. THE MATHEMATICS OF ' INVESTMENT BY WILLIAM L. HART, Pn.D, AS3O01ATH PBOTESSOR OF MATHEMATICS I2T THH OF MINWBSOTA D. 0. HEATH AND COMPANY BOSTON NEW YORK CHICAGO LONDON COPYRIGHT, 1934, BY D. C. HEATH AND COMPANY 2B0 PBINTBD D? U.S.A. PREFACE THIS book provides an elementary course in the theory and the application of annuities certain and in the mathematical aspects of life insurance. The book is particularly adapted to the needs of students in colleges of business administration, but it is also fitted for study by college students of mathematics who are not specializing in business. Annuities certain and their applications are considered in Part I, life insurance is treated in Part II, and a treatment of logarithms and of progressions IB given in Part III. The prerequisites for the study -of the book are three semesters of high school algebra and an acquaintance with progressions and logarithmic computation. Very complete preparation would be furnished by three semesters of \ high school algebra and a course in college algebra including logarithms. The material in the book has been thoroughly tested by the author through the teaching of it, in mimeographed form, for two years in ; classes at the University of Minnesota. It has been aimed in this book to present the subject in such a way that its beautiful simplicity & and great' usefulness will be thoroughly appreciated by all of the \ students to whom it is taught. Features of the book which will appeal to teachers of the subject are as follows : 1. Illustrative examples are consistently used throughout to illuminate new theory, to illustrate new methods, and to supply models for the solution of problems by the students. 2. Large groups of problems are supplied to illustrate each topic, and, in addition, sets of miscellaneous problems are given at the close of each important chapter, while a review set is placed at the end of 6ach of the major parts of the book. 3. Flexibility in the length, of the course is provided for ; the teacher can conveniently choose from this book either a one or a two semester, three-hour course, on account of the latitude afforded by (a) the large number of problems, (6) the segregation of optional methods and difficult topics into Supplementary Sections whose omission does not break the continuity of the remainder of the book, and (c) the possi-' bility of the omission of all of Part II, where life insurance is con- sidered. ui IV PREFACE 4. The concept of an equation of value is emphasized as a unifying principle throughout. 6. Formulas are simplified and reduced to as small a number as the author considers possible, if the classical notation of the subject is to be preserved. In Part I, a simplification is effected by the use of the interest period instead of the year as a time unit in a final pair of formulas for the amount and for the present value of an annuity certain. By use of these two formulas, the present values and the amounts of most annuities met in practice can be conveniently com- puted with the aid of the standard tables. In the applications of annuities certain, very few new formulas are introduced. The student is called upon to recognize all usual problems involving amortization, sinking funds, bonds, etc., as merely different instances of a single algebraic problem; that is, the finding of one unknown quantity by the solution of one of the fundamental pair of annuity formulas. 6. Interpolation methods are used to a very great extent and then 1 logical and practical completeness is emphasized. Some problems solved by interpolation are treated by other methods, as well, and such optional methods are found segregated into Supplementary Sections. 7. Practical aspects of the subject are emphasized throughout. 8. Very complete tables are provided, including a five-place table of logarithms, the values of the interest and annuity functions for twenty-five interest rates, tables of the most essential insurance functions, and a table of squares, square roots, and reciprocals. The tables may be obtained either bound with the book or bound separately. . 9. In the discussion of life annuities and life insurance, the em- phasis is placed on methods and on principles rather than on manip- ulative proficiency. It is aimed to give the student a clear conception of the mathematical foundations of the subject. No attempt is made to prepare the student as an insurance actuary, but the treatment in this book is an excellent introduction to more advanced courses in actuarial science. The interest and annuity tables prepared in connection with this book make possible the solution of most problems accurately to cents, if ordinary arithmetic is used. Results can be obtained with sufficient accuracy for most class purposes if the five-place table of logarithms is used in the computations. If the teacher considers it desirable to use seven-place logarithms, the author recommends the use of Glover's Tables from Applied Mathematics. These excellent tables contain the PREFACE V values, and the seven-place logarithms of the values, of the interest and annuity functions, a standard seven-place table of 'logarithms, and a variety of other useful tables dealing with insurance and statistics. The author acknowledges his indebtedness to Professor James Glover for his permission to publish in the tables of this book certain extracts from Glover's Tables which were published, for the first time, in that book. UNIVERSITY OP MINNESOTA, January 1, 1924. CONTENTS PART I ANNUITIES CERTAIN CHAPTER I SIMPLE INTEREST AND SIMPLE DISCOUNT IEOTION ' PAQH 1. Definition of interest 1 2. Simple interest 1 3. Ordinary and exact interest 3 4. Algebraic problems in simple interest 5 5. Simple discount 6 6. Banking use of simple discount ...... 9 7. Discounting notes 10 CHAPTER II COMPOUND INTEREST 8. Definition of compound interest . .. . * . . 14 9. Compound interest formula . 15 10. Nominal and effective rates . 18 11. Interest for parts of a period . 20 12. Graphical comparison of simple and compound interest . . 23* 13, Values of obligations . 24 14. Equations of value . 27 15. Interpolation of methods . 29 16.* Logarithmic methods . . , . 32 17.* Equated time ...-. . 33 18.* Interest, converted continuously . 35 Miscellaneous problems . 36 * Supplementary section. vu viii CONTENTS CHAPTER III ANNUITIES CERTAIN SECTION FAGH 19. Definitions 39 20. Special cases 41 21. Formulas in the most simple case 42 22. Annuities paid p times per interest period .... 45 23. Most general formulas 49 24. Summary of annuity formulas .50 25. Annuities due 56 26. Deferred annuities 59 27.* Continuous annuities , .62 28.* Computations of high accuracy 63 Miscellaneous problems 64 CHAPTER IV PROBLEMS IN ANNUITIES .29. Outline of problems 67 30. Determination of payment 67 31. Determination of term 70 32. Determination of interest rate 71 33.* Difficult cases and exact methods 74 Miscellaneous problems 76 CHAPTER V PAYMENT OF DEBTS BY PERIODIC INSTALLMENTS 34. Amortization of a debt 78 35. Amortization of a bonded debt 81 36. Problems where the payment is known 83 37. Sinking fund method 85 38. Comparison of the amortization and the sinking fund methods 88 Miscellaneous problems 89 39.* Funds invested with building and loan associations . . 92 40.* Retirement of loans made by building and loan associations 94 * Supplementary section. CONTENTS k CHAPTER VI DEPRECIATION, PERPETUITIES, AND CAPITALIZED COST PAOH 41. Depreciation, sinking fund plan ...... 96 42. Straight line method ........ 98 43. Composite life .......... 99 44. Valuation of a mine ........ 101 45. Perpetuities .......... 103 46. Capitalized cost ......... 105 47.* Difficult cases under perpetuities .' ..... 108 48.* Constant percentage method of depreciation .... 109 Miscellaneous problems ....... Ill CHAPTER Vn BONDS 49. Terminology .......... 113 50. Meaning of the investment rate ...... 113 51. Purchase price of a bond at a given yield .... 114 52. Changes in book value ........ 117 53. Price at a given yield between interest dates .... 121 54. Professional practices in bond transactions .... 124 55. Approximate bond yields ....... 126 66. Yield on a dividend date by interpolation .... 128 57. Special types of bonds ........ 130 58.* Yield of a bond bought between interest dates . . . .131 Miscellaneous problems ........ 133 Review problems on Part I ....... 135 PART II LIFE INSURANCE CHAPTER LIFE ANNUITIES 59. Probability .......... 147 60. Mortality Table ......... 148 61. Formulas for probabilities of living and dying ... 150 * Supplementary section. X CONTENTS SECTION PAGE 62. Mathematical expectation ....... 152 63. Present value of pure endowment 153 64. Whole life annuity 155 65. Commutation symbols 157 66. Temporary and deferred life annuities 159 67. Annuities due 162 Miscellaneous problems 163 CHAPTER IX LIFE INSURANCE 68. Terminology 165 69. Net single premium, whole life insurance .... 166 70. Term insurance 168 71. Endowment insurance 170 72. Annual premiums .171 73.* Net single premiums as present values of expectations . . 175 74.* Policies of irregular type 176 CHAPTER X POLICY RESERVES 75. Policy reserves 178 76. Computation of the reserve 180 Supplementary Exercise 183 Miscellaneous Problems on Insurance 184 PART III AUXILIARY SUBJECTS CHAPTER XI LOGARITHMS 77. Definition of logarithms ... ..... 187 78. Properties of logarithms .188 79. Common logarithms 190 80. Properties of the mantissa and the characteristic , . .191 81. Use of tables of mantissas 193 82. Logarithms of numbers with five significant digits , 194 * Supplementary section. CONTENTS xi SECTION . PAGH 83. To find the number when the logarithm is given . . . 196 84. Computation of products and of quotients .... 197 86. Computation of powers and of roots 198 86. Problems in computation 200 87. Exponential equations 201 88.* Logarithms to bases different from 10 203 CHAPTER XII PROGRESSIONS 89. Arithmetical progressions 204 90. Geometrical progressions 205 91. Infinite geometrical progressions 207 APPENDIX Note 1. Approximate determination of the time to double money at compound interest 211 Note 2. Approximate determination of the equated time . . 211 Note 3. Solution of equations by interpolation . ". . . 212 Note 4. Abridged multiplication 213 Note 5. Accuracy of interpolation for finding the time, under com- pound interest 214 Note 6. Accuracy of interpolation for finding the term of an annuity 214 INDEX 217 * Supplementary section. MATHEMATICS OF INVESTMENT PART I ANNUITIES CERTAIN CHAPTER I SIMPLE INTEREST AND SIMPLE DISCOUNT 1. Definition of interest. Interest is the income received from invested capital. The invested capital is called the prin- cipal ; at any time after the investment of the principal, the sum of the principal plus the interest due is called the amount. The interest charge is usually stated as a rate per cent of the principal per year. If $P is the principal, r the rate expressed as a decimal, and i the interest for 1 year, then by definition i = Pr, or -i en Thus, if $1000 earns $36.60 interest in one year, r - ^^ = .0366, or the 1000 rate is 3.66%. Also, if P = $1 in equation 1, then r = i, or the rate r equals the interest on $1 for 1 year. 2. Simple interest. If interest is computed on the original principal during the whole life of a transaction, simple interest is being charged. The simple interest on a principal P is propor- tional to the time P is invested. Thus, if the simple interest for 1 year is $1000, the interest for 5.7 years is $5700, Let / be the interest earned by P in t years, and let A be the amount due at the end of t years ; since amount = principal plus interest, A = P + I. (2) Since the interest earned by P in 1 year is Pr, the interest earned in t years is <(Pr) or j _ p r f (3) Hence P + I = P + Prt, so that, from equation 2, I 2 MATHEMATICS OF INVESTMENT It is important to realize that equation 4 relates two sums, P and A, which are equally desirable if money can be invested at, or is worth, the simple interest rate r. The possession of P at any instant is as desirable as the possession of A at a time t years later, because if P is invested at the rate r, it will grow to the amount A in t years. We shall call P the present value or pres- ent worth of the amount A, due at the end of t years. 3. Ordinary and exact interest. Simple interest is computed by equation 3, where t is the time in years. If the time is given in days, there are two varieties of interest used, ordinary and exact simple interest. In computing ordinary interest we assume one year to have 360 days, while for exact interest we assume 365 days. Example 1. Find the ordinary and the exact interest at 5% on $5000 for 59 days. Solution. For the ordinary interest I use equation 3 with f = ^, and for the exact interest I e use t = -gfc. I - 5000(.Q5) jfc = 840.97; I, = 5000(.05) Jft = $40.41. A relation exists between the ordinary interest I a and the exact interest I e on a principal P for D days at the rate r. From equations 5 it is seen that PrD = 360 I = 365 I e or Io _ 73 ,n\ J.-T? (6) which shows that ordinary interest is greater than exact interest. From equation 6, *, T '-^'-'-w- (8) Thus, if the exact interest 7. =* $40.41, we obtain from equation 7, /= 40.41 +^=$40.97. SIMPLE INTEREST AND SIMPLE DISCOUNT 3 EXERCISE I la the first five problems find the interest by use of equation 3. PBOB. PRINCIPAL RATH Turn INTHBBST 1. $ 6,570. 3.5% 75 days exact 2. 8,000. .045 93 days ordinary 3. 115,380. .0626 80 days exact 4. 4,838.70 7.5% 35 days ordinary 5. 2,500. .055 27 days exact s ' 6. The exact interest on a certain principal for a certain number of days is $60.45. Find the ordinary interest for the same period of time. 7. The ordinary interest on a certain principal for a certain number of days ia $35.67. Find the exact interest for the same period.- 8. In problem 1 find the ordinary interest by use of the result of problem 1. When the rate is 6%, the ordinary interest for 60 days is (.01)P, which is obtained by moving the decimal point in P two places to the left, while the interest for 6 days is (.001) P. These facts are the basis of the 6% method for computing ordinary interest at 6% or at rates conveniently related to 6%. 'Example 2. By the 6% method find the ordinary interest on $1389.20 for 83 days at 6%, and at 4.5%. Solution. $13.892 is interest at 6% for 60 days 4.168 ia interest at 6% for 18 days (3 times 6 days) 1.158 is interest at 6% for 5 days (^ of 60 days) $19.218 is interest at 6% for 83 days ' 4.805 is interest for 83 days at 1.5% (i of 6%) $14.413 is interest for 83 days at 4.5% - The extensive interest tables used in banks make it unnecessary to perform multiplications or divisions in computing simple in- terest. Table IV in this book makes it unnecessary to perform divisions. Example 3. Find the exact interest at 5% on $8578 for 96 days. Solution. From Table IV the interest at 1% on $10,000 for 96 days is $26.3013699. The interest on $8578 is (.8578) (5) (26.3014) - $112.81. 4 MATHEMATICS OF INVESTMENT To find the time between two dates, it is sometimes assumed that each month has 30 days. For example* February 23, 1922, is 1922 : 2 : 23 or 1921 : 14 : 23 (9) June 3, 1921, is 1921: 6: 3 Elapsed time - : 8 : 20 - 260 days The exact time can be found from Table III. February 23, 1922, is the 54th day of 1922 or the 419th day from January 1, 1921. June 3, 1921, is the 154th day from January 1, 1921. The elapsed time is (419 154) = 265 days. NOTE. In this book, for the sake of uniformity, proceed as follows, unless otherwise directed, in problems involving simple interest : (a) use ordinary interest if the time interval is given in days ; (b) in computing the number of days between dates, find the exact number of days; (c) if the time is given in months, reduce it to a fraction of a year on the basis of 12 months to the year, without changing to days. Methods in the business world are lacking in uniformity in these respects, and, in any practical application, ex- plicit information should be obtained as to the procedure to be followed. EXERCISE n Find the ordinary interest in the first five problems by use of the 6% method. 1. P = $3957.50, t = 170 days, r - .06. 2. P = $3957.50, t = 170 days, r = .07. 3. P = $4893.75, t = 63 days, r - .04. 4. P = $13,468.60, t = 41 days, r = .03. 6. P = $9836.80, t = 134 days, r = .05. 6. Find the exact interest in problem 4 by use of Table IV. 7. Find the ordinary interest on $8500 at 6% from August 11, 1921, to March 13, 1922. Use the approximate number of days, as in expression 9 above. 8. Find the ordinary interest in problem 7, but use the exact num- ber of days. 9. Find the ordinary and exact interest on $1750 at 5% from April 3, 1921, to October 13, 1921, using the exact number of days. 10. (a) Find the ordinary and exact interest in problem 9, using the approximate number of days. (&) Which of the four methods of problems 9 and 10 is the most favorable to a creditor? SIMPLE INTEREST AND SIMPLE DISCOUNT 5 4. Algebraic problems. If a sufficient number of the quantities (A, P, I, r, f) are given, the others can be determined by equations 2, 3, and 4, When the rate r, or the time t, is unknown, equation 3 should be used. When the present value P is unknown, equa- tion 4 is most useful. Example 1. If a $1000 principal increases to $1250- when in- vested at simple interest for 3 years, what is the interest rate ? Solution. P= $1000, A = $1250, t = 3. The interest 7 - $260. From 7 = Prt, 250 - 3000 r, or r = .0833. Example 2. What principal invested at 5.5% simple interest will amount to $1150 after 2 years, 6 months? Solution. Use equation 4. 1150 = P[l +2.5(.055)] = 1.1375 P; P _ 1150 _ J1010.99. An equivalent statement of Example 2 would be, 1.1375 " Find the present value of $1150, due at the end of 2 years, 6 months, if money is worth 5.5% simple interest." EXERCISE DI Find the missing quantities in the table below : PROS. A P / RATE TIME 1. $ 750. .04 3 yr., 6 mo. 2. $3500. .055 2yr. 3. 3500. .058 2yr. 4. $150 .075 6 mo. 6. 2500. .035 2 yr., 3 mo. 6. 1200. .06 2 yr., 6 mo. 7. 1800.60 300 .055 8. 650 .03 3 yr;, 9 mo. 9. 1680. .0375 11 mo. 10. 9850.50 .0725 1 yr., 6 mo. 11. Find the present value of $6000, due after 8 months, if money is worth 9%. 12. W borrowed $360 from B and agreed to repay it at the end of 8 months, with simple interest at the rate 5.25%. What did W pay at the end of 8 months? 6 MATHEMATICS OF INVESTMENT 13. Find the present worth of $1350, due at the end of 2 months, if money is worth 5% simple interest. ^ U. At the end of 3 months I must pay $1800 to B. To cancel this obligation immediately, what should I pay B if he is willing to accept payment and is able to reinvest money at 6% simple interest? Examples. A merchant is offered a $50 discount for cash payment. of a $1200 bill due after 60 days. If he pays cash, at what rate may he consider his money to be earning interest for the next 60 days? Solution. He would pay $1150 now in place of $1200 at the end of 60 days. To find 'the interest rate under which $1150 is the present value of $1200, due in 60 days, use I = Prf; 60 = , or r = .26087. Hia money earns in- 6 terest at the rate 26.087% ; he could afford to borrow at any smaller rate in order to be able to pay cash. 15. A merchant is offered a $30 discount for cash payment of a $1000 bill due at the end of 30 days. What is the largest rate at which he could afford to borrow money in order to pay cash? 16. A 3% discount is offered for cash payment of a $2500 bill, due at the end of 90 days. At what rate is interest earned over the 90 days if cash payment is made ? 17. The terms of payment for a certain debt are : net cash in 90 days or 2% discount for cash in 30 days. At what rate is interest earned if the discount is taken advantage of? HINT. For a $100 bill, $100 paid after 90 days, or $98 at the end of 30 days, would cancel the debt. 5. Simple discount. The process of finding the present value P of an amount A, due at the end of t years, is called discounting A. The difference between A and its present value P, or A P, is called the discount on A* From A = P + I, I A P; thus I, which is the interest on P, also is the discount on A, If $1150 is the discounted value of $1250, due at the end of 7 months, the discount on the $1250 is $100 ; the interest on $1150 for 7 months is $100. In considering I as the interest on a known principal P we com- puted I at a certain rate per cent of P, In considering 7 as the discount on a known amount A, it is convenient to compute / at a certain rate per cent per, year, of A, If i is the discount on SIMPLE INTEREST AND SIMPLE DISCOUNT 7 A, due at the end of 1 year, and if djs_thajiiscoiint-,rate,expreflafid^ as ajdecimal, then by (definition i = Ad, or Simple discount, like simple interest, is discount which is pro- portional to the time. If Ad is the discount on A, due in 1 year, the discount on an amount A due at the end of t years is t(Ad), or I=Adt. (11) From P = A- I, P = A- Adt, or P = A(l - df). (12) If the time is given in days, we may use either exact or ordinary simple discount, according as we use one year as equal to 365 or to 360 days, in finding the value of t. NOTB. In simple discount problems in this book, for the sake of uni- formity, proceed as follows unless otherwise directed : (a) if the time is given .orrlinnrjr dincouatj (ft) in computing the number of days between dates, findjthe^ exact number of dasa ; (c) if the time is given in months, duce'it to a fraction ojLa year^on the jgasisjDf J2^LQnEKB~to~tn"e yea Business practices are not uniform, and hence in any practical application of discount one should obtain explicit information as to the procedure to follow. Example 1 . Find the discount rate if $340 is the present value of $350, due after 60 days. ' Solution. Prom I - Adt, 10 = ; d = .17143, or 17.143%. 6 Example 2. If the discount rate is 6%, find the present value of $300, due at the end of 3 months. Solution. From /. = Adt, I = 300 (- 06 ) = 4.60. P = A - I - 300 - 4.60 = $295.50. NOTE, If A is known and P is unknown, it is easier to find P when the discount rate is given than when the interest rate is known. To appreciate this fact compare the solution of Example 2 above 'with that of Example 2 of . Section 4, where a quotient had to be commuted. This simplifying property of discount rates is responsible for their wide use in banking and 'business. The use of a discount rate in finding the present value P of a known amount A is equivalent to the use of some interest rate, which is always different from the discount rate. 8 MATHEMATICS OF INVESTMENT Example 3. (a) If a 6% discount rate is charged in discounting amounts due after one year, what equivalent interest rate could be used? (6) What would be the interest rate if amounts due after 3 months were being discounted? Solution. (a) Suppose A = $100, due after 1 year. Then / = 100(.06) = $6,. and P = 594. Let r be the equivalent interest rate, and use I = Prt. 6 = 94r; r -.063830. (&) If A 98.60 r . $100, due after 3 months, I = 1QO (- 06 ) 4 $1.60, and P = 898.60. From J = Prt, 1.50 .060914. Compare the results of Example 3. When a discount rate is being used, the equivalent interest rate is larger for long-term than for short-term transactions and in both cases is larger than the discount rate. The brief methods available for computing simple interest apply as well to the computation of simple discount because both opera- tions involve multiplication by a small decimal. Thus, we may use the 6% method for computing discount, and simple interest tables may be used as simple discount tables. EXERCISE IV 1. Find the discount rate if the discount on $1500, due after one year, is $82.50. 2. Find the discount rate if the present value of $1250, due after 8 months, is $1193.75. Find the missing quantities in the table by use of equations 11 and 12. PROS. DlSOOTFNTBD V A.IJ-CIH, P AMOUNT, A A IB DUB AJTBE /, DISCOUNT ON A OB INT. ON P DISCOUNT RATE, d 3. $1200 lyr. .05 4. $145.50 150 6mb. 6. 3 mo. $250 .07 6. ' 2000 90 da. .045 7. . 800 120 da. .06 8. 357.75 375 9 mo. 9. s * 5 mo. 300 .08 10. 750. 72 da. .0625 11. - 1500. 6 mo. 35 12. 7500 100 .06 SIMPLE INTEREST AND SIMPLE DISCOUNT 9 13. Write in words problems equivalent to problems 3, 4, 10, and 12 above. 14. (a) What simple interest rate would be equivalent to the charge made in problem 3 ; (&) in problem 7? 15. If d = .045, (a) what is the equivalent interest rate for a 1-year transaction ; (6) for one whose term is 4 months? r / 16. What discount rate would be equivalent to the use of a 6% interest rate in a 1-year transaction? HINT. Let P = $100; find A and I, and use I = Adt. , 17. What discount rate would be equivalent to the use of a 6% interest rate in a 6-month transaction? STTPPLEMENTABY NOTE. Formulas carx be obtained relating the discount rate d and the equivalent interest rate r on 1-year transactions. Suppose that $1 is due at the end of 1 year. Then, in equations 11 and 12, A = $1, I - d, and P = 1 - d. From I = Prt with t = 1, d - (1 d)r, or ' "rV as) From equation 13, r rd *= d, or r = d(l + r), so that d= iT7 . a*) It must be remembered that equations 13 and 14 connect the discount and the interest rates only in the case of 1-year transactions. SUPPLEMENTARY EXERCISE V 1. By use of equations 13 and 14, solve problems 14, 15, 16, and 17 of . Exercise IV. 6. Banking use of simple discount in lending money. If X asks for a $1000 loan for six months from a bank B which is charg- ing 6% discount, B will cause X to sign a note promising to pay B $1000 at the end of 6 months. Then, B will give X the present value of the $1000 which he has promised to pay. The bank computes this present value by use of its discount rate. P = 1000 30 = $970, which X receives. The transaction is - . equivalent to B lending X $970 for 6 months. The interest rate which X is paying is that which is equivalent to the 6% discount ; rate. The banker would tell X that he is paying 6% interest in \ I advance, but this would merely be a colloquial manner "of stating - V that the discount rate is 6%< In this book the phrase interest in advance is always used in this customary colloquial sense. 10 MATHEMATICS OF INVESTMENT Example 1. X requests a loan of $9000 for 3 months from a bank B charging 5% discount. Find the immediate proceeds of the loan and the interest rate which X is paying. Solution. X promises to pay $9000 at the end of 3 months. Discount on $9000 for 3 mouths at 5% is $112.50. Immediate proceeds, which X receives, are $9000 - $112.50 = $8887.50. To determine the interest rate, use I = Prt. 112.6 = 8887.6 r(*), or r = .060833. Example 2. X wishes to receive $9000 as the immediate proceeds of a 90-day loan from a bank B which is charging 5% interest payable in ad- vance. For what sum will X draw the note which he will give to B? Solution. P = $9000, t = 90 days, d = .05, and A is unknown. From equation 12, 9000 = A(l - .0126). A = $9113.92, for which X will draw the note. EXERCISE VI Determine how much X receives from the .bank B. In the first three problems, also determine the interest rate which X is paying. PBOB. LOAN REQUESTED BY X FOB DISCOUNT RATH OF B vl. $5,000 6 months .065 -A 2. 1,760 75 days .07' 3. 3,570 90 days .06 4. 190 30 days .05 5. 7,500 45 days .055 6. 3,800 3 months .0625 Determine the size of the loan which X would request from B if X desired the immediate proceeds given in the table. PBOB. IMMEDIATE PHOOBHDB THBM OF LOAN DISCOUNT RATH OF B 7. $ 3,500 30 days .06 8. 8,000 4 months .05 9. 1,300 3 months .07 10. 150,000 60 days .055 11. 4,300 90 days .045 12. 9,350 6 months .05 7. Discounting notes. The discounting of promissory notes gives rise to problems similar to those of Exercise VI. .Consider the following notes : <, / oiij-iiiiliiiiii SIMPLE INTEREST AND SIMPLE DISCOUNT 11 NOTE (a) Minneapolis, June 1, Six months after date I promise to pay to Y or order $5000 without interest. Value received. Signed X. NOTE (6) Chicago, June 1, 1922. One hundred and eighty days after date I promise to pay to Y or order $5000 together with simple interest from date at the rate 7%. Value received. Signed X. On August I, Y sells note (a) to a bank B. The sale is ac- complished by Y indorsing the note, transferring his rights to B, who -will receive the $5000 on the maturity date. The transaction is called discounting the note because B pays Y the present of discounted value of $5000, due on December 1, 1922. Example 1. If B discounts notes at 5%, what will Y receive on August 1? Solution. B is using the discount rate 5% in computing present values. The discount on $5000, due after 4 months, is $83.33 ; B will pay Y $4916.67. Example 2. On July 31, Y discounts note (6) at the bank B of Ex- ample 1. What are Y's proceeds from the sale of the note? . Solution. B first computes the maturity value of the note, or what X will pay on the maturity date, which is November 28. The transaction is equiva- lent to discounting this maturity value. Term of the discount is 120 days (July 31 to November 28). From equation 4, the maturity value of the note is 5000(1 + .036) = $5175.00. Discount for 120 days at 5% on $5175 is $86.25. Proceeds = $5175 - $86.25 = $5088.75. EXERCISE VH 1. X paid Y for an order of goods with the following note : Chicago, June 1, Sixty days after date I promise to pay to Y or order $375 at the Continental Trust Company. Value received. Signed X. 12 MATHEMATICS OF INVESTMENT Thirty days later, Y discounted this note at a bank charging 5.5% discount. Find Y's proceeds from the sale of the note. 2. Find the proceeds in problem 1, if the discount rate is 6%. 3. The bank B of problem 1, after buying the note from Y, immediately rediscounted it at a Federal Reserve Bank : whose rediscount rate, was .035. What did B receive for the note? 4. The holder of a non-interest-bearing note dated October. 1, 1911, payable 4 months, after date, discounted it at a bank on October 1, at the rate 4%. The bank's discount on the note was $20. What was the face of the note? 5. X owes a firm Y $800, due immediately. In payment X draws a 90- day non-interest-bearing note for such a sum that, if Y immediately dis- counts it at a bank charging 6% discount, the proceeds will be $800. What is the face value of the note? HINT. In equation 12, P ** $800 and A is unknown. 6. X draws a 60-day non-interest-bearing note in payment of a bill for $875, due now. What should be the face of the note so that the immediate proceeds to the creditor will be $875 if he discounts it immediately at a bank whose discount rate is 6.5%? Find the proceeds from the sale of the following notes : PBOD. FA.OD or NOTH DATE OF Norn THHM NOTHBlABfl INT. At SOLD OK Disc. RAfan . OF BUYER 7. $ 450 6/10/17 30 days .06 6/20/17 .07 8. 1200 6/12/18 ' 120 days .05 6/26/18 , .06 9. . 376 3/25/20 90 days .07 4/24/20 .04 10. 470 11/20/21 60 days .00 12/ 5/21 .065 11." 325 4/30/20 3 months .06 6/ 1/20 .08 12." 3000 8/14/19 6 months .08 12/16/19 .06 1 A Federal Reserve Bank discounts commercial notes brought to it by banks belonging to the Federal Reserve System. The rate of the Federal Reserve Bank is called a rediscount rate because all notes discounted by it have been discounted previously by some other bank. This previous discounting has no effect so far as the computation of the present value by the Federal Reserve Bank is concerned. 3 The note is due on 7/31/20, the last day of the third month from April. Find the exact number of days between 6/1/20 and 7/31/20. 1 Find the exaot number of days between 12/16/19 and 2/14/20. SIMPLE INTEREST AND SIMPLE DISCOUNT 13 J 13. Y owes W $6000 due now. In payment Y draws a 45-day non- interest-bearing note, which W discounts immediately at a bank charging 6% interest in advance. What is the face of the note if W's proceeds are $5000? 14. W desires $2500 as the immediate proceeds of a 6-month loan from a bank which charges 7% interest in advance. What loan .will W re- quest? 15. A bank B used the rate 6% in discounting a 90-day note for $1000. The note was immediately rediscounted by B at a Federal Reserve Bank whose rate was 4%. Find B's profit on the transaction. CHAPTER II COMPOUND INTEREST 8. Definition of compound interest. If, at stated intervals during the term of an investment, the interest due is added to the principal and thereafter earns interest, the sum by which the original principal has increased by the end of the term of the in- vestment is called compound interest. At the end of the term, the total amount due, which consists of the original principal plus the compound interest, is called the compound amount. We speak of interest being compounded, or payable, or con- verted into principal. The time between successive conversions of interest into principal is called the conversion period. In a compound interest transaction we must know (a) the conversion period and (&) the rate at which interest is earned during a con- version period. Thus, if the rate is 6%, compounded quarterly, the conversion period is 3 months and interest is earned at the rate 6% per year during each period, or at the rate 1.5% per con- version period. Example 1. Find the compound amount after 1 year if $100 is in- vested at the rate 8%, compounded quarterly. Solution. The rate per conversion period is ,02. Original principal is $100. At end of 3 mo. $2.000 interest is due j new principal is $102.000. At end of 6 nxo. $2.040 interest is due ; new principal is $104.040. At end of 9 mo. $2.081 interest is due ; new principal is $106.121. At end of 1 yr, $2.122 interest is due; new principal is $108.243. The compound interest earned in 1 year is $8.243. The rate of increase of O 0/tD principal per year is ^J22. = .08243, or 8.243%. EXERCISE Vm 1. By the method of Example 1 find the compound amount after 1 year if $100 is invested at the rate 6%, payable quarterly. What was the compound amount after 6 months? At what rate per year does principal increase in this case? 14 COMPOUND INTEREST 15 2. Find the annual rate of growth of principal under the rate .04, con- verted quarterly. NOTE. Hereafter, the unqualified word interest will always refer to compound interest. If a transaction extends over more than 1 year, compound interest should be used. If the tune involved is less than 1 year, simple interest generally is used. 9. The compound interest formula. Let the interest rate per conversion period be r, expressed as a decimal. Let P be the original principal and let A be the compound amount to which P accumulates by the end of k conversion periods. Then,. we shall prove that A = P(l + r)*. (15) The method of Example 1, Section 8, applies in estabhshing equation 15. Original principal invested is P. Interest due at end of let period is Pr. New principal at end of 1st period is P + Pr = P(l + r) . Interest due at end of 2d period is P(l + r)r. New principal at end of 2d period is P(l + r) + P(l + r)r = P(l + r) 2 . By the end of each period, the principal on hand at the beginning of the period has been multiplied by (1 + r). Hence, by the end of k periods, the original principal P has been multiplied k suc- cessive times by (1 + r) or by (1 + r)*. Therefore, the compound amount after k periods is P(l + r)*. If money can be invested at the rate r per period, the sums P .and A } connected by equation 15, are equally desirable, be- cause if P is invested now it will grow to the value A by the end of k periods. We shall call P the present value of A, due at the end of k periods. The fundamental problems under compound interest are the following : (a) The accumulation problem, or the determination of the amount A when we know the principal P, the interest rate, and the time for which P is invested. To accumulate P, means to find the compound amount A resulting from the investment of P. (6) The discount problem, or the determination of the present value P of a known amount A, when we know the interest rate and 16 MATHEMATICS OF INVESTMENT the date on which A is due. To discount A means to find its present value P. The discount on A is (A P). The accumulation problem is solved by equation 15. Exampk 1. Find the compound amount after 9 years and 3 months on a principal P = $3000, if the rate is 6%, compounded quarterly. Solution. The rate per period is r = .015 ; the number of periods is fc = 4(9.25) = 37. A = 3000(1.016)" = 3000(1.73477663) = $5204.33. (Using Table V) The compound interest earned is $6204.33 - $3000 - $2204.33. NOTE. If interest is converted m times per year, find the number fc of conversion periods in n years from the equation fc = mn. To solve the discount problem we first solve equation 15 for P, ob- taining A p - - = A Exampk 2. Find the present value of $5000, due at the end of 4 years and 6 months, if money earns 4%, converted semi-annually. Solution. Rate per period is r = .02; number of periods is fc = 2(4.5) = 9. P >= 6000(1.02)-' = 5000083675527) - $4183.78. (Using Table VI) The discount on A is $5000 $4183.78 = $816.22. NOTE. Recognize that Example 2 involves the formation of a product when solved by Table VI. A problem is solved incorrectly if available tables are not used to simplify the work. Since products are easier to compute than quotients, the following solution of Example 2 should be considered incorrect, although mathematically flawless, because a quotient is computed. F ~ In describing interest rates in the future, a standard abbrevia- tion will be used. When we state the rate to be (.05, m = 2), the " m = 2" signifies that interest is compounded twice per year, or semi-annually. The rate (.08, m = 1) means 8%, compounded annually; (.07, m = 12) means 7%, converted monthly; (.06, m = 4) means 6%, compounded quarterly. NOTE. The quantity (1 + r) in equation 16 is sometimes called the accumulation factor, while the quantity - or (1 + r)" 1 in equation 16 is called the discount factor. In many books the letter v is used to denote the discount factor, or v = (1 + r)" 1 . Thus, at the rate 7%, u 4 = (1.07)" 4 , COMPOUND INTEREST EXERCISE IX 17 1. By use of the binomial theorem verify all digits of the entry for (1.02) 4 in Table V. 2. In Table VI verify all digits of the entry for (1.02)- 6 by using the entry for (1.02) in Table V and by completing the long division involved. 3. Find the compound amount on $3,000,000 after 16 years and 3 months, if the rate is (.06, m = 4). 4. Accumulate a $40,000 principal for 15 years under the rate (.05, m = 4). What compound interest is earned? 6. Find the present value of $6000, due after 4J years, if money can earn interest at the rate (.08, m = 4). What is. the discount on the $6000? 6. Discount $5000 for 19 years and 6 months, at the rate (.05, m = 2). In the table below, find that quantity, P or A, which is not given. In the first four problems, before doing the numerical work, write equivalent problems in words. PROS. PRINCIPAL, P AMOUNT, A P ACCUMULATES FOB, OR A IB DUB APTHB RATE 7. $4000 6 yr., 6 mo. .04, m = 2 8. $1000. 10 yr., 3 mo. .07, m = 4 9. 3000. 12 yr. .06, m = 1 10. 6000 Syr.,. 6 mo. .03, 77i = 4 11. 2600. 13 yr., 9 mo. .08, m = 4 12. 1600 7 yr., 6 mo. .06, 77i = 2 13. 576.60 3 yr., 6 mo. .06, m - 12 14. 1398.60 16 yr., 3 mo. .05, m = 4 15. 8300 14 yr., 6 mo. ;056, 771 = 2 16. 0500 5yr. .045, TO = 1 17. 1300. 2 yr., 9 mo. .03, m = 4 18. 1. 76 yr. .05, m = 1 19. 100 100 yr. .035, m = 1 20. 1 100. 176 yr. .045, m = 1 21. 1 100 173 yr. .065, m = 1 22. 1 30 int. periods .04, per period 23. 1. 36 int. periods .06, per period 1 In problem 20 use (1.046) 1 " - (1.045) 1 (1.045). In. problem 21 use 18 MATHEMATICS OF INVESTMENT 24. (a) If the rate is i, compounded annually, and if the original principal is P, derive the formula for the compound amount after 10 years. (6) After n years. 26. If $100 had been invested in the year 1800 A.D. at the rate (.03, m = 1), what would be the compound amount now? 26. (a) If the rate is j, compounded m times per year, derive a formula for the compound amount of a principal P after 10 years. (&) After n years. _,i(, _ .._-;-' . , ? , - 10. Nominal and effective rates. Under a given type of compound interest _the rate per year at which principal grows is called the effective rate. The per cent quoted in stating a type of compound interest is called the nominal rate ; it is the rate per '/ year at which money earns interest during a conversion period. * In the illustrative Example 1 of Section 8 it was seen that, when the_"nommdl rg,te.,was 8%, converted quarterly, the effective rate was 8.24%. We shall say " tiitTrate is (j, m) " to abbreviate " the nominal rate is j, converted m times per year." Let i represent the effective rate. The effective rate i corresponding to the nominal rate j, converted m times per year, satisfies the equation + ff> ( &T) To prove this, consider investing P = $1 for 1 year at the rate (j, m). The rate per period is -^ and the number of periods in m 1 year is m. The amount A after 1 year and the interest I earned in that time are m The rate of increase of principal per year is i = = /, because P = $l. Hence i J Transpose the 1 in equation 18 and equation 17 is obtained. Example 1. What is the effective rate corresponding to the rate (.06, m- 4)? ,/ COMPOUND INTEREST Id Solution. Use equation 17. 1 + i = (1.0125) 4 = 1.05094634. i = .05094534. Example 2. What nominal rate, if converted 4 times per year, will . yield the effective rate 6%? i Solution. From equation 17, 1.06 = 1 + i = (1.06)* = 1.01467385, from Table X. (19) j = 4(.01467385) = .05869640." Table XI furniahes an easier solution. From equation 19, I - (1.06)* - 1 ; j = 4[(1.06)* - 1] - .05869538. (Table XI) Example 3. What nominal rate, converted quarterly, will give the same yield as (.05, m = 2) ? Solution. Let j be the unknown nominal rate. The effective rate i cor- responding to (.05, m = 2) must equal the effective rate corresponding to the nominal rate j, compounded quarterly. From equation 17, (1.025)'; 1 + i = (l + 0*. ... (1.025) 2 = (l + {)* (1.025)* = 1.01242284. j = 4(.01242284) = .04969136. 1 + i 1 + EXERCISE X 1. (a) In Table X verify the entry for (1.05)* by use of Table II. (6) In Table XI verify the entry f or p = 4 and i = .05, by using Table X. 2. Find .the effective rates corresponding to the rates (.06, m 2) and (.06, m = 4). 3. Find the nominal rate which, if converted semi-annually, yields the effective rate .05. (a) Solve by Table X. (6) Read the result out of Table XI. Solve the problems in the table orally by the aid of Tables V and XI. State equivalent problems in words. PROB. i ' m i- FqOB. 3 m i 4. .07 2 10. ' 12 .04 5. 2 .07 11. .03 2 6. .10 4 12. 2 .0275 7. 2 .035 13. 4 .026 * 8. .09 3 14. .05 1 9. .09 4 15. 1 .06- 12Q) MATHEMATICS OF INVESTMENT 16. Derive a formula for the nominal rate which, if converted p times per year, gives the effective rate i. NOTE. The resulting value of j is denoted by the notation j p , as in Table XI. 17. Which interest rate is the better, 5% compounded monthly or 6.5% compounded semi-annually? , 18. Which rate is the better, (.062, m = 1) or (.06, m = 2) ? 19. Determine the nominal rate which, if converted semi-annually, ./ may be used in place of the rate 5%, compounded quarterly. >f St>& ***S 20. What nominal rate compounded quarterly could equitably replace the rate (.04, m = 2) ? NOTE. When interest is compounded annually, the nominal and the effective rates are equal because in this case both represent the rate of increase of principal per year. This equality is seen also by placing m = 1 in equation 17. Hence, to say that money is worth the effective rate 5% is equivalent to saying that money is worth the nominal rate 5%, compounded annually. If m, the number of conversion periods per year of. a nomi- nal rate, is increased, the corresponding effective rate is also in- creased. If j = .06, the effective- rates i for different values of m are: m = 1 2 4 12 52 365 i f= .06000 .06090 .06136 .06168 .06180 .06183 We could consider interest converted every day or every moment or every second, or, as a limiting case, converted continuously (m = infinity). The more frequent the conversions, the more just is the interest method from the standpoint of a lender, so that interest, converted continuously, is theoretically the most ideal. The effective rate does not increase enormously as we increase the frequency of compounding. When j = .06, in the extreme case of continuous conversion (see Section 18, below), i = .06184, only slightly in excess of .06183, which is the-effective rate when m 365. 11. Interest for fractional parts of a period. In deriving equation 15 we assumed fc to be an integer. Let us agree as a new COMPOUND INTEREST 21 definition that^the compound a^ojjuil J <l.shaILbe_given Jby equation^ 15 also wEen k is not an integer. /' Exampk 1. Accumulate $1000 for 2 years and 2 months, at the rate '(.08, in = 4). ,- - ^ Solution. The rate per period is r = .02, and k =, 4(2J) = 8f . T A = 1000(1.02)'* = 1000(1.02)8(1.02)1- = 1000^:imS94JtL02)* ; f log 1.02 - 0.005734. | log 1171.66 - 3.068801. log A = 3.074635; A - $1187.23. Example 2. Find the present value of $3500, due at the end of 2 years and 10 months, if money is worth (.07, m = 2). / -^ "- 1 ,?-^, -- -- , - Solution. The rate per period is r = .035, and k = 2(2|) = 5f . P = 3500(1.035)^ - 3500(1.035)-(1.035)*. P = 3500(.81360064) (1.01163314) = $2880.09.1 (Tables VI and X) The methods of Examples 1 and 2 are complicated unless k is a convenient number. Approximate practical methods are described below. , Rule 1. To find the compound amount after k periods when k is not an integer, (a) compute the compound amount after the largest number of whole conversion periods contained in the given time. (6) Accumulate this amount for the remaining time at simple interest at the given nominal, rate, Example 3. Find the amount in Example 1 by use of Rule 1. Solution. Compound amount after 2 years is 1000(1.02)" = $1171.66. Simple interest at the rate 8% for 2 months on $1171.66 is $15.62. Amount at end of 2 years and. 2 months is 1171.66 + 15,62 = $1187.28, slightly greater than the result of Example 1. The use of Rule 1 is always slightly in favor of the creditor. Rule 2. To find the present value of A, due at the end of k periods, when k is not an integer, (a) discount the amount A for the smallest number of whole periods containing the given time. This gives the discounted value of A at a time before the present. 1 If five-place logarithms are used in multiplications or divisions, the results will be accurate to only four significant figures. Hence, in Example 2, if we desire P accurately to cents, ordinary multiplication must be used (unless logarithm tables with seven or more places are available). In performing the ordinary multiplica- tion, as in finding P in Example 2, the student is advised to use the abridged method described in the Appendix, Note 4. 22 MATHEMATICS OF INVESTMENT (&) Accumulate this result up to the present time at simple interest at the given nominal rate. Example 4. Solve Example 2 by use of Rule 2. Sol-uOon. The smallest number of conversion periods containing 2 years and 10 months is 6 periods, or 3 years. Discounted value of $3500, 3 years before due, or 2 months before the present, is SSOOfl.OSS)- 6 = $2847.25. Simple interest on $2847.25 at 7% for 2 months is $33.22. Present value is 2847.25 + 33.22 = $2880.47, which is greater than the result of Example 2. Results computed by use of Rule 2 are always slightly larger than the true present values as found from equation 16. NOTE. Unless otherwise directed, use the methods of Examples 1 and 2 when fc is not an integer. Compute the time between dates approximately, as in expression 9, of Chapter I, and reduce to years on the basis of 360 days to the year. EXERCISE XI Find P or A, whichever is not given. Use Table X whenever possible. "Pi* (TO PSBSHNT AMOUNT, P ACCUMULATES FOB, INTEREST STJMJJ3* VALUE, P A OR A IS DUE AFTER RATH 1. $2000 3 years, 3 mo. .06, m => 2 2. 1000 6 years, 1 mo. .07, m = 4 8. 8000 16 years, 8 mo. .05, m = 1 4. 4000 13 years, 7 mo. .08, m - 4 5. 5000 11 years, 5 mo. .04, m = 2 6. 1000 6 years, 4 mo. .05, m - 2 7. 1500 7 years, 10 mo. .06, m <= 4 8. 1500 7 years, 10 mo. .06, m = 4 9. Find the amount in problem 1 by use of Rule 1. 10. Find the present value in problem 5 by use of Rule 2. 11. Find the amount in problem 4 by use of Rule 1. 12. Find the present value in problem 7 by use of Rule 2. ^ia. On June 1, 1920, X borrows $2000, from Y and.agrees to pay the compound amount on whatever date ne settles his account, By use of Rule 1, determine what X shouldpay on August 1, 1922, if interest is at the* rate 6%, compounded quarterly? fl, 4, y t $ 14, At the end of 5 years and 3 mouths, $10,000 is due. Discount it to the present time if money is worth (.05, m = 2). Use Rule 2. COMPOUND INTEREST 23 12, Graphical comparison of simple and compound interest. In Figure 1 the straight line EF is the graph of the equation A = 1 + .06 t, the amount after t years if $1 is invested at simple in- terest at the rate .06. The curved line EH is the graph of the equation A = (1.06)*, the amount after t years if $1 is invested at the rate A .06 compounded annually. This curve was sketched through the points corre- sponding to the following table of values : FIG. i A 1 1.0147 1.0296 1.06 1.124 1.338 1.791 t i * 1 2 5 10 A=l,06 The entries f or t = % and t = are from Table X. In Figure 2, that part of the curves for which t = to t = 1 has been magnified (and distorted vertically, for em- phasis). Figure 2 shows that, when the time is less than one conversion period, the amount at simple interest is greater than the amount at compound inter- est. The two amounts are the same when t = 1, and, thereafter, the compound amount rapidly grows greater than the amount at simple interest* The ratio, (compound amount) -f- (amount at simple interest), approaches infinity as t approaches infinity. Fia, 2 24 MATHEMATICS OF INVESTMENT EXERCISE Xn 1. (a) Draw graphs on the same coordinate system, of the amount at simple interest, rate 5%, and of the amount at (.05, m = 1), for a principal of $1, from t = O'to t = 10 years. (6) Draw a second graph of that part of the curves for which t = to t = 1 with your original scales magni- fied 10 times. 13. Values of obligations. A financial obligation is a promise to pay, or, an obligation is equivalent to a promissory note. Con- sider the following obligations or notes : (a) Three years and 9 months after date, X promises to pay $1000 to Y or order. (&) Three years and 9 months after date, X promises to pay $1000 together with all accumulated interest at the rate 6%, com- pounded quarterly, to Y or order. Exampk 1. One year after date of note (a), what does Y receive on discounting it with a banker B to whom money is worth (.05, m = 4) ? Solution. B pays the present value of $1000, due after 2 years and 9 months, or 1000 (1.0 125) "" = $872.28. Example 2. One year after date of note (&), what is its value to a man W to whom money is worth (.07, m 4) ? Solution. Maturity value of obligation (6) is 1000(1.016) ll! - $1250.23. Its value to W, 2 years and 9 months before due, is 1250.23(1.0175)"" = $1033.03. Under a stipulated rate of interest, the value of an obligation, n years after its maturity date, is the compound amount which would be on hand if the maturity value had been invested for n years at the stipulated interest rate, Example 3. Note (6) was not paid when due. What should X pay at the end of 5 years to cancel the obligation if money is worth (.07, m =? 4) toY? Solution. Maturity value of note is lOOO(l.Olfi) 1 * - $1250.23. Value at the rate (.07, m = 4) to be paid by X, 1 year and 3 months after maturity date, is 1250.23(1.0175)* - $1363,52. NOTB. In all succeeding problems in compound interest, reckon elapsed time between dates approximately, as in expression 9 of Chapter I, If it is stated that a sum is due on & certain* date, the sum is understood to be due COMPOUND INTEREST %5 v -. . without interest. If a sum is dite with accumulated interest, this fact will be mentioned explicitly. EXERCISE 1. If money is worth (.07, m = 2) to W, what would he pay to Y for note (a), above, 3 months after date of the note? 2. If money is worth (.06, m = 2) to W, what should he pay to Y for note (&), above, 3 months after date of the note? 3. X borrows $1500 from Y and gives him the following note : BOSTON, July 15, 19SS, Three years and 6 months after date, I promise to pay to Y or order at the First National Bank, $1600 together with accumulated interest at the rate (.07, m = 2) Value received. Signed 5. On January 15, 1923, what does Y receive on selling this note to a bank which uses the rate (.06, m = 2) in discounting? 4. What would Y receive for the note in problem 3 if he discounted it on July. 15, 1922, at a bank using the rate (.055, m = 2) ? 6. X owes $300, due with accumulated interest at the rate (.04, m = 4) at the end of 5 years -and 3 months. What is the value of this obligation two years before it is due to a man to whom money is worth (.06, m = 1) ? 6. At the end of 4$ years, $7000 is due, together with accumulated interest at. the rate (.045, m = 2). (a) Find the value of this obligation 2i years before it is due if money is worth (.05, m = 2). (6) What is its value then under the rate (.045, m = 2) ? ^ 7. On May 15, 1918, $10,000 was borrowed. It was to be repaid on August 15, 1921, with accumulated interest at the rate (.08, m = 4). No payment was made until August 15, 1923. What was due then if money was considered worth (.07, m = 2) after August 15, 1921 T*/ 8.' On May 15, 1922, what was the value of the obligation of problem 7 if money was worth (.07, m = 4) after August 15, 1921 ? 9. Find the value of the obligation of problem 7 on November 15, 1923, if money is worth (.05, m = 4), commencing on August 15, 1921. 10. X owes Y (a) $2000, due in 2 years, and (6) $1000, due in 3 years with accumulated interest at the rate (.05, m = 2). At the end of one year what should X pay to cancel the obligations if money is worth (.04, m =* 2) to Y? HINT. X should pay the. sum of the values of his obligations. 26 MATHEMATICS OF INVESTMENT 11. At the end of 3 years and 3 months, $10,000 is due with accumulated interest at (.05, m = 4). (a) What is the value of this obligation at the end of 5 years if money is worth (.07, m = 4) ? (6) What is its value then if money is worth (.04, m = 4) ? 12. The note of problem 3 is sold by Y on October 16, 1924, to a banker to whom money is worth 6%, effective. By use of Rule 2 of Section 11 find the amount the banker will pay. The value of an obligation depends on when it is due. Hence, to compare two obligations, due on different dates, the values of the obligations must be compared on some common date. Example 4. If money is worth (.05, m - 1), which is the more valuable obligation, (a) $1200 due at the end of 2 years, or (&) $1000 due at the end of 4 years with accumulated interest at (.06, m = 2) ? Solution. Compare values at the end of 4 years under the rate (.05, m = 1). The value of (a) after 4 years is 1200(1.05)" = $1323.00. Tho value of (b) after 4 years is 1000(1.03)* = $1266.77. Hence, (a) is tho more valuable. NOTE. The value of an obligation on any date, the present for example, is the sum of money which if possessed to-day is as desirable as the payment promised in the obligation, If the present values of two obligations are the same, their values at any future time must likewise be equal, because these future values are the compound amounts of the two equal present values. Similarly, if the present values are equal, the values at any previous date must have been equal, because these former values would be the results obtained on discounting the two equal present values to the previous date. Hence, any comparison date may be used in comparing the values of two obligations, be- cause if their values are equal on one date they are equal on all other dates, both past and future. If the value of one obligation is greater than that of another on one date, it will be the greater on all dates. For instance, in Example 4 above, on comparing values at the end of 3 years, the value of (a) is 1200(1.05) = $1260.00; the value of (b) is 1000(1.03)(1.06)" a - $1206.45. Hence, as in the original solution, (a) is seen to be tho more valuable. The comparison date should be selected so as to minimize the computation required. Therefore, the original solution of Example 4 was the most desirable. EXERCISE XIV 1. If money ia worth (.04, m = 2), which obligation is the more valu- able : (a) $1400 due after 2 years, or (6) $1500 due after 3 years? 2. If money is worth (.05, m = 2), which obligation is the more valu- able: (a) $1400 due after 5 years, or (&} $1000 due after 4 yeajs with COMPOUND INTEREST 27 accumulated interest at (.07, m = 2) ? Use 4 years from now as the comparison date. 3. Solve problem 2 with 6 years from now as the comparison date. 4. If money is worth (.06, m = 2), compare the value of (a) $6000 due after 4 years with (&) an obligation to pay $4000 after 3 years with accumulated interest at (.05, m = 1). 6. Compare the set of obligations (a) with set (&) if money is worth (.06, m = 2) : (a) $1600 due after 3 years ; $1000 due after 2 years with accumu- lated interest at the rate (.04, m = 2). (&) $1200 due after 2 years ; $1400 due after 2 years with accumu- lated interest at the rate (.05, m = 2). 6. Which obligation is the more valuable if money is worth (.06, m = 4) : (a) $8000 due after 3 years with accumulated interest at (.05, m = 4), or (6) $8500 due after 3 years? 14. Equations of value. An equation of value is an equation stating that the sum of the values, on a certain comparison date, of one set of obligations equals the sum of the values on this date of another set. Equations of value are the most powerful tools available for solving problems throughout the mathematics of investment. NOTE. In writing an equation of value, the comparison date must be explicitly mentioned, and every term in the equation must represent the value of some obligation on this date. To avoid errors, make preliminary lists of the sets of obligations being compared. Exampk 1. W owes Y (a) $1000 due after 10 years, (&) $2000 due after 5 years with accumulated interest at (.05, m = 2), and (c) $3000 due after 4 years with accumulated interest at (.04, m = 1). W wishes to pay in full by making two equal payments at the ends of the 3d and 4th years. If money is worth (.06, m = 2) to Y, find the siae of Ws payments. Solution. Let $tc be the payment. W wishes to replace his old obligations by two new ones. Let 4 years from now be the comparison date. OLD OBLIGATIONS NEW OBLIGATIONS (a) $1000 due in 10 years. (6) 2000(1.025) 10 due in 6 years, (c) 3000(1,04)* due in 4 years. $z due in 3 years. $z due in 4 years. 2$ MATHEMATICS OF INVESTMENT In the following equation of value the left member is the sum of the values of the old obligations on the comparison date. This sum must equal the sum of the values of the new obligations given in the right member. 1000(1.03)~ w + 2000(1.025) 10 (1.03)~ a + 3000(1.04) 4 = 3(1.03)' + as. (20) 6624.16 - z(1.0609) + a? = 2.0609 as. x = $3214.21. If 5 years from the present were used as the comparison date, the equation would be 1000(1.03)" 10 + 2000(1.025) 10 + 3000(1.04)*(1.03) 3 = 3(1.03)* + aj(1.03) a , (21) from which, of course, the same value of re is obtained because equation 21 could be obtained by multiplying both sides of equation 20 by (1.03) 8 . All obligations were accumulated for one more year in writing, equation 21 as compared with equation 20. EXERCISE XV Solve each problem by writing an equation of value. List the obliga- tions being compared. 1. W owes Y $1000 due after 4 years and $2000 due after 3 years and 3 months. "What sum paid now will discharge these debts if money is worth (.08, m = 4) to Y? 2. W desires to discharge his obligations in problem 1 by two equal payments made at the ends of 1 year and of 1 year and 6 months, respec- tively. Find the payments if money is worth (.06, m = 4) to Y. 3. "W 'desires to pay his obligations in problem 1 by three equal pay- ments made after 1, 2, and 3 years. Find the payments if money is worth (.06, m = 4) to Y. 4. What payment made at the end of 2 years will discharge the fol- lowing obligations jf money is worth (.05, m 2) : (a) $10,000 due after 4 years, and (6) $2000 due after 3J years with accumulated interest ait (.07,m = 2)? 5. If money is worth (.06, m = 2), determine the size of the equal pay- ments which, if made at the ends of the 1st and 2d years, will discharge the obligations of problem 4, 6. What sum, paid at the end of 2 years, will complete payment of the obligations of problem 4 if twice that sum was previously paid at the end pf the first year? Money is worth (.08, m = 2). interest at (.06, m >- 2) after 4J years. ' W,paid $1500 after 2 years, v of thi 'I 7. W owed Y $1000 due after 3 years, and $3000 due with accumulated What should he pay at the end of 3 years to cancel his debts if money is worth 7%, compounded semVannually, to Y?jJ( COMPOUND INTEREST 29 8. A man, owing the obligations (a) and (6) of problem 4, paid $8000 at the end of 3 years. What single additional payment should he make at the end of 5 years to cancel his obligations if money is worth (.04, m = 2) to his creditor? 9. Determine whether it would be to the creditor's advantage in problem 8 to stipulate that money is worth (.05, m = 2) to him. 16. Interpolation methods. The usual problem in compound interest, where the rate or the time is the only unknown quantity, may be solved approximately by interpolating in Table V. The method is the same as that used in finding a number N from a logarithm table when log N is known. Example 1. Find the nominal rate under which $2350 will accumulate to $3500 by the end of 4 years and 9 months, if interest is compounded quarterly. Solution. Let r be the unknown rate per period. The nominal rate will be 4 r. From equation 16, 3600 = 2360(1 + r) 19 ; (1 + r) 19 - jj? = 1.4894. i (1 + i) 1B .02 i = r .0225 1.4568 1.4894 1.5262 1.4668 to 1.5262. Hence r = .02 + |H(-0025) = .0212. The first and third entries in the table are from the row in Table V for n = 19. In finding r by interpola- tion we assume that r is the same proportion of the way from .02 to .0225 as 1.4894 is of the way from 1.4568 to 1.5262. Since 1.5262 - 1.4568 = .0694, and 1.4894 - 1.4568 = .0326, then 1.4894 is f$$ of the way from The distance from .02 to .0225 is .0226 .02 = .0025. The nominal rate is 4 r = .0848. Interest rates per period determined as above are usually in error by not more than 1 J'fr.of the difference between the table rates used. Thus, the value of r above is probably in error by not more than ^(.0025) or about .0001. The error happens to be much less, because a solution by exact methods gives r = .02119. Results obtained by interpolation should be computed to one more than the number of decimal places wfiich are ex- pected to be accurate. * r NOTB. When interpolating, it is sufficient to use only four decimal places of the entries in Table V. Use of more places does not increase the accuracy of the final results and causes unnecessary computation. 1 The author gives no theoretical justification for this statement. He has verified its truth for- numerous examples distributed over the complete range of T,bi,v. 8*-. 4 30 MATHEMATICS OF INVESTMENT Example 2. How long will it take $5250 to accumulate to $7375 if invested at (.06, m = 4) ? Solution. Let k be the necessary number of interest periods. 7375 - 5250(1.015)*; (1.015)* - JJ^ - 1.4048. The first and third entries in the table are from Table V. Since 1.4084 - 1.3876 - .0208, and 1.4048 - 1.3876 = .0172,. then k is tff of the way from 22 to 23, or k = 22 + tfft = 22.83 periods of 3 months. The time n 22 n = Te 23 (1.015)" 1.3876 1.4048 1.4084 is = 5.71 years. A value of k obtained as above 4 is in error by not more than J of the interest rate J per period. The error hi Example 2 is much less, because an exact solution of the problem gives k = 22.831. Example 3. X owes Y $1000 due after 1 year, and $2000 due after 3 years with accumulated interest at (.05, m = 2) . When would the pay- ment of $4000 balance X'B account if money is worth (.06, m = 4) to Y? Solution. Let k be the number of conversion periods of the rate (.06, m = 4) between the present and the date when $4000 should be paid. With the present as a comparison date, the equation of value for the obligations is 4000(1.015)7* = 1000(1.015)"* + 2000(1.026)'(1.015)~ u = 2882.09. (1.015)~* = .72052. From interpolation in Table VI, k = 22 + yfrfo. = 22.02. X should pay $4000 after 22^ - 5.60 years. Example 4. How long will it take for money to double itself if left to accumulate at (.06, m = 2) ? Solution. Let P = $1 and A = $2. If k represents the necessary num- ber of conversion periods, a solution by interpolation gives k * 23.44; the time is 11.72 years. Another approximate method is furnished by the follow- ing rule. Rule I. 2 To determine the time necessary for money to double itself at compound interest : (a) Divide .693 by the rate per period. (6) Add .35 to this result. The sum is the time in conversion periods. The error of this approximate result generally is less than a few hundredths of a period. On solving Example 4 by this rule, Jb- ^ + .36 = 23.45. .03 1 For justification of this statement BQO Appendix, Note 5. A knowledge of the calculus is necessary in reading this note. * For a proof of this rule see Appendix, Note 1, COMPOUND INTEREST EXERCISE XVI 1 Solve all problems by interpolation unless otherwise directed. In each problem in the table, find the missing quantity. PBOB. AMOUNT, A PRINCIPAL, P P AOOUMTTLA.THB FOB, OB A IB DUB AFTEB NOMINAI RATE CONVBHSIONB PHB YBAB 1. $2735 $1500 .05 1 2. 2500 2000 .06 2 3. 2 1 15 years 1 4. 1000 750 3 years, 9 mo. 4 5. 5010 4250 .07 2 6. 6575 4270 7 years, 6 mo. 2 7. 3000 1000 .05 2 ^ 8. Find the nominal rate under which $3500 -is the present value of $5000, due -at the end of 12J years. Interest is compounded semi- annually. j? * V V *) * : j ^ 9. How long will it take for money to quadruple itself if invested at (.06, TO = 2)? % 3- H-$ f^ 10. (a) At what nominal rate compounded annually will money double in 14 years? (&) Solve by use of Rule 1. ' 11. If money is worth (.07, m = 1), when will the payment of $4000 cancel the obligations (a) $2000 due after 3 years, and (&) $2000 due after 7 years? 12. If money is worth (.05, m = 2), when will the payment of $3000 cancel the obligations (a) $1500 due after 3 years, and (&) $1000 due at the end of 2 years with accumulated interest at (.06, m = 4) ? 15. By use of Rule 1, determine how long it takes" for money to double itself under each of the following rates : (a) (.06, m = 4) ; (&) (.04, m = 2) ; (c) (.06, TO -2); (d) (.03, m = 1). 14. By use of the results of problem 13, determine how long it takes for money to quadruple itself under each of the four rates in problem 13. 16. If money is worth (.04, m = 2), when will the payment of $3500 cancel the liabilities (a) $1000 due after 18 months, and (6) $2000 due after 2J years? i The Miscellaneous Problems at the end of the chapter may be taken up im- mediately after the completion of Exercise XVI. 32 MATHEMATICS OF INVESTMENT SUPPLEMENTARY MATERIAL 16. Logarithmic methods. Problems may arise to which the tables at hand do not apply, or in which more accuracy is desired than is obtainable by interpolation methods. Logarithmic methods are available in such cases. Example 1. Find the present value of $350.75, due at the end of 6 years and 6 months, if interest is at the rate (.0374, m = 2). Solution. P = 350.75(1.0187)-" = log 350.75 = 2.54500 13 log 1.0187 = 13 (.0080463) = 0.10460 (Using Table H) (subtract) log P = 2.44040. P = $275.68. If Table I were used in obtaining log (1.0187) 131 , 13 log 1.0187 = 13(.00804) = 0.10452, in error by.. 00008. Example 2. If interest is converted quarterly, find the nominal rate under which $2350 is the present value of $2750, due after 4 years and 9 months. Solution. Let r be the unknown rate per period ; the nominal rate is 4 r. 2750 - 2.350(1 + r) ; 1 + r log 2750 =2.43933 log 2360 =2.37107 log quot. = 0.06826. ^ log quot. = 0.00359. .-. 1 + r = 1.0083, r = .0083. The nominal rate is 4 r = .0332, converted quarterly. Exampk 3, How long will it take for $3500 to accumulate to $4708 if interest is at the rate (.08, m = 4) ? Solution. Let k be the necessary number of conversion periods. 4708 - 3500(1.02)*; (1.02)* - JJ& ..% fc log 1.02 = log log 4708 - 3.67284 log 1.02 = 0.0086002 log 3600 - 3.54407 log quot. = 0.12877. .-. fc(0.0086002) = .12877. fc - 12877 lo g 12877 = 4.10982 860.02' log 860.02 - 2.93451 (subtract) log fe= 1.17531 The time is k = 14.973 periods of 3 months, or 3.743 years, , , COMPOUND INTEREST LJ ' EXERCISE XVH Use exact logarithmic methods in all problems on this page. Use Table II whenever advisable. 1. Find the compound amount after 3 years and 3 months, if $3500 is invested at the rate (.063, m - 4). 2. Find the present value of $3500 which is due at the end of 8 years and 6 months, if money is worth (.078, m = 2). 3. At what nominal rate, converted quarterly, is $5000 the present value of $7300, due at the end of 2 years and 9 months? 4. Find the length of time necessary for a principal of $2000 to ac- cumulate to $3600, if interest is at the rate (.05, m = 1). 6. Solve problems 2 and 5 of Exercise XVI by exact methods. 6. Solve problems 3 and 4 of Exercise XYI by exact methods. V 7. Find the nominal rate which, if converted semi-annually, yields the effective rate .0725. '/ * / $- */ % 8. Find the nominal rate which, if converted semi-annually, is equiva- lent to the rate .068, compounded quarterly. 9. (a) Determine how long it will take for money to double itself at the rate (.06, m - 1). (6) Compare your answer with the result you obtain on using Hule 1 of Section 15. 10. One dollar is allowed to accumulate at (.03, m - 2). A second dollar accumulates at (.06, m = 1). When will the compound amount on the second dollar be three times that on the first? HINT. Take the logarithm of both aides of the equation obtained. 17. The equated time. The equated date for a set of obligations is the date on which they could be discharged by a single payment equal to the sum of the maturity values of the obligations. The time between the present and the equated date is called the equated time, and it is found by solving an equation of value. Example 1. If money is worth (.05, m = 2), find the equated time for the payment of the obligations (a) $2000 due after 3 years, and (6) $1000 due after 2 years with accumulated interest at the rate (.04, m = 2). Solution, The sum of the maturity values of (a) and (b) is 2000 + 1000(1.02)* - $3082.43. Let the equated time be k conversion periods oi 34 MATHEMATICS OF INVESTMENT the rate (.05, m = 2). The value of the obligation $3082.43, due after k periods (on the equated date), must be equal to the sum of the values of the given obligations. With the present as the comparison date, the corresponding equation of value is 3082.4(1.025}-* = 2000(1.026)- 8 + 1000(1.02)(1.026)-* = 2705.2. (1.025)* = = 1.1394. By the method of Section 16, k = 5.286 six-month periods or, the equated time is 2.643 years. By the interpolation method, k = 5.28. The present was used as the comparison date above to avoid having k appear on both sides of the equation. To obtain the equated time approximately, the following rule is usually used. Rule I. 1 Multiply the maturity value of each obligation by the time in years (or months, or days) to elapse before it is due. Add these products and divide by the sum of the maturity values to obtain the equated time. On using this rule in Example 1 above, we obtain equated time = 3(2(100)^2(1082.4) - 2.65 years. Rule 1 is always used in finding the equated date for short-term commercial accounts. The equated date for an account is also called the average date and the process of finding the average date is called averaging the account. Since Rule 1 does not involve the interest rate, it is unnecessary to state the rate when asking for the equated date for an account. NOTE. Results obtained by use of Rule 1 are always a little too large, so that a debtor is favored by its use. The accuracy of the rule is greater when the interest rate is low than when it is high. The accuracy is greater for short- term than for long-term obligations. EXERCISE XVm 1. If money is worth (.05, m 1), find the equated time for the pay- ment of (a) $1000 due after 3 years, and (6) $2000 due after 4 years. Solve by Rule 1. 2. Solve problem 1 by the exact method of Example 1 above. 1 For derivation of the rule see Appendix. Note 2. COMPOUND INTEREST 3. (I) If money is worth (.07, m = 2), find the equated time for the payment of (a) $1000 due after 3 years and (&) $2000 due after 4 years with accumulated interest at (.05, m = 2). Solve by the exact method (II) Solve by Rule 1. * 4. Find the equated time for an account requiring the payment of $55 after 3 months, $170 after 9 months, and $135 after 7 months. Use Rule 1 . 6. (a) A man owes four 180-day, non-interest-bearing notes dated as follows: March 9, for $400; May 24, for $250; August 13, for $525; August 30, for $500. By use of Rule 1 find the equated time and the equated date for the payment of the notes, considering for convenience that March 9 is the present. (&) How much must be paid on the equated date to cancel these obligations, if no other payment is made? 6. If money is worth 6%, simple interest, what should be paid 30 days after the equated date in problem 5 in order to balance the account, if no other payment is made? 18. Continuously convertible interest ; the force of interest. 1 The compound amount on $1 at the end of one year, if inter- est is at the nominal rate j, converted m times per year, is A <= (\ + 3-\ - It was seen at the end of Section 10 that, as m \ m/ increases, the amount A increases. As m increases without bound, or in other words, as m approaches infinity, the amount A does not increase without bound but approaches a limiting value e', where e = 2,7182818 + is the base of the Naperian, or natural, system of logarithms. To prove this we use the theory of limits. lim A = lim (l + iy = lim |7l +^77 = flim (l + tn-oo m =fo\ m/ m=ooL\ m/ J l_m=co\ W It is known that 2 lim (l + ^V = e - Therefore, m=A m/ . (22) It is customary to say that this limiting value e 1 ' is the com- pound amount on $1 at the end of one year in the ideal case where 1 A knowledge of the theory of limits is advisable in reading this section. 1 See Granville's Calculus, Revised Edition, page 22. 3 ; MATHEMATICS OF INVESTMENT interest is converted continuously. For every value of w, the effective rate i corresponding to the nominal rate j is given by i = |Yi _|_ 1Y"_ i~l Hence, in the limiting case where interest is converted continuously, it follows from equation 22 that i = lim ( 1 + iV" - 1 - & - 1. 7n=co \ m/ J + z = ei. (23) Example 1. Find the effective rate if the nominal rate is .05, con- verted continuously. Solution. 1 + i = e- OH . log (1 + i) - -05 log e, where e - 2.71828. .05 log e = .05(0.43429) = 0.02171 = log (1 + i). .-. 1+i- 1.0513; i- .0513. The force of interest, corresponding to a given effective rate i, is the nominal rate which, if converted continuously, will yield the effective rate i. Hence, if 5 represents the force of interest, the value j = 8 must satisfy equation 23, or 1 + 1 - e. . (24) Example 2. Find the force of interest if the effective rate is .06. Solution. 1.06 = e. .'. 5 log e = log 1.06. log 1.06 .0253059 loge .43429 Under the effective rate i, the compound amount of a principal P at the end of n years is A = P(l + i) n . If the nominal rate is j, converted continuously, (1 + i} = e*', hence A P(e*) n , or A = Pert. To compute A we use logarithms ; log A = log P + nj log e. EXERCISE XIX 1. Find the effective rate if the nominal rate is .06, converted con- tinuously. - 9 - J 2. Find the force of interest if the effective rate is .05. "* ' / J 3. (a) Find the amount after 20 years if $2000 is invested at the rate .07, converted continuously, (6) Compare your answer with the com- pound amount in case the rate is (.07, m => 4) . F * <t COMPOUND INTEREST 37 MISCELLANEOUS PROBLEMS 1. A man, in buying a house, is offered the option of paying $1000 cash and $1000 annually for the next 4 years, or $650 cash and $1100 annually for the next 4 years. If money is worth (.06, m = 1), which method is the better from the purchaser's standpoint? 2. A merchant desires to obtain $6000 from his banker, (a) If the loan is to be for 90 days and if the banker charges 6% interest in advance, for what sum will the merchant make out the note which he will give to the banker? (&) What simple interest rate is the man paying? 3. A merchant who originally invested $6000 has $8000 capital at the end of 6 years. What has been the annual rate of growth of his capital if the rate is assumed to have been uniform through the 6 years? 4. If gasoline consumption is to increase at the rate of 5% per year, when will the consumption be double what it is now? Solve by two methods. 5. When will the payment of $5000 cancel the obligations $2000 due after 3 years, and $2500 due after 6 years? Money is worth (.05, m = 2) . 6. A certain life insurance company lends money to policy holders at 6% interest, payable in advance, and allows repayment of all or part of the loan at any time. Six months before the maturity of a $2000 loan, the policy holder A sends a check for $800 to apply on his loan. What additional sum will A pay at maturity? HINT. First find the sum, due in 6 months, of which $800 is the present value. 7. A must pay B $2000 after 2 years, and $1000 after 3 years and 6 months. At the end of 1 year A paid B $1500. If money is worth (.05, m = 2), what additional .equal payments at the ends of 2 years and 6 months and of 3 years will cancel A's liability? 8. (a) If you were a creditor, would you specify that money is worth a high or a low rate of interest to you, if one of your debtors desired to pay the value of an obligation on a date before it is due? Justify your answer in one sentence. (6) If a debtor desires to discharge an obligation by making payment on a date after it was due, what rate, high or low, should the creditor specify as the worth of money? 9. At the end of 4 years and 7 months, $3000 is due. Find its present value by the practical rule if money is worth (.08, m = 4). 10. A man has his money invested in bonds which yield 5%, payable semi-annually. If he desires to reinvest his money, what is the lowest rate, payable quarterly, which his new securities should yield? < 38 MATHEMATICS OF INVESTMENT 11. One dollar is invested at simple interest, rate 5%. A second dollar ." is invested at (.05, m = 1). When will the compound amount on the 1 second dollar be double the simple interest amount on the first? j HINT. Solve the equation by interpolation; see illustrative Example 1 in < Appendix, Note 3. 12. After how long a time will the compound amount on $1 at the rate (.06, m = 2) be double the amount on a second $1 at the rate (.035, m = l)? HINT. Use either interpolation or logarithmic methods. See problem 11. 13. If $100 is invested now, what will be the compound amount after 20 years if the effective rate of interest for the first 5 years will be 6%, whereas interest will be at the rate (.04, m = 2) for the last 15 years? 14. A man owes $2000, due at the end of 10 years. Find its present value if it is assumed money will be worth 4% effective, for the first 5 years, and 6% effective, for the last 5 years. IB. If $100 is due at the end of 5 years, discount it to the present time, (a) under the rate 5%, compounded annually ; (6) under the simple interest rate 6% ; (c) under the simple discount rate 5%. CHAPTER III ANNUITIES CERTAIN 19. Definitions. An annuity is a sequence of periodic pay- ments. An annuity certain is one whose payments extend over a fixed term of years. For instance, the monthly payments made in purchasing a house on the instalment plan, form an annuity certain. A contingent annuity is one whose payments -last for a period of time which depends on events whose dates of occurrence cannot be accurately foretold. For instance, a sequence of pay- ments (such as the premiums on an insurance policy) which ends at the death of some individual form a contingent annuity. In Part I of this book we consider only annuities certain. The sum of the payments of an annuity made in one year is called the antwl rent. The time between successive payment dates is the payment interval. The time between the beginning of the first 'payment interval and the end of the last, is called the term of the annuity. Unless otherwise stated, all payments of an annuity are equal, and they are due at the ends of the. payment inter- vals; the first payment is due at the end of the first interval, and the last is due at the end of the term. Thus, for an annuity of $50 per month for 15 years, the payment interval is 1 month, the annual rent is $600, and the term is 15 years ; the first payment is due after 1 month, and the last, after 15 years. Under a specified rate of interest, the present value of an annuity is the sum of the .present values of all .payments of the annuity. The amount of an annuity is the sum of the compound amounts that would be on hand at the end of the term if all payments should accumulate at interest until then from the dates on which they are due. NOOTB 1. Consider an annuity of $100, payable annually for 5 years, with interest at the rate 4%, effective. We obtain the present value A of this annuity by adding the 2d column in the table below, and the amount S by adding the 3d column. ^ -"" 39 Li 40 MATHEMATICS OF INVESTMENT PAYMENT OF $100 DUB AT END OF PRESENT VALUE OP PAYMENT COMPOUND AMOUNT AT END OF TERM IF PAYMENT is LBFT TO ACCUMULATE AT INTEREST 1 year 2 years 3 years 4 years 6 years 100C1.04)" 1 = 96.16385 100(1. 04) ""* = 92.46562 100(1.04) "* = 88.89964 100(1. 04) ^ = 86.48042 100(1.04)~ 5 - 82.19271 100(1.04)* = 116.98586 100(1.04) s = 112.48640 100(1.04) 9 = 108.16000 100(1.04) = 104.00000 100 = 100.00000 (add) A = $445.18224 (add) S = $641.63226 The present value A = $446.18 is as desirable as the future possession of all payments of the annuity. The amount S = $541,63, possessed at the end of 5 years, is as desirable as all of the payments. Hence, A should be the present value of S, due at the end of the term, or we should have S = A (1.04) 5 . This relation is verified to hold ; A(1.04)= = 446.182 X 1.21665290 = $541.632 = S. (25) NOTE 2. In the table below it is verified that, if a fund is formed by investing $445.182 at 4% effective, this fund will provide for all payments of the annuity of Note 1 and, in so doing, will become exactly exhausted at the time of the last payment. This result can be foreseen theoretically because $445.182 is the -sum of the present values of all of the payments. YEAR IN FUND AT BEGINNING OF YEAS INT. AT 4% ' Dux AT END OF YEAH IN FUND XT END OF YEAB BEFORE PAYMENT IB MADE PAYMENT AT END OF YEAR 1 $445.182 $17.807 $462.989 $100. 2 362.989 14.520 377.509 100 3 277.509 11.100 288.609 100 4 188.609 7.544 196.153 100 5 96.153 3.846 99.999 100 EXERCISE XX 1. (a) Form a table as in Note 1 above in order to find the present value and the amount of an annuity which pays $1000 at the end of each 6 months for 3 years. Money is worth 6%, compounded semi-annually. (6) Verify as in equation 25, that A is the present value of S, due at the end of the term, (c) Form a table as in Note 2, to verify that the present value A, if invested at (.06, m = 2), creates a fund exactly sufficient to provide the payments of the annuity. ANNUITIES CERTAIN 41 20. The examples below 1 illustrate methods used later to obtain fundamental annuity formulas. Example 1. If money is worth (.06, m = 4), find the present value A and the amount S of an annuity whose annual rent is $200, payable seml- annually for 15 years. Solution. Each payment is $100. The entries in the 2d and 4th columns below are verified by the principles of compound interest. PAYMENT OF $1OO Dms AT THE END OF PHHSENT VALUE OP PAYMENT TIME FROM DATE OF PAYMENT TO END OF TERM COMP. AMT. AT END op THHM IF PAYT. ifl LEFT TO ACCUMULATE AT INT. 6 months 1 year etc. 14 yr., 6 mo. 15 years 100(1.016)~ J 100(1.015) * etc. 100(1.015)~ 88 100(1.015) H10 14 yr., 6 mo. 14 years etc. 6 months months 100(1.015)" 100(1.015)" etc. 100(1.015) a 100 - Sum = A Sum = S Hence, S = 100[1 + (1.015) 3 + etc. + (1.015) + (1.015)"]. The bracket contains a geometrical progression of 30 terms for which the ratio is w - (1.015) 2 , the first term a = 1, and the last term L = (1.015) 68 . By the formula for the sum of a geometrical progression, 2 .2.44321978 - 1 100 : 84774.918. 1.03022500 - 1 On adding the 2d column in the table we obtain the present value A = lOOUl.QlS)^ + (1.015)- 58 + etc. + (1.015)-* + (1.015).-*]. The geometrical progression, in .the bracket has the ratio w = (1.015) 3 , while a = (LOIS)- 80 and L = (1.015)-*. Since wL - 1, L l - (1.015)-* = 1rtn 1 - .40929597 _ 1Q0 1.03022500 - 1 (1.015) 1 - 1 The present value of $4774.918, due at the end of 16 years, should equal A, orS = A(1.016) M . We verify that A (1.015)8 = (1954.356) (2.44321978) - $4774.920. (26) " l Geometrical Progressions in Part III, Chapter XII, should be studied if the student has not met them previously. Section 20 may be omitted without dis- turbing the continuity of the succeeding sections, but geometrical progressions are needed in Sections 21 and 22, > See Part III, Section 90. 42 MATHEMATICS OF INVESTMENT Example 2. Find the present value A of an annuity of $100 pei month for 3 years and 6 months, if money is worth (.05, m = 2) . Solution. A is the sum of the entries in the 2d row below. Payment of $100 due after 1 month 2 months etc. 3 yr., 5 mo. 3 yr.j 6 mo. Present value of payment 100(1.025)'^ 100(1 .025)'* etc. 100(1.025)-^ 100(1.025)-^ A - 100 [(1.025)-^ + (1.026)-* + - etc. - + (1.025)-* + (1.025)-*]. The ratio of the geometrical progression is w = (1.025)*; a = (1.025)~ T am L = (1.025)-*. 'Since wL ~ a = 1 - (1.025)- 7 , "~ 1QQ 1 - (1.025)-* 1Q0 ^- .84J.26S24 , ^^ (TablesVIajldX (1.025)i - 1.00412392 - I EXERCISE XXI In each problem derive formulas for A and S for the annuity describee using the method of Examples 1 and 2 above. *fr\ An annuity whose annual rent is $200, payable quarterly for 1 years. Money is worth (.08, m = 4) . Compute the formulas for A and and verify as in equation 26 that A is the present value of S, due at tt end of the term. /\ ~ S & * LrJL fi ~Jy a tf ^ "^ ^ ^ S . tf 6 ^ i 2. Fifteen successive annual payments of $1000, the first due aft< 1 year. Money is worth (.05, m = 2). Compute A and S and veril that A is the present value of S, due at the end of the term. 3. Payments of $100, made at the end of each 3 months for 15 year Money is worth (.05, m = 4). 4. (a) The annual rent of the annuity is $2000, the payment interv is 3 months, and the term is 12^ years. Money is worth (.06, m = 1 (6) Solve the problem if money is worth (.06, m = 2). 5. An annuity which pays $100 at the end of each interest period f 10 interest periods. Money is worth .045, per interest period. 21, Formulas for A and S in the most simple case. Co: sider the annuity paying $1 at the end of each year for n year Let (a/n\ at i) be the present value, and (Sn\ at i) be the amount this annuity when money is worth the rate i compounded annuall The entries in the table below are easily verified. ANNUITIES CERTAIN PAYMENT OP SI DUB AT THE END OP PHHSBNT VALUB OP THB PAYMENT TIME FBOM DATE OP PAYMENT TO END OFTBBM COMP. AMT. AT END OP THEM n- PAYT. IB LEFT TO ACCUMULATE AT INT. 1 year 2 years etc. etc. (n. 1) yr. (n - 2) yr. etc. (1 + i)"- J etc. (n - 1) yr. n years d + tr^ 1 - 1 year years. (1+0 1 Sum = (an\at i) Sum = (&n\ati) Hence, (85! 0*1) = 1 + (1 + i) + ..- etc. - + (1+ i)^ 2 + (1 + t)*~ x . This is a geometrical progression where the ratio 10= (1 + i), the first term a = 1, and the last term L = (1 + i) n-1 . Since (wL - a) = (1 + i)* - 1, and (w - 1) = i, the formula wL ~ a iw 1 = v + y - *. (27) On adding the 2d column of the table we obtain (o^oti) = (i+;r n +u+;r n+1 + - etc. . which is a geometrical progression with the ratio w = (1 + i), a = (1 + i)" n , and L = (1 + i)" 1 . Since (wL - a) = [(1 + i)(l + i)- 1 - - (1 + iT"], and (w - 1) = i, cfi) (28) If each payment of the annuity had been $# instead of $1, the present value A and the amount S would have been A = R(a^ati} and S = R(8n\at i). It is important to realize that formulas 27 and 28 may be used whenever the payment interval of the annuity equals the con- version period of the interest rate, In deriving the formulas, the interest period was called 1 year, merely for concreteness. Hence, if i is the interest rate per period, then R(Sn\ at i) represents the amount and R(a^ at i) the present value of an annuity which pays $R at the end of each interest period for n periods. Thus, 44 MATHEMATICS OF INVESTMENT at .025) is the present value of an annuity paying $100 at the end of each interest period for 18 periods if money is worth the rate .025 per period. Example 1. Find the amount and the present value of an annuity paying 1150 at the end of each 3 months for 15 years and 6 months, if money is worth 6%, compounded quarterly. Solution. Since the payment interval equals the interest period, formulas 27 and 28 apply with the number of payments n = 62, and with i = .016. Amount = 160(s m at .016) = 150(101.13773966) = $15170.66. Pr. val. = 150(0^ at .015) - 150( 40.18080408) = $6027.12. The value of s^ is from Table VII and that of ian is bom Table VIII. NOTE. - Recognize that the solution above makes' no use of the explicit expressions for 05^1 and s^ because their values are tabulated. The use of the explicit formulas for o^ or a^ in such a case would be a complicated, and therefore an incorrect method. EXERCISE XXE 1. (a) In Table VII verify the entry for (^ at .02) = (1 ' 2) J! ~ 1 .02 by use of Table V. (&) Verify the entry for (0^ at .04) in Table VTH by use of Table VI. 2. Find the present value and the amount of an annuity which pays $500 at the end of each year for 20 years, if money is worth (.05, m => 1). 3. If money is worth (.05, m = 2), find the present value and the amount of an. annuity whose annual rent is $240, payable semi-annually for 13 years and 6 months. Find the present values and the amounts of the annuities below. PBOB. EACH PAYMENT PAYMENT INTERVAL THEM ANNUAL RUNT INTEREST RATH 4. $ 50 3 mo. 14 yr., 9 mo. .06, m = 4 5. 10,000 1 yr. 18 yr. .065, m = 1 6. 500 6 mo. 19 yr., 6 mo. .07, m = 2 7. 6 mo. 15 yr. $1000 .055, m = 2 8. 300 , 1 vr. 25 yr. .04, m - 1 &, 6 mo. 23 yr. 2000 .03, m <- 2 10. 1 mo. 7yr. 2400 .06, TO 12 ANNUITIES CERTAIN In purchasing a house a man agrees to pay $1000 cash and at the end of each 6 months for the next 6 years. If money is worth (.07,. m - 2), what would be an equivalent cash valuation for the house? HINT. The cash price is the sum of the present values of all payments. The present value of the first payment is $1000. The remaining 12 payments come at the ends of the payment intervals and hence form a standard annuity .whose present value is W00(aj^ at .035). 12. The man of problem 11 has just paid the installment due at the end of 4 years and 6 months. What additional payment, if made immediately, would cancel his remaining indebtedness if money is worth (.08, m = 2) ? HINT. His remaining indebtedness at any time, or the principal out- standing, is the present value of all remaining payments. 13. If you deposit $50 at the end of each 3 months in a savings bank which pays interest quarterly at the rate 3%, how much will be to your credit after 20 years and 6 months, if you make no withdrawals? 14. A man in buying a house has agreed to pay $1000 at the beginning of each 6 months until 29 installments have been paid. If money is worth 6%, compounded semi-annually, what is an equivalent cash price for the house? 15. At the end of each year a corporation places $5000 in a depreciation fund which is to provide for plant replacement at the end of 12 years, (a) What sum will be in the fund at the end of 12 years if it accumulates at the effective rate 7% ? (&) What sum is in the fund at the beginning of the seventh year? 16. A man desires to deposit with a trust company a sufficient sum to provide his family with $500 at the end of each 3 months for the next 15 years. If the trust company credits interest at the rate 6%, quarterly, on all funds, what should the man deposit? HINT. See the table of Note 2 of Section 19. 22. Further annuity formulas. Consider the annuity whose annual.rent is $1, payable p times per year for n years. Each of the np payments is ^ ; the first is due at the end of - years, and the others are due at intervals of - years for the rest of the term. If money is worth the rate i, compounded annually, let (s at i} represent the amount of the annuity and (og, at i) its present 46 MATHEMATICS OF INVESTMENT we form the table value. To derive formulas for a^ and below. SI PAYMENT OF DUE AT THE END OF PRESENT VALUE OF THE PAYMENT Too, FBOM DATE OF PAYMENT TO END OF TERM COMF. AMT. AT END OF THBM IF PAYT. IB LEFT TO ACCUMULATE AT INTBBBST 1 - years P P (n-^yr. P 2 -years P P (o\ n ) yr. P' P etc. etc. etc. etc. ('-;)" P - years P P n years P years P Hence, on adding the fourth column we obtain The progression in the bracket has the ratio w = (1 + i)p, the first term a = 1, and the last term L = (1 + i) n ~5. Since wL a = (1+ i) n - 1, andw - 1 = (1 + i^ - 1, = ^ + V ~ ' . (29) P[(l + 1)3 - JT] The denominator of the last fraction is the expression we have previously called 1 (j p at ) . On multiplying numerator and denom- inator of the last fraction by i, we obtain Since, by formula 27, the last fraction is (s^ at i) t (30) 1 See Exercise X, Problem 16. Also see heading of Table XI. The fact that O'n at t) is the nominal rate which, if converted p times per year, yields the effective rate -i, is of importance in the applications of equation 29. We use j f merely as a convenient abbreviation for its complicated algebraic expression. ANNUITIES CERTAIN 47 From the second column of the table we obtain (off ati) = %l + t)- + (1 + i^* 1 * + - etc. - + (1 + i)-Jj.' p J The ratio of the geometrical progression in the bracket is w = (\ + 0*i the first term a = (1 + 0"*, and L = (1 -f 0~X Since wL - a = 1 - (1 + i)~*, From this expression we derive, as in formula 30, (<#>af lO-lfo of i). (32) fa If the sum of the payments made in 1 year, or the annual rent, had been $12 instead of $1, the present value of the annuity would have been R(a$ at i) and the amount, R(&j\ at {). In the discussion above, money was worth the rate i, compounded once per year, while the annuity was payable p times per year for n years and the sum of the payments made in 1 year was $1. The word year was used in this statement and in the proof of formulas 29 to 32 for the sake of concreteness. All of the reasoning remains valid if the word year is changed throughout to interest period. Thus, when money is worth the rate i, per interest period, if an annuity is payable p times per interest period for a term of n interest periods, and if the sum of the payments made in one in- terest period is $R, the present value A and the amount S are given by ? (33) S = R(s^ati) =R4-(s^ati). Jp Example 1. If money is worth (.05, m = 2), find A and S for an annuity of fifty quarterly payments of $100 each, the first due at the end of three months. 48 MATHEMATICS OF INVESTMENT Solution. Payments occur twice in each interest period. Hence, use formulas 33 with the data listed below. Tables XII, VIH, and VII are used in computing. n = 2(12.6) 25 int. periods, p =2, R - $200, i t = .025. A - 200(0^ at .025) = 200 ^(o^ .025), A= 200(1 .00621142) (18.42437642), A - $3707.76. -025 / B - 200(a^ at .025) - 200^(8^ of .026) = 200(1.00621142) (34.15776393), S = $6873.99. Example 2. If money is worth (.06, m = 4), find A and 5 for an annuity whose annual rent is $1000, payable monthly for 12 years and 3 months. Solution, Use formulas 33, because the payments occur three times in each interest period. 4(12^) = 49 int. periods, 3, R = $250, i - .015. A 250(ogai .015) = 250 ^(a^ at .015), A =250(1.00498346) (34.52468339). 8 - 260(g, of .015) - 250^(s^ of .015) - 250(1.00498346) (71.60869758). NOTHJ. When p = I, formulas 29 and 31 reduce to formulas 27 and 28, or (s^j* at i} = (s^ at i) and (aj^ erf i) = (o^ oi i) . We may obtain the same results on placing p = 1 in formulas 30 and 32, because (j\ at i) 1[ (1 + ^) - 1] - i and ^-= 1. These results could have been foretold because, when Ji p = 1, the payment interval equals the interest period, and hence formulas 27 and 28 apply as well as formulas 29 and 31. In the future think of (s^ erf i) as (a~? at i), with the value of j> left off and understood to be p = 1 (just as we omit the exponent 1 in algebra when we write x instead of x l ). Thus, formulas 30 and 32 express the present value and the amount of an annuity payable p tunes per interest period in terms of the present value and the amount of an annuity payable once per interest period. EXERCISE XXm 1. Verify the entry in Table XII for i = .06 and p = 2. HINT. ^ =s '^i. . from Table XI. Complete the division. 2, If money is w.orth (.06, m =2), find the present value and the amount of an annuity whose term is 9 years and 6 months, and whose rent is $1200, payable monthly. ANNUITIES CERTAIN (49O Compute the present values and the amounts of the annuities below. PHOB. ANNUAL RENT EACH PAYMENT PAYMENT INTERVAL TEBM INTEREST BATE 3. $1000 6 mo. 15 yr. t .05, TO = 1 4. 6000 1 mo. 12 yr. .06, m = 1 6. $500 6 mo. 9 yr., 6 mo. .07, m - 2 6. 226 3 mo. 19 yr. .05, TO = 1 7. - 200 3 mo. 8 yr., 6 mo. .08, TO = 2 8. 2000 3 mo. 10 yr., 6 mo. .055, TO = 2 9. 600 1 mo. 6 yr., 3 mo. .06, m = 4 10. 760 3 mo. Syr. .04, m = 1 i 11. In buying a farm it has been agreed to pay $100 at the end of each month for tHe next 25 years. If money is worth the effective rate 7%, what would be an equivalent cash valuation for the farm?*/^ <*/-_'?, * '"1 12. If $50 is deposited in a bank at the end of every month for the next 15 years and is, left to accumulate, what will be on hand at the end of 15 years if the bank pays 6%, compounded annually on deposits f )Jfa $ **-'. 6> 13. A sinking fund is being accumulated by payments of $1000, made at the end of each 3 months. Just after the 48th payment to the fund has been made, how much is in the fund if it accumulates at (.045, m 1) ? 14. An investment yields $50 at the end of each 3 months, and pay- ments will continue for 25i years. What is a fair valuation for the project if money is worth (.05, m 2) ? 16. How much could a railroad company afford to pay to eliminate a dangerous crossing requiring the attention of two watchmen, each re- ceiving $75 per month, if money is worth (.04, m = 1) ? Assume that the crossing will be used for 50 years. 16. Prove from formula 29, that (dfi at i) = -?-. Thus - is the sum which, if paid at the end of 1 year, is equivalent to p payments of made at equal intervals during the year. 23 , The most general annuity formulas . Consider the annuity whose annual rent is $1, payable p times per year for n years. To find the present value and the amount of this annuity when money is worth the nominal rate j, compounded m times per year, we might first compute the corresponding effective rate i and then use formulas 29 and 31. It is better to use equation 17 to obtain 50 MATHEMATICS OF INVESTMENT entirely new formulas in terms of the given quantities j and m. From equation 17, (i + i) = (l + } ; (l + i)-" = (l + \ m/ \ m On substituting these expressions in formulas 29 and 31 we obtain (34) If the annual rent of the annuity above were $R instead of $1, the present value would be R(a$ atj, m) and the amount would beRdJjgatj, m). NOTE. Formulas 34 include all previous formulas as special cases, because wnenwt - landj = i, formulas 34 reduce to formulas 29 and 31, from which we started. Thus, think of ( at $ as being ($> at j = i, m = 1) with the value of m left out and understood to be m - 1. Likewise, (a-, at i) = (a^? at 3 = Vm = 1). ^i ' x ^l 24. Summaiy. For an annuity under Case 1 below we usually may compute the present value A and the amount 8 by means of our tables. For an annuity under Case 2, the explicit formulas for A and S must be computed with much less aid from the tables. Case 1. The annuity is payable p times per interest period where p is an integer. The method of Section 22 applies, with additional simplification when p 1. If P = the number of payments per interest period, n = the term, expressed in interest periods, i = the rate, per interest period, and $R = the sum of the payments made in one interest period, then A = Z(a% at f) = R ati) S = *(,> at i) = R- at f. I ANNUITIES CERTAIN 51 When p = 1, $R is the annuity payment, n is the number of pay- ments, and ti). (H) The values of A and S in I and II can usually be computed by Tables VII, VIII, and XII. NOTE. One or more of Tables VII, VIII, and XII will not apply if i is not a table interest rate, or if n is not an integer. In that case the explicit formulas 29 and 31 for (a^ at i) and (^ at i) must be computed. Case 2. The annuity is not payable an integral number of times per interest period. The general formulas 34 must be used, and if n = the term in years, p = the number of payments per year, $R = the annual rent, j = the nominal rate, and m = the number of conversion periods per year, then S = tf atj, m) = *[(' + ) J - 'J STJPPLHMENTAHY NOTE. From formulas II and III it can be proved that / \ m These formulas can be used to simplify the computation of the present values and the amounts of many annuities coming under Case 2. Other sim- plifying formulas could be derived but they would not be of sufficiently gen- eral application to justify their consideration. Example 1. An annuity will pay $500 semi-annually for 8 years. Find the present value A if money is worth (.06, m = 4). Solution. The annuity comes under Case 2. A - 1000(a^ at .06, m - 4), Case 2 n = 8 years, p 2, j - .06, m *= 4, JB - $1000. 52 MATHEMATICS OF INVESTMENT Exampk 2. In buying a house a man has agreed to pay $1000 cash, and $200 at the end of each month for 4 years and 3 months. If money is worth (.06, m =.2), what would be an equivalent cash price for the property? Solution. First disregard the cash payment. The other payments form an annuity under Case 1 whose present value is Casel 7i = 8.5 int. periods, p = 6, i = .03, R - $1200. A = 1200(a^ at .03) = 1200 ^(Og^ a< -03), A = 1200(1.01242816) 1 A ~ (1.03)-"- B = (1.03)~ 9 (1.03)* = (.76641673) (1.01488916) = .7778280. A = 1200(1.01242816) (1 - .7778280) _ jg 997 33 .03 The equivalent cash price is $1000 + $8997.30 = $9997.33. Example 3. At the end of each 3 months a man deposits $60 with a building and loan association. What sum is to his credit at the end of 4 years if interest is accumulating at the rate (.075, m = 2), from the date of each deposit? Solution. The amount on hand is the amount of an annuity which comes under Case 1. Casel n 8 int. periods, p = 2, i - .0375, R = $100. S = 1000$' at .0375), S = 100 (1-0375) 8 - 1 . (Formula 29) 2[(1.0375)* - 1] } log (1.0375) = 0.0079940, from Table II. S (1.0375)* = 1.018577, from Table II. 8 log (1.0375) = 8C0159881) = 0.12790. S = = $921.8. .037154 (1.0375) 8 = 1.3425, from Table I. The answer is not stated to five digits because the numerator 34.25 was obtain- able only to four digits from Table I. NOTE. In every problem where the present value or amount of an annuity is to be computed, first list the case and the elements of the annuity as in the examples above. NOTE. To find A and S for an annuity we could always proceed as under Case 2, even though the annuity comes under Case 1. Thus, for the annuity of Example 3 above, the term is n = 4 years, the annual rent is R = $200, payable p = 4 times per year, j = .075, and m = 2. Hence, from formulas in of Case 2, S = 2000$ at .075, m = 2) = 200 P- 0375 ) 8 ~ 1 , 4[(1.0376) - 1] ANNUITIES CERTAIN 53 which is the same as obtained above. The only difference in method is that, under Case 2, the fundamental time unit is the year, whereas under Case 1 it is the interest period. The classification of annuity computations under two cases would not be advisable if we were always to compute A and S by the explicit formulas, as is necessary in Example 3. But, if we used the general formulas of Case 2, with the year as a time unit, in problems under Case 1 to which Tables VII, VIII, and XII apply, unnecessary computational confusion would result and other inconvenient auxiliary formulas would have to be derived. Hence, use the method of Case 1 whenever possible. EXERCISE XXIV Compute A and S for each annuity in the table. Use Table II when it is an aid to accuracy. PHOB. ANNUAL RENT EACH PAYMENT PAYMENT INTERVAL TERM INTEREST RATE 1. $10,000 1 month 15 years .05, m = 4 2. $ 400 1 month 12 years .06, m = 1 3. 2500 6 months 19 yr., 6 mo. .05, m = 4 4. 500 8 months 7 yr., 6 mo. .05, m = 2 6. 240 3 months 11 yr., 6 mo. .04, m = 4 6. 150 1 year 18 years .09, m = 4 7. 100 6 months 28 years .05, m = 1 8. 5,000 3 months 6 yr., 9 mo. .07, m =2 9. 125 6 months 10 years .0626, m = 1 10. 2,000 1 year 15 years .05, m = 2 11. 900 3 months 9 yr., 3 mo. .08, m = 4 12. 700 6 months 20 years .005, m = 1 13. 50 3 months 30 years .048, m =2 14. 100 4 months 9 years .04, m = 2 16. 3,000 3 months 12 years .04, m = 2 16. 500 1 year 35 years .07, m =2 17. 200 6 months 12 years .055, m = 2 18. 1,200 6 months 15 yr., 6 mo. .03, m = 4 19. 150 3 months 6 yr., 3 mo. .06, m = 2 20. 250 4 months 9 years .04, m = 2 21. ' 500 6 months 10 years .045, m = 2 22. 25 1 month 17 years .06, m = 1 23. To provide for the retirement of a bond issue at the end of 20 years, a city will place $100,000 in a sinking fund at the end of each 6 months, (a) If the fund accumulates at the rate (.05, m = 2), what sum will be available at the end of 20 years? (&) What sum is in the fund at the beginning of the 12th year? 54 MATHEMATICS OF INVESTMENT 24. An investment will yield $50 at the end of each month for the next 15 years. If money is worth (.05, m = 4), what would be a fair present valuation for the project? 25. A depreciation fund is being accumulated by semi-annual deposits of $250 in a bank which pays 5%, compounded quarterly. How much will be in the fund just after the 30th deposit? 26. A will decrees that X shall receive $1000 at the beginning of each 6 months until 10 payments have been made. If money is worth (.06, m = 2), on what sum should X's inheritance tax be computed, assuming that the payments will certainly be made ? 27. A certain bond has attached coupons for $5 each, payable at the end of each year for the next 25 years. If money is worth 5% effective, find the present value of the coupons. 28. The bond of problem 27 will be redeemed for $100 by the issuing corporation at the end of 25 years. What should an investor pay for the bond if he desires 5% effective on his investment? HINT. He should pay the present value of the coupons plus the present value of the redemption price. 29. A farm is to be paid for by 10 successive annual installments of $5000 in addition to a cash payment of $15,000. What is an equivalent cash price for the farm if money is worth (.05, m = 2) ? SO. (a) At the end of the 5th year in problem 29, after the payment due has been made, the debtor wishes to make an additional payment immediately which will cancel his remaining liability. The creditor is willing to accept payment if money is considered worth 4% effective. What does the debtor pay? (6) Why should the creditor specify the rate 4% effective instead of a higher rate, (.05, m = 1) for instance? HINT. Find the present value of the remaining payments. 81. A man has been placing $100 in a bank at the end of each month ' for the last 12 years. What is to his credit if his savings have been ac- cumulating at the rate 6%, compounded semi-annually from their dates \ of deposit? x 32. A man wishes to donate immediately to a university sufficient money to provide for the erection and the maintenance, for the next 50 years, of a building which will cost $500,000 to erect and will require $1000 at the end of each month to maintain. How much should he donate if the university is able to invest its funds at 5%, converted semi-annually ? ANNUITIES CERTAIN 55 33. A certain bond has attached coupons for $5 each, payable semi- annually for the next 10 years. At the end of 10 years the bond will be redeemed for $125. What should an investor pay for the bond if he desires 6%, compounded semi-annually, on his investment? HINT. See problem 28. 34. A man, who borrowed a sum of money, is to discharge the liability by paying $500 at the end of each 3 months for the next 8 years. What sum did he borrow if the creditor's interest rate is (.055, m 2) ? 36. (a) In problem 34, at the end of 4 years, just after the installment due has been paid, what additional payment would cancel the remaining liability if money is still worth (.055, m = 2) to the creditor? (&) What would be the payment if money is worth (.04, m = 4) to the creditor? 36. A man X agreed to pay $1000 to his creditor at the end of each 6 months for 15 years, but defaulted on his first 7 payments, (a) What should X pay at the end of 4 years, if money is worth (.06, m = 2) to his creditor? (6) What should he pay if money is worth (.05, m = 2) ? . 37. In problem 36, at the end of 4 years, X desires to make a single payment which will cancel his liability due to his previous failure to pay, and also will discharge the liability of the payments due in the future. (a) What should he pay if money is worth (.06, m = 2) to his creditor? . (6) Find the payment if the rate is (.05, m = 2). 38. A certain bond has attached coupons of $2 each, payable quarterly for the next 20 years, and at the end of that time the bond itself will be redeemed for $110. What should a man pay for the bond if he considers money worth 6%, -effective? 39. Prove by use of formulas 34 that the present value of an annuity, accumulated at the rate (j, m) for n years, will equal the amount of the annuity ; that is, prove algebraically that atj, m)l +- = B( atj, m) = S. Another statement of this result would 'be that "A is the present value of S, due at the end of the term of the annuity," 40. What is the amount of an annuity whose term is 14 years, and whose present value is $1575, if interest is at the rate (.06, m = 2) ? HINT. Use the result of problem 39. 41. What is the effective rate of interest in use if the present value of an annuity is $2500, the amount $3750, and the term 10 years? 56 MATHEMATICS OP INVESTMENT 42. A will bequeaths to a boy who is now 10 years old, $20,000 worth of bonds which pay 6% interest semi-annually. The will requires that half of the interest shall be deposited in a savings bank which pays 4%, compounded quarterly. The accumulation of the savings account, and the bonds themselves, are to be given to the boy on his 25th birthday. Find the value of the property received by him on that date. 43. A man desires to deposit with a trust company a sufficient sum to provide his family an annuity of $200 per month for 10 years. What should he deposit if the trust company will credit interest at the rate 5% compounded quarterly, on the unexpended balance of the fund? 44. If you can invest money at (.03, m = 2), what is the least sum you would take at the present time in return for a contract on your part to pay $100 at the end of each 6 months for the next 15 years? 45. If money is worth (.04, m = 1), is it more profitable to pay $100 at the end of each month for 3 years as rent on a motor truck, or to buy one for $3000, assuming that the truck will be useless after 3 years? Assume in both cases that you would have to pay the upkeep. 25. Annuities due. The payments of the standard annuities considered previously were made at the ends of the payment intervals. An annuity due is one whose payments occur at the beginning of each, interval, so that the first payment is due im- mediately. The definitions of the amount and of the present value of an annuity as given in Section 19 apply without change of word- ing to an annuity due. It must be noticed, however, that the last payment of an annuity due occurs at the beginning of the last interval, whereas the end of the term is the end of this interval. Hence, the amount of an annuity due is the sum of the compound amounts of the payments one interval after the last payment is ,made. For an annuity due whose annual rent is $100, payable quarterly for 6 years, the last $25 payment is made at the end of 5 years and 9 months. The amount of this annuity is the sum of the compound amounts of the payments at the end of 6 years, the end of the term. For the treatment of annuities due and for other purposes in the future, it is essential to recognize that, regardless of when a sequence of periodic payments start, they will form an ordinary annuity if judged from a date one payment interval before the first payment. Hence, one interval before the first payment, the ANNUITIES CERTAIN 57 sum of the discounted values of the payments is the present value of the ordinary annuity they form. Moreover, the sum of the accumulated values of the payments on the last payment date is the amount of this ordinary annuity. Example 1. If money is worth (.05, m = 2), find the present value A and the amount S of an annuity due whose annual rent is $100, payable quarterly for 6 years. Q > increasing time . 1 X X X K X X X X X X X X X )( X X X X )( )( )( ?( )( X -I N L represents 3 months N represents the present FIG. 3 Solution, Consider the time scale in Figure 3, where X represents a payment date, T is the end of the term, 6 years from the present, and L is the last pay- ment date, 3 months before T. Q is 3 months before the present. Considered from Q the payments form an ordinary annuity whose term ends at L and whose present value A' and amount S' are Case 1 12 int. periods, A' = 50(0^0^.025), Since A' is the sum of the discounted values of the payments at Q, 3 months before the present, we accumulate A' for 3 months to find A, the present value of the annuity due. A - ^'(1.025)* - 50(og,ai .025) (1.025)*. Since S' is the sum of the accumulated values of the payments at L, we ac- cumulate S' for 3 months to find S, which is the sum of the values at time T. S = S'(1.Q25)* = 50(3^ at .025) (1.025)*. Tables VII, VIII, X, and XII would be used to compute A and S. Second solution. The first $25 payment is cash and the remaining pay- ments form an ordinary annuity, as judged from the present. Its present value A ' is Case 1 n = 11.5 int. periods, p = 2, i - .025, R = $50. 4' = 50(0^ of .025). Hence, A = 25 + 60(ogL at .025). To find S, first consider a new annuity consisting of all payments of the an- nuity due, with an additional $25 due at tune T. Since T is the last payment date of the new sequence of payments, the sum of their values at time T is the amount S f of an ordinary annuity, or 58 MATHEMATICS OF INVESTMENT Case 1 n = 12.5 int. periods, p - 2, i = .026, R = $50. fl' - SOCsgj, a* .025). The value of the additional $25 payment at time T is included in S', or S' = S + 25. S - 60(8^, at .025) - 25. To find the nu- merical values of A and S, a^-Q and fij^ must be computed from formulas 29 and 31. Hence, the first solution was less complicated numerically. In some problems, however, the second solution would be the least complicated. Two rules may be stated corresponding, respectively, to the two methods of solution considered above. Rule 1. To find A and S for an annuity due, first find the present value A' and the amount S' of an ordinary annuity having the same term, a.nmifl.1 rent, and payment interval. Then : (a) A is the compound amount on A' after one payment interval. (6) S is the compound amount on S' after one payment interval. Rule 2. To find A for an annuity due, first find A', the 'present value of all payments, omitting the first. Then, if W is the annuity payment, A = A' + W. To find S first obtain S 1 , the amount of the ordinary annuity having a payment at the end of tho term in addition to the payments of the annuity due. Then, 8-S'-W. NOTE. It is customary in actuarial textbooks to use black roman type to indicate amounts and present values of annuities duo. Thus (s^i at j, m) represents the amount, and (a^ at j, m) the present value of an annuity duo whose annual rent is $1, payable p times per year for n years, if money is worth 0', ) EXERCISE XXV In each problem draw a figure similar to Figure 3. Find A and S for each annuity due in the table, by use of the specified rule. PHOB. TEBM PAYMENT iNTBHVALi ANNUAL RENT INTHRHBT RATH BUIiM 1. 10 yr. 3 mo. $ 300 .06, m - 4 2 2, 7 yr., 6 mo. 6 mo. 500 .05, m " 2 2 3. 12 yr., 6 mo. 6 mo, 3600 .03, m - 1 2 4. 12 yr. 3 mo. 1000 .06, m 4 1 ANNUITIES CERTAIN 6. Carry through, the solution of problem 2 by Rule 1 far enough to be able to state why it is inconvenient. 6. A man deposited $100 in a bank at the beginning of each 3 months for 10 years, (a) What sum is to his credit at the end of 10 years if the bank credits 6% interest quarterly from the date of deposit? (6) What sum is to his credit at the end of 9 years and 9 months, after the deposit hag been made at that time? J 7. In purchasing a house, a man has agreed to pay $100 at the beginning of each month for the next 5 years, (a) If money is worth 6% effective, find the present value of the payments. (6) If money is worth (.06, TO = 12), find their present value. { *\ l f / 7, ', ;', av V if $ * > 8. A man was loaned $75 on the 1st of each month, for 12 months each year, during the four years of his college course, (a) If his creditor con- siders money worth 3% effective, what is the liability of the debt at the end of the 4 years? (6) If the debtor makes no payment until four years after he graduates, .what should he pay then to settle in full? 9. If money is worth the effective rate i, prove that (a_, at i), the present value, and (s-, at i), the amount of an annuity due of $1 payable annually for n years, are given by (aj] at t) = l + (o^ at i), (s^ at t) = (^ at t) - 1. 10. Prove that (sjjy at j, w) = (1 + i)'(aj| at j, m) ; (a, at j, w) = (1 + t)(a^ atj, m). 26. Deferred annuities. A deferred annuity is one whose term does not begin until the expiration of a certain length of time. Thus, an annuity whose term is 6 years, deferred 4 years, and whose annual rent is $1000, payable semi-annually, consists of 12 pay- ments of $500, the first due after (4 years -+ 6 months) and the last, after (4 years + 6 years). Example 1. If money is worth (.05, m = 1), find A and S for the de- ferred annuity of the last paragraph. 2f J3 T 1 o o o o o o o o >( x x x x >< xx x x x > increasing time " " represents C months FIG. 4 In Fig. 4, X represents a payment date of the deferred annuity, N, the present, T, the end of 10 years, the end of the term, and B, the beginning of the term, 60 MATHEMATICS OF INVESTMENT Solution. Consider the time scale in Figure 4. The payments form an ordinary annuity when judged from B, 6 months before the first payment. S', the amount, and A', the present value (at B), of this ordinary annuity are p n = -2,t Case 1 6 int. periods, = .05, R = $1000. (2) S'-1000(BJjfoi.05). A' = 1000 (a^at .05). S' is the sum of accumulated values at time T. Since 8, the amount of the deferred annuity, is also equal to the sum of values at T, S = S' = 1000 (sjj at .05). Since A' is the sum of the discounted values at B, we must discount A' for 4 years to obtain the present value A. A - Second solution for A. Consider a new annuity having payments of $500 at the end of each 6 months for the first 4 years as well as for the last 6. The new payment dates are indicated by circles in Figure 4. The present value A of the deferred annuity equals the present value A ' of the new annuity over the whole 10 years minus the present value A" of the payments over the first 4 years, which are not to be received. Both A.' and A" are the present values of ordinary annuities. Case 1 n = 10, and 4, int. periods, p = 2, i = .05, R = $1000. A = A' -A 1 1000[(o^, 1000 ^[( Jt A' = 1000(ogj at .05). ^" = 1000(0^ at .05). a .05) - (aj? at .05)], ojol a* -05) - (aji at .05)]. From Example 1 it is clear that the amount of a deferred annuity equals the amount of an ordinary annuity having the same term. Corresponding to the two methods used above in obtaining A, we state the two rules below. Rule 1. To obtain A for an annuity whose term is deferred w years, first find A', the present value of the ordinary annuity having the same term. Then, A equals the value of A' discounted for w years. Rule 2. If the term of the annuity is n years, deferred w years, then ANNUITIES CERTAIN A = [present value of an ordinary annuity with term (w + n) years] (present value of an ordinary annuity with term w years), where these new annuities have the same annual rent and payment interval as the deferred annuity. NOTE. The present values and the amounts of deferred annuities are indicated in actuarial writings by the symbols for ordinary annuities with a number prefixed showing the time for which the term is deferred. Thus ( a-, atj, m) and ( a-, atj, m} x n\ " ' w n\ *" represent the present value and the amount when the term is deferred w years. EXERCISE XXVI 1 In each problem draw a figure similar to Figure 4. Find the present value of each deferred annuity in the table, by use of the specified rule. PBOH. TERM TERM DEFERRED PAYMENT INTERVAL ANNUAL RENT INTEREST RATE RULE 1. 6yr. 10 yr., 6 mo. 3 mo. $1000 .05, m => 4 2 2. 7yr. 8 yr., 6 mo. 1 mo. 200 .06, m = 2 1 3. 9 yr. 12 yr. 1 yr. 300 .07, m = 1 2 4. 13 yr. 10 yr., 6 mo. 1 mo. 1200 .05, m = 1 1 5. Carry through the solution of problem 4 by Rule 2 until you are able to state why it is inconvenient. I 6. Solve problem 3 if money is worth (.07, m = 2). 7. A man will receive a pension of $50 at the end of each month 10 years, first payment to occur 1 month after he is 65 years old. Assum- ing that he will live to receive all payments, find the present value of his expectation if money is worth (.04, m = 1), and if he is now 50 years old. 8. A certain mine will yield a semi-annual profit of $50,000, the first payment to come at the end of 7 years, and the last after 42 years, at which time the mine will become worthless. What is a fair valuation for the mine if money is worth 5%, effective? 9. A recently paved road will require no upkeep until the end of 3 years, at which time $3000 will be needed for repairs. After that, $3000 will be used for repairs at the end of each 6 months for 15 years. Find the present value of all future upkeep if money is worth (.05, m = 2). 1 The Miscellaneous Problems at the end of the chapter may be taken up im- mediately after the completion of Exercise XXVI. 62 MATHEMATICS OF INVESTMENT 10. By use of Rules 1 and 2 prove the relations below, for an annuity whose annual rent is $1, payable p times per year, and whose term is n years, deferred w years. . SUPPLEMENTARY MATERIAL :/ >."~' 27. 1 Continuous annuities. If money is worth the effective rate i, the present value of an annuity whose annual rent is $1, payable p times per year, is ti) = 1 (a-, ati). JP We may consider an annuity payable weekly, p = 52, or daily, p = 365, and we may ask what value does (o^| at i) approach as p becomes large without bound? We obtain lim (a-, at i) = i(a-, at i) lim ; P=OO v i w| ' z*-o (j p at i) But, from Section 18 we have lim j p = d, the force of interest p=00 corresponding to the effective rate i. Hence ,. , o , .v i(a-] at i) hm fe at t) = aL -- P-03 "' ' 5 In the same way it can be shown that g As p = oo, the annuity approaches the ideal case of an annuity whose annual rent is payable continuously. If we let (a^ at i) represent the present value and (s- at <) the amount of a contin- uous annuity, the results above show that /- , -N ^C a i at *) /- ^ ^ ^( s l at (a^, at i) = -i-aL - , (^ ai t ) -- -L -- (35) Recall that 1 + i - (* so that 6 = lo g-^ + ^- Since log e loge log 2.7182818 = 0.4342945, we obtain from equation 35, 1 Section 18 is a prerequisite for the reading of this section. ANNUITIES CERTAIN 63 i(a-i ati) (.4342945) tfa of a) (.4342945) (a-, ai t) = ^j - ,., , .. - , (s n a i) = , ,., , . N - . n| ' log (1 + 1) ' v nl log(l + &) The present value and the amount of an annuity, which is payable continuously, differ but slightly from the corresponding quantities for an annuity which is payable a very large number of times per year (see problem 1 below). Hence we may use o^ and I-| as approximations for a^ and s^ if p is very large. EXERCISE XXV3I 1. (a) If money is worth 6%, effective, find the present value and the amount of an annuity whose annual rent is $100, and whose term is 10 years, if the annuity is payable continuously. (6) Solve the problem if the annuity is payable monthly. 2. If one year equals 360 days, find approximately the amount of an annuity of $1 per day for 20 years if money is worth 4%, effective. HINT. Use a continuous annuity as an appropriation. 3. A member of a labor union has agreed to contribute $.20 per day to a benefit fund for 3 years. Under the rate (.05, m = 1), find approxi- mately the present value of his agreement if a year has 365 days. 4. An industrial insurance policy for $100 calls for a premium of 10 cents at the end of each week. Find approximately, by use of a contin- uous annuity, the equivalent premium which could be paid at the end of the year if money is worth 3$%. NOTE. If the conversion period of an interest rate is not stated, assume it to be 1 year. *-\ l*> k^^V*-*-"* 28. 1 Computations of high accuracy. The binomial theorem can be used in interest computations to which the tables do not apply. As a special case of the binomial theorem, 2 we have 21 o ! - 1) (n - r + l)s r + . . . When 7i is a positive integer, series 36 contains (n + 1) terms, the last of which is x n . If n is a negative integer or a fraction, the series contains infinitely many terms. In this case, if x lies 1 A knowledge of the binomial theorem ifl needed in this section. 1 See page 93 in Bietz and Crathprne's College Algebra. 64 MATHEMATICS OP INVESTMENT between 1 and + 1, the infinite series converges, and, if x is very small, the sum of the first few terms gives a good approxima- tion to the value of (1 + x) n . The proof of this statement is too difficult for an elementary treatment. Example 1. Find the value of (a^ at .033) accurately to six signifi- cant figures. Solution. From formula 31, (o .033)= l-ay>-. (37) From equation (36) with n = % and x = .033, (1.033)4= 1 +(.033) -TM-OSS^ + ^COS^-Te^OOSS)' + . - (38) 4=1(1.033)^ l] = .03300000- .00040838+. 00000786 -.00000018= .03269930. The next term in series 38, beyond the last one computed, is negligible in the 8th decimal place, and hence our result is accurate to the 7th decimal place with only a slight doubt as to the 8th. To compute (LOSS)"*, first compute (1.033) and then take the reciprocal; (1.033) 8 = 1.2150718; (1.033)- .8229966. Hence, "" The final 9 is not dependable because the final in the denominator was doubtful. EXERCISE XXVm /n\ /n\ 1. Compute (s^, at .0325) and (a^ at .0325) accurately to five sig- nificant figures. 2. Compute (3, at .02) accurately to the 7th decimal place. 3 . A man pays $50 to a building and loan association at the end of each week. If the deposits accumulate at the rate (.06, m = 2), how much will be to his credit at the end of 3,,years? Obtain the result correct to four significant figures. MISCELLANEOUS PROBLEMS 1. If $1750 is the present value of an annuity whose term is 12 years, what is its amount if money is worth (.05, m = 4) ? 2. If $100 is deposited in a hank at the beginning of each month for 10 years, what is the accumulated amount at the time of the last payment if interest is at the rate 6% compounded HRrni-n.nTinfl.ny on all money from the date of deposit? ANNUITIES CERTAIN 65 3. In problem 2, what is the amount at the end of 10 years? 4. A man W has occupied a farm for 5 years and, pending decision of a case in court, has paid no rent to B to whom the farm is finally awarded. What should W pay at the end of 5 years if rent of $100 should have been paid monthly, in advance, and if money is worth (.06, m = 12) ? 5. What should W pay in problem 4 if the rent is considered due at the end of each month and if money is worth 6%, compounded annually? 6. A building will cost $500,000. It will require, at the beginning of each year, $5000 for heat and light, $5000 for janitor service , and, at the end of each year, $3000 for small repairs. It is to be completely renovated, at a cost of $20,000 at the end of each 15 years. If the cost of the annual repairs is included in the cost of renovation at the end of each 16 years, and if the building is to be renovated at the end of 90 years, what present sum will provide for the erection of the building and for its upkeep for the next 90 years, if money is worth 6% effective ? 7. A young man, just starting a four-year college course, estimates his future earning power, in excess of living expenses, at $100 per month for the first 3 years after graduation from college, $200 per month for the next 7 years, and $300 per month for the next 20 years. What is the present value of this earning power, if money is worth 4%, effective? 8. If the man in problem 7 should place his surplus earnings in a bank, what will he have at the end of his working life if his savings earn interest at the rate (.05, m = 2), for the first 10 years, and at the rate (.04, m = 1) for the balance of the time? 9. In purchasing a homestead from the government, a war veteran has agreed to pay $100 at the end of 5 years, and monthly thereafter until the last payment occurs at the end of 9 years. What is the present value of his agreement if money is worth (.045, m = 1) ? 10. At the end of 3 years, the man of problem 9 decides to pay off his obligation to the government immediately. What should he pay if money is worth (.045, m = 1) ? 11. A savings bank accepts deposits of $1 at the beginning of each week during the year from small depositors who are creating Christmas funds. Just after the 52d payment, what will each fund amount to if the bank accumulates the savings at 6%, effective? 12. A contract provides for the payment of $1000 at the end of each 6 months for the next 25 years. What is the present value of the contract if the future liabilities are discounted at (.06, m = 2) over the last 15 years of the life of the transaction and at (.05, m = 2) over the first 10 years? 66 MATHEMATICS OF INVESTMENT 13. In purchasing a house it was agreed to pay $50 at the end of each month for a certain time. The purchaser desires to change to annual payments. What should he pay at the end of each year if money is con- sidered worth (.05, m = 2) ? 14. At what rate of interest compounded semi-annually will $1650 be the present value and $3500 the amount of an annuity whose term is 14 years, if the annuity is payable weekly? Would the result be any different if the annuity were payable annually ? 16. If money is worth 5%, effective, what is the least sum which you would accept now in return for a contract on your part to pay $50 at the end of each month for 20 years, first payment to occur at the end of 10 years and 1 month? CHAPTER IV PROBLEMS IN ANNUITIES 29. In every problem below, the payment interval of the annuity and the conversion period of the interest rate will be given or, in other words, the p and m of equations I and III of Section 24 will always be known. For an annuity under Case 1 there remain for consideration the five quantities (A, S, R, i, ri). If three of (S, R, i, ri) are known, we use S = R(s%\ at i} to find the fourth ; while, if three of (A, R, i, ri) are known, we use A = R(a ( Q at i). If an annuity comes under Case 2, similar remarks apply to the four quantities (S, R, j, ri) and to (A, R, j, ri). Problems in which A and S were unknown were treated in Chap- ter III. 30. Determination of the payment. For future convenience it is essential to know that I J -T\ ' /. _j '\ I * ~~ 7~ _J \ N""/ (s ; at i) (a, at i) (s at i) (a, at i) N n| ' x n| ' v n| ' x n| ' From formulas 27 and 28 we obtain i -!-.-* + *'d + fl" ~ * )V t \ ) n U - (1 + f)-*y (as] ai i) (1 + i) n - 1 and hence relation 39 is true. Exampk 1. Find the value of F) at - 67 68 MATHEMATICS OF INVESTMENT Solution. From Table IX, * ^ rN = .09634229. Therefore, from at .05) relation 39, - - - . = .09634229 - .05 = ,04634229. On doing this sub- traction mentally, we are able to read the result .04634229 directly from Table IX. NOTB. From Example 1 we see that, because of relation 39, Table IX gives the values of - - ^ as well as those of - - r It would be equally - - ^ - - r (a-, at i) (a^ at i) convenient to have a table of the values of - - r from which we could obtain those of - by adding the interest rate i. (a\ at i) Example 2. What annuity, payable quarterly for 20 years and 6 months, could be purchased for $5000, if money is worth (.05, m = 4) ? Solution. The present value of the annuity is $5000. Let $x be the quarterly payment. Casel n 82 int. periods, p = 1, t = .0125, A = $5000, R = $x. 5000 8a| x - 5000 1 MM . - 5000(.01956437) (ag2l at .0125) x = $97.822. The annual rent is 4 x = $391.29. Exam-pie 3. If money is worth (.06, m = 2), find the annual rent of an annuity, payable quarterly for Hi years, if its amount is $10,000. Solution. Let $x be the sum of the payments in one interest period. Case 1 n = 23 int. periods, p = 2, i = .03, S - $10,000, R = $x. 10000 = x(.crf .03) - x^ (jjj, c x 100000'j at .03) 100000'j at .03) it .03), 1 03(^0* .03) .03 ( s^ at .03) C03081390) . .03 Tables IX and XI were used. The annual rent is 2 x = $611.72. NOTE. Recognize that the solutions above wjere arranged so as to avoid computing quotients, except for the easy division by .03. Example 4. Find the annual rent if $3500 is the present value of an annuity which is payable semi-annually for 8 years. Interest is at the rate 5 %, compounded quarterly. PROBLEMS IN ANNUITIES Solution. Let $x be the B.nmifl.1 rent. Case 2 n = 8 yr., p = 2, j = .05, m = 4, R = $x, A = $3500. 3500 3500 = x -32801593 2(.02515625) (Tables V and VI) EXERCISE XXIX 1. Compute a< .05) 10.379658 Find the annual rents of the annuities below. to verify the entry in Table IX. PEOD. PATMHNT INTERVAL INTEREST RATH THEM AMOUNT PRESENT VALUE 2. 3 mo. .06, m = 4 12 yr., 3 mo. $ 6,500 3. 6 mo. .05, m = 2 17 yr., 6 mo. { 8,500 4. lyr. .04, m = 1 15 yr. 3,000 5. 6 mo. .05, m = 1 Syr. 15,000 6. 3 mo. .05, m = 2 6 yr., 6 mo. 3,750 7. 3 mo. .06, w = 1 15 yr. 4,000 8. lyr. .05, m = 1 17 yr. 7,000 9. 6 mo. .05, m = 4 12 yr., 6 mo. 10 3 000 10. 1 yr. .07, m = 2 9yr. 2,500 11. 6 mo. .07, m = 2 9yr. 2,500 ' J 12. If money is worth 6%, effective, find the annuity, payable annually for 25 years, which may be purchased for $1000. 13. If money is worth 6%, effective, find the annuity, payable annually for 10 years, whose amount is $1. 14. If money is worth 4%, compounded annually, what annuity, pay- able annually for 15 years, may be purchased for $1 ? 16. If money is worth the effective rate i, derive a formula for the payment of the annuity, payable annually for n years, which may be purchased for $1. 16. If money is worth the effective rate i, derive a formula for the pay- ment of the annuity, payable annually for n years, whose amount is $1. 17. In order to create a fund of $2000 by the end of 10 years, what must a man deposit at the end of each 6 months in a bank which credits interest semi-annually at the rate 3%? 70 MATHEMATICS OF INVESTMENT 18. The present liability of a debt is $12,000. If money is worth 5.5%, compounded semi-annually, what should be paid at the end of each year for 10 years to discharge the liability in full? 31. Determination of the term. If the term of an annuity is unknown, interpolation methods furnish the solution of the problem with sufficient accuracy for practical purposes. Example 1. For how long must a man deposit $175 at the end of each 3 months in a bank in order to accumulate a fund of $7500, if the bank credits interest quarterly at the rate G%? Solution. Tho deposits form an annuity whose amount is $7600. Let the unknown term be k interest periods. Case 1 n = k int. periods, p = 1, i => .015, R - $175, 8 - $7500. 7600 = 175(8^.015), at .015) = 7500 175 42.857. n fag, at .015) 33 42.299 k 42.857 34 43.933 From the column in Table VII for i = .015, we obtain the first and third entries at the left. . By interpolation, k - 33 + 658 1634 33.341 int. periods. k The term is - = 8.34 years. However, since an annuity whose term is not on integral number of payment intervals has not been defined, this result is useful only because it permits us to make tho following statement : The $175 pay- ments must continue for 8.5 years to create a fund of at least $7500 ; when the 33d payment occurs at tho end of 8.25 years, the fund amounts to less than $7500. Exampk 2. Find the term of an annuity whose present value is $8500 and whose annual rent is $2000, payable quarterly. Interest is at the rate (.06, m = 2). Solution. Let the unknown term be k interest periods. 8600 = 1000(0^ at .03) - 1000 '^(a* at .03), Case 1 n k int. periods, p 2, i .03, 12 - $1000, A - $8500. OCoff at .03) - 1000 '-^ *' Ja _8500(j a ot.Q3) (OB ol .03) - 8600(.Q2977 831) , 8 . PROBLEMS IN ANNUITIES 71 n (a^ at .03) 9 7.786 k 8.437 10 8.530 The first and third entries at the left are from the column hi Table VIII for i - .03. k k periods. The term is - 10 - - = 9.88 interest 744 4.94 years. Hence, for an annuity whose term is 4.75 years, the present value is less than $8500, while the present value would be greater than $8500 if the term were 5 years. NOTE. Values of k found by interpolation in Tables VII and VIII are in error by less than half of the interest rate per period. 1 In interpolating, use three decimal places of the table entries and compute the value of k to three decimal places. . EXERCISE XXX 2 Find the terms of the annuities below. PBOB. PAYMENT INTERVAL INTEREST RATE PRESENT VALUE AMOUNT ANNUAL RBNT 1. lyr. .05, m = 1 $5000 $ 500 2. 6 mo. .06, m = 2 7500 250 3. 1 yr. .03, m => 1 $8000 400 4. 3 mo. .08, m = 4 9000 1000 5. 6 mo. .05, m = I 6500 1300 6. 1 mo. .045, m =2 3500 600 7. 6 mo. .05, m = 2 8500 1000 8. 3 mo. .05, m = 2 8600 1000 9. 1 mo. .03, m = 1 4600 2500 10. 1 yr. .07, m = 1 7450 700 11. For how many full years will it be necessary to deposit $250 at the end of each year to accumulate a fund of at least $3500, if the deposits earn 5%, compounded annually? ^ 12. The cash value of a house is $15,000. In buying it on the install- ment plan a purchaser has agreed to pay $1000 at the end of each 6 months aa long as necessary. For how long must he pay if money is worth 6%, compounded semi-annually? 32. Determination of the interest. rate. Example 1. Under what nominal rate, converted quarterly, is $7150 the present value of an annuity whose annual rent is $880, payable quar- terly for 12 years and 6 months? 1 For justification of this statement see Appendix. Note 6. s See supplementary Section 33 for other problems in whioh the term ie unknown. 72 MATHEMATICS OF INVESTMENT Solution. Lot r be the unknown rate per period. Case 1 n = 50 iut. periods, p = 1, i = r, R $220, A = $7150. 7150 =220(a^ air), t nt r\ - 716 - *> finn a T l ~ l2(T ~ iz - ou U. .0176 r .0200 33.141 32.500 31.424 The first and third entries at the left were obtained from the row in Table VIII for n == 50. Since .0200 - .0175 = .0025, r =.0175 +^- 7 (.0025) = .0175 + .00093 = .01843. The nominal rate iaj = 4r = 4(.01843) = .07372, converted quarterly. NOTEJ. A value of r obtained as above usually is in error by not more than * Jjth of the difference of the table rates used in the interpolation. Hence, r = .01843 prolmbly is in error by not more than J&(.0025) = .0001, and the nominal rate j = .07372 is in error by not more than .0004. We are justified only in saying that the rate is approximately .0737, with doubt as to the lost digit. Supplementary Example 2. Determine the nominal rate in Example 1 accurately to hundredths of 1%. Solution. From Example 1, (0==-, at r) = 32.500, and r = .0184, approxi- mately. It is probable that ? is between .0184 and .0185, or else between .0184 and .0183. Since our tables do not use the rate .0184, we compute ~ (1.0184)- 1 ' 50 log 1.0184 = 50(.0079184) .0184' ' - .39592. .(a^ai.0184) .0184) .40186 .0184 = 32.507. log (1.0184)- BO = 9.60408 - 10. (1.0184)- 60 = .40186. Since 32.507 is greater than 32.500, r must be greater than .0184, and probably is between .0184 arid .0185. By logarithms, (o at. .0186) = 32.438. i (ogjy| at i) .0184 32.S07 T 32.500 .0186 32.438 From interpolation in the table at the left, r = .0184 + 1 (.0001) - .018410. The nominal rate is j = 4 r = .073640, which is cer- tainly accurate to hundredths of 1%, and is probably accurate to thousandths of 1%. 1 The author gives no theoretical justification for this statement. He has verified its truth for numerous examples scattered over the range of Tables VII and VIII. PROBLEMS IN ANNUITIES 73 NOTE. We could obtain the solution of Example 1 with any desired degree of accuracy by successive computations as in Example 2. Our accuracy would be limited only by the extent of the logarithm tables at our disposal. In Example 1, the most simple formulas (Case 1, with p = 1) applied because the conversion period equaled the payment interval. In more complicated examples, the solution may be obtained by first considering a new problem of the simple type met in Example 1. Example 3. Under what nominal rate, converted semi-annually, is $7150 the present value of an annuity whose annual rent is $880, payable quarterly for 12 years and 6 months? Solution. Let the unknown nominal rate be j. We could use the formulas of Case 1, with p = 2, in the solution, but the work would be slightly compli- cated. Instead, we first solve the following new problem: "Determine the nominal rate, w, converted quarterly, under which the present value of the annuity will be $7150." We choose quarterly conversions here because the annuity is payable quarterly. This new problem is the Example 1 solved above, so that w = .0737. The rate j, compounded semi-annually, must be equivalent to the rate .0737, compounded quarterly, because the present value of the annuity is $7150 under both of these rates. Hence, the effective rates corresponding to these two rates must be the same. 1 From equation 17, if i represents the effective rate, = (l +i)', 1 +< - ( (1.0184).. = (1.0184)*; 1 + 3 - = (1.0184)" = 1.03714. a Table II was used in computing 1.03714. The desired nominal rate is j = 2(.03714) = .07428, or approximately .0743, with doubt as to the last digit. EXERCISE XXXI 2 In the first ten problems find the nominal rates as closely 3 as is possible by interpolation in the tables. 1 For a similar problem see Section 10, illustrative Example 3. s The Miscellaneous Problems at the end of the chapter may be taken up im- mediately after the completion of Exercise XXXI. If the instructor desires, the students may be requested to obtain accuracy to hundredths of 1%, as in Example 2 above. 74 MATHEMATICS OF INVESTMENT PKOB. ANNUAL RUNT PAYMENT INTHBVAL INTEREST PERIOD AMOUNT PBHSHNT VALUE THHM 1. $1000 1 year 1 year $15,700 12 yr. 2. 100 1 year 1 year * 1,785 25 yr- 3. 600 1 year 1 year 5,390 15 yr. 4, 100 6 mo. 6 mo. 1,110 17 yr., 6 mo. 6. 400 3 mo. 3 mo. 2,500 5 yr., 3 mo. 6. 1000 1 year lyear 53,000 26 yr. 7. 1 200 3 mo. 1 year 2,750 9yr. 8. 1 200 3 mo. 6 mo. 2,750 9yr. 9. 2400 1 mo. 1 year 14,500 Syr. 10. 500 6 mo. 3 mo. 17,500 24 yr., 6 mo. "^ 11. A man has paid $100 to a building and loan association at the end of each 3 months for the last 10 years. If he now has $5500 to his credit, at what nominal rate; converted quarterly, does the association compute interest? 12. By use of the result of problem 11, find the effective rate of interest paid by the association of problem 11. 18. It has been agreed to pay $1100 at the end of each 6 months for 8 years. Under what nominal rate', compounded semi-annually, would this agreement be equivalent to a cash payment of $14,000? 14. A fund of $12,000 has been deposited with a trust company in order to provide an income of $400 at the end of each 3 months, for 10 years, at which time the fund will be exhausted. At what effective rate does the trust company credit interest on the fund? HINT. First find the equivalent nominal rate, payable quarterly. SUPPLEMENTARY MATERIAL 33. Difficult cases and .exact methods in finding the term. When the formulas of Case 2 apply to an annuity, it is necessary to use the explicit formulas III in finding the term if it is unknown, Example 1. The amount of an annuity is $8375, and the annual rent is $1700, payable semi-annually. What is the term if money is worth (.06, m = 4) ? 1 See illustrative Example 3. The same preliminary work should be used for both of problems 7 and 8. First determine the nominal rate, converted quarterly, under which. $2750 is the amount. PROBLEMS IN ANNUITIES 75 Solution. Case 2 applies. Hence, let the unknown term be k years. 8375 = 1700(8^ at .06, m = 4). Case 2 n = k yr., p - 2, j = .06, m = 4, R = $1700, A = $8375. From Table V, the denominator is 2 (.030225). (1.015)" - 1 = 8375(2) (.030225) = 2g781 1700 (1.015)" - 1 + .29781 = 1.29781. (40) (a) To solve equation 40 by interpolation, we use entries from the column in Table V for i = .015. We obtain 4 k = 17 + - . 17.507, or Jb = 4.377. 1932 (6) To obtain the exact value of k from equation 40, take the logarithm of both sides of the equation, using Table II for log 1.015. 4 k log 1.015 = log 1.29781 ; 4 fc(.0064660) = .11321. 11321 _ 11321 log 11321 = 4.05389 log 2568.4 = 3.41270 log k - 0.64119 k = 4(646.60) 2586.4 k = 4.3771. The solutions of problems, treated by interpolation in Section 31, may be obtained by solving exponential equations, as in solution (6) above. In these exact solutions it is always necessary to use the explicit alge- braic expressions for the present values and the amounts of the annui- ties concerned. Example 2. If the rate is (.06, m = 2), find the term of an, annuity whose present value is $8500 and whose annual rent is $2000, payable quarterly. Solution. Let the unknown term be k interest periods. Case 1 n = k int. periods, p = 2, i = .03, R - $1000, A = $8500. 8500 = 1000(a^ at .03) = 1000 1 - (1.03)-* 8600(.02Q77831) 1000 ,25312, 2[(1.03)* - 1] (Table XI) (By Table I) (76) MATHEMATICS OF INVESTMENT (1.03)-* = 1 - .25312 = .74688. /. - k log 1.03 = log .74688. - fc(.0128372) = 9.87325 - 10 = - .12675. Gog 1-03 from Table II) fc = .12675 g 873g periodfi of 6 mon tha. 1283.7 The term is 5 = 4.9369 years. Compare this with the result by interpolation 2 in Example 2 of Section 31. EXERCISE XXXII 1. At the end of each 6 months a man deposits $200 in a bank which credits interest quarterly at the rate 3%. For how many years must the deposits continue in order to create a fund of $3000? Use the exact method (&) of Example 1 above. 2. If money is worth 6%, compounded monthly, for how long must payments of $2000 be made at the end of each 6 months in order to dis- charge a debt whose present liability is $30,000? Solve by both an inter- polation and an exact method. 3. Solve illustrative Example 1 of Section 31 by the exact method. 4. Solve problem 9 of Exercise XXX by the exact method. B. To create an educational fund for a daughter, a father decides to deposit $500 at the end of each 6 months in a bank which credits interest annually, from the date of deposit, at the rate 4 %. When will the fund amount to at least $6000? Solve by the exact method. MISCELLANEOUS PROBLEMS 1. If money is worth 5%, effective, what equal payments should be made at the end of each year for 10 years in purchasing a house whose equivalent cash price is $5000? 2. If a man saves $200 at the end of each month, when will he be able to buy an automobile, worth $3000, if his deposits accumulate at the rate 5 %, compounded semi-annually? 3. A depreciation fund is being accumulated by equal deposits at the end of each month in a bank which credits 6 % interest monthly on deposits from date of deposit. What is the monthly deposit if the fund contains $7000 at the end of 5 years? ' 4. In purchasing a house, worth $20,000 cash, a man has agreed to pay $5000 cash and $1000 semi-annually for 9 years. What interest rate, compounded sew-annually, is being used in the transaction? PROBLEMS IN ANNUITIES (77 J. 6. An insurance policy, on maturing, gives the policy holder the option of an immediate endowment of $15,000 or an annuity, payable quarterly for 10 years. Under the rate 3.5%, effective, what will be the quarterly pay- ment of the annuity? 6. A fund of $50,000 has been deposited with a trust company which credits interest quarterly on all funds at the rate 5 %. For how long will this fund furnish a man payments of $1000 at the end of each 3 months? * 7. On the death of her husband, a widow deposited her inherited estate of $25,000 with -a trust company. If interest is credited semi-annually on the fund at the nominal rate 4 %, for how long will the widow be able to withdraw $800 at the end of each 6 months? 8. A certain loan bureau lends money to heads of families on the fol- lowing plan : In return for a $100 loan, $9 must be paid at the end of each month for 1 year. What effective rate of interest is being charged? HINT. First find the nominal rate, compounded monthly. 9. A certain homestead is worth $5000 cash. The government sold this to an ex-soldier under the agreement that he should pay $1000 at the end of each 6 months until the liability is discharged. If interest is at the rate (.04, m = 2), for how long must the payments continue? 10. The annual rent of an annuity is $50, payable annually. The present value of the annuity is $400 and the amount is $600. Find the effective rate of interest by use of the relation 39 of Section 30. 11. A certain farm has a cash value of $20,000. If money is worth (.05, m = 2), what equal payments, made at the beginning of each 6 months for 6 years, would complete the purchase of the farm? 12. A man borrowed $2000 under the agreement that interest should be at the rate (.06, m = 2) during the life of the transaction. He made no payments of either interest or principal for 4 years. At that time, he agreed to discharge all liability in connection with the debt by making equal payments at the end of each 3 months for 3 years. Find the quarterly payment. CHAPTER V THE PAYMENT OF DEBTS BY PERIODIC INSTALLMENTS 34. Amortization of a debt. A debt, whose present value is A, is said to be amortized under a given rate of interest, if all lia- bilities as to principal and interest are discharged by a sequence of periodic payments. When the payments are equal, as is usually the case, they form an annuity whose present value must equal A, the original liability. Hence, most problems in the amortization of debts involve the present value formulas for annuities. Many amortization problems have been solved in previous chapters. Example 1. A man borrows $15,000, with interest payable annually at the rate 5%. The debt is to be paid, interest as due and original principal included, by equal installments at the end of each year for 5 years, (a) Find the annual payment. (6) Form a schedule showing the progress of repayment (or amortization) of the principal. Solution. Let $x be the payment. The present value of the payment annuity, at the rate (.05, m = 1), must equal $16,000. 15000 = x(a at .05). Casel n = 5 int. periods, R 1, i .05, , A = $15,030. x - 15000 - $3464.622. AMORTIZATION SCHEDTJIJD YHAR OUTSTANDING PBINCIPAIi AT BEGINNING orYnAB INTEREST AT6 ?fc, DUB AT END OK YBAR ANNUAI, PAYMENT AT END OF YEAH FOB REPAYMENT OP PRINCIPAL AT END OF YEAR 1 2 3 4 5 $15,000.000 12,285.378 9,435,025 6,442.154 3,299.640 $ 750.000 614.269 471.751 322.108 164,982 $ 3,464.622 3,464.622 3,464.622 3,464.622 3,464.622 $ 2,714.622 2,850.353 2,992.871 3,142.514 3,299.640 Totals $46,462.197 $2323.110 $17,323.110 $15,000.000 78 PAYMENT OF DEBTS BY PERIODIC INSTALLMENTS 79 NOTH 1. The schedule shows that the payments satisfy the creditor's demands for interest and likewise return his principal in installments. If x = $3464.622 was computed correctly, we know, without forming the sched- ule, that these facts must be true because the present value of the five pay- ments is $15,000. The checks on the arithmetic done in the table are that the lust total should be $15,000, the sum of the second and the last should equal the third, and tho second should be interest on the first total for one year at 6%. Notice that tho repayments of principal increase from year to year, while tho interest, payments decrease. Amortization schedules are very useful in the bookkeeping of both debtor and creditor because the exact outstanding liability at every interest date is clearly shown. The outstanding principal, or liability at any date, is sometimes called the book value of the debt at that time. NOTE 2. Since money is worth 6%, in Example 1, we may assume that tho debtor invests tho $15,000 at 5% immediately after borrowing it. The accumulation of this fund should provide for all the annual payments, to be made to tho creditor, because then* present value is $15,000. A numerical veri- fication of this fact is obtained in the amortization table above if we merely alter the titles of tho columns, as below, leaving the rest of the table unchanged. YHAM IN FUND AT BHHINNINO op YSLMI iNTHnKBT RKGHIVHD AT END OF YBA.R PAYMENT TO CnBDiTon AT END OF YHAB TAKEN FROM FUND AT END OF YEAR 1 $15,000 $750 $3464.622 $2714.622 Thus, at the end of the first year, the debtor receives $760 from his invested fund and, in order to make the payment of $3464.622 to his creditor, he takes $2714.02 from tho principal. By the end of 6 years, the fund reduces to zero. Exampk 2. In Example 1, without using the amortization schedule, determine the principal outstanding at the beginning of the third year. Solution. The outstanding principal, or liability, is the present value of all payments remaining to bo made. These form an annuity whose term is three years. The outstanding principal is 3464.62(0^ at .05) = $9435.03. This is the third entry of the first column of tho amortization schedule. P Case 1 n 3 int. periods, 1, i - .05, R - $3464.62. Exampk 3. A debt whose present value ia $30,000, bearing interest at the rate 4.6%, compounded semi-annually, is to be amortized in 10 yearn by equal payments at the end of eaoh 3 months, (a) Find the quar- terly payment. (6) Find tho principal outstanding at the end of 6 years, after the payment due has been made. 80 ' MATHEMATICS OF INVESTMENT Solution. (a) Let $x be the quarterly payment. The present value of the payment annuity must equal $30,000. Case 1 n =*> 20 int. periods, p = 2, i = .0226, R = 2 x, A = $30,000. 30000 = 2 x(a at .0225) - 2 x (o at .0225). 15000 ji .0225 (ogyj at .0225) = 16006(.02237484) (OT294?q7) = $934.403. .0225 (b) At the end of the 6th year, or the beginning of the 6th, the outstanding liability, L, is the present value of payments extending over 5 years. L = 1868.81 (ajjai. 0225). Casel n = 10 int. periods, .0225, ,- 2, R. $1868.81. L . $16]69L95 . L = 1868.81 ^5(0 oi .0225). EXERCISE XXXffi ^ 1. A loan of $5000, with interest at 6%, payable semi-annually, is to be amortized by six semi-annual payments, the first due after 6 months, (a) Find the payment, to three decimal places. (&) Form the amor- tization schedule for the debt. 2. In problem' 1, without using the amortization table, find the prin- cipal unpaid at the end of 1 year and 6 months, just after the payment due has been made. 3. A man deposits $10,000 with a trust company which credits 5% interest annually. The fund is to provide equal payments at the end of each year for 5 years, at the end of which time the fund is to bo exhausted, (a) Find the annual payment to three decimal places. (&) Form a table showing the amortization of the fund. HINT. See Note 2, Section 34; think of the trust company as the debtor. 4. In problem 3, without using the table, find the amount remaining in the fund at the end of 2 years, after the payment due has been made. 6. A purchaser of a house owes $7500, and interest at 6% is payable semi-annually on all amounts remaining due. He wishes to discharge his debt, principal and interest included, by twelve equal semi-annual installments, the first due after 6 months. Find the necessary semi- annual payment. 6. A street assessment of $500 against a certain piece of real estate is to be amortized, with interest at 6%, by six equal annual payments, the first due after \ year. What part of the assessment will remain unpaid 8-t the beginning of the 4th year, after the payment due has been made? PAYMENT OF DEBTS BY PERIODIC INSTALLMENTS 81 7. A house is worth $25,000 and the owner, on selling, desires the equivalent of interest at the rate 5%, payable semi-annually. (a) What quarterly installment, for 8 years, in addition to a cash payment of $5000, would satisfy the owner? (&) How much of the principal of the debt remains unpaid at the end of 3 years and 6 months, after the payment due has been made ? 8. A debt of $12,000, with interest payable serai-annually at the rate 5 %, is to be amortized in 10 years by equal semi-annual installments, the first due after 6 months. What part of the debt will remain unpaid at the beginning of the 6th year, after the payment due has been made? 9. In problem S, what part of the llth payment is interest and what part is repayment of principal ? 10. A debt will be discharged, principal and interest, at 6% effective, included, by payments of $1200 at the end of each year for 12 years, (a) What is the original principal of the debt? (b) What principal will remain outstanding at the beginning of the 5th year? (c) What part of the 5th payment will be interest and what part repayment of principal? 11. A trust fund of $100,000 was created to provide a regular income at the end of each month for 20 years. If the trust company uses the interest rate 4%, converted semi-annually, what is the monthly payment, if the fund is to be exhausted by the end of 20 years? 12. It was agreed to amortize a debt of $20,000 with interest at 5%, by 12 equal annual payments, the first due in one year. The debtor failed to make the first four payments. What payment at the end of 5 years would bring the debtor up to date on liis contract? 13. A debt of 538,000 is to bo amortized by payments of $2000 at the end of each 3 months for 6 years, (a) Find the nominal rate, compounded quarterly, at which interest is being paid, (fo) What is the effective rate of interest? 14. A certain insurance policy on maturing gives the option of $10,000 cash or $345 at the end of each months for 20 years. What rate of interest is being used by the insurance company? 35. Amortization of a bonded debt. In amortizing a debt which is in the form of a bond issue, the periodic payments cannot be exactly equal. If the bonds are of $1000 denomination, for example, the principal repayments must be multiples of $1000, because any individual bond must be retired in one installment. 82' MATHEMATICS OF INVESTMENT Example 1. Construct a schedule for the amortization, by 10 annual payments as nearly equal as possible, of a $10,000 debt which is outstand- ing in bonds of $100 denomination, and which bears 4% interest payable annually. The first payment is due at the end of 1 year. SoMion. Let &c be the annual payment, which would be made if the payments were to be equal. 10000 - x(aat .04). Case 1 n = 10 int. periods, p = 1, i = .04, R = $x, A = $10,000. x = 10000 .04) $1232.91. The annual payments should be aa close as possible to $1232.91. Thus, at the ead of the 1st year, the interest due is $400, leaving 1232.91 400.00 = $832.91 available for repayment of principal. Hence, retire 8 bonds, or $800 of the principal on this date, making a total payment of $1200. At the end of the next year 1232,91 - 368.00 = $864.91 is available for retiring bonds. Therefore, pay 9 bonds or $900 of the principal. AMORTIZATION SCHEDULE FOR A BONDED DEBT YBAB PniNOIPAL OUT- STANDING AT BEGIN- NING OF YBAB INTEREST Dun AT END OF YHAR BONDS RHTIBBD AT END OF YHAB TOTAL PAYMENT AT END OF YEAH 1 $10,000 $400 8 $1,200 2 9,200 368 9 1,268 3 8,300 332 9 1,232 4 7,400 296 9 1,196 5 6,500 260 . 10 1,260 6 5,500 220 10 1,220 7 4,500 180 11 1,280 8 3,400 136 11 1,236 9 2,300 92 11 1,192 10 1,200 48 12 1,248 Totals $58,300 $2,332 100 $12,332 EXERCISE XXXTV ^ 1. A $1,000,000 debt is outstanding in the form of $1000 bonds which pay 6% interest annually. Construct a schedule for the retirement of the debt, principal and interest included, by five annual payments as nearly equal as possible, the first payment due at the end of 1 year. 2. A $1,000,000 issue of bonds, paying 5% interest annually, consists of 500 bonds of $100, 200 bonds of $500, 200 of $1000, and 130 of $5000 PAYMENT OF DEBTS BY PERIODIC INSTALLMENTS 83 denomination. Construct a schedule for the amortization of the debt by 10 annual payments as nearly equal as possible. HINT. In the schedule,, make a separate column for each class of bonds. 36. Problems in which the periodic payment is known. If the present liability of a debt, the interest rate, and the size and frequency of the amortization payments are known, the term of the payment annuity can be found as in Section 31. Bxampk 1. A house is valued at $10,000 cash. It is agreed to pay $1200 cash and $1200 at the end of each 6 months as long as necessary to amortize the given cash value with interest at 5%, payable semi-an- nually. (a) For how long must the payments continue? (6) Construct an amortisation schedule. Solution. (a) After the cash payment of $1200, $8800 remains due. Let k be the time in interest periods necessary to amortize it with interest at the rate (.05, m - 2). 8800 = 1200(0^ at .025) ; (fl at .025) = 7.333. Case 1 n = k int. periods, p = 1, i = .025, R = $1200, A = $8800. By interpolation in Table VIII, k = 8.20. Hence, 8 full payments of $1200 must be made in addition to the first cash payment. After the $1200 payment is made, at the end of 4 years, some principal is still outstanding because k is greater than 8. A partial payment will be necessary at the next payment date. These conclusions are verified in the schedule below. (6) AMORTIZATION SCHEDULE PAYMENT INTERVAL OtJTSTANDING PRIN- CIPAL AT BEGIN- NING OF INTERVAL INTEREST DUB AT END OP INTERVAL TOTAL PAYMENT AT END OF INTEBVAL PRINCIPAL RE- PAID AT END OP INTBBTAL 1 $8800.000 $220.000 $1200. $ 980.000 2 7820.000 195.500 . 1200. 1004.500 3 6815.500 170.388 1200. 1029.612 4 6785.888 144.647 1200. 1055.353 5 4730.535 118.263 1200. 1081.737 6 3648.798 91.220 1200. 1108.780 7 2540.018 63.500 1200. 1136.500 8 1403.518 35.088 1200. 1164.912 9 238.606 5.965 244.571 238.606 Totals $41,782.863 $1044.571 $9844.571 $8800.000 84 ; MATHEMATICS OF INVESTMENT Exampk 2. Without using the amortization table, find the principal still unpaid in Example 1 at the end of 2 years, after the payment due has been made. Solution. Let $M be the amount remaining due. The payment of $M at the end of 1\ years, in addition to the payments already made, would complete the payment of the debt whose original principal was $8800. Hence, this "Old Obligation" must have the same value as the "New Obligations " Hated below. OLD OBLIGATION NHW OBLIGATIONS $8800 due at the beginning of the transaction. (a) $M due at the end of 2$ years. (&) Payments of $1200 due at the end of each 6 months for 2| years. To find M, write an equation of value, under the rate (.06, m = 2), with the end of 2J years as the comparison date. The sum of the values of obligations (&) is the amount of the annuity they form. 8800(1. 025) B = M + 1200 (s^at .025). M = 8800(1.025) B - 1200(^0* .025) = 9956.39 - 6307.59 = $3648.80, (41) which checks with the proper entry in the table of Example 1. The debtor could close the transaction at the end of 2$ years by paying the regular in- stallment plus $3648.80 or (1200 + 3648.80) = $4848.80. NOTE 1. Recognize that 8800(1. 025) 5 is the amount the creditor should have at the end of 2i years if he invested $8800 at (.05, m = 2), whereas he actually has possession of only 1200 (s^ aL .025) as a consequence of the pay- ments received from the debtor. Hence, equation 41 shows that M is the difference between what the creditor should have and what he actually has. NOTE 2. By the method of Example 2 we can find the final installment in Example 1 without computing the amortization table. Let $ N be the amount remaining due just after the last full payment, at the end of 4 years. Then, N = 8800(1.025) 8 - 1200(85-, at .025) - 10721.946 - 10483.339 = $238.607. To close the transaction at the end of 4J years, the necessary payment is $238.607 plus interest for 6 months at 6%, or 238.607 + 5.966 $244.57. EXERCISE XXXV 1. (a) How long will it take to amortize a debt whose present value is $10,000 if payments of $2000 are made ai the end of each year and if these payments include interest at the rate 5%, payable annually. (6) Form an amortization schedule for this debt. PAYMENT OF DEBTS BY PERIODIC INSTALLMENTS 85 2. (a) Without using the table in problem 1, find the principal out- standing at the beginning of the 3d year. (&) Find the size of the final payment. 3. A debt of $50,000, with interest payable quarterly at the rate 8%, is being amortized by payments of $1500 at the end of each 3 months. (a) What is the outstanding liability just after the 10th payment? (6) Find the final installment. 4. A trust fund of $100,000 is invested at the rate 6%, compounded semi-annually. Principal and interest are to provide payments of $5000 at the end of each 6 months until the fund is exhausted, (a) How many full payments of $5000 will be made? (6) What will be the size of the final partial payment? 6. The purchaser of a farm has agreed to pay $1000 at the end of each 3 months for 5 years, (a) If these payments include interest at the rate 6%, payable quarterly, what is the outstanding principal at the be- ginning of the transaction? (&) Find the outstanding liability at the beginning of the 3d year. Notice that, since the exact number of the re- maining payments is known, part (6) should be done like illustrative Example 2 of Section 34 ; it would be clumsy to use the method of illus- trative Example 2 of the present section. 37. Sinking fund method. A sinking fund is a fund formed in order to pay an obligation falling due at some future date. In the following section, unless otherwise stated, it is assumed that the sinking funds involved are created by investing equal periodic payments. Then, the amount in a sinking fund at any time is the amount of the annuity formed by the payments, and examples involving sulking funds can be solved by use of the formulas for the amount of an annuity. Thus, to create a fund of $10,000 at the end of 10 years by investing $rc at the end of each 6 months for 10 years, at the rate (.06, m = 2), x must satisfy 10000 = .fa * - 03 > 5 - 1000 Suppose that $A is borrowed under the agreement that interest shall be paid when due and that the principal shall be paid in one installment at the end of n years. If the debtor provides for the future payment of $1 at the end of n years by the creation of a sinking fund, invested under his own control, his debt is said to be 86 MATHEMATICS OF INVESTMENT retired by the sinking fund method. Under this method, the expense of the debt to the debtor is the sum of (a) and (6) below : (a) Interest on $A, paid periodically to the creditor when due. (6) Periodic deposits, necessary to create a sinking fund of $A, to pay the principal when due. NOTE 1. Recognize that the sinking fund is a private affair of the debtor. The rate of interest paid by the debtor on $A bears no relation to the rate of interest at which the debtor is able to invest his sinking fund. Usually, the desire for absolute safety for the fund would compel the debtor to invest it at a lower rate than he himself pays on his debt. Example 1. A debt of $10,000 is contracted under the agreement that interest shall be paid semi-annually at the rate 6%, and that the principal shall be paid in one installment at the end of 2^ years, (a) Under the sinking fund method, what is the semi-annual expense of the debt if the debtor invests his fund at (.04, m = 2) ? (6) Form a table showing the accumulation of the fund. Solution. (a) Let $z be the semi-annual deposit to the sinking fund, whose amount at the end of 24 years is $10,000. 10000 = x 1 Case 1 n = 5 int. periods, p = 1, i = .02, R = $3, S - $10,000. x = 10000- at .02) ; $1921.584. expense is 300 + 1921.58 ( Sg] ai .02) Interest due semi-annually on the principal of the debt is (.03) (10000) = $300. Semi-annual = $2221.58. (6) TABLE SHOWING GROWTH OP SINKING FUND PAYMENT INTERVAL IN POND AT BEGINNING OF INTERVAL INTEBBST AT 4% RECEIVED ON FUND AT END OF INTERVAL PAYMENT TO FUND AT END OF INTERVAL IN FUND AT END op INTERVAL 1 $192L584 $1921.584 2 $1921.584 $ 38.432 1921.584 3881.600 3 3881.600 77.632 1921.584 6880.816 4 5880.816 117.616 1921.584 7920.016 5 7920.016 158.400 1921.684 10000.000 Nora 2. The book value of the debtor's indebtedness, or his net indebted- ness, at any time may be defined as the difference between what he owes and what he has in his sinking fund. Thus, at the end of 2 years, the book value of the debt is 10000 - 7920.016 - $2079.984. PAYMENT OF DEBTS BY PERIODIC INSTALLMENTS S7- NOTE 3. The amount in the sinking fund at any time is the amount of the payment annuity up to that date and can be found without forming the table (6). Thus, the amount in the fund at the end of 2 years is 1921.584(8^ at .02) = $7920.02. EXERCISE XXXVI * J 1. A debt of $50,000, with interest payable semi-annually at the rate 6%, is to be paid at the end of 3 years by the accumulation of a sinking fund, (a) If payments to the fund are made at the end of each 6 months and accumulate at the rate 3%, compounded semi-annually , what is the total semi-annual expense of the debt? (6) Form a table showing the accumulation of the fund. 2. (a) In problem 1, without using the table, determine the amount in the sinking fund at the end of 2 years. (6) What is the book value of the debtor's indebtedness at this time? 3. A loan of $10,000 bears 5% interest, payable serm-annually, and a sinking fund is created by payments at the end of each 6 months in order to repay the principal at the end of 4 years. Find the expense of the debt if the fund accumulates at the rate (.04, m = 2). 4. A city issues $100,000 worth of bonds bearing 6% interest, payable annually, and is compelled by law to create a sinking fund to retire the bonds at the end of 10 years. If payments to the fund occur at the end of each year and are invested at 6%, effective, what is the annual ex- pense of the debt? 6. How much is in the sinking fund in problem 4 at the beginning of the 7th year? 6. A loan of $5000 is made under the agreement that interest shall be paid semi-annually at the rate 5.5% on all principal remaining due and that the principal shall be paid in full on or before the end of 6 years. (a\ Find the semi-annual expense if the debt is amortised by equal pay- ments at the end of each 6 months for 6 years. (&) Find the semi-an- nual expense to retire the debt by the sinking fund method at the end of 6 years, if payments to the fund are made at the end of each 6 months and are invested at (.04, m = 2). (c) Find the semi-annual expense under the sinking fund method if the fund earns (.06, m = 2) . (d) Which method is most advantageous to the debtor? 7. A sinking fund is established by payments at the end of each 3 months in order to accumulate a fund of $300,000 at the end of 15 years. Find the quarterly payment if interest is at the rate (.06, m = 2). . (OBI at $) Since - - r = i -J -- -, 88 MATHEMATICS OF INVESTMENT 8. To accumulate a fund of $100,000, payments of $5000 are invested at the end of each 6 -months at the rate 5%, compounded semi-annually. (a) How many full payments must be made? (6) How much must be paid on the last payment date? 38. Comparison of the amortization and the sinking fund methods. To amortize a debt of $A, in n years, with interest payable annually at the rate i, by payments of R at the end of each year, we have A = R(ctn\ at i) or (42) R-Ai + Aj-' (43) (si at If the sinking fund method is used to retire this. debt, the amount in the fund at the end of n years is $A. If payments of $W are made to the fund at the end of each year and accumulate at the effective rate r, then A = W(sx\ at r) or W = A. l (44) (Sn\ at r) The annual interest on $A at the rate i is Ai, so that the total annual expense E under the sinking fund plan is E = Ai + W = Ai + A - ^-^ (45) fa\atr) When r = i in equation 45, E equals R, as given in equation 43. Thus, the amortization payment $R is sufficient to pay interest on $-4 at the rate i, and to create a sinking fund which amounts to &4 at the end of n years, if the fund also earns interest at the rate i. Hence, the amortization method may be considered as a special case of the sinking fund method, where the creditor is custodian of the sinking fund and invests it at the rate i. When r is less than i, (s^ at r} is less than (s^\ at i), so that A ,_ . ^ is greater than A -. - and E is greater than R. (Sn\atr) _ (%, at i) Similarly, when r is greater than i, the sinking fund expense is less than, the amortization expense. ft PAYMENT OF DEBTS BY PEEIODIC INSTALLMENTS 89 NOTE. The conclusions of the last paragraph are obvious without the use of any formulas. If the debtor is able to invest his fund at the rate r, greater than i, his expense will be less than under the amortization method because, under the latter, he is investing a sinking fund with his creditor at the smaller rate i. NOTE. Equation 42 is sometimes called the amortization equation, and equation 44 is called the sulking fund equation. Table IX for may be called a *\ a table of amortization charges for a debt of $1 (A <= 1 in equation 42), and a table of the values of would be a table of sinking fund charges for a debt of $1. MISCELLANEOUS PROBLEMS In solving the more difficult problems of the set below, the student should recall that the writing of an equation of value, for a conveniently selected comparison, date, furnishes a systematic method of solution, as in illustrative Example 2 of Section 36. 1. At the end of 5 years, a man will pay $15,000 cash for a house. (a) What equal amounts should he save at the end of each year to ac- cumulate the money if his savings earn 6%, effective? (6) What should he save at the beginning of each year in order to accumulate the money if the savings earn 6%, effective? 2. A loan of $5000 is made, with interest at 6%, payable semi-annually. Is it better to amortize the debt in 6 years by equal semi-annual in- stallments, or to pay interest when due and to retire the principal in one installment at the end of 6 years by the accumulation of a sinking fund by semi-annual payments, invested at (.04, m = 2) ? 3. A man, purchasing a farm worth $20,000 cash, agrees to pay $5000 cash and $1500 at the end of each 6 months, (a) If the payments in- clude annual interest at the rate 5%, effective, how many full payments of $1500 will be necessary? (6) What is the purchaser's equity in the house at the beginning of the 3d year? 4. A county has an assessed valuation of $50,000,000. The county borrows $500,000 at 5%, payable annually, and is to retire the principal at the end of 20 years through the accumulation of a sinking fund by annual payments invested at 4%, effective. By how much, per dollar of assessed valuation, will the annual taxes of the county be raised on account of the expense of the debt? V." MATHEMATICS OF INVESTMENT 5. A debt of $25,000, with interest payable semi-annually at the rate 6%, is to be amortized by equal payments, at the beginning of each 6 months for 12 years, (a) Determine the payment. (6) At the be- ginning of the 4th year, after the payment due has been made, what prin- cipal remains outstanding? 6. A debt of $12,000, with interest at 5%, compounded quarterly, is to be amortized by equal payments at the end of each 3 months for 8 years. At the end of 4 years, what payment, in addition to the one due, would cancel the remaining liability if the creditor should permit the future payments to be discounted, under the rate (.04, m = 4) ? 7. A debt of $100,000, with interest at 5% payable annually, will be retired at the end of 10 years by the accumulation of a sinking fund by annual payments invested at 4%, effective. Considering the total an- nual expense of the debtor as an annuity, under what rate of interest is the present value of this annuity equal to $100,000 ? The answer obtained is the rate at which the debtor could afford to amortize his debt, instead of using the sinking fund method described in the problem. 8. In order to accumulate a fund of $155,000, a corporation invests $20,000 at the end of each 3 months at the rate (.08, m = 4). (a) How many full payments of $20,000 must be made? (6) Three months after the last full payment of $20,000 is made, what partial payment will com- plete the fund? 9. The original liability of a debt was $30,000, and interest is at the rate 5%, effective. Payments of $2000 were made at the end of each year for 7 years. At the end of that time it was decided to amortize the remaining indebtedness by equal payments at the end of each year for 8 years. Find the annual payment. HINT. -Equations of value furnish a systematic method for solving prob- lems of this type. Let &c be the annual payment. Then, the value of the "Old Obligation" below must equal the sum of the values of the "New Obligations " on whatever comparison date is selected. OLD OBLIGATION NEW OBLIGATIONS $30,000 due at the begin- ning of the transaction. (a) $2000 due at the end of each year for 7 years, (b) Eight annual payments of $s, the first . due at the end of 8 years. The end of 7 years is the most convenient comparison date. PAYMENT OF DEBTS BY PERIODIC INSTALLMENTS 91 10. At the end of 5 years, $10,000 must be paid, (a) What equal amounts should the debtor deposit in a savings bank at the end of each 6 months in order to provide for the payment, if his savings accumulate at the rate (.04, m = 2) ? (6) How much must he deposit semi-annually if his first deposit occurs immediately and the last at the end of 5 years? 11. A debtor borrows $20,000, which is to be repaid, together with all accumulated interest at the rate (.05, m = 4), at the end of 6 years, (a) In order to pay the debt when due, what equal deposits must be made in a sinking fund at the end of each 6 months if the fund accumu- lates at the rate (.05, m = 2)? (6) At what rate, compounded semi- annually, could the debtor just as well have borrowed the $20,000, under the agreement that it be amortized in 6 years by equal payments at the end of each 6 months ? 12. A certain state provided for the sale of farms to war veterans under the agreement that (a) interest should be computed at the rate (.04, m = 2), and (6) the total liability should be discharged by 10 equal semi- annual installments, the first due at the end of 3 years. Find the nec- essary installment if the farm is worth $6000 cash. 13. Under what rate of interest will 25 semi-annual payments of $500, the first due immediately, amortize a debt of $9700? Determine both the nominal rate, compounded semi-annually, and the effective rate. 14. A sinking fund is being accumulated by payments of $1000 at the end of each year. For the first 10 years the fund earns 6%, effective, and, for the next 6 years, 4%, effective. What is the size of the fund at the end of 16 years? It is advisable, first, to find the amount at the end of: 16 years due to the payments during the first 10 years. 16. In order to create a fund of $60,000 by the end of 20 years, what equal payments should be made at the end of each 6 months if the fund accumulates at the rate (.04, m = 2) for the first 10 years and at 6%, effective, for the last 10 years? 16. A fund of $250,000 is given to a university. The principal and interest of this fund are to provide payments of $2000 at the beginning of each month until the fund is exhausted. If the university succeeds in investing the fund at 5%, compounded semi-annually, how many full payments of $2000 will be made? 17. Under what nominal rate, compounded semi-annually, would it be just as economical to amortize a debt in 10 years by equal payments at the end of each 6 months, as to pay interest semiTannually at the rate 92 MATHEMATICS OF INVESTMENT 6%, on the principal, and to repay the principal at the end of 10 years by the accumulation of a siring fund by equal payments at the end of each 6 months, invested at the rate (.04, m = 2). 18. A factory is worth $100,000 cash. At the time of purchase $25,000 was paid. Payments of $8000, including interest, were made at the end of each year for 6 years. The liability was completely discharged by six more annual payments of $9000, including interest, the first occurring at the end of the 7th year. What effective rate of interest did the debtor pay? HINT. Write an equation of value at the end of 6 years. Transpose all quantities in the equation to one side and solve by interpolation. Sec Appen- dix, Note 3, Example 2. SUPPLEMENTARY MATERIAL (} / 39. Funds invested with building and loan associations. A building and loan association is a cooperative enterprise whose main purpose is to provide funds from which loans may be made to members of the association desiring to build homes. Some members are investors only, and do not borrow from the as- sociation. Others are simultaneously borrowers and investors. Shares of stock are sold to members generally in units of $100. Each share is paid for by equal periodic installments called dues, payable at the beginning of each month. Profits of the association arise from investing the money received as clues. Members share in the profits in proportion to the amount they have paid on their shares of stock, and then- profits are credited as payments on their stock. When the periodic payments on a $100 share, plus the credited earnings, have reached $100, the share is said to mature. The owner may then withdraw its value or may allow it to remain invested with the association. Over moderate periods of time, the interest rate received by an association on its invested funds is approximately constant. The amount to the credit of a member, who has been making periodic payments on a share, is the amount of the annuity formed by his payments, -with interest at the rate being earned by the association. Example 1. A member pays $2 at the beginning of each month on a share in an association whose funds earn 6%, compounded monthly. What is to the credit of the member just after the. 20th payment? PAYMENT OF DEBTS BY PERIODIC INSTALLMENTS 93 Solution. The payments form an ordinary annuity of 20 payment inter- vals, if the term is considered to begin 1 month before the first payment is made. The amount on the 20th payment date is the amount of this ordinary annuity, or 2(a$Q\ at .005) = $41.96. Example 2. A member pays $1 at the beginning of each month on a $100 share. If the association is earning at the rate (.06, m = 12), when will the share mature? Solution. Let k be the number of installments necessary to bring the amount to the member's credit up to $100. - 100 = (a^of .005). By interpolation in Table VII, k = 81.3. Just after the 81st payment, at the beginning of the 81st month, the member is credited with (sg^ at .005) Case 1 n = k int. periods, p = 1, i - .005, R = $1, S = $100 $99.558. By the beginning of the 82d month, this book value has earned (.005) (99.558) = $.498, and the new book value is 99.568 + .498 = $100.056. Hence, no payment is necessary at the begin- ning of the 82d month. Properly speaking, the share matured to the value $100 at a time during the 81st month, but the member ordinarily would be informed of the maturity at the beginning of the 82d month. Example 3. A $100 share matures just after the 83d monthly pay- ment of $1. (a) What rate, compounded monthly, is the association earning on its funds ? (6) Wnat is the effective rate? Solution. (a) Let r be the rate per period of 1 month. The amount of the payment annuity is $100, Case 1 n s= 83 int. periods, p - 1, i = r, R = $1, S = $100. 100 = By interpolation in Table VII, r = = .441%. The nominal rate is 12 (.00441) = .0529, compounded monthly. (6) The effective rate i is obtained from 1 + i = (1.00441) 12 = 1.05422. (By Table n) Hence i = .0542, approximately. EXERCISE XXXVH 1. A member pays $1 at the beginning of each month on a share in an association which earns 5%, compounded monthly. What is to the mem- ber's credit just after the 50th payment? 2. When will the share of stock in problem I mature to the value $100, and what payment, if any, will be necessary on the maturity date? 3. If an association earns 6%, compounded monthly, when will a $100 share mature if the dues are $2 per month per share? 94 MATHEMATICS OF INVESTMENT 4. The monthly due on a $100 share is $.50, and the share will mature, approximately, just after the 131st monthly payment, (a) What rate, compounded monthly, does the association earn on its funds? (6) What is the effective rate? 5. An association earns approximately 6%, compounded quarterly, on its funds, (a) Find the date, to the nearest month, on which a $100 share will mature if the monthly due is $1, making your computation under the assumption that $3 is paid at the beginning of each 3 months. (&) Without any computation, tell why your result, as computed in (a), is smaller (or larger) than the actual result. 6. The monthly due on each $100 share is $1, in an association where the shares mature just after the 80th payment. What is the effective rate earned on the shares? 7. A man paid $30 per month for 80 months as dues on thirty $100 shares in an association, and, to mature his stock at the beginning of the 81st month, a partial payment of $6 was necessary. At what rate, com- pounded monthly, did his money increase during the 80 months? HINT. If he should pay $30 at the beginning of the 81st month, the book value of Ms shares would be $3024. 40. Retirement of loans made by building and loan associa- tions. If a man borrows $A from a building and loan associa- tion, he is usually caused, at that time, to become a member of the association, by subscribing for $A worth of stock. He must pay monthly interest (usually in advance) on the principal of his loan and dues on his stock. When his stock matures, with the value $^4., the association appropriates it as repayment of the principal of the loan. Thus, if a man borrows $2000 from an association which charges borrowers 7% interest, payable monthly in advance, the interest due at the beginning of each month is ^(.07) (2000) = $11.67. The borrower subscribes for twenty $100 shares of stock on which he pays $20 as dues at the beginning of each month if the due per share is $1. Thus, the monthly expense of the debt is $31.67. Payments continue until the twenty shares mature with the value $2000, at which time the association takes them as repayment of the principal of the debt, and closes the transaction. This method of retiring a debt is essentially a sinking fund plan, where the debtor's fund is invested in stock of the association. The debtor is benefited by this method because the rate received PAYMENT OF DEBTS BY PERIODIC INSTALLMENTS 95 on his stock investment is higher than could safely be obtained in the outside market. Example 1. A man borrows $2000 under the conditions of the pre- ceding paragraph. If shares in the association mature at the end of 82 months, at what nominal rate, compounded monthly, may the bor- rower consider that he is amortizing his debt ? Solution. The debtor pays $31.67 at the beginning of each month for 82 months. We wish the rate under which $2000 is the present value of this annuity due. Let r be the unknown rate per period of 1 month. The first $31.67 is paid cash and the remaining 81 payments form an ordinary annuity, .under Case 1, with p = 1. Hence, 2000 = 31.67 + 31.67(0^ at r), By interpolation in Table VIII, r = &% + (%) = .681%. The nominal rateis 12(.00681) = .082, approximately, compounded monthly. The effective rate i, if desired, can be obtained, with the aid of Table II, from 1 + i = (1.00681) 12 . EXERCISE 1. An association charges borrowers 6% interest payable monthly in advance and issues $100 shares on which the monthly dues are $1 per share, (a) At what rate of interest, compounded monthly, may a bor- rower consider his loan to be amortized, if shares in the association mature at the end of the 84th month, without a payment at that time? (6) What is the effective rate of interest? 2. Which would be more profitable, to borrow from the association of problem 1, or 'to pay 5% interest monthly in advance to some other lending source, and to repay the principal of the loan at the end of 84 months by the accumulation of a sinking fund, at the rate 6%, com- pounded monthly, if payments to the fund are made at the beginning of each month for 84 months? 3. An association charges borrowers 7% interest, payable monthly in advance, and issues $100 shares on which the monthly dues are $1 per share. If the shares mature at the end of 80 months, without a payment at that time, at what effective rate does a borrower amortize his debt? 4. An association charges 6% interest payable monthly in advance, and issues $100 shares on which the monthly due per share is $.50. If the shares mature at the end of 130 months, without a payment at that time, what is the effective rate paid by a borrower? CHAPTER VI DEPRECIATION, PERPETUITIES, AND CAPITALIZED COST 41. Depreciation ; .sinking fund plan. Fixed assets, such as buildings and machinery, diminish in value through use. Deprecia- tion is denned as that part of their loss in value which cannot be remedied by current repairs. In every business enterprise, the effects of depreciation should be foreseen and funds should be accumulated whose object is to supply the money needed for the replacement of assets when worn out. The deposits in these depreciation funds are called depreciation charges, 1 and are de- ducted periodically, under the heading of expense, from the cur- rent revenues of the business. NOTE 1. The replacement cost of an asset equals its cost when new minus its salvage or scrap value when worn out. Thus, if a machine costs $1000, and has a scrap value of $50 when worn out at the end of 10 years, its replace- ment cost is $950, the amount needed in addition to the scrap value in order to buy a new machine worth $1000. The replacement cost is also called the wearing value ; it is the value which is lost through wear during the life of the asset. A depreciation fund is essentially a sinking fund whose amount at the end of the life of the asset equals the replacement cost. Many different methods are in use for estimating the proper depreciation charge. Under the sinking fund method, the periodic depreciation charges are equal and are invested at compound inter- est at a specified rate. Under this plan, (a) the depreciation charges form an annuity whose amount equals the replacement cost, and (&) the depreciation fund is a sinking fund to which we may apply the methods for sinking funds used in Section 37. Exampk 1. A machine costs $1000 and it will wear out in 10 years. When worn out, its scrap value is $50. Under the sinking fund plan, i In this book, unless otherwise specified, we shall always assume that the charge for depreciation during each year is made at the end of the year, 96 DEPRECIATION AND CAPITALIZED COST 97 determine the depreciation charge which should be made at the end of each year for 10 years, if the fund is invested at 5%, effective. Solution. Let $2 be the annual charge. The replacement cost S = $950. Case 1 n = 10 int. periods, p = 1, i .05, R=$x,S = $950. 950 x = 950- at .05), . 05) $75.529. DEPRECIATION TABLE YEAH [NT. Dun ON FUND AT END OF YBAB PAYMENT TO FUND AT END OF YBAB IN Fmn> AT END OF YEAR BOOK VALUE AT END car YBAB 1 $0 $0 75.529 $ 75.529 $1000.00 924.47 2 3.777 75.529 154.835 845.16 3 7.742 75.529 238.106 761.89 4 11.906 75.529 325.540 674.46 . 5 16.277 75.529 417.346 582.65 6 20.867 75.529 513.742 486.26 7 25.687 75.529 614.958 385.04 8 30.748 75.529 721.235 278.76 9 36.062 75.529 832.826 16747 10 41.641 75.529 949.996 50.00 In Figure 5 the growth of the fund and the decrease in the book value are represented graphically. NOTE 2. A good depreciation plan is in harmony with the funda- mental principle of economics that capital invested in an enterprise should be kept intact. Thus, at the end of 2 years in Example 1, the $154.84 in the fund should be con- sidered as capital, originally in- vested in the machine, which has been returned through the revenues of the business. The book value of the machine, 1000 - 154.84 = $845.16, is the amount of capital still invested in the machine. IUUU onn \ / onn \ $> . / 7nn \A / snn ^ j, / S^ y jOO ^ r \ inn c- y \ onn X ^ \ ^ \ X s *_* 2 - .-Ye J ^ ars Ei [ . a. 5 ( i i 3 ) l( 98 MATHEMATICS OF INVESTMENT The condition per cent of an asset at any time is the ratio of its remaining wearing value to its wearing value when new. Example 2. In Example 1, find the condition per cent at the end of 6 years. Solution. Original wearing value is $950. Book value at end of 6 years is $486.26, and the remaining wearing value is 486.26 - 60 = $436.26. The condition per cent is 4 ^ 6 = .45923 or 45.923%. you EXERCISE XXXIX 1. In illustrative Example 1 above, find the amount in the deprecia- tion fund at the end of 4 years, without using the depreciation table. 2. A machine costs $3100 when new, it wears out in 12 years, and its final scrap value is $100. Under the sinking fund plan, determine the depreciation charge which should be made at the end of each 6 months if the fund accumulates at (.05, m = 2) . 3. (a) In problem 2, without forming a depreciation table, find the amount in the depreciation fund, and the book value of the asset, at the end of 5 years. (6) What is the condition per cent at this time? / 4. A building costs $100,000 and it will last 20 years, at the end of which time its salvage value is $5000. Under the pinking fund plan, when the fund earns 5%, effective, determine the size of the depreciation fund at the end of 6 years, if deposits in the fund are made at the end of each year. 6. A motor truck has an original value of $2500, a probable life of 6 years, and a final salvage value of $200. Its depreciation is to be covered by deposits in a fund at the end of each 3 months. Find the quarterly deposit if the fund earns (.055, m = 2). 6. A manufacturing plant is composed of part (a) whose post is $90,000, life is 15 years, and salvage value is $6000, and part (&) whose cost is $50,000, life is 20 years, and salvage value is $5000. If deprecia- tion charges are made at the end of each year and accumulate at 4%, effective, what is the total annual charge for the plant? 42. Straight line method. Consider an asset whose life is n years and whose replacement cost (wearing value) is $$, Sup- pose that annual depreciation charges are placed in a fund which does not earn interest. Then, in order to have $S at the end of cr n years, the annual charge must be -. The fund increases each n DEPRECIATION AND CAPITALIZED COST (99.) Sf year by -, and hence the book value decreases each year by - n n This method is called the straight line method, because we obtain straight lines as the graphs of the book value and of the amount in the fund (see problem 1 below). NOTE. An essential characteristic of the straight line method is that, under it, all of the annual decreases in book value are equal to -th of the total n wearing value S. It is usually stated, under this method, that is vrritten off n the book value each year for depreciation. Recall, from the table in Example 1 of Section 41, that the annual decreases in book value are not equal under the sinking fund plan ; they increase as the asset grows old. NOTE. The straight line method is the special case of the sinking fund method, where the rate earned on the depreciation fund is 0%. Hereafter, any general reference to the use of the sinking fund method should be understood to include the straight line plan as one possibility. In supplementary Section 48 below, another depreciation plan is considered which applies well to assets whose depreciation in early life is large compared to that in old age. The student is referred to textbooks on valuation and on accounting for many other special methods which are in use. Usually, each of these is particularly desirable for a certain type of assets. EXERCISE XL J 1. A building costs $50,000 and has a salvage value of $5000 when worn out at the end of 15 years, (a) Under the straight line method, form a table showing the amount in the depreciation fund and the book value at the end of each year. (6) Draw separate graphs of the book value and of the amount in the fund, using the years as abscissas. 2. In problem 1, find the annual depreciation charge under the sink- ing fund method, where the fund earns 4%, effective, and compare with the charge in problem 1. 43. Composite life. If the plant of an enterprise consists of several parts whose lives are of different lengths, it is useful to have a definition for the average or composite life of the plant as a whole. Let $S be the sum of the wearing values (replacement costs) and $D the sum of the annual depreciation charges for all parts. Under the sinking fund method let i be the effective rate earned on the depreciation fund. Then, the composite life is defined as the term of years, necessary for an annuity of $D, 100 MATHEMATICS OF INVESTMENT paid annually, to have the amount $& If k is the composite life in years, then 8 = D(SR at i), (46) which can be solved for k by interpolation. Under the straight line method of depreciation, fc is the number of times $D must be paid into a fund in order that the fund, with- out earning interest, should equal $S. Hence S = kD, k - |- (47) NOTE. In equation 46 place i = 0%. Then (s^ at 0) = k and equation 46 becomes 5 = kD, as found above. This is an obvious consequence of the fact that the straight line method is the sinking fund method with i = 0%. Example 1. A plant consists of part A, with life 20 years, original cost $55,000, and scrap value $5000 ; part B with life 15 years, original cost $23,000, and scrap value $3000 ; part C, with life 10 years, origi- nal cost $16,000 and scrap value $1000. Determine the composite Me, (a) under the sinking fund method, with interest at 4%, effective, and (&) under the straight line method. Solution. (a) The total wearing value is 50,000 -f 20,000 + 15,000 = $85,000. Under the sinking fund method, the annual charge for part A is 50000 t = $1679.09. Similarly, the charges for B and C are $998.82 (s^oi.04) and $1249.36, respectively. The total annual charge is $3927.27. Let k be the composite life. Then, the annuity of $3927.27, paid annually for k years, should have the amount $85,000, or Casel n = k int. periods, P = 1, * - .04, R - $3927.27, S = $85,000. 85000 - 8027.27(1)0 at .04) ; 4) =21.644. Table VII, k - 15.90 .(b) Under the straight line method, the -annual depreciation charges for the parts A, B, and C are, respectively, $2500, $1333.33, and $1500. The total annual charge is $5333.33. The composite lif e k is NOTE. Compare the results above. When the rate on the sinking fund is 0% (the straight line method), the composite life differs from the life when i = 4%, by only (15.94 15.90) = .04 year. Thus, under the sinking fund method, regardless of the rate earned on the fund, the composite life may be obtained approximately by finding the Me under straight line depreciation. DEPRECIATION AND CAPITALIZED COST EXERCISE XLI v 1. Fhid the composite life for the plant with parts A, B, and C below, under the sinking fund method (a) at 3%, effective, and (ft at 6%, effec- tive, and under (c) the straight line method. PABT LIFE COST SCHAP VALUE A 10 $20,500 $ 500 B 20 35,760 750 C 16 19,000 1000 44. Valuation of a mine. A mine, or any similar property, is a depreciable asset which becomes valueless when all of the ore is removed. Part of the net revenue of the mine should be used to accumulate a depreciation, or redemption fund, which will return the original invested capital when the mine is exhausted. The revenue remaining, after the depreciation charge, is the owner's net return on his investment. NOTE. We shall assume, in this book, that the annual revenue, or royalty, from a mine, or similar enterprise, is payable in one installment at the end of the year. Example 1. A mine, whose life is 20 years, costs $200,000 cash. "What should be the net annual revenue in order to pay 6% interest, annually, on the invested capital, and to provide an annual deposit for a redemption fund which accumulates at 4%, effective? Solution. Let $x be the annual deposit in the sinking fund to provide $200,000 at the end of 20 years. 200,000 - x(s at .04) ; x = $6,716.35. Case 1 n = 20 int. periods, p = 1, i = .04, R - $z, 8 = $200,000. Annual interest at 6% on $200,000 is $12,000. Annual revenue required is 12,000 + 6,716.35 = $18,716.35. Mining engineers furnish accurate estimates, for any given mine, of the life, k years, and of the annual revenue, $ JB. Suppose that a purchaser pays $P for a mine. If the annual revenue payments of $55 are exactly sufficient to amortize the original invested principal P at the effective rate i, then P is the present value of the annuity formed by the revenue installments, or P = R(an at i). (48) 1102) MATHEMATICS OF INVESTMENT Recall x that the amortization payments are exactly sufficient to pay interest at the rate i on P and to accumulate a sinking (redemp- tion) fund at the rate i to repay P at the end of k years. Therefore, the price paid under the assumption that P is amortized is the price we should obtain on assuming that the investor receives the rate i on his investment and places the surplus revenue in a redemption fund which accumulates at the rate i. Consider determining the purchase price P if the buyer desires the effective rate i on his investment and is able to invest his redemption fund at the effective rate r. Let $D be the deprecia- tion charge deposited annually in the redemption fund, which accumulates to the amount P at the end of k years. Then Annual interest on $ P at the rate i is PL Since revenue = (interest on capital) + (depreciation charge), (49) fi-Fi + D-Fi + JV-V -?(*+ 1_Y (50) (sji atr) \ (sj] at r)/ ^ ' P = - -- (51) Example 2. The annual revenue from a mine will be $30,000 until it becomes exhausted at the end of 25 years. What should be paid for the mine if 8% is to be earned on the invested capital while a redemption fund accumulates at 5%? Solution. From formula 51 P = 3000 . 3000 = $297,170. (Table IX) nft . 1 .10095246 ' V U8 + ( S2B1 oi.06) Nom In equation 61 place r = i, and use formula 39. One obtains p a jffi(ori at i), as obtained previously for this case in equation 48. EXERCISE XLII 1. A purchaser paid $300,000 for a mine which will be exhausted at the end of 50 years. What annual revenue from the mine will be required to pay 7% on the investment and to provide an annual deposit in a re- demption fund which accumulates at 5%, effective? 1 See Section 38. DEPRECIATION AND CAPITALIZED COST 103 2. The annual revenue from a mine will be $50,000 until the ore is ex- hausted at the end of 50 years. A purchaser desires 7% on his investment. What should he pay for the mine if his redemption fund accumulates (a) at 7%; (6) at 5%; (c) at 4%? 3. The privileges of a certain patent last for 10 years and the annual royalties from it will be $75,000. If a redemption fund can be accumu- lated at 5%, what should an investor pay for the patent rights if he de- mands 6% on his investment ? 4. An investor in an oil property desires 10% on his investment and assumes that he can accumulate a redemption fund at 5%. What should he pay for an oil field whose net annual revenue for its 10 years of life is estimated at $100,000? 5. A purchaser of a wooden ship estimates that the boat will be prac- tically valueless at the end of 6 years. If the net earnings for each of these years will be $50,000, what should the purchaser pay if he desires 8% on his investment and accumulates a depreciation fund at 4%? 45. Perpetuities. A perpetuity is an annuity whose payments continue forever. Present values l of perpetuities are useful in capitalization problems. Suppose that $1000 is invested at 6%, effective. Then, $60 interest is received at the end of each year, forever. That is, at 6%, the present value of a perpetuity of $60, paid annually, is $1000. Similarly, if $A is invested at the rate i, payable annually, it will yield R = Ai interest annually, forever. Hence, the present value $.A of a perpetuity of $72 paid at the end of each year is obtained from Ai = R ; or, A = - (52) NOTE 1. If, in the paragraph above, we change the word year to interest period, it is seen that, when the interest rate per period is i, the present value R of a perpetuity of $12, paid at the end of each interest period, is % Example 1. At the end of each 6 months, $50 is required to clean a statue. If money is worth (.04, m = 2), what is the present value of all future renovation? Solution. The future renovation costs form a perpetuity whose present value, by formula 62, is -^ = $2500. 02 1 The notion of the amount of a perpetuity is meaningless and useless, since the end of the term of a perpetuity does not exist. 104 MATHEMATICS OF INVESTMENT Consider a perpetuity of $1 paid at the end of each k years, and let (OK,, k at i) be its present value when money is worth the ef- fective rate i. In order to find a formula for a aa> t , first let us de- termine the installment $x which, if paid into .a fund at the end of each year for. k years, will accumulate to $1 at the end of k years. Then, $1 is the amount of the annuity of $x per annum and I = x(st\ at i) ; x ^- Therefore, a perpetuity of Ufcc per (St\ at i) annum will create a fund from which $1 can be paid at the end of each k years, forever. Hence, the present value of the perpetuity of $1 at the end of each k years is equal to the present value of the perpetuity of $#, per annum. Therefore, from formula 52 with R = x = i i i (sji at i) The present value $A of a perpetuity of $R paid at the end of each k years is :*") = ? 7^' (^) NOTE 2. Thus, if money is worth (.05, m = 1), the present value of a perpetuity of $90,000 paid at the end of each 20 years is A = PM- L_ . 9||p (.03024269) = $54,436.5. (Table IX) If i is not a table rate, formula 53 must be computed by inserting the explicit formula for (s^i ai i). NOTE 3. Recognize that formula 52, or formula 63, applies to a perpetuity whose first payment comes at the end of the first payment interval. The present value of a perpetuity due, or of a deferred perpetuity, can be obtained by the methods used for the corresponding type of annuity. EXERCISE XLHI 1. (a) Find the present value of a perpetuity which pays $100 at the end of each 3 months, if money is worth (.08, m = 4). (6) -What is the present value if the payments occur at the beginning of each 3 months ? v/2. An enterprise will yield $5000 net profit at the end of each year. At 4%, find the capitalized value of the enterprise, where the "capitalized value at 4%" is the present value of all future earnings. DEPRECIATION AND CAPITALIZED COST 105- 3. A bridge must be repainted each 5 years at a cost of $8000. If money is worth 5%, find the present value of all future repainting. 4. A certain depreciable asset must be replaced at the end of each 25 years at a cost of $50,000. At 6%, find the present value of all future replacements. 6. Find the present value of an annuity of $1000 paid at the end of each year for 75 years, if money is worth (.04, m = 1). Compare the result with the present value of a perpetuity of $1000, paid annually. 6. To repair a certain road, $1000 will be needed at the beginning of the 4th year and annually thereafter. Find the present value of all future repairs if money is worth 6%, effective. 46. Capitalized cost. The capitalized cost of an asset is de- fined as the first cost plus the present value of all future replacements, which it is assumed will continue forever. Let $C be the first cost and $# the replacement cost of an asset which must be renewed at the end of each k years. Then, the capitalized cost $X equals $C plus the present value of a perpetuity of $R paid at the end of each k years. When money is worth the effective rate i, we obtain, on using formula 53, K-C + ZL-. (54) i (s^ati) If the replacement cost equals the first cost, R C. Then, on changing C to -r and on placing R = C in formula 54, i = i + at i) i\ (SB at i \ (See formula 39) )J i (a^ati) Example 1. A machine costs $3000 new and must be renewed at the end of each 15 years, (a) Find the capitalized cost when money is worth (.05, m = 1), if the final scrap value of the machine is $500; (6) if the scrap value is zero. Solution. (a) Use formula 64 with C - $3000 and R = $2500. K - 3000 + ; - - $5317.12. .05 100 MATHEMATICS OF INVESTMENT (ft) When the scrap value is zero, C - R = $3000. From formula 65, NOTIS 1. If tho renewal coat of an asset is $1? and its life k years, the annual depreciation charge $D is given by R = D(s^ at i), or, (56) These future depreciation charges form a perpetuity of $D, paid annually, whoso present value is , or .1 D = l/fl 1A . 3 _J i i\ (s^jfrt i)/ * (a^oii) This renult is tlio same (see formula 54) as the present value of all future re- newal costs. Hence, the definition of capitalized cost may be restated to be " tho first cost plus tho present value of all future depreciation charges." NOTE 2. In formula 54, multiply both sides by i. Then Ki = Ci + R - - Ci + D. (See equation 56) (a ft at i) Thus, if an enterprise earns interest at the rate i on the capitalized cost K, tho revenue Ki provides for tho interest Ci at the rate i on the invested capital C and likewise for tho annual depreciation charge D. If two assets are available for serving the same purpose, that one should be used whoso capitalized cost is least. If their capitalized costs are the same, both assets are equally economical. Example 2. A certain type of pavement costs $12 per square yard, laid in place, and must be renewed at the same cost every 10 years. How much could a highway commission afford to pay to improve the pave- ment so that it would last 15 years, if money is worth 4%, effective? Solution. Let $* bo the cost per square yard of the improved type of pavement, whoso life is 15 years. If this type is just as economical na the old, its capitalized cost must bo tho same. The capitalized costs of the two types, as given by equation 66, are equated below : 12 1 .,_. 1 of .04) .04(0^0^.04) M) 12(11.1183874) $16 _ 450 . 4) 8.1108958 The commission could afford to pay anything less than (16.450 - 12) or $4.450 to improve the old pavement. DEPRECIATION AND CAPITALIZED COST EXERCISE i XLIV 1. Find the capitalized cost of a plant whose original cost is $200,000 and whose life is 25 years, if its final salvage value is $15,000. Money is worth 4%. 2. A section of pavement costing $50,000 has a lif e of 25 years. Find its capitalized cost if the renewal cost is $50,000, and if money is worth 3%. 3. If it costs $2000 at the end of each year to maintain a section of railroad, how much would it pay to spend, immediately, to improve the section so that the annual maintenance would be reduced to $500? Money is worth 5J%. 4. A bridge must be rebuilt every 50 years at a cost of $45,000. Find the capitalized cost if the first cost is $75,000, and if money is worth 3%. 6. One machine costs $15,000, lasts 25 years, and has a final salvage value of $1000. Another machine for the same purpose costs $18,000, lasts 28 years, and has a salvage value of $2000. If money is worth 5%, which machine should be used? 6. Would it be better to use tile costing $18 per thousand and lasting 15 years, or to use other material costing $22 per thousand and lasting 20 years, if money is worth 5% and if neither material has a scrap value? \f 7. A corporation is considering the use of motor trucks worth $5000 each, whose life is 4 years and salvage value is zero. How much would it pay to spend, per truck, to obtain other trucks whose life would be 6 years, and final salvage value zero ? Money is worth 5%. 8. The interior of a room can be painted at a cost of $10 and the paint- ing must be repeated every 2 years. If money is worth 6%, how much could one afford to pay for papering the room if the paper would need renewal every 3 years? 9. A certain manufacturing plant involves one part worth $100,000 new and needing replacement every 10 years at a cost of $90,000, and a second part costing $52,000 new and needing renewal every 12 years at a cost of $50,000. What should be the net operating revenue in order to yield 7% on the capitalized cost? 10. A certain dam will cost $100,000 and will need renewal at a cost of $50,000 every 10 years. If money is worth 4^%, how much could one afford to pay in addition to $100,000 to make the dam of permanent type? 1 After this exercise the student may proceed immediately to the Miscellaneous Problems at the end of the chapter, 108 MATHEMATICS OF INVESTMENT SUPPLEMENTARY MATERIAL 47. Difficult cases under perpetuities. Perpetuities are met to which formulas 52 and 53 do not apply. A systematic means for finding the present values of all perpetuities is furnished by infinite geometrical progressions. Example 1. If money is worth (.05, m = 2), find the present value of a perpetuity of $6 paid at the end of each 3 months. Solution. The present value $A of the perpetuity is the sum of the present values of all of the payments as listed below : Payment of $6 due at end of 3 mo. 6 mo. 9 mo. etc. to infinity . Present value of payment 6(1.025)~* GCl.CESr 1 6(1.025)"* etc. to infinitely many terms. A = 6[(1.025)-*+(1.026)^+ (1.025)-*+ . etc. to infinitely many terms}. The bracket contains an infinite geometrical progression whose first term a = (1.025)"*, and whose ratio w = (1.025)"*. The sum 1 of the series is - ; 6(1.025)"* _ 6(1.025)"* (1.025)* 1 - (1.025)-* 1 - (1.025)-* (1.025)* A 6 6 .0124228 $482.98. (Table X) (1.026)*- - 1 NOTE. The formulas 52 and 53 of Section 45 can be obtained by the method of Example 1 (see problems 3 and 5 below). EXERCISE XLV Use geometrical progressions unless otherwise directed. 1. . The annual rent of a perpetuity is $1000, payable in semi-annual installments. Find the present value when money is worth (.06, m = 1) . 2. Find the present value of the perpetuity in problem 1 if money is worth (.06, m = 4). 3. Derive the formula 52 for the present value of a perpetuity of $1 paid annually, when money is worth the effective rate i. This present value is generally denoted by the symbol a* ; that is, (a w at i) = T- 1 See Formula 22, Section, 91, DEPRECIATION AND CAPITALIZED COST 109/ 1 4. (a) By use of a geometrical progression, find the present value of a perpetuity of $100 paid at the end of each 10 years, if money is worth 5%, effective. (&) Compare with the result obtained by use of formula 53. 6. Derive formula 63 for R (a a0i fc at i), the present value of a perpetu- ity of $E paid at the end of each k years, with money worth the effective rate i. 6. At 6%, effective, find the capitalized value of an enterprise which yields a net revenue of $500 at the end of each month. 7. Let (a>at i) represent the present value, when money is worth i, effective, of a perpetuity whose annual rent is $1, paid in p installments per year. Prove that (a$at i) = = 1. (a^ at i). Jp Jp 8. If money is worth (.06, m = 4), find the present value of a perpetuity of $1000, paid semi-annually, by use of formula 53. 9. If money is worth (.05, m = 2), find the present value of a perpetu- ity of $100 paid monthly, by use of the result of problem 7. ^' 10. An irrigation system has just been completed. There will be no repair expense until the end of two years, after which $50 will be needed at the end of each 6 months. If money is worth 4%, effective, find the present value of the future upkeep. Solve by any method. 48. Constant percentage method of depreciation. Under the constant percentage method, the book value decreases each year by a fixed percentage of the value at the beginning of the year. If the life of the asset is n years, the constant percentage r, expressed as a decimal, must be chosen so that the original cost $C is reduced to the residual book value $22 at the end of n years. The decrease in the first year is Cr, and the value at the end of 1 year is C Cr = C(l r). Similarly, the book value at the end of each year is (1 r) tunes the value at the beginning of the year. By the end of n years, the original value C has been multiplied n times by (1 r) , or by (1 r) n , and the residual scrap value R is C(l r) n . Therefore, we may obtain r from , (57) Each annual reduction in book value is the depreciation charge for that year, and, if we consider all of these reductions in book value 110 MATHEMATICS OF INVESTMENT placed in a depreciation fund which does not earn interest, the fund will contain the replacement cost at the end of n years. Example 1. For a certain asset, the original cost is $3000, the life is 6 years, and the scrap value is $500. Find the annual percentage of de- preciatiou under the constant percentage method and form a table showing the changes in book value? Solution. From equation 67, log 5 = 9.22185 - 10. log (1 - r) = ilogi = 9.87031 - 10. V 3000 V o' 1 - r = .74183. r = .25817. The book values in the table below were computed by 5-place logarithms. Thus, at the end of 3 years, the book value is B = 3000(1 r) 8 ; log (1 - r) s = 3 log (1 - r) = 3(9.87031 - 10) - 9.61093 - 10 log 3000 = 3.47712 log B = 3.08805; B = $1224.7. DEPRECIATION TABLE YBAB BOOK VALUE AT END OF YBAB DEPRECIATION DUBINQ YEAB IN DHPR. PTTND AT END OF "XEAB 1 $2225.5 $774'.5 $ 774.5 2 1651.0 574.5 1349.0 3 1224.7 426.3 1775.3 4 908.6 316.1 2091.4 5 674.0 234.6 2326.0 6 500.0 174.0 2500.0 NOTE. When the scrap value R is relatively small, the method above gives ridiculously high depreciation charges in the early years. The method breaks down completely when R = 0. EXERCISE XLVI 1. (a) Find the annual percentage of depreciation under the constant percentage method, for a machine whose original cost is $10,000, life IB 5 years, and scrap value is $1000. (6) Form a depreciation table and draw a graph of the changes in book value. 2. In problem 1, find the annual depreciation charge under the sinking fund plan, where the fund earns 4%, effective, and compare with the result of problem 1, DEPRECIATION AND CAPITALIZED COST (ill A machine, whose life is 20 years, costs $50,000 when new and has a scrap value of $5000 when worn out. Find the annual rate of deprecia- tion under the constant percentage method. 4. For a certain asset, the depreciation in value during the early years of its life is known to be very great, as compared with the later years* Which of the three methods, straight line, sinking fund, or constant per- centage, would give a series of book values most in harmony with the actual values during the life of the asset? Justify your answer. MISCELLANEOUS PROBLEMS Depreciation, in problems below, is under the sinking fund method. 1. An automobile costs $3500 when new, and its salvage value at the end of 6 years is $400. (a) If the depreciation fund earns 4%, by how much is the book value decreased during the 4th year? (6) By how much is the book value decreased during the 4th year, under the straight line method? 2. A hotel has been built at a cost of $1,000,000 in an oil-boom city which will die at the end of 25 years. Assuming that the assets can be sold for $100,000 at that time, what must be the net annual revenue during the 25 years to earn 7% on the investment and to cover depreciation, where the depreciation fund earns 4%? 3. A syndicate will build a theater in a boom city which will die at the end of 30 years. For each $10,000 unit of net annual profit expected, how much can the syndicate afford to spend on the theater if 8% is desired on the investment while a redemption fund to cover the initial investment is accumulated at 5%? Assume that the theater will be valueless at the end of 30 years. 4. A machine, worth $100,000 new, will yield 12% net annual operating profit on its original cost, if no depreciation charges are made. If the life of the machine is 20 years, what annual profit will it yield if annual depreciation charges are made, where the depreciation fund accumulates at 4%, effective? Assume that the final salvage value of the machine is zero. 6. The life of a mine is 30 years, and its net annual revenue is $50,000. Find the purchase price to yield an investor 7%, if the redemption fund accumulates at 4%. 6. A mine will yield a net annual revenue of $25,000 for 20 years. It was purchased for $200,000. If, at this price, the investor considers 112 MATHEMATICS OF INVESTMENT that he obtains 10% on his investment, at what rate does he accumulate his redemption fund? 7. A certain railroad will cost $60,000 per mile to build. To maintain the roadbed in good condition will cost $500 per mile, payable at the be- ginning of each year. At the end of each 30 years, the tracks must be relaid at a cost of $30,000. What is the present value of the construction and of all future .maintenance and renewals, if money is worth 5%? 8. Find the capitalized value at 6%, effective, of a farm whose net annual revenue is $3000. 9. An automobile with a wearing value of $1200 has a life of 5 years. Upkeep and repairs cost the equivalent of $450 at the end of each year. What is the annual maintenance expense if the owner accumulates a de- preciation fund by annual charges invested at 5%? ^0. A certain piece of forest land will yield a net annual revenue of $25,000 for 15 years, at the end of which time the cut-over land will be sold for $15,000. (a) If money is worth 6% to an investor, what should he pay for the property? (&) If the investor desires 9% on his in- vested capital, and assumes that he can accumulate a redemption fund at 5% to return his original capital at the end of 15 years, find the price he should pay for the land, by use of the method which was used in deriving formula 51 for the valuation of a mine. CHAPTER VII BONDS 49. Terminology. A bond is a written contract to pay a definite redemption price $C on a specified redemption date and to pay equal dividends $D periodically until after the redemption date. The dividends are usually payable semi-annually, but may be paid annually or in any other regular fashion. The principal $F mentioned in the face of the bond is called the face value or par value. A bond is said to be redeemed at par if C = F (as is usually the case), and at a premium if C is greater than F. The interest rate named in a bond is called the dividend rate. The dividend $D is described in a bond by saying that it is the interest, semi-annual or otherwise, on the par value F at the dividend rate. Nona. The following is an extract from an ordinary bond : The Kansas Improvement Corporation acknowkdges itself to owe and, for value received, promises to pay to bearer FIVE HUNDRED DOLLARS on January 1st, 1926, with interest on said sum from and after January 1st, 1920, at the rate 6% per annum, payable semi-annually , until the said principal sum is paid. Furthermore, an additional 10% of the said principal shatt be paid to bearer on the date of redemption. For this bond, F = $500, C = $550, and the semi-annual dividend D = $15 is semi-annual interest at 6% on $500. A bond is named after its face F and dividend rate, so that the extract is from a $500, 6% bond. Corresponding to each dividend D there usually would be attached to the bond an individual coupon containing a written contract to pay $D on the proper date. 50. When an investor purchases a bond, the interest rate i which he receives 'on his investment is computed assuming that he will hold the bond until it is redeemed. It is important to recognize that the investment rate i is not the same as the dividend rate of the bond, except in very special cases, because i depends on all of the following : $P, the price paid for the bond ; $C, the redemption price ; the time to elapse before the redemption date ; the number of times per year dividends are paid, and the size of $1), the periodic dividend. 113 114 MATHEMATICS OF INVESTMENT In this chapter we shall solve two principal problems. First, the determination of the price $P which should be paid for a specified bond if we know the investment rate demanded by the buyer. Second, the determination of the investment rate if we know the price which the investor had to pay. 61. Purchase price to yield a given rate. The essential features of a bond contract are the promises (a) to pay $C on the redemption date and (6) to pay the annuity formed by the periodic dividends of SD. 1 If an investor desires a specified investment rate, the price $P he is willing to pay on purchasing the bond is p = (present value of $C due on the redemption date) (58) + (present value of the annuity formed by the dividends), where present values are computed under the investor's rate. Example 1. A $1000, 6% bond, with dividends payable semi-annually, will be redeemed at 105% at the end of 15 years. Find the price to yield an investor (.05, m = 1). Solution. " At 105% " means at a premium of 5% over the par value. F = $1000, C = 1000 + 50 = $1050. The semi-annual dividend D = (.03)1000 = $30. The redemption price $1050 is due at the end of 15 years. Hence, at the rate (.05, m = 1), Div. annuity, Case 1 n = 15 int. periods, p - 2, i = .05, R - $60. P = 1050(1.05)-" + 60(0^0* .05). F = 506.07 + 60 5 s . (a at .05) = $1136.54. Ja EXERCISE XLVII 1. A $1000, 5% bond, with dividends payable semi-annually, will be redeemed at 108% at the end of 7 years. Find the price to yield an in- vestor 6%, compounded semi-annually. The bonds in the table are redeemable at par. Find the purchase prices. The life is the time to the redemption date. 1 When a bond is sold on a dividend date, the seller takes the dividend $D which is due. The purchaser will receive the future dividends, which form an ordinary annuity whose first payment is due at the end of one dividend interval, and whose last payment is due on the redemption date. BONDS PBOB. PAB VALUE Lira DIVIDEND RATH DIVIDENDS PAYABLE INVESTMENT RATH 2. $ 1,000 10 yr., 6 mo. 5% semi-arm. (.06, m = 2) 3. 100 17 yr. 6% annually (.07, ro = 1) 4. 1,000 14 yr. 7% semi-ann. (.08, m = 1) 6. 500 9yr. 8% semi-ann. (.04, m = 2) 6. 2,000 7yr. 4% quarterly (.05, m = 4) 7. 1,000 8 yr., 6 mo. 3% semi-ann. (.06, m = 2) 8. 1,000,000 13 yr., 6 mo. 5J% semi-ann. (.04, m = 2) 9. 100,000 19 yr. 5% semi-ann. (.06, m = 1) 10. A $10,000, 5% bond, whose dividends are payable annually, will be redeemed at par at the end of 30 years. Find the purchase prices to yield (a) 5%, effective ; (6) 7%, effective ; (c) 4%, effective. Compare your results. 11. A $1000, 6% bond, whose dividends are payable semi-annually, is purchased to yield 5%, effective. Find the price if the bond is to be re- deemed at the end of (a) 5 years ; (6) 20 years ; (c) 75 years. Com- pare your results. s/ 12. A $100,000, 5% bond is redeemable at 110% at the end of 15 years, and dividends are payable annually. Find the price to yield (.06, m = 2). If a bond is redeemable at par (C = F) and if the investor's in- terest period equals the interval between successive dividends, it is easy to compute the premium (P F), the excess of the price P over par value F. Let k be the number of dividend periods to elapse before the bond matures, r the dividend rate per dividend interval, and i the investor's rate per interest period. Then, a dividend D = Fr is due at the end of each interest period and the redemption price F is due at the end of k periods. The equations below are easily verified. Div. annuity, Case 1 7i = k int. periods, p = 1, i = i, R = Fr. P = P - F = [ at i) From formula 28, - Fi (a^ at i) = - Fi Therefore, P - F - Fr(a^ at i)'- Fi * - F. = F(l + i)-* - F. i) = (Fr - Fi)(a m at i). Premium = P - F = F(r - 0(45-, at i). (59) 116 MATHEMATICS OF INVESTMENT NOTE 1. Formula 59 shows that, when r is greater than i, P F is positive, or the bond is purchased at a positive premium over par value F. When r is less than i, P F is negative or the bond is purchased at a negative premium, that is, at a discount from the par value F. Example 2. A $1000, 6% bond, with dividends payable semi- annually, is redeemable at par at the end of 20 years, (a) Find the price to yield an investor (.05, m = 2). (6) To yield (.07, m = 2). Solution. (a) From formula 59 with F = $1000, the premium is P F = 1000C03 - .025) (a^ at .025) = 5(0^ at .025) = $125.51. P - F+ 125.51 = $1125.51. (&) Premium = P - F = 1000(.03 - .035) (a^ at .035) - - 6(0^ at .035) = - $111.78. P = F - 111.78 = $882.22. In this case, we say the discount is $111.78. NOTE 2. Equation 59 could have been proved by direct reasoning. Sup- pose r is greater than i. Then, if an investor should pay $F for the bond, he would desire Fi as interest on each dividend date. Since each dividend is F r, he would be receiving (Fr Fi) = F(r i) excess income at the end of each interest period for k periods. Hence, he should pay, in addition to $F, a premium equal to the present value of the annuity formed by the excess income or F(r i)(a,j^at i). Similarly, when r is less than i, if the investor should pay $/*" for the bond, there would be a deficiency in income of Fi Fr = F(i r) at the end of each interest period. Hence, the present value of the defi- ciency or F (i r) (O F| at i) should be returned to the investor as a discount from the price F we supposed paid. VALUES TO THE NEAREST CENT OF A $100,000, 5% BOND WITH SEMI-ANNUAL DIVIDENDS INVEST. RATH WITH m 2 TIMB TO REDEMPTION D/LTB 101 YEARS 11 YEABS Hi YEABS 12 YEABS .0400 108505.60 108829.02 109146.10 109456.96 .0405 108060.01 108365.61 108665.14 108958.72 .0410 107616.62 107904.58 108186.75 108463.25 .0415 107175.43 107445.93 107710.93 107970.54 .0420 106736.43 106989.64 107237.65 107480.56 .0425 106299.59 106535.71 106766.91 106993.30 .0430 105864.92 106084.11 106298.69 106508.75 .0435 105432.40 105634.84 105832.97 106026.89 .0440 105002.01 105187.88 105369.74 105547.69 .0445 104573.75 104743.21 104908.99 105071.16 .0450 104147.61 104300.84 104450.70 104597.26 BONDS _ NOTE 3. To facilitate practical work with bonds, extensive tables have been computed showing the purchase prices of bonds redeemable at par. 1 The table on page 116 illustrates those found in bond tables. EXERCISE XLVm In the future use formula 59 to find P whenever F C and the in- vestor's interest period equals the dividend interval. Otherwise use the fundamental method involving formula 58. *! 1. Find the price to yield 4%, compounded serni-annually, of a $1000, 5% bond, with dividends payable semi-annually, redeemable at par at the end of 15^ years. 2. Verify all entries in the bond table on page 116, corresponding to the investment yields .04 and .045. 3. A $5000 bond, paying a $100 dividend semi-annually, is redeemable at par at the end of 11 years. Find the price to yield (.06, m = 2). 4. A man W signs a note promising to pay $2000 to M at the end of 5 years, and to pay interest semi-annually. on the $2000 at the rate 6|%. (a) What will M receive on discounting this note immediately at a bank which uses the interest rate 7%, compounded semi-annually? (6) What will M receive if the bank uses the rate 7% effective? 5. Find the price to yield 5%, effective, of a $10,000, 7% bond, with dividends payable annually, which is redeemable at par at the end of (a) 10 years ; (6) 15 years ; (c) 40 years, (d) Explain in a brief sen- tence how and why the price of a bond changes as the time to maturity increases, if the investor's rate is less than the dividend rate. 6. Find the price to yield 6%, effective, of a $10,000, 4% bond with annual dividends, which is redeemable at par at the end of (a) 5 years ; (6) 10 years; (c) 80 years, (d) Explain in one brief sentence how and why the price of a bond changes as the time to maturity is increased, if the investor's rate is greater than the dividend rate. 52. Changes in book value. On a dividend date, it is con- venient to use the term book value for the price $P at which a bond, would sell under a given investment rate i. Recall that this price $P, at which a purchaser could buy the bond, is the sum of the present values, under the rate i, of all payments promised in the bond. Hence, the dividends $D together with the redemption payment $(7 are sufficient to pay interest at rate i on the invested 1 Sprague's Complete Bond Tables contain the purchase prices to the nearest cent for a bond of $1,000,000 par value, corresponding to a wide range of investment rates. 118 MATHEMATICS OF INVESTMENT principal $P, and to return the principal intact. If a bond is pur- chased at a premium over the redemption price $C, only $C of the original principal $P is returned at redemption. Therefore, the remaining principal, which equals the premium (P C) originally paid for the bond, is returned in installments, or is amortized, through the dividend payments. Thus, each dividend $D, in addition to paying interest due on principal, provides a partial payment of principal. These payments reduce the invested principal, or book value, from $P on the date of purchase to $C on the redemption date. Example 1. A $1000, 6% bond pays dividends semi-annually and will be redeemed at 110% on July 1, 1925. It is bought on July 1, 1922, to yield (.04, m = 2). Find the price paid and form a table showing the change in book value and the payment for amortization of the premium on each interest date. Solution. =$1100, D=$30. P = 1100(1.02) ~ a +30(or l ai. 02) =$1144.811. TABLE OP BOOK VALUES FOR A BOND BOUGHT AT A PREMIUM DATH INT. AT 4% Dun ON BOOK VALTJH DIVIDEND RBOBIVBD FOR AMOBTIZATIOJ* OF PREMIUM FINAL BOOK VALUE July 1, 1922 $1144.811 Jan. 1, 1923 $22.896 $30.000 $7.104 1137.707 July 1, 1923 22.754 30.000 7.246 1130.461 Jan. 1, 1924 22.609 30.000 7.391 1123.070 July 1, 1924 22.461 30.000 7.639 1115.531 Jan. 1, 1925 22.311 30.000 7.689 1107.842 July 1, 1925 22.157 30.000 7.843 1099.999 On Jan. 1, 1923, for example, interest due on book value is .02(1144.81) = $22.896. Hence, the $30 dividend pays the interest due and leaves (30 22.896) =$7.104 for repayment, or amortization, of the premium; the new book value is 1144.811 - 7.104 = $1137.707. The check on the com- putation is that the fina.1 book value should be $1100, the redemption price. If a bond is purchased at a discount from the redemption price, that is, if P is less than C, the redemption payment C exceeds the original investment P by (C P). Hence, this excess must be the accumulated value on the redemption date of that part of the interest on the investment which the payments of D on the dividend dates .were insufficient to meet. Therefore, on each dividend date, the payment D is less than the interest due on BONDS 119 invested principal ; the interest which is not paid represents a new investment in the bond, whose book value is thereby increased. This writing up of the book value on dividend dates is called accumulating the discount because the book value increases from P on the date of purchase, to C on the redemption date, the total increase amounting to the original discount (C P). Example 2. A $1000, 4% bond pays dividends semi-annually and will be redeemed at 105% on January 1, 1924. It is purchased on Jan- uary 1, 1921, to yield (.06, m = 2). Find the price and form .a table showing the accumulation of the discount. Solution. C = $1050, and D = $20. P = 1060(1.03)-' + 20(ofi at .03) = $987.702. TABLE OP BOOK; VALUES FOB A BOND BOUGHT AT A DISCOUNT DATE INT. AT 8% Dura ON BOOK VA.LTJB DlVIDBND RBCBIVBD Fon AcouMuiiA.- TION OF DISCOUNT FINAL BOOK VALUE Jan. 1, 1921 $ 987.702 July 1, 1921 $29.631 $20.000 $ 9.631 997.333 Jan. 1, 1922 29.920 20.000 9.920 1007.253 July 1, 1922 30.218 20.000 10.218 1017.471 Jon. 1, 1923 30.524 20.000 10.524 1027.995 July 1, 1923 30.840 20.000 10.840 1038.835 Jan. 1, 1924 31.166 20.000 11.165 1050.000 In forming the row for July 1, 1921, for example, interest due at 6% is .03(987.702) - 29.631. Of this, only $20 is paid. The balance, 29.631 - 20 = $9.631, is considered as a new investment, raising the book value of the bond to 987.702 + 9.631 = $997.333. In his bookkeeping on July 1, 1921, the investor records the receipt of $29.631 interest although only $20 actually came into his hands. Also, his books show a new investment of $9.631 in the bond. Recognize that, when a bond is purchased at a premium, the dividend D is the sum of the interest / on the investment plus a payment for amortization of the premium, or I = D (amortization payment). (60) Thus, in illustrative Example 1 above on Jan. 1, 1923, the interest is 22.896 = 30 7.104. In accounting problems this fact is of importance. For instance, if a trust company purchases the bond of Example 1 for a trust fund, $1144.81 of the capital is invested. Suppose that the trust company considers all of each $30 dividend as interest and expends it for the beneficiary of the fund. Then, on July 1, 1925, the company faces (120* MATHEMATICS OF INVESTMENT? w' an illegal loss of $44.81 in the capital of the fund, because $1100 is re- ceived at redemption in place of $1144.81 invested. The company should consider only the entries in the 2d column of the table of Example 1 as income for the beneficiary. Similarly, when a bond is bought at a discount, / = D + (payment for accumulation of the discount) . (61) Thus, in illustrative Example 2 above on July 1, 1921, the interest is 29.631 = 20 + 9.631. From equations 60 and 61 we obtain, respectively, (amortization payment} D I, (payt. for accumulation of discount} = I D. Let P and PI be the book values on two successive dividend dates, at the same yield. Then, if the bond is at a premium, P\ equals P minus the amortization payment, or PI = P (D T) ; P! = PO+ (7-D). (62) If the bond is at a discount, Pi equals P plug the payment for the accumulation of the discount or Pi = P + (/ D), the same as found in equation 62. Hence, equation 62 holds true for all bonds. EXERCISE XLIX s | 1. A $1000, 8% bond pays dividends semi-annually on February 1 and August 1, and is redeemable at par on August 1, 1925. It is pur- chased on February 1, 1923, to yield (.06, m = 2) . Form a table showing the amortization of the premium. 2. A $1000, 5% bond pays dividends annually on March 1, and is redeemable at 110% on March 1, 1931. It is purchased on March 1, 1925, to yield (.07, m = 1). Form a table showing the accumulation of the discount. 3. By use of formula 58, find the book value of the bond of problem 2 on March 1, 1927, to yield (.07, m = 1) and thus verify the proper entry in the table of problem 2. Any book value in the tables of problems 1 and 2 could be computed in this way without forming the tables. 4. Under the investment rate (.04, m = 1), the book value of a $100, 5% bond on January 1, 1921, is $113.55, and dividends are payable an- nually on January 1. Find the amount of the interest on the investment, and of the payment for amortization on January 1, 1922. 4 1 BONDS 121 5. A $1000, 4% bond pays dividends annually on August 15 and is redeemable at par on August 15, 1935. An investor purchased it on August 15, 1923, to yield (.06, m = 1). (a) Without forming a table, find how much interest on invested capital should be recorded as re- ceived, in the accounts of the investor, on August 15, 1928. (6) How much new principal does the investor invest in the bond on August 15, 1928? 53. Price at a given yield between interest dates. The price of a bond on any date is the sum of the present values of all future payments promised in the bond. Let the investment rate be (i, m = 1), and suppose the last dividend was paid ~ th year ago. k At that time, the price P was the sum of the present values, at the rate (i, m = 1) of all future bond payments. To-day, the price P is the sum of the present values of these same payments because no more dividends have as yet been paid. Hence, P equals P accumulated at the rate (i, m = 1) for ^ years, or P = Po(l + i)*. (63) This price is on a strict compound interest yield basis. In practice, P is defined as PO, accumulated for - years at the rate i, simple K interest; P = Pof 1 + = i ), P = P -f Pofi (64) k That is, P equals PO plus simple interest on P from the last dividend date at the investment rate i. NOTB 1. The use of equation 64 favors the seller because it gives a slightly larger value of P than equation 63. The difference in price is neg* ligible except in large transactions. Use equation 64 in all problems in Ex- ercise L on page 123. Example 1. A $1000, 6% bond, with dividends payable July 1 and January 1, is redeemable at 110% on July 1, 1925. Find the price to yield (.05, m - 2) on August 18, 1922. Solution. July 1, 1922, was the lost dividend date. The price P then was Po = 1100(1.025)-" + 30(a u ~ja< .025) = $1113.77. Simple interest on 122 MATHEMATICS OF INVESTMENT $1113.77 from July 1 to Aug. 16 at the investment rate 5%, is $6.96. The price on Aug. 16 is P = 1113.77 + 6.96 - $1120.73. It is proper to consider that the dividend on a bond accrues (or is earned) continuously during each dividend interval. Thus, d days after a dividend date, the (accrued dividend) = (simple int. for d days on the face F at dividend rate). (65) Example 2. In Example 1, find the accrued dividend on August 16, 1922. Solution. From July 1 to Aug. 16 is 45 days. Accrued dividend is ^fc (.06) (1000) =$7.50. NOTE 2. In using equation 65, take 360 days as 1 year, and find the approximate number of days between dates, as in expression 9, Chapter I. When a bond is purchased at a given yield between interest dates, part of the price P is a payment to the seller because of the dividend accrued since the last dividend date. The remainder of P is the present value of future dividend accruals and of the future redemption payment. This remainder of P corresponds to what was defined as the book value of a bond in Section 52. Hence, between dividend dates, the price is P = (Accrued Dividend to Date) + (Book Value) ; (66) (Book Value) = P - (Accrued Dividend). (57) Equation 67 is also true on dividend dates ; the accrued dividend is zero because the seller appropriates the dividend which is due, and hence the book value and the purchase price are the same, as they were previously defined to be in Section 52. Exampk 3. For the bond of Example 1 above, find the book value on August 16, 1922, to yield (.05, m = 2). Solution. On Aug. 16, P - $1120.73, from Example 1. The accrued dividend to Aug. 16 is $7.50, from Example 2. Book value on AUK 16 is 1120.73 - 7.50 = $1113.23, from equation 67. NOTE 3. The accrued dividend, although earned, is not due-until the next chvidend date. Hence, theoretically, in equation 66 we should use, instead of the accrued dividend, its value discounted to date from the next dividend date. Thus, in Example 3 we should theoretically subtract the present value at COS, m - 2) on Aug. 16, 1922, of $7.50 due on Jan. 1, 1923, or 7.50(1.025)-$ - $7.35. The difference (in this case $.15) always is small unless a largo tranu- action is involved and, hence, it is the practice to use equation 67 as it standa. BONDS 123.) / EXERCISE L * 1. A $1000, 8% bond, with dividends payable January 16 and July 16, is redeemable at 110% on July 16, 1928. Find the purchase price and the book value on September 16, 1921, to yield (.04, m = 2) . Find the purchase prices and the book values of the bonds below on the specified dates. All bonds are redeemable at par. PHOB. PAH VALUE Div. RATH DIVIDEND DATES RBDBMP. DATE DATE OF PUBCHABH INVEST. RATE 2. $1000 5% June 1 6/1/1932 5/16/1922 (.07, m = 1) 3. 100 4% Jan. 1, July 1 7/1/1940 8/13/1931 (.04, m = 2) 4. 5000 44% May 1 6/1/1952 9/r /1924 (.06, m = 1) 5. 1000 6% June 1, Deo. 1 6/1/1937 8/16/1923 (.05, m = 2) 6. 100 64% May 1, Nov. 1 5/1/1934 3/1 /1927 (.07, m = 2) The book value between dividend dates may be found very easily by interpolation between the book values at the last and at the next dividend dates. This method is especially easy if a bond table is available. Example 4. A $100, 6% bond pays dividends on July 1 and January 1, and is redeemable at par on January 1, 1940. Find the book value and the purchase price on September 1, 1924, to yield (.04, m = 2). Solution. In the table below the book values to yield (.04, tn = 2) on 7/1/1924 and 1/1/1925 were computed from equation 69. Let B be the book value on Sept. 1, which is of the way from July 1 to Jan. 1. Henoe, sinoe 122.938 - 122.396 - .542, B = 122.938 - (.542) - $122.767. From equation 66, the purchase price P = 122.757 + 1 = $123.757. DATE BOOK VALUE 7/1/1924 9/1/1924 1/1/1925 $122.938 B $122.396 EXERCISE LI 1. A $1000, 4% bond pays dividends annually on July 1, and is re- deemable at par on July 1, 1937. (a) By interpolation find the book value on November 1, 1928, to yield (.06, m = 1). (&) Find the purchase price on November 1, 1928. 2. Find the book value in problem 1 by the method of illustrative example 3, Section 53, and compare with the result of problem 1. 3. A $5000, 6% bond pays dividends semi-annually on May 1 and November 1, and is redeemable at par on November 1, 1947. By use of interpolation find the book value and the purchase price to yield (.04, ?n - 2) on July 1, 1930, 124 MATHEMATICS OF INVESTMENT 4. A $1000, 5% bond, with dividends payable March 1 and September 1, is redeemable at par on March 1, 1935. By use of the bond table of Section 51, find by interpolation the book value to yield (.045, m = 2) on April 1, 1924. SUPPLEMENTARY NOTE. The interpolation method of illustrative Ex- ample 4, page 123, gives the same book value as is obtained by the method of Example 3, which uses equation 67. To prove this, lot the time to the present from the last dividend date be i th part of a dividend interval. Let Po be the /G book value on the last, and PI that on the next dividend date, P the purchase price to-day, D the periodic dividend, / the interest on P for a whole dividend interval at the investment rate, and B the book value of the bond to-day. First use the method of Example 3. Interest to date on Po at the investment rate is - (J), and the accrued dividend to date is - (D). From equation 64, K K Po -H rj and from equation 67, B /c l-D-p. 7 T^ * I - D (68) By interpolation, as in Example 4, since the present is ^th interval from the K last dividend date, B is -th part of the way /C DATE BOOK VALUH Last div. date Po Present B Next div. date Pi from Po to PI, or B From equation 02, PI PO hence B tion 68. Po+|(Pi-Po). I D, and Po + ~ -, the same as in equa- rC Equation 68 shows that, when a bond is selling at a discount, the accumula- tion of the discount in - th interval is f- th of the total accumulation for the k k interval, for, in equation 61 it is seen that I D is the accumulation for the whole interval. Similarly, if we write equation 68 as B => Po D - I k ' it is seen that the amortization of the premium on a bond in -th interval is -th of the amortization for the whole interval. 54. Professional practices in bond transactions. An investor buying a particular bond cannot usually demand a specified yield from his investment. He must pay whatever price is asked for that particular bond on the financial market. On bond exchanges, BONDS 125 and in most private transactions, the purchase price of a bond is described to a purchaser as a certain quoted price plus the accrued dividend. 1 That is, the market quotation on a bond is what we have previously called the book value of the bond, in equation 67. NOTE 1. The quotation for a bond is given as a percentage of its par value. That is, a $10,000 bond, quoted at 93, has a book value of $9325. Example 1. A $10,000, 6% bond, with dividends payable June 1 and December 1, is quoted at 93i on May 1. Find the purchase price. Solution. Quotation = book value = $9325.00. Accrued dividend since December 1 is $250. From equation 66, the price is 9325 + 250 = $9575. NOTE 2, Bond exchange methods are simplified by the quotation of book values instead of actual purchase a prices. If the yield at which a bond sells remains constant, the book value changes very slowly through the accumula- tion of the discount, or amortization of the premium, as the case may be. Hence, when the practice is to quote book values, the market quotations of bonds change very slowly and any violent fluctuation in them is due to a distinct change in the yields at which the bonds are selling. On the other hand, if the actual purchase price of a bond were the market quotation, the quotation would increase as the dividend -accrued and then, at each dividend date, a violent decrease would occur when the dividend was paid. Thus, even though the yield at which a bond were selling should remain constant, large fluctuations in its market quotation would occur . EXERCISE in 1. (a) A $1000, 5% bond, with dividends payable February 1 and August 1, is quoted at 98.75 on May 1 ; find the purchase price. (&) If the bond is purchased for $993.30 on April 1, find the market quotation then. 2. The interest dates for the 2d 4% Liberty Loan bonds are May 15 and November 15. Take their closing quotation on the New York Stock Exchange from the morning newspaper and determine the purchase price for a $10,000 bond of this issue. 3. A $1000, 6% bond whose dividend dates are January 1 and July 1 is quoted at 103& on October 16, Find the total price paid by a purchaser if he pays a brokerage commission of I % of the par value. 1 In bond market parlance, it is called accrued interest. The more proper word dividend has been consistently uaed in this book to avoid pitfalls which confront the beginner. As seen in Section 52, equations 00 and 61, the dividend is not the same as the interest on the investment. The terminology accrued interest in bond dealings must be learned by the student and appreciated to mean accrued dividend in the sonee of this chapter. 1 The purchase price is called the flat price in bond parlance, as contrasted with the price, and accrued interest quotation customarily used. 126 MATHEMATICS OF INVESTMENT 56. Approximate bond yields. On a given date the book value of a bond is quoted on the market and the problem is met of de- termining the yield obtained by an investor on purchasing the bond and holding it to maturity. We first consider an approxi- mate method of solution, using mere arithmetic. NOTE. A bond salesman, in speaking of the yield on a bond, usually refers to an investment rate compounded the same number of times per year as divi- dends are paid. Thus, by the yield on a quarterly bond, he means the invest- ment rate, compounded quarterly. We shall follow this customary usage in the future. Moreover, in computing yields it is usual to neglect the accrued dividend and brokerage fee paid at the time of purchase in addition to the book value. A yield is computed with reference to the book value of the bond. The justification of the following rules is apparent on reading them. Let $5 be the quoted book value of a bond, t the time in years before its maturity, and $C its redemption price. The in- vested principal changes from $5 at purchase to $C at redemption, so that the average book value $B is given by B Q = %(B + C). Even though a bond pays dividends quarterly or semi-annually, in using the rules below proceed as if the dividends were payable annually at the dividend rate and let $D be this annual dividend. Rule 1. When the quoted book value B is at a premium over C. Compute $A, the average annual amortization of the pre- mium from A = remium . Compute $1, the average annual t interest on the investment from I = D A. 1 Then, the ap- proximate yield r equals the average annual interest divided by the average invested capital or r = -^ -Do Rule 2. When B is at a discount from C. Compute $T, the average annual accumulation of the discount from T = un . t Compute I from I = D + T* Then, the approximate yield r equals the average annual interest divided by the average invested capital, or r = -- BQ * See equation 60. See equation 61, BONDS fiak Example 1. A $1000, 5% bond pays dividends semi-annuaUy and is redeemable at 110%. Eleven years before its maturity, the book value is quoted on the market at 93. Estimate the yield. Solution. Considering its dividends annual, D = $50, C = $1100, and the book value B = $930. Using Rule 2, the average accumulation of the discount is 1 ^ = $15.5, and I = 50 + 15.5 = $65.5. The average invested capital is 4(930 + 1100) = $1015. The approximate yield r = ffcV = -065, or 6.5%. Example 2. A $1000, 5 % bond pays dividends on July 1 and January 1 and is redeemable at par on January 1, 1961. Its quoted book value on May 1, 1922, is 113. Estimate the yield. Solution. Uae Rule 1 with B = $1130, t - 38f years, D = $50, and C = $1000. We find B = $1066. To find the average amortization of the premium we take t = 39, the nearest whole number, because the inaccuracy of our rule when t is large makes refinements in computation useless. A = -W = $3.3, I = 50 - 3.3 = $46.7, and therefore r = tffy = .044, or 4.4%. NOTE. -7- The author has experimentally verified that Rules 1 and 2 give estimated yields within .2% of the truth if : (a) the yield is between 4% and 8%, (b) the tune to maturity is less than 40 years, and (c) the difference between the dividend rate and the yield is less than 3%. Greater accuracy is obtained under favorable circumstances. For a bond whose term is more than 30 years, as in Example 2 above, take t as the whole number nearest to the time to maturity in years. In all other cases use the exact time to the nearest month. EXERCISE LIH Estimate the yields of the following bonds, by use of Rules 1 and 2. FBOB. PAGE To BB MARKET DrVlDHND TlMH TO i RATH PAID J * j $1000 par 107.24 5% semi-ann. 111 years 2. a 100 par 160.30 7% semi-ann. 25 years S.a 1000 par 84.28 4% semi-ann. 40 years 4. 100 110% 96.50 6% annually 231 years 5. 100 105% 115.00 5% annually 8 years 6. 100 116% 98.75 31% annually 20 years 1 Inspect the table of illustrative Example 2, Section 52. In that example, on computing the average semi-annual accumulation as in Rule 2, we obtain J(62.30) = $10.4, a result very close to all of the semi-annual accumulations. 3 The yields in the first three problems, determined by accurate means, are am follows: (1) 4.2%; (2) 3.4%; (3) 4.9%. Compare your results as found from Boles 1 and 2 in order to form an opinion of their accuracy. /128 MATHEMATICS OF INVESTMENT 7. A $10,000, 5% bond, -with dividends payable June 1 and December 1, is redeemable at par on December 1, 1950. On May 23, 1925, it is quoted at 89. Estimate the yield. 8. A Kingdom of Belgium 7|% bond, whose dividends are semi-annual, may be redeemed at 115% at the end of 8 years. Estimate its yield under the assumption that it will be redeemed then, if it is now quoted at 94. 66. Yield on a dividend date by interpolation. When the quoted value of a bond is given on a dividend date, the yield may be determined by interpolation. When annuity tables, but no bond tables, are available, proceed as follows : (a) Find the estimated yield r as in Section 55. (&) Compute the book value of the bond at the rate TI nearest to r for which the annuity tables may be used. (c) Inspect the result of (&) and then compute the book value for another rate r z , chosen so that the true yield is probably between TI and r 2 . Select r 2 as near as possible to TV (d) Find the yield i by interpolation between the results in (&) and (c). Example 1. A $100, 6% bond, with semi-annual dividends, is redeem- able at par. The quoted book value, 10| years before maturity, ia $111.98. Find the yield. Solution. (a) Average n.Tnrma.1 interest 1 = 6 ifj^a = $4.9 ; estimated yield r = &fc = 4.6%. (b) Book value 10J years before maturity to yield (.045, m = 2) is $112.44 (by equation 59). (c) Since $112.44 is greater than $111.98, the yield is greater than .046, and is prob- ably between .045 and .05. The book value at (.05, m = 2) is $108.09. Let (i, m - 2) be the yield. In the table, 112.44 - 108.09 - 4.35,' 112.44 - 111.98 = .46, and .06 - .045 = .005. Hence, * = -045 + jk (.005) - .0465, or the yield is ap- INVJEST. BATB .045, m = 2 i, m =* 2 .05, m = 2 BOOK VALUE 112.44 111.98 108.09 proximately 4.55%, compounded semi-annually. NOTE 1. A' more exact solution J may be obtained as follows :. At the yield (.0455, m = 2), found above, compute the book value P, using logarithms in equation 59 because the annuity tables do not apply ; P = 100 + .725 1 A solution as in Example 1 gives a result which is in error by not more than Ath of the difference between the table rates used in the interpolation. Wo ore, essen- tially, interpolating in Table VIII, and hence our result is subject only to the error we meet in using that table. BONDS (a^ } at .02275) = 111.998. Since $111.998 is greater than $111.980, the yield i is greater than .0455, and is probably between .0455 and .0456. By loga- rithms, the book value at (.0456, m = 2) is $111.910. From interpolation as in Example 1, i = .0455 + if (.0001) = .045520. The yield is 4.5520%, compounded semi-annually, with a possible error in the last decimal place. NOTE 2. The method of Example 1 is very easy if the desired book values can be read directly from a bond table (see problem 2 below) . If the bond table uses interest rates differing by $>&%, results obtained by interpolation in the table are in error by not more than a few .001%. Extension of the accuracy of a solution as in Note 1 is limited only by the extent of the logarithm tables at our disposal. NOTE 3. If the book value B is given on a day between dividend dates, the yield may be accurately obtained by the method of Section 68 below. An approximate result can be found by assuming B as the book value on the nearest dividend date and computing the corresponding yield. EXERCISE LIV Find the yield in each problem as in Example 1, page 128. If the in- structor so directs, extend the accuracy as in Note 1 above. 1. A $100, 4% bond pays dividends on January 1 and July 1 and is redeemable at par on January 1, 1932. (a) Find the yield if the quoted value on July 1, 1919, is 89.32. (6) Find the effective rate of interest yielded by investing in the bond. v 2. A $100, 5% bond pays semi-annual dividends and is, redeemable at par. By use of the bond table of Section 51, find the yield if the quoted value 11 years before maturity is 107.56. For each bond in the table, par value is $100. Find the yields. PBOB. TO BE REDEEMED AT DIVIDEND TIME TO MATDiuTr BOOK VALTJH RATH PAYABLE 3. 110% 6% annually 30 years $ 78.50 4. par 41% semi-ann. 15| years 110.76 6. par 3% semi-ann. 19 years 83.30 6, par 6% annually 12 years 121.00 7. 105% 6% semi-ann. 24 years 88.00 8. par 6% quarterly 10 years 107.00 9. On January 1, 1923, a purchaser paid $87.22, exclusive of brokerage, for a $100, 4% bond whose dividends are payable July 1 and January 1 130 MATHEMATICS OP INVESTMENT and which is redeemable at par on January 1, 1932. Find the yield ob- tained if the investor holds the bond to maturity. 10. A $100, 5% bond pays semi-annual dividends and is redeemable at par at the end of 9 years. If it is quoted at 83.20, find the effective rate of interest yielded by the investment. 67. Special types of bond issues. On issuing a set of bonds, a corporation, instead of desiring to redeem all bonds on one date, may prefer to redeem the issue in installments. The bonds are then said to form a serial issue. The price of the whole issue to net an investor a specified yield is the sum of the prices he should pay for the bonds entering in each redemption installment. Example 1. A $1,000,000 issue of 6% bonds was made on January 1, 1920, with dividends payable semi-annually, and the issue is redeemable serially in 10 equal annual installments. Find the price at which all bonds outstanding on January 1, 1927, could be purchased to yield an investor (M,m = 2). Solution. There is $300,000 outstanding. The price of the bonds for $100,000, which are redeemable at the end of 1 year, is 1000 (an of .02) + 100000 = $101,941.56; the prices of the bonds redeemable in the install- ments paid at the end of 2 years and of 3 years are $103,807.73 and $106,601.43, respectively. The total price of outstanding bonds is $311,350.72. An annuity bond, with face value $F, is a bond promising the payment of an annuity. The periodic payment $*S of the annuity is described as the installment which, if paid periodically during the life of the bond, is sufficient to redeem the face UPF in installments and to pay interest as due at the dividend rate on all of the face $F not yet redeemed. That is, the payments of $S amortize the face $F with interest at the dividend rate. When F and the dividend rate are known, S can be found by the methods of the amortization chapter. At a given investment yield, the price of an annuity bond is the present value of the annuity it promises. The annuity is always paid the same number of times per year as dividends are payable on the bond. Example 2. A certain ten-year, $10,000 annuity bond with the dividend rate 5% is redeemable in semi-annual installments of %8 each. (a) Find & (6) Find the purchase price of the bond, 5 years before maturity, to yield 6%, effective. BONDS .131 Solution. (a) The payments of $S amortize $10,000 at (.05, m 2). Bond annuity, Case 1 7i = 20 int. per., R = $5, A = $10,000, p - 1, i = .026. 10,000 = S(a m at .025) ; 8 = $641.47. (6) The price A at the yield (.06, m = 1) is the present value of semi- annual payments of S made for 5 years. Case 1 ,(2) n = 5 int. per., 33 = 2, i = 06, # = $1282.94. A = 1282.94(ojf.ri .06) = $5484.09. EXERCISE LV 1 1.- A $100,000 serial issue of 5% bonds, with dividends payable semi- annually, is redeemable in 5 equal annual installments. The issue was made July 1, 1927. On July 1, 1930, find the price of all outstanding bonds to net the investor (.06, m 2) . 2. For the bonds purchased in problem 1, form a table showing, on each dividend date, the dividend received, the installment (if any) which is paid, the interest due on the book value, and the final book value. 3. A house worth $12,000 cash is purchased under the following agree- ment : $2000 of the principal is to be paid at beginning of each year for six years ; interest at 6% is to be paid semi-annually on all principal outstand- ing. Two years later, the written contract embodying this agreement was sold to a banker, who purchased the remaining rights of the creditor !fco yield 7%, effective. What did the banker pay? J 4. A 5-year annuity bond for $20,000, with the dividend rate 6%, payable semi-annually, is issued on June 1, 1921. (a) Find the price on June 1, 1922, to yield (.03, m = 2). (&) Find the price on September 1, 1922, to yield (.03, m -- 2). 6. On June 1, 1924, find the price of the bond of problem 4 to yield (.06, m = 1). SUPPLEMENTARY MATERIAL 58. Yield of a bond between dividend dates. If the quoted value of a bond is given on a day between dividend dates, the yield may be found by interpolation by essentially the same procedure, with steps (a) , (&) , (c) and (d) , as used in Section 56 on a dividend date . After the completion of Exercise LV, the student may immediately proceed to the consideration of the Miscellaneous Problems at the end of the chapter. ( 132 ' MATHEMATICS OF INVESTMENT Example 1. A $100, 4% bond pays dividends annually on December 1 and is redeemable at par on December 1, 1931. Find the yield on Feb- ruary 1, 1926, if the book value is quoted at 95.926. Solution. (a) As in Section. 55, the average annual interest on the in- vestment is Z = 4 + .71 = $4.71 ; the estimated yield is ^5 = .048. (b) The nearest table rate is 5%. To find the book value at 5% on Feb. 1, 1926, first compute the values at 5% on Dec. 1, 1925, and Dec. 1, 1926, the last and the next interest dates. The results, 94.924 and 96.671, are placed in the first row of the table below, and from them we find by interpolation the book value TABLE OF BOOK VALUES YIELD DEO. 1, 1925 FEB. 1, 1020 Dno. 1, 1020 .05, m 1 i, m = 1 .045, m = 1 $94.924 97.421 $95.048 95.926 97.485 $95.671 97.805 on Feb. 1, 1926. Since 95.671 - 94.924 = .747, the book value on Feb. 1 at 5% is 94.924 + 1(.747) = 95.048. (c) Since 95.048 is less than 95.926 (the given book value), the yield is less than .05 and is probably between .046 and .05. Prices at .045 on Dec. 1, 1925, and on Dec. 1, 1926, were computed and from them the book value on Feb. 1, 1926, at .045 was obtained by interpolation. (d) The yield i is obtained by interpolation in the column of the table for Feb. 1. 97.485 - 95.048 = 2.437 ; 97.485 - 95.926 = 1.559 ; .05 - .045 = .005; hence i - .045 -f jjg .005 - .0482. The yield is approximately 4.82%, compounded annually, with a possible small error in the last digit. NOTE. The method above is extremely simple if the desired book values can be read from a bond table. The accuracy of the solution can be extended by the method of Note 1, Section 56. EXERCISE LVI N^ 1. By use of the bond table of Section 51, find the yield of a $100, 5% bond, with dividends payable on September 1 and March 1, and redeem- able at par on September 1, 1928, if the quoted book value December 1, 1917, is $106.78. 2. The interest dates of a $100, 4% bond are July 1 and January 1, and it is redeemable at par on January 1, 1930. (a) Find the yield if it is quoted at 83.25 on September 1, 1923. (6) Find the effective rate of interest yielded by the bond. BONDS 133 3. A man pays $87.22, exclusive of the brokerage commission, on Sep- tember 1, 1923, for a $100, 4% bond whose dividends are payable July 1 and January 1, and which is redeemable at par on January 1, 1930. De- termine the yield, if the bond is held to maturity. 4. A $1000, 5% bond pays dividends annually on June 16 and is re- deemable at 110% on. June 16, 1937. Find the yield if it is quoted at 112.06 on November 16, 1932. 6. The 3d Liberty Loan 4J% bonds are redeemable at par on Septem- ber 16, 1928. Interest dates are September 15 and March 15. If the bonds were quoted at 85 on May 15, 1921, what was the investment yield? MISCELLANEOUS PROBLEMS In the following problems, the word interest is used in the colloquial sense in connection, with bonds in place of the word dividend previously used. 1. A certain $1000, 5% bond pays interest annually. It is stipulated that, at the option of the debtor corporation, it may be redeemed at par on any interest date after the end of 10 years. The bond certainly will be redeemed at par by the end of 20 years. At what purchase price would a purchaser be certain to obtain 6% or more on his investment? 2. What is the proper price for the bond in problem 1 to yield 4%, or more? 3 . In return for a loan of $5000, W gives his creditor H the following note : Norfolk, June 1, 1915. For value received, I promise to pay, to H or order, $5000 at the end of 6 years and to pay interest on this sum semi-annually at the rate 6%. Signed, W. On December 1, 1916, H sold this note to an investor desiring (.07, m = 2) on his investment. "What did H receive? 4. What would H have received if he had sold the note to the same in- vestor as in problem 3 on February 1, 1917? 5. Two $1000 bonds are redeemable at par and pay 4% interest semi- annually. Their quoted prices on a certain date to yield (.05, m = 2) are $973 and $941.11, respectively. Without using annuity tables, and without computation, state which bond has the longer term to run and justify your answer. 134 MATHEMATICS OF INVESTMENT 6. Determine the term of the bond in problem 5 quoted at $941.11., 7. A $100, 4% bond, redeemable at par in 20 years, pays interest semi- annually. If it is quoted at $92.10, what is the effective rate of interest obtained by an investor? 8. On June 1, 1921, a corporation has its surplus invested in bonds which are redeemable at par on June 1, 1928, and which pay interest semi-annually at the rate 5%. If the bonds are quoted at 102.74, would it pay the corporation to sell the bonds and reinvest the proceeds in Government bonds which net 4.65%, effective? 9. A $1,000,000 issue of 5% bonds, paying interest annually, is to be redeemed at 110% in twenty annual installments. The first installmenl is to be paid at the end of 5 years and the last at the end of 24 years. II is desired that the annual payments (dividends on unpaid bonds and the redemption installment included)- at the end of, each year for the las- 20 years shall be equal. Determine the payment. 10. A house worth $12,000 is purchased under the following agreement $2,000 is to be paid cash and the balance of the principal is to be paid i\ four equal installments due at the ends of the 2d, 4th, 6th, and 8th years Interest at 6% ia to be paid semi-annually on all sums remaining due The note signed by the purchaser is sold after 3 years by the originf owner of the house. If the purchaser of the note demands 7%, compounde semi-annually, on his investment, what does he pay for the note? 11. A trust fund of $20,000 is invested in bonds which yield 5% at nually. The trust agreement states that $ of the income shall be give to the beneficiary each year and that the balance shall be re-invested in savings bank which pays 5%, compounded annually. The whole fun shall be turned over to the beneficiary after 10 years. If money is wori 6%, effective, to the beneficiary, what sum would he take now in place of h interest in the trust fund? Assume that he will live 10 years. 12. A corporation can sell at par a $1,000,000 issue of 6j% bon( redeemable at par in 20 years and paying interest annually. To pay the at maturity the corporation would accumulate a sinking fund by annu deposits invested at 4%, effective. Would it be better for the corpor tion to sell at par a $1,000,000 issue of 5% bonds y if these are redeemat in such annual installments during the 20 years that the total annual pa ments, dividends and redemption payments included, will be equa REVIEW PROBLEMS ON PART I 135 REVIEW PROBLEMS ON PART I 1. In purchasing a farm, $5000 will be paid at the end of each year for 10 years, (a) What is the equivalent cash price if money is worth 5%, effective? (6) What must be paid at the end of the 6th year to complete the purchase of the farm? 2. A depreciation fund is being accumulated by semi-annual deposits of $250 in a bank paying interest semi-annually at the rate 5%. What is in the fund just after the 30th payment? 3. A man wishes to donate to a university sufficient money to provide for the erection and the maintenance, for the next 50 years, of a building which will cost $500,000 to erect and will require $2000 at the end of each 3 months to maintain. What should he donate if the university is able to invest its funds at 5%, compounded semi-annually? 4. A debt of $100,000 bears interest at 6%, payable semi-annually. A sinking fund is being accumulated by payments at the end of each 6 months to repay the principal in one installment at the end of 10 years. If the sinking fund earns 4% interest, compounded semi-annually, what is the total semi-annual expense of the debt? 5. A debt of $100,000 is contracted under the agreement that interest at 6% shall be paid semi-annually on all sums remaining due. What payment at the end of each 6 months for 10 years will amortize this debt? 6. By use of a geometrical progression derive the expression for the amount of an annuity whose annual rent is $2000, payable in semi-annual installments for 10 years, if money is worth 6%, compounded quarterly. 7. Find the present value of an annuity whose annual rent is $3000, payable semi-annually for 20| years, if money is worth (.05, m = 4). 8. A merchant owes $6000 due immediately. For what sum should he make out a 90-day, non-interest-bearing note, so that his creditor may realize $6000 on it if he discounts it immediately at a bank whose discount rate is 8%? 9. If money is worth 5%, effective, find the equal payments which if made at the ends of the first and of the third years would discharge the liability of the following debts : (1) $1000 due without interest at the end of 3 years ; (2) $2000 due, with accumulated interest at the rate (06, m = 2), at the end of 4 years. 10. A trust fund of $100,000 is invested at 6%, effective. Payments of $10,000 will be made from the fund at the end of each year as long as pos- 136 f MATHEMATICS OF INVESTMENT sible. (a) Find how many full payments of $10,000 will be made, (b) How much will be left hi the fund just after the last full payment of $10,000? 11. Find the nominal rate of interest, compounded quarterly, under which payments of $1000 at the end of each 3 months for 20 years will be sufficient to accumulate a fund of $200,000. 12. Find the purchase price, to yield 6%, effective, of a $100, 5% bond with interest payable semi-annually, which is to be redeemed at 110% at the end of 10 yeara. 13. Estimate the yield on a bond which is quoted at 78, 10 years be- fore it is due, if it is to be redeemed at par and if its dividend rate is 6%, payable annually. 14. Find the capitalized cost of a machine, whose original cost is $200,000, which must be renewed at a cost of $150,000 every 20 years. Money is worth 5%, effective. '4 15. Find the yield of a $100, 6% bond, with semi-annual dividends, which is quoted at 93.70, 10 years before it is due, and is redeemable at par. 16. A $100, 5% bond, quoted at 86.33 on September 1, 1926, yields 6% if held to maturity. The last coupon date was July 1. What is the pur- chase price on September 1 ? 17. A man deposited $100 in a bank at the beginning of each 3 months for 10 years. What is to his credit at the end of 10 years if the bank pays 8%, compounded quarterly? 18. A man deposited $50 in a fund at the end of each month for 20 years, at which time deposits ceased. What will be in the fund 10 years later if it accumulated for the first 20 years at the rate 6%, effective, and at the rate 4%, effective, for the remainder of the time? 19. The cash price of a farm is $5000 and'money is worth (.06, m = 2) . What equal payments made quarterly will have an equivalent value, if the first payment is due at the end of 3 years and 9 months, and the last at the end of 134 years? 20. A corporation issues $200,000 worth of 6% bonds, redeemable at par at the end of 15 years, with interest payable semi-annually. The corporation is compelled by the terms of the issue to accumulate a sulking fund, to pay the bonds at maturity, by payments at the end of each 6 months, which are invested at (.04, m = 2). The bonds are sold by the corporation at 95 (95% of then* par value). Considering the total semi- annual expense as an annuity, under what rate of interest is the corpora- tion amortizing the loan it realizes from the bond issue? REVIEW PROBLEMS ON PART I 137 21. The present liability of a debt is $100,000. It is agreed that pay- ments of $5000 shall be made at the end of each 6 months for 10 years, and that, during this time, the payments include interest at the rate 6%, payable semi-annually. Then, commencing with a first payment at the end of 10J years, semi-annual installments of $10,000 shall be paid as long as necessary to discharge the debt, (a) After the end of 10 years, if the payments include interest at the rate 5%, payable semi-annually, how many full payments of $10,000 must be made? (&) What part of the payment at the end of lOf years is interest on outstanding principal and what part is principal repayment? 22. From whose standpoint, that of the debtor or of the creditor, is compound interest more desirable than simple interest? Tell why in one sentence. 23. A 90-day note whose face value is $2000 bears interest at 6%. It is discounted at a bank 30 days before due. What are the proceeds if the banker's discount rate is 8%? 24. A man borrows a sum of money for 72 days from a bank, charging 5% interest payable in advance, (a) What interest rate is he paying? (&) What interest rate would he be paying if he borrowed money for 1 year from this bank? 26. Estimate the yield of a bond whose redemption value is $135, whose dividends are each $10, and are paid annually, and whose purchase price 6 years before due is $147. 26. (a) Find the yield J of a $100, 6% bond bought for $103.53 on October 1, 1921. Coupons are payable semi-annually on February 1 and August 1 and the bond will be redeemed at par on February 1, 1928. (&) Find the effective rate of interest yielded by the bond. 27. The principal of a debt of $200,000 is to be paid after 20 years by the accumulation of a sinking fund into which 79 quarterly payments will be made, starting with the first payment in 6 months. Find the quarterly payment if the fund grows at 6%, compounded quarterly. 28. To amortize a certain debt at 6%, effective, 40 semi-annual pay- ments of $587.50 must be made. Just after the 26th payment, what principal will be outstanding? 29. (a) Find the capitalized worth at (.06, m = 12) of an enterprise which wnl yield a monthly income of $100, forever, first payment due now. (&) What is the present worth in (a) if the first monthly payment is due at the end of 6 months? 1 Find the yield as in Section 68, or, if that section has not been studied, use the method of Section 55. 138 MATHEMATICS OF INVESTMENT 30. A bridge will need renewal at a cost of $100,000 every 25 years. Under 5% interest, what is the present equivalent of all future renewals? 31. (a) By use of a geometrical progression determine a formula for the amount of an annuity whose annual rent is $20,000, which is paid quarterly for 30 years, if money is worth 7%, compounded annually. (6) Without a geometrical progression find the present value of the annuity. 32. A house is worth $50,000. In purchasing it $20,000 is paid cash and the remainder is to be paid, principal and interest at (.05, m = 2) included, by semi-annual installments of $2000, first payment to be made at the end of 2 years, (a) Determine by interpolation how many whole payments of $2000 will be necessary. (6) What liability will be out- standing just before the last full payment of $2000? 38. A debt of $50,000 is contracted and interest is at the rate 5%, compounded annually. The only payments (including interest) made were $5000 at the end of 2 years, and six. annual payments of $3000, starting with one at the end of 5 years. At the end of 10 years what ad- ditional payment would complete payment of the debt? 34. If money is worth (.05, m = 2), find the equal payments which must be made at the ends of the 3d and 4th years in order to discharge the following liabilities : (1) $5000 due at the end of 6 years, without in- terest ; (2) $4000 due at the end of 5 years with all accumulated interest at (.06, m = 1). 36. A debt of $100,000 is contracted and it is agreed that it shall be paid, principal and interest included, by equal payments at the end of each 6 months for 20 years. Interest is at the rate (4%, m = 2) for the first 10 years and at (5%, m = 2) for the next 10 years. What single rate of interest over the whole 20 years would have resulted in the same payments? 36. A man invested $100,000 in a certain enterprise. At the ends of each of the next. 10 years he was paid $4000 and, in addition, he received a payment of $25,000 at the end of 6 years. At the end of 10 years he sold his investment holdings for $80,000. 'Considering the whole period of 10 years, what was the effective rate of interest yielded by the invest- ment? HINT. Write an equation of value; solve by interpolation as in Note 3 in the Appendix. 37. Mr. A borrows $5000 from B to finance his college course and gives B a note, promising to pay $5000 at the end of 10 years, together with all accumulations at 3%, compounded semi-annually. (a) What will A pay REVIEW PROBLEMS ON PART I 139 at the end of 10 years? (&) At the end of 5 years, B sells A's promissory note to a bank, which discounts it, considering money as worth (.05, m = 1) . What does B realize from the sale ? 38. Find the price at which a $100, 5% bond would be quoted on the market on September 1, 1922, to yield the investor (.06, m = 2). The bond is to be redeemed at par on August 1, 1928, and interest dates of the bond are August 1 and February 1. 39. An industrial commission awards $10,000 damages to the wife of a workman killed in an accident, but suggests that this sum be paid out by a trust company in quarterly installments of $200, the first payment due immediately, (a) If the trust company pays (.04, m = 4) on money, for how long will payments continue ? (6) At the end of 10 years, the wife takes the balance of her fund. What amount does she receive? 40. Determine the capitalized cost of a machine worth $5000 new, due to wear out in 20 years, and renewable with a scrap value of $1000. Money is worth .05, effective. 41. Find the purchase price on December 1, 1920, of a $100, 6% bond with annual dividends, to yield at least 5%, if the bond may at the option of the issuing company be redeemed at 110% on any December 1 from 1930 to 1935, inclusive, or at par on any December 1 from 1943 to 1950. Justify your price. 42. A father wills to his son, who is just 20 years old, $20,000 of stock which pays dividends annually at the rate 6%. The will directs that the earnings shall be held to his son's credit in a bank paying 3%, effective, and that all accumulations as well as the original property shall become the direct possession of his son on his 30th birthday. Assuming that the market value of the stock on the 30th birthday will be $20,000, what is the present value of the estate for the son on his 20th birthday, assuming that money is worth 4|% and that the son will certainly live to age 30? 43. If money is worth (.06, m = 2), what equal installments paid at the ends of the 2d and 3d years will cancel the liability of the following obligations : (a) $1000 due without interest at the end of 5 years, and (6) $2000 due with accumulated interest at the rate 4%, compounded annually, at the end of 6 years? 44. Two years and 9 months ago X borrowed $2000 from Y, and has paid nothing since then, (a) If interest is at the rate 6%, payable semi- annually, determine the theoretical compound amount which X should pay to settle his debt immediately. (6) Determine the amount by the. practical rule, 140 MATHEMATICS OF INVESTMENT 45. At the end of each 6 months, $200,000 is placed in a fund which accumulates at the rate (.06, m = 2). (a) How many full payments of $200,000 will be necessary to accumulate a fund of $1,000,000 ? (6) What smaller payment will be needed to complete the fund on the next date of deposit after the last $200,000 payment? 46. Find the annual expense of a bond issue for $500,000 paying 5% annually, if it is to be retired at the end of 20 years by the accumulation of a sinking fund by annual payments invested at 4%, effective. 47. In problem 46, at what effective rate of interest could the bor- rower just as well borrow $500,000 if it is agreed to amortize the debt by equal payments made at the ends of the next 20 years? 48. How much is necessary for the endowment of a research fellowship paying $3000 annually, at the beginning of each year, to the fellow and supplying a research plant, whose original cost is $10,000, which requires $2000 at the beginning of each year for repairs and supplies? Money is worth 4%, effective. 49. A banker employs his money in 90-day loans at 6% interest, pay- able in advance. At what effective rate is he investing his resources? 50. Find the present value and the amount of an annuity of $50 per year for 20 years if money is worth 4%, payable annually. Use no tables and do entirely by arithmetic, knowing that (1.04) 20 = 2.191123. 51. $100,000 falls due at the end of 10 years. The debtor put $8000 into a sinking fund at the end of each of the first 3 years. He then decided to make equal a,nnua,1 deposits in his sinking fund for the remainder of the time in order to accumulate the necessary $100,000. If the fund earns (.04, m = T), what was the annual deposit? 52. A corporation is to retire, by payments at the end of each of the next 10 years, a debt of $105,000 bearing 5% interest, payable annually. The tenth annual payment, including interest, is to be $15,000. The other nine are to be equal in amount and are to include interest. Deter- mine the size of these nine payments. 53. Compute the purchase price to yield (,05, m = 4) of a $1000, 6% bond redeemable at 110% in 12J years, if it pays interest semi-annually. 54. Compute the present value of an annuity whose annual rent is $3000, payable quarterly for 6 years, if interest is at the rate 5.2%, effective. - 66. The maximum sum insured under the War Risk Insurance Act pays $57.50 at the beginning of each month for 20 years certain after death or disability. What would be the equivalent cash sum payable at death, or disability, at 3$% interest? REVIEW PROBLEMS ON PART I 141 66. A company issues $100,000 worth of 4%, 20-year bonds, which it wishes to pay at maturity by the accumulation of a sinking fund into which equal deposits will be made at the end of each year. The fund will earn 5% during the first five years, 4&% for the next 5 years, and 4% for the last 10 years. Determine the annual deposit. 67. The amount of a certain annuity, whose term is 7 years, is $3595 and the present value of the annuity is $2600. (a) Determine the effec- tive rate of interest. (&) Determine the nominal rate, if it is compounded quarterly. 58. How long will it take to pay for a house worth $20,000 if interest is at 5%, effective, and if payments of $4000, including interest, are made at the beginning of each year? Find the last annual payment which will be made, assuming that the debtor never pays more than $4000 at one time. 59. A sum of $1000 is due at the end of two years, (a) Discount it to the present time under the simple interest rate 6%. (6) Discount it under the simple discount rate 6%. (c) Discount it under the compound in- terest rate (.06, m = 1). 60. A concern issues $200,000 worth of serial bonds, paying 5% in- terest annually. It is provided that $30,000 shall be used at the end of each year to retire bonds at par and to pay interest. How long will it take to retire the issue ? Disregard the denomination of the bonds. 61. Find the value of a mine which will net $18,000 per year for 30 years if the investment yield is to be 6% and if the redemption fund is to be accumulated at 3%, compounded annually. 62. A man expects to go into business when he has saved $5000. He now has $2000 and can invest his savings at (5%, m 1) . How much must he save at the end of each year to obtain the necessary amount by the end of 5 years? 63. Find by interpolation the composite life on a 4% basis of a plant consisting of : Part (A), with life 10 years, cost new, $13,000, scrap value, $2000; Part (B), with life 16 years, cost new, $20,000, and scrap value, $3000. 64. How much could a telephone company afford to pay per $10 unit cost in improving the material in its poles in order to increase the length of life from 15 to 25 years? The poles have no scrap value when worn out, and money is worth (.05, m 1). 65. What are the net proceeds if a. 9Q-day n.gt;Q for $1000, bearing 6% interest, is discounted at 8%? 142 MATHEMATICS OF INVESTMENT 66. X requests a 60-day loan of $1000 from a bank charging 6% in- terest in advance. How much money does the bank give him and what interest rate is X paying on the loan? 67. A woman has funds on deposit in a bank paying (.04, m = 2). Should she reinvest in bonds yielding .0415, effective? 68. How long will it take for a fund of $3500 to grow to $4750 if in- vested at the rate 6%, compounded quarterly? 69. The sums $200, $500, and $1000 are due without interest in 1, 2, and 3 years respectively. When would the payment of $1700 equitably discharge these debts if money is worth (.06, m = 1) ? 70. A father has 3 children aged 4, 7, and 9. He wishes to present each one with $1000 at age 21. In order to do so he decides to deposit equal sums in a bank at the end of each year for 10 years. If it is assumed that the children will certainly live and that the bank pays (5%, m = 1), how much must the father deposit annually? 71. Which is worth more, if money is worth 6%, effective : (a) an in- come of 12 annual payments of $500, first payment to be made at the end of 2 years, or (&) 120 monthly payments of $50, first payment due at the end of 3 years and 1 month? 72. A $100, 5% bond pays interest quarterly and is redeemable at 110% at the end of 10 years. Find its price to yield 6%, effective. 73. Find the present value and the amount of an annuity of $3000 payable at the end of each 3 years for 21 years if interest is at the rate (.05, m = 2). 74. Find the nominal rate, converted quarterly, under which money will treble in 20 years. 76. (a) What effective rate is yielded by purchasing at par a $100, 4% bond, redeemable at par, which pays interest quarterly? (6) What rate, compounded semi-annually, does the investment yield ? 76. (a) In order to retire a $10,000 debt at the end of 8 years a sinking fund will be accumulated by equal semi-annual deposits, the first due im- mediately and the last at the end of 7 years. Find the semi-annual payment if the fund is invested at the rate (.04, m = 2) . (6) Find the size of the payments, under the same rate, if the first is made immediately and the last at the end of 8 years. 77. X lends $600 to B, who promises to repay it at the end of 6 years with all accumulated interest at (.06, m = 2). At the end of 3 years, B desires to pay in full. If X is now able to invest funds at only 4%, effective, what should the debtor pay? REVIEW PROBLEMS ON PART I 143 78. Find the nominal rate, converted quarterly, which yields the effective rate .0635. 79. (a) A house costs $23,000 cash. If interest is at the rate (.05, m = 1), what equal payments made at the beginning of each 6 months for 6| years will amortize the debt? (&) What liability is outstanding at the beginning of the 3d year before the payment due is made ? 80. How much must a man provide to purchase and maintain forever an ambulance costing $6000 new, renewable every 4 years at a cost of $4500 and requiring annual upkeep of $1500 payable at the beginning of each year? Money is worth 4%, effective. 81. A corporation was loaned $200,000 and, in return, made annual payments of $12,000 for 8 years in addition to making a final payment of $200,000 at the end of 9 years. What rate of interest did the corpora- tion pay? 82. A loan of $100,000 is to be amortized by equal payments at the end of each year for 20 years. During the first 10 years the payments are to include interest at 5%, effective, and, during the last 10 years, in- terest at 6%, effective. Determine the annual payment. 83. $10,000 is invested at 6%, effective. Principal and interest are to yield a fixed income at the end of each 6 months for 10 years, at the end of which time the principal is to be exhausted. Determine the semi-annual income. 84. A house worth $10,000 cash is purchased by B. A cash payment of $2000 is made and it is agreed in the contract to pay $500 of principal at the end of each 6 months until the principal is repaid and, in addition, to pay interest at the rate 6% semi-annually on all unpaid principal. Just after the payments are made at the end of two years, an investor buys the contract to yield 7%, compounded semi-annually, on the investment. What does the investor pay? 86. A farm worth $15,000 cash ia purchased by B, who contracts to pay $2000 at the beginning of each 6 months, these payments including semi-annual interest at 6%, until the liability is discharged. At the end of 4 years, just after the payments due are made, the contract signed by B is sold to an investor to yield him (.07, m = 2) on the investment. What does he pay? 86. A state, in making farm loans to ex-soldiers, grants them the fol- lowing terms: (a) interest shall be computed at the rate (.04, m = 2) throughout the life of the loan ; (&) no interest shall be paid, but it shall accumulate as a liability, during the first 4 years ; (c) the total indebted- 144 MATHEMATICS OF INVESTMENT ness shall be discharged by equal monthly payments, the first due at the end of 4 years and 1 month and the last at the end of 10 years. De- termine the monthly payment on a loan of $2000. 87. (a) A boy aged 15 years will receive the accumulations at 5%, effec- tive, of an estate now worth $30,000, when he reaches the age 21. What is the present value of his inheritance at 3$%, effective, assuming that he will certainly live to age 21? (6) Suppose that the boy is to receive, an- nually, the income at 5% from the estate and to receive the principal at age 21. Find the present value of the inheritance at 3?%, effective. 88. The quotation of a certain $100, 5% bond to-day (an interest date) is 88.37 and it yields 7% to an investor. Find the purchase price and market quotation 2 months later at the same yield. 89. A note signed by Y promises to pay $1000 at the end of 90 days with interest at 5%. (a) What would the holder X obtain on selling the note 30 days later to a banker whose discount rate is 6%? (&) What would he obtain if the note were discounted under the simple interest rate 6%? 90. A certain man invests $1500 at the rate (.04, m 1) on each of bis birthdays, starting at age 35 and ending at age 65. (a) At age 65, what does he have on hand ? (6) Suppose that at age 65 he decides to save no more and to spend all of his savings by taking from them an equal amount at the end of each month for 15 years, and suppose that he will certainly live that long. What can he take per month if the savings remain invested at (4%, m = 1) ? (c) If he desires to have $5000 left at the end of the 15 years, what will be his monthly allowance? 91. A depreciation fund is being formed by semi-annual deposits, to replace an article worth $10,000 new, when it becomes worn out after 6 years, (a) If money is worth (5%, m = 1), what is the semi-annual charge if the scrap value of the article is $1000? (6) How much is in the depreciation fund just after the third deposit? (c) Find the condition per cent of the article at the end of 3 years. 92. A debt of $50,000 is being amortized with interest at (.06, m = 2) by 24 equal Bemi-annual payments, the first payment cash. Find the payment and determine how much principal is outstanding just after the 12th payment. 93. Find the present value of a perpetuity of $1000, payable semi- annually, if interest is at the rate 6%, effective. 94. A man borrowed $10,000, which he agreed to amortize with interest at the rate 5%, payable annually, by equal payments, at the end of each. REVIEW PROBLEMS ON PART I 145 year for 12 years. Immediately after borrowing the money he invested it at 7%, payable semi-annually. In balancing his books at the end of 12 years, what is his accumulated profit on the transaction? 95. A loan agency offers loans to salaried workers under the following plan. In return for a $100 loan, payments of $8.70 must be made at the end of each month for 1 year. Determine the nominal rate, compounded quarterly, under which the transaction is executed. 96. A corporation can raise money by selling 6% bonds, with semi- annual dividends, at 95% of par value. To provide for their redemption at par at the end of 15 years, a sinking fund would be accumulated by in- vesting equal semi-annual deposits at (.04, m = 2). The corporation also can raise money by issuing, at par, 15-year, 7% annuity bonds redeemable in semi-annual installments, (a) Which method would entail the least semi-annual expense in raising $100,000 by a bond issue? (6) If money can be invested at (.04, m 2) by the corporation, what would be the equivalent profit, in values at the end of 15 years, from choosing the best method? 97. A corporation will issue $1,000,000 worth of 5% bonds, paying interest semi-annually and redeemable at par in the following amounts : $200,000 at the end of 5 years ; $300,000 at the end of 10 years ; $500,000 at the end of 15 years. A banking syndicate bids $945,000 for the issue. Under what interest rate is the corporation borrowing on the proceeds of the bond issue? 98. An investor paid $300,000 for a mine and spent $30,000 additional at the beginning of each year for the first 3 years for running expenses. Equal annual operating profits were received beginning at the end of the 3d year and ceasing with a profit at the end of 25 years, when the mine became exhausted. The investor reinvested all revenue from the mine at 5%, effective. What was the net operating profit for the last 23 years if, at the end of 25 years, he has as much as if he had received, and reinvested at 5%, effective, 8% interest annually on all capital invested in the mine and likewise had received back his capital intact at the end of 25 years? 99. A man who borrowed $100,000 under the rate 6%, payable semi- annually, is to discharge all principal and interest obligations by equal payments at the end of each quarter for 8 years. At the end of 2 years, his creditor agrees to permit him to discharge his future obligations by 4 equal semi-annual payments, the first due immediately, (a) What will be the semi-annual payment if the creditor, in computing it, uses the rate 146 MATHEMATICS OF INVESTMENT 5%, compounded semi-annually? (6) What will be the semi-annual payment if the rate (.07, m = 2) is used in the computation? 100. A contract for deed is the name assigned to the following type of agreement in real estate transactions : In purchasing a piece of property worth $2000 cash B agrees to pay $500 cash and to pay $25 at the end of each month, these payments to include interest at the rate 6%, payable monthly, until the property is paid for. The owner A agrees on his part to deliver the deed for the property to B when payment is completed. Six months after the contract above was made, A sells it to an investor, who obtains the rate 7%, compounded monthly, on his investment. What does he pay, if A has already received the $25 due on the contract on this date? PART II LIFE INSURANCE CHAPTER VIII LIFE ANNUITIES 69. Probability. The mathematical definition of probability makes precise the meaning customarily assigned to the words chance or probability as used, for example, in regard to the winning of a game. Thus, if a bag contains 7 black and 3 white balls and if a ball is drawn at random, the chance of a white ball being ob- tained is -^y because, out of 10 balls in the bag, 3 are white. Definition. If an event E can happen in h ways and fail in u ways, all of which are equally likely, the probability p of the event happening is 7, ' - ~b w and the probability q the event failing is NOTE 1 . In the ball problem above, the event E was the drawing of a white hall ; h = 3, h + u = 10, p = A- The probability of failure q = &. The denominator (u + A) in the formulas should be remembered as the total number of ways in which E can happen or fail. From formulas 1 and 2, it is seen that p and q are both less than 1. Moreover, , h , u _ u + h _ * P ^ q .h + u' r h + u u + h or the sum of the probabilities of failure and of success is 1. If an event is certain to happen, u = and p = - = 1. h j EXERCISE LVH / 1. An urn contains 10 white and 33 black balls. What is the probabil- ity that a ball drawn at random will be white? 2. A deck of 52 cards contains 4 aces. On drawing a card at random from a deck, what is the probability, that it will be an ace? 148 MATHEMATICS OF INVESTMENT ;| 3. Out of a class of 50 containing 20 girls and 30 boys, one member ia chosen by lot. What is the probability that a girl will be picked? 4. A cubical die with six faces, numbered from 1 to 6, is tossed. What is the probability that it will fall with the number 4 up ? t 6. A coin is tossed. What is the probability that it will fall head up ? -J 6. If the probability of a man living for at least 10 years is .8, find the probability of him dying within 10 years. 7. If the probability of winning a game is , what is the probability of losing? NOTE 2. It is important to recognize that when we say, as in problem 7, above, "the probability of winning is ," we mean : (a) if a very large number of games are played, it is to be expected that approximately $ of them will be won, and (&) if the number of games played becomes larger and larger with- out bound, it is to be expected that the quotient, of the number of them which are won divided by the total played, will approach as a limiting value. We do not imply, for instance, that out of 45 games played exactly $ of 45, or 27 games will be won. We must recognize that, if only a few games are played, it may happen that more, or equally well less, than of the total will be won. 8. As a cooperative class exercise, toss a coin 400 times and record at each trial whether or not the coin falls head up. How many were heads out of (a) the first 10 trials; (&) 50 trials; (c) 400 trials? Compare in each case the number of heads with of the number of trials so as to appreciate Note 2, above. NOTE 3. The assumption in the definition of probability that all ways of happening or failing are equally likely, is a very important qualification. For example, we might reason as follows : A man selected at random will either live one day or else he will die before to-morrow. Hence, there are only two possibilities to consider, and the probability of dying before to-morrow is $. This ridiculous conclusion would neglect the fact that he is mare likely to live than to die, and hence our definition of probability should not be applied. 60. Mortality Table. Table XIII was formed from the accu- mulated experience of many American life insurance companies. This table should be considered as showing the observed deaths among a group * of 100,000 people of the same age, all of whom 1 The actual construction, of a mortality table is a very difficult matter and cannot be considered here. It is, of course, impossible to obtain for observation 100,000 children of the same age, 10 years, and to keep a record of the deaths until all have died. However, data obtained by insurance companies, or census records of births and deaths, can be used to create a table equivalent to a death record of a repre- sentative group of 100,000 people of the same age, all of whom were alive at ago 10. LIFE ANNUITIES 149 were alive at age 10. In the mortality table, l x represents the number of the group still alive at age x, and d a the number of the group dying between ages x and x + 1. Thus, IK = 89,032, and das = 718 (= 89032 - 88314). In general, d x = l x - Z^+i. Out of l x alive at age x, Ix+n remain alive at age x + n, and hence l a lx+n die between ages x and x -\- n. Thus, Z 2 6 l& r 2154 die between ages 25 and 28. When the exact probability of the happening of an event is unknown, the probability may sometimes be determined by ob- servation and statistical analysis. Suppose that an event has been observed to happen h times out of m trials in the past. Then, as an approximation to the probability of. occurrence we may take v = This estimated value of p becomes increasingly reliable m as the number of observed cases increases. The statistical method is used in determining all probabilities in regard to the death or survival of an individual selected at random; our observed data is the tabulated record given in the mortality table. Exampk 1. A man is alive at age 25, (a) Find the probability that he will live at least 13 years. (6) Find the probability that he will die in the year after he is 42. Solution. (a) We observe Zst = 89,032 men alive at age 26. Of these, 79,611 ( = Z 38 ) remain alive at age 38. The probability of living to age 38 is B 796U A* 785 die in their 43d year. The probability of dying is p = ^i" EXERCISE LVm In the first nine problems find the probability : * 1. That a boy aged 10 will live to graduate from college at age 22. ^ 2. That a man aged 33 will live to receive an inheritance payable at age 45. 3. That a boy aged 15 will reach age 80. 4. (a) That a man aged 56 will die within 5 years. (6) That he will die during the 5th year. 6. That a man aged 24 will live to age 25. " 6. That a man aged 28 will die in his 38th year. 150 MATHEMATICS OF INVESTMENT 7. That a man aged 28 will die in the year after he is 38. 8. That a man aged 40 will live at least 12 years. 9. That a man aged 35 will live at least 20 years. ''-' 10. If a man is alive at age 22, between what ages is he most likely to die and what is his probability of dying in that year? NOTE. The problems in probability solved in the future in the theory of life annuities and of life insurance, so far as it is presented in this book, will be like those of' Exercise LVIII. None of the well known theorems on probabil- ity are needed in solving such problems ; the mere definition of probability is sufficient. Hence, no further theorems on probability are discussed in this text. The student is referred for their consideration to books on college algebra. 61. Formulas used with Table XTTT. In Table XIII, we verify that the number dying between ages 25 and 28 is Z 2 6 Ins = d%t + d 2 e + ^27, the sum of those dying in their 25th, 26th, or 27th years. Similarly, those dying between ages x and x + n are From Table XIII, Zg 8 = 0, and hence Z 87; !&, etc., are zero because all are dead before reaching age 96. The group of l x alive at age a are those who die in the future years, so that Z = d x + ds+i + + d w . It is convenient to use " (x} " to abbreviate a man aged x. Le' n p a represent the probability that (x) will live at least n years, or that (x) will still be alive at age x + n. Since Za; +n remain aliv< at age (x -\- n), out of la, alive at age x, * When n = 1, we omit the n = 1 on n p x and write p a for the prob ability that (jc) will live 1 year ; P. = -f 1 ' (6 la The values of p^ are tabulated in Table XIII. Let q x represen the probability that (x} will die before age (x + 1). Since a of the group of l x die in the first year, 2. = 150 MATHEMATICS OF INVESTMENT 7. That a man aged 2S will die in the year after he is 38. 8. That a man aged 40 will live at toast 12 years. 9. That a man aged 35 will live at least 20 years. 1 ' 10. If a man is alive at ago 22, between what ages is he most likely to die and what is his probability of dying in that year? NOTH. The problems hi probability solved in the future in the theory of life annuities and of life inmiranee, HO fur us it, in presented in this book, will bo like those of 'Exercise LV.III. None of the wnll known theorems on probabil- ity are needed in solving Huoh problems; the more definition of probability is sufficient. Honee, no further theorems on probability arc disuusHed in this text. The student is referred for their consideration to books on college algebra. 61. Formulas used with Table Xm. In Table XIII, we verify that the number dying botwoon ages 25 and 28 is l^ l& = rZ 2 r> + dQ -\- dvj, the sum of those dying in their 25th, 2Gth, or 27th years. Similarly, those dying between ages x and x -f- n are Z* L+ n = d x + d x+ i + + d+i-i. (3) From Table XIII, Zoo = 0, and hence 1&, Z 08 , etc., are zero because all are dead before reaching age 90. The group of l x alive at age x are those who die in the future years, so that lx = d x + d x+i + + <ZgG. (4) It is convenient to use " (x) " to abbreviate a man aged x. Let n ps represent the probability that (x} will live at least- n years, or, that (x) will still be alive at age x + n. Since l x+n remain alive at age (x + n), out of l x alive at ago x, When 7i = l,wa omit then = 1 on n p a and write p a far the prob- ability that (x) will live 1 year j p. = *-"- (6) Lfo The values of y) K are tabulated in Table XIII. Let q a represent tho probability that (x) will die before ago (x + 1). Since d x of tho group of l a die in tho first year, <1 - ( ' (7) LIFE ANNUITIES 151 The values of q a are tabulated in Table XIII. Let n \q a represent the probability that x will die in the year after reaching age (x + ri), between the ages (re + ri) and (x + n + 1). Of the original. group of l a alive at age x, d^+n die in the year after reaching age (x + ri). Hence Let n ([x represent the probability that x will die before reaching age x + n. Of the group of l e alive at age x, (l x Ix+n) will die before reaching age x + n, and hence the probability of dying is I n *+" ~ 8 - fQ~\ |n(Z* -- 7 -- (V) IB Example 1. State in words the probabilities denoted by the following symbols and find their values by the formulas above : (a) npas', (&) isl&a; (c) lug* Solution. (a) npas is the probability that a man aged 25 will be alive at age 42 ; npa 6 = ^ = L^r^n' W I 6 !? 22 ^ ^ e probability that a man aged 22 Its 89032 will die in the year after he reaches age 37; i&\q& == y =, j~ (c) lugaa ijj 91192 is the probability that a man aged 22 will die before he is 15 years older (before reaching age 22 + 16 - 37) ; | 18 <? M - lsL=JlL. EXERCISE LIX 1. Find the probability that a man aged 25 will live at least (a) 30 years ; (6) 40 years ; (c) 70 years. '' 2. Find the probability that a man aged 30 will die in the year after reaching age 40. 3. From formula 7 find the probability of a man aged 23 dying within 1 year, and verify the entry in Table XIII. 4. From formula 6 find the probability of a man aged 37 being alive at age 38, and verify the table entry. sj 5. Find the probability that a man aged 33 will die in the year after reaching age 55. 6. State in words the probabilities represented by the following sym- bols and express them as quotients by the formulas above: lap^J 15)242; wl?a; Mas; 270j* loPsaJ Pa', M I (Zsa- 152 MATHEMATICS OF INVESTMENT 7. From formulas 5 and 9 prove that nPx = 1 |n2 NOTE. If we consider the event of (x) living for at least n years, the failure of the event means that (x} dies within n years. Hence, the result of problem 7 should be true because, from Section 59 the sum of the probabilities of the success and of the failure of an event is 1, and p = 1 q. 8. What is the probability of a man aged 26 dying some time after he reaches age 45? 9. Verify formula 4 for x 90. 10. Verify formula 3 for x = 53 and n = 5. 62. Mathematical expectation; present value of an expecta- tion. If a man gambles in a game where the stake is $100, and where his probability of winning is .6, his chances are worth .6(100) = $60. Such a statement is made precise in the following Definition. If p is the probability of a person receiving a sum $S, the mathematical expectation of the person is pS. If the sum $S is due at the end of n years, the mathematical expectation at the end of n years is pS. If money is worth the ef- fective rate i, the present value $4 of the expectation is given by A = PS(1 + if". (10) NOTE. In the future, the arithmetical work in all examples will be per- formed by 5-place logarithms. Example 1. If money is worth 3$%, find the present value of the ex- pectation of a man aged 25 who is promised a payment of. $5000 at the end of 12 years if he is still alive. Solution. The probability p of receiving the payment is the probability of the man living to age 37, or p = upao- From equation 10, the present value of the expectation is A = ^55000(1.035)-" = fiOM(1.085)-*V ( Fo rmula 5) A = $2986.4. B (Tables VI and XIII) NOTE. In Example 1 it would be said that the payment of $6000 at the end of 12 years is contingent (or dependent) on the survival of the man. For brevity, in using formula 10, we shall speak of the present value of a contingent payment instead of, more completely, the present value of the expectation of this payment. EXERCISE LX 1. In playing a game for a stake of. $50, what is the mathematical expectation of a player whose probability of winning is .3 ? LIFE ANNUITIES 153' Norm. Suppose that a professional gambler should operate the game of problem 1 and charge each player the value of his mathematical expectation as a fee for entering the game. Then, if a very large number of players enter the game, the gambler may expect to win, or lose, approximately nothing. This follows from the facts pointed out in Note 2, Section 59, because, if a large number play, approximately .3 of them may be expected to win the stake, and the money won would, in this case, be approximately equal to the total fees collected by the gambler. If, however, the gambler should admit only a few players to the game, he might happen to win, or equally well lose, a large sum, because out of a few games he has no right to expect that exactly 3 of them will be won. The principle involved in this note is fundamental in the theory of insurance, and finds immediate application in problem 4, below. In any financial operation which is essentially similar to that of the professional gambler above, the safety of the operator depends on his obtain- ing a large number of players for his game. 2. At the end of 10 years a man will receive $10,000 if he is alive. At 5% interest, find the present worth of his expectation if his probability of living is .8. 1 3. A young man, aged 20, on entering college is promised $1000 at the end of 4 years, if he graduates with honors. At 5% interest, find the pres- ent value of his expectation. 4. Out of 1,000,000 buildings of a certain type, assume that the equiva- lent of 2500 total losses, payable at the end of the year, will be suffered through fire in the course of one year, (a) If an owner insures his build- ing for $20,000 for one year, what is the present value of his expectation, at 3% interest? (&) What is the least price that a fire insurance company could be expected to charge for insuring his building? 5. A certain estate will be turned over to the heir on his 23d birthday. If the estate will then be worth $60,000, what is the present worth of the inheritance if money is worth 4$% and if the heir is now 14 years old ? 6. A boy aged 15 has been willed an estate worth $10,000 now. The will directs that the estate shall be allowed to accumulate at the rate (.04, m = 2} until the heir is 21. If money is worth 3J% to the boy, find the present value of his expectation. 63. Present value of a pure endowment. If $1 ia to be paid to (x) when he reaches age (v + ri), we shall say he has (or is prom- ised) an n-year pure endowment of $1. Let n E x be the present value of this endowment when money is worth the effective rate i. The probability p of the endowment being paid equals the prob- 154 MATHEMATICS OF INVESTMENT ability of (x) living to age x + n, or p = n p x . Hence, from for- mula 10, with S 1, n E a = np*(l + i)~ n = Za+n(1 7 + *)"". (Formula 5) LX In the future we shall use u as an abbreviation for the discount factor (1 + O" 1 - Thus > v = (1 + 0~S w 2 = (1 + i)" 2 , etc., v n = (l+fl-. Hence, ^ _ ^ (u) *X The present value $A of an ?i-year pure endowment of $.R to a man aged x is given by (12) *x Norm 1. Remember the subscript (a; '+ n) on Z a+n in formula 12 as the age at which (x) receives the endowment. Example 1. A man aged 35 is promised a $3000 payment at age 39. Find the present value of this promise if money is worth 6%, effective. Solution. The man aged 35 has a 4r-year pure endowment of $3000. From formula 12, its present value is A = 3000(^6) = S 00 ?" 4 * 39 = SOQOft; 06 )"^ - $2290.3. (Tables VI, XIII) *36 ItS NOTE 2. Formula 11 may be derived by the following method. The present value JS^, is the sum which, if contributed now by a man aged x, will mate possible the payment of $1 to him at the end of n years, if money can be invested at the rate i, effective. Suppose that l a men of age x make equal contributions to a common fund with the object of providing all survivors of the group with $1 payments at the end of n years. Since l a + men will survive, the necessary payments at the end of n years total $i a+n . The present value of this amount at the rate i is l a+n (l + i)~" Z ffl+ u n , which is the sum needed in the common fund. Hence, the share which each of the Z people must con- tribute is v ni^ I. ' the same as obtained in formula 11. EXERCISE LXI 1. A man aged 31 is promised a gift of $10,000 when he reaches age 41. Find the present value of the promise at 3^% interest. 2. State in words what is represented by $2000 (lyJB'as) and find its value at 5% interest. f jf- LIFE ANNUITIES ('155.' v ' 3. A will specifies that the estate shall be turned over to the heir, now aged 23, when he reaches 30 years of age. If the estate will then amount to $150,000 find the present value of the inheritance at 4% interest. 4. A man aged 25 has $1000 cash. What pure endowment, payable at the end of 20 years, could he purchase from an insurance company which will compute the endowment at a 4% rate? Make use of equation 12, to determine the unknown quantity R. 6. If money is worth 3%, what endowment payable at age 45 could a man aged 30 purchase for $7500? / ..__. / i - . - - ., f\ t \ tf~ '*"' " ' f - -t-'jt'- - 64. Whole life annuity. A whole life annuity is an annuity whose periodic payments continue as long as a certain individual (or individuals) survives. We shall deal only with the case where one individual is concerned. In speaking of a life annuity we shall always mean a whole life annuity unless otherwise specified. NOTE 1. The periodic payments of all annuities will be supposed equal and will be due at the ends of the payment intervals unless otherwise stated. When no rate of interest is specified, it will be understood as the rate i, effectiye. Exampk 1. If money is worth 3%, find the present value of a life annuity of $1000 payable annually to a man aged 92. Solution. He is promised a payment, or endowment, of $1000 at age 93, another at 94, and a third at 95, which he will receive if lie is alive when they are due. No payment is possible after he is of age 95 because he is certainly dead at age 96. The present value A of his expectation from the annuity is the sum of the present values of the equivalent three endowments, due in 1, 2, and 3 years, foom formula 12, the present values of these endowments are 1000i#o2, lOOO^oa, and lOOOa^sa. Hence, by use of formula 12 and tables VI and XIII, we obtain A = 1000(1.002 wt / (1.036)^ 8 \ Let a x be the present value of a ife annuity of $1 payable at the end of each year to a man now aged x. This annuity is equivalent to pure endowments of $1 payable at ages (x -f 1), (x + 2), , to age 95. The present values of these endowments are tabulated below, and a x equals their sum. 156; MATHEMATICS OF INVESTMENT AQB AT WHICH SI ENDOW- MENT IB PAYABLE TIME FROM Now UNTIL ENDOW- MENT IB PAYABLB PRHSBNT VALUE OF THH ENDOWMBNT 3+2 1 yr. 2yr. ku A " t 95 95 x 95-3! EX On adding the last column we obtain a x = '95 (13) The present value $A of a life annuity of %R paid at the end of each year is given by A = Ra x . NOTE 2. IE contrast to the annuities certain considered in Part I, life annuities are called contingent annuities, because their payments are con- tingent (or dependent) on the survival of (x). The life annuity was inter- preted as heing paid to (x). Recognize that a, is the present value of pay- ments made at the end of each year during the life of (x), regardless of who receives the payments. EXERCISE LXH 1. By the process of Example 1 above, find the present value of a life annuity of $2000 payable at the end of each year to a man now aged 91, if money is worth 5%. 4 2. If money is. worth 6%, find the present value of a whole life pension of $1000 -payable at the end of each year to a man now aged 92. Use formula 13. 3. (a) By use of formula 13, write the explicit expression which would be computed in finding the present value of a life annuity of $500 paid at the end of each year to a man aged 65, if money is worth 6%. (6) How many multiplications would be necessary in computing the nu- merator? 4. By the method used in deriving the formula for , find the present value of a life annuity of $1000 payable at the end of each 3 years to a man aged 35, if money is worth (.04, m = 1). Do not compute the ex- pression obtained. LIFE ANNUITIES 157 66. Commutation symbols. Auxiliary symbols (such as D h and Nb below), called commutation symbols, are used in life an nuity and insurance formulas. From formula 11 for nE X) we obtain n E a = vn ^ n = l a V l x Let Dh be an abbreviation for y^*, or D k - irt*. ' (14) Thus, DBO = v 60 l5Q. Hence, v*l x = D x , v^^lg+n = Dx+n, and A = %* (15) UK The present value $A of an Ti-year pure endowment of $J2 is A - R(JBJ - ^5=*=- (16) Z/B Example 1. Compute AB if money is worth 3%. Solution. D S8 - ii = (1.035)- se (89032) = 37674. NOTE. It is very customary for insurance companies to use 3i% as the rate in annuity computations. The values of DI O , DH, to DOB at 3J% are tabulated in Table XIV, and the result of Example 1 above is seen to check the proper table entry. Formulas 15 and 16 may be used, in connection with Table XTV, only when the rate is 3J%. Tables of the values of D* at, other rates 1 are found in collections of actuarial tables. In problems in this book, when the rate is not 3J%, formulas 11 and 12 must be used. To simplify formula 13 for a a , .multiply numerator and de- nominator by V. We obtain, Since v*l 9 = D a , V+H^i = >+!, etc., v 96 ^ = D 9 B, Introduce Nk as an abbreviation for the sum of all JD's from Db to D 96 : Nk = Dk + Dk+i + + DM. (18) Thus, JVoo = DM 4- D n + D 92 + D 93 + D M + D 9B . Since the numerator in formula 17 is Na+i, a ^*H. - ^IQ'i Q x ^ i \-*-"/ 1 See Tables of Applied Mathematics, by Glover. 158 ' MATHEMATICS OF INVESTMENT The present value $A of a life annuity of $R per year to (x) is A = Ra x = *^i- (20) MX NOTE. The values of Nk are tabulated in Table XIV for the rate 34%. For this rate, formulas 19 and 20 may be used in connection with Table XIV. For all other rates the previous formula 13 must be used. The values of Nk for a few other interest rates are found in actuarial tables. Exampk 2. If money is worth 3%, what life annuity, payable at the end of each year, can a man aged 50 purchase for $10,000? Solution. Let R be the payment of the annuity. From formula 20, 10000 = .8(050) = R^, DM p 10000D60 10000(12498.6) 9.70000 R = -Jfc 169166 * 738 ' 83 - EXERCISE LXTTT 1. Compute the value of D& for i = .035 and verify the entry in Table XIV. 2. By use of formula 18 and Table XIV for the D's, find the value of (a)N 66 -, (&) tf M ; (c) N m . 3. Find the present value of a life annuity of $1000 at the end of each year for a man aged 24, at 3J%. 4. (o) Find the present value of a pure endowment of $3500 at the end of 12 years for a man aged 33, at 3|%. (6) Find the present value of the endowment at the rate 4%. 5. A man aged 65 is promised a pension of $2000 at the end of each year as long as he lives, (a) If money is worth 3|%, find the present value of his pension. (&) What is the present value if $2000 is to be paid at the beginning of each year? 4 6. An estate is worth $100,000 and is invested at 5%, effective. The annual income is willed to a woman, aged 30, for the rest of her life. Find the present value of her inheritance if money is worth 3J%. 7. A man aged 45 has agreed to pay a $75 insurance premium at the end of each year as long as he lives. At 3$% interest, what is the present value of his premiums from the standpoint of the insurance company? 8. A man aged 26 has agreed to pay $50 insurance premiums at the end of each year for the rest of his life. At 3J%, what is the present value of his premiums? LIFE ANNUITIES 159 9. A man aged 60 gives $10,000 to an insurance company in return for an annuity contract promising him payments at the end of each year as long as he lives. If money is worth 3|% to the company, what annual payment does he receive ? 10. From the formulas previously developed, prove that a x = vp x (l NOTE. Recognize that this formula would make the computation of a table of the values of a, very simple. First, we should- compute at&, which is zero ; then, 094 = vpu(l + ctt&) gives the value of OM, etc., for a ra , <ZM, , down to au>. 66. Temporary and deferred life annuities. A temporary life annuity of $R per year for n years to (x) furnishes payments of $jffi at the end of 1 year, 2 years, etc., to the end of n years, if (x) continues to live. The payments cease at the end of n years, even though (x) remains alive. Let a^\ represent the present value of a temporary Me annuity of $1 paid annually for n years to (Jt). This annuity promises n pure endowments whose present values are tabulated below. AQB AT WHIOH SI ENDOWMENT is PAYABLE TlMB FHOM NOW UNTIL ENDOWMENT is PAYABLE PBESENT VALUE OF THE ENDOWMENT X+ 1 'z + 2 x + n 1 yr. 2 yr. nyr. 77 _ W ^B+I lJ2/ a - va E v*U z ZJCia - 7 L-e IP _ n k+n n jG/ai ~ - LX The sum of the present values is a an\ ^ or vx (21) Formula 21 applies for all interest rates. To obtain a formula in terms of N and D (which, with our tables, will be useful only 160 MATHEMATICS OF INVESTMENT when the rate is &%), multiply numerator and denominator in equation 21 by v x . From formula 18, Hence, #3+1 - 2V I+n+1 = >*+! + D^+a + + D x+n , and there- fore a*n = N ** -">** (22) t'x If the annual payment of the temporary annuity is $E, the present value $A is given by A = R(a x *\) = R ( N ** ~ ^ + " +l) (23) "* The definition of a deferred life annuity is similar to that for" a deferred annuity certain (see Section 26, Part I). A life .annuity of $1 per year, whose term is deferred 10 years, to a man aged 30, promises the first $1 payment at the end of (10 + 1) or 11 years, and $1 annually thereafter. Let n \Q>x be the present value of a life annuity of $1 per year, whose term is deferred n years, to a man aged x. The first payment of the deferred annuity is due at the end of (n + 1) years. It is clear that a whole life annuity of $1 per year to a man aged x pays him $1 at the end of each year for the first n years, and also at the end of each year after that, provided that he lives. The payments during the first n years form a temporary life annuity whose present value is a^. The payments after the nth. year are those of the deferred annuity, whose present value we are representing by nja,. Hence, the pres- ent value a x of the whole life annuity is the sum of the other two present values or a x Jo, + a*i; (24) n|a* - o, - a^\. (25) LIFE ANNUITIES / 161 , On using formulas 19 and 22 in formula 25, (27) The present value $A of a life annuity of $72 per year, deferred n years, for a man aged x, is - A = *(!<**) = *%5i- (28) l) x EXERCISE LS3V 1. At 3?%, find the present value of a life annuity of $1000 per annum, deferred 20 years, to a man aged 23. 2. At 3%, find the present value of a life annuity of $2000 paid an- nually for 25 years to a man aged 45. 3. If money is worth 5%, find the present value of a life annuity of $1000 paid annually for 3 years to a man aged 27. 4. A man aged 25 will pay 20 annual premiums of $50 each on a life insurance policy, if the man remains alive. If the first premium is cash, find their present value, at 3%. ' 5. A man aged 50 gives an insurance company $10,000 in return for a contract to pay him a fixed income at the end of each year for 20 years, if he lives. If money is worth 3i% to the company, what is the annual income? Use formula 23. v/ 6. A man aged 40 pays an insurance company $20,000 in return for a contract to pay him a life annuity whose first annual payment will be made when his age is 65. Find the annual payment, if money is worth 3&% to the insurance company, by use of formula 28. 7. A corporation has promised to pay an employee, now aged 48, a pension of $1000 at the end of each year, starting with a payment on his 61st birthday. At 3%%, what is the present value of this obligation? NOTE. Any pension system instituted by a company constitutes a definite present obligation whose value can be determined by finding, as in the prob- lem above, the present value of the pension promised to each employee. \/ 8. A man aged 43 estimates his future earnings at $5000 at the end of each year for the next 25 years. At 3f%, find the capitalized (present) value of his earning power. 162 MATHEMATICS OF INVESTMENT 67. Annuities due. A life annuity due is one whose payments occur at the beginnings of the payment intervals, so that the first payment is cash. Let a z be the present value of a life annuity due of $1 paid annually to (x). A cash payment of $1 is due and the remaining payments of $1 at the end of each year form an ordinary life annuity whose present value is a x . Hence, the pres- ent value of the annuity due is given by a x = 1 + a x . (29) From formula 19, _ i _i_ N x+ i _ D x +N x+ i _ P.+ (IWi+ At+H ----- hflro) /cm a a -l+ -- - (30) .-* (3D MX The present value $A of a life annuity due of $.R paid annually is A = *(a) = $& (32) L>x Let a.\ be the present value of a temporary life annuity due, whose term is n years,' paying $1 annually, to a man aged x. The first $1 is paid cash and the remaining payments form an ordi- nary temporary life annuity whose term is (n 1) years. Hence, asSi = 1 -h Oxt=i\> (33) From formula 22, a x Hence, a^ = 1 + 1 D a D a Since D x + N x+ i = N x , * X n\= N *~ N * +n ' (34) = Nx + l ~ -n AWi- D x N- x+n Dx + D x N .-, N , 1.1 (B+l JV ffi+n. The present value $A of a temporary annuity due of $# payable annually for n years to (x) is A = *(a^|) = * ~ *+ . (35) "X Example 1. In a certain insurance policy, the present value of the benefits promised to the policyholder is $3500. If the polioyholcler is of age 27, what equal premiums should he pay to the insurance company at the beginning of each year for 10 years, in payment of the policy, if money is worth 3J% to the company? LIFE ANNTJITIES < 163 Solution. Let $R be the annual premium. The premiums form a tem- xary life amiuity due whose present value equals $3600. Prom formula , with A = $3600, x = 27, and n = 10, 3500 = V- ^) R = 3600 ' R = 3500(3^01) = 287510 NOTE 1, In discussing premiums on life insurance policies, formulas 32 d 35 are of great use. Formulas 16, 28, 32, and 35 of this chapter are the as we shall use moat frequently in the future. Summary of present value formulas Pure endowment : A = #(*) = R ^ - (16) DX Whole life annuity: A = R(a x )=^^- (20) DX Temporary life annuity : A = R(a^ = R Nx+1 ~ Nx ^^- (23) D x Deferred life annuity : A = R( n \ a*) = R ^i . (28) D x Whole life annuity due : A = R(eL x ) = R^' (32) * Temp, life annuity due : A = /?(a^) = R N * ~ N ^ n - (36) MISCELLANEOUS PROBLEMS L. A man aged 40 pays $10,000 to an insurance company in return for ontract to pay him a fixed annual income for life, starting with a pay- nt on his 60th birthday. Find the annual income if money is worth & to the company. I. At age 65 a man considers whether he should (a) pay his total ings of $20,000 to an insurance company for a life annuity whose first mal payment would occur in 1 year, or (6) invest his savings at 6%, ictive. Find the difference in his annual income under the two methods, uming that money is worth 3|% to the insurance company. 1. In problem 2, what will be received by the heirs of the man at his ,th if he adopts plan (a) ? What will they receive under plan (6) ? 164 MATHEMATICS OF INVESTMENT 4. A certain insurance policy taken out by a man aged 28 calls for premiums of $200 at the beginning of each year as long as he lives. Find the present value of these premiums at 3%. 5. A certain insurance policy matures when the policyholder is of age 35 and gives Him $2000 cash or the optipn of equal payments at the be- ginning of each year for 10 years as long as he lives. If money is worth 3&%, find, the annual payment under the optional plan. 6. A boy of 16 has been left an estate of $100,000, which is invested at 5%, effective. If he lives, he will receive the income annually for the next 10 years and the principal of the estate when he reaches age 26. If money is worth 3|%, find the present value of his inheritance. 7. Derive formula 13 for a f by the mutual benefit fund reasoning used in Note 2, Section 63. Thus, at the end of 1 year, $k+! will be needed for payments ; $Z I+a at the end of 2 years, etc., $Ze 6 at the end of (95 x) years. Discount all of these payments and divide by l a , 8. Derive formula 21 for a^ by the mutual fund method of reason- ing. '( 9. A man aged 22 agrees to pay $50 as the premium on an insurance policy at the beginning of each year for 10 years if he lives. Find the present value of his premiums at 3% interest. 10. The present value of the benefits promised in a certain insurance policy is $8000. If the policyholder is aged 30, what equal premiums should he agree to pay at the beginning of each year for 15 years, provided he lives, if money is worth 3$% to the insurance company? 11. A man is to receive a life annuity of $2000 per year, the first pay- ment occurring on his 55th birthday. If he postpones the annuity so that the first annual payment will occur on his 65th birthday, what will be the annual income, if the new annuity has the same present value as the former one, under 3|% interest? 12. A certain professor at age 66 enters upon a pension of the Carnegie Foundation which will pay $2000 at the end of each year for life. In order to have, at age 65, an amount equal to the present value at 3J% of the pension he is to receive, what equal sums would the professor have had to have invested annually at 5%, assuming that his first investment would have occurred at age 41 and his last at age 65? CHAPTER DC LIFE INSURANCE 68. Terminology. Insurance is an indemnity or protection against loss. The business of insuring people against any variety of disaster is on a scientific basis only when a large number of in- dividuals are insured under one organization, so that individual losses may be distributed over the whole group according to some scientific principle of mutuality. That is, each of the insured should pay in proportion to what he is promised as an insurance benefit. In this chapter we shall discuss the principles and most simple aspects of the scientific type of life insurance furnished by old line, or legal reserve companies. When an individual is insured by a company, he and the com- pany sign a written contract, called a policy. The individual is called a policyholder, or the insured. In the contract the com- pany promises to pay certain sums of money, called benefits, if certain events occur. The person to whom the benefits are to be paid is called the beneficiary. The insured agrees to pay cer- tain sums called gross or office premiums in return for the con- tracted benefits. The policy date is the day the contract was entered into. The successive years after this date are called policy years. The fundamental problem of a company is to determine the pre- miums which should be charged a policyholder in return for speci- fied benefits. Every insurance company adopts a certain mor- tality table and an assumed rate of earnings on invested funds as the basis for its computations. We shall use the American Experience Table and 3%, as is the custom among many com- panies. The net premiums for a policy are those whose present value is equal to the present value of the policy benefits under the following assumptions : (a) the benefits from the policy mil be paid at the ends of the years in which they fall due; (&) the company's funds will earn interest at exactly the specified rate (3% in our case) ; 166 166 MATHEMATICS OP INVESTMENT (c) the deaths among the policyholders will occur at exactly the rate given by the mortality table (Table XIII in our case). Under these assumptions, if a company were run without profit or administra- tive expense, it could afford to issue policies in return for these net premiums. The actual gross premiums for a policy are the net premiums plus certain amounts which provide for the adminis- trative expense of the company and for added expense due to vio- lations of the theoretical conditions (a), (6), and (c) assumed above. In computing gross premiums, insurance companies use their own individual methods. Our discussion is concerned entirely with net premiums and related questions. Nona. In the future, if the interest rate in a problem is not given, it is understood to be 3%. 69. Net single premium ; whole life insurance. If a policy- holder agrees to pay all premium obligations in one installment, it is payable immediately on the policy date and is called the single premium for the policy. The net single premium is the present value on the policy date of all benefits of the policy. A whole life insurance of $R on the life of (x) is an agreement by the company to pay $R to the beneficiary at the end of the year in which (x) dies. A policy containing this contract is called a whole life policy. Example 1. Find the net single premium for a whole life policy for $1000 for a man aged 91. Solution. Suppose that the company issues whole life policies for $1000 to ZDI, or 462 men of age 91. During the first year, d 9 i - 246 men will die; $246,000 in death claims will be payable to beneficiaries at the end of 1 year. The present value of this payment is 246,000(1. 035) ~ l = 246,000 v. The other entries below are easily verified. Pouor YEAB DEATHS DURING YBAB BENEFITS DBB AT END OP YHAB PIIBBTIINT VALOT op BmNHfflTH 1 d 9 i = 246 $246,000 246,000 v 2 d 9 a - 137 137,000 137,000 w 3 dgs = 58 58,000 58,000 it 4 dw-lS 18,000 18,000 v* 5 d 8B =3 3,000 3,000 V 6 - LITE INSURANCE 167 Hence, on the policy date, the insurance company should obtain through the net single premiums from the lai men, a fund equal to the sum of the last col- umn. Thia sum, divided by 462, is the share or net single premium paid by each of the ZBI men. By use of Table VI, we find that each pays 246000 v + 137000 + 58000 t> + 18000 v* + 3000 v 5 462 $943.93. Let lAs be the net single premium for a whole life insurance of $1 on the life of (x). To obtain A a by the method of Example 1 above, suppose that a company issues whole life policies for $1 insurance to each of l a men of age x. During the first policy year, d x will die ; $d ffl in death benefits is payable to beneficiaries at the end of 1 year. The present value of these benefits at the rate i is d x (1 + i}~ 1 = vd a . The other entries below are easily verified. POLICY YHAB DEATHS DUBINQ YBAB BENEFITS Dim AT END op YEAR ___^^_ PBUBBNT VALtra os BEND PITS 1 d, ^ wd, 2 dx+i 9d x +i t^dz+i 3 d : + * V.* "^ 96-s ,;. ^., In the last row, notice that when the group reaches age 95, the policies have been in force (95 re) years, or the (96 :c)th year is just entered on. Honce, dw is due at the end of (96 x) years. From the net single premiums paid on the policy date, the company must obtain a fund equal to the sum of the values in the last column. The share of each of the l a men, or his net single premium A x > is A =1 v( ** + y2rfg+1 + y3 ^ g+2 "^ ---- *" &*~* d u . (36) On multiplying numerator and denominator above by v*, + v*+*d f+l + + 168 MATHEMATICS OF INVESTMENT Introduce a new symbol C* = v k+1 dk. Thus, C& = v 94 cZ B a, etc., v^^dj, = C X ) V^dx+i == CWij and ir^d^ = C B 5. Hence ri \ fi I ... ' ^ 4 _ ^a "t" v^g+l ~T T ^96. A ~ D x Introduce a new symbol M k = C k + Cft+i H h C 9B - (37) Thus, M 92 = C B2 4- Cgg + C B4 + C B5 ; M s - C a + + C^. Hence 4 X = ^- (38) X The net single premium A for a whole life policy of $72 for (x) is (39) NOTE. The values of Mb for the rate 3%, v = (1.035) ~ l , ore given in Table XIV. EXERCISE LXV Use formulas 38 and 39 unless otherwise directed. 1. Compute the values of (7 9 4 and of Ceo at 3$% ; verify the entries for M 94 and for M 9B in Table XIV. 4 2. By the method of illustrative Example 1, page 166, find the net single premium for a whole life insurance for $1000 for a man aged 93, if interest is at the rate 5%. 3. Find the net single premium for a whole life insurance of $1000 on the life of a man, (a) aged 90 ; (6) aged 50 ; (c) aged 30 ; (d) aged 10. g 4. How much whole life insurance can a man aged 50 purchase from a company for $1500 cash? 6. How much whole lif e insurance can a man aged 35 purchase from a company for $2000 cash? 70. Term insurance. An n-year term insurance for $R on the life of (x) promises the payment of $R at the end of the year' in which (x) dies, only on condition that his death occurs within n years. Thus, a 5-year term insurance gives no benefit unless (jc) dies' within 5 years. Let A 1 ^ represent the present value of an Ti-year term insurance for $1 on the life of (x}. To obtain the value of A, assume that the company issues n-year term in- LIFE 169 surance policies for $1 to each of Z a men aged x. The present values of the benefits which will be paid are tabulated below ; the policy has no force after n years. POLICY YEAR DEATHS DURING YEAH BENEFITS DUB AT END OP YHAH PBBBENT VALUE OF BENEFITS 1 2 d, d a +i 94. $da+I vd x &d a+ i n dj;+n-l $dx+n-l f^s+n-l The net single premium paid by each man is the sum of the last column, divided by l a , or [ .i2J I ... I .nJ (40) in t 3711 z a On multiplying numerator and denominator above by if and on using the symbol C k = -i + -\ \- C 9B , Since M x = C x + C I+i + + and M x+n = it is seen that the numerator in equation 41 is M x M^ n ; hence ^ ^ ~ M ^+". (42) *n\ D; ^ The net single premium $A for T^year term insurance of $B on the lifo of (x) is A = J R4* = g(^ ~ M x+ n). (43) *n| ^ The not fiinglo premium for a 1-year term insurance for (x) is called the natural premium at age x. The natural premium for $1 insurance is obtained from bquation 42, with n = 1 : Natural Premium - 4*-, *** ~ M * +1 - ^-, (44) **x *^< where M - C s because of formula 37. 170- MATHEMATICS OF INVESTMENT EXERCISE LXVI Use formulas 42 and 43 unless otherwise specified. 1. By use of the method used in deriving formula 40, find the expres- sion for the net single premium for a 3-year term insurance for $1000 on the life of a man aged 25 and compute its value at 5% interest. '- r 2. Find the net single premium for a 10-year insurance for $2000 on the life of a man* aged 31. 3. Find the natural premium for $1 insurance at age 22 ; at age 90. 4. (a) Find the net single premium for a whole life insurance of $1000 at age 50. (6) Find the net single premium at age 50 for a 10-year term insurance for $1000. J 6. How much term insurance for 10 years can be purchased for $2000 cash by a man aged 35? 6. How much term insurance for 10 years can be purchased for $2000 cash by a man aged 55? 71. Endowment insurance. An n-year endowment insur- ance of $R on the life of a man aged x furnishes (a) a payment of $R at the end of the year in which (x) dies, if he dies within n years, and (b) a pure endowment of $J5 to (x) at the end of n years if (x) is alive at that time. Thus, a 20-year endowment insurance of $1000 pays $1000 at death, if it occurs within 20 years ; or, if (x) is alive at the end of 20 years, he receives the endowment of $1000. Let A X n\ repre- sent the net single premium (or present value) of an n-year en- dowment insurance of $1 on the life of (x}. The present valuo Axft is the sum of the present values of (a) the n-year term in- surance for $1 on the life of (x}, and of (6' the n-yoar pure endow- ment of $1 to (x}. Hence, on using formulas 16 and 42, (45) If the endowment insurance is for $A, the net single premium $A is given by Af* - Af*+n + *+n) LIFE INSURANCE , 171 EXERCISE LXVH 1. (a) Compute the net single premium for a $1000, 20-year endow- ment insurance on the life of a man aged 23. (&) Compute the present value of ti pure endowment of $1000 payable to the man at age 43. (c) Find the net single premium for a 20-year term insurance for $1000 on the life of the man aged 23, by using (a) and (&). '2. Find the net single premium for a 10-year endowment insurance for $5000 on the life of a inun aged 30. 3. (a) Find the net single premium for a 10-year endowment insurance for $3000 on the life of a man aged 26. (6) Find the net single premium for a 10-year term insurance for $3000 on his life, (c) From the results of (a) and (&), find the present value of a 10-year pure endowment of $3000 for the man. f /4. How much 20-year endowment insurance can a man aged 33 pur- chase for $3000 cash? 6. How much 10-year endowment insurance can a man aged 45 pur- chase for $2500 cash? 72, Annual premiums. If the net premiums for a policy are payable annually, instead of in one installment (the net single premium), they must satisfy the condition that the (pr. val. of annual premiums) = (net single premium), (47) because the net single premium is the present value of the policy benefits. When paid annually, the premiums for a policy are always equal and are payable at the beginnings of the years, as long as the policyholdor lives. Example 1. (a) Find the net single premium for a 10-year term in- surance for $10,000 on the life of a man aged 46. (&) Find the equivalent annual premium which the man might agree to pay for 10 years, if he lives. Solution, (a) From formula 43, the not single premium is 10000(4*^) = 1QQ 00(M4 fl " MM) = $1119.30. (Table XIV) 4010 1 , JJlft That is, the present value of the insurance benefits is $1119.30. (b) Let P be the annual premium. The 10 premiums form a 10-year life annuity due paid by a man aged 46. Their present value ie P(a 46 iji), and it must equal $1119.30. Hence, 1119.30 - P(a) - pN * Ntt ' (Formula 35) 1119.30 Dq m $137.33, (Table XIV) 172 MATHEMATICS OF INVESTMENT The annual payments of $137.33 have a value equivalent to $1119.30 paid cash. NOTE 1. The solution of (a) was not necessary in order to solve (6) above. Thus, we may write, immediately, from equation 47, 10000 W6- - N In insurance practice the most simple forms of insurance policies are those tabulated below. Their names, policy benefits, and manner of premium payment should be memorized. All premiums are payable in advance, at the beginning of the year. .The num- bers assigned are merely for later convenience in this book. NUMBBB NAME OF POLICY POLICY BHNBFITS PBBMIUMB PAID I Ordinary life Whole life insurance Annually for life II w-payment lif e Whole life insurance Annually for n years III ?i-year term 7i-year term insurance Annually for n years rv Tfc-year endowment (a) Ti-year pure endowment (6) 7i-year term insurance Annually for n years To determine the net annual premiums for these policies, we use the fundamental equation 47, and the method x of Note 1 above. Consider an ordinary life policy for $1 for a man aged x. Let P a be the net annual premium. The premiums paid by the man aged x form a whole life annuity due whose present value is P a ,(a a ). The net single premium for the policy is A& Hence, from equation 47, JP.(a,} -A.; Px~~ - 7T 2 - (Formulas 32, 38) 1 The student ia advised to solve problejn 1 of reading the rest of the section. (48) we &XVIH below before LIFE INSURANCE 173, Let n Px be the net annual premium for an n-payment life policy for $1 for a man aged x. The premium payments by (x) form an ?>-year Me annuity due whose present value is nPsCa^). Hence, from equation 47, n P*(a.*nd - A*', nP* Nx ~ Nx + n = ^=- (Formulas 35, 38) Uy U x nPx = M * (49) It is left as an exercise (problem 3, below) for the student to prove that the net annual premium P^ for an n-year term insur- ance policy for $1 for (x) is given by M x+n _ -- ,.. (60) Let P.^ be the net annual premium for an n-year endowment policy for $1 for (x). The premiums paid by the man aged x form an w-year life annuity due whose present value is P a -ni(a a ^ 1 ). The net single premium for the policy is A^. From equation 47, - M,+ n + D^ t (Formulas 35, 45) , -*1* ^. , (51) P*x JVjc+n If the policies I to IV are for $J? instead of $1, the annual pre- miums are found from formulas 48 to 51 by multiplying by R. EXERCISE LXVIE 1. By the method of Note 1, page 172, find the net annual premium for a 5-payment life policy for $1000 for a man aged 45. J 2. A man aged 29 has agreed to pay 15 annual premiums of $100 for a certain policy. Find the net single premium for the policy, 3. Establish formula 50 for the net annual premium for an n-year term insurance policy for $1 for a man aged &. 4, The net single premium for a certain policy for a man aged 26 is $3500, (a) Wbftt is the net annual premium if hQ agress to pay premiums 174/ MATHEMATICS OF INVESTMENT annually for life? (&) What is the net annual premium if he agrees to pay annually for 12 years? 6. For a man aged 25-, find the net annual premium (a) for an ordinary life policy for $2000 ; (&) for a 20-payment life policy for $2000. 6. Find the net annual premium for a 10-year term policy for $1000 for a man (a) aged 32 ; (&) aged 42 ; (c) aged 62. 7. Find the net annual premium for a 20-year endowment policy for $1000 for a person of your own age. 8. (a) Find the net annual premium for a 5-year term policy for $1000 for a man aged 40. (&) Find the natural premiums for a $1000 insurance at each of the ages 40, 41, 42, 43, and 44. (c) Compare the result of (a) with the five results in (6) . 9. A whole life insurance policy for $1000 taken at age 30 states that the annual premiums were computed as if : (a) term insurance of $1000 were given for the first year, and (&) an ordinary whole life policy were then written when the man reaches age 31. Find the net premium (a) for the first year, and (&) for the subsequent years. NOTE. Such a policy is very common and is said to be written on tho 1-year term plan. The advantages from an insurance company's standpoint are apparent after reading the next chapter. 10. A certain endowment policy for $1000 taken at age 23 provides that the net premium for the 1st year is that for 1-year torm insurance and that the net premiums for the remaining 19 years are those for a 19- year endowment policy for $1000, taken at age 24. Find the net annual premium (a) for the first year ; (&) for subsequent years. This policy is .another example of the 1-year term plan. 1 11. How much insurance on the 20-payment life plan can a man aged 32 purchase for a net annual premium of $75 ? It is advisable, first, to find the equivalent net single premium. 12. How large a 20-year endowment policy can a man aged 23 pur- chase for net annual premiums of $100? 13. (a) For a boy aged 16, find the not annual premium for a $1000 endowment policy, which matures at age 85. (/;) Find the- not annual premium for an ordinary life policy for $1000 tulam at ago l(i. (e) Ex- plain the small difference between the results. 14. A man aged 30 takes out a policy which provides him with $10,000 insurance for the first 10 years, $8000 for the next 10 yours, and $5000 for the remainder of his life. He is to pay premiums annually for 10 years. Find the net annual premium. LIFE INSURANCE 175 15. A certain policy on maturing at age 55 offers the option of a pure endowment of $2000, or an equivalent amount of paid up whole life in- surance, that is, as much insurance as the $2000, considered as a net single premium, will buy. Find the amount of paid up insurance. NOTE ON GHOSH PHRMIUMB. Premiums previously discussed were net premiums, or present values of the benefits to be paid under the policy. In conducting on insurance company there is expense due to the salaries paid to administrative officials, the commissions paid to agents for obtaining new policyholders, the expense of the medical examination of policyholders, book- keeping expense, etc. To provide for these items and for unforeseen con- tingencies, it is necessary for the company to add to the net premium an amount called the loading. The not premium plus the loading is the gross or actual premium paid by the policyholdor. Sometimes the loading is de- termined as a certain percentage of the not premium plus a constant charge independent of the nature of the policy. Sometimes the loading may be determined simply as n percentage of the net premium, the percentage either being independent of the policyholder's age, or varying with it. Each com- pany uses its own method for loading, but the resulting gross premiums of all large, well-managed companies are essentially the same. SUPPLEMENTARY MATERIAL 73. Net single premiums as present values of expectations. A whole life policy on a life aged x promises only one payment, due at the end of the year in which (x) dies. However, we may think of the policy as promising payments at the end of each year up to the man's 96th birthday, the payment at each date being contingent on his death during the preceding year. Then, the method used in deriving formulas for life annuities may be used to obtain the present value, or net single premium, for the policy. Consider obtaining the net single premium A a for a whole life insurance of $1 on the life of (*). At the end of 1 year, $1 will be paid if (*) dies during the preceding year. The probability of (x) dying in this year is ~^j from formula 10 with S 1, the IK present value of the expectation of this payment is ~ (1 + i)~ l l a or ^2. The other present values below are verified similarly, la 176 MATHEMATICS OF INVESTMENT PAYMENT OF SI WILL BE MADE AT END OF IF MAN DEBS BETWEEN AGES PROBABILITY OF PAYMENT BEING MADE PRESENT VAT/CH OB- THE PAYMENT lyr. x and (x + 1) d, I, vd x I, 2yr. 96 - x yr. (a; + 1) and (x + 2) 95 and 06 "Ir I, i x I, The expression obtained for A x on adding the present values in the last column is the same as previously obtained in formula 36. NOTE. From the present point of view, an insurance company may be likened to a gambler who plays against all of the beneficiaries of the policies. The net single premiums are the present values of the expectations of the beneficiaries. So many players are involved as opponents of the company that the probabilities of winning and of losing as given by the mortality table will be practically certain to operate. Hence, the company will neither lose nor win in the long run. EXERCISE L33X 1. By the method which was used above to obtain A x , find the expres- sion for the present value of a 10-year term insurance policy for $1000 on a life aged 20. 2. By the method above, derive the formula 40 for A 1 ^. 74. Policies of irregular type. Equation 47 enables us to find the premiums for any policy for which the present value of the benefits is known. Example 1. A policy written for a man aged 32 promises the follow- ing benefits : (a) Term insurance for $5000 for 28 years ; (ft) a life annuity of $1000 paid annually, first payment due at age 60. It is agreed that premiums shall be paid annually for 28 years. Find P, the net annual premium. Solution. The present value of benefit (a) is 6000(A3 33ffl ) ; benefit (b) is a life annuity, term deferred JJ7 years, whose present value is 1000(jff|aa). LIFE INSURANCE 177 The annual premiums form a 28-year temporary annuity due, whose present value is P-,). Hence, from equation 47, Aa 1000 N 60 + 500Q(M a - ^) = $234.43. (Formulas 35, 42, 28) (Table XIV) EXERCISE LXX Find the periodic premium payment for each policy described. The age of policyJwlder is the age at the time the policy was written. Pnon. BBXEFITS OP POLICY A an OF POLIOY- HOLDBB METHOD OP PATINO PREMIUMS 1. (a) 10-year term insurance for $1000. (6) A pure endowment of $2000 at the end of 10 years. 27 10 annual premiums 2. (a) Term insurance of $10,000 for 20 years. (6) Life annuity of $1000 paid annually, first payment at age 06. 46 20 annual premiums 3. Life annuity of $1000 paid annually, first pay- ment at age 65. 30 25 annual premiums 4. Life annuity of $1000 paid annually, first pay- ment at age 70. 45 10 annual premiums 6. (a) Term insurance of $10,000 for first 10 years, (b) Term insurance for $6000 for next 20 years, (c) Life annuity of $2000, paid annually with first payment at age 65. 45 20 annual premiums 6. In what way does the policy in problem 1 differ from a 10-year endowment policy? NOTE. The policy of problem 4 is called an annuity policy and is a familiar form for those wishing protection in thoir old age. This same feature of protection is present in the policy of problem 2. CHAPTER X POLICY RESERVES 75. Policy reserve. At age 30, the natural premium for $1000 insurance, that is, the net single premium for 1-year term insur- ance for $1000, is found to be $8.14. This is the sum which each of Zao men of age 30 should pay in order to provide benefits of $1000 in the case of all of the group who will die within 1 year. The $8.14 premium is the actual expense of an insurance company in insuring a man aged 30 for $1000 for 1 year. The expense of insurance for 1 year increases continually during life, after an early age, being $17.94 per $1000 insurance at age 55 and $139.58 at age 80. Consider a man aged 30 who takes out a $1000, ordinary life policy. Throughout life he pays a net level premium (that is, a constant premium) of $17.19, as obtained from formula 48, and is insured for $1000 all during life. The expense of the company in insuring him during the first year is the natural premium, $8.14. Hence, in the first year the man pays (17.19 8.14) = $9.05 more than the expense. The insurance company may be consid- ered to place this unused $9.05 in a reserve fund which will accu- mulate at interest for future needs. Up to age 54, each annual premium of $17.19 is more than the expense of insurance and the company places the excess over expense in the reserve fund. At age 55 the $17.19 premium is less than the insurance expense, which is $17.94. The deficiency, (17.94 - 17.19) = $.75, IB taken from the reserve fund. From then on until the end of life, the expense of insurance is met more and more largely from tho re- serve fund. Thus, at age 80, the expense is $130,58 (tho natural premium, as given above) so that (139.58 - 17.19) - $122,39 comes from the reserve. For every insurance policy (except a 1-year term policy) where a net level premium P is paid, tho annual expense of insurance 178 POLICY RESERVES 179 during the early policy years is less than the premium P. Hence, the insurance company should place the unused parts of the pre- miums in a reserve fund and accumulate it at interest to answer the future needs of the policy. When the expense of insurance, in later years, becomes greater than the level premium, the defi- ciency is made up by contributions from the reserve fund. The reserve funds should be regarded as a possession, of the policy- holders, merely held and invested by the insurance company. The reserve on a policy at the end of any policy year, before the next premium due is paid, is called the terminal reserve for that year. In this chapter we consider the determination of the ter- minal reserve for a given year. Example 1. Form a, table showing the terminal reserves for the first 6 years for a 5-payment life policy for $1000 written at age 40. Solution. From formula 49, the not annual premium is $89.4674. Assume that the company issues the same policy to each of Z*o = 78,106 men. The following table shows the disposition of the funds received as premiums. POUOT YHAB PREMIUMS PAID AT START off YEAH RBsHirvm FUND AT STAHT OF YBAB ACCOM. FUND AT 34%, AT END on 1 YEAH RBBIJIIVII FUND AVTEB DEATH RMNHPITS AT END ov YEAH RBSHRVH ran SUB- vivon AT END aw YBAII 1 $6,987,100 $ 6,987,160 $ 7,231,710 S 6,466,710 $ 84 2 0,918,725 18,385,435 13,848,925 13,074,925 171 3 6,849,485 19,924,410 20,021,764 19,836,704 262 4 6,779,201 20,01(5,025 27,547,586 26,750,586 357 5 6,707,903 33,458,549 34,029,508 33,817,598 456 33,817,608 35,001,224 34,173,224 466 For example, at tho beginning of the first year, (78, 106) (89.4674) is received in premiums. At tho and of the your, interest at 3J% added to the premium fund gives $7,231,710. During tho your, (k 70S deaths ooeurred so that $706,000 is payable to beneficiaries, leaving $6,460,710. There are Zi = 77,341 survivors; the total fund $0,400,710, divided by 77,341, gives $84 as tho share, or reserve, per policy. At the beginning of the 2d year, 77,341 men pay premiums, etc. After tho 6th year, no more premiums will be received, HO that all death benefits in the future come from the fund, $33,817,698, on hand at tho end of 5 years, and it future accumulations at interest, 180 MATHEMATICS OF INVESTMENT EXERCISE LXXI 1. Assume that an insurance company issues $1000 ordinary life poli- cies to each, of ^ men of age 92. Compute a table showing the disposi- tion of funds received as premiums and the total reserve per policy at the end of each year. 76. Remaining benefits of a policy; computation of the re- serve. At any time after a policy is written, the remaining benefits of a policy are the promised payments of the policy as they affect the policyholder at his attained age. Example 1. A certain insurance policy, written for a man aged 32, promises Kim (a) temporary life insurance for $1000 for 25 years ; (&) a pure endowment of $1000 payable at the end of 25 years ; (c) a lif e annuity of $1000 payable annually, first payment at age 60. (1) Describe the re- maining benefits, 8 years later. (2) Find the present value of the re- maining benefits, 8 years later. Solution. (1) Eight years later, the attained age of the man is 40 years. The policy promises (a) term insurance for $1000 for 17 years to a man aged 40 ; (&) a pure endowment of $1000 payable at the end of 17 years to a man now aged 40 ; (c) a deferred life annuity, for a man aged 40, of $1000 paid annually. Since the first payment of the annuity is due at age 60, which is 20 years later, the annuity is deferred 19 years. (2) The present value of the remaining benefits at age 40 is the sum of the present values or net single premiums for the three individual benefits or 1000 A + lOOOiT^o + 1000i.]o4o, which can be computed by use of the proper formulas. Consider the conditions in regard to a policy, written for a man aged x, n years after the policy date. The attained age of the policyholder is x + n, and the reserve fund for the policy contains a certain amount $"7, the terminal reserve at the end of n years. The company is liable for the remaining benefits of the policy, and the policyholder is liable for the future premiums, Since all future benefits must be. paid from the reserve and from the future Dremiums, the following equation is satisfied : -ingle premium for \ /Pr.val.atagex + n\ /Terminal \ . val. of) remaining ) = (' of .net premiums J-H reserve at ) (52) is at age x + n / \due in the future/ \age x + n' POLICY RESERVES 181 To find the terminal reserve on a policy, first find the net annual premium and then use equation 52. Example 2. Find the terminal reserve at the end of 6 years on a 20- year endowment policy for $1000 written at age 24. Solution. (a) The net annual premium is 1000 (P z ^\) = $39.085, from formula 51. (&) The remaining benefits at the attained age of 30 years are a pure endowment of $1000 payable at the end of 14 years to a man now aged 30 and term insurance for $1000 for 14 years on a life aged 30 ; in other words, the benefits form a 14-year endowment insurance for $1000 for a man aged 30. The remaining premiums form a 14-year temporary annuity due. Let V be the reserve at the end of 6 years. From equation 62, 1000(Jfao -M + Z>) _ 39.086(^0 - AT). (Formulas 46, 35) Dao D$o v 1000(Jf 30 - M u + Pit) - 39.Q85(J\r 30 - Jfa) Pn V = 66 - $217.9. (Table XIV) The method used in Example 2 may be applied in the case of any standard poli cy to obtain a general formula for the reserve at the end of a given number of years. For example, consider an ordinary life policy for $1 written for a man aged x. Let n V a represent the terminal reserve at the end of n years. The net annual premium is P a = ^- ? - The remaining benefit at the at- rf x tained age (x + n) is whole life insurance for $1 for a man aged (x + n). The remaining premiums form a whole life annuity due of P m payable annually by a man now aged (x + ri). From equation 52, n y a = Axn NOTB 1. For the advantage of the insurance actuary, who has occasion to compute the reserves on numerous policies, it is advisable to develop general formulas and convenient numerical methods for the computation of reserves. In the case of a student meeting the subject for the first time, it is more im- portant to appreciate thoroughly the truth of equation 62. Such appreciation is attained only by direct application of the equation. The problems of Exorcise LXXII below should be solved by direct application of equation 52, as was done in illustrative Example 2 above, MATHEMATICS OF INVESTMENT EXERCISE LXXn 1 J l. If the net single premium for the remaining benefits of a policy is $745, and if the present value of the future premiums is $530, what is the reserve? i ~i 2. At an attained age of 42, the net single premium for the remaining benefits of a policy is $750. There are six annual premiums of $50 remain- ing to be paid, the first due immediately. Find the policy reserve. 3. At the attained age of 44, the reserve on a certain policy is $500. Annual premiums of $25, the first due immediately, must be paid for the remainder of life. Find the present value of the remaining policy benefits. " 4. A $1000, 10-payment life policy is written at age 34. (a) Find the reserve on the policy at the end of 6 years. (6) Find the reserve at the end of 10 years. 6. A $1000, 5-payment life policy is written at age 40. (a) Take the premium, as computed in illustrative Example 1, Section 75; compute the reserve at the end of 3 years and compare with the result given in the table of that example. (6) Find the terminal reserve at the end of 5 years and compare with the table. 6. A $2000, 20-year endowment policy is written at age 33. (a) Find the terminal reserve at the end of 15 years. (&) What is the terminal reserve at the end of 20 years, before the endowment is paid? 7. In the case of a 1-year term policy, why is the reserve zero at the end of the year? i 8. An ordinary life policy for $5000 is written at age 25. Find the terminal reserve at the end of 15 years. 9. Derive a formula for the terminal reserve at the end of n years for an w-payment life policy written at age x. (6) Derive a formula for the reserve at the end of m years, where m is greater than n. 10. Find the reserve at the end of 5 years for a 10-year term policy for $10,000 written at age 35. 11. (a) Find the reserve at the, end of 5 years for an ordinary life policy for $10,000 written at age 35. (&) Compare your answer with that in problem 10 and give a brief explanation of the difference. 12. A man aged 25 pays the net single premium, for a 10-year term in- rance for $1000. What is the policy reserve, 5 years later? L3. A man aged 30 pays the net single premium for a whole life insur- ance for $1000. Ten years later, what is the policy reserve? 1 After the completion of Exorcise LXXII, tho student may proceed immediately to the Miscellaneous Problems at tho end of tho chapter. POLICY RESERVES 183 14. Derive a general formula, as in equation 53, for the reserve at the end of m years for an n-year endowment policy for $1 written at age x. NOTE 2. The method for computing reserves, furnished by equation 52, is called the prospective method because the future history of the policy is the basis for the equation. Retrospective methods also are used. NOTE 3. Insurance companies are subject to legal regulation. It is usually specified by state law that, at periodic times, an insurance company must show net assets equal to the sum of the reserves for all of its outstanding policies. The law specifies a standard mortality table and interest rate to be used. A company is insolvent if it cannot show net assets equal to the neces- sary reserve. It is likewise recognized by law that a company's reserve be- longs to its policyholders as a whole. Hence, the reserve on a policy is the basis for its cash surrender value, the amount which the company must pay to a policyholder if he decides to withdraw from the company and surrender his policy. The cash surrender value equals the reserve, minus a surrender charge. The surrender charge in most states is specified by law and may be considered as a charge by the insurance company on account of the expense entailed in finding a new policyholder to take the place of the one withdrawing. This charge is legitimate because the theoretical reserve was computed by the company on the assumption that it had so many policyholders that the laws of averages, as dealt with in using the mortality table, would hold. Hence, the number of policyholders must be maintained and any one withdrawing should pay for the expense of obtaining a new policyholder in his place. NOTE 4. It should be recognized that the discussion in the preceding three chapters is merely an introduction to the subject of life annuities and of life insurance. We have not considered joint life, or survivorship annuities and insurance. Moreover, the subject of reserves requires a thorough treatment, beyond what we have given, from both the theoretical and the computational standpoint. The surplus of a company, its manner of declaring dividends to policyholders, and many other practical questions connected with the account- ing and business methods of insurance companies have not even been men- tioned. The student who wishes to pursue the subject farther is referred to the Text Book of the Institute of Actuaries, and to the courses of study described by the Educational Committees of the Actuarial Society of America and of the American Institute of Actuaries, SUPPLEMENTARY EXERCISE LXXni Students working the problems below should have previously completed Supplementary Section 74 of Chapter IX. 1. A policy written for a person aged 38 promises whole life insur- ance for $10,000, and a life annuity of $1000 payable annually, with the 184 MATHEMATICS OF INVESTMENT first payment at age 65. Premiums are payable annually for 20 years, (a) Find the reserve at the end of 10 years. (6) Find the reserve at the end of 20 years. 2. A policy written at ago 27 promises $1000 term insurance for 20 years and a pure endowment of $5000 at the end of 20 years. Premiums are payable annually for 10 years. Find the reserve at the end of G years. 3. A policy written at age 15 promises 20-year endowment insurance for $1000, and the premiums are payable annually for 10 years. () De- termine the reserve at the end of 10 years. (6) Determine the reserve at the end of 9 years. . 4. A certain pure annuity policy written at age 40 promises a life annuity of $1000 with the first payment at age 61, The premiums are payable annually for 21 years. Find the reserve (a) at the ond of 5 years ; (&) at the end of 20 years. NOTE, When a corporation or association promises a pension to a person, its act is equivalent to writing a pure annuity policy for the person involved. Hence, a pension association should be considered solvent only whon its reserve fund is equal to the sum of the reserves on each of its pension contracts. As judged by this standard, there are an unfortunately largo numbor of insolvent pension associations in operation. Their insolvency does not become apparent until after they have been operating long enough so that the theoretical reserve (which they do not possess) becomes necessary in order to moot liabilities falling due. 5. A group of workers of the same age entered a pension association which promised $500 annual payments for life, starting with payments at age 61. At ago 55, 10,000 workers remain alivo. They aro required to pay $50 at the beginning of each year up to and including their 00th birthdays. How much should tho asHociation have on hand as a reserve before the $50 payments due at ago 55 have boon made? MISCELLANEOUS PROBLEMS ON INSURANCE* 1. "Write a sample of each of tho following tyjxis of insurance policies, stating the age of the policyholder, tho bencfitH ho will reooivo, and how he is required to pay premiums : (a) 20-payment life ; (b) 10-year endow- ment; (c) ordinary whole life ; (d) 10-year term, 1 Insurance companies mentioned in those problems aro aeaumed to operate under assumptions (a), (b), and (o) of Section, 68, POLICY RESERVES 185 2. (a) A man aged 47 desires to set aside a sufficient sum which he can invest at 5%, effective, to pay him an annual income of $1000 for 10 years, starting with a payment on his 61st birthday. Find the amount set aside, assuming that ke will certainly live to age 70. (6) At age 47 what would he have to pay to an insurance company for a contract to, pay him $1000 at the end of each year for life, with the first payment at age 61, with the understanding that the company would compute the charge in accordance with the principles of scientific life insurance, at 3^%? 3. A woman offers $3000 to a benevolent organization on condition that the organization pay her 5% interest thereon at the end of each year for life. If the organization can purchase the required annuity for her from an insurance company, which uses the rate 3|%, will it pay to accept her offer if she is 55 years old ? 4. According to a will, a trust fund of $200,000 will go to a charity at the death of a girl who is now aged 19, and she is to receive the income at 4% for the remainder of her life. On a 3|% basis, find the present value of (a) her inheritance and of (6) the bequest to the charity. 5. A man borrows $200,000, on which he pays 5% interest annually. The principal is due at the end of 8 years. To protect his creditor he is compelled to take out an 8-year term insurance policy for $200,000. Assume that the man will certainly live to the end of 8 years, and find the present value at 6%, effective, of all of his payments on account of the debt, assuming that he pays merely the net premiums for his insurance as computed by a company which uses the rate 3i%. His age is 40 years. 6. A man aged 35 pays the net single premium on a whole life insurance for $1000. What is the policy reserve 10 years later? 7. A man aged 30 took out a 10-payment life policy. At the end of 10 years he desires to convert it into a 20-year endowment insurance as of that date. How much paid up endowment insurance will he obtain if the company permits all of his reserve to be used for that purpose? Notice that his reserve is the net single premium for the new insurance. 8. (a) Find the net annual premium at age 43 for an ordinary life policy for $2000. (6) Suppose that the man is alive at the end of 25 years. Find the reserve on his policy and compare it with the sum he would have on hand if he had invested all of his annual premiums at 5%, effective, 9. A man aged 42 borrows $100,000 and agrees to pay 4% interest annually. He agrees to provide for the payment of the principal at his death, or at the end of 10 years if he lives, by taking out a 10-year endow- ment policy for $100,000, with the creditor as beneficiary. The debtor 186 MATHEMATICS OF INVESTMENT considers his future payments, assuming (1) that he will pay merely the net premiums at 3% for his policy, (2) that he will certainly live to the end of 10 years, and (3) that he is able to invest his money at 7%, effective. He asks if it would pay to borrow $100,000 elsewhere at 6%, payable annually, with the agreement that the principal may be re- paid at the end of 10 years through the accumulation of a sinking fund, (a) Which method is best? (6) In terms of present values, how much could the debtor save by selecting the best method? 10. Compare the net single premiums for whole life insurance for $1000 (a) at ages 25 and 26; (6) at ages 75 and 76. (c) For which pair is the change in cost greatest? PART III AUXILIARY SUBJECTS CHAPTER XI LOGARITHMS 77. Definition of logarithms. Logarithms are exponents. The logarithm of a number N with respect to a base a, where a is > 0, 7* 1, is the exponent of the power to which a must bo raised to obtain N. That is, by definition, if #*-*, (1) then, the logarithm of N with respect to the base a is re ; or, in abbreviated form, l oga N = x. (2) Thus, since 49 = 7 2 / then Iog 7 49 = 2 ; since 1000 = 10 3 , then logio 1000 = 3. Also, if logs 'N = 2, then, from equation 1, N = 5 2 ; if Iog 6 N = 4, then N = 6 4 = 1296. In the future, whenever we talk of the logarithm of a number we shall be referring to a positive number N. This is necessary because, in the definition of a logarithm, the base a is positive, and hence only positive numbers N have logarithms as long as the x in equation 1 is a real number. EXERCISE LXXIV 1. Since 3 a - 9, what is logs 9? 2. Since 5 4 = 625, what is logG 625? 3. Since 100 10 2 , what is logm 100? 4. Since 2 3 8, what is log* 8? 6. Since 10 = 1, what is Iog 10 l? Since 17 1, what is logn 1? Since every number to the power zero is 1, what is the logarithm of 1 with respect to every base ; that is, since a = 1, what is loga 1 ? 6. What is Ipg 8 36? 8. What is log a 16? 10. .What is Iog 6 25? 7. What is logio 10,000? 9. What is Iog 7 7? 11. What is log a? 187 188 MATHEMATICS OF INVESTMENT /12. If Iog4# = 2, findiV. 19. Find log 10 1. 13. If logs N = 4, find N. 20. Find log lfl 4. 14. H logic N = 5, find N. 21. Find logioo 10. 15. If lo&W = i, find N. 22. If 10 1 - 6 = 31 .62, find Iog 10 31.62. 16. If lo&N = i find N. 23. If 10- 009 = 5, find logio 5. 17. If loga# = 6.5, find N. 24. If 10 2 - 4814 = 303, find Iog 10 303. 18. Find Iog 9 81. 25. If 10- 4771 = 3, find log w 3. Express in another way the fact that : 26. Iogio86.6 = 1.9370. 29. Iog 10 4730 = 3.6749. 27. logio 684 = 2.8351. 30. 343 = 7 3 . 28. logio 6.6 = .8195. 8 1. V = 1.732. 32. If N = i, find log* N. HINT. i = (4)" 1 . 33. Iftf = 4,findlog 2 tf. HINT. ^ - 3 2" 1 . lo lo A 34. If N = .1, find logio JV. HINT. .1 = A- 36. Find logio .001. Find logio .00001/: Find logio .0000001. 78. Properties of logarithms. Logarithms have properties which make them valuable tools for simplifying arithmetical computation. Property I. [The logarithm of., the product of two numbers M and N is equal to the sum of the logarithms of M and N : log tt AflV = log* Jf + log, tf. (3) Proof. Let logo M = x and logo N ** y. Then, since logo-W = x, then M = a*, (Def. of logarithms) and since logo AT = y, then N = a". (Def. of logarithms) Hence, MN = o a a = a a+v . (Law of oxpozionte) Since . MN=a a+v , thenlog a MN^x+y. (Def . of logarithms) Hence, logo MN log a M + logo N. (Subst. x = log M',y** logo ^V) Property n. The logarithm of the quotient of two numbers, M divided by N, is equal to the logarithm of the numerator minus the logarithm of the denominator : lOga j - lOga M- loga N. (4) Proof. Let logo M = x and logo N = y. * Then, since log a M = x, then M = a", (Def. of logarithms) and since log a N = y, then N = a". (Def. of logarithms) LOGARITHMS 189 TUf n* Hence, 41 = <L = a*-//. (La W O f exponents) Since = a""", then logo ^ = x y. (Def. of logarithms) Hence, logo ^ = Iog M logo N. (Subst. x = logo M;y^ Iog 2V) Property HE. The logarithm of a number N, raised to a .power Jc, is k times the logarithm of N: lOga-JV* = felOgaW. (5) Proof. Let logo N = x; then N = a*, by the definition of logarithms. Hence, . N k = (a") 6 = a* x . (Law of exponents) Since N h a* 1 , then logo N h = kx. (Def. of logarithms) Hence, logo N h = k logo N. (Subst. x = logo N) NOTE. In the future we shall deal entirely with logarithms to the base 10. Hence; for convenience, instead of writing logio N we shall write merely log N, understanding that the base always is 10. Logarithms to the base 10 are called Common Logarithms ; the name Briggs' logarithms is also used, in honor of an Englishman named Henry Briggs (1556-1630), who computed the first table of Common Logarithms. Example 1. Given that: log 2 = .3010, log 5 = .6990, log 17 = 1.2305,_fuid the logarithms of each of the following numbers: 34, 85, , Vl7, 25. Solution. log 34 - log 2(17) - log 2 + log 17 - .3010 + 1.2305= 1.5315. log 85 = log 5(17) - log 5 + log 17= .6990 + 1.2306 = 2.2295. (Prop. I) log jr. = i og 17 _ fog 5 =* 1.2305 - .6990 = .5315. (Prop. II) log Vl7 - log 17* = i log 17 = J(1.2305) = .61525. (Prop. Ill with k = *) log 25 - log 6 = 2 log 5 = 2(.6990) - 1.3980. (Prop. Ill with k - 2) EXERCISE LXXV In the problems below find the logarithms of the given numbers, given that : log 2 = .3010 log 3 = .4771 ' log 5 = .6990 log 7 - .8451 log 11 - 1.0414 log 13 = 1.1139 log 17 - 1.2305 log 23 1.3617 log 29 = 1.4624 1. 6 2, 9 3. 46 4. 61 B. -^ 6. A 7. 20 r 8. V 9. -^5 10. V7 11. 49 12. 16 13. 5 14. ^17 IB. Jf 16. 50 17. 55 18. 154 19. 20. U 21. 10 22. 100 23, 1000 24. 10,000 190 'MATHEMATICS OF INVESTMENT 26. 230 26. 2300 27. 23,000 28. 230,000 29. .1 = A 30. .01 = ^ 31. .001 32. .0001 33. .5 = A 34. .05 36. .005 36. .0005 79. Common logarithms. If one number N = 10* is larger than another number M = 10", then x must be larger than y. Since x = log N and y = log M, it follows that, if N is larger than M, then log N is larger than log M . Thus, since 9 is larger than 7, log 9 must be larger than log 7. The table below gives the logarithms of certain powers of 10. SINOB: THHN: 10000 = 10* log 10000 = 4 1000 = 10" log 1000 = 3 100 = 10* log 100 = 2 10 = 10 1 log 10 = 1 .1 = A = lo- 1 log .1 = - 1 .01 - T*IF = 10-* log .01 = - 2 .001 - T*W = lo- 3 log .001 = - 3 Consider the number 7, or any other number between 1 and 10. Since 7 is greater than 1 and less than 10, log 7 is greater than log 1, which is 0, and is less than log 10, which is 1. That is, since 7 is between 1 and 10, log 7 lies between and 1. Hence, log 7 = + (a proper fraction). From a table of logarithms, as described later, log 7 = .84510, approximately, so that the fraction men- tioned above is .84510. Similarly, since 750 is between 100 and 1000; log 750 lies between 2 and 3 ; therefore, log 750 = 2 + (a proper fraction) ; since 5473 is between 1000 and 10,000, log 5473 = 3 + (a proper fraction). In the same manner, since .15 lies between .1 and 1, log .15 lies between 1 and 0, and hence 1 log .15 = 1 + (a proper fraction). In general, the logarithm of every positive number can be expressed as an integer) either positive or negative, plus a positive proper fraction. The integral part of a logarithm is called its characteristic. When a number N is greater than 1, the characteristic of log N is positive ; when N is less than 1, the characteristic of log N is negative, 1 Any number between 1 and oan be expressed as 1 + (a proper fraction). Thus, - .67 - -* 1 + .43 ; - .88 - - 1 + .12, etc, LOGAEITHMS 191 The fractional part of a logarithm is called its mantissa. Thus, given that log 700 = 2.84510, the characteristic of log 700 is 2, and the mantissa is .84510 ; given that log .27 = 1 -|- .43136, the characteristic of log ,27 is 1 and the mantissa is .43136. 80. Properties of the mantissa and the characteristic. Given that log 3.8137 = .58134, then, by use of Properties I and II of Section 78, and from the logarithms of powers of 10 given in Sec- tion 79, we prove the following results : log 3813.7 = log 1000(3.8137) = log 1000+log 3.8137=3+.58134=3.68134. log 381.37 = log 100(3.8137) = log 100 +log 3.8137 =2 +.58134 =2.58134. log 38.137 = log 10(3.8137) = log 10 +log 3.8137 = 1 +.58134 = 1.58134. log 3.8137 = 0.58134. log .38137 = log = log 3.8137- log 10 - .68134- 1 = - 1+. 58134. log .038137 = log ^j|? = log 3.8137- log 100 = .58134- 2 = - 2 +.58134. log .0038137 = log = log 3.8137- log 1000 = .58134- 3 = - 3+.58134. NOTE. The characteristics of the logarithms above could have been obtained as in Section 79. Thus, since 3813.7 lies between 1000 and 10,000, log 3813.7 lies between 3 and 4 ; log 3813.7 = 3 + (a proper fraction). There- fore, the characteristic of log 3813.7 is 3, as found above. From inspection above, we see that .58134 is the mantissa of all of the logarithms. This result, which obviously would hold for any succession of digits as well as it does for the digits 3, 8, 1, 3, 7, may be summarized as follows : Rule 1. The mantissa of the logarithm of a number 2V de- pends only on the succession of digits in N. If two numbers have the same succession of digits, that is, if they differ only in the posi- tion of the decimal point, their logarithms have the same mantissa. The logarithms above also illustrate facts about the character- istic. Rule 2, The characteristic of the logarithm of a number greater than 1 is positive and is 1 less than the number of digits in the number to the left of the decimal point. NOTE. Thus, in accordance with Rule 2, 3 is the characteristic of log 3813.7 j 2 is the characteristic of log 381.37, etc, Eule 2 is justified in general 192 MATHEMATICS OF INVESTMENT by recognizing that, if a number N has (k + 1) digits to the left of the decimal point, then N is between 10* and lO* 41 ; hence log N is between k and (k + 1) and log N = k + (a proper fraction). That is, k is the characteristic of log N. Rule 3. If a number N is less than 1, the characteristic of log AT is a negative integer ; if the first significant figure of N appears in the fcth decimal place, then the characteristic of log N is ft. Thus, the first significant figure of .38137 is 3 and appears in the first decimal place, and, in accordance with Rule 3, the characteristic of log .38137 is 1. The first significant figure of .038137 appears in the 2d decimal place, while the characteristic of log .038137 is - 2, etc. NOTE. It may appear strange to the student that we write, for example, log .0038137 = 3 + .58134, instead of performing the subtraction. For every number N which is less than 1, log N is a negative number ; thus, log .0038137 = 3 + .58134 = - 2.41866. Written in this way, the mantissa .58134 and the characteristic 3 are lost sight of. We write the logarithm in the form 3 + .58134 to keep the characteristic and the mantissa in a prominent position. Since the mantissa depends merely on the succession of digits in the number, it is customary. to speak of a mantissa as corre- sponding to a given succession of digits without thinking of any decimal point being associated with the digits. Thus, above, we would say that the mantissa for the digits 38137 is .58134. Example 1. Given that the mantissa for the digits 5843 is .76664, find log 5843 ; log 584.3 ; log 58,430,000; log 5.843 ; log .0005843. Solution. The characteristic of log 6843 is 3 ; hence, log 5843 = 3.76664. Similarly, log 584.3 =2.76664; log 58,430,000=7.76664; log 6.843 - 0.76664 ; log .0005843 = - 4 + .76664. EXERCISE LXXVI 1. Given that .75101 is the mantissa for. the digits 56365, find log 5636.5; log 56365; log 563.65; log 56,365,000 ; log .0056365; log .56365. 2. Given that .93046 is the mantissa for 85204, find log 85.204 ; log 852,040,000; log 8.5204; log "85204; log .085204; log .0000085204, ' 3. Given that .39863 is the mantissa for 2504, find log 2504 ; ' log 2.504 ; log 25,040; log .2504; log .00000000002504. 4. Given that log 273.7 = 2.43727, find log 2.737; log 27.37; log 27,370; log .02737; log .002737. Moke use of Rule 1. LOGARITHMS 193 5. Given that log 68,025 = 4.83267, find log 68.025; log 6.8025; log .68025 ; log 6802.5 ; log .00068025. 6. What is the mantissa of log 1 ; of log 10; of log 10,000; of log .1 ; of log .00001? 81. Tables of mantissas. The mantissa for a given succession of digits can be computed by the methods of advanced mathematics. The computed mantissas are then gathered in tables of logarithms which, more correctly, should be called tables of mantissas. Ex- cept in special cases, mantissas are infinite decimal fractions. Thus the mantissa for 10705. is .02958667163045713486 to 20 deci- mal places. In a 5-place table of logarithms, this mantissa would be recorded correct to 5 decimal places, giving .02959. In an 8- place table, it would be recorded as .02958667, correct to 8 deci- mal places. NOTB. Table I in this book is a 5-place table of logarithms. A decimal point is understood hi front of each tabulated mantissa. To find the mantissa for N = 3553, for example, go to the sixth page of Table I. Find the digits 355 in column headed N ; the mantissa for 3553 is entered in the corresponding row under the column headed 3. The entry is " 060," but the first two digits of tlie mantissa are understood to be " 65," the same as for the first entry in the row. Thus, the mantissa for 3553 is .55060. From Table I the student should now verify that : FOR THE DIGITS BHLOW THE MANTISSA is 3630 .55991 3947 .59627 4589 .66172 9331 .96993 9332 .96997 9333 1 .97002 Example 1. Find log 38570 ; log .008432. Solution. By inspection, the characteristic of log 38570 is 4 ; the mantissa as found in Table I is .58625. Hence, log 38570 = 4.58626. The character- istic of log .008432 is - 3; log .008432 = - 3 + .92593. 1 In Table I, for 9333, wo find the entry " *002. ' ' The asterisk (*) on the " 002 " means that tho first two digits are to be changed from 96, as at the beginning of the 1 row, to 97. 194 MATHEMATICS OF INVESTMENT' In order to obtain greater convenience in computation, it is customary to write negative characteristics in a different manner than heretofore. Thus, in log .008432 = - 3 + .92593, change the - 3 to (7 - 10). Then log .008432 = - 3 + .92593 = 7 - 10 + .92593 = 7.92593 - 10. Recognize cle 4 arly that log .008432 = - 3 + .92593 = - 2.07407. We verify that 7.92593 - 10 = 2.07407. The two ways introduced for writing log .008432 are merely two different ways of writing the negative number 2.07407, which is the actual logarithm involved. Similarly, log .8432 = - 1 + .92593 = 9.92593 - 10; log .000'000'000'ob8432 = '- 12 + .92593 = 8 - 20 + .92593 = 8.92593 - 20, etc. NOTE. The change from the new form to the old or vice versa is oasy. Thus, given that log .05383 = 8.73102 - 10, we see that the characteristic is (8 - 10) or - 2; given that log .006849 = - 3 + .76708, then log .006849 = 7.76708 - 10. EXERCISE LXXVH 1. What are the characteristics of the following logarithms : 9.8542 10 ; 7.7325 - 10; 6.5839 - 10; 4.3786 - 10? 2. Write the following logarithms in the other form : 3 + .5678 ; - 5 + 7654; - 7 + .8724; - 1 + .9675. 3. Write the following logarithms as pure negative numbers : 3 + .5674; - 1 + .7235; 9.7536 - 10; 7.2539 - 10. 4. By use of Table I verify the logarithms given below : N Loo N N Loo# 3616. 3.54593 35.88 1.65486 .01832 8.26293 - 10 1.170 0.06819 889,900 5.94934 .0008141 0.91008 - 10 .6761 9.83001 - 10 . 27,770 4.44358 621.8 2.79365 .00004788 5.08015 - 10 NOTE. When the characteristic of log 2V is 0, log N is equal to Us man- tissa. Thus, log 1.578 = 0.19811. Hence, a table of mantissas is a tiiblo of the actual logarithms of all numbers between 1 and 10. 82. Logarithms of numbers with five significant figures. If a number N has five significant digits, log N cannot be read di- rectly from the table. We must use the process of interpolation as described in the following examples. LOGARITHMS 195 Exampk 1. Find log 25.637. Solution. The characteristic is 1. To find the mantissa, recognize that 25.637 is between 25.630 and 25.640 ; the mantissas for 2563 and for 2564 were read from Table I and the logarithms of 25.630 and 25.640 are given in the table below. Since 25.637 is .7 of the way from 25.630 toward 25.640, we assume l that log 25.637 is .7 of the way from 1.40875 toward 1.40892. The total way, or difference, is .40892 - .40875 = .00017; .7 of the way is .7(.00017) = .000119. We reduce this to .00012, the nearest number of five decimal places. Hence, NUMB an LOGABrrHM 25.630 25.637 25.640 1.40875 ? ? 1.40892 log 25.637 = 1.40875 + .00012 = 1.40887. NOTE. At first, the student should do all interpolation in detail as in Example 1 above. Afterward, he should aim to gain speed by doing the arithmetic mentally. The small tables in the column in Table I headed PP, an abbreviation for proportional parts, are given to reduce the arithmetical work. Exampk 2. Find log .0017797. Solution. The characteristic is 3 or (7 10). The digits 17797 form a number between 17790 and 17800. The tabular difference between the corre- sponding "mantissas is (.25042 .25018) = .00024, or 24 units in the 6th decimal place. Since 17797 is .7 of the way from 17790 to 17800, we wish .7(24). By multiplication, ,7(24) = 16.8. This should be found without multiplication from the small table headed 24 under the column PP. From this table we read .1(24) = 2.4, .2(24) - 4.8, etc., .7(24) = NUMBER MANTISSA 17790 .25018 17797' ? ? 17800 .25042 16.8. Hence, the mantissa for 17797 is .25018 + .17 = .25035, and log .0017797 - 7.25035 - 10. NOTE. The following situation is sometimes met in interpolating. Sup- pose that .6(15) 7.5 is the part of the tabular difference which-we must add. Wo may, with equal justification, call 7.5 cither 7 or 8. As a definite rule in this book, whenever such' an ambiguity is met, we agree to choose the even number. Hence, we choose 8 above. Similarly, in using .7(15), or 10.5, we should call it 10, because we have a choice between 10 and 11. 1 This assumption is justified by the first paragraph of Section 70, Since 25,037 ia between 26.030 and 25.640, log 25,637 must be between log 25.030 and log 25.040. In interpolating as in Example 1, we merely go ono step farther than, tliis admitted fact when we assume that the change in the logarithm is proportional to the change in the number. This assumption, although not exactly true, is sufficiently accurate for all practical purposes. 196 MATHEMATICS OF INVESTMENT EXERCISE LXXVIH 1. Verify the following logarithms : log 256.32 = 2.40878 log 8966.1 = 3.95211 log 13.798 = 1.13982 log 931.42 = 2.96915 log .073563 = 8,86666 - 10 log 33.581 = 1.52609 log .59834 - 9.77695 - 10 log .00047178 = 6.67374 - 10 log 1.1675 = 0.06725 log 676.93 = 2.83064 2. Find the logarithms of the following numbers : 18.156 .31463 .061931 151.11 5321.7 83196 48.568 6319.1 67.589 113.42 384.22 9.3393 .031562 .92156 .52793 .000031579 .009567 5.6319 1.1678 83.462 83. To find the number when the logarithm is given. Example 1. Find N if log# = 7.67062 - 10. Solution. Since the characteristic is (7 10) = 3, the first significant figure of 2V will appear in the 3d decimal place ; N = .00 . . . . To find the digits of N, we must obtain the number whose mantissa is .67062. We search for this mantissa, or those nearest to it, in Table I ; we find .67062 as the mantissa of 4684. Hence, N = .004684. Example 2. Find N if log N = 5.41152. Solution. We wish the 6-figure number whose mantissa is .41152. On inspecting Table I we find the tabular mantissas .41145 and .41162 between which .41162 lies. The total way, or tabular difference, between .41145 and .41162 is .00017, or 17 units in the 5th decimal place. The partial difference .41152 - .41146 > .00007, or 7 units hi the 6th decimal place. Hence, .41152 is fr of the way from .41145 to .41162. Wo then assume that the number x, whose mantissa is .41162, is -^ of the way from 25790 to 25800. The total way, or dif- ference, is 10 units in the 6th place; ^(10) 4.1; the nearest unit is 4. Hence, .41152 is the mantissa of 25790 + 4 - 25794. Since the characteristic of log N is 5, N = 267,940. Nora. The arithmetic in Example 2 above is simplified by use of the table headed 17 under the column of proportional parts. In Example 2 wo desire ' />1 .0), which we can easily obtain if we know -fr oorroot to tho nearest tenth. m the table headed 17, we read .4(17) - 6.8, or ^ - .4; .6(17) - 8.6, P? - .5. Since 7 is between 6.8 and 8,5, ^ is between .4 and .5, but is rest to .4. Thus, ^(10) - 4, to the nearest unit. With practice, this NUMBEH MANTISSA 25790, X 25800 .41145 .41152 .41162 LOGARITHMS 107 result should be obtained almost instantaneously. Thus, look under the table headed 17 for the number nearest to 7 ; we find 6.8 ; at the left it is shown tfiat this is A of 17 ; hence ^(10) = 4. EXERCISE LXXIX 1. Find the numbers corresponding to the given logarithms and verify the answers given : log N = 3.21388; N = 1636.4. log N - 3.75097; N - 5636. log N = 8.40415 - 10; N = .02536. log N = 0.46839; AT = 2.9403. log N - 2.16931; N - 144.31. log N = 3.33590; N - 2167.2. log N = 9.52163 - 10; N - .33238. log N - 8.66267 - 10; N = .044944. log N =0.89651; N - 7.8797. log N - 0.36217; N - 2.2499. 2. Find the numbers corresponding to the following logarithms : log N = 5.21631 log N - 3.19008 log N - 9.64397 - 10 log N = 1.39876 log N - 7.56642 - 10 log N - 2.57938 - 10 log N = 8.95321 - 10 . log N = 0.89577 log N = 1.77871 log N = 4.32111 - 10 log N - L21352 log JV - 7.77853 log N = 2.15678 log N - 8,45673 - 10 log AT = 3.15698 84. Computation of products and of quotients. Exampk 1. Compute P - 787.97 X .0033238 X 14.431. Solution. From Property I of Section 78, log P is the sum of the logarithms of the factors. From Table I, log 787.97 =2.89661 log .0033238 = 7.52163 - 10 log 14.431 -1.15031 (add) log P -11.57746 -10 = 1.67746 . From Table I, P - 37.796. Solution. From Property II of Section 78, log Q equals the logarithm of the numerator minus the logarithm of the denominator. Both numerator and denominator are products whose logarithms are determined by Property I. log 4.8031 - 0.68152. log 78797 - 4.89651 log 269.97 - 2.43131 log 253.6 =2.40415 log 1.6364 0.21389 (add) log Denom, 7.30066 (add) log Numer. - 3.32672 log Denom. 7.30066 (subtract) log Q = ? 198 MATHEMATICS OF INVESTMENT We recognize that the result on subtracting will be negative. To obtain log Q in standard form, we add and also subtract 10 from the log numerator. log Numer. - 8.32672 = 13.32672 - 10 log Denotn. = 7.30066 (subtract) logQ = 0.02606 - 10; Q = .00010618. NOTE. Before computing any expression by logarithms, a computing form should be made. Thus, the first operation in solving Example 2 above was to write down the following form : log 4.8031 - log 78797 = log 269.97 = log 253.6 = log 1.6364 = (add) logDenom. = (add) log Numer. = log Denom. = (subtract) log Q => A systematic form prevents errors and makes it easy to repeat the work if it is desired to check the computation. EXERCISE LXXX Compute by logarithms : 1. 563.7 X 8.2156 X .00565. 2. 4.321 X 21,98 X .99315. 675.31 4 66.854. 13.215' ' 2356.7 .008315 6 783.12 X 11.325 .0003156' ' 8932 86 X 73 X 139.68 fi 9.325X631.75. 3215.7 X .4563 ' ' .8319 X .5686 ft .42173 X .21667 1Q 5.3172 X .4266 .3852 X. 956 ' ' 18.11X31.681 85. Computation of powers and of roots. Exampk 1. Find (.3156) 4 . Solution. From Property III of Section 78 with fc 4, log (.3166)* - 4 log .3156 - 4(9.49914 - 10) - 37.99600 - 40 - 7.99656 - 10. From Table I, (.3156)* - .009921. Example 2. Find ^856.31. SoMion. -^856.31 - (856.31)*. From Property III with k -*, log ^ - * log 866.31 2 ' 93 263 - 0.97754 j hence, ^856\3T - 9.4960. 3 LOGARITHMS 199 Exampk 3. Find ^08361; ^.08351. Solution. Since #.08351 = (.08351)*, we obtain from Property III, log ^08351 - Jlog .08351 - 8 ' 9217 f ~ 1Q . If we divide this as it stands, D we obtain 1.48696 J& a most inconvenient form. Hence, we add and, at the same time, subtract 50 from log .08351 in order that, after the division by 6, the result will be in the standard f orm for logarithms with negative character- istics. Hence, log v'.'oWl = 8 -92174 - 10 _ 50 + 8.92174 - 10 - 50 _ 58.92174 - 60 log ^ From Property III, log 6 6 9.82029 - 10 ; hence, from Table I, 6 .66113. log .08361 8.92174 - 10 28.92174 - 30 9.64058 - 10; hence, ^.08361 - .43710. EXERCISE LXXXI Compute by logarithms : 1. (175) 2 . 2. (66.73) 8 . 4. V53T2. 6. ^.079677. 7. (353.3 X 1.6888) 2 . 8. ^199^62. 10. (1.06)*. 11. (1.03)". 13. 1 16 85.75 56.35 X 4.3167 14. (45.6) 2 8. (.013821) 4 . 6. (.38956)*. 9. (1.05) 7 . 12. (1.06) 29 . 16. (1.03)" 8 - 1 (1.03)' V '21.36 X V52L9 HINT. For this problem, the computing form is : 17. 19. log 56.36 = log 4.3157 - log 621.9 - f i log 521.9 = 1 log 21 .36 - (add) log Numer. = log Denom. (add) log Denom. = (subtract) log fract. = 4 log frftct. 535 X 831,6 X (1.03) 8 Result = 18. (189.5)*. ' 20. V896.33. M .03166 X 75.31 475 X 938 ^00356. (163.2) a X 257.3 1893.2 X 35830 221.38 X (.3561) a 28. 24, (1.035)"", 200 MATHEMATICS OF INVESTMENT 86. Problems in computation. It is very important to realize tbat the properties I, II, and III of logarithms may bo used in com- puting products, quotients, and powers, but that they may not bs used in computing differences or sums except in the auxiliary Banner illustrated below, _ v Tin 4 n V896+ (.567) (35.3) Example 1. Compute Q = 532 - (15 31)* -- Solution. By logarithms, ve perform each of the three computations below. log V896 - * tag 896 - 2 ' 9 ^ 231 = 1.47616; V896 = 29.934. 2 log .567 - 9,76358 - 10 log 35.3 = 1,84777 log prod. - 11,30135 - 10; (.567) (35.3) = 20.016. leg (15.31)' = 2 log 16.31 = 3(1.18498) = 2.36996; hence, (15.31) 3 234.40. e *. above, log 49.949 = 1.69853 = 11.69853 - 10 log 297.60 = 2.47363 - 2.47363 (subtract) log Q = 9.22490 - 10; hence, Q = .16784. NOTE. A computation cUne with a 5-place table of logarithms will give results which are accurate to 4 significant figures, but the 5th figure always tvill be open to question. Each mantissa in the table, and each of those we determine by interpolation is subject to an error of part of 1 unit hi the 5th (fecimal place, even though all of our interpolation is done correctly. During a Jong computation, these accumulated errors hi the logarithms, together with tie allowable error due to oil* final interpolation, cause an unavoidable error in the 5th significant figure <vf our final result. Therefore, if a number with BUore than 5 significant figures, such as 2,986,633, is mot in a computation with A,5-place table, we should reduce this number to 2,986,500, tho nearest number having 5 significant figures, before finding its logarithm. To retain more than & significant digits is fictitious accuracy, since our final results will bo accurate fa only 4 digits. For the sanw reason, in looking up tho number corresponding fa a given logarithm, the interpolation should not be carried beyond the near- est unit in the 5th significant place. NOTE. Logarithmic computation of products, quotients, and powers must d&al entirely with positive ntwnbers, according to the statements of Section 77. Hence, if negative numbew a?e involved, we first compute the expression by logarithms as if all numbers vere positive, and then by inspection determine tie proper sign to be assigned to the result. Thus, to compute (- 75.3) X ( - 8.392) X (- 32.15) we firsb find 75.3 X 8.392 X 32.15 - 20316 ; then, we ft&te that a negative sign rnngt be attached, giving 20316 as the result* LOGARITHMS 201 EXERCISE LXXXH Compute by logarithms : (35.6) 2 + 89.53 . 1.931 X 5.622 - V11L39 - 2.513 ' 5.923 3 dOS) 5 ~ 1- 4 1 ~ (1.Q4)- 4 . (1.03)* -1 ' -04 6 (1-07) 8 ~ 1. 6 251 + 63.95 X 41.27 .07 ' 787 7. 395X856. 8. ||||. 9. (Iog395)(log856). 10. |^|^|- That is, compute That is, compute (2.59660) (2. 93247). 11 log 88-2 - 3 log (1.04) 654 log 2 ' 13 Io 6 6 - 532 14. log 8.957 ' log 1.04' log 1.06' 16. 153.5(1.025) 10 . 16. 35.285(1.04)- B . 17. (1.05)*. 18. (1.035)*. 19. 12[(1.02)A - 1]. 20. log <85 + 3 . , 87. Exponential equations. An equation in which the unknown is involved in an exponent is called an exponential equation. Thus, 3* 7 = 27 is an exponential equation for 2. In this sec- tion we shall treat exponential equations of the type that can be solved by use of the following rule : Rule, To solve a simple exponential equation, take the log- arithm of both sides of the equation and solve the resulting equation. <- Exampk 1. Solve the equation IS 2 ** 2 = (356)5*. Solution. Toko tho logarithm of both sides of the equation, making use of Property I. Then (2 x -f 2) log 13 - log 366 + * log 6, or 202 MATHEMATICS OF INVESTMENT (2 x + 2) (1.11394) - 2.66146 + s(.69897). (Table I) 2.22788 x + 2.22788 - 2.66146 + .69897 x. 1.62891 x = .32367; .32357 011*4 log -32367 = 9.60997 - 10 1.62891 ' log 1.5289 = 0.18438 (subtract) log x = 9.32659 - 10 The exponential equations met in applications to the mathe- matics of investment are of the form A' = B, (6) where A and B are constants, and where v is a function of the un- known quantity. Example 2. Solve (1.07) 2n = 4.57. Solution. Taking the logarithm of both sides, we obtain 2 n.]og 1.07 = log 4.57; n = lo ? 4 ' 57 . = ' 659 ' 92 = *&&. 2 log 1.07 2(.02938) .05876 log -.85082 = 9.81949 - 10 log .05876 = 8.76908 - 10 (subtract) logn = 1.05041; n =11.231. EXERCISE LXXXm Solve the following equations : 1. (1.05)" = 6.325. 2. 15* = 95. 3. 12 a+1 = 38. 4. (1.025) 2 " = 3.8261. 6. S3* = 569. 6. 2 n = 31. 7. 5* = 27(2-). 8. 25(6*) - 282. HINT. Clear the equation of fractions and reduce to the form of equation 6, obtaining (1.035) n = 1.06875. 10. (1.045)"" = .753. HINT. The equation becomes - n log 1.045 = 9.87679 - 10 = - .12321. 11. (1.03)-* - .8321. 12. 850(1.05)" = 1638. 13. 65.30(1.025)-" - 52.67, 14. 750 CLOg)*-l = 3500> -02 LOGARITHMS 203 SUPPLEMENTARY MATERIAL 88. Logarithms to bases different from 10. To avoid confu- sion we shall explicitly denote the bases for all logarithms met in this section. From Section 77, x = log JV satisfies the equation a* = N. By solving this exponential equation, we can find x when N and a ane given. Thus, taking the logarithm to the base 10 of both sides of a x = N, we obtain logio N or x = log, N = - . logiotf. (7) NOTE. Equation 7 enables us to find the logarithm of any number with respect to a given base a, provided that we have a table of logarithms to the base 10. The quantity logio a is called the modulus of the system of logarithms to the base 10 with respect to the system to the base a. The natural system of logarithms is that system where the base is the number e = 2.718281828 ---- The number e is a very im- portant mathematical constant and logarithms to the base e are useful in advanced mathematics.- From an 8-place table, we find logio e = 0.43429448; log .43429448 = 9.63778431 -10. Exampk 1. Find log, 35. Solution, Let x log. 35. Then, e a = 36 ; taking the logarithm of both sides to the base 10, x logio - logio 35 ; x a iQRio 35 = 1.54407 log 1.5441 - 10.18868 - 10 logio e 0.43429* log .43429 9.63778 - 10 x 3.5555. (subtract) log x = 0.55090 EXERCISE LXXXIV 1. Find log. 76; logs 10; log, 830; log, 657. 2. Find the natural logarithm of 4368. 3. Find logo 353; logs 10; Iogg895; Iogi 5 33. 4. If a and 6 arc any two positive numbers, prove that logs N <=> log a N - logs a. HINT. Let x Iog N and y - log& N. Then N = a* = b. Take the logarithm with respect to the base & of both siclee of the equation ft" = a*, CHAPTER XII PROGRESSIONS 89. Arithmetical progressions. A progression is a sequence of numbers formed according to some law. An arithmetical pro- gression is a progression in which each term is obtained from the next preceding term by the addition of a fixed constant called the common difference. Thus, 3, 6, 9, 12, , etc., is an arith- metical progression in which the common difference is 3. Simi- larly, 3, f, 2, f , , etc., is an arithmetical progression in which the common difference is ( ^-). Let a represent the first term of an arithmetical progression, d the common difference, and n the number of terms in the progres- sion. Then, in the progression, a = 1st term, a + d = 2d term, a + 2d = 3d term, a + 3d = 4th term, - etc. (8) a + (n l)d = nth term. If we let I represent the last, or the nth, term, we have proved that I = a + (n -1 ) d. (9) If we start with the last term, the next to the last term is formed by subtracting d, the second from the last by subtracting 2 d, etc. That is, in going backward, we meet an arithmetical progression with the common difference ( d). Thus, Z = last term, Z d = 1st from last term, Z 2 d = 2d from last term, i! 3 d = 3d from last term, etc. - (10) a = Z - (n l)d = (n l)st from last. Let s represent the sum of the terms of the progression. Then, we obtain equation 11 below by using the terms aj3 given in equa- tions 8, and equation 12 by using equations 10. s = a +a da 2d ..-etc. . + [a + (n - l)fl. (11) etc,,-.-+[Z-(n-l)4 (12) 304 PROGRESSIONS 205 On adding equations 11 and 12, we obtain 2 a - (a + I) + (a + I) + ( + J) + ' ' ' etc. + (a + I). (13) There are n terms in equation 13, one corresponding to each term of the progression. Hence, 2 s = n(a + I), or s=2( a + Z). (14) 2 If any three of the quantities (a, d, n, I, s) are given, the equa- tions 9 and 14 enable us to find the other two. We call (a, d, n, I, s) the elements of the progression. Example 1. In an arithmetical progression with the first term 3, the 6th term is 28. Find the common difference and the intermediate terms. Solution. We have a = 3, n = 6, and I = 28. Hence, from equation 9, 28 = 3 + 5 d ; 5 d = 25 ; d = 5. The terms of the progression are 3, 8, 13, 18, 23, 28. EXERCISE LXXXV 1. Find the last term and the sum of the progression 3, 5, 7, 9, ... to twelve terms. 2. Find the sum of the progression 5, 4, 3, 2, . . . , to eighteen 'terms. 3. Find the last term and the sum of the progression 1000(.05), 950(.05), 900(.05), . . . etc., to twenty terms. 4. If 10 is the first term and 33 is the 20th term of an arithmetical progression, find the common difference and the sum of the progression. 5. If 15 is the 4th term and 32 is the 10th term of an arithmetical progression, find the intermediate terms. 90. Geometrical progressions. A geometrical progression is a progression in which each term is formed by multiplying the preceding term by a fixed constant r. The number r is called the common ratio of the progression because the ratio of any term to the preceding term is equal to r. Thus, 4, 12, 36, 108, - etc., is a geometrical progression with the common ratio r => 3, The sequence (1.05), (1.05) 2 , (1.05) 8 , (1.05)*, etc., is a geometrical progression with the ratio r * (1.05). 206 MATHEMATICS OF INVESTMENT Let a represent the first term, r the common ratio, and n the number of terms in a geometrical progression. Then, ar 6 = 5th term, . . '. etc., = (73, __ i) B t term, a = 1st term, ar = 2d term, ar 2 = 3d term, ar 8 = 4th term, If we let I represent the last, or nth, term, we have proved that Let s represent the sum of the terms of the progression. Then s = a + ar + ar 2 + etc. + ar n ~ 2 + ar"" 1 , (16) rs = ar 4- ar 2 + ar 8 + etc. + ar"" 1 + * (17) On subtracting equation 17 from equation 16, all terms will cancel except a from equation 16 and ar n from equation 17. Thus, s T-S = 5(1 r) = a ar n . Hence, s = a -f^ = 5^1 ( 18 ) Since I ar n-1 , then rl = ar n ; on substituting this in the first fraction of equation 18, rl a s = _ T (19) Example 1. Find the sum of 1 + -J- + i + etc ---- to six terms. Solution. Use formula 18 with a = 1, r ~ 4, and n = 6. S= l-* = } = 24' Exampk 2. Find an expression for the sum of 1 + (1.05) " + (1-05) + (1.05) 1 - 6 + etc. . . . + (1.05) 38 - 5 . , Solution. The terms form a geometrical progression for which o 1, r = (1.05)- B , and I - (1.05) W - B . From formula 19, ' - 1 ra (l.Q5) M - 1 . (1.05) - B - 1 (1.05) - 1 EXERCISE LXXXVI 1. Find the last term and the sum of 25, 5, 1, t, A, etc. to seven terms. 2. Find the last term and the sum of 2, 4, 8, . . . etc, to eighteen terms. 3. Find the ratio, the number of terms, and the sum for the progres- sion 3, 9, 27, ... t? t ; to 729. PROGRESSIONS 207 4. Find the sum of 2, 1, , etc. to eight terms. 5. Find the sum of 1 + * + H ----- j-rk- Find expressions for the following sums : 6. (1.05) + (1.Q5) 2 + (1.05) 8 + etc. + (1.05) 28 . 7. (1.04) 2 + (1.04) 4 + (1.04) 8 + etc. + (1.04) 38 . 8. (1.06) - 215 + (LOG)" 24 + (1.06)' 28 + . . . etc. + (1.06) "\ 9. (1.03) - 1 + (1.03)' 2 + (LOS)' 8 + - - etc. + (1.03)-". 10. (1.02) + (1.02) J + (1.02) 8 + etc. + (1.02) 80 . 91. Infinite geometrical progressions. Consider the follow- ing hypothetical example. A certain jar contains two quarts of water. One quart is poured out; then, ^ of the remainder, or ^ quart, is poured out ; then, % of the remainder, or quart, is poured out, etc., without ceasing. The amounts poured out are 1, , i, -g-, etc. to infinitely many terms. The sum of the amounts poured out up to and including the nth pouring is , , 1 , 1 , , 1 Sn = 1+ 2 + 4 + ' ' + 2^i' Since the amount originally in the jar was 2 quarts, s n can never exceed 2. Also, it is clear intuitionally that, as n increases with- out bound, s n must approach the value 2 because the amount of water left in the jar approaches as the process continues. We can prove this fact mathematically ; from formula 19, 2) = 2 - (20) 1 As n grows large without bound, continually decreases and approaches zero. Thus, from equation 20 we prove that, as n increases without bound, s n approaches the limit 2, as was seen intuitionally above. Hence, we may agree, by definition, to call this value 2 the sum of the infinite geometrical progression, or to say 2 = l + -f-+''- etc. to infinitely many terms. This example shows that a sensible definition, in accordance with our intuitions, may be given for the sum of an infinite geometrical progression, 208 MATHEMATICS OF INVESTMENT In general, consider any infinite geometrical progression for which the ratio r is numerically less than 1, that is, for which r lies between 1 and + 1. The terms of the progression are a, ar, ar 2 , ar 3 , etc. to infinitely many terms. Let s n represent the sum of the first n terms of the progression : s n = a + ar + ar 2 + + ar n ~ l . The statement as n approaches infinity will be used as an abbre- viation for the statement as n increases wthout bound. The sum S of an infinite geometrical progression is defined as the limiting value, if any exists, approached by s n as n approaches infinity. From formula 18, a ar n a ar n As n approaches infinity, it is evident that r n approaches zero l because r is numerically less than 1. Hence, from equation 21 it is seen that, as n approaches infinity, s n approaches _ as a limiting value, because the other term in equation 21 approaches zero. Since, by definition, this limiting value of s n is the value we assign to the sum S = a + ar + ar 2 + etc. to infinitely many terms, we have proved that a 1 r Example 1. Find the sum of the progression (1.04) "* + (1.04) " + (1.04)" 8 H ---- etc. to infinitely many term. Solution. The ratio of the infinite geometrical progression is r = (1.04)~ a ; a = (1.04) ~*. From formula 22, the sum is S (1.04)-* 1 - (1.04)-' Exampk 2. Express the infinite repeating decimal .08333 as a fraction. Solution. We verify that .08333 equals .08 plus .003 + .0003 + .00003 + etc. to infinitely many terms. 1 For a rigorous proof of this intuitional fact the student is referred to the theory pf limits as presented, for example, in books on the Calculus. PROGRESSIONS 209 These terms form an infinite geometrical progression with a = .003, and r = .1. Their sum is ~^- = :92. Hence, .08333 = .08 + = JL + JL = .?A = -L. .9 100 900 300 12 NOTE. By the method of Example 2 above, any infinite repeating decimal can be shown to represent a fraction whose numerator and denominator are integers. EXERCISE LXXXVH Find the sums of the following progressions : 1. 2 + 1 + i + to infinitely many terms. 2. 5 + l+i + -jfr+---to infinitely many terms. 8 - (W + (Ti5y' + (i^ + ^ 4. (1.04)- 1 + (1.04)" 8 + (1.04) " 3 H to infinitely many terms.- 6. (1.03)~* + (1.03)" 4 + (1.03)~ H to infinitely many terms. 6. (l.Oir 1 + (1.01)~ 2 + (l.Oir 8 + ... to infinitely many terms. Express the following infinite decimals as fractions : 7. .333333 -. 8. .66666 9. .11111 . 10. .41111 -. 11. .5636363 -. 12. .24222222 . APPENDIX Note 1 Proof of Rule l; Section 16, Part I. Consider the equation 2 = (1 + r). The solution of this equation for n is the time required for money to double itself if r is the rate per period. On taking the logarithms, with respect to the base e = 2.71828 . . . , of both sides of the equa- tion, we obtain l og 2 71 " log (1 + r) f where " log " means " log fl ." From textbooks on the Calculus, we find that log (1 + r) =r-^+^ ----- r(l - | + ~ -), and from a table of natural logarithms we obtain log 2 = .693. Hence .693 .693. ,r r 2 , * On 1 neglecting the powers of r in the parenthesis from r 2 on, we obtain as an approximate solution .693 , .693 .693 , , n = -- = -- r '5O- r 2 r Note 2 Proof of Rule 1, Section 17, Part I. Consider three obligations whose maturity values are Si, 89, and 83, which are due, respectively, at the ends of n\, n a , and nj years. We shall prove Rule 1 for this special case. The reasoning and the details of proof are the same for the case of any number of obligations. Let i be the effective rate of interest, and let n bo the equated time. By definition, n satisfi.es the equation iT" 1 + &(1 + ff"" 1 + A(l + i)-" 3 . (1) 211 212 MATHEMATICS OF INVESTMENT By use of the binomial theorem, we obtain the following infinite series "(I . 2 Since i is small, we make only a slight error if we use only the first two terms of each infinite series as an approximate value for the corresponding power of (1 + i). On using these approximations in equation 1, we obtain (1 - ni)GSi + S + &) = Si(l - ni*) + S 2 (l - n*i) + &(1 - *) On expanding both sides and on solving for n, we obtain ni + n a + na which establishes Rule 1 for the present case. Note 3 Solution, of equations by interpolation. The method of inter- polation which the student has used in connection with logarithm and compound interest tables can be used in solving equations whose solution would otherwise present very great difficulties. Example 1. Solve for n in the equation 5(1.06)" = 7.5 + 7.6(.00)n. (1) Solution. On rewriting the equation and on using the abbreviation F(n) fov the left member, wo obtain F(n) - 5(1.06)" - 7.6 - .45 n - 0. We desire a value n fc such that F( - 0. If we find a value n - n t such that F&I) is negative, and another value n - ni such that F<j) is positive, then it will follow that there is a value n k, between n\ and nt t such that /?(*) - 0. That is, there must be a solution n h between Wi and n*. From a rough inspection of Table V we guess that the solution is greater than n - 19. With the aid of Table V wo compute Fw for n - 19, 20, and 21. F(iw _ .922; F(ao) - - .464; J^OD - -H .048. Hence, there is a solution n k of the equation between n * 20 and n - 21, We find fc by interpola- APPENDIX 213 0. The total dif- The partial differ- tion in the table below where we use the fact that Fw = ference in the tabular entries is .048 ( .464) = .512. ence is ( .464) = .464. Hence, since is $ff = .91 of the way from .464 to +.048, we assume that the solution k is .91 of the way from 20 to 21, or that k = 20 + .91 = 20.91. Of course, this is only an approximate solution of the equation, but such a one is extremely useful in practical applications. An in- spection of the equation shows that there cannot be any other solution because 5(1.06)" increases much more rapidly than .45 n, and hence F(n) will be positive for all values of n greater than 21. Example 2. A man invests $6000 in the stock of a corporation. He receives a $400 dividend at the end of each year for 10 years. At the end of that time lie sells his holdings for $5000. Considering the whole 10-year period, at what effective rate may the man consider his investment to have been made ? Solution. Let r be the effective rate. With the end of 10 years as a comparison date, we write the following equation of value : 6000(1 + r) 10 - 5000 + 400^ at r}, Fw - 6000(1 + r) 10 - 5000 - 400( aini at r) = 0. (1) We shall solve equation 1 by interpolation. If the $1000 loss in capital had been uniformly distributed over the 10 years, the loss per year would have been $100. Hence, under this false (but ap- proximately true) condition, the net annual income would have been $300. The average invested capital would have been $(6000 + 5000) = $5500. Hence, since -^nnr = -055, we guess 1 .055 as an approximation to the solution of the equation. When r .055, F(.osS) <=> + 98.73. Since this is positive, the solution must be less than .055. We find ^(.oe) 257.80. Hence, the solution r = k of equation 1, for which F(K) = is between r => .056 and r = .05. We interpolate in the table below. 98.7 - (- 267.8) - 356.5; - (- 257.8) 267.8; .055 - .06 - .005. Hence, - .05 + .0036, or. approxi- The solution could be obtained ac- r *V) .05 - 257.S r = k .055 98.7 ' 366.5 mately, k <= .054. curatcly to hundredths (or to thousandths, or less) of 1%, if desired, by the method used in Example 2, Sec- tion 32, Part I. Note 4 Abridged multiplication. Consider forming the product (11.132157) X (893.214) . We decide in advance that wo desire the result accurately 1 Notice the uimilaiity between this reasoning and that employed in Section 55 of Parti, 214 MATHEMATICS OF INVESTMENT to the nearest digit in the second decimal place. The ordinary mul- tiplication would proceed as at the left below, while the abridged method proceeds as at the right. OHDINABY METHOD ABIUDOBD METHOD 11.132157 893.214 xxxxx 11.132157 893.214 Multiply by 44528628 11132157 22264314 33396471 100189413 89057256 8905.7256 1001.8935 33.3963 2.2264 .1113 .0444 800 90 3 .2 .01 .004 9943.398481598 9943.3975 Add ResuU = 9943.40 Result = 9943.40 In multiplying by the abridged method we proceed as follows : Since we desire the result to be accurate in the 2d decimal place, we carry two extra places, or four decimal places, for safety. To multiply by 893.214 we multiply in succession by 800, 90, 3, .2, .01, and .004 and then add the results (this is the same as is done in the ordinary method of multiplication, except that the multiplications are performed in the reverse order). We first multiply by 800, that is, we multiply by 8 and then move the decimal point. All digits of 11.132157 are used in this operation in order to obtain four significant decimal places in the result. To obtain four decimal places when multiplying by 90 we need one less digit of 11.132157 ; we put X over the "7" to indicate that we multiply 11.13215 at this time. Wo put X over the "5" and then multiply 11.1321 by 3; we put X over the last " 1 " and then multiply 11.132 by .2 ; etc. The advantages of this method are obvious. Less labor is involved, the decimal point is accurately located, and fewer mis- takes will occur in the final addition, Note 6 Accuracy of the interpolation method in solving for the time in the compound interest equation. Consider the equation A = (1 + r) n , ' (1) where A and r are known. To determine the value of n by interpola- tion, we first find from our interest table (Table V if A > 1, Table VI if A < 1) two integers ni and n a , n a n\ 1, such that the oorre- APPENDIX 215 spending values A\ = (1 + r) n i and A z = (1 + r)" 1 include A between them. That is, AI< A< A*. Then, as obtained by interpolation, the solution of equation 1 is A A , "AT I **- * 1 J\ =s n\ -\ The exact solution of equation 1 is obtained by taking the logarithm of both sides ; log A = n log (1 + r), where log means log,. _ log A log (1 + r) From equation 2 we obtain dn _ 1 dA Alo g (l+r: Hence since dn/dA is positive, n is an increasing function of A. Moreover, since dtn/dA 2 is negative, the graph of n as a function of A, with the A-&XJB horizontal, will be concave downward, as in Figure 6, dis- torted for illustration. It is seen graphically that the dif- ference between n, as given in equation 2, and N is given by the line EF in the figure. This error is less than DH, where H is the point in which the tangent (drawn at P) in- tersects the ordinate at J.a. Since CD - 1, tfn dA* - 1 4 2 log(l (2) (3) TlAi A Fia. 6 , -. , Ailog(l+r) - 1. Since A 2 = (1 + r)" 1 = (1 + r)(l + r)" 1 ** Ai(l + r}, A t - Aj. = rAi, Hence, on inserting the infinite series for log (1 + r), as obtained from any textbook on the Calculus, if _rf '_, r DH ft* M r 2 , r 3 - 1 = 216 MATHEMATICS OF INVESTMENT If r < .10. as is the case in the tables of this book. DH <-(}, -2\.95/ which is approximately r, if computed to only two decimal places. Hence, if we are computing results to only two decimal places, a solu- tion of equation 1, obtained by interpolation, is in error by at most Jr. Note 6 Accuracy of the interpolation method in solving for the time in the annuity equations. Consider the equation 0^ at r) = S, (1) where S and r are known. From equation 1, on inserting the explicit algebraic expression for (s| at r), we obtain t 1 + r)n - 1 = 8; (1 + 70" - Sr + 1. T If we solve equation 1 for n by interpolation in Table VII, our solution is the same as. we should obtain in solving the equivalent equation (1 + r) n = A (2) (where A = Sr + 1) for n, by interpolation in Table V. For, the solution of equation 1 by interpolation would be while that for equation 2 would be -- _ L oV + 1 - Sir - which is the same as the result for equation 1. Hence, it follows from Note 6 of the Appendix that the error in the solution of equation 1 obtained by interpolation in Table VII is at most i of the interest rate r. Similarly, it follows that, if we should solve for n in the equa- tion (ctn\ at r) = A, by interpolating in Table VIII, the error of the result would be at most r. INDEX Numbers refer to pages Abridged multiplication, 213 Accrued dividend, on a bond, 126 Accumulation factor, 16 Accumulation of diacount, on a bond, 119 Accumulation problem, 16 American Experience Table of Mor- tality, 148 Amortization, of a debt, 78; of the premium on a bond, 118 Amortization equation, 89 Amortization plan, 78; bonded debt retired by a, 81 ; comparison of sinking fund method with the, 88 ; final payment under the, 84 Amortization schedule, for a debt, 78; for the premium on a bond, 118 Amount, at compound interest, 15; at simple interest, 1; in a sinking fund, 87 ; of an annuity certain, 39 Annual premium; see net annual premium Annual rent of an annuity, 39; de- termination of the, 68 Annuities certain, 39; formulas for, 43, 47, 60; interpolation methods for, 70, 72; summary of formulas for, 60 Annuity; see annuity certain, and life annuity Annuity bond, 130 Annuity certain, 39; amount of an, 39; annual rent of an, -39; con- tinuous, 02; deferred, 69; determi- nation of the annual rent of an, 68 ; interest rate borne by an, 71 ; term of an, 70; due, 56; payment in- terval of an, 39 ; present value of an, 39; term of an, 39 Annuity due, certain, 56 ; life, 162 Annuity policy, 177 Approximate bond yield, 126 Arithmetical progression, 204 Asset, scrap value of an, 96; wearing value of an, 96; condition per cent of an, 98 Average date, 34 Averaging an account, 34 Bank discount, 9 , Base of system of logarithms, 187 Beneficiary, 165 Benefit of a policy, 166 Binomial theorem, 63 Bond, 113 ; accumulation of discount on a, 119; accrued dividend on a, 126; amortization of premium on a, 118; approximate yield on a, 126; book value of a, 117, 122; changes m book value of a, 117; dividend on a, 113; face value of a, 113; flat price of a, 125; pur- chase price of a, 114, 121; quoted price of a, 125 ; redemption price of a, 113; the yield of a, 126; yield on a, by interpolation, 128, 131 Bond table, 116 Book value, of a debt, 86; of a de- preciable asset, 97 Book value of a bond, on an interest date, 117; between interest dates, 122 Briggs' system of logarithms, 189 Building and loan associations, 92; dues of, 92; interest rates earned by, 93; loans made by, 94; profits in, 92; shares in, 92; time for stock to mature in, 62 Capitalized cost, 105 Cash surrender value, 183 Characteristic of a logarithm, 190 Common logarithms, 189 217 218 INDEX Numbers refer to pages Commutation symbols, 157 Comparison date, for comparing values, 26; for writing an equa- tion of value, 27 Composite life, 99 Compound amount, 14; for a frac- tional period, 20 Compound interest, 14; accumula- tion problem under, 15; amount under, 14; continuous conversion under, 35 ; conversion period under, 14; discount problem under, 15; effective rate under, 18; nominal rate under, 18 Concluding payment, under amortiza- tion process, 84 Condition per cent, 98 Contingent annuity, 39 Contingent payment, present value of a, 152 Continuous annuity, 62 Continuously converted interest, 35 Conversion period, 14 Deferred annuity, certain, 59 Deferred life annuity, 159 Depreciation, 96; constant percen- tage method for, 109; valuation of mining property under, 101; sinking fund plan for, 96; straight line method for, 98 Discount, banking use of, 9 ; problem of, under compound interest, 15; bond purchased at a, 119; rate of, 7; simple, 7 Discount factor, 16 Discounting of notesj under simple discount, 10; under compound interest, 24 Dividend, on a bond, 113 Dues of a building and loan associa- tion, 92 . Effective rate of interest, 18 Endowment insurance, 170; also see pure endowment Equated date, 33 Equated time, 33 ; equation for, 33, 212 Equation of value, 27; comparison date for, 27 Exact simple interest, 2 Exponential equation, 201; use of, in annuity problems, 75 Force of interest, 36 Geometrical progressions, 205; use of, under annuities certain, 41, 43, 46 ; use of, under perpetuities, 108 Glover's tables, 167 Graphical representation, of accu- mulation under interest, 23; of a deferred annuity, 59 ; of an annuity due, 57 ; of depreciation, 97 Gross premium, 165 Infinite geometrical progressions, 207 ; use of, under perpetuities, 108 Insurance, 165 ; endowment, 170 ; gross premium for, 165, 175 ; net annual premium for, 171; net premiums for, 165; net single premium for, 166; policy of, 165; term, 168; whole lif e, 166 Insurance policy, 165; beneficiary of an, 165; benefits of an, 165; policy date of an, 165; endowment, 172; w-paymcnt life, 172; n-yoar term, 172 ; ordinary life, 172 ; reserve on an, 178; whole life, 166 Insurance premium ; seo premium Interest, 1 ; compoxind, 14 ; converted continuously, 35 ; effective rate of, 18 ; exact, 2 ; force of, 36 ; graphi- cal representation of accumulation under, 23 ; in advance, 9 ; nominal rate of, 18; ordinary, 2; rate of, 1; simple, 1 Interest period, 14 Interpolation, annuity problems solved by, 70, 72, 213 ; book valuo of bond between interest dates by, 123; compound interest problems solved by, 29; use of, in logarith- mic computation, 194; yield of bond by, 128, 131 INDEX 219 Numbers refer to pages Investment yield of a bond, 113, 126; by approximate method, 126; by interpolation, 128, 131 Legal reserve insurance company, 166 Level premium, 178 Life annuity, 155 ; deferred, 159 ; due, 162; present value of a, 165, 159, 163; temporary, 159; whole, 155 Life insurance ; see insurance Loading, 175 Logarithms, 187; base of a system of, 187; Briggs 1 system of, 189; change of base of, 203 ; . characteris- tics of, 190; common, 189; man- tissas of, 190; Napierian, 203; natural, 203; properties of, 188; use of tables of, 194, 196 Mathematical expectation, 152; net single premium as a, 175; of a con- tingent payment, 152 Mantissa, 190 Mining property, valuation of, 101 Modulus, of a system of logarithms, 203 Mortality, American Experience Table of, 148 n-payment endowment policy, 172 Tirpaymont life policy, 172 7i-year term policy, 172 Natural premium, 169 Net annual premium, 171; endowment policy, 173 ; for for an an irregular policy, 176; for an w- payment life policy, 173 ; for an n- year term policy, 173 ; for ordinary life policy, 172 Net premiums, 165 Net single premium, 166, 175; for endowment insurance, 170; for irregular benefits, 176; for term insurance, 168; for whole, life in- surance, 166 Nominal rate of interest, 18 Notes ; sec discounting of notes Office premium, 165 Old lino insurance company, 165 Drdinary life policy, 172 Ordinary simple interest, 2 Par value of a bond, 113 Payment of a debt, amortization process for, 78 ; amortization sched- ule for, 78; building and loan association arrangement for, 94; comparison of amortization and sinking fund methods for, 88; sinking fund method for, 85 Pensions, present value of, 184 Perpetuities, 103; infinite geometri- cal progressions applied to, 108; present values of, 103, 104; use of, in capitalization problems, 105 Policy ; see insurance policy Policy date, 165 Policyholder, 165 Policy year, 165 Premium, annual, 171; gross, 165; level, 178; natural, 169; net, 165; net annual, 171 ; net single, 166 Premium on a bond, formula for the, 115 ; amortization schedule for the, 118 Present value, of a contingent pay- ment, 152; of a pure endowment, 153; of a life annuity, 155, 159, 162 ; of an annuity certain, 39 ; of life insurance, 165; under com- pound interest, 15; under simple discount, 7 ; under simple interest, 2 Principal, 1 ; amortization of, 78 Probabilities of life, 150 Probability, 147 Progressions, 204; arithmetical, 204; geometrical, 205 ; infinite geometri- cal, 207 Proportional parts, 195 Prospective method of valuation, 183 Pure endowment, 153 ; present value of a, 153, 157 Rate of discount, 7 Rate of interest, 1 ; borne by an an- nuity, 71; effective, 18; nominal, 18 ; paid by a borrower of a build- 220 INDEX Numbers rofer to pages ing and loan association, 95; yielded by a bond, 126, 128, 131 Redemption fund, for a mine, 101 Reserve, terminal, 178; table show- ing growth of a, 179; formula for the, 180 Scrap value, 96 Serial bonds, 130 Simple discount, 6 Simple interest, 1; exact, 2; ordi- nary, 2 ; six per cent rule for, 3 Sinking fund, 85; amount in a, 87; table showing growth of a, 86 Sinking fund equation, 89 Sinking fund plan, for depreciation, 96 ; for retiring a debt, 85 Six per cent rule, 3 Straight line method for depreciation, 98 Temporary life annuity, 159 Term, of an annuity certain, 39 ; de- termination of the, 70 Terminal reserve, 178; for an ordi- nary life policy, 181; prospective method for obtaining the, 183 Term insurance, 168 Time, to double money, 30, 211 Valuation, of a mine, .101 ; of an insurance reserve, 180 Value, cash surrender, 183; of an obligation, 24 Values, comparison of, 26 Wearing value, 96 Whole life annuity, 155 Whole life insurance policy, 166 Yield of a bond, by approximate method, 126; by interpolation, 128, 131 TABLE I COMMON LOGARITHMS OF NUMBERS TO FIVE DECIMAL PLACES Pages 2 to 19 TABLE II COMMON LOGARITHMS OF NUMBERS FROM 1.00000 to 1.10000 TO SEVEN DECIMAL PLACES Pages 20 to 21 N 1 9 3 4 5 ' 6 7 8 9 PP 100 00000 043 087 130 173 217 260 303 340 88'J 01. 02 03 432 860 01284 475 903 326 518 945 368 561 988 410 604 +030 452 647 +072 494 089 +116 536 732 +167 678 775 +199 020 817 +242 002 44 43 42 04 05 06 07 08 09 703 02119 531 938 03342 743 745 160 672 979 383 782 787 202 612 *019 423 822 828 243 653 *060 463 862 870 284 . 694 +100 503 902 912 325 736 +141 543 941 953 306 776 +181 583 981 005 407 816 +222 023 +021 +036 449 857 +202 003 +000 +078 490 898 +302 703 +100 1 2 3 4 5 G 7 8 4.4 4.3 4.2 8,8 8.0 8.4 13.2 12.0 12.0 17.0 17.2 10.8 22.0 21.5 21,0 20.4 2C.8 2fl.2 30.8 30.1 20.4 3G.2 34.4 33.0 30.0 38.7 37.8 110 04139 179 218 258 297 336 376 415 454 493 11 12 13 532 922 05308 671 961 346 610 999 385 650 *038 423 689 +077 461 727 +115 600 760 +154 638 805 +102 676 844 +231 . 014 883 +200 652 41 40 30 14 15 16 17 18 19 690 06070 446 819 07188 555 729 108 483 856 225 591 767 145 521 893 262 628 805 183 568 930 298 664 843 221 505 987 335 700 881 258 633 +004 372 737 918 206 670 +041 408 773 950 333 707 *078 445 809 094 371 744 +116 482 840 +032 408 781 +151 518 882 1 3 4 5 7 8 4.1 4.0 3.1) 8.2 8.0 7.H 12.3 12.0 11.7 10.4 10.0 15.0 20.S 20.0 10.0 24.0 24.0 23.4 28.7 28.0 27.3 32.8 32.0 31.2 36,9 30.0 36.1 120 918 954 990 +027 +063 *099 +135 +171 +207 +243 21 22 23 08279 636 991 314 072 *026 360 707 *061 386 743 +006 422 778 +132 458 814 +167 493 849 +202 529 884 *237 505 020 +272 600 066 +307 38 87 8ft 24 25 26 .27 28 29 09342 691 10037 380 721 11059 377 726 072 415 766 093 412 760 106 449 789 126 447 795 140 483 823 160 482 830 176 517 857 193 617 864 209 561 890 227 552 899 243 585 924 201 587 934 278 010 058 204 621 908 312 053 002 327 656 +003 340 087 +02. 1 ) 361 1 3 6 6 7 B 3.8 3,7 3.0 7.0 7.4 7.2 11.4 11.1 10.8 15.2 14.8 14.4 10.0 18.5 18.0 22.8 22.2 21.0 20.0 20.0 20.2 30.4 20.6 28.8 34.2 33.3 32.4 130 394 428 461 494 528 561 504 628 601 094 31 32 33 727- 12067 385 760 090 418 793 123 450 826 156 483 860 189 516 893 222 648 926 254 581 959 287 613 992 320 046 *024 352 678 35 84 33 34 35 36 37 38 39 710 13033 354 672 988 14301 743 066 386 704 +019 333 776 098 418 735 *051 364 808 130 460 767 *082 395 840 162 481 799 +114 426 872 194 513 830 +145 457 905 226 545 862 +176 489 937 258 577 803 +208 520 969 290 609 025 +239 551 *001 322 640 056 +270 582 2 I 6 7 1 3.0 3.4 3.3 7.0 0.8 0.6 10,5 10.2 0.0 14.0 13,6 13.2 17.5 17.0 10.0 21.0 20.4 10.8 24.5 23.8 23.1 28.0 27.2 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531 536 542 547 652 657 562 567 572 578 43 583 588 593 508 603 600 614 619 624 629 44 634 630 646 650 655 660 665 670 675 681 45 686 691 696 701 706 711 716 722 727 732 46 737 742 747 752 758 763 768 773 778 783 47 788 703 799 804 800 814 819 824 829 834 48 840 845 850 855 860 865 870 875 881 886 40 891 896 901 906 911 016 921 927 932 937 850 942 947 952 067 062 967 973 978 983 088 MT 1 2 3 4 5 8 7 8 9 PP 16 N 1 a 3 4 5 G 7 8 V P 850 02942 947 052 957 962 967 073 078 083 088 51 52 53 54 55 56 57 58 993 93044 095 146 107 247 298 340 008 040 100 151 202 252 303 354 *003 054 105 156 207 258 308 350 *008 050 110 161 212 263 313 364 *013 064 115 166 217 268 318 369 *018 060 120 171 222 273 323 374 *024 075 125 176 227 278 328 370 *029 080 131 181 232 283 334 384 *034 085 136 186 237 288 330 380 *039 000 141 102 242 293 344 304 6 59 309 404 400 414 420 425 430 435 440 445 1 0.6 860 450 455 460 465 470 476 480 485 400 495 3 1.8 61 62 63 64 65 66 67 68 60 600 551 601 651 702 752 802 852 002 505 556 606 056 707 757 807 857 907 510 661 611 661 712 762 812 862 912 515 506 616 666 717 767 817 807 917 520 571 621 671 722 772 822 872 922 526 576 626 676 727 777 827 877 027 531 681 631 682 732 782 832 882 932 536 586 636 687 737 787 837 887 937 641 501 641 692 742 702 842 802 942 646 696 646 697 747 707 847 897 047 5 e I 3.0 3.0 4.2 4.8 6.4 870 952 057 002 967 072 077 982 987 092 007 71 72 73 74 75 76 77 78 70 94002 002 101 151 201 250 300 349 390 007 057 106 156 206 255 305 354 401 012 062 111 161 211 260 310 359 409 017 067 116 166 216 265 315 364 414 022 072 121 171 221 270 320 369 419 027 077 126 176 226 275 325 374 424 032 082 131 181 231 280 330 379 420 037 086 136 186 236 285 335 384 433 042 091 141 191 240 290 340 380 438 047 090 146 100 245 205 345 394 443 2 3 5 a 7 9 5 0.6 1.0 1.5 2.0 2.5 3.0 3.6 4.0 4.6 880 448 453 458 463 468 473 478 483 488 498 81 82 83 84 85 86 87 88 89 408 547 596 645 694 743 702 841 800 503 552 601 650 000 748 707 846 806 607 657 606 655 704 753 802 851 900 512 562 611 660 709 758 807 856 905 617 667 616 665 714 763 812 861 010 522 571 621 670 719 708 817 866 916 527 576 626 675 724 773 822 871 910 532 581 630 680 729 778 827 876 024 537 586 635 685 734 783 832 880 020 542 591 640 689 738 787 836 885 934 2 3 4 0,4 0.8 1.2 800 039 914 949 954 050 963 968 073 978 983 5 1.0 2.0 91 92 93 94 95 96 97 98 90 088 05036 085 134 182 231 270 328 376 003 041 000 130 187 236 284 332 381 008 046 005 143 102 240 280 337 386 *002 051 100 148 197 245 294 342 300 *007 056 105 153 202 250 209 347 305 *012 061 109 158 207 255 303 352 400 *017 066 114 163 211 260 308 357 405 *022 071 no 168 216 265 313 361 410 *027 076 124 173 221 270. 318 366 416 *032 080 120 177- 226 274 323 371 419 7 8 B 2.4 3.8 3.2 3.0 000 424 429 434 439 444 448 453 458 463 468 N 1 % 3 4 S 6 7 8 9 1 P 17 N 1 2 3 * 5 6 7 8 9 PP 900 95424 429 434 439 444 448 453 458 403 408 01 472 477 482 487 492 497 501 500 611 516 02 521 525 530 535 540 545 650 554 550 564 03 569 574 678 ,583 588 593 598 002 607 612 04 617 622 626 631 636 641 646 650 055 060 05 665 670 674 679 684 689 694 098 703 708 06 713 718 722 727 732 737- 742 740 751 756 07 761 766 770 775 780 785 789 794 790 804 08 809 813 818 823 828 832 837 842 847 852 09 856 861 866 871 876 880 885 890 805 899 910 904 909 914 918 923 028 933 938 942 947 11 952 967 901 960 971 076 980 985 000 005 12 999 +004 +009 +014 +019 *023 *028 +033 +038 +042 13 96047 052 057 061 066 071 076 080 085 090 5 1 00 14 005 099 104 100 114 118 123 128 133 137 2 1.0 15 142 147 152 150 161 166 171 176 180 185 3 1.5 16 190 194 199 204 209 213 218 223 227 232 4 2.0 5 2.0 17 237 242 246 251 250 261 265 270 275 280 3.0 7 3.5 '18 284 289 294 298 303 308 313 317 322 327 19 332 336 341 340 360 355 360 305 369 374 4lfl 920 379 384 388 393 398 402 407 412 417 421 21 426 431 435 440 445 450 454 459 464 468 22 473 478 483 487 492 497 501 500 511 516 23 520 525 530 534 539 544 548 553 558 502 24 567 572 577 581 586 591 695 000 605 600 25 614 619 624 628 633 G38 642 047 652 050 26 661 666 670 675 680 685 689 094 609 703 27 708 713 717 722 727 731 730 741 745 750 28 755 759 704 709 774 778 783 788 792 797 29 802 806 811 810 820 825 830 834 830 844 930 848~ 853 868 862 867 872 870 881 880 800 31 895 900 904 909 914 918 923 928 032 937 32 942 946 951 956 960 965 970 974 979 984 4 . 33 988 993 997 +002 +007 +011 +016 +021 +025 +030 34 97035 039 044 049 053 058 063 067 072 077 2 Oil 3 1.2 35 081 086 090 095 100 104 109 114 118 123 4 1.0 36 128 132 137 142 140 151 155 160 165 169 5 2.0 6 2.4 37 174 179 183 188 102 197 202 206 211 216 7 2.8 8 32 38 220 225 230 234 239 243 248 253 257 202 o a!o 39 267 271 276 280 285 290 294 209 304 308 910 313 317 322 327 331 330 340 346 350 354 41 359 364 368 373 377 382 387 391 390 400 42 405 410 414 419 424 428 433 437 442 447 43 451 466 460 465 470 474 470 483 488 493 44 497 502 506 611 510 620 525 629 534 530 45 543 548 552 667 562 666 571 575 580 585 46 589 694 598 603 607 612 017 621 626 630 47 635 640 644 649 663 658 663 667 072 676 48 49 681 727 685 731 690 736 695 740 699 745 704 749 708 754 713 759 717 763 722 768 950 772 777 782 786 791 7.95 800 804 809 813 N 1 * 8 A 5 6 7 8 9 PP 18 N 1 2 3 4 A 6 7 8 PP 050 51 52 C3J 54 55 50 57 58 50 060 61 (12 G3 04 (15 00 07 08 OOj 970 71 72 73 74 75 70 77 78 70 080 81 82 83 84 85 80 87 88 80 000 01 02 03 04( 05 00 07 08 00 1000 97772 777 782 780 701 705 800 804 809 813 1 0.5 2 1.0 3 1.5 4 2.0 B 2.6 3.0 7 8.5 8 4.0 4.6 4 T-5T" 2 0.8 3 1.2 4 1.6 6 2.0 6 2.4 7 2.8 8 3.2 3.0 818 864 000 055 08000 040 001 137 182 823 868 914 050 005 050 000 141 180 827 873 018 904 000 055 100 140 101 832 877 023 008 014 050 105 150 105 830 882 028 073 010 064 100 165 200 841 886 032 078 023 068 114 159 204 845 801 037 082 028 073 118 164 209 850 896 041 987 032 078 123 108 214 855 900 946 901 037 082 127 173 218 859 905 960 996 041 087 132 177 223 227 232 230 241 246 260 254 250 263 268 272 318 363 408 463 498 543 CSS 032 677 277 J22 307 412 457 502 547 502 037 281 327 372 417 402 507 552 597 041 280 331 376 421 466 511 550 001 046 200 330 381 426 471 510 561 605 050 295 340 385 430 476 520 505 610 655 200 345 300 435 480 525 670 614 050 304 349 304 439 484 529 674 610 064 308 354 300 444 480 534 579 623 668 313 368 403 448 403 538 583 628 673 082 080 001 605 700 704 700 713 717 722 707 811 856 900 046 080 00034 078 iiif 720 771 810 860 006 040 004 038 083 731 770 820 865 000 054 008 043 087 736 780 825 800 014 058 *003 047 002 740 784 820 874 018 063 *007 052 006 744 789 834 878 023 007 012 056 100 740 703 838 883 027 972 *016 061 105 753 798 843 887 032 076 +021 065 100 758 802 847 802 036 981 *026 069 114 762 807 851 896 941 985 *020 074 118 127 131 136 140 145 140 154 158 162 107 211 255 300 344 388 432 470 520 eoT 171 210 200 304 348 302 430 480 524 176 220 204 308 352 300 441 484 528 180 224 209 313 357 401 445 480 533 185 220 273 317 361 406 440 403 637 189 233 277 322 360 410 464 408 542 103 238 282 326 370 414 458 602 640 108 242 286 330 374 419 403 606 650 202 247 201 336 379 423 407 511 555 207 251 295 339 383 427 471 615 550 508 572 577 581 585 600 594 509 603 007 051 095 730 782 820 870 013 957 612 050 000 743 787 830 874 917 901 016 060 704 747 701 835 878 922 065 621 004 708 752 795 830 883 026 070 625 660 712 750 800 843 887 930 074 620 673 717 700 804 848 801 935 078 034 677 721 705 808 852 800 030 083 638 082 726 769 813 866 900 044 087 642 686 730 774 817 861 904 948 901 647 691 734 778 822 806 909 052 096 00000 004 009 013 017 022 026 030 036 030 N 1 a 8 4 5 6 7 8 PP 19 N i 2 3 4 5 6 7 8 9 1000 1001 1002 1003 0000000 0434 0869 1303 1737 2171 2605 3030 3473 3007 4341 8677 001 3009 4775 9111 3442 5208 9544 3875 5642 9977 4308 6076 *0411 4741 6610 *0844 6174 6943 +1277 5607 7377 *1710 6039 7810 *2143 6472 8244 +2670 6005 1004 1005 1006 7337 002 1661 5980 7770 2093 6411 8202 2625 6843 8636 2957 7275 9067 3389 7706 9499 3821 8138 9932 4253 8560 *0384 4685 0001 *0796 5116 0432 +1228 5S48 0803 1007 1008 1000 1010 1011 1012 1013 003 0295 . 4605 8912 0726 5036 9342 1157 5467 9772 1588 6898 *0203 2019 6328 *0033 2451 6760 *1063 2882 7190 *1403 3313 7620 *1024 3744 8051 *2364 4174 8481 +2784 0043214 3644 4074 4504 4033 5363 5703 6223 6052 7082 7612 0051805 6094 7941 2234 6523 8371 2663 6952 8800 3092 7380 9229 3521 7809 9659 3960 8238 *0088 4370 8666 *0617 4808 0004 *0047 5237 9523 *137fl 6000 0(151 1014 1016 1016 006 0380 4660 8937 0808 5088 9365 1236 5516 9792 1664 6944 +0219 2092 6372 *0647 2521 6799 *1074 2940 7227 *1601 3377 7656 *1928 3806 8082 *2355 4233 8. r >lQ *2782 1017 1018 1019 1020 1021 1022 1023 007 3210 7478 0081742 3637 7904 2168 4064 8331 2594 4490 8767 3020 4017 9184 3440 5344 9010 3872 5771 *0037 4208 0198 +0403 4724 0024 +0880 5160 7061 *131fl 6,170 6002 6427 6853 7279 7704 8130 8550 8981 0407 ~3669~ 7007 *2151 0832 009 0257 4509 8758 0683 4934 9181 1108 6359 9605 1533 5784 *0030 1959 6208 +0454 2384 6633 +0878 2800 7058 *1303 3234 7483 +1727 4084 81192 +2675 1024 1026 1026 010 3000 7239 Oil 1474 3424 7662 1897 3848 8086 2320 4272 8510 2743 4696 8933 3166 5120 9357 3590 5544 9780 4013 6967 *0204 4436 0301 +0027 4859 681/5 +1050 5282 1027 1028 1029 1080 1031 1032 1033 5704 9931 012 4154 6127 *0364 4576 6550 *0776 4998 6973 *1198 5420 7396 +1621 5842 7818 *2043 0264 8241 *2465 0685 8064 +2887 7107 0086 +3310 7520 0600 +3732 7051 8372 8794 9215 9637 +0059 *0480 *0901 +1323 +1744 +2106 013 2587 6797 014 1003 3008 7218 1424 3429 7639 1844 3850 8059 2264 4271 8480 2685 4692 8901 3105 5113 9321 3625 5534 9742 3945 5055 *0102 4305 0370 +0683 4785 1034 1035 1036 5205 9403 015 3598 5625 9823 4017 6046 *0243 4436 6465 *0662 4866 6885 *1082 5274 7305 *1501 5693 7725 +1920 6112 8144 +2340 0531 8564 *2760 0950 8084 *3178 7300 1037 1038 1039 1040 1041 1042 1043 7788 016 1974 6155 8206 2392 6573 8625 2810 6991 9044 3229 7409 9462 3647 7827 9881 4065 8245 *0300 4483 8663 *0718 4001 9080 +1137 5319 0408 *1556 5737 0010 017 0333 0751 1168 1586 2003 2421 2838 3256 3673 4000 4507 8677 0182843 4924 9094 3259 5342 9511 3676 5759 9927 4092 6176 *0344 4508 6593 +0761 4926 7010 +1177 5341 7427 *1594 6757 7844 *2010 0173 8200 +2427 6580 1044 1045 1046 7005 019 1163 5317 7421 1578 5732 7837 1994 6147 8253 2410 6562 8669 2825 6977 9084 3240 7392 9500 3656 7807 9916 4071 8222 +0332 4480 8037 +0747 4002 0062 1047 1048 1049 1000 9467 020 3613 7765 9882 4027 8169 *0296 4442 8583 *0711 4856 8997 *1126 5270 9411 *1540 6684 9824 +1955 6099 *0238 +2309 6513 +0652 *2784 0927 +1066 *3108 7341 +1470 021 1893 2307 2720 3134 3547 3961 4374 4787 5201 6614 N e 1 2 3 A 5 6 V 8 20 N 1 a 8 4 5 6 7 8 1050 1061 1052 1053 021 1893 2307 2720 3134 3547 3001 4374 4787 5201 5614 0027 022 0157 4284 0440 0570 4606 0854 0983 6109 7207 1390 5621 7680 1808 5933 8093 2221 0345 8600 2034 6758 8919 3046 7170 9332 3459 7582 9745 3871 7994 1064 1055 1056 8406 023 2525 0030 8818 2036 7060 9230 3348 7402 9042 3759 7873 +0054 4171 8284 *0406 4582 8095 +0878 4994 9106 *1289 5405 9517 +1701 5817 9928 *2113 6228 +0339 1057 1058 1059 1000 1001 1002 1003 024 0750 4857 8960 1101 5207 9370 1672 6678 9780 1982 6088 +0190 2393 0498 +0600 2804 0900 +1010 3214 7310 *1419 3625 7720 *1829 4030 8139 +2230 4446 8540 *2649 026 3059 3408 3878 4288 4097 5107 5510 5020 0336 6744 7164 026 1245 6333 7603 1054 5741 7972 2003 6150 8382 2472 6558 8701 2881 0967 9200 3289 7375 9009 3098 7783 *0018 4107 8192 +0427 4515 8600 +0836 4924 9008 1004 1005 1000 9416 027 3496 7572 9824 3904 7979 *0233 4312 8387 *0641 4719 8794 *1049 6127 9201 +1457 5535 9009 +1865 5942 +0010 1-2273 6350 *0423 +2680 0757 +0830 *3088 7166 +1237 1007 1008 1009 1070 1071 1072 1073 0281044 5713 9777 2051 0119 +0183 2458 0520 +0500 2865 0932 *0990 3272 7339 *1402 3679 7746 *1808 4086 8162 "2214 4492 8558 *2020 4899 8904 +3020 6306 9371 +3432 029 3838 4244 4049 5065 5401 5867 0272 6078 7084 7489 7895 030 1948 6997 8300 2353 0402 8706 276S 6807 9111 ,3163 7211 9516 3568 7610 9922 3973 8020 *0327 4378 8425 +0732 4783 8830 +1138 51S8 9234 +1543 5592 9638 1074 1075 1070 0310043 4085 8123 0447 4489 8520 0851 4893 8930 1256 5290 9333 1000 5700 9737 2064 0104 *0140 2468 6508 +0544 2872 0912 +0947 3277 7315 *1360 3681 7719 +1754 1077 1078 1079 1080 1081 1082 1083 032 2157 0188 033 0214 2560 0590 0017 2903 0993 1010 3367 7396 1422 3770 7799 1824 4173 8201 2220 4676 8004 2029 4979 9007 3031 5382 9400 3433 5785 9812 3835 4238 4640 5042 5444 5846 6248 6050 7052 7453 7855 8267 0342273 0285 8659 2074 0680 9000 3075 7087 9402 3477 7487 9864 3878 7888 +0266 4279 8289 +0607 4080 8690 +1008 5081 9091 +1470 5482 0491 *1871 6884 9892 1084 1085 1080 035 0293 4297 8298 0003 4608 8008 1004 6098 9098 1495 5498 9498 1895 5898 9898 2290 0208 +0297 2696 6098 "0007 3096 7098 +1097 3407 7498 +1490 3807 7808 *1806 1087 1088 1089 1000 1001 1092 1093 030 2295 0289 037 0279 2606 0688 0078 3004 7087 1070 3494 7480 1476 3893 7885 1874 4203 8284 2272 4602 8083 2071 5091 9082 3070 5491 9481 3408 6800 9880 3867 4205 8248 038 2220 6202 4003 5062 5400 5868 0257 0655 7053 7451 7849 8040 2024 0500 9044 3022 0090 9442 3419 7393 0839 3817 7791 +0237 4214 8188 +0035 4612 8585 +1033 6000 8982 +1431 5407 9370 +1829 6804 9770 1094 1095 1096 039 0173 4141 8100 0570 4538 8502 0967 4934 8898 1304 6331 9294 1701 6727 0690 2158 0124 *008Q 2554 ,6520 10482 2051 0917 +0878 3348 7313 +1274 3746 7709 +1070 1097 1098 1090 1100 040 2060 0023 9977 2462 0410 *0372 2858 0814 "0707 3254 7210 +1102 3660 7005 +1G57 4045 8001 +1952 4441 8306 +2347 4837 8701 +2742 5232 9187 *3137 5028 9682 +3B32 041 3927 4322 4710 51U 5506 6900 6295 0690 7084 7479 N 1 9 3 4 5 6 7 8 9 21 TABLE HI The Number of Each Day of the Year fa hi Q n Ss d 4 i | 1 I 1 ! 1 i i 1 1 E 31 I 1 32 00 91 121 152 182 213 244 274 305 335 i 2 2 33 61 92 122 153 183 214 245 275 306 336 2 3 3 34 62 93 123 154 184 215 246 276 307 337 3 4 4 35 63 94 124 155 185 216 247 277 308 338 4 ft 6 36 64 95 126 156 180 217 248 278 309 339 & 6 6 37 65 9G 126 157 187 218 240 279 310 340 7 7 38 66 97 127 158 188 219 250 280 311 341 7 8 8 39 67 98 128 150 189 220 261 281 312 342 8 9 S 40 68 99 120 160 190 221 252 282 313 343 9 10 10 41 69 100 130 161 191 222 253 283 314 344 10 11 11 42 70 101 131 162 192 223 254 284 315 345 11 12 12 43 71 102 132 163 193 224 255 285 316 340 12 18 13 44 72 103 133 164 194 225 256 286 317 347 13 14 14 45 73 104 134 165 195 226 257 287 318 348 14 15 15 46 74 105 135 166 196 227 258 288 310 340 15 16 16 47 75 106 136 167 197 228 260 280 320 350 10 17 17 48 76 107 137 " 168 198 . 229 260 200 321 351 17 18 18 49 77 108 138 169 199 230 261 201 322 352 18 19 19 50 78 109 139 170 200 231 202 202 323 353 19 20 20 51 79 110 140 171 201 232 203 203 324 354 20 21 21 52 80 111 141 172 202 233 264 204 325 355 21 22 22 53 '81 112 142 173 203 234 265 205 320 356 22 23 23 54 82 113 143 174 204 235 260 206 327 357 23 24 24 55 83 114 144 175 205 236 267 207 328 358 24 25 25 56 84 115 145 176 206 237 268 208 320 350 25 2G 20 57 85 116 146 177 207 238 269 200 330 360 26 7 27 58 86 117 147 178 208 239 270 300 331 361 27 28 28 50 87 118 148 179 209 240 271 301 332 362 28 29 29 88 119 149 180 210 241 272 302 333 363 29 30 30 89 120 150 181 211 242 273 SOU 334 364 30 31 31 90 151 212 243 304 365 31 NOTB. la leap years, alter February 28, add 1 to the tabulated number. TABLE IV Ordinary and Exact Interest at 1% on $10,000 Ezact Interest for 1 to 865 Days Ordinary Interest for 1 to 360 Days Days Interest For days below add to interest column Days Interest For days below add to interest column 980 40 60 $80 120 940 160 980 1 $ .2739726 74 147 220 293 1 $ .2777778 73 146 217 280 2 .5479452 75 148 221 294 2 .5555556 74 140 218 290 3 .8219178 76 140 222 295 3 .8333333 76 147 219 291 4 1.0958904 77 150 223 296 4 1.1111111 76 148 220 292 5 1.3098030 78 161 224 297 5 1.3888880 77 140 221 293 1.0438356 79 152 225 298 1.0006607 78 150 222 294 7 1.9178082 80 163 220 299 7 1.0444444 70 151 223 296 8 2.1917808 81 154 227 300 8 2.2222222 SO 152 224 296 e 2.4057534 82 155 228 301 9 2.5000000 81 153 225 297 10 2.7397200 83 150 229 302 10 2.7777778 82 154 226 298 11 3.0130980 84 157 230 303 11 3.0566656 83 155 227 299 12 3.2870712 85 158 231 304 12 3.3333333 84 150 228 300 13 3.5010438 86 159 232 305 13 3.0111111 85 157 229 301 14 3.8350104 87 100 233 300 14 3.8888880 80 158 230 302 15 4.10Q5890 88 101 234 307 15 4.1060007 87 150 231 303 10 4.383501G 89 102 235 308 10 4.4444444 88 160 232 304 17 4.0575342 90 103 236 300 17 4,7222222 89' 101 233 305 IS 4.9315008 91 104 237 310 18 5.0000000 90 102 234 306 19 5.2054795 92 105 238 311 19 5.2777778 91 163 235 307 20 6.4794521 93 100 230 312 20 5.5555556 92 164 236 308 21 5.7534247 94 107 240 313 21 5.8333333 03 105 237 309 22 0.0273973 95 108 241 314 22 6.1111111 94 106 238 310 23 0.3013099 90 109 242 315 23 6.3888889 05 167 239 311 24 0.5753425 97 170 243 310 24 0.6000007 00 108 240 312 25 0.8493151 98 171 244 317 25 0.0444444 07 109 241 313 20 7.1232877 99 172 245 318 26 7.2222222 98 170 242 314 27 7.3972003 100 173 246 319 27 7.5000000 99 171 243 315 28 7.0712320 101 174 2-17 320 28 7.7777778 100 172 244 310 29 7.9452055 102 175 248 321 20 8.0555566 101 173 246 317 30 8.2191781 103 170 249 322 30 8.3333333 102 174 246 318 31 8.4931507 104 177 250 323 31 8.6111111 103 175 247 319 32 8.7071233 105 178 251 324 32 8.8888889 104 176 248 320 33 9.0410959 106 179 252 325 33 9.1006007 105 177 249 321 34 9.3150085 107 180 253 326 34 9.4444444 100 178 260 322 3C 9.5890411 108 181 254 327 35 9.7222222 107 179 251 323 30 9.8630137 109 182 255 328 30 10.0000000 108 ISO 252 324 37 10.1309803 110 183 260 329 37 10.2777778 109 181 253 325 38 10.4109580 111 184 257 330 38 10.5565650 110 182 254 326 39 10.0849315 112 185 268 331 30 10.8333333 111 183 255 327 40 10.9589041 113 186 259 332 40 11.1111111 112 184 250 328 41 11.2328707 114 187 200 333 41 11.3888889 113 185 257 329 42 11.5008493 115 188 201 334 42 11.0060607 114 186 258 330 43 11.7808210 11C 189 202 335 43 11.0444444 115 187 269 331 44 12.0547945 117 190 203 330 44 12.2222222 110 188 260 332 45 12.3287071 118 191 204 337 45 12.5000000 117 189 261 333 40 12.0027397 119 192 205 338 40 12.7777778 118 190 262 334 47 12.8707123 120 193 206 339 47 13.0555650 119 191 263 335 48 13.1500849 121 194 207 340 48 13.3333333 120 192 264 330 49 13.4240575 122 195 208 341 40 13.0111111 121 193 205 337 60 13.0980301 123 190 269 342 50 13.8888889 122 194 266 338 51 13.9720027 124 197 270 343 51 14.1006607 123 195 267 339 52 14.2405753 125 198 271 344 52 14.4444444 124 196 268 340 53 14.5205470 120 199 272 345 53 14.7222222 126 197 269 341 54 14.7945206 127 200 273 340 64 15.0000000 126 198 270 342 56 15.0084932 128 201 274 347 55 15.2777776 127 190 271 343 50 15.3424058 120 202 275 . 348 60 15.5555550 128 200 272 344 57 15.0104384 130 203 270 349 57 15.8333333 129 201 273 345 58 15.8904111 131 204 277 350 58 16.1111111 130 202 274 340 59 16.1043830 132 205 278 351 69 10.3888889 131 203 275 347 00 10.4383602 133 200 279 352 00 10.0660007 132 204 276 348 01 10.7123288 134 207 280 363 01 10.0444444 133 205 277 349 02 10.9803014 136 208 281 354 02 17.2222222 134 300 278 350 03 17.2002740 136 209 282 355 03 17.5000000 135 207 279 351 04 17.5342406 137 210 283 350 64 17.7777778 136 208 280 352 05 17.8082192 138 211 284 367 05 18.0665566 137 209 281 353 60 18.0821918 139 212 285 368 06 18.3333333 138 210 282 351 07 18.3501044 140 213 280 369 67 18.6111111 139 211 283 355 08 18.6301370 141 214 287 300 08 18.8888889 140 212 284 350 09 18,9041090 142 215 288 301 00 10.1060067 141 213 286 357 70 19.1780822 143 210 280 302 70 10.4444444 142 214 286 358 71 19.4520548 144 217 200 363 71 10.7222222 143 215 287 350 72 19.7260274 145 218 291 364 72 20.0000000 144 216 288 360 73 20.0000000 140 219 292 306 TABLE V COMPOUND AMOUNT OF 1 (1 + 0" . n 5% 1% % !% 1% 1 2 3 4 5 1.0041 6667 1.0083 5069 1.0125 5210 1.0167 7112 1.0210 0767 1.0050 0000 1.0100 2500 1.0160 7513 1.0201 5050 1.0252 6125 1.0058 3333 1.0117 0069 1.0170 0228 1.0235 3830 1.0296 0894 1.0075 0000 1.0150 6025 1.0220 0017 1.0303 3910 1.0380 6073 1.0100 0000 1.0201 0000 1.0303 0100 1.0400 0401 1.0510 1005 6 7 8 9 10 1.0252 6187 1.0205 3370 1.0338 2352 1.0381 3111 1.0424 5666 1.0303 7761 1.0355 2940 1.0407 0704 1.0460 1058 1.0511 4013 1.0355 1440 1.0415 6400 1.0470 3064 1.0537 4182 1.0598 8805 1.0458 5224 1.0530 0613 1.0015 9885 1.0606 6084 1.0775 8255 1.0015 2015 1.0721 3635 1.0828 5671 1.0936 8527 1.1046 2213 11 12 13 14 15 1.0468 0023 1.0C11 0100 1.0555 4174 1.0690 3983 1.0643 5025 1.0663 9683 1.0616 7781 1.0060 8620 1.0723 2113 1.0770 8274 1.0660 7133 1.0722 0008 1.0785 4511 1.0848 3002 1.0011 6483 1.0866 0441 1.0038 0000 1.1020 1045 1.1102 7553 1.1186 0259 1.1166 0835 1.1268 2503 1.1380 9328 1.1494 7421 1.1600 0806 16 17 18 10 20 1.0087 9100 1.0732 4430 1.0777 1621 1.0822 0070 1.0867 1589 1.0830 7115 1.0884 8051 1.0030 2894 1.0003 9868 1.1048 9658 1.0075 2000 1.1030 3222 1.1103 7182 1.1168 4890 1.1233 6395 1.1260 0211 1.1354 4455 1.1430 6039 1.1525 4000 1.1611 8414 1.1726 7864 1,1843 0443 1.1061 4748 1.2081 0895 1.2201 9004 21 22 23 24 25 1.0912 4387 1.0967 9072 1.1003 5662 1.1049 4134 1.1095 4520 1.1104 2006 1.1169 7216 1.1215 5202 1.1271 6078 1.1327 9558 1.1290 1600 1.1365 0808 1.1431 3771 1.1408 0002 1.1505 1322 1.1008 9302 . 1.1780 6722 1.1875 0723 1.1964 1353 1.2053 8663 1.2323 9104 1,2447 1680 1.2571 6302 1.2607 3465 1.2824 3200 26 27 28 20 30 1.1141 6836 1.1188 1073 1.1234 7244 1.1281 5358 1.1328 5422 1.1384 5055 1.1441 5185 1.1498 7261 1.1566 2197 1.1614 0008 1.1032 5055 1.1700 4523 1.1708 7040 1.1837 3657 ' 1.1006 4000 1.2144 2703 1.2235 3523 1.2327 1175 1.2410 5700 1.2612 7176 1.2052 5631 1.3082 0888 1.3212 9007 1.3345 0388 1.3478 4802 81 32 33 34 35 1.1375 7444 1.1423 1434 1.1470 7308 1.1518 5346 1.1566 5284 1.1672 0708 1.1730 4312 1.1780 0833 1.1848 0288 1.1907 2G89 1.1076 8610 1.2045 7202 1.2115 0860 1.2186 6034 1.2267 7523 1.2006 5G30 1.2701 1122 1.2796 3706 1.2802 3434 1.2980 0350 1.3613 2740 1.3749 4068 1.3880 0000 1.4025 7600 1.4160 0270 : 36 37 38 30 40 1.1014 7223 1.1063 1170 1.1711 7133 1.1700 5121 1.1809 5142 1.1966 8052 1.2026 6303 1.2080 7725 1.2147 2063 1.2207 0424 1.2320 2550 1.2401 1705 1,2473 5107 1.2546 2780 1.2610 4055 . 1.3086 4537 1.3184 6021 1.3283 4866 1.3383 1128 1.3483 4801 1.4307 6878 1.4450 7647 1.4505 2724 1.4741 2251 1.4888 6373 41 42 43 44 45 1.1858 7206 1.1008 1319 1.1067 7491 1.2007 5731 1.2057 0046 1.2268 0821 1.2330 3270 1.2301 978(5 1.2453 0385 1.2510 2082 1.2003 0701 1.2707 1220 1.2841 6960 1.2010 5062 1.2001 8525 1.3584 0123 1,3080 4060 1,3780 1460 1.3802 5642 1.3906 7684 1.B037 5237 1.5187 8089 1.5330 7779 1.5403 1757 1.5048 1075 46 47 48 4 50 1.2107 844Q 1.2158 2940 .1.2208 9530 1.2259 8242 1.2310 0068 1.2578 7802 1.2641 6832 1.2704 8016 1.2768 4161 1.2832 2681 1.3067 6383 1.3143 8662 1.3220 5383 1.3207 6686 1.3375 2283 1.4101 7341 1.4207 4071 1.4314 0533 1.4421 4087 1.4520 5603 1.5804 5885 1.5902 6344 1.0122 2608 1.0283 4834 1.6446 3182 TABLE V COMPOUND AMOUNT OF 1 n 5 , 12% |% i% !% 1% 51 62 3 1 55 1.2302 2002 1.2413 7114 1.2405 4352 1.2517 3745 1.2509 5302 1.2890 4194 1.2900 0015 1.3025 70(50 1.3090 8340 1.3150 2887 1.3453 2504 1.3531 7277 1.3010 6028 1.3090 0583 1.3700 9170 1.4038 5411 1.4748 3301 1.4858 9420 1,4970 3847 1.6082 002G 1.6010 7814 ,1.0770 8892 1.0944 0581 1.7114 1047 1.7285 2457 56 57 58 59 00 1.2021 0033 1.2074 4040 1.2727 3050 1.2780 3354 1.2833 5868 1.3222 0702 1.3288 1805 1.3354 0214 1.3421 3940 1.3488 5016 1.3850 2415 1.3931 0340 1.4012 2000 1.4004 0374 1.4170 2520 1.5106 7826 1.5300 7009 1.5424 6740 1.5540 2583 1.5660 8103 1.7458 0082 1.7032 (5702 1.7800 0000 1.7087 0001) 1.8160 0670 61 02 63 G4 65 1.2887 0001 1.2940 7561 1.2994 0700 1.3048 8204 1.3103 1905 1.3556 9440 1.3023 7238 1.3001 8424 1.3700 3010 1.3S29 1031 1.4258 0474 1.4342 1240 1.4425 7870 1.4500 9374 1.4694 6787 1.5774 2303 1.6892 5431 1.0011 7372 l.Giai 8252 1.0252 8130 1.8348 0307 1.8532 1230 1.8717 4443 1.8904 0187 1.0003 0040 66 07 68 69 70 1.3167 7872 1.3212 0113 1.3267 6038 1.3322 0458 1.3378 4580 1.3808 2480 1.3007 7399 1.4037 5785 1.4107 7604 1.4178 3053 1.4679 7138 1.4705 3454 1.4851 4700 1.4938 1102 1.5025 2492 1.0374 7100 1.0407 6203 1.0021 2517 1.0745 9111 1.0871 5055 1.0284 0015 1.9477 4475 1.0072 2220 1.9808 9442 2.0Q07 0337 71 78 73 74 75 1.3434 2010 1.3490 1774 1.3546 3805 1.3002 8298 1.3659 5082 1.4249 1908 1.4320 4428 1.4392 0450 1.4464 0052 1.4536 3252 1.5112 8906 1.5201 0550 1.6280 7279 1.5378 0170 1.5408 6283 1.0098 0418 1.7125 5271 1.7253 9085 1.7383 3733 1.7513 7480 2.0208 3100 2.0470 0931 2.0075 7031 2.0882 4001 2,1001 2847 70 77 78 79 80 1.3710 4220 1.3773 5746 1.3830 0045 1.3888 5035 1.3046 4627 1.4009 0069 1.4082 0519 1.4755 4022 1.4829 2395 1.4903 3857 1.5558 8620 1.5040 0220 1.5740 0115 1.5832 7334 1.5925 0910 1.7045 1017 1.7777 4400 1.7010 7708 1.8045 1015 1.8180 4308 2.1302 1076 2.1515 2195 2.1730 3717 2.1947 0754 2.2107 1522 '81 89 83 81 85 1.4004 6729 1.4062 9253 1.4121 6209 1.4180 3005 1.4230 4454 1.4977 9026 1.5052 7921 1.5128 0561 1.5203 6964 1.6279 7148 1.6017 0874 1.0111 4257 1.0205 4090 1.6290 0405 1.0396 0236 1.8310 7031 1.8454 1001 1.8502 5763 1.8732 0100 1.8872 5008 2.2388 8237 2.2012 7119 2.2838 SHOO 2.3007 2274 2.3297 8007 86 87 88 89 90 1.4298 7704 1.4358 3540 1.4418 1811 1.4478 2568 1.4638 5820 1.5350 1134 1.6432 8040 1.5510 0585 1.5587 6087 1.5665 6408 1.0490 0012 1.8586 8667 1.6683 6134 1.6780 0344 1,6878 8232 1.9014 0536 1.0150 0500 1.0300 3330 1.9446 0865 1.9590 9240 2.3630 8787 2.3700 1875 2.4003 8404 2.4248 8879 2.4480 3207 91 92 93 M 95 1.4590 1003 1.4050 0002 1.4721 0735 1.4782 4113 1.4844 0047 1.6743 8745 1.5822 6930 1.5901 7069 1.5981 2154 1.6001 1216 1.0977 2830 1.7070 3172 1.7175 9290 1.7270 1210 1.7376 8993 1.0737 8505 1.0886 8905 2.0035 0340 2.0186 2974 2.0330 6871 2.4731 1000 2.4978 6019 2.5228 2801) 2.5480 5098 2,6735 3765 96 97 98 99 100 1.4905 8547 1.4967 9024 1.5030 8289 1.6092 0653 1.5155 8420 1.6141 4271 1.6222 1342 1.6303 2449 1.6384 7011 1.6406 6849 1.7478 2040 1.7580 2211 1,7082 7724 1.7786 0219 1.7880 6731 2.0489 2123 2.0642 8814 2.0797 7030 2.0953 0858 2.1110 8384 2.5002 7203 2.0252 0506 2.0515 1831 2.0780 3340 2.7048 1383 26 TABLE V COMPODTTD AMOUNT OF 1 (1 + 0" n i% 1% s% ! 1% 101 102 103 104 105 1.0218 9019 1.5262 4044 1.6346 0811 1.6410 0231 1.6474 2315 1.G549 0183 1.0631 7634 1.6714 9223 1.0708 4000 1.0882 4894 1.7094 0295 1.8098 0047 1.8204 5722 1.8310 7065 1.8417 5783 2.1269 1097 2.1428 6885 2.1589 4036 2.1751 3242 2.1014 4591 2.7318 6197 2.7591 8059 2.7807 7239 2.8146 4012 2.8427 8052 106 107 108 109 110 1.5538 7075 1.5003 4521 1.5608 4006 1.6733 7518 1.5700 3001 1.6060 9018 1.7051 7303 1.7136 0950 1.7222 6800 1.7308 7034 1.8525 0142 1.8633 0768 1.8741 7097 1.8851 0067 1.8061 0014 2.2078 8175 2.2244 4087 2.2411 2417 2.2579 3200 2.2748 0710 2.8712 1438 2.8999 2653 2.9289 2579 2.9582 1605 2.9877 9720 111 112 113 114 115 1.5805 1396 1.5031 2443 1.5007 0246 1.0004 2812 1.0131 2167 1.7395 3373 1.7482 3140 1.7500 7266 1.7667 5742 1.7746 8621 1.9071 0070 1.9182 9100 1.0204 8104 1.9407 3726 1.9620 5822 2.2019 2860 2.3001 1807 2.3204 3645 2.3438 8472 2.3614 6386 3.0170 7517 3.0478 5192 3.0783 3044 3.1001 1375 3.1402 0489 110 117 118 119 120 1.0108 4201 1.6205 0220 1.0333 0973 1.0401 7543 1.0470 09GO 1.7834 6014 1.7023 7644 1.8013 3832 1.8103 4501 1.8103 9073 1.0034 4522 1.0748 0866 1.9804 1800 1.9080 0634 2.0006 0138 2.3791 7484 2.3070 1865 2.4149 0620 2.4331 0876 2.4513 5708 3.1716 0093 3.2033 2300 3.2353 5623 3.2677 0980 3.3003 8689 121 122 123 124 125 1.0538 7204 1.6007 0317 1.0070 8302 1.0746 3170 1.6816 0933 1.8284 0372 1.8376 3619 1.8468 2437 1.8560 5849 1,8653 3878 2.0213 8440 2.0331 7581 2.0450 3600 2.0500 6538 2.0080 6434 2.4607 4226 2.4882 6532 2.5009 2731 2.6257 2927 2.6446 7224 3.3333 9076 3.3607 2407 3.4003 9192 3.4343 0584 3.4087 3980 120 127 128 129 130 1.0886 1603 1.6056 5193 1.7027 1715 1.7008 1181 1.7100 3002 1.8746 6548 1.8840 3880 1(8034 5000 1.9020 2629 1.9124 4002 2.0810 3330 2.0931 7260 2.1053 8284 2.1170 6424 2.1300 1728 2.6637 6728 2.5829 8540 2.6023 5785 2.0218 7653 2.0415 3060 3.5034 2719 3.5384 6147 3.5738 4008 3.6005 8454 3.6456 8030 131 132 138 134 135 1.7240 8002 1.7312 7303 1.7384 8727 1.7467 3097 1.7530 0485 1.9220 0313 1.0316 1314 . 1.9412 7121 1.9500 7757 1.0607 3245 2.1424 4238 2.1640 3900 2.1076 1044 2.1801 5425 2.1028 7182 2.0613 6115 2.6813 1128 2.7014 2112 2.7216 8177 2.7420 0430 3.0821 3719 3.7180 6856 3.7501 4815 3.7037 0903 3.8310 4673 130 137 138 139 140 1,7003 0903 1.7670 4305 1.7760 0884 1.7824 0471 1.7808 3130 1.9705 3012 1.9803 8880 1.0002 9074 2.0002 4210 2.0102 4340 2.2056 0357 2.2185 2904 2.2314 7137 2.2444 8828 2.2576 8113 2.7626 0000 2.7833 8005 2.8042 6640 2.8252 8731 2.8404 7007 3.8009 6319 3.0086 6282 3.0477 4045 3,9872 2695 4.0270 9922 141 142 143 144 145 1.7072 8902 1.8047 7773 1.8122 0763 1.8108 4837 1.8274 3158 2.0202 0402 2.0303 9609 2.0405 4808 2.0507 5082 2.0010 0457 2.2707 6036 2.2830 0640 2.2073 1071 2.3107 2074 2.3241 0905 2.8678 2554 2.8803 3424 2.9110 0424 2.9328 3077 2.0548 3305 4.0073 7021 '4.1080.4391 4.1401 2435 4.1000 1660 4.232G 2175 146 147 148 14 150 1.8350 4688 1,8426 0100 1.8503 6978 1.8580 7906 1.8668 2106 2.0713 0950 2.0816 0614 2.0020 7447 2.1026 3484 2.1130 4762 2.3377 6778 2.3513 9470 2.30B1 1117 2.3780 0705 2.3927 8461 2,0709 0430 2.0003 2176 3.0218 1067 3.0444 8020 3.0673 1389 4.2748 4697 4.3175 9544 4.3607 7139 4.4043 7910 4.4484 2290 27 TABLE V COMPOUND AMOUNT OP 1 (1 + i)" n l|% . l\% 1|% l|% 2% l 2 3 4 5 1.0112 6000 1.0226 2666 1.0341 3111 1.0467 6600 1.0676 2094 1.0126 0000 1.0261 6626 1.0370 7070 1.0609 4634 1.0640 8216 1.0160 0000 1.0302 2600 1.0460 7838 1.0013 0355 1.0772 8400 1.0176 0000 1.0353 0026 1.0534 2411 1.0718 5003 1.0000 1050 1.0200 0000 1.0404 0000 1.0012 0800 1.0824 3210 1.1040 8080 6 7 8 9 10 1.0694 2716 1.0814 6821 1.0036 2462 1.1060 2780 1.1183 6068 1,0773 8318 1.0908 6047 1.1044 8610 1.1182 0218 1.1322 7083 1.0934 4320 1.1098 4401 1.1204 9250 . 1.1433 8098 1.1006 4083 1.1097 0235 1.1291 2215 1.1488 8178 1.1089 8721 1.1894 4440 1.1201 0242 1.1480 8607 1.1718 5938 MOW) 0207 1.2189 0442 11 is 13 14 15 1.1300 6124 1.1436 7444 1.1666 4078 1.1606 6186 1.1827 0032 1.1464 2422 1.1607 6462 1.1762 6306 1.1800 6476 1.2048 2018 1.1779 4804 1.1960 1817 1.2135 5244 1.2317 5573 1.2602 3207 1.2102 6077 1.2314 3031 1.2529 8050 1.2740 1082 1.2972 2780 1.2433 7431 1.2082 4170 1.2080 0003 1.3104 787(1 1.3458 081)4 16 17 18 IB 90 1.1060 1480 1.2004 6007 1.2230 7660 1.2368 3611 1.2607 6062 1.2198 8066 1.2361 3817 1,2605 7739 1.2662 0961 1.2820 3723 1.2080 8555 1.2880 2033 1.3073 4004 1.3200 6075 1.3408 5601 1.3100 2035 1.3430 2811 1.3005 3111 1.3004 4540. 1.4147 7820; 1.3727 8571 1.4002 4142 1.4282 4025 1.4508 1117 1.4859 4740 21 22 28 34 85 1.2648 2146 1.2700 6071 1.2034 4003 1.3070 0123 1.3227 0613 1.2980 6270 1.3142 8848 1.3307 1709 1.3473 5106 1.3641 9294 1.3070 6783 1.3876 0370 1.4083 7715 1.4206 0281 1.4509 4635 1.4305 3081 1.4647 2871 1.4003 0140 1.5104 4,279 1.5420 8054 1.5150 0634 1.5450 7007 1.57(18 0920 1.0084 3725 1.0400 0500 J86 27 28 29 80 1.3376 8667 1.3626 3442 1.3678 6166 1.3832 3080 1.3088 0134 1.3812 4636 1.3986 1002 1.4160 0230 1.4336 0221 1.4616 1336 1.4727 0953 1.4948 0018 1.5172 2218 1.5300 8051 1.6630 8022 1.5009 8200 1.5974 5739 1.0254 1200 1.0538 5702 1.6828 0013 1.0734 1811 1.7068 8048 1.7410 2421 1.7758 4400 1.8113 0168 31 32 33 84 ; 35 1.4146 3786 1.4304 6140 1.4466 4308 1.4628 1760 1.4792 7430 1.4607 6863 1.4881 3051 1.6067 3214 1.5265 6020 1.6446 3687 1.6806 2642 1.6103 2432 1.0344 7918 1.6580 0637 1.6838 8132 1,7122 4913 1.7422 1349 1.7727 0223 1.8037 2452 1.8362 8070 1.8475 8882 1.8846 4069 1.9222 3140 1.9000 7003 1.0908 8055 36 87 88 39 40 1.4060 1613 1.6127 4610 1.6297 6367 1.6460 7341 1.6643 7687 1.6630 4382 1.6834 0312 1.6032 8678 1.6233 2787 1.6436 1046 1.7001 3954 1.7347 7603 1.7007 0828 1.7872 1025 1.8140 1841 1.8074 0727 1.0000 8689 1.0333 3841 1,9071 7184 2.0015 0734 2.0308 8734 2.0800 8500 2.1222 0879 2.1047 4477 2.2080 3000 41 42 48 44 45 1.6810 7611 1.5097 7334 1.6177 7079 1.6360 7071 1.6643 7638 1.6641 6471 1.6840 6677 1.7060 2885 1.7273 6421 1.7480 4614 1.8412 2808 1.8088 4712 1.8968 7082 1.9263 3302 1.0542 1301 2.0300 2530 2,0722 0024 2.1086 3090 2.1454 3010 2.1829 7522 2.2522 0040 2.2972 4447 2.3431 8936 2.3900 5314 2.4378 5421 46 47 48 49 50 1.6720 8710 1.6018 0821 1.7108 4106 1.7300 8801 1.7406 6160 1.7708 0707 1.7929 4306 1.8153 6485 1.8380 4679 1.8610 2237 ' 1.0835 2021 2.0132 7010 2.0434 7820 2.0741 3046 2.1052 4242 2.2211 7728 2.2600 4780 2.2005 0872 2.3308 4170 2.3807 8803 2.4800 1120 2.5303 4361 2.15870 7039 2.0388 1170 2.0016 8803 TABLE V COMPOUND AMOUNT OF 1 (1 + *)" n *I% i;% 1|% 1|% 2% 51 53 58 54 55 1.7602 3395 1.7891 3784 1.8092 0504 1.8206 1988 1.8602 0310 1.8842 S615 1.0078 3872 1.0310 8670 1.0558 3279 1.0802 8070 2.1308 2100 2.1088 7337 2.2014 0047 2.2344 2757 2.2679 4398 2.4224 5274 2.4048 4006 2.5070 8046 2.5518 7012 2.5065 2785 2.7454 1079 2.8003 2810 2.8503 3475 2.0134 0144 2.9717 3007 6 57 58 5 ' 60 1.8710 1788 1.8020 6684 1.0133 5260 1.0348 7780 1.0506 4518 2.0050 3420 2.0300 9713 2.0554 7335 2.0811 6070 2.1071 8135 2.3010 6314 2.3304 0269 2.3715 3008 2.4071 1308 2.4432 1078 2.0410 0708 2.0882 0161 2.7362 4503 2.7831 1182 2.8318 1028 3.0311 0529 3.0917 8859 3.1530 2430 3.2166 0085 3.2810 3070 61 62 63 04 65 1.9780 6744 2.0000 1733 2.0234 2765 2.0461 9121 2.0002 1087 2.1335 2111 2.1001 0013 2.1871 0250 2.2145 3241 2.2422 1407 2.4708 0807 2.5170 0009 2.5548 2208 2.5031 4442 2.0320 4168 2.8813 7306 2.0317 9709 2.9831 0354 3,0343 0785 3.0884 2574 3.3466 5140 3.4135 8443 3.4818 5612 3.5514 9324 3.6225 2311 66 67 68 . 69 70 2.0024 8940 2.1160 2990 2.1308 3533 2.1030 0848 2.1882 6245 2,2702 4174 2.2980 1976 2.3273 5251 2.3504 4442 2.3858 9097 2.6715 2221 2.7115 9504 2.7522 0890 2.7035 6300 2.8354 5629 3.1424 7319 3.1974 0647 3.2534 2213 3.3103 5702 3.3682 8827 3.6040 7357 3.7088 7304 3.8442 5050 3.9211 3551 3.9995 5822 71 72 78 74 75 2.2128 7020 2.2377 0508 2.2020 3904 2.2833 0801 2.3141 4249 2.4167 2372 2.4450 2027 2.4704 0427 2.6074 6045 2.5387 0358 2.8770 8814 2.9211 5706 2.0049 7533 3.0004 4906 3.0546 9171 3.4272 3331 3.4872 0900 3.6482 3607 3.6103 3020 3.6736 1008 4.0795 4930 4.1011 4038 4.2443 6318 4.3292 5045 4.4158 3540 76 77 78 79 80 2.3401 7650 2.3666 0358 2.3031 2070 2,4200 4042 2.4472 7498 2.5705 2850 2.6020 6011 2.6351 0336 2.6681 3327 2.7014 8404 3.1004 1060 3.1469 1674 3.1041 2050 3.2420 3230 3.2006 6279 3.7377 0742 3.8032 0888 3.8697 6503 3.0374 8502 4.0063 9102 4.5041 5216 4.5042 3521 4.0801 1001 4.7798 4231 4.8754 3016 81 82 83 84 85 2.4748 0082 2.5020 4840 2.5308 0310 2.5502 7473 2.5880 0657 2.7352 6350 2.7004 4417 2.8040 6222 2.8301 1300 2.8746 0101 3.3400 2273 3.3901 2307 3.4409 7402 .3.4925 8954 3.5449 7838 4.0765 0378 4.1478 4260 4.2204 2984 4.2942 8737 4.3694 3740 4.0729 4704 5,0724 0600 5.1738 5504 5.2773 3214 5.3828 7878 86 87 88 89 90 2.0171 8232 2.0466 2502 2.6764 0016 2.7066 0966 2.7300 5780 2.0105 3444 2.0409 1612 2.0837,6257 , 3.0210 4048 3.0688 1260 3.5981 5306 3.6521 2535 3,7069 0723 3.7025 1084 3.8189 4851 4.4450 0255 4.5237 0584 4.6028 7070 4.6834 2003 4.7653 8080 5.4005 3636 5.6003 4708 5.7123 5402 5.8266 0110 5.9431 3313 91 92 93 8 2.7677 4867 2.7988 8584 2.8303 7331 2.8022 1501 2.8044 1402 3.0070 4776 3.1367 0085 3.1740 6786 3.2146 4483 3.2548 2789 3.8702 3273 3.9343 7022 3.9933 0187 4.0532 0275 4.1140 9214 4.8487 7406 4.0336 2853 5.0199 0703 6.1078 1646 5.1972 0324 6.0619 0579 6.1832 3570 6.3069 0042 0.4330 3843 6.6616 9920 96 97 98 99 100 2.9269 7700 2.0500 QB50 2.0932 0462 3.0268 7807 3.0600 3045 3.2065 1324 3.3367 0710 3.3784 1600 3.4206 4020 3.4634 0427 4.1758 0362 4.2384 4067 4.3020 1718 4.3066 4744 4.4320 4565 5.2881 5429 5.3800 9009 5.4748 5919 5.5706 6923 5,6681 5504 6.6929 3318 6.8267 9184 6.0633 2708 7.1026 9423 7.2440 4612 TABLE V COMPOUND AMOUNT OP 1 (1 + 0" n 2 -Of t'O 2 -or 2/0 2|% 3% 3|% 1 2 3 4 5 1.0225 0000 1.0465 0625 1.0600 3014 1.0930 8332 1.1176 7769 1.0250 0000 1.0506 2500 1.0768 9063 1.1038 1280 1.1314 0821 1.0276 0000 1.0557 5625 1.0847 8955 1.1146 2126 1.1452 7334 1.0300 0000 1.0609 0000 1.0027 2700 1.1255 0881 1.1592 7407 1.0350 0000 1.0712 2500 1,1087 1788 1.1476 2300 1,1876 8631 e 7 8 9 10 1.1428 2644 1.1085 3001 1.1048 3114 1.2217 1484 1.2492 0343 1.1506 9342 1.1886 8575 1.2184 0290 1.2488 6297 1.2800 8454 1.1767 6830 1.2091 2049 1.2423 8055 1.2765 4602 1.3110 6103 1.1040 6230 1.2208 7387 1.2067 7008 1.3047 7318 1.3430 1638 1.2202 5533 1.2722 7926 1.3168 0004 1.3028 0735 1.4105 9870 11 12 13 14 15 1.2773 1050 1.3060 4900 1.3354 3611 1.3654 8343 1.3962 0680 1.3120 8666 1.3448 8882 1.3785 1104 1.4129 7382 1.4482 9817 1.3477 2144 1.3847 8378 1.4228 6533 1.4019 9413 1.5021 0890 1.3842 3387 . 1.4257 6080 1.4685 3371 1.5125 8972 1,5579 6742 1.4590 6072 1.5110 0866 1.5639 5606 1.6186 9452 1.6753 4883 16 17 18 19 20 1.4276 2146 1.4597 4294 1.4925 8716 1.5261 7037 1.5605 0920 1.4845 0562 1.5216 1826 1.5506 5872 1.6986 5010 1.6386 1644 1.5435 0944 1.6859 5595 1.6296 6973 1.6743 8200 1.7204 2843 1.0047 0644 1.6528 4763 1.7024 3300 1.7535 0005 1.8061 1123 1.7339 8604 1.7046 7555 1.8574 8020 1.9225 0132 1.0897 8886 31 99 23 24 25 1.5956 2066 1.6316 2212 1.8682 3137 1.7057 6658 1.7441 4632 1.6795 8185 1.7215 7140 1.7646 1068 1.8087 2695 1.8539 4410 1.7677 4021 1.8163 6307 1.8663 0278 1.9176 2610 1.9703 6082 - 1.8002 9467 1.9161 0341 1.0735 8651 2.0327 0411 2.0037 7703 2.0694 3147 2,1316 1168 2.2061 1448 2.2833 2849 2.3632 4408 26 27 28 29 30 1.7833 8962 1.8235 1588 1.8645 4499 1.9064 9725 1.0493 9344 1.9002 9270 1.9478 0002 1.0964 9502 2.0464 0739 2.0976 6758 2.0245 4575 2.0802 2075 2.1374 2682 2.1962 0600 2.2566 0173 2.1506 9127 2.2212 8901 2.2879 2708 2.3565 6551 2.4272 6247 2.4459 6856 2.6315 6711 2.6201 7100 2.7118 7708 2.8067 9370 31 32 33 34 35 1.9D32 5479 2,0381 0303 2.0839 6034 2.1308 4945 2.1787 9356 2.1500 0677 2.2037 5604 2.2588 5086 2.3153 2213 2.3732 0619 2.3186 5828 2.3824 2138 2.4479 3797 2.5152 5620 2.5844 2581 2.5000 8035 2.5760 8270 2.6523 3524 2.7319 0530 2.8138 6246 2.0060 3148 3.0067 0750 3.1119 4235 3.2208 6033 3.3336 9045 36 37 38 39 40 2.2278 1642 2.2779 4229 2.3291 9599 2.3810 0290 2.4351 8897 2.4325 3532 2.4033 4870 2.5556 8242 2.6195 7448 2.6850 6384 2.6554 0752 2.7286 2370 2.8035 5810 2.8806 5595 2.0508 7399 2.8982 7833 2.0852 2608 3.0747 8348 3.1670 2608 3.2620 3779 3.4502 6611 3.8710 2543 3,6900 1132 3.8263 7171 3.9592 5072 41 42 43 44 45 2.4809 8072 2.5460 0528 2.6032 0040 2.6618 6444 2.7217 6639 2.7521 9043 2.8209 0520 2.8015 2008 2.9638 0808 3.0379 0328 3.0412 7052 3.1249 0546 3.2108 4036 3.2991 3847 3.3898 6478 3.3598 9893 3.4606 9689 3.6645 1077 3.8714 6227 3.7815 9584 4.0978 3381 4.2412 5799 4.3807 0202 4.5433 4160 4.7023 5865 46 47 48 4 50 2.7829 0590 2.8456 1331 2.9096 3061 2.9751 0650 3.0420 4640 3.1138 6086 3.1916 9713 3.2714 8966 3.3532 7680 3.4371 0872 3.4830 8606 3,5788 7093 3.6772 8088 3.7784 1535 3.8823 2177 3.8960 4372 4.0118 0503 4.1322 5188 4.2562 1044 4.3839 0602 4.8669 4110 5.0372 8404 5.2136 8898 5.3960 6459 5.5849 2680 30 TABLE V COMPOUND AMOUNT OP 1 n 2l% 2% 2|% 3% 3 lor a > 61 52 53 51 55 3.1104 0244 3.1804 7862 3.2520 3020 3.3262 1017 3.4000 2740 3.5230 3044 3.0111 1235 3.7013 0010 3.7030 2401 3.8887 7303 3.0800 8502 4.0087 8547 4.2115 0208 4.3273 1838 4.4463 1004 4.5154 2320 4.6508 8500 4.700-1 1247 4.0341 2485 5.0821 4850 5.7803 0030 5.0827 1327 0. 1921 0824 0.4088 3202 0.0331 4114 5G 57 5S 5ft 60 3.4766 2802 3.6547 4000 3.6347 3177 3.7106 1324 3.8001 3470 3.0850 0230 4.0850 4217 4.1877 8322 4.2024 7780 4.3997 8975 4.5685 0343 4.0042 2075 4.8233 2107 4.0550 0230 5.0922 5130 5.2340 1305 5.3010 5144 5.5534 0008 5.7200 0301 5.8010 0310 0.8053 0108 7.1055 8002 7.3542 8215 7.61 1U 8203 7.8780 0090 61 02 63 64 65 3.8860 3782 3.0730 6407 4.0624 6802 41538 0304 4.2473 2688 4.5007 8440 4.0225 2010 4.7380 0233 4.8565 4404 4.9770 5826 5.2322 8827 5.3761 7020 5.5240 2105 5.0750 3102 5.8320 1074 0.0083 5120 0.2504 0173 6.4379 1370 0.6310 5120 6.8299 8273 8.1538 2408 8.4392 0703 8.7345 8020 9.0402 9051 0.3507 0008 66 67 68 69 70 4.3428 9071 4.4400 0676 4.6406 1030 4.0426 8107 ,4.7471 4140 5.1024 0721 5.2200 6739 5.3007 1668 5.4047 3440 5.0321 0280 5.0924 0020 0.1571 0130 0.3205 1400 0.5004 0310 6.0702 6076 7.0343 8222 7.2450 2808 7.4033 0054 7.0872 0674 7.9178 2191 0.6841 8520 10.0231 3168 10.3730 4129 10.7370 2924 11.1128 2520 71 72 73 74 75 4.8630 6208 4,0631 C(iOO fi.0748 3723 5.1890 2107 C.3057 7405 5.7729 0543 5.0172 2806 0.0651 6876 0.2107 8773 0.3722 0743 0.8020 3032 7.0516 6700 7.2455 8701 7.4448 4158 7.0496 7472 8.1553 5657 8.4000 1727 8.0520 1778 8.0115 7832 0,1780 2607 11.5017 7414 11.9043 3G24 12.3200 8801 12.7622 2250 13.1985 5038 76 77 78 79 80 6.4251 6300 5.5472 1093 5.6720 3237 5.7000 5310 5.0301 4530 0.6315 1261 0.0948 0043 0.8021 7044 7.0337 2470 7.2005 0782 7.8500 3802 8.0700 8032 8.2981 7800 8.5203 7801 8.7008 5402 9.4542 0344 9.7370 2224 10.0300 6001 10.3300 0171 10.0408 0050 13.0604 0004 14.1380 1713 14.0334 0873 15.1450 4013 16.6767 3754 81 . 82 83 84 85 6.0035 7357 6.2000 0307 0.3395 0406 6,4821 4290 Q.6270 0112 7.3898 0701 7.5745 5210 7.7039 1609 7,0680 13S9 8.1560 0424 9.0017 7761 0.2403 2030 0.5030 8286 0.7050 3414 10.0335 7268 10.0001 1727 11.2880 2070 11.6275 8842 . 11.0704 1607 12.3357 0855 16.2243 8835 16.7922 4195 17.3790 7041 17.9882 6038 18.0178 5S81 86 87 88 80 00 0.7771 2002 6.0200 0014 7.0866 2228 7.2440 4053 7.4070 6782 8.3008 8834 8.5000 1055 8.7841 5832 0.0037 0228 0.2288 5033. 10.3004 0583 10,5930 0000 10,8843 1405 11.1836 3331 11.4911 8322 12.7057 7081 13.0800 6320 13.4705 0180 13.8830 4805 14.3004 6711 10.2604 8387 10.0430 IfiSO 20.0410 5285 21.3044 2120 22.1121 7595 01 02 02 04 05 7.6740 3088 7.7460 OC21 7.0103 3020 8.0076 1B12 8.2707 OW21 0.4505 7774 9.6060 0718 0.0384 0886 10.1800 3068 10.4410 0385 11.8071 0070 12.1318 8861 12.4055 1544 12.8083 1711 13.1006 4684 14.7204 8112 15.1713 6556 15.6265 0652 10.0053 0172 10.6781 6077 22.8801 0210 23.6871 150H 24.5101 6473 25.3742 3040 20.2023 2856 OG 07 08 00 100 8.4000 0267 8.6664 8773 8:8B12 f>871 0.0504 1203 0.2540 4030 10.7020 4305 10.0702 1004 11.2444 0530 11.5265 7003 11,8137 1035 13.5224 0085 13.8043 2852 14,2704 2255 14.0000 2417 10.0724 2234 17.0765 0550 17.5877 7070 18.1154 0388 18.0588 0600 19.2180 3108 27.1815 1000 28.1328 0201 20.1175 1311 30.1360 2807 31.1914 0798 31 TABLE V COMPOUND AMOUNT OF 1 (l + 0" n 4% 4%' 6% 5|% 6% 1 2 3 4 5 1.0400 0000 1.0816 0000 1.1248 6400 1.1008 5850 1.2166 6200 1.0450 0000 1.0920 2500 1.1411 6613 1.1025 1860 1.2461 8104 1.0500 0000 1.1025 0000 1.1576 2600 1.2155 0625 1.2762 8150 1.0560 0000 1.1130 2500 1.1742 4138 1.2388 2465 1.3000 6001 1.0600 0000 1.1236 0000 1.1910 1600 1.2624 7606 1.3382 2558 6 7 8 9 10 1.2663 1002 1.3150 3178 1.3685 6905 1.4233 1181 1.4802 4428 1.3022 6012 1,3608 6183 1.4221 0061 1.4860 0514 1.6520 6042 1.3400 0564 1,4071 0042 1.4774 6644 1.5513 2822 1.0288 9463 1.3788 4281 1.4546 7916 1.5346 8651 1.0100 0427 1.7081 4446 1.4186 1011 1.6036 3026 1.5038 4807 1.6804 7806 1.7008 4770 11 12 13 11 15 1.5304 6406 1.0010 3222 1.6650 7351 1.7316 7645 1.8000 4351 1.6228 6305 1.6058 8143 1.7721 9610 1.8510 4402 1.9352 8244 1.7103 3936 1.7958 5633 1.8856 4914 1.9709 3160 2.0780 2818 1.8020 0240 1.0012 0740 2.0067 7390 2.1160 9146 2.2324 7040 1.8082 0856 2.0121 0647 2.1320 2826 2,2600 0306 2.3065 5810 16 17 18 19 20 1.8720 8125 1.0470 0050 2.0258 1652 2.1068 4018 2.1011 2314 2.0223 7015 2.1133 7681 2.2084 7877 2.3078 6031 2.4117 1402 2,1823 7459 2.2020 1832 2.4066 1923 2.5209 5020 2.6532 0771 2.3662 6270 2.4848 0216 2.0214 6627 2.7650 4601 2,9177 6749 2.6403 5168 2.6027 7270 2.8643 3015 3.0255 0050 3.2071 3647 21 22 23 24 25 2.2787 6807 2,3690 1879 2.4647 1554 2.5633 0416 2.6658 3633 2.5202 4116 2.6336 5201 2.7521 6635 2.8760 1383 3.0054 3446 2.7850 6250 2.0252 0072 3.0716 2376 3.2250 9994 3.3863 6494 3.0782 3415 3.2475 3703 3.4261 6157 3.6145 8990 3.8133 9236 3.3096 6360 3.6035 3742 3.8107 4960 4.0480 3464 4.2918 7072 26 27 28 29 30 2.7724 6078 2.8833 6858 2.9087 0332 3.1186 6145 3.2433 9761 3.1406 7001 3.2820 0956 3.4296 9000 3.5840 3640 8.7453 1813 , 3.5556 7269 3.7334 5632 3.9201 2914 4.1161 3660 4.3219 4238 4.0231 2803 4.2444 0102 4.4778 4307 4.7241 2444 4.0839 5120 4.6493 8296 4.8223 4594 6.1110 8670 5.4183 8790 5.7434 0117 31 32 33 94 35 3.3731 3341 3.5080 5876 3.6483 8110 3.7043 1634 3.0460 8800 3.9138 6745 4.0800 8104 4.2740 3018 4.4603 6154 4.6673 4781 4.5380 3949 4.7649 4147 5,0031 8854 5.2633 4797 6.5160 1537 5.2680 6861 6.6472 6238 5.8623 6181 6.1742 4171 6.5138 2501 6.0881 0064 6,4533 '8068 6.8405 8088 7.2610 2528 7.6860 8679 36 37 38 39 40 4.1030 3255 4.2680 8086 4,4388 1345 4.0163 6590 4.8010 2063 4.8773 7846 5.0968 6040 5.3262 1921 6.5658 9908 5.8163 6454 5.7918 1614 6.0814 0604 6.3854 7729 6,7047 6115 7,0399 8871 6.8720 8538 7.2500 6008 7.6488 0283 8.0604 8600 8.5133 0877 8.1472 5200 8.6360 8712 0.1542 5235 9.7035 0749 10.2857 1794 41 42 43 44 45 4.9930 6145 5.1927 8391 5.4004 9527 5.6165 1508 5.8411 7568 6.0781 0094 6.3616 1648 6.6374 3818 6.9361 2290 7.2482 4843 7.3919 8815 7.7615 8766 8.1496 6693 8.5571 5028 8.9850 0770 8.0815 4076 0.4755 2550 9.9066 7040 10.6464 9677 11.1265 5400 10.9028 6101 11.5570 3267 12.2504 5463 12.0854 8191 13.764(3 1083 46 47 48 49 50 6.0748 2271 6.3178 1562 6.5705 2824 6.8333 4937 7.1066 8335 7.5744 1961 7.9152 6849 8.2714 5567 . 8.6436 7107 9.0326 3627 9.4342 5818 9.0059 7109 10.4012 6065 10.0213 3313 11.4673 9970 11.7385 1466 12.3841 3287 13.0652 6017 13.7838 4048 14.6410 6120 14.5004 8748 16,4069 1673 16.3938 7173 17.3775 0403 18.4201 5427 32 TABLE V COMPOUND AMOUNT OP 1 n 4% 4j% 6% 65% 6% 51 52 53 54 55 7.3009 5068 7.6865 8871 7.0040 5220 8.3138 1435 8.6403 6002 0.4301 0490 0.8638 0403 10.3077 3863 10.7715 8077 11.2563 0817 12.0407 0978 12.6428 0826 13.2740 4868 13.0386 9011 14.0350 3002 16.3417 6007 10.1865 0037 17.0767 7252 18.0140 4001 10.0057 0171 10.5253 0353 20.0968 8634 21.0386 9840 23.2650 2037 24 0503 2150 56 57 58 59 60 8.9022 21*30 0.3510 1040 0.7250 8088 10.1150 2035 10.5106 2741 11.7028 4204 12.2021 0093 12.8453 1758 13.4233 5687 14.0274 0703 15.3074 1240 16.1357 8309 16.0426 7224 17.7897 0085 18.6701 8580 20.0510 7860 21.1538 8703 22.3173 6170 23.5448 0611 24.8307 7045 26.1203 4080 27.6B71 0134 20.3680 2742 31.1204 0307 32.9876 0085 61 62 63 64 65 10.0404 1250 11.3780 2000 11.8331 5010 12.3004 7017 12.7087 3522 14.0586 4129 15.3182 8014 16.0076 0275 10.7270 4487 17.4807 0230 10.6131 4619 20.5038 0245 21.0234 0257 22.7040 6720 23.8300 0056 26.2059 5782 27.0472 8650 20.1678 8620 30.7721 1004 32.4645 8654 34.0609 5230 37.0649 6944 30.2888 0761 41.0461 0067 44.1440 7165 60 67 68 69 70 13.3106 8403 13.8431 1201 14.3068 3640 14.0727 0905 15.5716 1835 18.2073 3400 10.0893 0403 10.9483 8541 20.8460 6276 21.7841 3558 25.0318 9659 26.2834 0037 27.5076 6488 28.0775 4813 30.4264 2554 34.2501 3880 30.1338 0643 38.1212 0074 40.2170 3008 42.4200 1023 40.7036 0094 40.0012 0014 , 62.5773 6766 65.7320 0060 50.0750 3018 71 . 72 73 74 75 16.1044 8308 16.8422 6241 17.5150 5200 18.2165 0102 18.0452 5460 22.7644 2168 23.7888 2006 24.8503 1769 25.0770 8088 27.1460 0620 31.9477 4681 33.5451 3415 35.2223 0080 36.0835 1040 38.8320 8502 44.7635 6163 47.2255 5761 40.8229 6318 52.6632 2615 55.4542 0350 62.0204 8609 00.3777 1516 70.3003 7806 74.6820 0074 70.0669 2070 76 77 78 79 80 10.7030 6485 20.4011 8744 21.3108 3404 22.1032 6834 23.0407 0007 28.3686 1112 20.6451 0862 30.0702 3250 32.3732 0802 33.8300 0643 40.7743 2022 42.8130 3023 44.0630 8804 47.2013 7244 40.5014 4107 68.6041 8470 61.7219 1495 65.1166 2027 68.0980 3430 72.4764 2628 83.8003 3603 88.8283 5620 94.1680 6757 00.8075 4102 105.7950 0348 81 82 83 84 85 23.0717 0103 24.0306 6207 25.0278 8018 26.0650 0475 28,0436 0404 35.3524 5077 36.0433 1106 38.6057 0000 40.3430 1026 42.1584 6513 52.0305 1312 64.6414 8878 67.3735 0322 00.2422 4138 03,2543 6344 76.4626 2973 80.6680 7436 85.1048 1845 80.7855 8347 04.7237 0066 112.1437 5309 118.8723 7828 126.0047 2007 133.5650 0423 141,6780 0440 86 87 88 89 99 20.1053 4014 30.3310 6310 31.5452 4103 32,8070 5120 34.1103 3334 44.0555 8601 46.0380 8006 48.1008 0087 50.2747 410.1 62.6371 0630 60.4170 7112 60.7370 2467 73.2248 2001 70.8860 0106 80.7303 0605 00.0336 0004 105.4209 4098 111.2285 0407 117.3461 0674 123.8002 0601 160.0736 3875 160.0780 6708 168.6227 4060 178.7401 0403 180.4646 1123 91 93 93 94 95 35.4841 0068 36.0034 7004 38.3706 0078 30.9147 0417 41.5113 8504 64.9012 7603 67.3718 3241 60.0536 0487 62.0514 7520 66.4707 0108 84.7668 8330 80.0052 2747 93.4554 8884 08.1282 0328 103.0346 7645 130.0002 1724 137.7927 2419 145,3713 2402 163.3667 4684 101.8019 1791 200,8323 8100 212.8823 2482 225.6652 0431 230.1046 8017 263.6462 408 96 97 98 99 100 43.1718 4138 44.8087 IfiOS 46.6046 6363 48.5624 6018 50.5040 4818 68.4169 7730 71.4957 4128 74.7130 4904 78.0751 3087 81.6885 1803 108.1864 1027 113.5957 3078 119.2785 1732 125.2302 0319 131.5012 5786 170.7010 2340 180.0806 7909 189.9046 0057 200.4442 0443 211.4686 3667 268.7600 3028 284.8846 7200 301.9776 4042 320.0963 0620 330,3020 8351 33 TABLE V COMPOUND AMOUNT OF 1 (1 + i) n 6|% 7% 7|% 8% 8l% 1 2 3 4 5 1.0650 0000 1.1342 2600 1.2070 4963 1.2864 0636 1.3700 .8066 1.0700 0000 1.1449 0000 1.2250 4300 1.3107 9601 ' 1.4025 6173 1.0750 0000 1.1556 2500 1.2422 0888 1.3354 0914 1.4350 2933 1.0800 0000 1.1604 0000 1.2697 1200 1.3004 8SUG 1.4093 2808 1.0850 0000 1.1772 2500 1.2772 8013 1.3868 6870 1.5036 0000 6 7 8 9 10 1.4501 4230 1.5539 865C 1.0549 9507 1.7025 7039 1.8771 3747 1.5007 3035 1.6057 8148 1.7181 8018 1.8384 5921 1.9671 6136 1.5433 0153 1.6590 4914 1.7834 7783 1.9172 3860 2.0010 3156 1.5868 7432 1.7138 2427 1.8509 3021 1.0990 0463 2.1589 2500 1.6314 6751 1.7701 4225 1.0206 0434 2.0838 5571 2.2609 8344 11 12 13 - 14 15 1.9901 5140 2.1200 9024 2.2074 8750 2.4148 7418 2.5718 4101 2.1048 5195 2.2521 9159 2.4098 4500 2.5785 3415 2.7590 3154 2.2156 0893 2.3817 7900 2.6004 1307 2.7624 4406 2.9588 7736 2.3316 3900 2.5181 7012 2.7196 2373 2.9371 9362 3.1721 6011 2.4531 6703 2.6616 8623 2.8879 2956 3.1334 0367 3.3997 4288 16 17 18 19 20 2.7390 1067 2.9170 4637 3.1066 5438 3.3085 8691 3.5236 4506 2.9521 6375 3.1588 1521 3.3799 3228 3.6165 2764 3.8696 8446 3.1807 9315 3.4193 5204 3.6758 0409 3.9514 8940 4.2478 5110 3.4259 4264 3.7000 1806 3.9960 1050 4.3157 0100 4.6609 5714 3.6887 2102 4.0022 6231 4.3424 5461 4.7116 6325 5.1120 4612 21 22 23 24 25 3.7526 8199 3.9066 0632 4.2563 8573 4.5330 5081 4.8276 9911 4.1405 6237 4.4304 0174 4.7406 2986 5.0723 6695 5.4274 3264 4.5664 3993 4.9089 2293 5.2770 9215 5.6728 7400 6.0983 3961 5.0338 3372 5.4305 4041 5.8714 0365 6.3411 8074 6.8484 7520 6.5465 7005 6.0180 2850 6.5205 6002 7.0845 7360 7.6867 6236 28 27 28 29 30 5.1414 9955 5.4756 9702 5.8316 1733 8.2106 7245 6.6143 6616 5.8073 6292 6.2138 6763 6.6488 3836 7.1142 6705 7.6122 5504 6.5657 1508 7.0473 9371 7,5759 4824 8.1441 4436 8.7549 5519 7.3963 5321 7.0880 6147 8.0271 0039 9.3172 7490 10.0026 5089 8.3401 3716 9.0490 4881 9.8182 1796 10.6527 6649 11.5582 5164 31 32 38 34 35 7.0442 9996 7.5021 7946 7.9808 2113 8.5091 5950 9.0622 5487 8.1451 1290 8.7162 7080 9.3253 3975 9.9781 1354 10.6765 8148 9.4116 7683 10.1174 4509 10.8762 5347 11.6919 7248 12.5688 7042 10.8676 6944 11.7370 8300 12.6760 4964 13.6901 3361 14.7853 4429 12.5407 0303 13.6060 6279 14.7632 2913 10.0181 0300 17.3796 4241 36 37 38 39 40 9.6513 0143 10.2786 3603 10.9467 4737 11.6582 8596 12.4160 7453 11.4239 4210 12.2236 1814 13.0792 7141 13.9948 2041 14.9744 5784 13.5115 3570 14.5249 0088 15.6142 6844 16.7853 3858 18.0442 3897 15.9681 7184 17.2456 2558 18.6252 7563 20.1152 9768 21.7246 2150 18.8560 1201 20.4597 4053 22.1088 2824 24.0857 2800 26.1330 1558 41 42 43 44 45 13.2231 1938 14.0826 2214 14.9979 9258 15.9728 6209 17.0110 9813 16.0226 6989 17.1442 6678 18.3443 5475 19.6284 6959 21.0024 5176 19.3975 5689 20.8523 7306 22.4163 0168 24.0975 2431 25.9048 3863 23.4624 8322 25.3394 8187 27.3666 4042 29.5659 7160 31.9204 4939 28,3543 2190 30.7644 3927 33.3794 1000 36.2166 6702 39.2950 8371 46 47 48 49 N 18.1168 1951 19.2044 1278 20.5485 4961 21.8842 0533 23.3066 7868 22.4726 2338 24.0457 0702 25.7289 0651 27.5299 2997 29.4570 2506 27.8477 0163 20.9362 7915 32.1815 0008 34.5951 1259 37.1897 4603 34.4740 8534 37.2320 1217 40.2106 7314 43.4274 1899 46.9016 1261 42.6351 6583 46.2591 5402 50.1911 8309 54.4574 3365 59.0863 1551 34 TABLE VI PRESENT VALUE OF 1 " = (! + i)~ n n a% \% 5% !% 1% 1 2 3 4 5 0.0068 5002 0.9917 1846 0.9876 0345 0.0835 0551 0.9704 2457 0.9050 2488 0.9900 7450 0.9851 4876 0.9802 4752 0.9763 7067 0.9942 0060 0.0844 3463 0.0827 0220 0.0770 0302 0.0713 3G88 0.0025 5583 0.0861 0708 0.0778 3333 0.0706 5417 0.0033 2020 0.0900 9001 0.0802 0005 0.9705 9015 0.9000 8034 0.9514 6500 6 7 8 9 10 0.0753 6057 0.9713 1343 0.0672 8308 0.9632 6946 0.9692 7240 0.0706 1808 0.9656 8963 0.0(108 8520 0.0561 0468 0.9513 4794 0.0657 0301 0.0601 0301 0.0545 3480 0.0480 0007 0.0434 0534 0.0501 5802 0.0400 4022 0.9419 7540 0.0349 6318 0.0280 0315 0.0420 4524 0.9327 1805 0.0234 8322 0.0143 3082 0.9052 8605 11' 12 13 14 15 0.9562 9211 0.9513 2824 0.9473 8082 0,9434 497S 0.9396 3505 0.0466 1489 0.0419 0634 0.0372 1024 0.0325 5046 0.0270 1688 0.9380 2354 0.0326 8347 0.0271 7406 0.9217 0780 0.0164 5183 0.9210 9404 0.9142 3815 0.9074 3241 0.0008 7733 0.8030 7254 0.8063 2372 0.8874 4023 0.8786 6260 0.8600 6297 0.8613 4947 IG 17 18 19 20 0.9356 3656 0.9317 6425 0.9278 8805 0.9240 3789 0.9202 0371 0.9233 0037 0.9187 0084 0.0141 3016 0.0005 8822 0.0050 0290 0.0111 3686 0.0058 5272 0.9005 9023 0.8953 7620 0.8001 3346 0.8873 1700 0.8807 1231 0.8741 6014 0.8676 4878 0.8611 8985 0.8528 2120 0.8443 7749 0.8300 1731 0.8277 3992 0.8195 4447 21 22 23 24 25 0.9163 8544 0.0125 8301 0.9087 9036 0.9060 2542 0.0012 7012 0.9005 6010 0.8000 7071 0.8016 2160 0.8871 8567 0.8827 7181 0.8850 2084 0.8708 8816 0.8747 8525 0.8607 1103 0.8046 6803 0.8547 7901 0.8484 1680 0.8421 0014 0.8358 3140 0.8206 0933 0.8114 3017 0.8033 0621 0.7954 4170 0.7875 6013 0.7707 6844 20 27 28 29 30 . 0.8976 3041 0.8938 0022 0.8000 0748 0.8864 0413 0.8827 2610 0.8783 7001 0.8740 0086 0.8096 6155 0.8063 3488 0.8010 2973 0.8506 5330 0.8546 6782 0.8497 1118 0.8447 8327 0.8398 8305 0.8234 3358 0.8173 0330 0.8112 1000 0.8051 8080 0.7901 8600 0.7720 4706 0.7644 0392 0.7508 3657 0.7493 4215 0.7419 2202 31 82 33 34 35 0.8700 6334 0.8754 1677 0.8717 8334 0.8681 0599 0.8646 6364 0.8507 4000 0.8524 8358 0.8482 4237 0.8440 2226 0.8308 2314 0.8350 1304 0.8301 7038 0.8263 5681 0.8205 6915 0.8158 1020 0.7032 3762 0.7873 3262 0.7814 7158 0.7756 5418 0.7608 8008 0.7345 7715 0.7273 0411 0.7201 0307 0.7129 7334 0.70J30 1420 30 37 38 39 40 0.8600 7024 0.8574 0372 0.8538 4603 0.8603 0310 0.8407 7487 0.8356 4492 0.8314 8748 0.8273 5073 0.8232 3455 0.8101 3880 0.8110 7807 0.8003 7511 0.8010 9854 0.7070 4008 0.7924 2660 0.7641 4800 0.7684 6061 0.7528 1440 0.7472 1032 0.7416 4700 0.0089 2405 0.6020 0490 0.6861 3337 0.6783 6067 0.6710 6314 41 42 43 44 45 0.8432 0128 0.8397 6227 0.8362 7778 0.8328 0776 0.8203 6211 0.8150 6354 0.8110 0850 0.8009 7363 0.8020 5884 0.7989 6402 0.7878 3002 0.7832 0189 0.7787 1936 0.7742 0317 0.7607 1318 0.7301 2701 0.7306 4716 0.7262 0809 0.7108 0052 0.7144 6114 0.6660 0311 0.6584 1802 0.6518 0092 0.0464 4546 0.6390 6402 46 47 48 49 50 0.8259 1082 0.8224 8380 0.8190 7100 0.816B 7237 0.8122 8784 0.7940 8907 0.7910 3300 0.7870 0841 0.7831 82SO 0.7702 8607 0.7652 4023 0,7608 1116 Q.7G03 9884 0.7C20 1210 0.7476 6080 0.7001 3264 0.7038 5374 O.H086 1414 0.6034 1353 0.6882 C165 0.6327 2704 0.6204 6301 0.0202 0041 0.6141 1021 0.6080 3882 TABLE VI PRESENT VALUE OF 1 n &* 1% ** 12/o !% 1% 51 52 53 54 55 0.8089 1735 0.8055 6084 0.8022 1827 0.7088 8066 0.7955 7467 0.7764 0002 0.7716 5127 0.7677 1270 0.7638 9324 0.7600 9277 0.7433 1480 0.7390 0394 0.7347 1809 0.7304 6700 0.7262 2080 0.6831 2819 0.6780 4286 0.6720 9540 0.6079 8651 0.6030 1291 0.6020 1864 0.5960 5800 0.5001 5649 0.5843 1330 0.5785 2808 56 57 58 59 60 0.7922 7363 0.7880 8608 0.7857 1228 0.7824 5207 0.7792 0538 0.7663 1122 0.7525 4847 0.7488 0445 0.7450 7906 0.7413 7220 0.7220 0008 0.7178 2179 0.7136 6878 0.7096 1991 0.7054 0505 0.6680 7733 0.6531 7849 0.6483 1612 0.6434 8995 0.6386 0970 0.5728 0008 0.6071 2879 0.5015 1305 0.5559 6411 0.5604 4962 61 62 63 64 65 0.7759 7216 0.7727 6236 0.7695 4591 0-7063 6278 0.7631 7289 0.7376 8378 0.7340 1371 0.7303 6190 0.7267 2826 0.7231 1209 0.7013 1405 0.6972 4678 0.6932 0310 0.6891 8286 0.6861 8694 0.6339 4511 0.6292 2692 0.6245 4185 0.6198 9260 0.6162 7807 0.5449 9962 0.5300 0358 0.5342 0097 0.6289 7120 0.5237 3392 66 67 68 69 70 0.7600 0620 0.7568 5265 0.7537 1218 0.7605 8474 0.7474 7028 0.7195 1512 0.7159 3644 0.7123 7357 0.7088 2943 0.7053 0291 0.6812 1221 0.6772 0151 0.6733 3373 0.6694 2873 0.6655 4038 0.6106 9784 0.6061 6170 0.6016 3940 0.5971 6070 0.5927 1533 0.5185 4844 0.5134 1429 0.5083 3099 0.5032 9801 0.4983 1480 71 72 73 74 75 0.7443 6874 0.7412 8008 0.7382 0423 0.7361 4114 0.7320 9076 0.7017 9394 0.6983 0243 0.6048 2829 0.6913 7143 0.6879 3177 0.6616 8654 0.6578 4909 0.6540 3389 0.6502 4082 0.6464 6975 0.6883 0306 0.5839 2363 0.5796 7681 0.5752 6234 0.6709 7999 0.4933 8105 0.4884 9009 0.4830 5949 0.4788 7078 0.4741 2949 76 77' 78 79 80 0.7290 5304 0.7260 2792 0.7230 1536 0.7200 1529 0.7170 2768 0.0845 0923 0.6811 0371 0.6777 1513 0.6743 4342 0.6709 8847 0.6427 2064 0.6389 9308 0.6352 8724 0.6316 0289 0.6279 3991 0.6667 2052 0.5026 1009 0.5583 2326 0.5541 6701 0.5500 4170 0.4004 3514 0.4047 8726 0.4001 8541 0,4566 2912 0.4511 1794 81 82 83 84 85 0.7140 6246 0.7110 8959 0.7081 3901 0,7062 0067 0.7022 7453 0.6676 5022 0.6643 2868 0.6610 2346 0.6577 3479 0,6544 6248 0.6242 9817 0.6206 7755 0.6170 7793 0.6134 9919 0.6099 4120 0.5459 4710 0.5418 8297 0.5378 4911 0.6338 4527 0.5298 7123 0.4400 5142 0.4422 2913 0.4378 5003 0.4335 1547 0.4292 2324 86 87 88 89 90 0.6993 6052 0.6964 5861 0.6935 6874 0.6906 9086 . 0.6878 2493 0.6512 0644 0.6479 6661 0.6447 4290 0.6415 3622 0.6383 4350 0.6064 0384 0.6028 8700 0,5993 9066 0.5969 1439 0.5924 6838 0.5259 2678 0.5220 1169 0.5181 2675 0.6142 6873 0.5104 4043 0.4249 7350 0,4207 0585 0.4165 9985 0.4124 7510 0.4083 0119 91 92 93 94 95 0.6849 7088 0.6821 2868 0.6792 9827 0.6764 7960 0.6736 7263 ' 0,6351 6766 Q.6320 0763 0.6288 6331 0.6257 3464 0.6226 2163 0.5890 2242 0.5866 0638 0.5822 1015 0,5788 3303 0.5764 7668 0.5066 4063 0,5028 6911 0.4991 2667 0.4964 1009 0.4917 2217 0.4043 4771 0.4003 4427 0.3903 8040 0.3924 5590 0.3885 7020 96 97 98 99 100 0.6708 7731 0.6680 9359 0.6663 2141 0.6626 6074 0.6598 1163 0.6196 2391 0.6164 4170 0.6133 7483 0.6103 2321 0.6072 8678 0.5721 3020 0.5088 2108 0,5656 2220 0.5622 4245 0.5589 8172 0.4880 6171 0.4844 2860 0.4808 2233 0.4772 4301 0.4736 9033 0.3847 2297 0.3809 1383 0.3771 4241 0.3734 0832 0.3697 1121 TABLE VI PRESENT VALUE OP 1 n H |% le, la % !% 1% 101 102 103 104 105 0.6570 7372 0.0643 4727 0.0510 3214 0.0489 2827 0.6402 3502 0.6042 0545 0.6012 5015 0.5082 6781 0.6052 0130 0.5023 2071 0.5557 3901 0.5525 1080 0.5403 1257 0.5461 2083 0.5420 5957 0.4701 0410 0.4000 6412 0.4031 0019 0.4597 4213 0.4503 1073 0.3660 6071 0.3624 2644 0.3588 3806 0.3552 8521 0.3517 6753 106 107 108 109 110 0.0435 415 0.6408 8380 0.0382 2453 0.6355 7030 0.6320 3006 0.5803 8279 0.58G4 5054 0.6836 3288 0.5806 2973 0.5777 4102 0,5308 1067 0.5300 8004 0.5335 6756 0.5304 7313 0.5273 9065 0.4520 2281 0.4405 5117 0.4402 0404 0.4428 8302 0.4305 8012 0.3482 8400 0.3448 3632 0.3414 2210 0.3380 4168 0.3346 9474 111 112 113 114 115 0.0303 1276 0.0270 9734 0.0250 9270 0.0224 9004 0.6109 1006 0.5748 6669 0.5720 0600 0.5601 0085 0.5603 2021 0.5635 1105 0.5243 3801 0.5212 9711 0.5182 7385 0.5162 0812 0.5122 7982 0.4303 1377 0.4330 6677 0.4298 4190 0.4200 4124 0.4234 6616 0.3213 8003 0.3280 0003 0.3248 5141 0.3210 3500 0.3184 5056 116 117 118 119 120 0.0173 4379 0.0147 8220 0.0122 3123 0.6090 9080 0.6071 6102 0.5G07 0811 0.6579 1852 0.5551 4280 0.5523 8000 0.6400 3273 0.5003 0885 0.5003 5512 0.5034 1851 0,6004 9893 0.4975 9029 0.4203 1379 0.4171 8491 0.4140 7031 0.4100 0083 0.4079 3730 0.3152 0758 0.3121 7582 0.3000 8407 0.3000 2473 0.3029 9478 121 122 123 124 125 0.0040 4168 0.0021 327G 0.5990 3431 0.5971 4620 0.5946 0842 0.5408 0824 0.6441 7730 0.5414 7001 0.5387 7012 0.6300 0565 0.4947 1047 0.4018 4140 0,4889 8896 0.4801 5307 0.4833 3363 0.4040 0055 0.4018 8040 0.3988 9409 0.3050 2625 0.3929 7702 0.2009 9483 0.2970 2459 0.2940 8375 0.2011 7203 0,2882 8014 126 127 128 129 130 0.5922 0091 0.5897 4366 0.5872 9658 0.5848 5060 0.5824 3286 0,5334 2850 0.5307 7463 0.5281 3300 0.5266 0643 0,5228 9107 0.4805 3063 0.4777 4369 0.4740 7302 0.4722 1841 0.4604 7978 0.3900 5252 0.3871 4801 0.3842 6601 0.3814 0630 0.3785 6711 0.2864 3470 0.2820 0870 0.2798 1060 0.2770 4019 0.2742 0722 131 132 133 134 135 0.6800 1613 0.5770 0942 0.5762 1270 0.5728 2503 0.5704 4906 0.5202 9052 0.6177 0201 0.6151 2637 0.5125 0350 0.5100 1349 0.4667 5703 0.4640 6007 0.4613 6881 0.4686 8310 0.4560 2303 0.3757 4800 0.3729 6185 0.3701 7553 0.3674 1988 0.3646 8475 0.2716 8141 0.2688 0248 0.2662 3018 0.2636 9424 0.2609 8430 136 137 138 139 140 0,5630 8205 0.5657 2486 0.5633 7746 0.5010 3979 0.5587 1182 0.5074 7611 0.6040 5136 0.5024 3016 0.4009 3040 0.4074 5220 0.4633 7832 0.4607 4805 0.4481 3483 0.4455 3587 0.4420 5108 0.3619 G997 0.3502 7541 0.3666 0000 0.3539 4630 0.3513 1147 0.2584 0030 0.2658 4197 0.2533 0888 0.2508 0087 0.2483 1770 141 142 143 144 145 0.5503 9351 0.5540 8483 0.6617 8672 0.6494 0015 0.6472 1009 0.4049 7731 0.4025 1474 0.4000 6442 0.4870 2628 0.4862 0028 0.4403 8308 0.4378 2908 0.4352 8980 0.4327 0642 0.4302 5560 0.3480 0625 0.3401 .0040 0.3435 2406 0.3400 0681 0.3384 2800 0.2458 5011 0.2434 2486 0.2410 1471 0.2380 2843 0.2362 6577 146 147 148 149 150 0.5449 4548 0,5420 8429 0,5404 3249 0.5381 0003 0.6360 6688 0.4827 8635 0.4803 8443 0.4779 9446 0.4760 1637 0,4732 5012 0.4277 0033 0,4252 7953 0,4228 1312 0.4203 0102 0.4170 2313 0,3350 0028 0.3334 0871 0.3300 2670 0.3284 6320 0.3200 1815 0.2330 2050 0.2310 1040 0.2293 1723 0.2270 4070 0.2247 0877 37 TABLE VI PRESENT VALUE OP 1 n 1|% l\% 1|% 1|% 2% i 2 3 4 5 0.9888 7615 0.9778 7407 0.9669 9537 0.9662 3770 0.9455 9970 0.9876 5432 0.9754 6106 0.9634 1833 0.9515 2428 0.9397 7706 0.9852 2107 0.9706 6176 0.9563 1099 0.9421 8423 0.9282 6033 0.9828 0098 0.9058 9777 0.9402 8528 0.9329 5851 0.9109 1254 0.9803 0216 0.0011 0878 0.9423 2233 0.9238 4543 0.0057 3081 6 7 8 9 10 0.9350 8005 0.9248 7743 0.9143 9054 0.9042 1808 0.8941 5881 0.9281 7488 0.0107 1693 0.9053 9845 0.8942 2069 0.8831 8093 0.9145 4219 0.9010 2070 0.8877 1112 0.8745 9224 0.8610 6723 0.0011 4254 0.8856 4378 0.8704 1167 0.8554 4136 0.8407 2800 0.8870 7138 0.8705 0018 0.8534 9037 0.8307 5527 0.8203 4830 11 12 13 14 15 0.8842 1142 0.8743 7470 0.8646 4742 0,8550 2835 0.8455 1629 0.8722 7740 0.8015 0860 0.8508 7269 0.8403 6809 0.8299 9318 0.8489 3323 0.8363 8742 0.8240 2702 0.8118 4928 0.7998 5150 0.8262 0889 0.8120 5788 0.7980 9128 0.7843 6490 0.7708 7469 0.8042 0304 0.7884 0318 0.7730 3253 0.7578 7502 0.7430 1473 10 17 18 19 20 0.8361 1005 0.8268 0846 0.8176 1034 0.8085 1455 0.7995 1995 0.8197 4635 0.8096 2002 0.7996 3064 0.7897 6860 0.7800 0856 0.7880 3104 0.7763 8620 0.7649 1159 0.7536 0747 0.7424 7042 0.7576 1631 0.7445 8605 0.7317 7090 0.7191 9401 0.7008 2468 0.7284 4581 0.7141 0250 0.7001 5937 0.0864 3070 0.0720 7133 21 22 23 24 25 0.7906 2542 0.7818 2983 0.7731 3210 0.7645 3112 0.7560 2583 0.7703 7881 0.7608 0796 0.7614 7453 0.7421 9707 0.7330 3414 0.7314 9795 0.7206 8703 0,7100 3708 0.0995 4392 0.0892 0683 0.0040 6789 0.6827 2028 0.6709 7817 0.0594 3800 0.6480 9032 0.0507 7682 0.0408 3904 0.0341 5602 0.0217 2149 0.6095 3087 26 27 28 29 30 0.7476 1616 0.7392 9806 0.7310 7348 0.7229 4040 0.7148 9780 0.7239 8434 0.7150 4028 0.7062 1853 0.6974 9978 0.6888 8867 0.0790 2052 0.6689 8674 0.0590 0925 0.6493 6887 0.6397 6243 0.0369 4070 0.6259 9479 0.6152 2829 0.0040 4697 0.5042 4764 0.5976 7928 0.6858 6204 0.6743 7465 0.6631 1231 0,5620 7080 31 32 33 34 35 0.7069 4467 0.6990 8002 0.6913 0287 0.6836 1223 0.6760 0716 0.6803 8387 0.6719 8407 0.6630 8797 0.6564 9429 0.6474 0177 0.6303 0781 0.6209 9292 0.6118 1668 0.6027 7407 0.5938 6608 0.5840 2710 0.5739 8247 0.5641 1063 0.5544 0830 0.6448 7311 0.5412 4507 0.5300 3330 0.5202 2873 0.6100 2817 0.6000 2701 30 37 40 0.6684 8667 0.6610 4986 1 0.6536 9578 0.6464 2352 0.6392 3216 0.6394 0910 0.6316 1622 0.6237 1873 0.6160 1850 0.6084 1334 0.6850 8074 0.6764 4309 0.6679 2423 0.6595 3126 0.6612 6232 0.5356 0183 0.5202 9172 0.5172 4002 0.5083 4400 0.4996 0098 0.4902 2318 0.4806 1003 0.4711 8719 0.4619 4822 0.4528 9042 41 42 43 44 45 0.6321 2080 0.6250 8855 0.6181 3454 0.6112 789 0.6044 5774 0.6009 0206 0.5934 8352 0.6881 5656 0.5789 2006 0.6717 7290 0.6431 1560 0.6350 8Q25 0.6271 8163 0.5193 9067 0.5117 1494 0.4910 0834 0.4826 6348 0.4742 6386 0.4661 0690 0.4580 0040 0,4440 1021 0.4353 0413 0.4207 0875 0.4184 0074 0.4101 0680 40 47 48 49 00 0.6977 3324 0.6910 8356 0.5845 0784 0.6780 0528 0.6715 7506 0.6647 1397 0,5577 4219 0.5608 5649 0.5440 5579 0.6373 3906 0.5041 6266 0.4967 0212 0.4893 6170 0.4821 2976 0.4760 0468 0.4502 1170 0.4424 6850 0.4348 5848 0.4273 7934 0.4200 E883 0.4021 5373 0.3942 0830 0.3865 3701 0.3780 6844 0.3716 2788 38 TABLE VI PRESENT VALUE OF 1 IX* = (1 + ~ n n 1|% ll% 1|% lf% 2% 51 58 53 54 55 0.5662 1637 0.5580 2843 0.5527 1044 0.5405 (1102 0.5404 8120 0.5307 0624 0.5241 5332 0.517G 8220 0.5112 9115 0.5040 7802 0.4679 8401 0.4010 0887 0.4542 5605 0.4475 4192 0.4400 2800 0.4128 0475 0.4057 0492 0.3087 2719 0.3018 6047 0.3861 2970 0.3642 4302 0.3571 0100 0.3500 9902 0.3432 3433 0.3305 0425 56 57 58 50 60 0.5344 0843 0.5285 2250 0.5220 4282 0.5168 2850 0.5110 7887 0.4087 4401 0.4925 8727 0.4S05 0594 0.480-1 0970 0.474C 0700 0.4344 1182 0.4270 0104 0.4210 0094 0.4154 3541 0.4092 9607 0.3785 0685 0.3710 9592 0.3655 0796 0.3593 1003 0.3531 3025 0.3200 0613 0.3234 3738 0.3170 9547 0.3108 7791 0.3047 8227 61 03 63 64 65 0.5053 0319 0.4997 7077 0.4942 1000 0.4887 1288 0.4832 7002 0.4087 0874 0.4020 2222 0.4572 0713 0.4515 (1250 0.4450 8775 0.4032 4720 0.3972 8704 0.3014 1000 0.3S5G 3221 0.3799 3321 0.3470 5676 0.3410 8772 0.3352 2135 0.3204 5687 0.3237 8056 0.2988 0014 0.2920 4720 0.2872 0314 0.2816 7170 0.2700 5009 66 67 68 69 70 0.4778 9905 0.4725 8300 0.4673 2568 0.4021 2076 0.4500 8500 0.4404 8173 0.4350 4303 0.4200 7277 0.4243 0817 0.4101 2005 0.3743 1843 0.3687 8003 0.3033 3058 0.3670 0708 0.3520 7002 0.3182 2009 0.3127 4701 0.3073 6866 0.3020 8222 0.2008 8070 0.2706 3703 0.2053 3130 0.2001 2873 0.2550 2817 0.2500 2701 71 78 73 74 75 0.4610 0177 0.4408 7443 0.4410 0302 0.4300 8G02 0.4321 2551 0.4130 5402 0.4088 4407 0.4037 9001 0.3988 1147 0.3038 8787 0.3474 0495 0.3423 3000 0.3372 7093 0.3322 8003 0.3273 7600 0.2917 8064 0.2867 0221 0.2818 3018 0.2709 8208 0.2722 1914 0.2451 2611 0.2403 1874 0.2350 0061 0.2300 8087 0.2204 6771 76 77 78 70 80 0.4273 1818 0.4225 0433 0.4178 0337 0.4132 1470 0.4080 1775 0.3800 2500 0.3842 2228 0.3704 7870 0,3747 9387 0.3701 0079 0.3225 3793 0.3177 7130 0.3130 7C23 0.3084 4860 0.3038 9015 0.2675 3724 0.2620 3686 0.2584 1362 0.2539 0916 0.2490 0114 0.2220 1737 0.2170 0408 0,2133 0616 0.2092 1192 0.2051 0073 81 88 83 84 85 0.4040 7104 0.3005 7070 0.3061 3148 0.3907 3570 0,3803 8882 0.3055 0083 0.3010 8329 0.3500 2547 0.3522 2208 0.3478 7420 0.2993 9010 0.2949 7454 0.2900 1631 0.2803 2Q50 0.2820 8017 0.2463 0825 0.2410 8010 0.2309 4269 0.2328 6761 0.2288 0242 0.2010 8707 0.1971 4607 0.1032 7048 0.1894 8068 0.1867 7420 86 87 88 89 00 0.3820 0031 0.3778 3001 0.3730 3621 0.3004 70(50 0.3053 0010 0.3436 7051 0.3303 3770 0.3351 4843 0.3310 1080 0.3200 2425 0.2770 2036 0.2738 1310 0.2007 6000 0.2057 7907 0.2018 5218 0.2240 2621 0.2210 5770 0.2172 6672 0.2135 1014 0.2008 4082 0.1821 3167 0.1786 6036 0.1760 5018 0.1716 2065 0.1682 6142 91 08 03 94 95 0.3013 0448 0.3572 8503 0.3533 1020 0.3403 7070 0.3454 0207 0.3228 8814 0.3180 0187 0.3140 0481 0.3110 7030 .0.3072 3601 0.2579 8245 0.2541 0000 0.2504 1300 0.2407 1300 0.2430 0000 0.2002 3766 0.2026 0067 0.1002 0450 0.1957 7837 0.1024 1118 0.1040 0217 0.1017 2762 0.1585 5049 0.1564 4754 0,1623 0055 96 07 08 90 100 0.3410 4041 0.3878 4801 0.3340 9010 0.3303 7340 0.32GO 0805 0.3034 4287 0.2000 0006 0.2059 0070 0.2023 4242 0.2887 8320 0.2304 7487 0.2350 3583 0.2324 4000 0.2200 1389 0.2260 2044 . 0.1801 OlflO 0.1868 4053 0.1820 6310 0.1705 1105 0.1764 2422 0,1494 1132 0,1404 8109 0.1436 0060 0.1407 0363 0.1380 3297 TABLE VI PRESENT VALUE OP 1 n 2j% * 2 -Of t' 3% s|% i 2 3 4 5 0.9779 9511 0.9564 7444 0.9354 2732 0.9148 4335 0.8047 1232 0.0750 0970 0.0518 1440 0.9285 9941 0.0050 5004 0.8838 5420 0.0732 3001 0.0471 8833 0.9218 3770 0.8971 0573 0.8731 5400 0.0708 7379 0,0425 0601 0.9151 4100 0.8884 8705 0.8026 0878 0.0001 8357 0.9335 1070 0.0010 4271 0.8714 4223 0.8410 7317 7 8 9 10 0.8750 2427 0.8557 0040 0.8300 3835 0.8185 2101 0.8005 1013 0.8022 0687 0.8412 0524 0.8207 4657 0.8007 2830 0.7811 0840 0.8407 8401 0.8270 4128 0.8040 0035 0.7833 0385 0.7023 0791 0.8374 8426 0.8130 0151 0.7804 0923 0.7004 1073 0.7440 0801 0.8135 0004 0.7850 0000 0.7894 1156 0.7337 3007 0.7080 1881 11 12 13 14 15 0.7828 0400 0.7650 6748 0.7488 1905 0.7323 4137 0.7102 2028 0.7021 4478 0.7435 5589 0.7254 2038 0.7077 2720 0.0904 0550 0.7410 9310 0.7221 3440 0.7028 0720 0.0830 0728 0.0060 9078 0.7224 2128 0.7013 7988 0.0800 5134 0.0011 1781 0.0418 0105 0.0840 4671 0.0017 8330 0.0394 0415 0.0177 8170 0.5008 9002 10 17 0.7004 6580 0.6850 5212 0.0730 2493 0.6571 0500 0*0478 7424 O.G305 3464 0.0231 0094 0.0050 1046 0.5707 0501 5572 0378 18 10 20 0.6699 7763 0.6552 3484 0.6408 1047 0.6411 6591 0.6255 2772 0.0102 7004 0.0130 6802 0.6072 3400 0.6812 6057 0.5873 0401 0.6702 8003 0.6536 7675 (X5383 0114 0.6201 5609 0.5026 0688 21 0.0207 1538 0.5953 S020 0.5656 0308 0.5375 4928 0,4855 7090 22 28 24 0.6129 2457 0.6994 3724 0.6802 4008 0.6808 0407 0.5000 0724 0.5528 7535 0.5505 5376 0.5358 1874 0.5214 7800 0.5218 0260 0.6066 9175 0.4019 3374 0*4691 5003 0.4532 8503 0,4379 5713 25 0.5733 4039 0.5303 0050 0.5076 2120 0.4770 0667 0*4231 4099 20 27 0.5007 2997 0.5483 9117 0.5202 3472 0.5133 0073 0.4030 3700. 0.4807 1821 0.4630 0473 0.4501 8006 0.4088 3767 0.3050 1224 28 0.5363 2388 0.5008 7778 0.4078 6227 0.4370 7075 0*3816 5434 20 80 0,5245 2213 0.5129 8008 0,4880 0125 0.4707 4209 0.4563 3008 0.4431 4421 0.4243 4030 0.4110 8676 0*3087 4816 0.3502 7841 81 0.5010 9201 .0.4651 1481 0,4312 8301 0.3009 8715 0.3442 3035 32 33 34 35 0.4000 5233 0.4708 5558 0.4692 0641 0.4580 0900 0.4537 7055 0,4427 0208 0.4310 0534 0.4213 7107 0.4107 4103 0,4086 0708 0.3075 7380 0.3800 3314 0.3883 3703 0.3770 2625 0.3600 4490 0.3553 8340 0.3325 8071 0.3213 4271 0.3104 7005 0.2009 7080 30 0.4488 7002 0.4110 0372 0.3705 7727 0.3460 3243 0.2898 3272 37 0.4389 9208 0.4010 0705 0.3064 0860 0.3340 8294 0.2800 3101 88 80 0.4293 3270 0.4108 8528 0.3012 8402 0,3817 4139 0.3506 8059 0.3471 4310 0.3262 2016 0.3167 5355 0.2705 0194 0.2014 1250 40 0,4100 4575 0.3724 3002 0.3378 5222 0.3005 5084 0.2525 7247 41 0.4016 0954 0.3633 4005 0.3288 0005 0.2076 2800 0,2440 3137 42 0.3927 7210 0.3544 8483 0.3200 0008 0.2880 6022 0,2357 7910 43 0,3841 2925 0.3458 3880 0.3114 4405 0.2805 4294 0.2278 0590 44 0.3750 7053 0.3374 0370 0.3031 0044 0.2723 7178 0.2201 0231 45 0.3074 0981 0.3291 7440 0.2040 0702 0.2644 3862 0.2126 6924 40 0.3593 2500 0,3211 4670 0.2871 0172 0.2507 3663 0.2054 6787 . 47 0.3514 1809 0.3133 1204 0.2704 1773 0.2492 5870 0.1085 1068 48 0.3430 8518 0.3056 7116 0.2710 3040 0,2410 9880 0.1018 0045 40 0.3301 2242 0.2082 1579 0.2040 0122 0.2340 6020 0.1863 2024 50 0.3287 2008 0.2000 4221 0.2576 7783 0.2281 0708 0.1790 6337 TABLE VI PRESENT VALUE OF 1 n 2|% 2 -Of a % 2 -or i% 3% s|% 51 53 53 54 55 0.3214 9250 0.3144 1810 0.3074 0930 0.3007 3287 0.2041 1528 0.2838 4600 0.2709 2208 0.2701 0870 0.2335 7028 0.2671 6052 0.2600 8402 0.2439 7471 0.2374 4497 0.2310 0000 0.2240 0(511 0.2214 6318 0.2150 1280 0.2087 5029 0.2020 7010 0.1967 0717 0.1720 9843 0.1671 4824 0.1014 9689 0.1580 3407 0.1507 6814 50 57 58 59 00 0.2870 4330 0.2813 1374 0.2731-2347 0.2600 0040 0.2031 4850 0.2508 7855 0.2447 5056 0.2387 8082 0.2329 0568 0.2272 8350 0.2188 8575 0,2130 2740 0.2073 2003 0.2017 7710 0.1003 7070 0.1910 3600 0.1854 7103 0.1800 0084 0.1748 2508 0.1007 3309 0.1456 0004 0.1407 3433 0.1369 7520 0.1313 7701 0.1260 3431 61 02 03 64 05 0.2573 C801 0.2516 0487 0.2401 6035 0.2407 3971 0.2354 4220 0.2217 4009 0.2103 3170 0.2110 5541 0.2059 0771 0.2008 8557 0.1911 2097 0.1860 0581 0.1810 2755 0.1701 8263 0.1714 0718 0.1647 8941 0.1509 8072 0.1553 2082 0.1508 0505 0.1464 1325 0.1220 4184 0.1184 0453 0.1144 8747 0.1100 1501 0.1008 7628 00 07 08 09 70 0.2302 0138 0.2261 9450 0.2202 3012 0.2153 9278 0.2106 5300 0.1050 8593 0.1012 0578 0.1865 4223 0.1810 0241 0,1775 6368 0.1068 7804 0.1624 1172 0.1680 6403 0.1538 3448 0.1407 1720 0.1421 4879 0.1380 0853 0.1330 8887 0.1300 8028 0.1262 0730 0.1032 0114 0.0097 0022 0.0963 0638 0.0031 3503 0.0890 8012 71 78 78 74 75 0.2060 1700 0.2014 8420 0.1070 5066 0.1027 1458 0.1884 7301 0.1732 2300 0.1089 0806 0.1048 701(5 0.1008 5478 0.1500 3140 0.1457 1023 0.1418 1044 0.1380 1603 0.1343 2119 0.1307 2622 0.1220 1880 0.1100 4737 0.1155 7098 0.1122 1357 0.1080 4521 0.0860 4311 0.0840 0300 0.0811 6232 0.0784 1770 0.0767 6590 70 77 78 79 80 0.1843 2657 0.1802 7048 0.1703 0365 0.1724 2411 0.1686 2093 0.1531 0380 0.1403 6065 0.1457 2040 0.1421 7218 0.1387 0457 0.1272 2747 0.1238 2235 0.1206 0837 0.1172 8309 0.1141 4412 0,1057 7205 0.1026 9131 0.0907 0030 0.0007 0641 0.0939 7710 0.0732 0376 0.0707 2827 0.0683 3650 0.0600 2060 0.0037 9285 81 82 83 84 85 0.1640 1025 0.1012 0022 0.1577 4105 0.1542 6097 0.1508 7528 0.1353 2153 0.1320 2101 0.1288 0008 0.1256 5040 0.1225 9463 0.1110 8017 0.1081 1608 0.1052 2237 0.1024 0620 0.0006 6540 0.0012 3000 0.0885 8243 0.0800 0230 0.0834 9743 0.0810 6547 0.0610 3561 0.0596 5131 0.0576 3750 0.0555 0178 0.0537 1187 80 87 88 89 90 0.1475 5528 0.1443 0835 0.1411 3286 0.1380 2724 0.1340 8007 0.1106 0452 0,1106 8733 0.1138 4130 0.1110 6468 0.1083 5579 0.0069 9706 0.0044 0100 0.0018 7533 0.0804 1038 0.0870 2324 0.0787 0434 0.0764 1108 0.0741 8630 0.0720 2502 0.0000 2770 0.0518 9553 0.0501 4000 0.0484 4J503 0.0408 0079 0.0452 2395 91 918 93 94 95 0.1320 1053 0.1201 1445 0.1262 7331 0.1234 9468 0.1207 7719 0.1057 1200 0.1031 3460 0.1006 1012 0.0981 0600 0.0057 7073 0.0846 9415 0.0824 2740 0.0802 2131 0.0780 7427 0,0750 8400 0.0678 9105 0.0059 1364 0.0039 9383 0.0621 2903 0.0603 2032 "0.0430 9464 0.0422 1704 0.0407 8941 0.0304 1000 0.0380 7735 90 97 98 99 100 0.1181 1050 0.1155 2029 0.1120 7828 0.1104 9221 0.1080 6084 0.0034 3486 0.0011 5506 0.0880 3264 0.0867 6356 0.0846 4737 0.0730 6104 0.0710 7181 0.0700 4550 0.0081 7080 0.0663 4634 0.0685 6342 0.0668 5769 0.0552 0104 0.0685 9383 0.0620 3284 0.0307 8071 0,0356 4562 0.0343 4350 0.0331 8221 0,0320 6011 41 TABLE VI PRESENT VALUE OP 1 n 4% 4f% 5% 5|% 6% 1 0.9615 3846 0.0569 3780 0.9523 8005 0.0478 0730 0.0433 002,'i % 0.0245 5621 0.91 57 2995 0.0070 2048 0.8984 5242 0.8800 9044 3 0.8889 9636 0.8762 0060 0.8638 3760 0.8510 1300 O.&'iOO 1028 4 0.8548 0410 0.8385 6134 0.8227 0247 0.8072 1074 0.71)20 03UI1 5 0.8219 2711 0.8024 5105 0.7835 2017 0.7051 3435 0.7472 5817 6 0.7903 1463 0.7678 9574 0.7402 1640 0.72C2 4683 0.7040 (IOM 7 0.7699 1781 0.7348 2840 0.7106 8133 0.0874 3081 0.0(150 5711 8 0.7306 9021 0.7031 8513 0.6708 393(1 0.0615 0887 0.0274 12H7 9 0.7026 8674 0.6729 0443 0.6440 0892 0.0170 2H20 0.51)18 1)84(1 10 0.6765 6417 0.6439 2768 0.6139 1325 0.5854 3058 0.5683 0478 11 0.6495 8093 0.6161 9874 0.5846 7029 0.5540 1050 0.5207 876:1 12 0.6245 9705 0.5896 6380 0.6568 3742 0.5250 81. r )2 . 0,41)00 Ol);t(! 13 0.6005 7409 0.5642 7164 0.6303 2135 0.4085 (1008 0.4088 31)02 14 0.5774 7508 0.5390 7286 0.5060 0795 0.4725 0037 0.4423 001)1) 15 0.5552 6450 0.5187 2044 0.4810 1710 0.4470 3305 0.4172 OfiOO 16 0.5339 0818 0.4944 6932 0.4581 1152 0.4245 8100 0.3030 4028 17 0.5133 7325 0.4731 7639 0.4362 9669 0.4024 453 0,3713 0442 18 0.4936 2812 0.4528 0037 0.4155 2005 0.3814 0500 0.3608 4371) 19 0.4746 4242 0.4333 0179 0.3057 3300 0.3016 7900 0.3805 1111)1 20 0.4563 8695 0.4146 4286 0.3768 3948 0.3427 280G 0.3118 0473 21 0.4388 3360 0.3967 8743 0.3589 4230 0.3248 6158 0.2041 6540 22 0.4219 5539 0.3797 0089 0.3418 4087 0.3070 2507 0.2775 0610 23 0.4057 2633 0.3633 5013 0.3265 7131 0.2918 7207 0.2017 0720 24 0.3901 2147 0.3477 0347 0.3100 0701 0.2700 5050 0.2400 7865 25 0.3751 1680 0.3327 3060 0.2953 0277 0.2022 3370 0.2320 9803 26 0.3606 8923 0.3184 0248 0.2812 4073 0.2485 6275 0.2108 1003 27 0.3468 1657 0.3046 9137 0.2678 4832 0.2350 0450 0.2073 0705 28 0.3334 7747 0.2915 7069 0.2560 9364 0.2233 2181 0.1050 3014 29 0.3206 5141 0.2790 1502 0.2429 4032 0.2116 7044 0.1845 6074 80 0.3083 1867 0.2670 0002 0.2313 7746 0.2000 4402 0.1741 1013 81 0.2964 6026 0.2555 0241 0.2203 5047 0.1001 8300 0.1042 5484 32 33 0.2850 5794 0.2740 0417 0.2444 9991 0.2339 7121 0.2008 6017 0.1998 7264 0.1802 0910 0.1708 7119 0.1640 5740 0.1401 8022 34 0.2635 5209 0.2238 9589 0.1003 5480 0.1010 0321 0.1370 11(53 35 0.2534 1547 0.2142 5444 0.1812 9029 0.1535 1903 0.1801 0522 86 37 38 39 40 0.2436 6872 0.2342 9685 0.2252 8543 0.2166 20G1 0.2082 8904 0.2050 2817 0.1001 0921 0.1877 5044 0.1706 6549 0.1719 2870 0.1726 5741 0.1644 3563 0.1566- 0536 0.1401 4707 0.1420 4568 0.1465 1024 0.1370 3008 0.1307 3041 0.1230 2302 0.1174 0314 0,1227 4077 0.1167 0318 0.1002 3K8fi 0.1030 6552 0,0072 2210 41 42 43 44 45 0.2002 7703 0.1925 7493 0.1851 6820 0.1780 4635 0.1711 0841 0.1645 2507 0.1574 4026 0.1506 6054 0.1441 7276 0.1379 6437 0.1352 8100 0.1288 3962 0.1227 0440 0.1168 0133 0.1112 9061 0.1113 3047 0.1055 3504 0.1000 3322 0.0048 1822 0.0898 7500 0.0017 1005 0.0805 2740 0.0810 2002 0.0770 0908 0.0720 15007 46 47 48 49 50 0.1546 1386 0.1582 8256 0.1521 8476 0.1463 4112 0.1407 1262 0.1320 2332 0.1263 3810 0.1208 S771 0.1166 9158 0.1107 0965 0.1050 9668 0.1009 4921 0,0961 4211 0.0915 6391 0.0872 0373 0.0861 8905 0.0807 4840 0.0765 3885 0.0725 4867 0.0687 0052 0.0085 37H1 0.0640 5831 0.0600 0840 0.0575 4500 0.0542 8830 TABLE VI PRESENT VALUE OF 1 n = (1 + f)-n n 4% 4% 6% 6l% 6% 51 52 53 54 55 0.1353 0059 0,1300 0072 0.1250 9300 0.1202 8173 0.1150 5551 0.1050 4225 0.1013 8014 0.0070 1440 0.0928 3(583 0.0888 3007 0.0830 5117 0.0790 0036 0.0753 2080 0.0717 4272 0.0683 2040 O.OG51 8153 0.0617 8344 0.0585 6250 0.0565 0948 0.0526 1562 0.0512 1544 0.0483 1046 0.0455 8150 0.0430 0147 0.0405 6742 56 57 58 59 60 0.1112 0722 0.10GO 3002 0.1028 1733 0.0088 6282 0.0050 0040 0.0850 1347 0.0813 5200 0.0778 4938 0.0744 9701 0.0712 8001 0.0060 7270 0.0619 7400 0.0500 2291 0.0562 1230 0.0535 3552 0.0498 7263 0.0472 7263 0.0448 0818 0.0424 7221 0.0402 6802 0.0382 7115 0.0301 0480 0.0340 6119 0.0321 3320 0.0303 1434 61 62 03 64 65 0.0014 0423 0.0878 8808 0.0845 0835 0.0812 5803 0.0781 3272 0.0082 1915 0.0652 8148 0.0024 7032 0.0507 8021 0.0572 0594 0.0600 8021 0.0485 5830 0.0402 4600 0.0440 4381 0.0419 4648 0.0381 6026 0.0301 6092 0.0342 8428 0.0324 9695 0.0308 0270 0.0285 9843 0.0269 7UU5 0.0254 5250 0.0240 1179 0.0226 5264 66 67 68 69 70 0.07C1 2702 0.0722 3809 0.0004 5070 0.0067 8818 0.0042 1040 0.0547 4253 0.0523 8510 0.0501 2037 0.0479 7069 0.0459 0497 0.0300 4003 0.0380 4670 0.0302 3495 0.0345 0048 0.0328 6017 0.0201 0600 0.0270 7485 0.0262 3208 0.0248 0453 0.0235 6828 0.0213 7041 0.0201 0077 0.0100 1050 0.0170 4301 0.0160 2737 71 72 73 74 75 0.0617 4042 0.0603 7445 0.0570 0081 0.0548 0501 0.0527 8367 0.0430 2820 0.0420 3055 0.0402 2637 0.0384 9413 0.0368 3640 0.0313 0111 0.0208 1058 0.0283 0103 0.0270 3008 0.0267 5150 0.0223 3960 0.0211 7498 0.0200 7107 0.0190 2471 0.0180 3200 0.0150 0021 0.0150 6530 0.0142 1254 0.0134 0800 0.0126 4011 76 77 78 79 80 0,0507 5353 0.0488 0147 0.0460 2440 0.0451 1070 0.0433 8433 0.0352 5023 0.0337 3228 0.0322 7960 0.0308 8005 0.0205 5048 0.0245 2524 0.0233 5737 0.0222 4512 0.0211 8582 0.0201 7098 0.0170 9279 0.0102 0170 0.0163 6706 0.0145 5646 0.0137 9759 0.0119 3313 0.0112 6767 0.0106 2044 0.0100 1928 0.0094 5216 81 82 83 84 85 0.0417 1570 0.0401 1125 0.0385 6861 0.0370 8510 0.0350 5875 0.0282 8058 0.0270 0850 0.0250 0287 0.0247 8744 0.0237 2003 0.01TB2 1017 0.0183 0111 0.0174 2003 0.0165 0005 0.0158 0019 0.0130 7828 0.0123 0648 0.0117 5022 0.0111 3706 0.0106 5701 0.0089 1713 0.0084 1238 0.0070 3021 0.0074 8609 0.0070 6320 86 87 88 89 90 0.0342 8720 0.0320 6852 0,0317 0050 0.0304 8125 0.0203 0800 0.0220 9800 0.0217 2115 0.0207 8579 0.0198 0070 0.0100 3417 0.0150 6037 0.0143 3940 0.0130 5657 0,0130 0020 0.0123 8(501 0.0100 0004 0.0004 8407 0.0080 0040 0.0085 2180 0.0080 7763 0.0000 6340 0.0062 8022 0.0050 3040 0.0055 0472 0.0062 7803 91 92 93 M 95 0.0281 8163 0.0270 9772 0.0200 5550 0.0260 6337 0.0240 8078 0.0182 1451 0.0174 3016 0.0100 7058 O.OlfiO-0132 0.0162 7399 0.0117 9706 0.0112 3530 0.0107 0028 0.0101 9074 0.0007 0547 0.0076 5043 0.0072 6728 0.0008 7804 0.0066 2032 0.0001 8040 0.0040 7028 0.0040 9743 0.0044 3154 0.0041 8070 0.0030 4405 96 97 98 99 190 0.0231 63215 0.0222 7235 0,0214 1572 0.0205 0204 0.0108 0004 0.0146 1626 0.0139 8085 ' 0.0183 8454 0.0128 0817 0.0122 5063 0.0002 4331 0.0088 0315 0.0083 8306 0.0079 8471 0.0070 0440 0.0058 5820 0.0065 5279 0.0052 6331 0.0049 8802 0.0047 2883 0.0037 2081 0.0035 1019 0.0033 1160 0.0031 2400 0.0029 4723 43 TABLE VI PRESENT VALUE OP 1 n 6|% 7% 7j% 8% 8|% 1 9 3 4 ff 0.0389 0714 . 0.8810 5928 0.8278 400fl 0.7773 2309 0.7298 808-1 0.9345 7944 0.8734 3873 0.8162 9788 0.7028 9531 0.7129 8018 0.9302 3250 0.8053 3201 0.8040 0057 0.7488 (H163 0.0005 5803 0.0251) 2503 0.8573 3882 0.7038 3221 0.7350 20S5 0.0805 Site!) 0.1)2111 fiHOil 0.8-Ul.l flflao 0.7S211 (IS10 0.7l!1.1 7-128 (l.tlOfil) 4. r >11! C 7 8 9 10 0.0853 3412 0.0435 0021 0.0042 3110 0.5073 5323 0.5327 2004 0.0003 4222 0.0227 497-1 0.r>S20 0010 0.5430 337-t 0.5083 4020 0.0-17!) (1162 0.0027 5400 0.5007 0223 0.5215 8347 0.4851 0303 0. llUDt (111(13 0.5S3-1 iioio 0./540S! 0888 0.5002 -IS'.)? o,-i<m '.wit) 0.15] 21) .15011 o./in-iii 2ii3. r > o.fwoii on ir. 0.-179H 70(18 0.-M22 85-ia 11 12 13 11 15 O.C002 1224 0.4000 8286 0.4410 1070 -0.4141 0025 0.3888 2052 0.4750 9280 0.4440 110(1 0.4149 0445 0.3878 1724 0.3024 4002 0.4513 431!) 0.4108 5-113 0.3006 ClOH 0.3033 1347 0.3370 0002 0.428S 82SO 0.31)71 1370 0.307(1 0702 0.3401 0101 0.3152 4170 o.d07 3033 0.3757 0108 0.3-102 0883 0.31SI1 4178 O.tllMl 3UKO 16 17 18 10 20 0.3050 9533 0.3428 1251 0.3218 8909 0.3022 4384 0.2837 9703 0.3387 3-100 0.3106 7-130 0.2958 0302 0.2705 0832 0.2584 1000 0.3143 8000 0.2024 5302 0.2720 4032 0.2530 00i;t 0.2354 1315 0.2018 0017 0.2703 (18SI5 0.2502 40o;< 0.2317 1200 0.2145 -1N21 0.2710 UU07 0.2-11IM 58110 0.2302 8-160 0.2122 -1378 0.105(1 1I33U 21 22 23 24 25 0.2004 7008 0.2502 1228 0.2349 4111 0.2200 0198 0.2071 3801 0.2415 1300 0.2257 1317 0.2100 4088 0.1971 4002 0.1842 4018 0.2180 8807 0.2037 1007 0.1804 0830 0.17(52 7740 0.1039 700H 0.1080 6575 0.1831) ll)fll 0.1703 1C28 0.1570 ODH'l- ().14() 1700 0.1802 OHIO 0.10111 (t738 0.1531 4005 O.H11 5I7I 0.1300 0378 26 27 28 20 30 0.1944 9570 0.1826 2515 0.1714 7902 0.1610 1316 0.1511 8607 0.1721 9540 0.1009 3037 0.1504 0221 0.1405 0282 0.1313 0712 0.1525 3800 0.1418 0043 0.1310 0008 0.1227 87(11 0.1142 2103 0.1362 0170 0.1251 8(182 0.1151) 1372 0.1073 2762 0.0003 7733 0.1101) 0210 0.1105 0885 0.1018 51-18 0.0038 7233 O.OKlin 182H 31 32 33 34 85 0.1419 5875 0.1332 9400 0.1251 5925 0.1175 2042 0.1103 4781 0.1227 7301 0.1147 4113 0.1072 3470 0.1002 1934 0.0930 6204 0.1002 5212 0.0988 3018 0.0010 4343 0.0855 2H77 0.0795 0104 0.0920 1006 0.0852 0005 0.0788 8N03 0.07,40 -IfiUl 0.0070 34M 0.0707 40fi 0.0734 OiMl 0.0077 :)58il 0.0(12-1 2(13(1 0.0575 ,'18.18 86 37 38 30 40 0,1030 1207 0.0972 8917 0.0913 5134 0.0857 7590 0.0805 4075 0.0875 3540 0.0818 0884 0.0764 5080 0.0714 5501 0.0607 8038 0.0740 1083 0.0088 4720 0.0040 4300 0.0505 7580 0.0554 1035 0.0020 2458 0.0670 H572 0.0630 1)018 0.041)7 1,'M1 o.o4m> 3003 0.0630 3005 0.0-188 7Mfi 0.0460 4742 0.0-11 fi 1830 0.0382 0677 41 42 43 44 45 0.0760 2612 0.0710 0950 0.0060 7550 0.0026 0610 0.0587 8515 0.0624 1157 0.0583 2867 0.0545 1268 0.0509 4043 0.0476 1349 0.0515 5288 0.0479 5017 0.0440 1030 0.0414 9804 0.0380 0283 0.0420 2123 0.0304 (till 0.0305 4084 0.0338 15411 0.0313 27S8 0.0362 07UO 0.0326 0500 0.0200 6868 0.0271J 1100 0,0264 4H4H 40 47 48 40 50 0.0551 9733 0.0518 2848 0.0480 0524 0.0466 9506 0.0429 0610 0,0444 9869 0.0415 8747 0.0388 0079 0.0303 2410 0.0339 4770 0.0359 0901 0.0334 0438 0.0310 7375 0.0289 0682 0.0268 8913 0.0200 07.10 0.0208 6801 0.0248 0008 0.0230 2(103 0.0213 2123 O.OarU 5-1K2 o.oairt 17;M o.oioo a:ma 0.018.1 02U7 0.0100 24311 TABLE VH AMOUNT OF ANNUITY OF 1 PER PERIOD n 5% 1% H% S* 1% 1 2 3 4 5 1.0000 0000 2.0041 6667 3.0125 1738 4.0260 0052 5.0418 4064 1.0000 0000 2.0050 0000 3.0150 2500 4.0301 0013 5.0502 5003 1.0000 0000 2.0058 3333 3.0175 8403 4.0351 3031 5.0580 7460 1.0000 0000 2.0075 0000 3.0225 5625 4.0452 2542 5.0755 6461 1.0000 0000 2.0100 0000 3.0301 0000 4.0604 0100 5.1010 0601 6 7 8 10 0.0628 4831 7.0881 1018 8.1176 43fl7 9.1G14 6749 10.1805 0860 0.0755 0188 7.1058 7939 8.1414 0870 9.1821 1583 10.2280 2641 6.0881 8364 7.1230 9794 8.1052 5284 9.2128 8349 10.2660 2531 6.1136 -3136 7.1594 8358 8.2131 7971 0.2747 7866 10.3443 3940 6.1620 1506 7.2135 3621 8.2856 7050 9.3685 2727 10.4622 1254 11 12 13 14 15 11.2320 5520 12.2788 5549 13.3300 1739 14.3855 5013 15.4454 0896 11.2791 6654 12.3355 6237 13.3972 4018 14.4642 2039 15.5305 4752 11.3265 1398 12.31)25 8529 13.4048 7537 14.5434 2048 15.6282 5710 11.4219 2194 12.6075 8036 13.0013 9325 14.7034- 0370 15.8136 7923 11.5668 3467 12.6825 0301 13.8093 2804 14.0474 2132 10.0968 9664 16 17 18 19 20 10.5098 5520 17.5780 4027 18.6518 9063 19.7290 0684 20.8118 1353 16.0142 3020 17.0973 0141 18.7867 8791 19.8797 1685 20.9791 1544 16.7194 2193 17.8109 5189 18.9208 8411 20.0312 5593 21.1481 0493 16.0322 8183 18.0692 7304 19.1947 1849 20.3386 7888 21.4912 1897 17.2678 0449 18.4304 4314 19.6147 4767 20.8108 9504 22.0190 0399 21 22 23 24 25 21.8985 2942 22.9897 7330 24.0855 6402 25.1859 2054 26.2908 6187 22.0840 1101 23.1944 3107 24.3104 0322 25.4319 5524 26.5591 1502 22.2714 6887 23.4013 8577 24.5378 9386 26.6810 3157 26.8308 3759 22.6524 0312 23.8222 9614 25.0009 0336 26.1884 7059 27.3848 8412 23.2391 9403 24.4715 8598 25.7163 0183 26.0734 6485 28.2431 9950 26 27 28 29 30 27.4004 0713 28.5145 7549 29.6333 8622 30.7508 6867 31.8850 1224 27.6919 1059 28.8303 7015 29.9745 2200 31.1243 9461 32.2800 1658 27.987a 5081 29.1500 1036 30.3206 6558 31.4976 2607 32.0812 6164 28.5902 7075 29.8046 9778 31.0282 3301 32.2009 4476 33.6029 0184 29.5256 3150 30.8208 8781 32.1200 '9609 33.4503 8760 34.7848 9153 31 32 33 34 36 33.0178 6646 34.1554 4090 35,2977 5524 30.4448 2922 37.5960 8268 33.4414 1666 34.6086 2375 35.7816 6086 36.9605 7520 38.1463 7807 33.8719 0233 35.0094 8843 36.2740 6045 37.4860 5913 38.7043 2648 34.7541 7381 36.0148 2991 37.2849 4113 38.5645 7819 39.8538 1253 36.1327 4045 37.4940 6785 38.8690 0853 40.2576 9802 41.6602 7560 36 87 38 30 40 38.7533 3552 39.9148 0775 41.0811 1945 42.2522 9078 43.4283 4199 39.3361 0496 40.5327 8549 41.7354 4042 42.9441 2066 44.1588 4730 39,9301 0071 41.1630 2030 42.4031 4395 43.6504 9502 44.0051 2352 41.1527 1612 42.4613 6149 43.7798 2170 45.1081 7037 46.4404 8164 43.0768 7830 44.5076 4714 45.9527 2361 47.4122 6086 48.8863 7336 41 42 48 44 45 44.6092 0342 45.7961 6548 46.9859 7866 48.1817 5358 49.3825 1088 45.3706 4153 46.6065 3974 47.8395 7244 40.0787 7030 50.3241 6415 46.1670 7007 47.4363 7798 48.7130 9018 49.9972 4988 61.2889 0050 47.7948 3026 49.1532 9148 60.5210 4117 61.9008 5573 63.290X 1215 50.3762 3709 51.8789 8946 63.3077 7936 54.9317 5715 56.4810 7472 46 47 48 49 60 50.5882 7134 51.7990 5581 53.0148 8521 54.2357 8056 55.4017 6298 51.5767 8407 52.8336 6390 54.0978 3222 55.3683 2138 56.6451 6299 52.5880 8575 53.8048 495D 55.2092 3621 66.6312 9009 57.8610 5596 54.6897 8799 56.0990 6140 57.5207 1111 58.9521 1644 60.3942 5732 58.0458 8547 50.6263 4432 61.2226 0777 62.8348 3385 64.4631 8218 TABLE Vn AMOUNT OF ANNUITY OF 1 PER PERIOD , a + f) n - 1 i n sk* 1% 5% ! 1% 51 52 53 54 55 56.8928 5366 57,9290 7388 69.1704 4603 60.4189 8855 61.6687 2600 57.0283 8880 59.2180 3075 60.5141 2090 61.8166 9150 63.1257 7496 59.1985 7877 60.5430 0381 61.8070 7059 63.26S1 4287 64.6271 4870 01.8472 1424 03.3110 (1835 04.7850 0130 8H.2717 052 67.7088 3400 60.1078 1401 07.7BHS 1)215 00.4405 8107 71.1410 4088 72.8524 5735 56 57 58 59 60 62.9256 7902 64.1878 6935 65.4563 1881 66.7280 4930 68.0060 8284 64.4414 0384 65.7636 1080 67.0924 2891 68.4278 9105 69.7700 3051 86.0041 4040 67.3891 6465 68.7822 6801 70.1834 0701 71.6929 0165 00.2771 0035 70.7900 7800 72.3270 5301) 73.8701 1100 75.4241 3003 74./380I) ll2 70.3207 174 78.0000 fiOOfl 7U.870I) 0025 81.0800 0080 61 62 63 64 65 69.2894 4162 70.5781 4753 71.8722 2314 73.1716 9074 74.4765 7278 71.1188 8086 72.4744 7607 73.8368 4744 76.2060 3168 76.6820 6184 73.0105 2001 74.4364 2165 75.8706 3411 77.3132 1281 78.7642 0666 70.9898 1705 78.5072 4159 80.1564 9690 81.7670 0062 83.3708 5214 83.480.3 flflfifl 85.3212 3022 87.1744 4852 80.0401 809/5 00.0300 4882 66 67 68 68 70 75.7868 9184 77.1026 7055 78.4239 3168 79.7506 9806 81.0829 9264 77.0649 7215 79.3547 9701 80.7615 7099 82.1553 2885 83.5661 0549 80.2236 6442 81.8916 3579 83.1681 7034 84.6533 1800 88.1471 2902 84.0001 3353 80.0336 0453 88.2833 5657 80.0454 8174 91.0200 7286 02.8400 1531 04.7744 7540 00.7222 2021 08.0804 4242 100.G73 3084 71 72 73 74 75 82.4208 3844 83.7642 5860 85.1132 7634 86.4679 1500 87.8281 9797 84,9839 3602 86.4088 5570 87.8408 9908 80.2801 0448 Q0.7265 0500 87.6406 5394 89.1609 4359 90.6810 4000 92.2100 2188 93.7479 1367 03.3072 2340 05.0070 2768 96.7105 802S 98.4440 7714 100.1833 1440 102.0831 0021 104.7000 3121 100.7570 3052 108.8246 008 110.9128 4084 76 77 78 79 80 89.1941 4880 90.6657 9109 91.9431 4855 93.3262 4500 94.7151 0436 92.1801 3762 93.0410 3821 95.1092 4340 96.5847 8962 98.0677 1357 95.2947 7650 96.8506 6270 98.4166 2490 99.9897 1604 101.5729 8038 101.0346 8032 103.0991 9940 105.4760 4340 107.2080 2066 100.0725 3072 113.0210 7530 115.1521 9500 117.3037 1701 110.47(37 5418 121.0715 2172 81 82 83 84 85 96.1097 5062 97.6102 0792 98.9165 0046 100.3286 6254 101.7466 8859 90.5680 5214 101.0568 4240 102.5611 2161 104.0739 2722 105.5942 9685 103.1654 9849 104.7672 9723 106.3784 3980 107.9989 8070 109.8286 7475 110.8905 7470 112.7222 5401 114.5676 7091 116.4209 2845 118.3001 3041 123.8882 3004 120.1271 1031 128.3883 0050 130.0722 7440 132.0780 0718 86 87 88 89 90 103.1706 3312 104.6005 1076 106.0363 4622 107.4781 6433 108.9259 9002 107.1222 6834 108.0578 7968 110.2011 6908 111.7621 7492 113.3109 3580 111.2884 7710 112.9175 4322 114.5762 2889 110.2445 9022 117.0226 8367 120.1873 8130 122.0887 8675 124.0044 5206 125.9344 8004 127.8789 0400 135.3087 8712 137.0018 74flO 140.0384 0874 142.4388 7808 144.8032 0740 91 93 93 94 95 110.3708 4831 111.8397 6434 113.3057 6336 114.7778 7071 116.2661 1184 114.8774 9048 116.4618 7793 118.0341 3732 119.6243 0800 121.2224 2964 119.6106 6699 121.3082 9429 123.0159 2001 124.7335 1891 126.4611 3110 129.8380 8715 131.8118 7280 133.8004 6186 136.8039 6631 137.8224 9506 147.3110 0014 140.7850 1014 152,2828 0033 164,8056 080S 157.3537 6501 96 97 98 99 100 117.7406 1230 119.2310 9777 120.7278 9401 122.2309 2690 123.7402 2243 122.8285 4169 124.4426 8440 126.0648 9782 127.6952 2231 129,3336 9842 128.1988 2103 129.9466 4749 131.7046 6960 133.4729 4684 135.2616 3903 139.8561 6377 141.0060 8499 143.0693 7313 146.0491 4343 148.1446 1201 160.0272 0200 102.5285 0548 165.1518 3114 167.8033 4045 170.4813 8204 46 TABLE VH AMOUNT OF ANNUITY OF 1 PER PERIOD ( S at fl - <* + *)" " 1 n] n 3% i% 5% > 1% 101 102 103 104 105 125.2558 0000 120.7777 0580 128.3060 4033 120.8405 5444 131.3815 5075 i:i0.0803 0602 132.0352 0875 134.20S4 4501) 135.9H09 3732 137.0407 8701 137.0405 0034 138.8300 0020 140.0408 0870 142.4702 0508 144.3013 4253 150.2555 0585 162.3825 1281 154.5253 8100 150.6843 2202 158.8504 5444 173.1801, 0677 175.9180 5874 178.0772 3033 181.4040 1172 18-1.2780 6184 100 107 108 100 110 132.0280 7000 13-1.4828 5005 130.0431 0580 137.0100 4251 130.1834 1709 130.3380 3594 141.0347 2012 142.7308 0076 144.4536 0025 140.1758 0725 140.1431 0030 147.0050 0178 140.8580 0940 151.7330 8043 153.0181 0010 101.0500 0035 103.2587 8210 105.4832 2200 167.7243 4714 160.0822 7974 187.1214 3830 189.0020 5274 102.8025 7927 105.8215 0500 108.7707 2011 111 112 113 114 115 140.7033 4800 142.3408 6255 143.0420 8008 145.5427 4942 147.1401 7754 147.0007 4058 140.0-102 8032 151.3SJ45 1172 153.1514 8428 154.0172 4170 155.5143 0225 157.4214 0001 150.3307 0001 101.2002 4285 103.2000 8010 172.2571 4084 174.5490 7544 176.8581 9351 170.1846 2006 181.5285 1408 201.7075 1731 204.7851 0248 207.8330 4441 210.9113 7485 214.0204 8800 110 117 118 119 120 148.7022 9012 150.3821 4203 152.0087 3420 153.0421 0401 155.2822 7045 15(1.0018 2701 158.4752 8704 100.2070 0348 102.0000 0180 103.8703 4081 105.1020 3832 107.1254 8354 109.1003 8210 171.0808 0109 173.0848 0743 183.8800 7854 186.2001 6338 188.0001 7203 101.0811 0832 193.5142 7708 217.1006 0340 220.3323 0042 223.5350 2343 226.7709 7000 230.0386 8046 121 122 123 124 125 150.0202 8805 158.5831 0098 100.2430 2415 101.0110 0717 103.5802 3887 105.0087 4354 107.5272 3720 109.3048 7344 171.2110 9781 173.0077 5030 176.0944 0881 177.1158 5321 170.1490 2002 181.1940 0502 183.2510 3040 105.9066 3410 198.4363 7042 200.9236 4174 203.4305 0905 206.9662 9832 233.3300 7035 236.6724 6712 240.0301 0179 243.4305 8370 246.8739 7054 120 127 128 129 130 105.2078 4810 100.0504 0423 108.6521 1010 170.3548 3331 172.0040 4512 174.9330 0508 170.8077 0060 178.0017 0030 180.5S52 5830 182.4881 8405 185.3199 0474 187.4010 2805 180.4042 0071 101.5005 8355 103.7172 4778 208.5009 7050 211.0047 2784 213.0477 1330 210.2500 7115 218.8710 4668 250.3427 1034 253.8401 4053 257.3840 0800 200.0584 5408 204.5080 3862 181 132 133 134 135 173.7816 8114 176,5050 7106 177.2360 4400 178.0764 3100 180.7211 0203 184.4000 2567 180.3220 2870 188.2542 4184 100.1055 1305 102.1404 0002 105.8472 0500 107.0807 0744 200.1446 4740 202.3121 5785 204.4023 1210 221.5134 8628 224.1748 3743 226.8561 4871 220.5575 6082 232.2792 5100 208.2137 1000 271.8958 6619 275.6148 1476 279.3700 0200 283,1046 7263 130 137 138 139 140 182.4741 0777 184.2344 7081 180.0021 2040 187.7771 2020 180.5505 3400 104.1072 2307 100.0777 5010 108.0681 4708 200.0484 3872 202.0480 8002 200.0851 8302 208.8008 4740 211.1093 7744 213.3408 4881 215.5853 3709 235.0213 4598 237.7840 0608 240.5073 8012 243.3710 4152 240.1960 2883 286.0003 1020 200.8062 8245 204.7740 4527 208.7226 9473 302.7009 2107 141 142 143 144 145 191.3403 0530 198.1400 5441 104.0514 3214 100.7037 2077 108.5836 7805 204.0580 2432 200.0702 1804 208.1000 1504 210.1501 (1311 212.2000 1303 217.8429 1822 220.1130 0858 222.3070 0408 224.0940 8400 227.0057 0544 249.0434 0580 251.9112 3134 254.8005 0558 257.7115 6082 200.04-44 0659 300.7370 2089 310.8043 0110 314.0124 3501 310.0615 5030 323.2521 7405 146 147 148 149 150 200.4110 1023 202.2400 5010 20-1.0887 4800 205.0301 1770 207.7071 0744 214.2010 1850 2111,3332 2K09 218.4148 0423 220.5000 (1870 222.0005 0354 220.3290 0538 231.0070 0317 234.0100 5787 230.3841 0004 238.7030 7000 203.5092 3064 200.5702 3304 209.5755 5569 272.6073 7236 275,6418 6205 327.4846 0070 331.7695 4367 330.0771 3011 340.4379 1050 344,8422 8000 47 TABLE VH AMOUNT OF ANNUITY OF 1 PER PERIOD . (1 + 0" - 1 i n ll% lj% l|% lj% 2% 1 2 3 4 5 1.0000 0000 2.0112 5000 3.0338 7066 4.0080 0767 6.1137 7278 1.0000 0000 2.0125 0000 3.0376 5625 4.0756 2605 5.1265 7229 1.0000 0000 2.0150 0000 3.0462 2500 4.0000 0338 5.1522 6693 1.0000 0000 2.0175 0000 3.0528 0625 4.1002 3030 5.1780 8038 1.0000 0000 2.0200 0000 3.0004 0000 4.1210 0800 5.2040 4010 6 7 8 9 10 6.1713 0270 7.2407 2980 8.3221 8807 0.4168 1260 10.6217 4068 6.1906 6444 7.2680 3762 8.3688 8809 0.4633 7420 10.6810 6637 6.2295 6003 7.3229 0410 8.4328 3011 9.5593 3160 10.7027 2107 6.2087 0590 7.3784 0831 8.5075 3045 0.0504 1224 10.8253 0945 6.3081 2090 7.4342 8338 8.6829 0905 0.7540 2843 10.0407 2100 11 12 13 14 15 11.6401 1016 12.7710 6140 13.9147 3684 16.0712 7062 10.2408 2848 11.7130 3720 12.8603 6142 14.0211 1604 16.1003 7088 10.3803 3463 11.8632 0240 13.0412 1143 14.2368 2960 15.4603 8205 10.0821 3778 12.0148 4304 13.2251 0371 14.4505 4303 15.7095 3253 16.9844 4935 12,1687 1542 13.4120 8973 14.0803 3152 15.0730 3815 17.2934 1002 16 17 18 19 20 17.4236 3780 18.0106 '5260 19.8290 2267 21.0620 9907 22.2889 3619 17.5911 0382 18.8110 5336 20.0461 0153 21.2067 6893 22.5629 7864 17.0323 6984 10.2013 5630 20.4893 7672 21.7967 1030 23.1236 6710 18.2816 7721 19.0010 0056 20.0446 3408 22.3111 0578 23.7016 1110 18.6302 8525 20.0120 7096 21.4123 1238 22.8405 5803 24.2973 0080 21 22 23 24 25 23.6396 8671 24.8046 0717 26.0836 6788 27.3769 0790 28:6849 8913 23.8460 1577 26.1430 7847 20.4573 6605 27.7880 8403 29.1364 3508 24.4705 2211 25.8376 7904 27.2251 4364 28.0335 2080 30.0630 2361 25.1103 8038 20.5559 2620 28.0206 5400 29.5110 1037 31.0274 6015 26.7833 1719 27.2089 8354 28.8449 0321 30.4218 6247 32.0302 9972 26 27 28 29 80 30.0076 9626 31.3462 8183 32.6979 1625 34.0667 6781 36.4490 0769 30.4996 2802 31.8808 7337 33.2793 8420 34.6963 7069 36.1290 6880 31.5139 0896 32.9866 7850 34.4814 7867 35.9987 0085 37.6386 8137 32.5704 3900 34.1404 2238 35.7378 7977 37.3032 0267 30.0171 5020 33.6709 0572 35.3443 2383 37.0612 1031 38.7022 3451 40.6680 7021 31 32 33 34 85 30.8478 0903 38.2623 4688 39.6927 9829 41.1393 4227 42.0021 5987 37.5806 8210 39.0604 4069 40.5385 7120 42.0463 0334 43.5708 0903 39.1017 6159 40.0882 8801 42.2986 1233 43.0330 9152 45.5020 8780 40,6099 5042 42.4121 9955 44.1544 1305 45.0271 1527 47.7308 3979 42.3794 4070 44.2270 2961 46.1115 7020 48.0338 0100 40.0044 7703 36 87 38 39 40 44.0814 3417 46.6773 6030 47.0000 9649 48.6198 5906 60.1668 3248 45.1155 0550 46.6794 4932 48.2926 4243 49.8862 2921 51.4895 6708 47.2750 6921 48.9861 0874 50.7108 8638 52.4806 8306 54.2678 9301 40.5661 2049 51.4335 3675 53.3336 2305 55.2669 6200 57.2341 3300 51.9943 6710 54.0342 5453 56.1149 3002 58.2372 3841 60.4019 8318 41 42 43 44 45 61.7312 0934 53.3131 8546 64.9129 5879 66.5307 2957 58.1007 0028 53.1331 7654 54.7973 4125 50,4823 0801 .68.1883 3087 50.9156 9108 56.0819 1232 57.9231 4100 50.7919 8812 61.6888 6794 63.6142 0096 59.2357 3124 01.2723 5654 63.3446 2278 65.4531 5367 67.5985 8386 62.6100 2284 64.8022 2330 67.1594 0777 69.5020 6712 71.8927 1027 46 47 48 49 50 59.8210 7566 61.4940 6276 63.1858 7097 64.8967 1201 66.6268 0002 61.6646 3721 63.4354 4518 65.2283 8824 67.0437 4310 08.8817 8980 65.5084 1398 67.5519 4018 69,5662 1920 71.6086 9758 73.6828 2804 69,7815 5908 72.0027 3637 74.2627 8426 76.5623 8298 78.9022 2468 74.3305 0447 76.8171 7676 79.3535 1027 81.9405 8906 84,5704 0145 48 TABLE Vn AMOUNT OF ANNUITY OF 1 PER, PERIOD . a + o n - 1 f n 1|% l|% 1|% 1*0, J- 4 % 2% si 53 53 54 55 68.3703 5152 70.1455 8548 71.0347 2332 73.7439 8895 76.5736 0883 70.7428 1226 72.6270 9741 74.5340 3613 76.4000 2283 78.4224 5502 75.7880 7046 77.0248 9152 80.0937 6489 82.2951 7136 84.5295 0803 81.2830 1301 83.7054 6635 86.1703 1201 88.0782 9247 91.2301 6259 87.2700 8948 90.0164 0927 02.8107 3746 95.6730 7221 98.5865 3365 56 57 58 59 60 77.4238 1193 70.2048 2081 81.1868 0065 83.1002 4023 85.0361 2704 80.4027 3631 82.4077 7052 84.4378 6765 80.4933 4000 88.5745 0776 88.7075 4202 80.0005 0600 91.4359 0865 03.8075 3863 06.2140 6171 03.8266 9043 96.4680 5752 00.1508 5002 101.8921 0405 104.0752 1588 101.6582 6432 104.6894 2961 107.6812 1820 110.8348 4257 114.0515 3042 61 62 63 64 65 86.0917 7222 88.0704 2066 00.0713 4009 91.0047 7464 95.0409 6586 90.0810 8010 92.8152 1022 94.0754 0034 97.1625 0285 99.3771 2520 08.6578 7149 101.1377 3956 103.6548 0565 100.2006 2774 108.8027 7216 107.5070 3216 110.3884 0522 113.3202 0231 116.3033 0585 119.3386 1370 117.3326 7021 120.6702 2161 124.0928 0604 127.5740 6216 131.1201 5541 66 67 68 69 70 07.1101 7672 90.2026 6021 101.3186 0021 103.4585 3154 105.0224 4002 101.6103 3933 103.8896 8107 100.1882 0083 108.5155 6334 110.8719 0776 111.4348 1374 114.1063 3504 110.8179 3098 119.5701 0005 122.3037 5205 122.4270 3944 125.5605 1263 128,7669 7010 132.0204 0124 135.3307 5826 134.7480 7852 138.4430 5209 142.2125 2513 140.0507 7663 149.9779 1114 71 72 73 74 75 107.8106 9247 110.0235 6276 112.2013 2784 114.6242 6778 110.8126 6570 113.2578 9773 116.6730 2145 118.1106 4172 120.5050 3600 123.1034 8644 126.1092 0024 128.0771 0738 130.9083 6534 133.9033 3007 136.9727 8003 138.6990 4653 142.1262 7084 145.6134 8074 140.1817 2581 152.7720 6601 163.9774 6937 158.0570 1875 162.2181 5913 166.4625 2231 170.7917 7276 76 77 78 79 80 110.1268 0828 121.4600 8487 123.8334 8845 120.2260 1520 128.0466 6462 125.6422 8002 128.2128 0852 130.8154 6863 133.4506 6100 136.1187 0526 140.0273 7234 143.1277 8202 140.2740 0067 140.4088 2010 162.7108 5247 156.4455 6609 160.1833 6441 103.0865 7329 167.8563 3832 171,7938 2424 176.2070 0821 179.7117 6038 184.3059 9558 188.0921 1649 193,7719 6780 81 82 83 84 85 131.0039 3060 133.5687 4642 130.0713 0481 138.0021 0801 141.1614 7273 138.8202 8020 141.5555 3370 144.3249 7787 147.1200 4010 149.0681 5310 150.0015 1525 150.3415 3708 102.7310 0106 160.1720 3507 160.6652 2651 176.8002 1017 179.8707 1996 184.0245 6255 188.2440 9239 192.5302 7976 198.6473 9696 203.6203 4400 208,6927 6180 213.8666 0683 210.1439 3807 86 87 88 89 90 143.7495 3930 140.3607 2102 140.0133 4724 151.0807 4739 154.3962 5705 152.8427 5501 155.7632 8045 168.7002 0557 161.0830 5814 164.7050 0702 173.2102 0380 170.8083 5605 180.4004 8230 184.1073 8054 187.0200 0038 196.9087 1716 201.3646 1971 205.8783 2556 210.4811 0625 215.1646 1718 224.5268 1775 230.0173 6411 235.6177 0110 241.3300 5521 247.1506 5632 91 92 93 94 95 157.1332 1404 150.0000 6301 102.0098 4045 165.6302 2276 168.3024 3776 167.7038 2021 170.8608 6700 173.0066 2881 177.1715 8067 180.3802 3151 191.7488 4880 195.6260 8102 100.5694 5784 203.5528 4071 207.0061 4240 210.9209 0708 224.7787 7205 229.7124 0148 234.7323 6850 239.8401 8496 253.0097 8944 259.1617 8523 205.3450 2094 271.0610 2136 278.0840 5078 96 97 98 99 100 171.2808 5269 174.2138 2078 177.1737 3537 180.1069 3089 183.1038 1706 183.6410 6040 186.0305 7264 100.2732 7080 103.6510 0580 107.0723 4200 211.7202 3459 215.8000 3811 220.1344 7808 224.4364 9680 228.8030 4330 245.0373 8819 250.3255 4248 255.7062 3947 201.1810 0866 266.7517 6780 284.6400 6898 201.3305 0210 298.1603 8400 305.1207 1108 312.2323 0501 TABLE VH AMOUNT OF ANNUITY OF 1 PER PERIOD n 2|% 2ior 2% 2f% 3% 3|% i 2 3 4 5 1.0000 0000 2.0225 0000 3.0680 0025 4.1370 3839 6.2301 1971 1.0000 0000 2.0250 0000 3.0756 2600 4.1525 1603 5.2503 2852 1.0000 0000 2.0276 0000 3.0832 5625 4.1080 4580 5.2826 6706 1.0000 0000 2.0300 0000 3.0000 0000 4.1836 2700 5.3091 3581 1.0000 0000 2.0350 0000 3.1062 2500 4.2140 4288 5.3024 0588 6 7 8 9 10 6.3477 9740 7.4900 2284 8.6591 0180 0.8539 9300 11.0757 0784 6.3877 3873 7.5474 3016 8.7301 1690 9.9545 1880 11.2033 8177 0.4279 4040 7.0047 0870 8.8138 3825 10.0602 1880 11.3327 0482 6.4684 0088 7.6024 0218 S.S923 3005 10.1591 0613 11.4038 7931 0.5501 5218 7.7704 0761 0.0516 8077 10.3084 0681 11.7313 0316 11 13 13 14 Iff 12.3249 1127 13.6022 2177 14.9082 7176 16.2437 0788 17.6091 9130 12.4834 6831 13.7956 5297 15.1404 4170 16.5189 5284 17.9310 2680 12.0444 1585 13.0921 3720 15.3709 2107 10.7007 8039 18.2017 8062 12.8077 0669 14.1020 2950 16.0177 0045 17.0803 2410 18.5089 1380 13.1410 0102 14.0010 0104 10.1130 3030 17.0700 8030 19.2956 8088 16 17 18 19 20 19.0053 9811 20.4330 1957 21.8927 0251 23.3853 4966 24.9116 2003 19.3802 2483 20.8047 3045 22.3863 4871 23.0460 0743 25.5446 5701 10.7039 7948 21.3074 8892 22.8934 4487 24.5230 1460 26.1073 9750 20.1568 8130 21.7015 8774 23.4144 3537 25.1108 0844 20.8703 7449 20.9710 2971 22.7060 1575 24.4996 9130 26.3671 8050 28.2700 8181 21 22 23 24 25 26.4720 2923 28.0676 4989 29.0991 7201 31.3074 0338 33.0731 0996 27.1832 7405 28.8628 5500 30.5844 2730 32.3490 3798 34.1577 0393 27.9178 2593 29.0856 0015 31.5010 1921 33.3082 2190 35.2858 4810 28.6704 8672 30.5367 8030 32.4528 8370 34.4204 7022 30.4592 0432 30.2094 7068 32.3280 0215 34.4604 1373 30.6605 2821 38.9498 5660 26 27 28 29 30 34.8173 1628 36.0007 0590 38.4242 2178 40.2887 0077 42.1952 0402 36.0117 0803 37.9120 0073 39.8598 0075 41.8662 9577 43.9027 0316 37.2502 0802 39.2807 5407 41.3609 7542 43.4984 0224 45.0940 0830 38.3530 4225 40.7000 3352 42.0300 2252 45.2188 5020 47.5754 1671 41.3131 0168 43.7500 6024 46.2906 2734 48.9107 0930 51.8226 7728 31 32 33 34 35 44.1446 5746 46.1379 1226 48.1760 1528 50.2599 7503 52.3908 2508 46.0002 7074 48.1502 7761 50.3540 3445 52.6128 8531 54.9282 0744 47.9612 1003 50.2608 6831 52.0522 8900 55.1002 2706 57.0154 8391 60.0020 7818 52.5027 5852 66.0778 4128 57.7301 7062 60.4020 8181 54.4294 7008 57.3345 0247 60.3412 1005 63.4631 6240 00.6740 1274 86 37 38 39 40 54.5696 1804 56.7974 3500 59.0763 7735 61.4045 7334 63.7861 7024 57.3014 1263 59.7339 4794 02.2272 9064 04.7829 7900 07.4025 5354 00.1900 0972 02,8554 0724 65.5839 3094 88.3874 8904 71.2681 4490 83.2760 4427 60.1742 2259 60.1594 4927 72.2342 3275 75.4012 5973 70.0070 0318 73,4678 6930 77.0288 0472 80.7240 0004 84.5602 7775 41 42 43 44 45 60.2213 6521 68.7113 4592 71.2673 6121 73.8606 4161 76.5225 0605 70.0876 1737 72.8398 0781 75.6008 0300 78.5523 2308 81.5101 3116 74.2280 1898 77.2092 8950 80.3941 9496 83.0060 3532 86.9041 7379 78.6632 9753 82.0231 9645 85.4838 9234 89.0484 0911 92.7198 0130 88,5005 3747 92.6073 7128 90.8486 2028 101.2383 3130 105.7816 7290 46 i 47 48 49 50 79.2442 6243 82.0272 5834 84.8728 7165 87.7826 1126 90.7576 1776 84.6640 3443 87.6678 8530 90.8695 8243 94.1310 7199 97.4843 4879 90.2940 3867 93.7771 2463 07.3559 9566 101,0332 8544 104.8117 0079 06.6014 5723 100.3905 0095 104.4083 0698 108.5406 4785 112.7968 6729 110.4840 3145 115.3500 7265 120.3882 6650 125.6018 4557 130.9979 1016 50 TABLE VH AMOUNT OF ANNUITY OF 1 PER PERIOD fe fit iT = (1 H~ *')" - 1 n 2-w i/o 2|% 2|% 3% 3|% 51 58 53 54 55 03.7096 6416 06.0101 5661 100.0006 3513 103.3426 7442 106.6678 8460 100.0214 57S1 104.4444 9305 108.0558 0629 111.7560 0645 115.5500 2136 108.6940 2256 112.6831 0818 110.7818 0305 120.0933 0573 125.3207 1411 117.1807 7331 121.6001 0651 126.3470 8240 131.1374 0488 130.0716 1072 136.5828 3702 142.3632 3831 148.3459 4058 154.6380 5782 100.0468 8084 50 57 58 50 60 110.0670 1200 113.5444 4002 117.0001 8002 120.7330 2160 124.4504 3403 110.4306 9440 123.4256 8676 127.6113 2893 131.6901 1215 135.0915 8005 129.7070 3375 134.3356 2718 139.0208 5602 143.8531 7700 148.8091 4038 141.1537 6831 140.3883 8136 151.7800 3280 157.3334 3379 163.0534 3680 107.5800 3009 174.4463 3207 181.6500 1809 188.9052 0085 100.5108 S288 61 en 03 04 65 128.2505 6072 132.1362 0764 136.1002 7221 140.1717 3083 144.3255 9477 140.3013 7070 144.0011 6419 140.5236 9330 154.2617 8503 150.1183 3027 153.0013 0174 150.1336 8002 104.5098 5022 170.0338 7726 176.7098 0880 168.9450 3001 175.0133 9110 181.2037 9284 187.7017 0662 104.3327 5782 204.3949 7378 212.5487 9780 220.0880 0570 229.7225 8609 238.7028 7650 66 67 68 69 70 148.5720 2066 ] 52.0168 1137 157.3664 1713 161.8060 3651 166.6396 1758 164.0062 8853 109.1986 0574 174.4286 6314 170.7893 7971 185.2841 1421 181.5418 2863 187.5342 2892 103.6014 2021 200.0179 3427 206.5184 2746 201.1027 4055 208.1970 2277 215.4435 5145 222.0068 5800 230.5040 6374 248.1105 7718 257.8037 8238 267.8268 9400 273.2008 3535 288.0378 6450 71 72 73. 74 75 171.2867 5898 176.1407 1100 181.1038 7705 180.1787 1420 191.3677 3536 100.0162 1706 106.6891 2240 202.0063 6055 208.6715 0031 214.8882 9705 213.1076 8422 220.0006 2064 227.1122 8700 234.3578 7561 241.8027 1709 238.5118 8565 246.0072 4222 255.0672 5049 263.7102 7727 272.0308 6569 300.0508 8085 311.5524 6400 323.4568 0024 335.7777 8824 348.5300 1083 76 77 78 79 80 100.6735 0041 202.0986 6337 207.6458 8320 213.3179 1667 210.1175 0877 221.2605 0447 227.7020 1709 234.4868 1751 241.3489 8706 2-18.3827 1265 249.4522 9181 257.3122 2083 265.3883 1615 273.0804 0485 282.2128 7345 281.8007 8126 291.2040 7460 301.0019 9603 311.0320 5084 321.3630 1855 361.7288 8121 375.3890 6085 389.5276 7708 404.1011 4071 410.3067 8086 81 82 83 84 85 226.0477 1407 231.1112 8703 237.3112 0160 243.0607 9667 250.1329 3867 255,5022 8047 262.0820 8748 270.5560 3906 278.3205 5566 286.2785 6955 200.9737 2747 200,0755 0408 300.2248 3137 318.7285 1423 328.4935 4837 332.0030 0010 342.9640 2038 354.2529 4717 365.8805 3558 377.8560 5168 434.0825 2430 451.2009 1274 407.0091 5409 485.3791 2610 603.3873 9448 86 87 88 89 90 256.7609 2060 263.5380 5060 270.4676 6674 277.5531 7902 284.7081 2555 294.4355 3379 302.7064 2213 311.3663 3268 320.1604 9100 329.1542 5328 338.5271 2005 348.8366 1678 350.4200 2374 370.3139 3830 381.4976 7170 390.1026 6020 402.8084 4001 416.9853 9321 420.4040 5500 443.3489 0365 521.9852 5320 541.2547 3715 501.1080 5295 581.8400 0581 603.2060 2701 91 92 93 94 95 202.2060 8337 299.7807 2025 307.5267 8045 315,4451 1066 323.6420 3177 338.3831 0961 347.8420 8735 357.5387 6453 387.4772 2339 377.6041 5308 302.0887 5492 404.7059 4568 416.0278 3418 420.3933 4962 442.2016 0674 457.6493 7076 472.3788 5189 487.5602 1744 503.1767 2397 510.2720 2669 625.3172 0206 648.2033 0506 071,8904 2073 696.4065 8540 721.7808 1506 96 97 98 99 100 331.8223 4090 340.2883 4360 348.9448 3130 357.7060 9010 366.8465 0213 388.1057 5783 398.8084 0177 400.7786 1182 421.0230 7711 432.6486 5404 455.3622 1257 468.8846 7342 482.7790 0104 497.0554 2449 511,7244 4867 535.8501 8645 552.9250 0205 670.5134 3281 588.0288 6000 607.2877 3270 748.0431 4451 775.2246 5457 803.3575 1748 832.4750 3059 862.6116 5666 61 TABLE YD AMOUNT OF ANNUITY OF 1 PER PJJRIOD ' (r-fo-Ci + y-i "I i n 4% 4% 5% 6|% 6% l 9 3 4 5 I.OOOO 0000 2.0400 0000 3.1216 0000 4.2464 61400 .4163 2266 1.0000 0000 2.0450 0000 3.1370 2500 4.2781 9113 5.4707 0973 1.0000 0000 2.0500 0000 3.1525 0000 4.3101 2500 5.5256 3125 1.0000 0000 2.0550 0000 3.1080 2500 4.3422 6638 5.5810 9103 1.0000 0000 2.0600 0000 3.1830 0000 4.3740 1600 5.6370 9290. 6 7 8 9 10 0.6329 7540 7.8982 0448 9.2142 2020 10.6827 0631 12.0001 0712 6.7108 9166 8.0191 6170 9.3800 1362 10.8021 1423 12.2882 0937 6.8019 1281 8.1420 0845 9.5491 0888 11.0205 6432 12.5778 9254 6.8880 5103 8.2608 0384 9.7215 7300 11.2502 5051 12.8753 5370 G.0753 1864 8.3038 3705 9.8974 0791 11.4013 1508 13.1807 9494 11 It 13 14 15 13.4803 5141 15.0268 0546 16.0268 3768 18.2919 1110 20.0235 8764 13.8411 7879 15.4650 3184 17.1599 1327 18.9321 0937 20.7840 5429 14.2067 8716 15.9171 2652 17.7129 8285 19.5986 3199 21.5785 6359 14.5834 0825 10.3855 0065 18.2807 9814 20.2925 7203 22.4080 0350 14.9710 4264 16.8600 4120 18.8821 3767 21.0150 0503 23,2750 0988 16 17 18 19 20 21.8245 3114 23.0975 1239 25.6454 1288 27.0712 2940 29.7780 7858 22.7103 3073 24.7417 0089 26.8550 8370 29.0635 6246 31.3714 2277 23.0674 9177 25.8403 6636 28.1323 8467 30.5390 0391 33.0650 5410 24.6411 3099 26.9904 0269 29.4812 0483 32.1020 7110 34.8683 1801 25.6725 2808 28.2128 7970 30.9050 5255 33.7590 9170 36.7855 9120 21 23 23 24 25 31.0092 0172 34.2479 6979 36.6178 8858 30.0820 0412 41.6459 0820 33.7831 3080 36.3033 7796 38.9370 2996 41.0891 9631 44.6652 1015 35.7192 5181 38.6052 1440 41.4304 7512 44.5019 9887 47.7270 9882 37.7860 7560 40,8643 0965 44.1118 4609 47.5379 9825 51.1525 8816 39.0927 2068 43.3022 9028 46.9058 2700 50.8155 7735 64.8645 1200 28 27 28 20 30 44.3117 4462 47.0842 1440 40.0075 8298 52.9602 8630 66.0840 3775 47.5706 4460 50.7113 2301 53.9933 3317 57.4230 3316 61.0070 6966 61.1134 5376 54.0691 2645 68.4025 8277 62.3227 1191 66.4388 4750 54.0659 8051 68.0891 0943 63.2335 1046 67.7113 6353 72.4364 7797 69.1563 8272 63.7057 0608 68.5281 1102 73.0397 9832 79.0581 8022 31 32 83 34 35 59.3283 3526 62.7014 6807 66.2095 2742 60.8579 0851 73.6522 2486 64.7623 8779 68.6602 4524 72,7562 2628 77.0302 6640 81.4066 1800 70.7607 8988 75.2988 2937 80.0637 7084 85,0009 5938 90.3203 0735 77.4104 2926 82.6774 9787 88.2247 6026 04.0771 2207 100.2513 6378 84.8016 7739 90.8897 7803 97.3431 0471 104,1837 5460 111.4347 7987 86 37 38 39 .40 77.5083 1386 81.7022 4640 85.0703 3026 90.4091 4971 96.0255 1570 86.1630 6581 91.0413 4427 96.1382 0476 101.4644 2398 107.0303 2306 05.8363 2272 101.6281 3886 107.7095 4580 114,0950 2309 120.7907 7424 106.7661 8879 113.6372 7417 120.8873 2425 128.6301 2708 136.6056 1407 119.120S 0000 127.2681 18(10 135.0042 0678 145,0584 5813 154.7619 6502 41 42 43 44 45 90.8265 3033 104.8195 9778 110.0123 8169 115.4128 7696 121.0293 9204 112.8466 8760 118.9247 8854 126.2764 0402 131.9138 4220 138.8490 6510 127.8397 6295 135.2317 6110 142.9933 3866 151.1430 0550 159.7001 5587 145,1189 2285 154.1004 6360 163.5759 8910 173.5726 6850 184.1191 6527 165.0470 8356 176.0605 4457 187.5075 7724 199,7680 3188 212.7435 1370 46 47 48 49 50 126.8705 6772 132.9453 9043 139.2632 0604 145.8337 3429 152.8670 8366 146.0982 1363 153.6726 3314 161.5879 0163 109.8503 5720 178.5030 2828 168.6851 6366 178.1194 2186 188.0253 0204 108.4266 6259 209.3479 9572 105.2457 1936 206.9842 3302 219.3683 6679 232.4336 2696 246.2174 7645 226.5081 2462 241.0086 1210 256.5645 2882 272.0584 0055 290,3369 0458 52 TABLE VH AMOUNT OF ANNUITY OF 1 PER PERIOD n 4% 4% 5% 6|% 6% 51 52 53 54 55 169.7737 6700 107.1647 1708 174.8513 0630 182.8453 5805 101.1501 7200 187.6356 0455 190.9747 0040 200.8380 3408 217.1403 7262 227.0170 5038 220.8163 0550 232.8501 6528 245.4980 7354 258.7739 2222 272.7120 1833 200.7604 3706 270.1012 0072 292.2807 7309 300.3025 4601 327.3774 8562 308.7500 5880 328.2814 2230 348.0783 0773 370.9170 0020 394.1720 2057 50 57 58 59 60 100.8055 3001 208.7077 Olfil 218.1406 7197 227.8756 5885 237.0006 8520 230.1742 0750 250.9371 0080 203.2202 7053 276.0745 9711 280.4970 5398 287.3482 4924 302.7150 0171 318.8514 4470 335.7040 1703 353.5837 1788 340.3832 4733 360.4343 2503 387.5882 1386 409.9056 0502 433.4503 7173 418.8223 4810 444.0510 8905 472.0487 0040 602.0077 1782 533.1281 8080 01 02 63 04 65 248.5103 1261 250.4507 2511 270.8287 6412 282.6610 0428 204.0683 8045 303.5253 0190 318.1840 0319 333.5022 8333 349.5098 8008 366.2378 3096 372.2029 0378 391.8700 4807 412.4008 5141 434.0933 4308 458.7080 1118 458.2001 4217 484,4900 9999 512.1433 8540 541.3112 7170 572.0S33 0104 500.1168 7174 001.0828 2405 038.1477 9349 077.4360 0110 719.0828 6070 06 07 08 09 70 307.7671 1507 321.0778 0030 334.0200 1231 340.3177 4880 364.2004 5876 383.7185 3335 401.9858 0735 421.0752 3138 441.0236 1670 461.8696 7055 480.6379 1174 505.6698 0733 531.9532 9770 650.5500 0258 588.5285 1071 004.6470 7818 638.7081 1608 674.0320 1341 713.0532 7415 753.2712 0423 703.2278 3241 810.0215 0230 850.6227 9250 012.2001 0005 007.0321 0065 71 72 73 74 75 370.8620 7711 396.0505 6010 412.8088 2200 430.4147 7550 448.0313 6052 483.6538 1513 606.4182 3081 630.2070 6747 555.0663 7505 581.0443 0103 018.0540 3025 050.9020 8300 084.4478 1721 710.0702 0807 750.6537 1848 705.7011 2046 840.4640 8209 887.0002 3000 937.6132 0278 000.0764 2803 1027.0030 0083 1089.0285 8582 1150.0063 0097 1220.3060 7903 1300.0480 7077 76 77 78 . 79 80 467.6760 2118 487.2706 3003 507.7708 7347 520.0817 0841 551.2440 7675 008.1913 6822 030.5599 0934 000.2051 6790 607.1844 0052 720.6570 9864 795.4864 0440 830.2007 2402 879.0737 0085 024.0274 4889 971.2288 2134 1045.5300 3252 1104.0348 1731 1165.7567 3226 1230.8733 5264 1299.5713 8093 1380.0060 0055 1403.8050 3059 1552.0342 9278 1040.7023 5035 1746.5998 9137 81 82 83 84 85 574.2947 7582 508.2005 6085 623.1972 2052 640.1251 1870 076.0001 2345 703.3877 9497 708.7402 4576 835.0835 5680 874.2893 1680 914.0323 3012 1020.7002 6240 1072.8207 7552 1127.4712 0430 1184.8448 2752 1245.0870 6880 1372.0478 1321 .1448.5104 4204 1520.1786 1730 1014.2833 3575 1704.0639 1921 1852.3058 8485 1964.5306 3794 2083.4120 1622 2200.4167 3719 2342.0817 4142 80 87 88 89 00 704.1337 2839 733.2090 7753 763.0310 4063 796.1762 8226 827.0833 3354 960.7007 0125 1000.8463 7086 1046.8844 0381 1004.0042 0468 1145.2600 0050 1308.3414 2234 1374.7684 0345 1444.4064 1812 1617.7212 3903 1504.6073 0008 1708.7927 0977 1808.7203 0881 2004.1502 5570 2115.3848 4980 2232.7310 1000 2484.5000 4591 2634,6342 8400 2793.7123 4174 2002.3360 8225 3141.0761 8718 91 92 93 94 95 862.1020 0688 807.5807 7356 034.4002 4450 972.8608 5428 1012.7840 4846 1107.8061 1180 1252.7073 8692 1310.0792 1933 1370.0327 8420 1432.0842 5049 1676.3376 0003 1760.1045 4033 1849.1007 7080 1042.5052 6564 2040.0036 2892 2350.5312 2252 2487.1404 3970 2024.9331 0304 2770.3044 8796 2923.0712 3480 3330.5390 9841 3531.3720 8032 3744,2644 0514 3069.0000 0044 4209.1042 4901 90 97 98 99 100 1054.2060 3430 1097.4678 7677 1142.3005 0080 1180.0012 5443 1237.6237 0461 1498.1660 5117 1600.6720 2847 1038.0677 0976 1712.7808 1930 1700.8559 5027 2143.7282 0537 2251.9140 1504 2305.6103 4642 2484.7858 0374 2010.0251 5003 3085.4731 5271 3250.1741 7011 3430.2037 5580 3020.2582 6237 3820.7024 0080 4402.6506 0469 4731.4095 3486 5010.2041 0090 6318.2717 6337 6638.3080 5857 63 TABLE VH AMOUNT OF ANNUITY OF 1 PER PERIOD : (1 '+ 0" ~ 1 i n 6|% 7% 7|% 8% 8l% 1 2 3 4 1.0000 0000 2.0060 0000 3.1992 2500 4.4071 7463 5.0930 4098 1.0000 0000 2.0700 0000 3.2149 0000 4.4399 4300 5.7507 3901 1.0000 0000 2.0750 0000 3.2300 2500 4.4729 2188 6.8083 9102 1.0000 0000 2.0800 0000 3.2404 0000 4.5001 1200 5.8800 0006 1.0000 0000 2.0850 0000 3.2022 2500 4.6305 1413 6.0253 7283 7 8 9 10 7.0637 2764 8.6228 0094 10.0768 5648 11.7318 6215 13.4944 2254 7.1532 9074 8.0540 2109 10.2598 0267 11.9779 8876 13.8104 4706 7.2440 2034 8.7873 2187 10.4403 7101 12.2298 4883 14.1470 8750 7.3350 2004 8.9228 0330 10.0300 2703 12.4875 5784 14.4806 0247 7.4290 2052 0.0004 0702 10.8300 3027 12.7512 4301 1-1.8360 0032 11 12 13 14 Iff 15.3715 8001 17.3707 1141 19.4998 0765 21.7672 9515 24.1821 6933 15.7835 9932 17.8884 5127 20.1406 4286 22.5504 8780 25.1290 2201 10.2081 1900 18.4237 2799 20.8055 0759 23.3659 2000 20.1183 0470 10.6454 8740 18.9771 2040 21.4902 0058 24.2149 2030 27.1521 1303 17.0000 8270 10.6402 4079 22.2100 3003 25.0988 0559 28.2322 0016 16 1? 18 10 20 26.7540 1034 20.4930 2101 32.4100 0738 35.5167 2176 38.8253 0807 27.8880 5355 30.8402 1730 33.9990 3251 37.3789 6479 40.9954 0232 29.0772 4200 32.2580 3621 35.0773 8786 39.3531 0194 43.3046 8134 30.3242 8304 33.7502 2509 37.4502 4374 41.4402 0324 45.7019 6430 31.0320 1204 35.3207 3300 39.3220 0638 43.0054 4908 48.3770 1323 21 22 23 24 2S 42.3489 5373 46.1016 3573 00.0982 4205 54.3546 2778 58.8876 7859 44.8651 7878 49.0057 3910 53.4361 4090 68.1706 7070 63.2490 3772 47.5525 3244 52.1189 7237 57.0278 0530 62.3049 8744 67.0778 0150 50.4220 2144 65.4507 6610 00.8932 0557 00.7047 5022 73.1060 3996 53.4800 5R30 59.0360 2940 05.0530 6700 71.6832 1882 78.0077 0242 26 27 28 29 30 63.7153 7769 68.8588 7725 74.3325 7427 80.1041 9159 86.3748 6405 68.6764 7030 74.4838 2328 80.6976 9091 87.3466 2927 94.4007 8632 74.0702 0112 80.0319 1020 87.0703 0901 06.2552 5810 103.3994 0252 70.0544 1515 87.3507 0836 96.3388 2083 103.0659 3622 113.2832 1111, 80.3545 6478 04.0040 0103 103.7437 4075 113.5010 5871 124.2147 2520 31 32 33 34 35 92.9892 3021 100.0335 3017 107.5357 0903 115.5255 3076 124.0340 9026 102.0730 4137 110.2181 5426 118.9334 2506 128.2587 0481 138.2308 7835 112.1543 5771 121.6050 3454 131.6833 7003 142.5590 3310 154.2510 0558 123.3458 0800 134.2136 3744 145.0506 2044 158.626(5 7007 172.3168 0308 136.7720 7084 148.3130 7087 161.0203 4261) 170.08JJ6 7170 103.7010 71530 30 37 38 39 40 133.0969 4513 142.7482 4650 153.0268 8259 103.9736 2995 175.8319 1590 148.9134 5084 100.3374 0202 172.5010 2017 185.6402 9158 109.0351 1199 100.8204 7000 180,3320 1170 194.8669 1258 210.4711 8102 227.2585 1060 187.1021 4707 203.0703. 1081 220.3150 4540 238.0412 2103 259.0666 1871 210.0813 1780 228.0382 2081 2411.31)79 7035 271-.5008 0750 205.0825 3024 41 42 43 44 45 188.0479 9044 201.2711 0981 215.3537 3195 230,3517 2463 240.3245 8602 214.6006 6983 230.6322 3972 247.7764 9050 268,1208 5125 286,7493 1084 245.3007 5857 204.6983 1546 285.5600 8012 307.9009 0080 332.0(345 1511 280.7810 4081 304.2435 2342 320.5830 0530 356-9406 4572 386.5050 1788 321.8155 5182 350,1008 7372 380.0343 1200 414.3137 2959 450,5303 0061 46 47 48 49 50 263,3356 8475 281.4525 0426 300.7489 1704 321.2954 6606 343.1796 7198 300,7617 0260 329,2243 8598 353.2700 9300 378.9989 9051 406.6289 2047 367.0603 5376 385.8170 5528 415.7533 3442 447.9348 3451 482.6209 4709 418.4280 8677 452.9001 0211 490.1321 6428 530,3427 3742 573.7701 5642 480.8254 8032 632.4008 4015 678.7108 0107 028.9100 8418 6B3.3884 1782 TABLE Vm PRESENT VALUE OF ANNUITY OF 1 PER, PERIOD 1 - (1 + f)- z n 5% !% 5* 1% 1% 1 2 3 4 5 0.0058 5062 1.0875 0908 2.9761 7253 3.0580 7804 4.9381 0201 0.0060 2488 1.0850 9038 2.0702 4814 3.0504 0560 4.0258 6633 0.9042 0050 1.0820 3513 2.0063 3733 3.0423 4034 4.0130 7723 0.9925 5583 1.0777 221)1 2.0555 5024 3.9261 1041 4.8894 3901 0.9900 0901 1.9703 9500 2.9409 8521 3.9019 6555 4.8634 3124 7 8 10 C.9134 6318 0.8847 7061 7.8520 5969 8.8163 2915 0.7740 0104 6.8963 8441 0.8620 7404 7.8229 5924 8.7790 6302 9.7304 1180 5.8703 8084 0.8394 8385 7.7040 1876 8.7430 1781 9.6805 1316 5.8455 0703 0.7946 3785 7.7366 1325 8.6715 7042 9.5005 7058 5.7954 7047 6.7281 0453 7.6610 7775 8.5600 1768 0.4713 0463 11 12 13 11 15 10.7298 9374 11.0812 2198 12.0280 0280 13.6720 5267 14.6115 8702 10.0770 2073 11.0189 3207 12.5501 5131 13.4887 0777 14.4166 2466 10.6245 3009 11.5571 2010 12.4842 0511 13.4060 9201 14.3226 4473 10.5200 7452 11.4340 1207 12.3423 4608 13.2430 2242 14.1360 9405 10.3078 2825 11.2650 7747 12.1337 4007 13.0037 0304 13.8050 5252 16 17 18 19 20 15.4472 2418 16.3789 7843 17.3068 G048 18.2300 0438 19.1611 0809 15.3309 2602 10.2586 3186 17.1727 6802 18.0823 5624 18.9874 1915 15.2336 8100 10.1305 3432 17.0401 3354 17.0365 0974 18.8266 0320 15.0243 1201 15.0050 2402 10.7701 8107 17.0468 2084 18.5080 1000 14.7178 7378 15.5622 5127 16.3082 6858 17.2260 0850 18.0456 5297 21 22 23 24 25 20.0674 0352 20.9800 7053 21.8888 7289 22.7938 9831 23.6051 6843 19.8879 7025 20.7840 5800 21.0750 8056 22.5628 0622 23.4450 3803 10.7107 1404 20.5906 0220 21.4053 8745 22.3350 0038 23.1997 0741 10.3027 9870 20.2112 1459 21.0533 1473 21.8891 4614 22.7187 5547 18.8569 8313 10.0603 7934 20.4558 2113 21.2433 8720 22.0231 5570 26 27 28 29 30 24.5020 0884 25.4805 0500 20.3766 0254 27.2630 0008 28.1457 3278 24.3240 1704 25.1980 2780 26.0076 8033 26.0330 2423 27,7040 5307 24.0594 2070 24.0140 8802 25.7037 9970 26.6085 8307 27.4484 0702 23.6421 8005 24.3504 0280 25.1707 1251 26.0758 0331 26.7760 8021 22.7952 0300 23.5590 0759 24.3104 4310 25.0057 8530 25.8077 0822 31 32 33 34 35 20.0247 9612 20.9002 1189 30.7710 0524 31.6401 0122 32.5047 2-180 28.0/507 9907 20.5032 8355 30.3615 2502 31.1955 4818 32.0353 7132 28.2a34 8000 20.1130 5044 20.0300 0025 30.759C 7540 31.5753 8506 27.5083 1783 28.3550 5045 29.1371 2203 20.0127 7021 30.6826 5020 26.5422 8537 27.2605 8047 27.9800 9256 28.7020 6580 29.4085 8000 36 37 38 39 40 33.3057 0109 34.221)1 0181 35.0760 5084 36.9272 53U4 30.7740 2881 32.8710 1024 33.7025 0372 34.5298 5445 35.3630 8000 30.1722 2780 32.3804 0463 33.1028 3074 33.00.15 3828 34.7915 8736 35.5840 1300 31.4408 0525 32.2062 6570 32.0580 8010 33.7052 9048 34.4460 3344 30.1075 0504 30.7905 0094 31.4840 6330 32.1630 3208 32.8340 8611 41 42 43 44 45 37.0172 0009 38.4570 5236 30.2033 3013 40.1261 3788 40.9554 8090 30.0872 0141 37.7082 0091 38.0052 7364 30.4082 3238 40.2071 0640 36.3718 4487 37.1561 0670 37.0338 2012 38.7080 2020 30.4777 4248 35.1830 6646 35.0137 1200 30.6380 2070 37.3587 3022 38.0731 8130 33.4090 8922 34.1581 0814 34.8100 0800 35.4554 5352 30.0045 0844 40 47 48 40 50 41.7814 0081 42.0038 8461 43.4220 5562 44.2380 2700 46.0500 1682 41.0021 8547 41.7032 1037 42.6803 1778 43.3035 0028 44.1427 8635 40.2429 0170 41.0038 0287 41.7602 0170 42.5122 1380 43.2508 0460 38.7823 1401 39.4801 0774 40.18-17 8189 40.8781 0642 41.5804 4707 36.7272 3008 37.3530 0000 37.0730 5040 38.5880 7871 30.1001 1753 TABLE Vm PRESENT VALUE OF ANNUITY OF 1 PER PERIOD n 55* ! 3% i* !% 1% 51 S& 64 55 46.8598 3317 46.6363 0401 47.4676 1228 48.2665 0184 49.0620 7661 44.9181 9537 46.6897 4664 46.4674 5934 47.2213 5258 47.9814 4635 44.0031 7940 44.7421 8336 45.4709 0144 46.2073 6853 46.0335 7933 42.2403 7525 42.0270 1S12 43.6000 13C1 44.2685 0002 44.0310 1103 30.7081 3617 40.3041 0423 40.1)843 5072 41.5(580 (1408 42.1471 0210 50 57 58 59 60 49.8643 6003 60.6433 3612 61.4290 4840 62.2116 0046 52.9907 0684 48.7377 5057 49.4903 0605 50.2391 0950 50.9841 8855 61.7256 6075 47.6555 8841 48.3734 1020 40.0870 6898 40.7005 8880 50.6010 9304 45.5890 8020 46.2428 0770 46.8011 8388 47.6340 7382 48.1733 7352 42.7100 9224 43.2871 2102 43.8480 3408 44.40-15 8870 44.0550 3841 01 69 63 M 65 53.7668 7800 64.6394 3036 55.3089 7627 56.0753 2905 56.8386 0194 52.4632 4453 53.1972 5824 53.9276 2014 54.6643 4839 55.3774 6109 51.2033 0800 51.0005 5478 62.5937 6787 53.2829 4073 63.0681 2668 48.8073 1863 49.4365 4455 60.0610 8040 50.6809 7006 51.2962 5713 46.6000 3803 40.0306 4161 46.5730 0258 47.1028 7385 47.0200 0777 60 67 68 69 70 67.6985 0814 68.3653 6078 59.1090 7296 69.8696 5770 60.6071 2798 56.0969 7621 56.8129 1166 57.5252 8522 58.2341 1405 58.9394 1766 54.3403 3888 55.3266 0040 55.0900 3413 56.0693 6287 57.3349 0925 51.0060 5407 62.5131 0007 63.1147 4007 63.7110 0077 64.3040 2210 48.1451 5021 48.0585 7060 40,1660 0140 40.0701 0040 60.1086 1435 71 n 73 71 75 61.3614 9672 62.0027 7680 62.8309 8103 63.5661 2216 64.2982 1292 59.6412 1151 60.3395 1394 61.0343 4222 61.7267 1366 62.4136 4643 67.9965 0579 58.6644 4488 59.3084 7877 50.0587 1959 60.6061 8934 64.8920 2516 55.4708 4880 56.0564 2501 66.6316 8706 57.2020 6704 60.6018 OG39 61.1603 0148 51.0340 6007 62.1120 2175 52.5870 5124 70 77 78 79 80 66.0272 6696 65.7632 9388 66.4763 0924 67.1963 2453 67.9133 6221 63.0981 5460 63.7792 5836 64.4569 7350 65.1313 1691 65.8023 0638 61.2479 0988 61.8860 0207 62.5221 9021 63.1537 9310 63.7817 3301 57.7603 9740 58.3310 0815 68.8002 3141 50.4443 0842 50.0044 4012 63,0504 8637 53.6212 7304 53.0814 6006 64.4370 8817 54.8882 0011 81 89 83 84 85 68.6274 0467 69.3384 9426 70.0466 3326 70.7518 3393 71.4541 0846 66.4699 5561 67.1342 8419 67.7963 0705 68.4630 4244 69.1075 0491 64.4060 3118 65.0267 0874 65.6437 8667 66.2572 8585 66.8672 2705 60.5403 8722 61.0822 7010 61.6201 1030 62.1530 6460 62.0838 3570 55.3348 6753 56.7770 8008 50.2140 3720 60.0484 6270 67,0776 7000 80 87 88 89 90 72.1634 6898 72.8499 2759 73.5434 9633 74:2341 8720 74.9220 1212 69.7687 1136 70.4066 7796 71.0514 2086 71.6929 5608 72.3312 9058 67.4736 3080 68.0766 1789 68.6760 0845 69.2718 2283 69.S642 8121 63.2007 0257 63.7317 7427 64.2400 0002 64.7041 6875 65.2746 0018 57.5020 4961 57.9234 1636 58.3400 1520 58.7524 9030 50,1008 8148 91 92 98 94 95 75.6069 8300 76.2891 1168 76.9684 0995 77.6448 8955 78.3185 6218 72.9664 6725 73.5984 7487 74.2273 3818 74.8530 7282 75.4756 9434 70.4533 0363 71.0380 1001 71.6211 2017 72.1000 6370 72.7754 3047 65.7812 4081 66.2841 1802 66.7832 4458 67.2780 5467 67.7703 768B fiO,50C2 2010 50.0055 7340 60.3610 5302 60.7644 0082 01.1420 8002 90 97 98 99 160 78.9894 3960 70.6575 3308 80.3228 5450 80.9854 1524 81.6452 2677 76.0952 1825 76.7116 6995 77.3250 3478 77,9353 6799 78.5426 4477 73.3475 6967 73.9163 9076 74.4810 1204 75.0441 5530 75.6031 3712 68.2584 3856 68.7428 6705 69,2236 8038 60.7009 3230 70.1746 2272 61.5277 0200 61.0086 1082 62.2857 5023 62.6591 6756 63.0288 7877 56 TABLE Yin PRESENT VALUE OF ANNUITY OF 1 PER PERIOD 1 - (1 + f)-" z (a^ati) n or .|% 4% i% 1% i 82.3023 0049 82.9600 4777 83.6082 7991 84.2572 0818 84.0034 4381 79.1409 1021 79.7481 0937 80.3404 3718 80.9417 2864 81.5340 5825 70.1588 7702 70.7113 9392 77.2007 0048 77.8008 3331 78.3497 9288 70.6447 8682 71.1114 6094 71.6746 4113 72.0343 8325 72.4907 0298 63.3949 2947 63.7673 6691 64.1101 9397 64.4714 7918 04.8232 4071 100 107 108 100 110 85.5409 9796 80.1878 8175 80.8201 0028 87.4610 8258 88.0040 2163 82.1234 4104 82.7098 9158 83.2934 2440 83.8740 5410 84.4517 9522 78.8896 0355 79.4202 8350 79.9598 5115 80.4903 2428 81.0177 2093 72.0436 2670 73.3931 7696 73.8393 8160 74.2822 6461 74.7218 5073 66.1715 3140 66.5163 6772 66.8577 8983 06.1968 3151 60.5305 2025 111 112 113 114 115 88.7249 3437 89.3520 3171 89.9777 2460 00.0002 2364 91.2201 3059 85.0206 6191 85.5980 0850 80.1078 2942 80.7341 5802 87.2976 7027 81.5420 5895 82.0033 5600 82.5810 2991 83.0908 0803 83.0091 7785 76.1581 6450 76.6012 3027 70.0210 7223 70.4477 1437 76.8711 8052 00.8619 0718 67.1900 0710 07.5148 6852 07.8364 9358 68.1540 4414 116 11? 118 119 120 91.8374 8338 92.4522 0658 03.0644 9081 03.0741 8767 04.2813 4800 87.8683 7838 88.4162 9G90 88.9714 3970 89.5238 2059 90.0734 5333 84.1184 8071 84.0248 4182 85.1282 6033 85.0287 6920 86.1263 6554 77.2914 9431 77.7086 7922 78.1227 5863 78.5337 6536 78.9410 9207 68.4702 4172 08.7824 1755 00.0015 0252 00.3075 2725 09.7005 2203 121 123 124 125 94.8859 9030 95.4881 2315 98.0877 5747 96.6849 0367 97.2795 7209 00.6203 6167 01.1045 2892 91.7059 9893 92.2447 7505 02.7808 7070 86.6210 6602 87.1129 0742 87.6018 0038 88.0880 4946 88.5713 8308 79,3405 9322 79,7484 7962 80.1473 7432 80.5432 9957 80.9362 7749 70.0005 1080 70.2975 4145 70.5010 2520 70.8827 9722 71.1710 8036 120 127 128 120 130 97.8717 7301 98.4015 1666 99.0488 1324 99.0336 7290 100.2161 0670 03.3142 0920 03.8450 7384 04.3732 0780 04.8087 1422 05.4216 0619 80.0510 1361 89.5296 5731 00.0040 3032 00.4708 4873 00.9463 2851 81.3203 3001 81.7134 7802 82.0977 4583 82.4701 6219 82.8577 1929 71.4506 2115 71.7391 2985 72.0189 4045 72.2959 8064 72.6702 7786 131 133 133 , 134 135 100.7901 2189 101.3737 3131 101.0480 4401 102.5217 6994 103.0922 1800 95.0418 9071 96.4595 9872 90.9747 2509 97.4872 8865 97.0973 0214 01.4130 8564 01.8771 3661 92,3384 9442 02.7971 7758 03.2532 0000 83.2334 6828 83.6064 2013 83.9706 0506 84.3440 1554 84.7087 0029 72.8418 5927 73.1107 6175 73.3709 8193 73.0405 7017 73.0015 6058 130 137 138 130 140 103,0003 0104 104.2260 2590 104.7894 0335 105.3504 4314 105.0091 6490 08.5047 7825 00.0097 2900 99.5121 0875 100.0121 0821 100.5095 0041 03.7005 7892 04.1673 2787 04.0054 0270 05.0500 0857 05,4030 5050 86.0706 7020 85.4299 4667 85.7805 4657 80.1404 0288 80.4018 0434 74.1599 6095 74.4158 0293 74.0601 1181 74.0199 1268 75.1682 3038 141 143 143 144, 145 100.4655 4847 107.0190 3330 107.5714 1902 108,1209 1517 108,0081 3120 101,0045 3772 101,4970 6246 101.9871 1088 102.4747 4310 102.9500 4344 05.9343 3304 00.3721 0272 00.8074 6201 07.2402 1804 07.0704 7304 86.8405 0059 87.1806 0108 87.5301 2514 87.8710 0195 88.2005 2055 75.4140 8948 75.6676 1434 75.8985 2905 76.1371 5747 70.3734 2324 140 147 148 140 150 109.2130 7074 100.7657 0103 110.2901 9363 110.8343 8350 111.3703 4044 103.4427 2979 103.0231 1422 104.4011 0808 104.8707 2505 105.3499 7518 08.0082 3307 08.5235 1360 08,0403 2003 90.3006 8765 00.7840 1078 88.5464 2082 88.8788 3864 80.2007 0630 80.6382 2858 80.8642 4073 70.6073 4974 70.8389 0014 77.0082 7737 77.2053 2413 77.6201 2290 67 TABLE VD3 PRESENT VALUE OF ANNUITY OP 1 PER PERIOD .!-(!+ Q-" n 1|% l\% 1|% ' lf% 2% l 2 3 4 5 0.9888 7515 1.9667 4923 2.9337 4460 3.8890 8230 4.8355 8200 0.0876 5432 1.9631 1538 2.0265 3371 3.8780 5708 4.8178 3504 0.9852 2107 1.0558 8342 2.0122 0042 3.8543 8465 4.7826 4497 0.0828 0008 1.9486 0875 2.8079 8403 3.8300 4254 4.7478 5508 0.0803 0216 1.9415 6004 2.8838 8327 3.S077 2870 4.7134 5061 G 7 8 9 10 6.7706 6205 6.6953 3048 7.6097 3002 8.5130 4810 9.4081 0690 5.7460 0902 6.6627 2585 7.5681 2420 8.4623 4408 0.3455 2501 5.0071 8717 0.5982 1306 7.4859 2508 8.3605 1732 0.2221 8455 5.0489 9702 0.5346 4139 7.4050 6297 8.2604 0432 0.1012 2291 5.0014 3080 0.471!) 0107 7.3254 8144 8.1622 3071 8.9825 8501 11 13 IB 14 15 10,2923 1832 11.1666 9302 12.0313 4044 12.8863 6880 13.7318 8509 10.2178 0337 11.0793 1197 11.0301 8466 12.7706 6276 13.6005 4502 10.0711 1770 10.0075 0521 11.7315 3222 12.5433 8150 13.3432 3301 0.9274 0181 10.7396 4000 11.5370 4097 12.3220 0587 13.0028 8046 9.7808 4805 10.6753 4122 11.3483 7375 12.1002 4877 12.8402 0350 10 17 18 IB 20 14.C679 0514 15.3948 0360 16.2124 1395 17.0209 2850 17.8204 4845 14.4202 9227 15.2290 1829 16.0295 4893 16.8193 0769 17.6093 1613 14.1312 6405 14.9070 4031 15.6725 6080 10.4261 6837 17.1680 3879 13.8504 0677 14.5950 8282 15.3268 6272 10.0400 6073 16.7528 8130 13.5777 0031 14.2918 7188 14.0920 3125 16.6784 6201 16.3514 3334 21 22 23 24 25 18.0110 7387 10.8920 0371 20.1660 3580 20.0305 6603 21.6865 0276 18.3696 0495 19.1305 6291 19.8820 3744 20.6242 3451 21.3572 6865 17.9001 3073 18.0208 2437 19.3308 6145 20.0304 0537 20.7190 1120 17.4475 4010 18.1302 0948 18.8012 4764 10.4006 8505 20.1087 8106 17.0112 0010 17.0580 4820 18.2922 0412 18.0130 2600 19.5234 5047 20 27 30 22.4342 0792 23.1735 0508 23.9045 7940 24.6275 1986 25.3424 1766 22.0812 5290 22.7062 9025 23.5025 1778 24.2000 1756 24.8889 0623 21.3980 3172 22.0670 1746 22.7207 1071 23.3700 7668 24.0168 3801 20.7457 3106 21.3717 2644 21.9860 6474 22.5910 0171 23.1858 4034 20.1210 3570 20.7068 0780 21.2812 7230 21.8443 8400 22.3004 6555 31 32 33 34 35 26.0493 6233 26.7484 4236 27.4307 4522 28.1233 5745 28.7003 6460 26.5692 9010 26.2412 7418 26.9049 6215 27.5604 5644 28.2078 5822 24.6461 4582 25.2671 3874 25.8789 6442 26.4817 2840 27.0755 9468 23.7698 7650 24.3438 5807 24.0079 0951 25.4023 7780 20.0072 6100 22.0377 0162 23.4083 3482 23.0885 0355 24.4086 0172 24.0986 1933 38 37 38 39 40 20.4378 127 30.1289 0114 30.7825 0692 31,4290 2044 82.0682 5260 28.8472 6737 29.4787 8259 30.1026 0133 30.7185 1983 31.3269 3316 27.6606 8431 28.2371 2740 28.8050 5103 29.3645 8288 20.0158 4520 26.6427 6283 27.0690 4456 27.5862 8457 28.0046 2867 28.5942 2055 25.4888 4248 25.0604 5341 26.4406 4060 26.0025 8883 27.3554 7024 41 42 43 44 45 32.7903 7340 33.3254 6195 33.0435 9649 34.5548 5438 35.1593 1212 31.9278 3622 32.5213 1874 33.1074 7530 33.6863 9536 34.2581 6825 30.4580 0079 30.9040 5004 31.5212 3157 32.0406 2223 32.6623 3718 20.0862 3780 20.6078 0135 30.0420 6522 30.5081 7221 30.0602 0261 27.7994 8045 28.2347 0358 28.0615 0233 20.0790 0307 20.4901 6987 46 47 48 49 M 36.7570 4536 36.3481 2891 36.0326 3674 37.5106 4202 38.0822 1708 34.8228 8222 36.3806 2442 35.9314 8091 36.4755 3670 37.0128 7574 33.0564 8083 33.5531 0195 34.0425 6305 34.5246 8339 34.9006 8807 31.4164 7431 31.8589 4281 32.2038 0120 32.7211 8063 33.1412 0046 29.8023 1360 30.2805 8100 30.6731 1067 31.0620 7801 31.4236 0589 68 TABLE VHI PRESENT VALTTE OF ANNUITY OF 1 PER PERIOD n 1|% 1|% 1|% l|% 2% 51 52 53 51 55 38.0474 3345 30.2003 01 88 30.7CflO 7232 40.3050 3304 40.8401 1514 37.5435 8000 38.0077 3431 38.5854 1000 39.0007 0770 30.0010 8007 35.4076 7208 35.0287 4185 30.3820 0000 30.8305 3882 37.2714 0081 33.5540 1421 33.0597 1013 34.3584 4033 34.7503 1579 35.1354 4550 31.7878 4892 32.1449 4992 32.4950 4894 32.8382 8327 33.1747 8752 5G 57 58 59 00 41.3805 8358 41.0001 0013 42.4317 4806 42.0485 7740 43.4500 5033 40.1004 3128 40.6030 1855 41.0705 2449 41.5000 2410 42.0345 9170 37.7058 7803 38.1338 7058 38.6565 3761 38.9700 7202 30.3802 0880 35.5130 5135 35.8850 4727 30.2515 4623 30.6108 5520 30.9030 8552 33.5040 0305 33.8281 3103 34.1452 2050 34.4501 0441 34.7008 8008 01 ca 03 01 05 43.0050 4052 44.4048 2020 44.0600 3110 43.4477 4407 45.0310 2000 42.5033 0054 42.0002 2275 43.4234 2088 43.8749 0247 44.3200 8022 30.7835 1614 40.1808 0408 40.5722 2077 40.0678 5298 41.3377 8018 37.3110 4228 37.0521 3000 37.9873 5135 38.3108 0723 38.0405 9678 35.0590 0282 35,3520 4002 35.0398 4310 35.9214 1480 30.1974 0555 00 07 08 00 70 40.4080 1075 40.8815 0284 47.3488 2852 47.8100 5627 48.2070 4004 44.7014 0105 45.1005 0503 45.0201 7840 40.0505 4050 40.4006 7602 41.7121 0401 42.0808 0125 42.4442 2783^' 42.8021 0400 43,1548 7183 38.0588 1748 39.2715 0509 30.5780 3375 30.8810 1597 40,1779 0207 30.4081 0348 30.7334 3478 30.9035 0351 37.2485 9108 37.4986 1020 71 78 73 74 75 48.7108 4270 40,1007 1714 40.0086 2010 50.0450 0708 50.4777 32BO 40.8830 3024 -47,2024 7431 47.0002 7003 48.0050 8240 48.4880 7027 43.5023 3078 43.84-10 0077 44.1810 3771 44.5142 2434 44,8410 0034 40.4000 8321 40.7504 4542 41.0382 7500 41.3162 5867 41.5874 7771 37.7437 4441 37.9840 0314 38.2100 0975 38.4506 5002 38.0771 1433 7G 77 78 79 80 50.0050 5077 51.3270 1510 51.7464 78-17 52.1580 0317 52.5073 1002 48.8770 0533 40.2022 1701 40.0410 0040 50.0104 0027 50.3800 5700 45.1041 3820 45.4810 0002 45.7040 8485 40.1034 3335 40.4073 2340 41.8550 1495 42.1179 5081 42.3703 0443 42.0303 3359 42.8709 3474 38.8091 3170 39.1107 9578 39.3301 0194 30.5304 0380 39.7445 1359 81 88 83 84 85 52.0713 8280 63.3709 5057 53.7000 0104 54.1508 2074 54.5432 1557 50,7522 5380 51.1133 3717 51,4099 0204 51.8221 8532 62.1700 5058 40,7007 2205 47.0010 0720 47.2023 1251 47.5780 3301 47.8007 2218 43,1252 4298 43.3003 3217 43,0032 7480 43.8301 4237 44.0060 0470 30.0450 0160 40.1427 40B3 40.3300 2011 40.5255 1570 40.7112 8090 80 87 88 89 90 54.0253 0588 55.3031 4540 65.0707 8100 50.0402 0120 50.4110 30-11 62.5130 3000 52.8520 7088 53.1881 2631 53.5101 3011 53.8400 0035 48.1380 4254 48.4124 5571 48.0822 2237 48.0480 0234 40.2008 5452 44.2809 3090 44.5109 8869 44.7282 4441 44.9417 0355 45.1510 1037 40.8034 2156 41.0710 8192 41.2470 4110 41.4180 0774 41.5860 2010 91 99 93 94 95 50.7720 3400 57.1302 1002 57.4835 3021 57.8320 0007 58.1784 0204 54.1080 4850 54.4878 5037 54.8028 151S 55.1138 0154 55.4211 2744 40.4078 3000 40.7220 0080 40,0724 2065 50.2101 3365 50.4022 0054 45.3578 4803 45.6005 3800 45,7507 4310 45.0655 2147 40.1479 3205 41.7618 9133 41.9130 1895 42.0721 7645 42.3270 2200 42. .'1800 2264 90 97 98 90 190 58.5200 5235 58.8579 0000 50.1010 0100 50.5223 0440 50.8400 0251 55.7245 7031 60.0242 0008 50.3202 0308 60.0126 0010 60.0013 3030 50.7010 7541 50.0370 1124 61.1700 0034 61.3000 7422 61.0247 03(17 40.3370 3455 40.5228 8408 40.7065 3718 40.8850 4882 47.0014 7304 42.5204 3380 42.0760 1555 42.8105 2506 42.9003 1807 43.0983 5104 59 TABLE vm PRESENT VALUE OF ANNUITY OF i PER PERIOD .!-(!+ i)-* n 2 -or 4% 2|% Q?0/ "l'0 3% 8l% l 2 3 4 5 0.9779 9511 1.9344 6955 2.8698 9687 3.7847 4021 4.6794 5253 0.9766 0976 1.9274 2415 2.8660 2356 3.7019 7421 4.6458 2850 0.9732 3001 1.9204 2434 2.8422 6213 3.7394 2787 4,6125 8136 0.9708 7379 1.9134 6970 2.8280 1135 3.7170 9840 4.6797 0719 0.9661 8357 1.8996 0428 2.8010 3608 3.6730 7921 4.6150 5238 6 7 8 9 10 6.5544 7680 6.4102 4626 7.2471 8461 8.0657 0622 8.8602 1635 6.5081 2536 6.3493 9060 7.1701 3717 7.9708 6653 8.7520 6393 5.4623 6078 6.2894 0800 7.0943 1441 7.8770 7820 8.6400 7616 5.4171 9144 6.2302 8200 7.0196 9219 7.7861 0892 8.5302 0284 C.3285 5302 0.1146 4398 0.8730 G6G4 7.0070 8051 8.3106 0532 ii 12 13 14 15 9.6491 1134 10.4147 7882 11.1635 9787 11.8959 3924 12.6121 6551 9.5142 0871 10.2577 6460 10.9831 8497 11.6909 1217 12.3813 7773 0.3820 6920 10.1042 0360 10.8070 1080 11.4910 0814 12.1500 9892 9.2520 2411 9.9540 0399 10.0349 6633 11.2900 7314 11.9379 3509 9.0015 5104 9.0033 3433 10,3027 3849 10.9205 2028 11.6174 1000 16 17 18 19 20 13.3126 3131 13.9976 8343 14.6676 6106 15.3228 9590 15.9637 1237 13.0550 0206 13.7121 0772 14.3633 6363 14.9788 9134 15.5891 6229 12.8046 7316 13.4361 0769 14.0487 0061 14.0400 0157 15.2272 5213 12.6011 0203 13.1061 1847 13.7535 1308 14.3237 9911 14.8774 7486 12.0941 1681 12.6513 2059 13.1896 8173 13.7098 3742 14.2124 0330 21 22 23 24 25 16.6904 2776 17.2033 6232 17.8027 8955 18.3890 3624 18.9623 8263 16.1845 4857 16.7654 1324 17.3321 1048 17.8849 8583 18.4243 7642 16.7929 4612 16.3434 9987 10.8793 1861 17.4007 9670 17.9083 1795 15.4150 2414 15.9369 1664 16.4430 0839 10.9355 4212 17.4131 4709 14.0979 7420 15.1071 2484 16.0204 1047 10,0683 0760 10.4815 1450 26 27 28 29 30 19.5231 1260 20.0716 0376 20.0078 2764 21.1323 4977 21.6453 2985 18.9506 1114 19.4640 1087 19.9648 8866 20.4536 4991 20.9302 9269 18.4022 5592 18.8829 7413 19.3508 2040 19.8061 5708 20.2493 0130 17.8768 4242 18.3270 3147 18.7641 0823 19.1884 5459 19.6004 4135 10.8903 5220 17.2853 6451 17.6070 1886 18.0367 6700 18.3620 4541 31 82 33 34 35 22.1470 2186 22.6376 7419 23.1175 2977 23.5868 2618 24.0467 9577 21.3954 0741 21.8491 7796 22.2918 8094 22.7237 8628 23.1451 5734 20.6805 8520 21.1003 2023 21.5088 3332 21.9064 0712 22,2933 4026 20.0004 2849 20.3887 0553 20.7067 9178 21.1318 3068 21.4872 2007 18.7302 7670 19.0088 6547 19.3902 0818 19.7006 8423 20.0000 6110 36 37 38 89 40 24.4946 6579 24.9336 5848 25.3629 9118 25.7828 7646 26.1936 2221 23.5662 5107 23.9573 1812 24.3486 0304 24.7303 4443 26.1027 7505 22.6099 1753 23.0364 1009 23.3931 0568 23.7402 4884 24.0781 0106 21.8322 5250 22.1672 3544 22.4924 6159 22.8082 1513 23,1147 7197 20.2004 0381 20.5705 2542 20.8410 8736 21.1024 0087 21.3550 7234 41 42 43 44 45 26.5951 3174 26.9879 0390 27.3720 3316 27.7477 0969 28.1161 1950 25.4661 2200 25.8206 0683 26.1604 4569 26.5038 4945 26.8330 2386 24,4009 1101 24.7269 2069 25.0383 0563 25.3414 7607 25.0304 7209 23.4123 9997 23.7013 5920 23.9819 0213 24.2542 7392 24.5187 1254 21.6901 0371 21.8348 8281 22.0626 8870 22.2827 0102 22.4964 5026 46 47 48 49 50 28.4744 4450 28.8258 6259 29.1696 4777 29.5066 7019 29.8343 9627. 27.1541 6962 27.4674 8256 27.7731 6371 28.0713 6947 28.3623 1168 26.9236 7381 26.2029 9154 26.4749 3094 26.7396 9215 26.9971 6998 24.7764 4907 25.0247 0783 25,2667 0664 26,5016 5693 25.7297 6401 22.7000 1813 22.8994 3780 23.0912 4425 23.2766 0450 23.4550 171J7 60 TABLE Vm PRESENT VALUE OF ANNUITY OF 1 PER PERIOD . 1 - (I + z)- n n 2j% 2|% 2-07 4/0 3% 3|% 51 52 53 54 55 30.16fi8 8877 30.4703 0687 30.7778 0023 31.0786 3910 31.3726 6438 28.6461 5774 28.0230 8072 29.1032 4048 20.4508 2876 20.7130 7928 27.2478 5400 27.4018 2871 27.7202 7368 27.9603 6368 28.1852 0870 25.9512 2710 26.1602 3990 26.3740 0028 20.5776 6047 26.7744 2764 23.6280 1630 23.7957 6454 23.9572 6043 24.1132 9510 24.2040 5323 50 57 58 59 60 31.6602 0708 31.0416 1142 32.2167 3489 32.4868 0420 32.7480 5285 20.9048 5784 30.2090 1740 30.4484 0722 30.6813 7200 30.0086 5649 28.4041 5454 28.0171 8203 28.8245 0800 20.0262 8522 20.2226 6201 26.0654 6373 27.1500 3506 27.3310 0540 27.5058 3058 27.6765 0307 24.4097 1327 24.5504 4700 24.6864 2281 24.8177 9981 24.9447 3412 61 02 03 04 65 33.0063 1086 33.2680 0673 33.5041 6208 33.7440 0170 33.0803 4406 31.1303 0667 31.3467 2836 31.5577 8377 31.7636 0148 31.9045 7705 20.4137 8298 20.5007 8870 20.7808 1634 20.0560 9887 30.1284 6605 27.8403 5307 28.0003 4270 28.1656 7201 28.3064 7826 28.4528 0152 25.0073 7596 25.1858 7049 25.3003 5700 25.4109 7388 25.5178 4010 00 07 08 09 70 34.2106 0643 34.4367 0003 34.6560 3005 34.8714 3183 35.0820 8402 32.1605 0208 32.3517 6876 32.5383 1009 32.7203 0340 32.8078 6608 30.2053 4400 30.4577 5581 30.0168 2074 30.7000 5522 30.0193 7247 28.5050 4031 28.7330 4884 28.8670 3771 28.9071 2390 29.1234 2135 25.6211 1030 25.7208 7051 25.8172 7480 25.9104 1052 26.0003 9604 71 72 73 74 75 35.2881 0261 35.4805 8001 35.6866 3750 35.8703 5214 36.0678 2605 33.0710 7098 33.2400 7803 33.4049 5417 33.5658 0805 33.7227 4044 31.0060 8270 31.2068 9314 31.3440 0816 31.4792 2036 31.0099 5568 20.2400 4015 20.3650 8752 29.4806 6750 20.5028 8106 29.7018 2628 28.0873 3975 20.1713 4275 26.2525 0508 26.3309 2278 26.4066 8868 70 77 78 79 80 36.2521 5262 36.4324 2310 36.6087 2675 36.7811 5085 30.0407 8070 33.8758 4433 34.0252 1398 34.1700 4047 34.3131 1205 34.4518 1722 31.7371 8304 31.8010 0540 31.0815 1377 32.0087 0085 32.2120 4008 29.8076 9833 29.9102 8004 30.0099 8004 30.1067 8035 30.2007 6346 26.4708 9244 26.5506 2072 26.6189 5721 26.6840 8281 20.7487 7667 81 82 83 84 85 37.1147 0004 37.2750 0020 37.4337 3130 37.6880 0127 37.7388 7655 34.6871 3876 34.7191 5070 34.8479 6074 34.0730 2023 35.0062 1486 32.3240 3015 32.4321 4613 32.6373 6850 32.6307 7469 32.7304 4000 30.2020 0335 30.3805 8577 30.4666 8813 30.5600 8556 30.6311 5103 26.8104 1127 26.8600 6258 20.9275 0008 26.9830 9186 27.0368 0373 80 87 88 89 90 37.8864 3183 38.0307 4018 38.1718 7304 38.3000 0028 38.4448 0025 35.2158 1938 35.3325 0671 35.4463 4801 35.5674 1269 35.6657 6848 32.8364 3804 32.0308 3094 33.0227 1627 33.1121 3165 33.1901 5480 30.7008 5637 30.7862 6735 30.8604 5374 30.9324 7036 31.0024 0714 27.0886 0020 27.1388 3986 27.1872 8480 27.2340 9108 27.2703 1664 91 92 93 94 95 38.5760 0078 38.7060 2423 38.8322 0754 38.0557 0221 30,0765 6040 35.7714 8144 35.8746 1604 35.0752 3510 36.0734 0010 36.1601 7080 33.2838 4005 33.3602 7644 33,4404 9776 33.6245 7202 33.6005 5671 31.0702 9820 31.1362 1184 31.2002 0567 31.2623 3560 31.3220 5502 27.3230 1028 1 27.3652 2732 27.4000 1673 27.4454 2680 27.4835 0415 90 97 98 99 too 30.1046 8800 30.3102 0020 30.4231 8748 30.6336 7068 30.6417 4052 30.2626 0574 38.3637 6170 36.4420 0434 36.6294 5790 36.6141 0520 33.674B 0775 33.7404 7056 33.8105 2612 33.8846 0508 33.0510 4232 31.3812 1034 31.4380 7703 31.4032 7807 31.6408 7250 31.5089 0534 27.6202 9387 27.5558 3048 27.5001 8308 27.0233 6529- 27.6554 2540 TABLE Vm PRESENT VALUE OP ANNUITY OP 1 PER PERIOD n 4% 4|% 5% 6 1% 6% 1 a 3 4 5 0.0615 3846 1.8860 9467 2.7760 0103 3.6298 9622 4.4518 2233 0.0560 3780 1.8726 6775 2.7480 6435 3.5876 2570 4.3800 7674 0.0523 8005 1.8594 1043 2.7232 4803 3.5450 5050 4.3204 7607 0.9478 0730 1.8403 1071 2.0970 3338 3.5051 6012 4.2702 8448 0.0433 0623 1.8333 9207 2.0730 1105 3.4051 0501 4.2123 0370 6 7 8 9 10 5.2421 3686 6.0020 5407 6.7327 4487 7.4353 3161 8.1108 9578 5.1678 7248 5.8927 0094 6.5958 8007 7.2687 9050 7.0127 1818 5.0750 0206 5.7863 7340 6.4632 1276 7.1078 2108 7.7217 3493 4.0955 3031 6,6820 0712 6.3345 6500 6.9521 0525 7.5376 2583 4.0173 2433 5.5823 8144 0.2097 0381 0.8010 0227 7.3000 8705 11 in IS 11 16 8.7604 7671 9.3850 7376 9.0856 4785 10.5631 2293 11.1183 8743 8.5280 1692 9.1185 8078 9.6828 5242 10.2228 2528 10.7305 4573 8.3064 1422 8.8632 5164 0.3935 7299 9.8986 4094 10.3706 5804 8.0925 3033 8.0185 1786 9.1170 7853 9.5806 4790 10.0375 8004 7.8808 7458 8.3838 4304 8.8520 8290 9.2040 8303 0.7122 4800 16 17 18 19 99 11.6522 9561 12.1656 6885 12.6592 9607 13.1339 3940 13.5903 2634 11.2340 1505 11.7071 9143 12.1599 9180 12.5032 9359 13.0070 3645 10.8377 6050 11.2740 6626 11.6805 8690 12.0853 2086 12.4622 1034 10.4621 6203 10:8046 0856 11.2400 7447 11.6070 5352 11.0503 8240 10.1058 0527 10.4772 6000 10.8270 0348 11.1581 1040 11,4609 2122 21 22 23 24 26 14.0201 5095 14.4611 1533 14.8568 4167 15.2460 6314 15.6220 7904 13.4047 2388 13.7844 2476 14.1477 7489 14.4954 7837 14.8282 0896 12.8211 5271 13.1630 0268 13.4885 7388 13.7080 4170 14.0030 4457 12.2752 4400 12.5831 0073 12.8760 4240 13.1516 9806 13.4139 3206 11.7640 7002 12.0415 8172 12.3033 7808 12.6503 5763 12.7833 5010' 28 27 28 29 30 15.9827 6918 16.3295 8575 16.6630 6322 16.0837 1463 17.2020 3330 15.1466 1145 15.4513 0282 15.7428 7361 16.0218 8853 16.2888 8854 14.3751 8530 14.6430 3362 14.8981 2726 15.1410 7358 15.3724 5103 13.6624 9541 13.8080 9091 14.1214 2172 14.3331 0116 14.5337 4517 13.0031 6019 13.2105 3414 13.4061 0428 13.5007 2102 13.7048 3115 31 32 38 34 36 17.5884 9356 17.8735 5150 18.1476 4687 18.4111 9776 18.6646 1323 16.5443 9095 16.7888 9086 17.0228 6207 17.2467 5796 17.4610 1240 15.5028 1050 15.8026 7667 16.0025 4921 10.1020 0401 16.3741 0420 14.7230 2907 14.0041 0817 15.0750 6936 15.2370 3257 15.3005 5220 13.0200 8509 14.0840 4330 14.2302 2001 14,3081 4114 14.4082 4(530 39 37 38 39 10 18.9082 8195 19.1426 7880 19.3678 6423 19.5844 8484 19.7927 7388 17.6660 4058 17.8622 3970 18.0400 0023 18.2296 5572 18.4015 8442 16.5468 5171 16.7112 8734 10.8678 0271 17.0170 4067 17.1590 8635 15.5360 6843 15.6739 0851 15.8047 3793 15.0286 6154 16.0461 2469 14.0200 8713 14.7307 8031 14.8460 1010 14.9400 7408 15.0402 0087 41 42 48 44 46 19.9930 5181 20.1856 2674 20.3707 9494 20.5488 4120 20.7200 3970 18.5661 0040 18.7235 4075 18.8742 1029 19.0183 8305 19.1563 4742 17.2943 6706 17.4232 0758 17.5450 1108 17.6627 7331 17.7740 6082 16.1574 6416 16.2620 0920 16.3630 3242 16.4578 6003 16.5477 2572 15.1380 1592 15.2245 4332 16.3001 7294 15.3831 8202 15.4558 3200 4ft 47 48 49 50 20.8846 5356 21.0429 3612 21.1951 3088 21.3414 7200 21.4821 8462 19.2883 7074 10.4147 0884 10.5356 0654 10.6512 0813 19.7620 0778 17.8800 6650 17.9810 1571 18.0771 6782 18,1687 2173 18.2550 2546 16.6329 1537 16.7136 6386 16.7902 0271 16.8027 5139 16.9315 1700 15.5243 0000 15.5800 2821 15.6500 2061 15.7075 7227 15.7618 6064 62 TABLE Vm PRESENT VALUE OF ANNUITY OF 1 PER PERIOD n 4% 4|% 6% 6-<y "3 JO 6% 51 52 63 54 55 21.6174 8521 21.7476 8103 21.8720 7403 21.0029 6007 22.1080 1218 10.8070 5003 10.0003 3017 20.0003 4400 20.1501 8140 20.2480 2057 18.3380 7663 18.4180 7298 18.4034 0284 18.5051 4566 18.6334 7100 10.0066 0943 17.0584 8287 17.1170 4538 17.1725 5486 17.2251 7048 16.8130 7007 15.8013 0252 15.9009 7408 15.9409 7554 15.9006 4207 50 57 58 59 60 22.2180 1040 22.3267 4043 22,4205 0076 22.5284 2057 22.0234 8007 20.3330 3404 20,4143 8004 20.4922 3002 20.5007 3303 20.6380 2204 18.6085 4473 18.7006 1870 18.8195 4170 18.8767 5400 18.0292 8952 17.2760 4311 17.3223 1575 17.3071 2393 17.4006 9614 17.4498 5416 10.0288 1412 16.0649 1808 16.0080 8017 10.1311 1337 10.1614 2771 61 62 63 64 65 22.7148 0421 22.8027 8280 22.8872 0124 22.9086 4027 23.0400 8109 20.7062 4118 20.7715 2266 20.8330 0298 20.8037 7310 20.0500 7913 18.9802 7674 19.0288 3404 19.0750 8003 10.1101 2384 10.1010 7033 17.4880 1343 17.5241 8334 17.6684 0762 17.5900 6457 17.0217 0737 16.1000 2614 10.2170 0579 10.2424 5829 16.2604 7009 10.2801 2272 66 67 68 69 70 23.1218 0001 23.1040 4770 23.2035 0740 23.3302 0558 23.3045 1408 21.0057 2106 21.0581 0084 21.1082 3021 21.1502 0000 21.2021 1187 19.2010 1930 19.2300 0000 10.2753 0101 10.3008 1048 10.3420 7005 17.6600 6433 17.6780 3017 17.7048 7125 17.7297 3570 17.7533 0400 10.3104 0314 16.3306 5300 10.3406 7349 16.3076 1650 16.3845 4387 71 72 73 74 75 23.4502 6440 23.5150 3885 23.6727 2060 23.0270 2408 23.0804 0834 21.2460 4007 21.2880 7662 21.3283 0298 21.3067 0711 21.4030 3360 10.3730 7776 19.4037 8834 10.4321 7037 10.4502 1845 19.4840 0005 17.7750 4300 17.7008 1804 17.8168 8970 17.8350 1441 17.8539 4731 16.4005 1308 10.4155 7838 10.4207 0093 16.4431 0899 10.4558 4810 76 77 78 79 80 23.7311 0187 23.7700 0333 23.8268 8782 23.8720 0752 23.0163 0185 21.4388 8383 21.4720 1011 21.5048 0579 21.5357 8545 21.5003 4403 19.5094 0510 19.5328 5257 19.5560 0708 10.5762 8361 10.5004 0048 17.8710 4010 17.8872 4180 17.9026 9887 17.9171 6532 17.0300 6291 16.4077 8123 16.4790 3889 16.4800 6033 16.4096 7802 16.5091 3077 81 82 83 84 85 23.0571 0754 23.0072 1870 24.0357 8730 24.0728 7240 24.1085 3110 21.6936 3151 21.0207 0001 21.6466 0288 21.6713 0032 21.0951 1035 10.6156 7606 10.0339 7776 10.0614 0730 19.0080 0704 10.0838 1623 17.9440 3120 17.9564 2708 17.0081 7789 17.9703 1554 17.0808 7256 16.5180 4700 16.5264 6028 16.5343 0640 16.6418 8348 16.5480 4608 86 87 88 80 90 24.1428 1842 24.1767 8094 24,2074 8746 24,2370 6870 24.2872 7750 21.7178 0806 21.7306 3000 21.7003 1588 21.7802 0658 21.7002 4075 19.0088 7200 10.7132 1200 19.7208 6857 19.7308 7483 10.7522 6174 17.0008 7910 18.0003 6410 18.0183 5400 18.0208 7645 18.0340 5308 10.6550 1008 16.5018 9030 10.5678 2670 16.5734 2141 16.5780 0944 91 92 93 94 95 24.2954 5023 24,3226 6005 ' 24.3480 1245 24.3730 0682 24.3077 5650 21.8174 5626 21.8348 8542 21.8515 0400 21.8075 2631 21.8828 0030 10.7040 5880 10.7752 0410 10.7850 0438 10.7001 8612 10.8058 0059 18.0426 1041 18.0498 0700 18.0507 4062 18.0632 0094 18.0604 4734 10.5830 7872 16,5883 7615 16.5028 0700 16.6060 8830 16.0000 3244 96 97 98 99 100. 24.4200 1884 24.4431 0110 24.4640 0002 24.4851 08tt 24,5049 0000 21.8074 1065 21.0114 0340 21.0247 8704 21,0375 0012 21.0408 5274 10.8151 3300 10.8230 3705 10.8323 2100 10.8403 0571 10,8470 1020 18.0753 0553 18.0808 5833 18.0801 2104 18.0911 1055 18.0058 3039 16.0040 5325 16.6081 0344 10,6114 7494, 16.0145 9000 10.6175 4623 TABLE VET PRESENT VALUE OP ANNUITY OF 1 PER PERIOD z) 1 - (1 + n B|% 7% 7|% 8% 8|% 1 3 3 4 5 0.9389 6714 1.8200 2642 2.0484 7551 3.4257 0800 4,1556 7044 0.9345 7944 1.8080 1817 2.6243 1604 3.3872 1120 4.1001 9744 0.0302 325(5 1.7056 0517 2.6005 2574 3.3403 2027 4.0458 8490 0.0250 2503 1.7832 0475 2.5770 9090 3.3121 2084 3.0027 1004 0.9210 5890 1.7711 1427 2.6640 2237 3.2756 0606 3.0406 4208 6 7 8 10 4.8410 1356 5.4845 1977 0.0887 5000 0.6501 0419 7.188S 3022 4.7065 3900 5.3802 8040 6.9712 9851 6.5152 3225 7.0235 8154 4.6038 4042 5.2066 0132 5.8673 0355 6.3788 8703 6.8640 8000 4.0228 7006 5.20U3 7006 5.7400 3804 0.2468 8701 0.7100 8140 4.6535 8717 5.1185 1352 5.6301 8207 6.1100 6264 6.5613 4800 11 13 13 14 15 7.6800 4240 8.1587 2532 8.5907 4208 9.0138 4233 9.4026 6S85 7.4086 7434 7.9426 8630 8.3676 5074 8.7464 0799 9.1079 1401 7.3164 2415 7.7352 7827 8.1258 4020 8.4801 5373 8.8271 1074 7.1389 6426 7,5300 7802 7.0037 7694 8.2442 3098 8.5604 7800 6.0680 8430 7.3440 8607 7.6000 6400 8.0100 0668 8.3042 3658 16 17 18 19 20 9.7677 6418 10.1105 7670 10.4324 6638 10.7347 1022 11.0185 0725 9.4466 4860 9.7032 2299 10.0590 8691 10.3355 0524 10.5940 1425 0.1415 0674 9.4330 5976 9.7000 0008 0.0590 7821 10.1044 0130 8.8513 6016 0.1216 3811 9.3718 8714 0.6035 0020 0.8181 4741 8.5753 3326 8.8251 0104 0,0554 7044 0.2077 2022 0.4633 3001 21 22 23 24 25 11.2840 8333 11.5351 9562 11.7701 3673 11.9007 3871 12,1078 7672 10.8355 2733 11.0612 4050 11.2721 8738 11.4693 3400 11.6535 8318 10.4134 8033 10.0171 0101 10.8060 8931 10.0829 6680 11.1460 4586 10.0168 0310 10.2007 4300 10.3710 5805 10.6287 5828 10.6747 7019 0,6436 2821 0.8097 0550 0.0629 4524 10.10-10 0700 10.2341 0078 26 Vt 28 29 30 12.3023 7251 12./5749 9766 12.7464 7668 12.9074 8084 13.0580 7591 11.8257 7867 11.9867 0904 12.1371 1125 12.2776 7407 12.4000 4118 11.2094 8452 11.4413 8005 11.6733 7703 11.6961 6524 11.8103 8627 10.8090 7706 10.9361 0477 11.0510 7840 11.1684 0001 11.2577 8334 10.3540 0288 10.4646 0174 10.5004 5321 10.0003 2564 10.7408 4382 81 82 33 84 35 13.2006 3465 13.3339 2025 13.4590 8850 13.6766 0892 13.0869 5673 12.5318 1410 12.6465 5532 12.7537 0002 12.8540 0936 12.9470 7230 11.9166 3839 12.0154 7767 12.1074 2009 12.1929 4076 12.2725 1141 11.3407 0030 11.4340 0944 11.5138 8837 11.5800 3307 11.6545 0822 10,8205 8410 10.0000 7767 10.0078 1343 11.0302 4270 11,0877 8137 36 37 38 39 40 13.7905 6970 13.8878 5887 13.9792 1021 14.0049 8611 14.1455 2687 13.0352 0776 13.1170 1660 13.1934 7345 13.2649 2846 13.3317 0884 12.3465 2224 12.4153 6063 12.4704 1361 12.5380 8031 12.6044 0860 11.7171 0270 11.7751 7851 11.8288 6809 11.8785 8240 11.0246 1333 11.1408 1233 11.1806 8878 11.2347 3020 11.2702 6467 11.3145 2034 41 42 43 44 45 14.2211 5190 14.2921 6149 14.3588 3708 14.4214 4327 14.4802 2842 13.3941 2041 13.4524 4898 13.5069 6167 13.5579 0810 13.6055 2159 12.6459 6155 12.6939 1772 12.7885 2811 12.7800 2015 12.8186 2898 11.0672 3467 12.0000 0867 12.0432 3951 12.0770 7302 12.1084 0160 11.3407 8833 11.3822 0339 11.4122 5107 11.4308 0367 11.4653 1206 46 47 48 50 14.5354 2575 14.5872 5422 14.6350 1046 14.0816 1451 14.7246 2067 13.6500 2018 13.6916 0764 13.7304 7443 13.7667 9853 13.8007 4629 12.8645 3858 12,8879 4287 12.0190 1662 12.9479 2244 12.9748 1167 12.1374 0880 12.1642 0741 12.1801 3649 12.2121 6341 12.2334 8464 11.4887 6086 11,5103 8420 11.6303 0802 11.6486 7099 11.6656 0538 64 TABLE IX PERIODICAL PAYMENT OF ANNUITY WHOSE PRESENT VALUE IS 1 1 = 1 = ; , 1 (a-, ati) 1 - (1 + i)- (s-, at 2) n| * n/ n fi% |% % 1% 1% 1 9 3 4 5 1.0041 6687 0.6031 2717 0.3361 1406 0.2626 0068 0.2026 0693 1.0050 0000 O.C037 6312 0.3360 7221 0.2531 3279 0.2030 0097 1.0058 3333 0.5043 7024 0.3372 2976 0.2536 5044 0.2035 1367 1.0075 0000 0.5050 3200 0.3383 4579 0.2547 0501 0.2045 2242 1.0100 0000 0.5075 1244 0.3400 2211 0.2502 8109 0.2060 3980 G 7 8 10 0.1601 0564, 0.1452 4800 0.1273 5512 0.1134 3876 0.1023 0506 0.1605 0646 0.1457 2864 0.1278 2886 0.1139 0738 0.1027 7057 0.1700 8504 0.1462 0086 0.1283 0351 0.1143 7608 0.1032 3632 0.1710 6891 0.1471 7488 0.1202 5552 0.1153 1020 0.1041 7123 0.1725 4837 0.1486 2828 0.1306 0020 0.1167 4037 0.1056 8208 11 12 13 14 15 0.0031 9767 0.0866 0748 0.0791 8532 0.0736 8082 0.0680 1045 0.0038 5903 O.OSOO 6643 0.0700 4224 0.0741 3BOO 0.0003 6436 0.0941 2176 0.0865 2675 0.0801 0004 0.0746 0205 0.0608 1990 0.0950 5094 0.0874 5148 0.0810 2188 0.0755 1146 0.0707 3639 0.0964 6408 0.0888 4870 0.0824 1482 0.0769 0117 0.0721 2378 16 17 18 19 20 0.0647 3055 0.0010 5387 0,0677 8063 0.0548 5191 0.0522 1630 0.0651 8037 0.0015 0579 0.0582 3173 0.0553 0253 0.0520 6645 0.0656 4401 0.0610 5960 0.0586 8409 0.0657 5532 0.0531 1889 0.0665 6879 0.0628 7321 0.0595 0768 0.0608 6740 0.0540 3063 0.0679 4460 0.0642 5808 0.0600 8205 0.0580 5175 0.0564 1632 21 22 23 24 25 0.0408 3183 0.0476 6427 0.0456 8531 0.0438 7130 0.0422 0270 0.0502 8163 0.0481 1380 0.0461 3465 0.0443 2061 0.0420 5180 0.0507 3383 0.0485 6585 0.0405 8603 0.0447 7258 0.0431 0388 0.0616 4543 0.0494 7748 0.0474 9846 0.0466 8474 0.0440 1650 0.0630 3076 0.0608 6371 0.0488 8684 0.0470 7347 0.0464 0676 26 27 28 29 30 0.0406 6247 0.0302 3646 0,0379 1230 0.0366 7074 0.0355 2036 0.0411 1163 0.0308 8565 0.0383 6107 0.0371 2014 0.0350 7802 0.0415 6376 0.0401 3793 0.0388 1415 0.0375 8186 0.0304 3191 0.0424 7093 0.0410 6176 0.0307 2871 0.0384 0723 0.0373 4816 0.0438 6888 0.0424 4553 0.0411 2444" 0.0308 9502 0.0387 4811 31 32 33 34 35 0.0344 5330 0.0334 4458 0.0324 9708 0.0316 0540 0.0307 6476 0.0349 0304 0.0338 0453 0.0320 4727 0.0320 5580 0.0312 1550 0.0353 5633 0.0343 4816 0.0334 0124 0.0326 1020 0.0316 7024 0.0362 7352 0.0352 6834 0.0343 2048 0.0334 3053 0.0326 0170 0.0376 7673 0.0306 7080 0.0367 2744 0.0348 3907 0.0340 0308 36 37 38 39 40 0.0299 7000 0.0202 2003 0.0286 0875 0.0278 3402 0.0271 9310 0.0304 2104 0.0206 7130 0.0280 6045 0.0282 8607 0.0276 4552 0.0308 7710 0.0301 2698 0.0294 1649 0.0287 4258 0.0281 0251 0.0317 0973 0.0310 5082 0.0303 4157 0.0298 0893 0.0200 3016 0.0332 1431 0.0324 6805 0.0317 6150 0.0310 0160 0.0304 6560 41 42 43 44 45 0.0263 8352 0.0260 0303 0.0254 4061 0.0249 2141 0.0244 1675 0.0270 3631 0,0264 5622 0.0250 0320 0.0253 7541 0.0248 7117 0.0274 0379 0.0260 1420 0.0203 6170 0.0258 3443 0.0253 3073 0.0284 2276 0.0278 4452 0.0272 0338 0.0207 0761 0.0262 6521 0.0298 5102 0,0202 7563 0,0287 2737 0.0282 0441 0.0277 0505 46 47 48 49 50 0.0239 3409 0.0234 7204 0.0230 2020 0.0226 0468 0.0221 9711 0.0243 8804 0.0239 2733 0.0234 8503 0.0230 0087 0.0220 5376 0.0248 4005 0.0243 8708 0.0230 4024 0.0235 2205 0.0231 Iflll 0.0257 8405 0.0253 2532 0.0248 8504 0.0244 0202 0,0240 5787 0.0272 2775 0.0207 7111 0.0203 3384 0,0260 1474 0.0255 1273 TABLE IX PERIODICAL PAYMENT OF ANNUITY WHOSE PRESENT VALUE IS 1 1 * _.- , 1 n n% |% h% !% 1% 51 0.0218 0657 0.0222 6260 0.0227 2663 0.0230 0888 0.0251 2680 Kt 0.0214 2010 0.0218 8075 0.0223 5027 0.0232 0503 0.0247 5603 53 0.0210 6700 0.0215 2507 0.0210 8910 0.0220 3546 0.0243 0050 54 0.0207 1830 0.0211 7086 0.0210 4167 0.0225 8038 0.0240 5058 56 0.0203 8234 0.0208 4130 0.0213 0071 0.0222 6005 0.0237 2637 56 0.0200 5843 0.0205 1707 0.0200 8300 0.0210 3478 0.0234 0823 57 0.0107 4503 0.0202 0508 0.0206 7261 0.0216 2400 0.0231 0156 58 0.0104 4426 0.0100 0481 0.0203 7100 0.0213 2507 0.0228 0573 50 0.0101 5287 0.0100 1302 0.0200 8170 0.0210 3727 0.0225 2020 00 0.0188 7123 0.0103 3280 0.0108 0120 0.0207 6836 0.0222 4445 61 0.0185 0888 0.0100 6006 0.0105 2000 0.0204 8873 0.0219 7800 62 0.0183 8636 0.0187 0706 0.0192 0762 0.0202 2706 0.0217 2041 63 0.0180 8026 0.0185 4337 0.0100 1368 0.0100 7500 0.0214 7125 64 0.0178 3316 0.0182 0681 0.0187 0773 0.0107 3127 0.0212 3013 65 0.0175 0371 0.0180 5780 0.0185 2046 0.0194 0460 0.0200 0667 66 0.0173 6166 0.0178 2027 0.0182 0848 0.0102 0524 0.0207 7062 67 0.0171 3639 0.0176 0163 0.0180 7440 0.0100 4280 0.0205 6130 68 0.0169 1788 0.0173 8366 0.0178 6710 0.0188 2716 0.0203 3888 69 0.0167 0674 0.0171 7206 0.0176 4622 0.0186 1786 0.0201 3280 70 0.0164 0971 0.0160 6657 0.0174 4138 0.0184 1404 0.0100 3282 71 0.0162 0052 0.0167 6603 0.0172 4230 0.0182 1728 0.0107 3870 72 0.0161 0403 0.0165 7280 0.0170 4901 0.0180 2554 0.0106 5010 73 0.0150 1572 0.0163 8422 0.0168 6100 0.0178 3017 0.0103 6700 74 0.0157 3166 0.0162 0070 0.0166 7814 0.0176 6706 0.0191 8010 75 0.0155 5253 0.0160 2214 0.0165 0024. 0.0174 8170 0.0100 1600 76 0.0163 7816 0.0158 4832 0.0163 2709 0.0173 1020 0.0188 4784 77 0.0152 0836 0.0156 7008 0.0161 5851 0.0171 4328 0.0186 8416 78 0-0150 4206 0.0156 1423 0.0150 0432 0.0160 8074 0.0185 2488 79 0.0148 8177 0.0153 5360 0.0158 3436 0.0108 2244 0.0183 6084 80 0.0147 2464 0.0151 0704 0.0156 7847 0.0106 6821 0.0182 1886 81 0.0145 7144 0.0150 4430 0.0165 2650 0.0105 1700 0.0180 7180 82 0.0144 2200 0.0148 0562 0,0153 7830 0.0163 7136 0.0170 2851 88 0.0142 7620 0.0147 5028 0.0152 3373 0.0102 2847 0.0177 8880 84 0.0141 3301 0.0140 0855 0.0160 0268 0.0100 8008 0.0176 6273 85 0.0130 0600 0.0144 7021 0.0149 5501 0.0150 5303 0.0175 1008 86 0.0138 6035 0.0143 3513 0.0148 2000 0.0158 2034 0.0173 0060 87 0.0137 2685 0.0142 0320 0.0146 8036 0.0156 0076 0.0172 6417 89 0.0135 0740 0.0140 7431 0.0145 6115 0.0155 0423 0.0171 4080 89 0.0134 7088 0.0130 4837 0.0144 3688 0.0154 4064 0.0170 2050 90 0.0133 4721 0.0138 2527 0.0143 1347 0.0153 1080 0.0100 0300 91 0.0132 2620 0.0137 0403 0.0141 0380 0.0152 0100 0.0167 8832 92 0.0131 0803 0.0136 8724 0.0140 7679 0.0150 8657 0.0166 7624 98 0.0120 0234 0.0134 7213 0.0130 6236 0.0140 7382 0.0165 0673 94 0.0128 7016 0.0133 5060 0.0138 5042 0.0148 0356 0.0164 6071 95 0.0127 6837 0.0132 4030 0.0137 4000 0.0147 6571 0.0163 5511 96 0.0126 5002 0.0131 4143 0.0136 3372 0.0146 5020 0.0102 5284 97 0.0125 5374 0.0130 3583 0.0135 2880 0.0145 4606 0.0161 5284 98 0.0124 4076 0.0120 3242 0.0134 2608 0.0144 4602 0.0160 5603 99 0.0123 4700 0.0128 3115 0.0133 2540 0.0143 4701 0.0150 5030 100 0.0122 4811 0.0127 3104 0.0132 2606 0.0142 6017 0.0158 6574 TABLE IX PERIODICAL PAYMENT OF ANNUITY WHOSE PRESENT VALUE IS 1 , f . 1 C^tfQ n % |% -% 12 10 !% 1% 101 102 103 104 105 0.0121 5033 0.0120 6440 0.0119 6054 0.01 IS 0842 0.0117 7800 0.0128 3473 0.0125 3947 0.0124 4611 0.0123 6457 0.0122 6481 0.0131 3045 0.0130 3587 0.0120 4310 0.0128 5234 0.0127 6238 0.0141 5633 0.0140 0243 0.0130 7143 0.0138 8226 0.0137 0487 0.0157 7413 0.0156 8440 0.0155 0608 0.0155 1073 0.0154 2066 10G 107 108 100 110 0.0110 8948 0.0110 0250 0,0115 1727 0.0114 3358 0,0113 5143 0.0121 7679 0.0120 9045 0.0120 0575 0.0119 2264 0.0118 4107 0.0120 7594 0.0125 0020 0.0125 0028 0.0124 2385 0.0123 4208 0.0137 0022 0.0130 2524 0.0136 4201 0.0134 6217 0.0133 8296 0.0153 4412 0.0152 6336 0.0151 8423 0.0151 0609 0.0150 3009 111 112 113 114 115 0.0112 7079 0,0111 9161 0.0111 1386 0.0110 3750 0.0100 6240 0.0117 6102 0.0116 8242 0.0116 0626 0.0115 2048 0.0114 6500 0.0122 6301 0.0121 8671 0.0121 0023 0.0120 3414 0.0110 6041 0.0133 0527 0.0132 2905 0.0131 5425 0.0130 8084 0.0130 0878 0.0149 5320 0.0148 8317 0-0148 1166 0.0147 4133 0.0146 7245 116 117 118 110 120 0.0108 8880 0,0108 1030 0.0107 4524 0.0106 7630 0.0106 0655 0.0113 8105 0.0113 1013 0.0112 3060 0.0111 7021 0.0111 0205 0.0118 8709 0.0118 1680 0.0117 4098 0.0110 7832 0.0116 1085 0.0120 3803 0.0128 6857 0.0128 0037 0.0127 3338 0.0120 0758 0.0146 0488 0.0145 3860 0.0144 7356 0.0144 0973 0.0143 4709 131 122 123 124 125 0.0105 3800 0.0104 7261 0.0104 0715 0,0103 4288 0.0102 7965 0.0110 3505 0,0109 0918 0.0100 0441 0.0108 4072 0.0107 7808 0.0115 4464 0.0114 7030 0.0114 1528 0.0113 5228 0.0112 0033 0.0120 0294 0.0125 3042 0.0124 7702 0.0124 1568 0.0123 6540 0.0142 8501 0.0142 2525 0.0141 0590 0.0141 0780 0.0140 5065 126 127 128 120 130 0.0102 1746 0.0101 5625 0.0100 9603 0.0100 3077 0.0090 7844 0.0107 1047 0.0106 5586 0.0105 0023 0.0105 3755 0.0104 7081 0.0112 2040 0.0111 6948 0.0111 1054 0.0110 5265 0.0100 9560 0.0122 9614 0.0122 3788 0.0121 8000 0.0121 2428 0.0120 0888 0.0130 9462 0.0130 3030 0.0138 8624 0.0138 3203 0.0137 7975 131 132 133 134 135 0.0090 2102 0.0008 0440 0.0008 0883 0.0007 5403 0.0007 0005 0.0104 2208 0,0103 6704 0.0103 1107 0.0102 677B 0.0102 0430 0,0100 3035 0.0108 8410 0.0108 2072 0.0107 7619 0,0107 2340 0.0120 1440 0.0110 0080 0.0110 0808 0.0118 6621 0.0118 0510 0.0137 2837 0.0130 7788 0.0130 2825 0.0135 7947 0.0135 3151 136 137 138 130 140 0.0006 4080 0.0005 9463 0,0005 4205 0.0004 0213 0.0094 4205 0.0101 5179 0.0101 0002 0.0100 4002 0.0009 0870 0.0099 4030 0,0100 7101 0.0106 2052 0.0105 7021 0.0105 2007 0.0104 7187 0.0117 5493 0,0117 0550 0.0110 5684 0-0116 0804 0.0116 0170 0.0134 8437 0.0134 3801 0.0133 9242 0.0133 4759 0.0133 0340 141 142 143 144 145 0,0003 9271 0,0003 4408 0,0002 0015 0.0002 4800 0.0002 0233 0-0099 0056 0.0008 5250 0.0008 0610 0.0007 6860 0.0007 1252 0.0104 2380 0.0103 7044 0.0103 2078 0.0102 8381 O.OL02 3851 0.0115 1536 0,0114 6966 0.0114 2464 0.0113 8031 0.0113 3064 0.0132 6012 0.0132 1746 0,0131 7640 0.0131 3419 0.0130 0366 146 147 148 140 ISO 0.0001 5Q41 0.0001 1114 0.0000 0050 0.0000 2247 0.0089 7005 0.0000 0710 0.0000 2250 0.0005 7844 0.0005 3500 0.0004 0217 0.0101 0380 0,0101 4086 0.0101 0040 0.0100 6373 0.0100 2150 0.0112 9364 0.0112 5127 0.0112 0063 0.0111 0841 0.0111 2700 0.0130 5358 0.0130 1423 0.0120 7651 0,0120 3739 0.0128 0088 67 TABLE IX PERIODICAL PAYMENT OF ANNUITY WHOSE PRESENT VALUE IS 1 1 (a^ati) l-(l+i)- n . , n 1|% 1|% ll% lf% 2% i 3 3 4 5 1.0112 5000 0.6084 5323 0.3408 0130 0.2670 7058 0.2068 0034 1.0126 OOOO 0.5003 0441 0.3417 0117 0.2678 8102 0.2076 0211 1.0150 0000 0.5112 7792 0.3433 8206 0.2504 4478 0.2000 8032 1.0175 0000 0.6131-0295 0.3460 0746 0.2010 3237 0.2100 2142 1.0200 0000 0.6150 4950 0.3407 5407 0.2620 2375 0.2121 5830 6 7 8 9 10 0.1732 9034 0.1403 6762 0.1314 1071 0,1174 6432 0.1002 9131 0.1740 3381 0.1500 8872 0.1321 3314 0.1181 7055 0.1070 0307 0.1765 2521 0.1516 5010 0.1336 8402 0.1106 0082 0.1084 3418 0.1770 2250 0.1530 3059 0.1350 4202 0.1210 5813 0.1098 7634 0.1785 2681 0.1545 1106 0.1365 0080 0.1225 1644 0.1113 2063 11 12 13 14 15 0.0071 5084 0.0805 5203 0.0831 1020 0.0770 0138 0.0728 2321 0.0978 6839 0.0002 5831 0.0838 2100 0-0783 0616 0.0735 2646 0.0002 9384 0.0018 7090 0.0852 4030 0.0797 2332 0.0749 4430 0.1007 3038 0.0931 1377 0.0800 7283 0.0811 5602 0.0703 7730 0.1021 7704 0.0045 6060 0.0881 1835 0.0820 0107 0.0778 2547 16 17 13 10 20 0.0086 4363 0.0040 5008 0.0610 8113 0.0587 5120 0.0561 1531 0.0603 4072 0.0650 6023 O.OG23 8479 0.0504 5648 0.0508 2039 0.0707 6508 0.0670 7006 0.0638 0578 0.0008 7847 0.0682 4574 0.0721 9058 0.0685 1023 0.0652 4402 0.0623 2061 0.0590 0122 0.0730 5013 0.0000 0984 0.0067 0210 0.0637 8177 0.0011 5072 21 22 23 24 25 0.0537 3145 0.0515 6626 0.04Q5 8833 0.0477 7701 0.0461 1144 0.0544 3748 0.0522 7238 0.0502 0666 0.0484 8005 0.0468 2247 0.0558 0550 0.0537 0331 0.0517 3075 0.0400 2410 0.0482 0346 0.0573 1404 0.0551 5038 0.0581 8706 0.0513 8665 0.0497 2052 0.0587 8477 0.0566 3140 0.0540 0810 0.0528 7110 0,0512 2044 26 27 28 20 30 0.0445 7479 0.0431 5273 0.0418 3209 0.0406 0408 0.0304 5063 0.0452 8729 0.0438 6677 0.0425 4863 0.0413 2228 0.0401 7854 0.0407 3196 0.0463 1627 0.0440 0108 0.0427 7878 0.0416 3010 0.0482 0200 0.0407 0070 0.0454 8151 0.0442 0424 0.0431 2076 0.0406 9923 0.0482 9309 0.0460 8907 0.0457 7830 0.0440 4902 31 32 33 34 35 0.0383 8866 0.0373 8535 0.0364 4349 0.0355 5763 0.0347 2200 0.0391 0942 0.0381 0791 0.0371 6786 0.0362 8337 0.0354 5111 0.0405 7430 0.0305 7710 0.0380 4144 0.0377 0180 0.0309 3303 0.0420 7005 0.0410 7812 0.0401 4770 0.0302 7303 0.0384 5082 0.0435 0635 0.0426 1001 0.0416 8053 0,0408 1807 0.0400 0221 36 37 38 3ft 40 0.0330 3529 0.0331 0072 0.0324 8589 0.0318 1773 0.0311 8349 0.0340 0533 0.0330 2270 0.0332 1083 0.0326 6365 0.0310 2141 0.0301 5240 0.0364 1437 0.0347 1613 0.0340 5463 0.0334 2710 0.0370 7507 0.0360 4257 0.0362 4090 0.0356 0399 0.0349 7200 0.0302 3286 0.0386 0078 0.0378 2067 0.0371 7114 0.0305 6575 41 42 43 44 45 0.0305 8009 0.0300 0709 0.0294 0064 0.0289 3949 0.0284 4197 0.0313 2068 0.0307 4906 0.0302 0466 0.0290 8557 0.0291 0012 0.0328 3106 0.0322 0420 0.0317 2405 0.0312 1038 0.0307 1970 0.0343 8170 0.0338 2057 0.0332 8606 0.0327 7810 0.0322 9321 0.0359 7188 0.0354 1720 0.0348 8003 0.0343 8704 0.0339 0062 46 47 48 49 50 0.0279 6652 0.0275 1173 0.0270 7032 0.0266 5010 0.0262 5898 0.0287 1676 0.0282 6406 0.0278 3075 0.0274 1563 0.0270 1763 0.0302 5125 0.0208 0342 0.0203 7500 0.0280 6478 0.0286 7108 0.0318 3043 0.0313 8830 0.0300 0569 0.0305 6124 0.0301 7391 0.0334 5342 0.0330 1702 0.0320 0184 0.0322 0306 0.0318 2321 TABLE PERIODICAL PAYMENT OF ANtttJITY "WHOSE PRESENT VALUE IS 1 1 i 1 n ll% ll% l|% lf% 2% 61 6% 63' 64 66 0.0258 7404 0.026C 0006 0.0251 6149 0.0248 1043 0.0244 8213 0.0266 3571 0.0202 6807 0.0259 1653 0.0266 7760 0.0252 5146 0.0281 0460 0.0278 3287 0.0274 8537 0.0271 6138 0.0268 3018 0.0208 0209 0.0204 4665 0.0201 0492 0.0287 7672 0.0284 6120 0.0314 6856 0.0311 0909 0.0307 7392 0.0304 5226 0.0301 4337 66 67 68 69 60 0.0241 6502 0.0238 6116 0.0236 6726 0.0232 8366 0.0230 0085 0.0240 3730 0.0246 3478 0.0243 4303 0.0240 6158 0.0237 8003 0.0265 2106 0.0262 2341 0.0250 3661 0.0250 6012 0.0253 9343 0.0281 5705 0.0278 0606 0.0275 8503 0.0273 1430 0.0270 5336 0.0298 4656 0.0295 6120 0.0202 8667 0.0200 2243 0.0287 6707 61 63 63 64 66 0.0227 4534 0.0224 8060 0.0222 4247 0.0220 0320 0.0217 7178 0.0235 2758 0.0232 7410 0.0230 2004 0.0227 0203 0.0225 6268 0.0251 3604 0.0248 8751 0.0246 4741 0.0244 1534 0.0241 0004 0.0268 0172 0.0265 5892 0.0263 2455 0.0260 9821 0.0258 7952 0.0285 2278 0.0282 8643 0.0280 5848 0.0278 3855 0.0276 2624 66 67 68 60 70 0.0215 4758 0.0213 3037 0.0211 1085 0.0200 1671 0.0207 1760 0.0223 4065 0.0221 2560 0.0210 1724 0.0217 1627 0.0215 1041 0.0230 7386 0.0237 6370 0.0235 6033 0.0233 6320 0.0231 7235 0.0266 6813 0.0254 8372 0.0252 6506 0.0260 7459 0.0248 8030 0.0274 2122 0.0272 2316 0.0270 3173 0.0268 4665 0.0260 6766 71 73 78 74 75 0.0205 2552 0.0203 3806 0.0201 6770 0.0100 8177 0.0198 1072 0,0213. 2041 0.0211 4501 0.0200 6600 0.0207 9215 0.0206 2325 0.0229 8727 0.0228 0779 0.0226 3368 0.0224 6473 0.0223 0072 0.0247 0985 0.0245 3600 0.0243 6750 0.0242 0413 0.0240 4670 0.0264 0440 0.0263 2683 0.0261 6454 0.0260 0736 0.0268 5608 76 77 78 79 80 0.0100 4442 0.0104 8260 0.0103 2536 0.0101 7226 0.0100 2323 0.0204 5010 0.0202 0053 0.0201 4435 0.0100 0341 0.0108 4652 0.0221 4146 0.0210 8676 0.0218 3045 0.0216 0036 0,0215 4832 0.0238 0200 0.0237 4284 0.0236 0806 0.0234 5748 0.0233 2003 0.0257 0751 0.0255 6447 0.0254 2576 0.0252 0123 0.0251 6071 81 82 83 84 85 0.0188 7812 0.0187 3678 0.0185 0008 0.0184 6480 0.0183 3400 0.0107 0356 0.0106 6437 0.0104 2881 0.0102 0675 0.0101 6808 0.0214 1019 0.0212 7583 0.0211 4500 0.0210 1784 0.0208 0306 0.0231 8828 0.0230 5036 0.0220 3403 0.0228 1223 0.0226 0375 0.0250 3405 0.0240 1110 0.0247 9173 0.0246 7681 0.0245 6321 86 87 88 80 00 0.0182 0654 0,0180 8215 0.0179 6081 0.0178 4240 0.0177 2684 0.0100 4207 0.0180 2041 0.0188 0110 0.0180 8400 0.0185 7140 0.0207 7333 0.0206 6584 0.0205 4138 0,0204 2084 0.0203 2113 0.0225 7850 0.0224 6636 0.0223 5724 0.0222 5102 0.0221 4760 0.0244 5381 0.0243 4760 0.0242 4416 0.0241 4370 0.0240 4602 01 99 93 04 06 0.0176 1403 0.0176 0387 0.0173 0020 0.0172 0110 0.0171 8851 0.0184 6076 0.0183 5271 0.0182 4724 0.0181 4426 0.0180 4366 0.0202 1516 0.0201 1182 0,0200 1104 0.0190 1273 0.0198 1681 O.d220 4600 0.0219 4882 0.0218 6327 0.0217 0017 0.0216 6944 0.0230 6101 0.0238 5850 0.0237 6868 0.0236 8118 0.0235 0602 06 07 08 00 100 0.0170 8810 0.0160 0007 0.0168 0418 0.0168 0041 0.0167 0870 0.0170 4540 0.0178 4041 0.0177 5660 0.0170 6391 0.0176 7428 0.0107 2321 0.0196. 3186 0.0105 4268 0.0104 5560 0.0103 7057 0.0216 8101 0.0214 9480 0.0214 1074 0.0213 2876 0.0212 4880 0.0235 1313 0.0234 3242 0,0233 6383 0.0232 7720 0.0232 0274 69 TABLE IX PERIODICAL PAYMENT OF ANNUITY WHOSE PRESENT VALUE IS 1 1 _ i _.-, 1 n 2-07 4% 2l% 2|% 3% 3l% 1 2 3 4 5 1.0225 0000 0.5100 3768 0.3484 4458 0.2042 1803 0.2137 0021 1.0260 0000 0.5188 2716 0.3501 3717 0.2668 1788 0.2152 4086 1.0275 0000 0.6207 1825 0.3518 3243 0,2074 2059 0.2107 0832 1.0300 0000 0.5220 1084 0,3535 3030 0.2000 2706 0.2183 5457 1.0350 0000 0.5264 0049 0.3500 3418 0.2722 5114 0.2214 8137 6 7 8 10 0.1800 3400 0.15GO 0026 0.1370 8402 0.1230 8170 0.1127 8768 0.1815 4007 0.1574 9543 0.1394 0735 0.1254 5089 0.1142 6876 0.1830 7083 0.1589 9747 0.1409 6706 0.1260 4005 0.1157 3972 0.1845 0750 0.1005 0035 0.1424 5030 0.1284 3380 0.1172 3051 0,1870 0821 0.1035 4440 0.1464 7005 0.1314 4001 0.1202 4137 11 12 18 14 15 0.1030 3049 0.0900 1740 0.0805 7080 0.0840 6230 0.0702 8852 0.1051 0596 0,0974 8713 0.0910 4827 0.0855 3053 0.0807 6640 0.1005 8029 0,0089 6871 0.0025 3252 0.0870 2457 0.0822 5017 0.1080 7745 0.1004 6200 0.0040 2054 0.0885 2034 0.0837 0068 0.1110 0107 0.1034 8305 0.0070 0167 0.0015 707H 0.0808 2507 1ft 17 18 1 20 0.0751 1063 0.0714 4039 0.0681 7720 0.0052 0182 0.0026 4207 0.0765 9899 0.0720 2777 0.0000 7008 0.0607 0002 0.0041 4713 0.0780 9710 0.0744 3180 0.0711 8003 0.0082 7802 0,0066 7173 0.0700 1086 0.0750 6253 0.0727 0870 0.0098 1388 0.0072 1571 0.0820 8483 0.0700 4313 0.0758 1084 0.0720 4033 0.0703 0108 21 22 28 24 25 0.0002 7672 0.0681 2821 0.0661 7097 0.0543 8023 0.0527 3590 0.0617 8733 0.0506 4061 0.0570 9638 0.0550 1282 0.0542 7592 0.0033 1041 0.0011 8040 0,0592 4410 0.0574 0803 0.0568 3907 0.0048 7178 0.0027 4730 0.0008 1300 0.0500 4742 0.0574 2787 0.0080 3060 0.0060 3207 0.0640 1880 0.0022 7283 0.0600 7404 20 27 28 30 80 0.0512 2134 0.0408 2188 0.0485 2625 0.0473 2081 0.0401 9934 0.0527 6875 0.0513 7087 0.0500 8793 0.0488 9127 0.0477 7764 0.0543 4110 0.0529 5770 0.0616 7738 0.0604 8936 0.0493 8442 0.0559 3820 0.0545 0421 0,0532 9323 0.0521 1407 0,0510 1020 0.0592 0540 0.0578 5241 0.0500 0205 0.0564 4538 0.0543 7133 31 32 83 34 35 0.0451 5280 0.0441 7415 0.0432 5722 0.0423 9665 0.0416 8731 0.0467 3000 0.0457 6831 0.0448 5938 0,0440 0075 0.0432 0558 0.0483 5463 0,0473 9203 0.0464 0253 0.0450 4875 0.0448 6046 0.0400 9803 0.0400 4002 0.0481 6012 0.0473 2106 0.0405 3920 0.0533 7240 0.0624 4160 0.0516 7242 0.0607 50(50 0.0400 0835 36 37 88 80 40 0.0408 2622 0.0401 0043 0.0394 2753 0.0387 8543 0.0381 7738 0.0424 5158 0.0417 4090 0.0410 7012 0.0404 3015 0.0398 3623 0,0441 1132 0.0434 0063 0.0427 4704 0.0421 2260 0.0416 3161 0.0468 0379 0,0451 1102 0.0444 5934 0.0438 4385 0.0432 0238 0.0403 8416 0.0480 1325 0.0479 8214 0.0473 8775 0.0408 2728 41 42 43 44 45 0.0376 0087 0.0370 5364 0.0305 3364 0.0360 3901 0.0355 0805 0.0392 6786 0.0387 2870 0.0382 1088 0,0377 3037 0.0372 0752 0.0409 7200 0.0404 4175 0.0300 3871 0,0394 6100 0.0300 0093 0,0427 1241 0.0421 0107 0.0410 9811 0.0412 2085 0.0407 8518 0.0462 0822 0.0467 9828 0.0453 2630 0.0148 7708 0.0444 5343 46 47 48 40 50 0.0351 1921 0.0340 9107 0.0342 8233 0.0338 9179 0,0335 1836 0.0368 2676 0.0364 0609 0.0300 0590 0.0356 2348 0.0352 6806 0,0385 7403 0,0381 6358 0.0377 7158 0.0373 9773 0.0370 4092 0.0403 0264 0.0390 0051 0.0395 7777 0.0302 1314 0.0388 6550 0.0440 5108 0,0430 0010 0.0433 0040 0,0420 0107 0.0426 3371 70 TABLE IX PERIODICAL PAYMENT OP ANNUITY WHOSE PRESENT VALUE IS 1 1 i (a^ati) l-(l+f)- f n 2|% Q-Q7 -4 a % 2|% 3% 3i% 51 58 53 51 55 0.0331 0102 0.0328 1384 0.0324 9004 0.0321 7064 0.0318 7480 0.0340 0870 0.0345 7440 0.0342 5440 0.0330 4790 0.0336 5410 0.0307 0014 0.0363 7444 0.0360 0207 0.0367 0491 0.0354 7063 0.0386 3382 0.0382 1718 0.0379 1471 0.0370 2568 0.0373 4907 0.0423 2166 0.0420 2429 0.0417 4100 0.0414 7000 0.0412 1323 50 57 58 50 60 0.0315 8530 0.0313 0712 0.0310 3977 0.0307 8268 0.0305 3533 0.0333 7243 0.0331 0204 0.0328 4244 0.0325 0307 0.0323 5340 0.0352 OG12 0.0340 4404 0.0346 0270 0.0344 5153 0.0342 2002 0.0370 8447 0.0368 3114 0.0365 8848 0.0363 5503 0.0361 3200 0.0409 6730 0.0407 3245 0.0405 0810 0.0402 9366 0.0400 8862 61 6% 63 64 65 0.0302 0724 0.0300 0705 0.0208 4704 0.0206 3411 0.0204 2878 0.0321 2204 0.0310 0126 0.0316 8700 0.0314 8240 0.0312 8403 0.0330 0707 0.0337 8402 0.0335 7860 0.0333 8118 0.0331 0120 0.0350 1008 0.0357 1386 0.0356 1682 0.0363 2760 0.0351 4581 0.0308 0249 0.0397 0480 0.0395 2513 0.0303 6308 0.0391 8826 60 67 68 00 70 0.0202 3070 0.0200 3055 0.0288 5500 0.0286 7077 0.0285 0458 0.0310 0308 0.0300 1021 0.0307 3300 0.0305 6206 0.0303 0712 0.0330 0837 0.0328 3236 0.0326 6285 0.0324 0055 0.0323 4218 0.0340 7110 0.0348 0313 0.0346 4150 0.0344 8618 0.0343 3663 0.0300 3031 0.0388 7892 0.0387 3376 0.0385 0463 0.0384 6005 71 73 78 74 75 0.0283 3810 0.0281 7728 0.0280 2160 0.0278 7118 0.0277 2654 0.0302 3790 0.0300 8417 0.0200 3508 0.0207 0222 0.0206 5358 0.0321 0048 0.0320 4420 0.0310 0311 0.0317 6008 0.0316 3560 0.0341 9266 0.0340 6404 0.0330 2053 0.0337 9191 0.0336 0706 0.0383 3277 0.0382 0973 0.0380 9160 0.0379 7810 0.0378 6010 76 77 78 79 80 0.0275 8467 0.0274 4808 0.0273 1680 0.0271 8784 0.0270 0370 0.0205 1960 0.0203 8007 0.0202 0403 . 0.0201 4338 0.0200 2605 0.0316 0878 0.0313 8633 0.0312 6800 0.0311 5382 0.0310 4342 0.0336 4849 0.0334 3331 0.0333 2224 0.0332 1610 0.0331 1176 0.0377 6460 0.0376 6300 0.0375 0721 0.0374 7420 0.0373 8480 81 83 83 84 85 0.0200 4350 0.0268 2062 0.0207 1387 0.0200 0423 0.0204 0787 0.0280 1248 0.0288 0254 0.0286 0608 0.0285 0208 0.0284 0310 0.0309 3074 0.0308 3361 0.0307 3380 0.0306 3747 0.0305 4420 0.0330 1201 0.0329 1570 0.0328 2284 0.0327 3313 0.0320 4660 0.0372 0804 0.0372 1628 0.0371 3070 0.0370 6025 0.0360 8062 86 87 88 80 00 0.0203 0467 0.0202 0452 0,0261 0730 0.0201 0201 0.0200 1120 0.0283 0033 0.0283 0255 0.0282 linn 0.0281 2363 0.0280 3809 0.0304 5307 0.0303 6007 0.0302 8210 0-0302 0041 0.0301 2126 0.0326 6284 0.0324 8202 0.0324 0303 0,0323 2848 0.0322 6560 0.0360 1576 0.0308 47C6 0.0367 8190 0.0367 1868 0.0366 6781 91 99 92 94 05 0.0250 2224 0.0258 3577 0,0267 5176 0,0256 7012 0.0255 0078 0,0270 5523 0.0278 7486 0.0277 9000 0.0277 2120 0.0276 4786 0.0300 4400 0.0299 7038 0.021)8 0860 0.0208 2887 0.0207 6141 0.0321 81)08 0.0321 1604 0.0320 6107 0.0319 8737 0.0310 2677 0.0365 0019 0.0365 4273 0.0364 8834 0.0364 3594 0.0363 8546 06 97 98 99 100 0,0265 1306 0.0254 3868 0.0253 6678 0.0262 9480 0,0252 2594 0.0275 7602 0,0276 0747 0.0274 4034 0.0273 7617 0.0273 1188 0.0200 0605 0.0206 3272 0.0206 7134 0.0206 1185 0.0294 5418 0.0318 6610 0.0318 0856 0.0317 6281 0.0316 9886 0.0310 4667 0.0363 3682 0.0362 8905 0.0362 4478 0.0362 0124 0.0361 6927 TABLE IX PERIODICAL PAYMENT OF ANNUITY WHOSE PRESENT VALUE IS 1 1 _ i 1 (a^ati) l- n 4% *!% 6% 5|% 6% 1 2 8 4 5 1.0400 0000 0.5301 9608 0.3603 4864 0.2754 9005 0.2246 2711 1.0450 0000 0.5339 9756 0.3637 7336 0.2787 4365 0.2277 9164 1.0600 0000 0.5378 0488 0.3672 0856 0.2820 1183 0.2309 7480 1.0650 0000 0.5416 1800 0.3706 5407 0.2852 9449 0.2341 7644 1.0600 0000 0.5454 3689 0.3741 0981 0.2885 9149 0.2373 0640 6 7 8 9 10 0.1907 6190 0.1666 0961 0.1485 2783 0.1344 9299 0.1232 9094 0.1938 7839 0.1697 0147 0.1516 0965 0.1375 7447 0.1263 7882 0.1970 1747 0.1728 1982 0.1647 2181 0.1406 9008 0.1295 0458 0.2001 7895 0.1759 6442 0.1678 6401 0.1438 3946 0.1326 6777 0.2033 0263 0.1701 3502 0.1010 3604 0.1470 2224 0.1368 6706 11 12 13 U 15 0.1141 4904 0.1065 5217 0.1001 4373 0.0946 6807 0.0899 4110 0.1172 4818 0.1096 6619 0.1032 7635 0.0978 2032 0.0931 1381 0.1203 8889 0:1128 2541 0.1064 5577 0.1010 2397 0.0963 4229 0.1235 7006 0.1160 2923 0.1096 8426 0.1042 7012 0.0996 2500 0.1207 0294 0.1192 7703 0.1120 0011 0.1075 8401 0.1020 0276 18 17 18 19 20 0.0858 2000 0.0821 9852 0.0789 9333 0.0761 3862 0.0736 8176 0.0890 1537 0.0854 1758 0.0822 3690 0.0794 0734 0.0768 7614 0.0022 6991 0.0886 9914 0.0855 4622 0.0827 4501 0.0802 4259 0.0955 8254 0.0020 4197 0.0889 1092 0.0861 5006 0.0830 7033 0.0080 5214 0.0054 4480 0.0023 5654 0.0800 2080 0.0871 8456 21 22 23 24 25 0.0712 8011 0.0691 9881 0.0673 0906 0.0655 8683 0.0640 1190 0.0740 0057 0.0725 4565 0.0706 8249 0.0689 8703 0.0674 3903 0.0779 9611 0.0759 7061 0.0741 3082 0.0724 7090 0.0709 5246 0.0814 6478 0.0794 7123 0.0776 6066 0.0760 3580 0.0745 4035 0.0850 0455 0.0830 4557 0.0812 7848 0.0790 7000 0.0782 2072 26 27 28 29 80 0.0625 6738 0.0612 3854 0.0600 1298 0.0588 7993 0.0678 3010 0.0660 2137 0.0647 1946 0.0635 2081 0.0624 1461 0.0613 9154 0.0095 6432 0.0082 0186 0.0671 2253 0.0600 4551 0.0660 5144 0.0731 9307 0.0710 5228 0.0708 1440 0.0607 6867 0.0088 0539 0.0700 0436 0.0750 0717 0.0745 0255 0.0736 7901 0.0726 4801 31 82 83 34 35 0.0568 5535 0.0559 4859 0.0551 0357 0.0543 1477 0.0535 7732 0.0604 4345 0.0595 6320 0.0587 4453 0.0570 8191 0.0572 7046 0.0641 3212 0.0632 8042 0.0624 9004 0.0617 5645 0.0610 7171 0.0079 1065 0.0670 0519 0.0663 3469 0.0056 2958 0.0649 7493 0.0717 9222 0.0710 0234 0.0702 7203 0.0605 0843 0.0680 7380 36 37 38 39 40 0.0528 8688 0.0522 3957 0.0516 3192 0.0510 6083 0.0505 2349 0.0566 0578 0.0550 8402 0.0554 0169 0.0548 5567 0.0643 4315 0.0004 3446 0.0598 3070 0.0592 8423 0.0587 6462 0.0582 7816 0.0643 603G 0.0637 0903 0.0032 7217 0.0027 7901 0.0623 2034 0.0683 0483 0.0078 5743 0.0673 5812 0.0068 9377 0.0604 0154 41 42 43 44 45 0.0500 1738 0.0495 4020 0.0490 8989 0.0486 0454 0.0482 6246 0.0538 0158 0.0534 0868 0.0520 8235 0.0625 8071 0.0522 0202 0.0578 2220 0.0573 0471 0.0569 0333 0.0566 1025 0.0562 6173 0.0618 0000 0.0614 8027 0.0611 1337 0.0607 0128 0.0604 3127 0.0000 5880 0.0066 8342 0.0653 3312 0.0650 0000 0.0047 0050 46 47 48 49 50 0.0478 8205 0.0475 2189 0,0471 8065 0.0468 5712 0.0466 6020 0.0518 4471 0.0615 0734 0.0611 8858 0.0508 8722 0.0606 0215 0.0559 2820 0.0656 1421 0.0653 1843 0.0550 3065 0.0547 7074 0.0601 2175 0.0598 3120 0.0595 5854 0.0503 0230 0.0590 6145 0.0044 1485 0.0041 4708 0.0038 9706 0.0630 0356 0,0634 4429 72 TABLE IX PERIODICAL PAYMENT OF ANNUITY WHOSE PRESENT VALUE IS 1 1 _ i . = i -I ^ n 4% 4% 5% ftt 6% 51 52 53 54 55 0.0402 5885 0.045D 8212 0.0457 1015 0.0454 0010 0.0452 3124 0.0503 3232 0.0500 7070 0.0498 3400 0.0490 0510 0.0493 8754 0.0545 2807 0.0542 0450 0.0540 7334 0.0538 (M38 0.0530 0080 0.0588 3405 0.058(1 218(1 0.05H4 21HO 0.0582 3245 0.0580 5458 0.0032 3880 O.OIJ30 4(117 0.0028 (1551 0.0020 IXJ02 0.0(125 3(100 56 57 58 59 00 0.0450 0487 0.0447 8032 0.0445 8401 0.0443 8830 0.0442 0185 0.0401 8105 0.0489 8500 0.0487 0807 0.0480 2221 0.0484 5420 0.0534 8010 0.0533 03-13 0.0531 3020 0.0520 7802 0.0528 2818 0.0678 8008 0.0677 21100 0.0675 8000 0.0674 3051) 0.0673 0707 0.0023 8705 0.0(122 4744 0.0021 1574 0.0010 0200 0.0018 7572 61 62 03 0.0440 2308 0.0438 5430 0.0436 0237 0.0482 0402 0.0481 4284 0.0479 0848 0.0620 8027 0.0625 5183 0.0524 2442 0.0671 8202 0.0670 (1400 0.05110 5258 0.0(117 (1042 o.ooio 0:100 0.0015 (1704 04 05 0.0435 3780 0.0433 0019 0.0478 0115 0.0477 3047 0.0523 0806 0.0521 8015 0.0608 4737 0.0607 4800 0.0014 7016 0.0013 9000 66 67 08 0.0432 4921 0.0431 1451 0.0420 8578 0.0470 0008 0.0474 8705 0,0473 7487 0.0520 8057 0.0519 7757 0.0518 7080 0.0600 6-113 0,0605 (1544 0.05(1.1 8103 0.0013 1022 0.0012 3454 0.0(111 (11130 09 79 0.0428 6272 0.0427 4500 0.0472 0745 0.0471 0611 0.0617 8716 0.0510 91G 0.06(14 0242 0.0603 2754 0.0010 0025 0.0010 3313 71 0.0420 3253 0.0470 0750 0.0510 1503 0.0602 5076 0.0000 7370 72 0.0425 2480 0.0400 7405 0,0515 8033 O.Omil 8082 0.0000 1774 73 0.0424 2100 0.0408 8600 0.0514 0103 0.05(11 2(152 0.0008 0505 74 0.0423 2334 0.0408 0159 0.0513 8053 0.0500 0(1(15 0.0008 1642 75 0.0422 2000 0.0407 2104 0.0613 2101 0.06(10 1002 0.0007 0807 76 0.0421 3809 0.0400 4422 0.0512 5700 0.0650 5(145 0.0007 2403 77 0.0420 5221 0.0405 7004 0.0511 0580 0.0550 0677 0.000(1 an 5 78 0.0419 0930 0.0405 0104 0.0611 3750 0.0668 6781 0.0000 4407 79 0.0418 0007 0.0404 3434 0.051O 8222 0.0568 1243 0.0000 0724 89 0.0418 1408 0.0403 7069 0.0510 2002 0.0557 0048 0.0005 7254 81 0.0417 4127 0.0403 0995 0.0509 7003 0,0557 2884 0.0005 3084 82 0.0410 7160 0.0402 5107 0.0500 3211 0.0650 008(1 0.0005 0003 83 0.0410 0403 0.0401 0063 0.0508 801)4 0,0550 5:il)5 0.0(104 70JI8 84 0.0415 4054 0.0401 4379 0.0608 4800 0.0650 1047 0.0(104 52(11 85 0.0414 7909 0.0400 0334 0.0508 0810 0.0655 8(183 0.0004 20H1 86 0.0414 2018 0.0400 4510 0.0507 0433 0.0666 6503 0.0004 0240 87 0.0413 0370 0.0450 0915 0,0507 2740 0.0556 20(17 0,(H!()3 706(1 88 0.0413 0953 0.0459 6622 0.0500 Oa28 0.0664 0800 0.0003 6706 89 0.0412 5V58 0,0450 1325 0.0500 5888 0.0664 727.1 0.0003 8767 90 0.0412 0776 0.0468 7310 0.0500 2711 0,0554 4788 0.0003 183(1 91 0.0411 5005 0.0458 3480 0.0505 0081) 0.0554 2485 0.0(103 0025 92 0.0411 1410 0.0457 9827 0.0505 OH15 0,0554 0207 0.0002 a*m 93 0.0410 7010 0.0457 0331 0.0505 4080 0.056S 800(1 0.0(102 0708 94 0,0410 2789 0.0457 2001 0.0505 1478 0,0653 0007 0.0002 5100 95 0.0409 8738 0.0450 0709 0.0504 0003 0.05511 4204 0.0002 3768 96 0.0409 4800 0.0450 0740 0.05O4 0048 0.0653 2410 0.0(102 2408 97 0.0409 1119 0.0450 3834 0.0504 4407 0.0561) 0711 0.01102 1135 98 0.0408 7538 0.0450 1048 0.0504 2274 0.0652 0101 0,0001 0035 99 0.0408 4100 0.0465 8385 0.0504 0245 0.0552 7fl77 0.0001 8803 100 0.0408 0800 0.0455 6839 0,0503 8314 0.0662 0132 0.0001 7730 73 TABLE IX PERIODICAL PAYMENT OF ANNUITY WHOSE PRESENT VALUE IS 1 1 _ i 1 (a^ati) !-(!+*)- (s^ati) n 6|% 7% 7f% 8% 8|% i 2 3 4 5 1.0050 0000 0.5402 6150 0.3775 7G70 0.2010 0274 0.2400 3454 1.0700 0000 0.5530 9170 0.3810 5100 0.2952 2812 0.2438 0000 1.0760 0000 0.5500 2771 0.3845 3703 0.2085 0751 0.2471 6472 1.0800 0000 0.5007 0023 0.3880 3351 0.3010 2080 0.2504 5045 1.0850 0000 0.6040 1031 0.3915 3925 0.3052 8781) 0.2537 0575 6 7 8 9 10 0.2005 0831 0.1823 3137 0.1042 3730 0.1502 3803 0.1301 0409 0.2097 0580 0.1855 5322 0.1074 0770 0.1534 8647 0.1423 7750 0.2130 4489 0.1888 0032 0.1707 2702 0.1507 0710 0.1450 8593 0.2103 1539 0.1920 7240 0.1740 1470 0.1000 7071 0.1490 2040 0.2100 0708 0.1053 0022 0.1773 3005 0.1034 2372 0.1524 0771 11 12 13 14 15 0.1300 5521 0.1226 6817 0.1162 82CO 0.1109 4048 0.1063 6278 0.1333 5090 0.1250 0199 0.1100 5085 0.1143 4494 0.1007 9462 0.1300 9747 0.1292 7783 0.1230 0420 0.1177 0737 0.1132 8724 0.1400 7034 0.1320 9502 0.1206 2181 0.1212 0085 0.1108 2954 0.1434 0293 0.1301 5280 0.1300 2287 0.1248 4244 0.1204 2040 16 17 18 19 20 0.1023 7767 0.0089 0033 0.0058 5401 0.0931 5676 0.0907 5040 0.1058 5706 0.1024 2610 0.0094 1200 0.0007 5301 0.0943 0293 0.1003 9110 0.1000 0003 0.1030 2890 0.1004 1090 0.0080 0219 0.1120 7087 0.1090 2943 0.1007 0210 0.1041 2703 0.1018 5221 0.1100 1354 0.1133 1108 0.1104 3041 0.1079 0140 0.1050 7007 21 22 23 24 25 0.0880 1333 0.0800 0120 0.0840 6078 0.0833 9770 0.0819 8148 0.0022 8900 0.0004 0677 0.0887 1393 0.0871 8002 0.0858 1052 0.0060 2937 0.0941 8087 0.0025 3528 0.0010 5008 0.0807 1067 0.0908 3225 0.0980 3207 0.0004 2217 0.0049 7700 0.0030 7878 0.1030 0541 0.1019 3802 0.1003 7193 0.0989 0975 0.0077 1108 26 27 28 29 30 0.0806 9480 0.0705 2288 0.0784 5305 0.0774 7440 0.0765 7744 0.0845 6103 0.0834 2573 0.0823 9193 0.0814 '4866 0.0805 8040 0.0884 0001 0.0874 0204 0.0804 0520 0.0854 9811 0.0840 7124 0.0025 0713 0.0914 4809 0.0004 8801 0.0896 1854 0.0888 2743 0.0005 8010 0.0055 0025 0.0046 3014 0.0038 0577 0.0030 5058 31 32 33 34 35 0.0757 5393 0.0740 9005 0,0742 0924 0.0736 6010 0.0730 6220 0.0797 9001 0.0700 7292 0.0784 0807 0.0777 0074 0.0772 3300 0.0839 1028 0.0832 2699 0.0825 9307 0.0820 1401 0.0814 8291 0.0881 0728 0.0874 5081 0.0808 BIOS 0.0803 0411 0.0858 0320 0.0023 0524 0.0017 4247 0.0011 7588 0.0000 60K4 0.0001 8037 36 37 38 39 40 0.0725 1332 0.0720 0634 0.0715 3480 0,0710 9854 0.0706 9373 0.0707 1631 0.0762 3086 0.0757 0606 0.0753 8070 0.0760 0914 0.0809 9447 0.0805 4533 0.0801 3197 0.0797 6124 0.0794 0031 0.0853 4467 O.OU40 2440 0.0846 8HH4 0.0841 8513 0.0838 0010 0.0807 0000 0.0803 (1790 0.0800 0066 0.0886 8103 0,0883 8201 41 42 48 44 AS 0.0703 1779 0.0699 6842 0.0696 4352 0.0693 4119 0.0690 5068 0.0746 5962 0.0743 3691 0.0740 3600 0.0737 6760 0.0734 9957 0.0790 7663 0.0787 7789 0.0785 0201 0.0782 4710 0.0780 1146 0.0835 6149 0.0832 8084 0.0830 3414 0.0828 0152 0.0825 8728 0.0881 0737 0.0878 6570 0.0876 2512 0.0874 1363 0.0872 1961 46 47 48 49 50 0.0687 9743 0.0685 5300 0.0683 2606 0.0081 1240 0.0679 1393 0.0732 5996 0.0730 3744 0.0728 3070 0.0726 3863 0.0724 5985 0.0777 9353- 0.0775 9190 0.0774 0827 0.0772 3247 0.0770 7241 0.0823 8991 0.0822 0799 0.0820 4027 0.0818 8567 0.0817 4886 0.0870 4154 0.0808 7807 0.0867 2796 0.0866 9005 0.0864 6334 74 TABLE X COMPOUND AMOUNT OP 1 FOR FRACTIONAL PERIODS p ' n% \% S* ! 1% 2 3 4 e 12 18 20 1.0020 8117 1.0013 8(100 1.0010 4004 1.0000 0324 1.0003 4050 1.0003 1000 1.0001 6004 1.0024 OOS8 1.001(1 0300 1.0012 4700 1.0008 3100 1.0004 1571 1.0003 8373 1.0001 0185 1.0020 1243 1.0010 4068 1.0014 5515 1.0000 01)87 1.0004 8482 1.0004 4751 1.0002 2373 1.0037 4200 1.0024 0378 1.0018 0075 1.0012 4011 1.0000 22SO 1.0005 7404 1.0002 8743 1.0040 8750 1.0033 2228 1.0024 0008 1.0010 5077 1.0008 2054 1.0007 0570 1.0003 8270 P lj% lj% 1|% ll% 2% 2 3 4 6 12 13 26 1.0050 0027 1.0037 3002 1.0028 0081 1.0018 0027 1.0001) 3270 1.0008 0092 1.0004 3037 1,0002 3050 1.0041 4043 1.0031 1040 1.0020 7257 1.0010 3575 1.0000 5004 1.0004 7700 1.0074 7208 1.0040 7521 1,0037 2900 1.0024 8452 1.0012 4149 1.0011 4504 1.0005 7280 1.0087 1205 1.0057 0003 1.0043 4058 1.0028 0502 1.0014 4077 1.0013 3540 1.0000 0748 1.0009 5050 - : 1.0000 2271 1.0040 0203 1.0033 0580 1.0010 5168 1.0015 2444 1.0007 0103 P 2|% 2|% 2-V 4/0 3% 3|% 2 3 4 G 12 26 52 1.0111 8742 1.0074 1444 1.00H5 7816 1.0037 1532 1.001S 5S04 1.0008 5016 1.0004 2709 L0124 2284 1.0082 0484 1.0001 0225 1,0041 231)2 1.0020 5084 1.0000 5017 1.0004 741)7 1.0130 5076 1.0(100 8300 1.0008 0522 1.00.16 3108 1.0022 0328 1.0010 4300 1.0005 2184 1,0148 8010 1.0090 0103 1.0074 1707 1.00-10 3862 1.0024 0027 1.0011 3752 1.0005 6800 1.0173 4060 1.0115 3314 1.0080 8745 1.0067 5004 1.0028 7000 1.0013 2401 1.0000 0170 P 4% 4-<5f, *JJ fO 5% 5\% 6% 2 3 4 12 20 52 . 1.01518 0300 1,0131 5H41 1.0008 5341 1.0005 5K20 1.0032 7374 1.0015 0003 1.0007 5453 1.0222 5242 1,0147 8040 1.0110 0400 1.0073 0312 1.0030 7481 1.0010 1)430 1.0008 4084 1.0240 0508 1,0103 0030 1.0122 7224 1.0081 04H5 1. 00-10 7412 1.0018 7831 1.0000 3871 1.0271 3103 1.0180 0713 1,0134 7518 1.0080 0340 1.0044 7170 1.0020 0138 1.0010 3010 1.0206 Q302 1.011)0 1282 1.014(5 7386 1.0007 6880 1.004S 0755 1.0022 43(13 1.0011 2118 P 6% 7% 7-% 1 2 8% 8|% a 4 12 2 52 1.0310 8837 1.0212 1347 1.0108 0828 1.0105 5107 1.0052 6109 1.0024 2504 1.0012 1179 1.0344 0804 1.0228 0012 1.0170 5R53 1.0113 4020 1,0050 5415 l.OOSfl 0564 1.0013 0107 1.0308 2207 1.0243 0081 1.0182 4400 1.0121 2038 l.OflflO 4402 1.0027 8544 1.0013 0176 1.0302 3048 1.0250 8557 1.0104 2056 1.0120 0040 1.0004 3403 1.0020 0443 1.0014 8112 1,0410 3333 1.0275 6044 1,0200 0440 1,0130 8052 1.0008 2140 1.0031 4262 1.0016 7008 75 TABLE XI NOMINAL RATE j WHICH IF CONVERTED TIMES PER YEAR GIVES EFFECTIVE RATE i P a% 1% 5% !% 1% 2 3 4 6 12 13 2ffi .0041 6234 .0041 6089 .0041 6017 .0041 5945 .0041 6873 .0041 6808 .0041 834 .0049 9377 .0049 9169 .0049 9005 .0049 8962 .0049 8858 .0049 8850 .0049 8802 .0058 2485 .0058 2203 .0068 2062 .0068 1921 .0058 1780 .0068 1709 .0058 1704 .0074 8599 .0074 8133 .0074 7900 .0074 7007 .0074 7434 .0074 7416 .0074 7309 .0099 7612 .0099 0085 .0090 0272 .0099 5850 .0090 5440 .0099 5414 .0009 5224 P 1|% 1|% 1|% l|% 2% 2 3 4 6 12 13 26 .0112 1854 .0112 0807 .0112 0285 .0111 9763 .0111 9241 .0111 9200 .0111 8960 .0124 0118 .0124 4828 .0124 4183 .0124 3530 .0124 2895 .0124 2846 .0124 2540 .0149 4417 .0149 2562 .0149 1636 .0149 0710 .0148 9785 .0148 9714 .0148 9288 ,0174 2410 .0173 9890 .0173 8631 .0173 7374 .0173 0119 .0173 0022 .0173 5443 .0109 0099 .0198 6813 .0198 5173 .0198 3634 .0108 1898 .0198 1772 .0198 1017 ^ 2;% 2|% 2 -or 4 % 3% 3*% 2 3 4 6 12 26 52 .0223 7484 .0223 3333 .0223 1261 .0222 9192 .0222 7125 .0222 6013 .0222 5537 .0248 4667 .0247 9451 .0247 6809 .0247 4349 .0247 1804 .0247 0434 .0240 9848 .0273 1349 .0272 5170 .0272 2087 .0271 9009 .0271 5936 .0271 4283 .0271 3575 .0297 7831 .0297 0490 .0296 0829 .0290 3173 .0205 9524 .0295 7561 .0295 6721 .0346 0809 .0345 0043 .0345 4078 ,0345 0024 .0344 5078 .0344 2420 .0344 1281 P 4% 4|% 6% 6|% 6% 2 3 4 6 12 26 52 .0396 0781 .0394 7821 .0394 1363 .0393 4918 .0392 8488 .0392 5031 .0392 3561 .0445 04S3 .0443 4138 .0442 5996 .0441 7874 .0440 9771 .0440 5417 .0440 3552 .0493 9015 .0491 8907 .0490 8894 .0489 8908 .0488 8949 .0488 3597 . .0488 1300 .0542 3380 .0540 2139 . .0539 0070 .0537 8036 .0536 6039 .0535 9593 .0535 0834 .0601 2003 .0588 3847 .0580 0538 .0585 5277 .0584 1001 .0583 3426 .0583 0157 P 6|% 7% 7|% 8% 8|% 2 3 4 6 12 28 2 .0639 7674 .0636 4042 .0634 7314 .0633 0644 .0631 4033 .0630 5113 .0630 1295 .0688 1009 .0684 2737 .0682 3410 .0680 4166 .0678 4974 .0677 4676 .0677 0268 .0736 4414 .0731 9942 .0729 7840 .0727 5827 .0725 3903 .0724 2134 .0723 7098 .0784 0097 .0779 6070 .0777 0619 .0774 5074 ,0772 0830 .0770 7606 .0770 1802 .0832 0007 , .0826 9033 .0824 1768 .0821 3712 .0818 5702 .0817 0811 .0816 4401 76 TABLE XH THE VALUE OP THE CONVERSION FACTOR - f '-, tf-o p 35% 8% n% !% 1% % 3 4 12 13 26 1.0010 4058 1.0013 8701 1.0015 811G 1.0017 3471 1.0010 0821) 1.0010 2164 1.0020 0170 1.0012 4844 1.0010 0482 1.0018 7305 1,0020 8131 1.0022 8000 1.0023 050 1.0024 2182 1.0014 5B21 1.0010 4103 1.0021 8485 1.002-1 2781 1.0020 7080 1.0028 8050 1.0028 0100 1.0018 7150 1.0024 0585 1,0028 0812 1.0031 2040 1.0034 3280 1.0034 50UO 1.0036 0111 1.0024 0378 1.0033 2500 1.0037 4223 1.0041 f>8fll 1.0045 7510 1.004W 0714 1.0047 1)041 / 1|% lj% i-;-% l|% 2% 2 3 4 6 12 13 88 1.0028 0403 1.0037 4008 1.0042 0802 1.0040 7730 1.0051 4583 1.0051 8188 1.0053 0818 1.0031 1520 1.0041 5510 1.0046 7537 1.00C1 0575 1.0057 1032 1.0057 5037 1.0050 0000 1.0037 3004 1.0040 8340 1.0050 0755 1.0002 3101 1,0008 5052 1.0000 0458 1.0071 0290 1.0043 8176 1.0058 1084 1.0005 3878 1.0072 0707 1.0070 0571 1.0080 5177 1.0083 8820 1,0040 7525 l.OOHO 3733 1.0074 H850 1.0083 0125 1.0001 3380 1.0001 H700 1.0005 8243 /> 2|% 2 1% 2|% 3% 3|% 2 3 4 6 12 588 52 LOOSE 0371 1,0074 0202 1.0083 0830 1,0003 3444 1.0102 7107 1.0107 7505 1.0109 0195 1.0002 1142 1.0082 8701 1.0003 2677' 1.0103 6006 1.0114 0726 1.0110 0781] 1.0122 0810 1.0008 2837 1.001)1 1141 1.0102 5422 1.0113 0780 1.0125 4243 1.0131 5008 1.0134 2343 1.0074 4458 1.0000 3431 1.0111 8072 1.0124 2S10 1.0136 7002 1.0143 4020 1.0140 3757 1.0080 7475 1.0115 7748 1.0130 3004 1.0144 8578 1.0151) 4203 1.0107 2074 1.0170 0310 P 4% 4j% 5% 6f% 6% 2 3 4 G 12 20 52 1.0000 0105 1.0132 1713 1.0148 7744- 1,0165 3057 1.0182 0351 1.0101 0023 1.0194 8470 1.0111 2021 1.0148 5328 1.0107 2020 1.0185 8053 1.0204 0100 1.0214 0080 1.0210 0231 1.0123 4754 1.0184 8507 1.0185 5042 1.0200 3570 1.0227 1470 1.0238 3548 1.0243 1002 1.0135 R/506 1.0181 1522 1.0203 405 1.0226 7810 1.0240 0400 1.0201 0720 1.0207 2580 1.0147 8151 1.0107 4104 1,0222 2GH8 1.0247 1070 1.0272 1070 1.0385 5520 1.0201 318(i J 6|% 7% 1-% I j 70 8% 8*% 2 3 4 6 12 20 S2 1.0159 0419 1.0213 0348 1,0240 6523 1.0267 5172 1,0204 5204 1,0300 0941 1.0315 3404 1.0172 0402 1.0220 8254 1.0258 8002 1.0287 8208 1.0316 9143 1.0332 5078 1.0330 3242 1.0184 1103 1,0245 0820 1,0277 0120 1.0308 1059 1.0330 2617 1.0356 0640 1.0363 2706 1.0100 1524 1.0262 1065 1.0)205 1004 1.0328 3450 1,0301 5721 1.0379 4927 1.0387 1704 1.0208 1067 1.0278 1074 1.0313 3332 1.0348 5402 1.0383 8456 1.0402 8846 1.0411 Ofill 77 TABLE Xm AMERICAN EXPERIENCE TABLE OF MORTALITY Ag X Num- ber living , Num her of death * Yearly proba- bility of dying ?* Yearly proba- bility of living P, Ag X Num- ber living ' Num bor of deatbfl ** Yearly proba- bility of dying Vx Yearly proba- bility of living Px 10 100,000 749 0.007 490 0.002 510 53 66,797 1,091 0.010 333 0.083 007 11 99,251 746 0-007 516 0.002 484 54 05,706 1,143 0.017 300 0.082 604 12 98,605 743 0.007 543 0.992 457 55 64,563 1,100 0.018 571 0.081 420 13 97,762 740 0.007 669 0.992 431 50 63,364 1,200 0.019 885 0.080 115 14 07,022 737 0.007 596 0.002 404 57 62,104 1,325 0.021 335 0.078 005 15 96,286 735 0.007 634 0.002 366 58 60,779 1,394 0.022 936 0.077 064 16 95,550 732 0.007 661 0.992 330 59 50,385 1,468 0.024 720 0.075 280 17 94318 729 0.007 688 0.992 312 60 57,917 1,640 0.020 603 0.973 307 18 94,080 727 0.007 727 0.902 273 61 68,371 1,628 0.028 880 0.071 120 19 93,362 725 0.007 766 0.902 235 62 54,743 1,713 0.031 202 0.068 708 20 92,637 723 0.007 805 0.002 105 63 53,030 1,800 0.033 043 0.006 067 21 91,914 722 0.007 855 0.992 145 64 51,230 1,880 0.030 873 0.003 127 22 01,102 721 0.007 906 0.992 094 65 49,341 1,980 0,040 120 0.050 871 23 90,471 720 0.007 968 0.002 042 66 47,361 2,070 0.043 707 0.056 203 21 89,751 719 0.008 Oil 0.991 989 67 45,291 2,168 0.047 047 0.052 353 25 80,032 718 0.008 065 0.991 935 68 43,133 2,243 0.052 002 0.047 008 26 88,314 718 0.008 130 0.991 870 69 40,800 2,321 0.056 762 0.043 238 27 87,596 718 0.008 197 0.991 803 70 38,560 2,391 0.061 993 0.038 007 28 86,878 718 0.008 264 0.001 736 71 36,178 2,448 0.007 665 0.032 335 29 86,160 719 0.008 345 0.901 655 72 33,730 2,487 0.073 733 0,020 207 30 85,441 720 0.008 427 0.991 573 73 31.243 2,505 0.080 178 0.010 822 31 84,721 721 0.008 610 0.991 490 74 28,738 2,501 0.087 028 0.012 972 32 84,000 723 0.008 607 0.901 303 75 26,237 2,470 0.094 871 0.005 620 33 83,277 726 0.008 718 0.001 282 76 23,761 2,431 0.102311 0,897 080 34 82,651 729 0.008 831 0.991 169 77 21,330 2,360 0.111 004 0,888 030 36 81,822 732 0.008 046 0.001 054 78 18,061 2,201 0.120 827 0.870 173 30 81,090 737 0.000 089 0.900 911 79 10,670 2,190 0.131 734 0.868 20(1 37 80,353 742 0.009 234 0.090 776 80 14,474 2,091 0.144 400 0.855 534 38 79,611 749 0.000 408 0.900 592 81 12,383 1,964 0.158 005 0.841 306 39 78,862 756 0.009 686 0.900 414 82 10,410 1,816 0.174 297 0,825 703 40 78,106 765 0.000 704 0.090 206 83 8,603 1,048 0.191 501 0.808 430 41 77,341 774 0.010 008 0.980 092 84 0,055 1,470 0.211 359 0.788 641 42 76,667 785 0.010 252 0.989 748 85 5,485 1,292 0.236 652 0.704 448 43 75,782 707 0.010 617 0.980 483 86 4,193 1,114 0.205 081 0,734 310 44 74,985 812 0.010 820 0.989 171 87 3,070 033 0.303 020 0.000 080 45 74,173 828 0.011 163 0.988 837 88 2,140 744 0.340 602 0.053 308 46 73,345 848 0.011 562 0.988 438 89 1,402 555 0.305 863 ).0()4 137 47 72,407 870 0.012 000 0.988 000 90 847 385 0.454 645 0.54.1 4515 48 71,627 896 0.012 600 0.987 491 91 462 240 0.532 468 0.467 534 49 70,731 927 0.013 106 0.980 804 92 210 137 0.634 250 0.305 741 50 51 52 69,804 68,842 67,841 962 1,001 1,044 0.013 781 0.014 641 0.015 380 0.086 210 0.085 450 0.084 611 93 94 05 70 21 3 58 18 3 0.734 177 0.857 143 1.000000 0.265 823 0,142857 0.000 000 78 TABLE XIV COMMUTATION COLUMNS, AMERICAN EXPERIENCE TABLE, 8& Age X *>* NX M x Age X *, N x M x 10 70 801.0 1 575 535.3 17 012.01 53 10 787.4 146 915.7 5 853.005 11 67 081.6 1 504 043.4 17 000.89 54 10 252.4 135 128.2 5 682.801 12 05 180.0 1 430 601.0 16 606.20 55 9 733.40 124 875.8 5 510.644 13 02 00.4 1 371 472.0 10 131.12 50 229.00 116 142.4 5 335.808 11 50 038.4 1 308 003,5 15 073.00 57 8 740.17 105 012.8 6 168.C73 15 54 471.0 1 240 025,0 16 234.05 58 8 204.44 97 172.64 4 078.406 10 55 104.2 1 101 653.4 14 810.17 50 7 801.83 88 008.20 4 705.206 17 52 S32.0 1 136 440.2 14 402.30 00 7 351.65 81 100.38 4 008.020 18 GO 063.0 1 033 G10.2 14 000.83 01 013.44 73 764.73 4 410.322 10 48 562.8 1 032 902.4 13 031.08 62 480.75 66841.28 4 226.413 20 46 550,2 084 300.6 18 207.32 63 071.27 60 364.64 4 030.200 21 44 630,8 037 843.4 12 016.25 64 5 600.85 64 283.27 3 331.187 22 42 782.8 803 212.6 12 577.63 65 6 273.33 48 010.41 3 020.300 23 41 000.2 850 420.0 12 260.71 GO 4 800.66 43 343.08 3 424.843 24 39 307.1 809 420.0 11 035,38 67 4 518,05 38 462.53 3 218.321 25 37 073.6 770 113.0 11 031.14 08 4 167.82 33 033.88 3 010.290 20 30 106.1 732 430.0 11 337.50 60 3 808.32 29 776.0(5 2 SOl.SOfl 27 34 001.5 600 333.8 11 063.07 70 3 470.67 25 067.74 2 502.538 28 33 167.4 001 732.4 10 770.04 71 3 145.43 22 497.07 2384.657 28 31 771.3 628 575.0 10 515.18 72 2 833,42 19 351.64 2 170.018 30 30 440.8 500 803.6 10 250.02 73 2 535.75 10 618.22 1 077.107 31 20 103.5 506 302.0 10 011.17 74 2 253.57 13 082.47 1 780.731 32 27 937.5 637 199.3 9 771.375 75 1 087.87 11 728.00 1 501.240 33 26700.5 609 261.8 539.044 76 1 730.39 741.028 1 400.088 31 25 630.1 482 501.3 313.638 77 1 508.03 8 001.633 1 238.047 35 24544.7 460 871.2 004.065 78 1 205.73 6 492.000 1 070.158 36 23 502.6 432 320.5 8 882,708 79 1 100.65 5 197.271 024.803 7 37 22 501.4 408 824.0 8 070,415 80 023.338 4 000.024 784,804 38 21 630.7 380 322.6 8 475.068 81 703.234 3 173,280 655.024 5 39 20 015.5 364 782.0 8 270.860 82 620.405 2 410.052 538.005 7 40 10 727.4 344 107.4 8 088.915 83 404.995 1 780.587 434.477 41 18 873.0 324 440.0 7 902.231 84 386.641 1 204.592 342.862 4 42 18 052.0 305 506.3 7 710.738 85 204.610 007.051 3 203.005 43 17 203.0 287 513,4 7 540.910 86 217.598 613.341 7 100.850 44 10 504.4 270 240.8 7 365.480 87 154.383 305.743 8 141.000 3 45 15 773.0 253 745,5 7 102.800 88 103,903 241.360 05.801 07 40 15 070.0 237 071.0 7 022.083 80 65.023 1 137.307 8 00.070 82 47 14 302.1 222 001.0 fl 854.337 90 38.304 7 71.774 70 35.S77 52 48 13 738.5 208 500,8 087.406 91 20.180 33.470 01 10.056 00 40 13 107.0 194 771.3 6 621.410 92 0,11880 13,283 00 8.000 605 50 12408.0 181 063.4 355.436 98 3.222 30 4.164 21 3.081 545 51 11 000.6 100 104.7 6 180,012 94 0.827 Oil 0.041 84 .705 7G2 &9 11 330.5 167 256.2 021.006 95 0.114 232 0.114 23 .110 300 79 TABLE XV SQUARES SQUARE ROOTS RECIPROCALS n if Vi VlOn 1/n n R> V^ VlOn l/n 1.00 1.0000 1.00000 3.16228 1.000000 1.60 2.2500 1.22474 3.87298 .666667 1.01 1.0201 1.00499 3.17805 .990099 1.51 2.2801 1.22882 3.S8587 .662262 1.02 1.0404 1.00095 3.19374 .080392 1.52 2.3104 1.23288 3.89872 .657896 1.03 1.0009 1.01489 3.20936 .970874 1.53 2.3409 1.23003 3.91152 .653595 1.04 1.0S16 1.01980 3.22490 .961538 1.54 2.3716 1.24097 3.02428 .649351 1.06 1.1025 1.02470 3.24037 .052381 1.65 2.4025 1.24400 3.03700 .645161 1.06 1.1236 1.02966 3.25576 .043306 1.56 2.4336 1.24900 3.04068 .641020 1.07 1.1440 1.03441 3.27100 .034579 1.57 2.4649 1.26300 3.00232 .036943 1.08 1.1604 1.03923 3.28634 .025920 1.58 2.4964 1.25008 3.07402 .632011 1.09 1.1881 1.04403 3.30151 .017431 1.59 2.5281 1.26005 3.08748 .628031 1.10 1.2100 1.04881 3.31662 .900001 1.60 2.6600 1.20401 4.00000 .625000 1.11 1.2321 1.05357 3.33167 .000901- 1.61 2.5921 1.26886 4.01248 .621118 1.12 1.2544 1.05830 3.34664 .892857 1.62 2.6244 1.27279 4.02402 .617284 1.13 1.2769 1.06301 3.36155 .884956 1.63 2.6569 1.27071 4.03733 .613407 1.14 1.2906 1.06771 3.37639 .877193 1.64 2.6806 1.28062 4.04060 .600756 1.15 1.3225 1.07238 3.39116 .860665 1.65 2,7225 1.28452 4.06202 .606061 1.16 1.3456 1.07703 3.40588 .862069 1.66 2.7550 1.28841 4.07431 .602410 1.17 1.3689 1.08167 3.42053 .854701 1.67 2.7889 1.29228 4.08650 .508802 1.18 1.3924 1.08628 3.43511 .847458 1.68 2,8224 1.29616 4.09878 .505238 1.10 1.4161 1.09087 3.44964 .840336 1.69 2.8561 1.30000 4.11006 .501716 1.20 1.4400 1.09545 3.46410 .833333 1.70 2.8900 1.30384 4.12311 .588235 1.21 1.4041 1.10000 3.47851 .826446 1.71 2.9241 1.30707 4.13621 .584706 1.22 1.4884 1.10454 3.49285 .819672 1.72 2.9584 1.31149 4.14729 .581305 1.23 1.5129 1.10905 3.50714 .813008 1.73 2.0029 1.31529 4.16033 .578035 1.24 1.5376 1.11355 3.62136 .806452 1.74 3.0276 1.31909 4.17133 .574713 1.25 1.5625 1.11803 3.53653 .800000 1.76 3.0625 1.32288 4.18330 .671429 1.26 1.5876 1.12260 3.64966 .793651 1.76 3.0976 1.32666 4.19624 .568182 1.27 1.6129 1.12694 3.56371 .787402 1.77 3.1320 1.33041 4.20714 .564072 1.28 1.6384 1.13137 3.67771 .781250 1.78 3.1684 1.33417 4.21900 .561708 1.20 1.6641 1.13578 3.59166 .776194 1.79 3.2041 1.33791 4.23084 .568669 1.30 1.6900 1.14018 3.60555 .769231 1.80 3.2400 1.34164 4.24264 .555556 131 1.7161 1.14455 3.61939 .763350 1.81 3.2761 1.34536 4.26441 .552486 1.32 1.7424 1.14891 3.63318 .757676 1.82 3.3124 1.34907 4.26615 .540451 1.33 1.7689 1.15326 3.64692 .751880 1.83 3.3489 1.35277 4.27786 .546448 1.34 1.7956 1.15768 3.66060 .746269 1.84 3.3856 1.35647 4,28052 .643478. 1.35 1.8225 1.16190 3.67423 .740741 1.85 3.4225 1.36015 4.30116 ..640541 1.36 1.8496 1.16619 3.68782 .735294 1.86 3.4596 1.36382 4.31277 .537634 1.37 1.8769 1.17047 3.70135 .729927 1.87 3.4969 1.36748 4.82435 .534750 1.38 1.0044 1.17473 3.71484 .724638 1.S8 3.5344 1.37113 4.33590 .531015 1.30 1.9321 1.17898 3.72827 .719424 1.89 3.5721 1.37477 4.34741 .529101 1.40 1.9600 1.18322 3.74166 .714288 1.90 3.6100 1.37840 4.36890 .526316 1.41 1.9881 1.18743 3.76600 .700220 1.91 3.0481 1.38203 4.37036 .523560 1.42 2.0164 1.10164 3.76829 .704225 1.Q2 3.6864 1.38564 4.38178 .620833 1.43 2.0449 1.19583 3.78153 .699301 1.03 3.7249 1.38024 4.30318 .518136 1.44 2.0736 1.20000 3.79473 .694444 1.04 3.7636 1.39284 4.40464 .515464 1.45 2:1025 1.20416 3.80780 .689656 1.95 3.8025 1.39642 4.41588 .612821 1.46 2.1316 1.20830 3.82099 .684932 1.96 3.8416 1.40000 4.42719 ,510204 1.47 2.1609 1.21244 3.83406 .680272 1.07 3.8809 1.40357 4,43847 .507614 1.48 2.1004 1.21655 3.84708 .675676 1,08 3.0204 1.40712 4.44972 , .505051 1.49 2.2201 1.22066 3.86005 .671141 1.09 3.9601 1.41067 4.46094 .502513 1.SO 2J2500 1.22474 3.87298 .666667 9,00 4,0000 1.41421 4.47214 .600000 n rt> v Vlon Vn n n V VfiF/i l/n 80 TABLE XV SQUARES SQUARE ROOTS RECIPROCALS n n V vlon l/ n n? VS VlOn 1/n 2.00 4.0000 1.41421 4.47214 .600000 2.50 6.2500 1.68114 6.00000 .400000 2.01 4.0401 1.41774 4.48330 .407512 2.51 6.3001 1.68430 6.00090 .398400 2.02 4.0804 1.42127 4.40444 .406060 2.52 6.3604 1.68746 5.01090 .396825 2.03 4.1200 1.42478 4.50555 .402611 2.53 6.4000 1.50060 5.02091 .396257 2.04 4.1616 1.42820 4.51664- .400106 2.64 6.4516 1.50374 5.03084 .303701 2.05 4.2025 1.43178 4.52760 .487805 2.55 6.5025 1.50087 5.04075 .392157 2.06 4.2436 1.43527 4.53872 .485437 2.50 0.6536 1.00000 5.05064 .390025 2.07 4.2840 1.43875 4.54073 .483002 2.57 6.0049 1.60312 5.00052 .380105 2.08 4.3264 1.44222 4.50070 .480760 2.68 6.0504 1,60024 5.07937 .387597 2.00 4.3681 1.44568 4.67165 .478400 2.50 6.7081 1.60935 6.08020 .386100 2.10 4.4100 1.44914 4.68258 .470190 2.00 0.7000 1,61245 6.00002 .384015 2.11 4.4521 1.45258 4.50347 .473034 2,61 0.8121 1.61555 6.10882 .383142 2.12 4.4044 1.45602 4.60435 .471698 2.62' 6.8644 1.61804 5.11850 .381679 2.13 4.5369 1.46045 4.61610 .400434 2.63 6.0160 1.62173 5.12835 .380228 2.14 4.5706 1.40287 4.62601 .407290 2.04 6.0606 1.02481 6.13800 .378788 2.15 4.6225 1.46620 4.63081 .486118 2.65 7.0225 1.62788 5.14782 .377368 2.10 4.6656 1.46060 4.04758 .432963 2.66 7.0750 1.03095 5.15752 .375040 2.17 4.7080 1.47300 4.66833 .400820 2.07 7.1289 1.03401 6.10720 .374532 2.18 4.7524 1.47048 4.00905 .458710 2.G8 7.1824 1.63707 6.17687 .373134 2.10 4.7061 1.47086 4.67974 .456621 2.60 7.2301 1.64012 5.18652 .371747 2.20 4.8400 1.48324 4.60042 .454545 2.70 7.2900 1.64317 6.10015 .370370 2.21 4.8841 1.48661 4.70108 .452489 2.71 7.3441 1.04021 5.20577 .360004 2.22 4.0284 1.48097 4.71169 .450450 2.72 7.3984 1.64024 5.21530 .367047 2.23 4.0720 1.40332 4.72220 .448430 7.73 7,4529 1.05227 5.22404 .300300 2.24 5.0176 1.40660 4.73286 .446429 2.74 7.5070 1.05520 5.23450 .364064 2.25 5.0625 1.50000 4.74342 .444444 2.75 7.6625 1.66831 5.24404 .303036 2.26 5.1070 1.50333 4.75305 .442478 2.76 7.0170 1.66132 5.26357 .302310 2.27 5.1520 1.50605 4.70445 .440529 2.77 7.0729 1.60433 5.20308 .301011 2.28 5.1084 1.60007 4.77403 .438500 2.78 7.7284 1.66733 5.27257 .350712 2.20 5.2441 1.51327 4.78530 .430681 2.70 7.7841 1.67033 5.28205 .358423 2.80 5.2000 1.51068 4.70583 .434783 3.80 7.8400 1.67332 5.20150 .357143 2.31 5.3301 1.51087 4.80625 .432000 2.81 7.8901 1.07031 5.30004 .365872 2.32 5.3824 1.62315 4.81664 .431034 2.82 7.9524 1.67920 6.31037 ,354010 2.33 5.4280 1.52043 4.82701 .429185 2.83 8.0089 1.68226 5.31077 .353367 2.34 5.4760 1.62071 4.83735 .427350 2.84 8.0060 1.08523 6.32017 .352113 2.35 5.5225 1.53207 4.84708 .426632 2.85 8.1225 1.08810 5.33854 .350877 2.30 6.5606 1.63G23 4.85708 .423729 2.86 8,1796 1.09116 6.34700 .349050 2.37 5.6100 1.53048 4.86826 .421941 2.87 8.2369 1.60411 5.35724 .348432 2.38 5.6644 1.54272 4.87852 .420168 2,88 8.2044 1.69700 5.30056 .347222 2.30 5.7121 1.64500 4.88876 .418410 2.80 8.3621 1.70000 6.37687 .346021 2.10 5.7000 1.64019 4.80808 .410067 2.00 8.4100 1.70204 6.38516 .344828 2.41 5.8081 1.55242 4.00918 .414038 2.01 8,4681 1.70587 5.30444 .343043 2.42 5.8564 1.65563 4.01935 .413223 2.02 8.5204 1.70880 6.40370 .342400 2.43 5,9049 1.56885 4.02050 .411523 2.93 8.6849 1.71172 5.41295 .341207 2.44 5.0536 1.56205 4.93004 .400836 2.94 8.6430 1.71404 5.42218 .340130 2.45 6.0026 1.66525 4,04075- .408163 2.05 8.7026 1.71756 5.43130 .338983 2.46 6.0610 1.50844 4,05984 .406504 2.06 8,7616 1.72047 6.44050 .337838 2.47 6,1009 1.67162 4.06901 .404858 2.07 8.8209 1.72337 5.44077 .336700 2.48 6.1504 1.67480 4.97906 .403220 2.08 8.8804 1.72627 5.45804 .335570 2.4Q 6.2001 1.67797 4.08009 .401600 2.99 8,0401 1,72916 6.40809 .334448 2.50 6.2500 1.68114 6.00000 .400000 3.00 0.0000 1.73205 6.47723 .333333 n If V^ ViOn 1/n n rt> V^ Vlbln Vn 81 TABLE XV SQUARES SQUARE ROOTS RECIPROCALS n ri> Vn" V5fn Vn n n Vn VlOn Vn 3.00 0.0000 1.73205 5.47723 .333333 3.50 12.2500 1.87083 6.91008 .286714 3.01 0.0601 1.73404 5.48635 .332226 3.51 12.3201 1.87360 6.92453 .284000 3.02 9.1204 1.73781 5,49545 .331126 3.52 12.3904 1.87017 5.93206 .284001 3.03 9.1809 1.74060 5.50454 .330033 3.63 12.4609 1.87883 6.04138 .283280 3.04 9.2416 1.74356 5.51362 .328047 3.54 12.5316 1.88140 6.94070 .282486 3.05 9.3025 1.74642 5.52268 .327869 3.55 12.6026 1.88414 5.05810 .281600 3.06 9.3636 1.74029 6.53173 .326707 3.56 12.6730 1.88680 5.00657 .280800 3.07 0.4240 1.75214 5.64076 .325733 3.57 12.7440 1.88044 6.07406 .280112 3.08 9.4864 1.75400 6.64977 .324875 3.58 12.8164 1.80200 5.08331 .270330 3.00 9.5481 1.75784 5.55878 .323625 3.50 12.8881 1.80473 5.00100 .278552 3.10 9.6100 1.76068 5.56776 .322581 3.00 12.9000 1.80737 0.00000 .277778 3.11 9.6721 1.76362 5.57674 .321543 3.01 13.0321 1.00000 0.00833 .277008 3.12 0.7344 1.76636 6.58570 .320513 3.62 13.1044 1.90203 0.01064 .276243 3.13 0.7969 1.76918 5.50464 .310480 3.03 13.1700 1.00526 6.02405 .276482 3.14 9.8506 1.77200 5,60357 .318471 3.64 13.2400 1.00788 6.03324 .274725 3.15 9.0225 1.77482 5.61249 .317460 3.65 13.3226 1.91050 0.04152 .273973 3.10 9.0856 1.77764 5.62130 ,316456 3.66 13.3950 1.01311 6.04070 .273224 3.17 10.0489 1.78045 5.63028 .315467 3.67 13.4689 1.91672 6.05805 .272480 3.18 10.1124 1.78326 5.63015 .314465 3.68 13.5424 1.91833 0.00030 ,271731) 3.19 10.1701 1.78606 5.64801 .313480 3.60 13.6161 1.02004 6.07454 .271003 3.20 10.2400 1.78885 6.65685 .312500 3.70 13.6000 1.02354 6.08276 .270270 3.21 10.3041 1.70165 5.66560 .311526 3.71 13.7641 1.02014 6.00008 ,260542 3.22 10.3684 1.79444 5.67450 .310659 3.72 13.8384 1.92873 6.00018 .268817 3.23 10.4329 1.7Q722 6.68331 .309598 3.73 13,9120 1.03132 6.10737 .208007 3.24 10.4976 1.80000 6.69210 .308042 3.74 13.9876 1.03301 6.11555 .287380 3.25 10.5626 1.80278 5.70088 .307092 3.75 14.0625 1.93640 6.12372 .200067 3.26 10.6276 1.80555 5.70964 .308748 3.76 14.1376 1.03007 6.13188 .266057 3.27 10.6929 1.80831 6.71839 .305810 3.77 14.2129 1.94105 6.14003 .265252 3.28 10.7584 1.81108 5.72713 .304878 3.78 14.2884 1.94422 6.14817 .264650 3.20 10.8241 1.81384 6.73685 ,303951 3.79 14.3641 1.94079 0.15630 .263852 3.30 10.8900 1.81650 5,74456 .303030 3.80 14.4400 1.04936 6.16441 .263158 3.31 10,0561 1,81034 6.75326 .302115 3.81 14.6161 1.96102 0.17252 .262467 3.32 11.0224 1.82209 5.76104 .301205 3.82 14.5024 1.95448 0.18081 .261780 3.33 11.0880 1.82483 6.77002 .300300 3.83 14.6680 1.05704 6.18870 .261007 3.34 11,1556 1.82757 6.77027 .209401 3.84 14.7450 1.05060 6.19677 .260417 3.35 11.2225 1.83030 5.78702 .298507 3.85 14.8225 1.06214 6.20484 .250740 3.36 11.2896 1.83303 5.79655 .207619 3.86 14.8006 1.06460 6.21280 .250067 3.37 11.3560 1,83576 6.80517 .206736 3.87 14.9769 1.06723 6.22003 .258308 3.38 11.4244 1.83848 5.81378 .295858 3.88 15.0544 1.06077 6.22896 .257732 3.39 11.4921 1.84120 5.82237 .294985 3.89 15.1321 1.07231 0.23000 .257000 3.40 11.5600 1.84391 5.83005 .294118 3.00 16.2100 1.07484 0.24500 .266410 3.41 11.6281 1.84662 5.83052 .293255 3.01 15,2881 1.97737 6.25300 .255754 3.42 11.6064 1.84932 5.84808 .292308 3.02 15.3064 1.07900 6.20000 .255102 4.43 11.7640 1.85203 5.85662 .291545 3.93 16.4449 1.98242 6.28897 .264453 3.44 11.8336 1.85472 5.86615 .200608 3.94 16.6236 1.98494 6.27694 .253807 3,45 11.9025 1.85742 5.87367 .289855 3.95 15.6026 1.08746 6.28400 .263165 3.46 11.9718 1.86011 5,88218 .280017 3.90 16.6816 1.98097 6.20285 .252525 3.47 12.0409 1.86270 6.89067 .288184 3.97 15.7609 1.99249 0.30070 .261880 3.48 12.1104 1.86548 6.89015 .287356 3.98 15.8408 1.99400 6.30872 .261256 ; 3,49 12,1801 1.86815 6.00762 ,286533 3.99 15,9201 1.99750 6.31064 .260027 3.50 12.2500 1.87083 6.01608 .285714 4.00 16.0000 2.00000 6.32456 .250000 '. n n Vn" Vl0n Vn n rt> Vn Vion Vn 82 TABLE XV SQUARES SQUARE ROOTS RECIPROCALS It if V^ VSTn Vn n a V^ VlOn Vn 4.00 lfi.0000 2.00000 6.32450 .250000 4.50 20.2500 2.12132 0.70820 .222222 4.01 10.0801 2.00250 0.33240 .249377 4.61 20.3401 2.12368 6.71665 .221729 4.02 10.1004 2.00400 6.34035 .248750 4.52 20.4304 2.12603 6.72309 .221230 4.03 10.2400 2.00749 6.34823 .248130 4.53 20.5200 2.12838 0.73063 .220751 4.04 10.3210 2.00008 0.35610 .247525 4.54 20.0116 2.13073 6.73796 .220264 4.05 10.4025 2.01240 0.36300 .240014 4.56 20.7025 2.13307 6.74537 .21078O 4.00 10.4830 2.01404 0.37181 .240305 4.50 20.7030 2.13542 6.76278 .210298 4.07 10.6040 2.01742 0.37000 .245700 4.57 20.8840 2:13776 6.70018 .218818 4,08 10.0404 2.01000 0,38740 .245008 4.58 20.9764 2.14000 6.70767 218341 4.00 10.7281 2.02237 0,30531 .244400 4.50 21.0681 2.14243 6.77496 .217866 4.10 10.8100 2.02485 0.40312 .243002 4.GO 21.1000 2.14476 6.78233 .217391 4.11 10.8021 2.02731 0.41003 .243300 4.01 21.2521 2.14709 6.78970 .21692O 4.12 10.0744 2.02978 0.41872 .242718 4.02 21.3444 2.14042 6.79706 .21645O 4.13 17.0500 2.03224 6.42051 .242131 4.03 2i:4300 2.16174 6.80441 .215983 4.14 17.1390 2,03470 6.43428 .241540 4.04 21.6206 2.15407 6.81176 .215617 4.15 17.2225 2.03715 6.44205 .240004 4.05 21.6225 2.15630 0.81009 .215054 4,10 17.30flO 2.03001 0.44081 .240385 4.06 21.7166 2.15870 6.82642 .214592 4.17 17.3880 2.04200 0.45755 .230808 4.07 21.8080 2.16102 6.83374 .214133 4.18 17.4724 2.04450 G.40520 .230234 4.08 21.9024 2.16333 6.84105 .213675 4,10 17.5501 2.04605 6.47302 .238063 4.00 21.9961 2.16604 6.84836 .213220 4.20 17.0400 2.04030 0.48074 .238005 4.70 22.0000 2.16795 6.85665 .212766 4.21 17,7241 2.05183 6.48845 .237630 4.71 22.1841 2.17025 6.86294 .212314 4.22 17.8084 2.05426 6.40615 .230067 4.72 22.2784 2.17256 6.87023 .211864 4.23 17.8920 2.05070 6.50384 .230407 4.73 22.3720 2.17486 6.87750 .211416 4.24 17.0770 2.05013 6.51153 .235840 4.74 22.4076 2.17715 6.88477 .21097O 4.25 18.0025 2.00155 6.51020 .235204 4.75 22.5025 2.17945 6.89202 .210526 4.20 18.1470 2.00308 6.62687 .234742 4.70 22.0676 2.18174 6.89928 .210084 4.27 18.2320 2.00040 6.5U52 ,234102 4.77 22.7629 2.18403 6.90652 .209644 4.28 18.3184 2.0G882 6.S4217 .233045 4.78 22.8484 2.18632 6,01376 .2092O5 4.20 18.4041 2.07123 6.54081 .233100 4.79 22.9441 2.18801 6.92008 .208768 V 4.30 18.4000 2.07304 6.55744 .232558 4.80 23.0400 2.10080 6.92820 .208333 4.31 18.5701 2.07005 0.50506 .232010 4.81 23.1301 2.19317 6.93642 .2070OO 4.32 18.0024 2.07840 8.67207 .231481 4.82 23.2324 2.19545 0.94262 .207469 4.33 18.7480 2.08087 6.C8027 .230047 4.83 23.3280 2.10773 6.04982 .207039 4.34 18.8356 2.08327 6.58787 .230415 4.84 23.4250 2.20000 6.95701 .206612 4.35 18.0225 2.08507 0.60545 .220885 4.85 23.5226 2.20227 0.96410 .206186 4.30 10.0006 2.08806 6.60303 .220358 4.80 23,0100 2,20454 6.97137 .205761 4.37 10.0060 2.00045 0,01000 .228833 4.87 23.7109 2.20081 6.97864 .206339 4,38 10.1844 2.00284 6.01810 .228311 4.88 23.8144 2.20907 6.08670 ,204918 4.30 10.2721 2.00523 0.82571 .227700 4.80 23.9121 2,21133 6.90285 .204490 4.40 10.3600 2,00762 6.63325 .227273 4.90 24,0100 2.21360 7.00000 .204082 4.41 10.4481 2.10000 6.04078 .226757 4.01 24.1081 2,21585 7.00714 .203666 4.42 10.5304 2.10238 fl.64831 .220244 4.02 24.2064 2.21811 7.01427 .203252 4.43 10.0240 2.10470 0.65582 .226734 4.03 24.3049 2.22036 7,02140 .202840 4.44 10.7136 2.10713 0.00333 .225225 4.04 24.4086 2.22261 7.02851 .202429 4.45 10.8025 2.10050 0.07083 .224710 4.05 24.6026 2.22486 7.03662 .202020 4.4U 10.8010 2.11187 0.67832 .224215 4.00 24.0016 2.22711 7.04273 .201613 4.47 10.0800 2.11424 0.68B81 .223714 4.07 24.7000 2.22935 7,04982 .201207 4.48 20.0704 2.11000 0.00328 .223214 4.08 24,8004 2,23159 7-06691 .200803 4.40 20.1601 2.11896 0.70070 .222717 4.00 24.0001 2.23383 7.06399 ,200401 4.50 20.2500 2.12132 0.70820 .222222 5.00 26.0000 2,23607 7,07107 .200000 n n vS VlOn 1/n n 7 V VlSn Vn 83 TABLE XV SQUARES SQUARE ROOTS RECIPROCALS n n ^ Vlfln Vn n n> ^ ViOn Vn 5.00 25.0000 2.23607 7.07107 .200000 5.50 30.2500 2.34521 7.41620 181818 6.01 25.1001 2.23830 7.07814 .190601 5.61 30.3801 2.34734 7.42294 181488 5.02 25.2004 2.24054 7.08520 .100203 5.52 30.4704 2.34047 7.42067 181150 5.03 25.3000 2.24277 7.09225 .198807 5.53 30.5809 2.35100 7.43640 180832 5.04 25.4016 2.24490 7.00930 .108413 6.54 30.6916 2.35372 7.44312 .180505 6.05 25.5026 2.24722 7.10034 .108020 5.55 30.8025 2.35584 7.44983 .180180 5.06 25.6036 2.24944 7.11337 .107628 5.56 30.9136 2.36707 7.45054 .179856 5.07 25.7040 2.26167 7.12030 .107239 5.57 31.0249 2.30008 7.40324 .179533 6.08 25.8064 2.25389 7.12741 .196850 5.58 31.1364 2.36220 7.40994 .170211 5.09 25.0081 2.25610 7.13442 .196464 5.50 31.2481 2.36432 7.47003 .178801 5.10 20.0100 2.25832 7.14143 .106078 5.00 31.3600 2.30643 7.48331 .178571 5.11 26.1121 2.26053 7.14843 .195095 5.61 31.4721 2.36854 7.48999 .178253 5.12 20.2144 2.26274 7.15542 .195312 5.02 31.5844 2.37065 7.49667 .177936 6 13 28.3160 2 26405 7.16240 .104032 5.63 31.0000 2.37276 7.50333 .177020 6.14 26.4106 2.26716 7.16938 .104553 5.64 31.8096 2.37487 7.60999 .177306 5.15 6.10 5.17 6.18 5.19 26.5225 26.6266 26.7280 26.8324 26.0361 2.26936 2.27166 2.27376 2.27696 2.27816 7.17636 7.18331 7.10027 7.19722 7.20417 .194175 .193798 .193424 .193050 .192078 5.06 5.00 5.67 5.08 5.69 31.0225 32.0350 32.1489 32.2024 32.3761 2.37607 2.37908 2.38118 2.38328 2.38537 7.51Q65 7.52330 7.52994 7.53858 7.64321 .178991 .178678 .176387 .176056 .175747 5.20 5.21 5.22 5.23 5.24 27.0400 27.1441 27.2484 17.3620 27.4576 2.28035 2.28254 2.28473 2.28692 2.28910 7.21110 7.21803 7.22400 7.23187 7.23878 .192308 .191939 .101571 .191205 .190840 5.70 5.71 6.72 5.73 5.74 32.4900 32.6041 32.7184 32.8329 32.0470 2.38747 2.38056 2.30105 2.39374 2.39583 7.64983 7.55645 7.6Q307 7.50968 7.57628 .175439 .17C131 ,174825 .174620 .174216 5.26 5.26 6.27 5.28 5.29 27.6025 27.6676 27.7729 27.8784 27.0841 2.20120 2.29347 2.29565 2.29783 2.30000 7.24600 7.26259 7.25048 7.26636 7.27324 .100476 .190114 .180753 .180304 .180036 5.75 5.76 5.77 5.78 6.79 33.0625 33.1770 33.2020 33.4084 33.5241 2.39702 2.40000 2.40208 2.40410 2.40624 7.58288 7.68947 7.59005 7.60263 7.60920 .173013 .173611 .173310 .173010 .172712 5.30 5.31 5.32 5.33 5.34 28.0000 28.1001 28.3024 28.4080 28.5150 2.30217 2.30434 2.30651 2.30808 2.31084 7.28011 7.28697 7.20383 7.30068 7.30753 .188079 .188324 .187070 .187017 .187266 5.80 5.81 5.82 5.83 5.84 33.6400 33.7561 33.8724 33.0880 34.1050 2.40832 2.41039 2.41247 2.41454 2.41661 7.01577 7.62234 7.02889 7.63644 7.04199 .172414 .172117 .171821 .171527 .171233 5.35 5.30 6,37 5.38 5.30 28.6225 28.7206 28.8360 28.0444 20.0521 2.31301 2.31517 2.31733 2.31048 2.32164 7.31437 7.32120 7.32803 7.33486 7.34100 .186016 .180567 .136220 .185874 .185529 5.85 5.86 5.87 5.88 5.89 34.2225 34.3306 34.4560 34.5744 34.6021 2.41868 2.42074 2.42281 2.42487 2.42003 7.04863 7.0S500 7.00169 7.00812 7.67403 .170940 .170040 .170358 .170008 .160770 5.40 5.41 6.42 5.43 5.44 20.1600 20.2081 29.3764 20.4840 20.5036 2.32379 2.32594 2.32800 2.33024 2.33238 7.34847 7.35527 7.36200 7.30886 7.37604 .185185 .184843 .184502 ,184102 .183824 5.90 6.01 5.02 5.03 5.04 34.8100 34.9281 35.0464 35.1040 35.2836 2.42899 2.43105 2.43311 2.43510 2.43721 7.08115 7.08765 7.00416 7.70006 7.70714 .160402 .100205 .108019 .168034 .168350 5.45 5.40 5.47 5.48 5.40 20.7026 20.8116 20.0200 30.0304 30.1401 2.33452 2.33060 2.33880 2.34004 2.34307 7.38241 7.38918 7.39594 7.40270 7.40045 .183480 .183150 .182815 .182482 .182149 5.06 5.06 5.07 5.08 5.00 35.4025 35.5216 35.6400 35.7604 35.8801 2.43020 2.44131 2.44336 2.44540 2.44745 7.71362 7.72010 7.72858 7,73305 7.73051 .168067 .107785 .107504 .107224 .160045 5.50 30.2500 2.34521 7.41620 .181818 0.00 30.0000 2.44040 7.74697 .160607 n n* V^ VlOn 1/n n n V^ VlOn 1/n 84 ABLE XV SQUARES SQUARE ROOTS RECIPROCALS * v; VBT 1/n n 71" A VlOn 1/n uo.oooo 2.44040 7.74B07 .100007 G.50 42.2600 2.54051 8.00226 .163846 M0.1201 2.451 fi3 7.75242 .160380 0.51 42.3801 2.55147 8.00846 .153610 .'HI.2404 2.45H. r .7 7.75887 .106113 6.52 42.5104 2.55343 8.07465 .163374 30.3000 2.455(11 7.70531 .106837 6.53 42.0400 2.55539 8.08084 .163139 30.4810 2.45704 7.77174 .100501) 6.64 42.7710 2.55734 8.08703 .152005 30.0025 2.45007 7.77817 .105280 0.55 42.0025 2.55030 8.09321 .152072 30.723(1 2.40171 7.78460 .106017 0.50 43.0330 2.50125 8.00938 '. 152439 :ill.H14l) 2.40374 7.70102 .104745 0.57 43.1040 2.50320 8.10555 .152207 30.0004 2.40577 7.70744 .104474 0.58 43.2004 2.50515 8.11172 151076 37.0881 2.40770 7,80385 .104204 0.50 43.4281 2.50710 8.11788 .161745 37.2100 2.40082 7.81026 .103034 0.60 43.5000 2.60005 8.12404 .151515 37.3321 2.471 H4 7.81006 .103060 0.01 43.0021 2.67000 8.13010 .151286 37.4544 2.47:18(1 7.82304 .103300 0.02 43.8244 2.57294 8.13634 .151067 37.6700 2.475H8 .103132 0.03 43.0560 2.57488 8.14248 .160830 37.0000 2.47700 7'.83682 .102800 0.04 44,0800 2.57082 8.14802 .160602 H7.8225 2.47002 7,84210 .102(502 0.05 44,2225 2.57870 8.15475 .160370 37,045(1 2.4X103 7.84857 .102338 0.00 44.355U 2.58070 8.10088 .150160 38.00KD 2.4K305 7.85403 .102075 0.07 44.4880 2.58263 8.10701 .149925 38.1024 2.48.100 7.80130 .101812 O.OS 44.0224 2.58-157 8.17313 .140701 38.3101 2.48707 7.80700 .101551 0.00 4-1.7601 2.58060 8.17824 .149477 38.4400 2.48008 7.87401 .101200 6.70 44.8000 2.58844 8.18635 .149264 3K.5041 2.41)100 7.88030 .101031 0.71 45.0241 2.50037 8.10146 .149031 38.0884 2.40300 7.8S070 . .100772 0,72 45.1684 2.50230 8.19758 .148810 38.8120 2.40000 7.80303 .100514 0.73 45.2020 2,60422 8.20306 .148588 3S.0370 2.40800 7.80037 ,100250 0.74 45.4270 2.50015 8.20075 .148368 30.0025 2.50000 7.00fiOO .100000 0.76 45.6025 2.50808 8.21684 .148148 30.1870 2.60200 7.01202 .160744 0.70 45.6070 2.00000 8.22192 .147929 iiO.3120 2.5O400 7.91&13 .150400 0.77 45.8320 2.00192 8.22800 .147710 30 4384 2.60.WO 7.02406 .159230 0.78 45.0084 2.00384 8.23408 .147493 8!SQ41 2.50700 7.03005 .158983 0.70 40.1041 2.00070 8.24016 .147276, 30.0000 2.50008 7.03726 .168730 6.80 46.2400 2.00708 8.24021 .147050 30.8101 2.51107 7.94.'155 .158470 0.81 40.3761 2.00000 8.25227 .140843 30.0424 2.51.-IOO 7.0408-1 ,158228 6.82 40.6124 2.01151 8.25833 .146028 40.0080 2.61fiOfi 7.05013 .157078 0.83 46,0480 2.01348 8.20438 .146413 40.105G 2.51704 7.0(1241 ,157720 6.84 40.7866 2.01634 8.27043 .146109 40.3225 2.61002 7.00809 .157480 0.85 40.0225 2.01725 8.27047 .145085 40.4400 7.07400 .157233 0.80 47.0500 2.01010 8.28251 .146773 40.5700 rt j|O'JU() 7.08123 .150080 0,87 47.1000 2.02107 8.28856 .145560 40.7044 o '*" I |U7 7.0H740 .150740 0.88 47.3,'J44 2.02208 8.20458 .146349 40.8321 2.527K4 7.00375 .150405 0.80 47.4721 2.02488 8.30060 .145138 40.0000 2.52082 8.00000 .150250 6.00 47.0100 2.02079 8.30662 .144928 41.08K1 2.531 HO 8.00025 .150000 0.01 47.7481 2.02800 8.31264 .144718 41.2154 2.6UH77 8.01240 .155703 0.02 47,8804 2.03050 8.31806 .144600 41.3440 2 63fi74 8.01 873 .155521 0.03 48.0240 2,63240 8.32406 .144300 41.4730 2i6772 8.02400 .165280 0.04 48.1030 2.03439 8.33067 .144092 41.0025 2 53001) 8,03110 .155030 0,06 48.3025 2.03020 8.33067 .143885 41.7310 2>)41n 8.03741 .154700 0.00 48.4410 2.03818 8.34200 .143678 41.8000 8.04303 .154600 0.07 48.5800 2.04008 8.34805 .143472 41.0004 B^45.TH 8.04084 .154321 O.OS 48.7204 2,04107 8.36404 .143266 42.1201 2.54708 8.05(106 .154083 6.00 48.8001 2,04880 8.36002 .143062 42.2500 2.54001 8,00220 .153846 7.00 40.0000 2.04675 8.30660 .142857 n' Vn vB5 1/n n tfi vS vlon Vn 85 TABLE XV SQUARES SQUARE ROOTS RECIPROCALS n **\ _ V " N/lOn 1/n n na v VlOn 1/n 7.00 *&*&_' 8.36060 142857 7.50 6.2500 2.73801 8.66025 133333 7.01 IIWKD4 8.37257 142653 7.51 6.4001 2.74044 8.60603 133166 7.02 ''.QffX 8.37854 142460 7.52 6.5504 2.74226 8.67179 132979 7.03 9.4209 irifcyBL B.38451 142248 7.53 56.7009 2.74408 8.07756 132802 7.04 9.5G1G 2)Q5^m JL39047 142046 7.54 6.8510 2.74591 8.68332 132026 7.05 49.7025 2.GrilC s ^.30843 141844 7.55 67.0026 2.74773 8.68907 132450 7.00 49.8430 2.65707 8.41238 141643 7.56 57.1530 2.74955 8.69483 132275 7.07 49.0849 8.41*33 141443 7.57 57.3049 2.75130 8.70057 .132100 7.08 60.1204 2.60093 SA&7 .141243 7.58 57.4504 2.75318 8.70032 .131920 7.09 60.2081 2.60271 A ,832021 .141044 7.59 67.0081 2.75500 8.71206 .131752 7.10 60.4100 2.60458 8.42615 .140845 7.60 67.7000 2.75081 8.71780 .131579 7.11 60.5521 2.60646 8.43208 .140047 7.61 67.9121 2.75802 8.72363 .131400 7.12 60.6944 2.60833 8.43801 .140449 7.62 58.0644 2.70043 8.72920 .131234 7.13 60.8369 2.67021 8.44393 .140252 7.63 58.2109 2.76226 8.73499 .131002 7.14 60.9796 2.67208 8.44985 .140050 7.64 68.3090 2.70405 8.74071 .130890 7.16 61.1226 2.67306 8.46577 .139860 7.65 68.6225 2.70580 8.74043 .130719 7.10 51.2056 2.67682 8.40108 .139065 7.06 68.0750 2.70707 8.75214 .130548 7.17 51.40S9 2.07769 8.40759 .139470 7.67 68.8289 2.76948 8.75786 .130378 7.18 51.5524 2.67955 8.47349 .139276 7.68 58.9824 2.77128 8.70356 .130208 7.19 51.6961 2.68142 8.47939 .139082 7.69 69.1361 2.77308 8.70920 , .130039 7.20 51.8400 2.68328 8.48528 .138889 7.70 59.2900 2.77489 8.77490 .129870 7.21 51.9841 2.68514 8.49117 .138690 7.71 59.4441 2.77669 8.78000 .129702 7.22 52.1284 2.68701 8.49700 .138604 7.72 59.5984 2.77849 8.78036 .129534 7.23 52.2729 2.68887 8.50294 .138313 7.73 69.7529 2.78029 8.79204 .129366 7.24 52.4170 2.69072 8.50882 .138122 7.74 59/9076 2.78209 8.79773 .129199 7.26 52.5026 2.69258 8.51469 .137931 7.75 60.0625 2.78388 8.80341 .129032 7.26 52.7076 2.69444 8.52050 .137741 7.76 60.2170 2.78568 8.80909 .128800 7.27 52.8529 2.69629 8.52643 .137562 7.77 60.3729 2.78747 8.81470 .128700 7.28 52,9984 2.69815 8.53229 .137363 7.78 00.6284 2.78927 8.82043 .128635 7.29 63.1441 2.70000 8.53815 .137174 7.79 00.6841 2.79100 8.82010 .128370 7.30 63.2900 2.70185 8.54400 .136986 7.80 00.8400 2.79285 8.83170 .128205 7.31 53.4361 2.70370 8.54985 .136799 7.81 00.9901 2.79464 8.83742 .128041 7.32 53.5824 2.70655 8.56570 .136612 7.82 61.1624 2.79043 8.84308 .127877 7.33 53.7289 2.70740 8.56154 .136426 7.83 61.3089 2.79821 8.84873 .127714 7.34 53.8750 2.70924 8.50738 .136240 7.84 61.4666 2.80000 8.85438 .127651 7.36 54.0225 2.71109 8.57321 .136054 7.86 61.6226 2.80179 8.80002 .127389 7.30 54.1600 2.71293 8.57904 .135870 7.86 61.7796 2,80357 8.80500 .127220 7.37 54.3169 2.71477 8.58487 .135685 7.87 01.9309 2.80535 8.87130 .127005 7.38 54.4644 2.71602 8.59009 .135601 7.88 62.0944 2.80713 8.87694 .126904 7.39 54.612 2.71846 8.59651 .135318 7.89 62.252 2.80891 8.88257 .120743 7.40 54.760 2.72029 8.60233 .135135 7.0 62.4100 2.81069 8.S8819 .120582 7.41 54.00S 2.72213 8.60814 .134953 7.91 02.568 2.81247 8.89382 .126422 7.42 55.0564 2.72307 8.61394 .13477 7.92 62.7204 2.81425 8.89944 .126263 7.43 55.204 2.72580 8.61974 .134590 7.93 02.884 2.81603 8.90506 -.126103 7.44 55.353 2.72764 8.02554 .13440 7.94 63.043 2.81780 8.91067 .125945 7.46 55.502 2.72947 8.63134 .13422 7.96 63.202 2.81967 8.91628 .125780 7.40 55.651 2.73130 8.63713 .13404 7.96 93.361 2.82135 8.92188 .125628 7.47 66.800 2.73313 8.64292 .13386 7.97 63.520 2.82312 8.92749 .125471 7.48 55.0504 2.73496 8.64870 .13369 7.98 63.6804 2.82489 8.93308 .125313 7.40 56.100 2.73679 8.65448 .13351 7.99 03.840 2.82066 8.93868 .126150 7.50 56.250 2.73861 8.60025 .13333 8.00 64.000 2.82843 8.04427 .125000 n n V^ v^On .1/n n n V^ VlOn 1/n TABLE XV SQUARES SQUARE ROOTS RECIPROCALS n n a v VlOn 1/n n n* v^ VlOn 1/n 8.00 8.01 8.02 8.03 8.04 04.0000 04.1001 04.3204 04.4800 04.0410 2.82843 2.83019 2.83100 2.83373 2.83549 8.04427 8.04080 8.05545 8.00103 8.06000 .125000 .124844 .124088 .124533 .124378 8.50 8.51 8.52 S.53 8.54 72.2500" 72.4201 72.6004 72.7009 72.9310 2.01548 2.01719'. 2.01890- 2.02002 2.02233 9.21954 0.22/07 9.23038 0.23580 9.24121 .117647 .1J7609 .117371 .117233 .117000 8.05 8.00 8.07 8.08 8.09 04.8025 04.0030 05.1249 05.2804 ,05.4481 2.83725 2.83901 2.84077 2.84253 2.84429 8.07218 8.97775 8.08332 8.08SSS 8.00444 .124224 .124000 .123910 .123702 .123009 8.55 8.56 8.57 8.58 8.50 73.1025 73.2730 73.4440 73.0104 73.7881 2.92404 2.92C76 2.92746 2.02910 2.03087 0.24602 9.25203 0.25743 0.20283 0.20823 .110050 .110822 .110080 .110550 .116414 8.10 8.11 8.12 8.13 8.14 05.0100 05.7721 05.9344 00.0900 06.2590 2.84005 2.84781 2.84050 2.85132 2.85307 0.00000 0.00555 0.01110 0.01006 0.02210 .123457 .123305 .123153 .123001 .122850 8.60 8.01 8.02 8.03 8.04 73.0000 74.1321 74.3044 74.4709 74.0490 2.03258 2.03428 2.93508 2.03700 2.93039 0.27362 0.27901 0.28440 0.2897S 9.20510 .116270 .110144 .110000 .115875 .115741 8.1S 8.10 8.17 8.18 8.10 00.4225 00.5850 00.7480 00.0124 07.0701 2.85482 2.85657 2.85832 2.80007 2.80182 9.02774 9.03327 0.03881 0.04434 0.04080 .122009 .122540 .122300 .122240 .122100 S.05 8.00 8.07 8.08 8.00 74.8225 74.0050 75.1080 75.3424 75.6101 2.94109 2.94279 2.04449 2.94018 2.94788 0.30054 0.30591 0.31128 9.31605 0.32202 .115607 .115473 .115340 .115207 .116076 8.90 8.21 8.22 8.23 8.24 07.2400 07.4041 Q7.S084 07.7329 07.8070 2.80350 2.80531 2.80706 2.80880 2.87054 0.0/3530 0.00001 0.00042 9.07103 0.07744 .121951 .121803 .121055 .121507 .121359 8.70 8.71 8.72 8.73 8,74 75.0000 75.8041 70.0384 70.2129 76.3870 2.94058 2.95127 2.05200 2.05400 2.05035 0.32738 9.33274 0.33800 0.34345 9.34880 .114043 .114811 .114679 .114548 .114410 8.25 8.20 8.27 8.28 8.29 08.0625 08.2270 08.3920 68.5584 68.7241 2.87228 2.87402 2.87570 2.87750 .2.87024 0.08205 0.08845 9.00305 9.00945 0.10404 .121212 .121005 .120010 .120773 .120027 8.75 8.70 8.77 8.78 8.70 70.5025 70.7370 76.0129 77.0884 77.2041 2.95804 2.05973 2.00142 2.06311 2.00479 9.35414 9.35940 9.30483 0.37017 9.37560 .114280 .114165 .114025 .113806 .113700 8.30 8.31 8.32 8.33 8.34 68.8000 00.0501 00.2224 09.3880 00.5556 2.88007 2.88271 2.88444 2.88017 2.88701 0.11043 fl.11502. 9,12140 0.12088 0.13230 .120482 .120337 .120102 .120048 .110004 8.80 8.81 8.82 8.83 8.84 77.4400 77.8101 77,7024 77.0089 78.1456 2.90048 2.00810 2.00086 2.07163 2.07321 0.38083 0.38010 0.30140 0.30081 0.40213 .113030 .113507 .113370 .113250 .113122 8.35 8.30 8.37 8.38 8.30 09.7225 09,8800 70.0500 70.2244 70.3921 2.88904 2.80137 2.80310 2.80482 2.80055 0.13783 0.14330 0.14877 0.15423 9.15000 .110700 .110017 .110474 .110332 .110100 8.85 8.80 8.87 8.88 8.80 78.3226 78.4000 78.0700 78.8544 70,0321 2.07480 2.07658 2.07825 2.07903 2,08101 0.40744 0.41270 0.41807 0.42338 0.42808 .112094 .112807 .112740 .112613 .112480 8.40 8.41 8,42 8.43 8.44 70.5000 70.7281 70.8064 71.0040 71.2330 2.80828 2.90000 2.00172 2.00345 2.00517 0.10,11/5 0.17001 0.17000 0.18150 0.18006 .110048 .118000 .118705 .118024 .118483 8.00 8.01 8.02 8.03 8.04 70.2100 70.3881 70.5004 70.74-10 70.0230 2.08320 2.08-100 2.08004 2.08831 2.08098 0.43398 9.43028 0.44468 0.44087 0.45510 .112300 .112233 .112108 .111082 .111857 8.45 8.40 8.47 8.48 8.40 71.4025 71.5710 71.7409 71.9104 72.0801 2,00080 2.00801 2.01033 2.01204 2.91370 0.10230 0.10783 0,20320 0.20800 0.21412 .118343 .118203 .118004 .117926 .117786 8.0i> 8.00 8.07 8.08 8.00 80.1025 80.2810 80,4000 80.0404 80.8201 2,00100 2.90333 2.99500 2,00006 2,00833 9.40044 0.40673 9.47101 0.47620 0,48160 .111732 .111007 .111483 .111369 .111236 8.50 72.2500 2.01548 0.21054 .117047 0.00 81.0000 3.00000 0.48083 .111111 n n V VlOn t/n n n Vn- VlOn Vn 87 TABLE XV SQUARES SQUARE ROOTS RECIPROCALS n n> v vTOn 1/n n n V/z N/lOn 1/n 9.00 9.01 0.02 9.03 0.04 81.0000 81.1801 81.3604 81.5409 81.7216 3.00000 3.00167 3.00333 3.00500 3.00666 9.48083 9.49210 9.49737 9.50263 9.50789 .111111 .110088 .110865 .110742 .110010 9.50 0.51 0.52 0.53 9.54 90.2500 90.4401 90.6304 00.8209 91.0116 3.08221 3.08383 3.08545 3.08707 3.08860 0.74079 0.76192 9.75705 0.70217 0.76720 .106263 .105162 .106042 .104932 .104822 0.05 9.06 9.07 0.08 0.00 81.9025 82.0836 82.2640 82.44G4 82.6281 3.00832 3.00998 3.01164 3.01330 3.01406 9.51315 9.51840 0.62365 9.52890 0.53415 .110497 .110376 .110254 .110132 .110011 9.55 9.56 9.57 9.58 9.50 91.2025 91.3930 01.6840 91.7704 91.0681 3.09031 3.00102 3.09354 3.00616 3.09677 0.77241 0.77753 0.78264 0.78776 0.79285 .104712 .104003 .104493 .104384 .104275 9.10 0.11 0.12 0.13 9.14 82.8100 82.9021 83.1744 83.3560 83.5396 3.01662 3.01828 3.01993 3.02150 3.02324 9.53939 9.54463 9.54087 9.56510 9,56033 .109890 ,109769 .109649 .109520 .109400 9.60 9.61 9.62 9.63 9.64 02.1600 02.3521 92.5441 02.7369 02.0206 3.00830 3.10000 3.10161 3.10322 3.10483 0.70790 9.80306 0.80816 0.81326 0.81835 . .104167 .104058 .103050 .103842 .103734 0.15 0.16 0.17 0.18 0.10 83.7225 83.9056 84.0889 84.2724 84.4561 3.02490 3.02655 3.02820 3.029&5 3.03150 9.56556 9.67079 9.57001 9.58123 0.68645 .100200 .100170 .109061 .108032 .108814 9.65 9.66 9.07 9.68 9.69 03.1225 03.3150 03.5080 03.7024 93.8961 3.10644 3.10805 3.10966 3.11127 3.11288 0.82344 0.82853 0.83362 0.83870 0.84378 .103627 .103520 .103413 .103306 .103100 9.20 0.21 0.22 0.23 0.24 84.6400 84.8241 85.0084 85.1929 85.3776 3.03315 3.03480 3.03645 3.03809 3.03974 9.50166 9.50687 9.60208 0.60729 9.61249 .108606 .108578 .108400 .108342 .108225 9.70 9.71 9.72 9.73 9.74 94.0900 04.2841 04.4784 94.0729 94.8676 3.11448 3.11609 3.11700 3.11020 3.12000 0.84886 9.85393 0.85001 0.80408 0.86014 .103003 .102087 102881 .102775 .102669 0.25 9.26 9.27 9.28 9.29 85.5625 85.7476 85.9320 86.1184 86.3041 3.04138 3.04302 3.04467 3.04631 3,04795 9.61769 9.62289 9.62808 9.63328 9.63846 .108108 .107001 .107876 .107750 .107643 9.75 9.78 9.77 9.78 9.79 95.0025 95.2576 95.4620 95.6484 95.8441 3.12250 3.12410 3.12570 3.12730 3.12800 0.87421 9.87027 0.88433 0.88039 9.89444 .102564 .102450 .102364 .102240 .102145 9.30 9.31 0.32 0.33 0.34 86.4900 86,6761 86.8624 87.0489 87.2356 3.04060 3.05123 3.05287 3.05450 3.05614 9.64365 9.64883 9.65401 9.65919 9.66437 .107527 .107411 .107206 .107181 .107060 9.80 9.81 9.82 0.83 0.84 90.0400 96.2361 06.4324 06.0289 06.8256 3.13050 3.13209 3.13369 3.13528 3.13088 9.80049. 9.00454 0.00050 0.01464 0.01008 .102041 .101037 .101833 ,101720 .101020 9.35 9.36 9.37 9.38 9.39 87.4225 87.6090 87.7909 87.9844 88.1721 3.05778 3.05041 3.06105 3.00268 3.06431 0.66954 9.67471 0,67988 9.68504 0.60020 .106052 .100838 .100724 ,100010 .100406 0.85 0.86 9.87 0.83 9.89 07.0226 97.2106 97.4109 97.6144 97,8121 3.13847 3.14006 3.14166 3.14326 3.14484 0.92472 0.02075 0.03470 0.93082 0.04485 .101523 .101420 .101317 .101216 .101112 9.40 9.41 9.42 0.43 0.44 88.3600 88.5481 88.7364 88.9243 89.1136 3.06504 3.00757 3.00920 3.07083 3.07246 0.09536 0.70052 0.70567 0.71082 0.71597 .100383 .100270 .100167 .100045 .105032 9.90 9.91 0.92 0.03 9.94 98.0100 08.2081 08.4064 98.0040 98.8036 3.14643 3.14802 3.14000 3,16110 3.16278 0.94087 9.95490 9.95992 0.00404 9.96096 .101010 .100008 .100800 .100705 .100604 0.45 9.46 9.47 9.48 .9.49 89.3026 89.4916 89.6809 89.8704 90.0601 3.07409 3.07571 3.07734 3.07896 3.08058 9,72111 9.72625 9.73139 9.73653 0.74166 ,105820 .106708 .105607 .405486 .105374 9.95 9.06 9,97 9.98 9.00 99.002C 00.2016 09.4000 90,0004 00.8001 3.15430 3,15595 3.15763 3.15011 3.16070 0.97497 0.07008 9.98499 0.08000 9.99500 .100503 .100402 400301 .100200 .100100 9.50 90.2500 3.08221 9.74670 .105263 10.00 100.000 3.16228 10.0000 .100000 n n* V v'lOn 1/n n n> V^ VWn 1/n 88 13841 156