.
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,
threehour 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 fiveplace table
of logarithms, the values of the interest and annuity functions for
twentyfive 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 fiveplace table of logarithms
is used in the computations. If the teacher considers it desirable to
use sevenplace logarithms, the author recommends the use of Glover's
Tables from Applied Mathematics. These excellent tables contain the
PREFACE V
values, and the sevenplace logarithms of the values, of the interest and
annuity functions, a standard sevenplace 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 longterm than
for shortterm 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.IJCIH, 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 1year
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 1year 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 6month transaction?
STTPPLEMENTABY NOTE. Formulas carx be obtained relating the discount
rate d and the equivalent interest rate r on 1year 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 1year 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
90day 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 : <, /
oiijiiiiliiiiii
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 noninterestbearing 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 noninterestbearing 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 60day noninterestbearing 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 45day non
interestbearing 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 6month 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 90day 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 semiannually.
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 semiannually. 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 semiannually, 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 semiannually?
, 18. Which rate is the better, (.062, m = 1) or (.06, m = 2) ?
19. Determine the nominal rate which, if converted semiannually, ./
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 theeffective
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 fiveplace 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.
SoluOon. 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 today 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 semiannually, yields the
effective rate .0725. '/ * / $ */ %
8. Find the nominal rate which, if converted semiannually, 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 sixmonth 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 shortterm
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 longterm 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 180day, noninterestbearing 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 +
tnoo 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
semiannually. 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 semiannually.
(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) n1 . 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 semiannually 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 semiannually, 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 lOlfo 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 semiannually 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 (10375) 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 semiannual 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 semiannually 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 semiannually ?
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 semiannually, 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 semiannually. 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,
8S'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 semiannually, 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 semiannual 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 
P03 "' ' 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) = iaL  , (^ 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(ai 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)= lay>. (37)
From equation (36) with n = % and x = .033,
(1.033)4= 1 +(.033) TMOSS^ + ^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 HRrnin.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 fouryear 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 semiannually 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.
Exampie 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 semiannually 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 semiannually 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 semiannually, 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 semiannually?
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 semiannually, 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 semiannually, 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 semiannually, 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 semiannually. 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 103 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 semiannually?
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 semiannually for 9 years. What interest rate,
compounded sewannually, 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 semiannually 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 exsoldier 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 semiannually, 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 semiannually, is to be
amortized by six semiannual 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
semiannually on all amounts remaining due. He wishes to discharge
his debt, principal and interest included, by twelve equal semiannual
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
8t 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 semiannually. (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 seraiannually at the rate
5 %, is to be amortized in 10 years by equal semiannual 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 semiannually, 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 semian
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
semiannually. 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 semiannually 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 semiannual 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 semiannual 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 semiannually on the principal of
the debt is (.03) (10000) = $300. Semiannual
= $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 semiannually 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 semiannually , what is the
total semiannual 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 sermannually, 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 semiannually 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 semiannual expense if the debt is amortised by equal pay
ments at the end of each 6 months for 6 years. (&) Find the semian
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 semiannual 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 semiannually.
(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)
RAi + 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 semiannually.
Is it better to amortize the debt in 6 years by equal semiannual 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 semiannual 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 semiannually 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 semiannually
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 semiannual payments of
$500, the first due immediately, amortize a debt of $9700? Determine
both the nominal rate, compounded semiannually, 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 semiannually, how many full
payments of $2000 will be made?
17. Under what nominal rate, compounded semiannually, 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.
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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)
fiFi + DFi + JVV ?(*+ 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_ . 9p (.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,
KC + 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 semiannual
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 semiannually, 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 5place 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 oilboom 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 cutover 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 semiannually, 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,
semiannual 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 semiannually , 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 semiannual dividend D = $15
is semiannual 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 semiannually,
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 semiannual 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 semiannually, will be
redeemed at 108% at the end of 7 years. Find the price to yield an in
vestor 6%, compounded semiannually.
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%
semiarm.
(.06, m = 2)
3.
100
17 yr.
6%
annually
(.07, ro = 1)
4.
1,000
14 yr.
7%
semiann.
(.08, m = 1)
6.
500
9yr.
8%
semiann.
(.04, m = 2)
6.
2,000
7yr.
4%
quarterly
(.05, m = 4)
7.
1,000
8 yr., 6 mo.
3%
semiann.
(.06, m = 2)
8.
1,000,000
13 yr., 6 mo.
5J%
semiann.
(.04, m = 2)
9.
100,000
19 yr.
5%
semiann.
(.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 semiannually, 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
SEMIANNUAL 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 serniannually, of a $1000,
5% bond, with dividends payable semiannually, 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 semiannually, 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 semiannually. 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 semiannually?
(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 semiannually 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 semiannually 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+ (7D). (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 semiannually 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. Today, 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 dueuntil 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 semiannually 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 today, 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 today.
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
lDp.
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+(PiPo).
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 semiannually,
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 semiannuaUy 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%
semiann.
111 years
2. a
100
par
160.30
7%
semiann.
25 years
S.a
1000
par
84.28
4%
semiann.
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 semiannual accumulation as in Rule 2, we obtain J(62.30)
= $10.4, a result very close to all of the semiannual 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 semiannual,
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 semiannual 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 semiannually.
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 semiannually, 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 semiannual 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%
semiann.
15 years
110.76
6.
par
3%
semiann.
19 years
83.30
6,
par
6%
annually
12 years
121.00
7.
105%
6%
semiann.
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 semiannual 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 semiannually, 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 tenyear, $10,000 annuity bond with the
dividend rate 5% is redeemable in semiannual 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 semiannually 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 5year annuity bond for $20,000, with the dividend rate 6%,
payable semiannually, 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 semiannually 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
semiannually 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 semiannually 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
semiannually, 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 reinvested 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 semiannual deposits
of $250 in a bank paying interest semiannually 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 semiannually?
4. A debt of $100,000 bears interest at 6%, payable semiannually.
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 semiannually, what is
the total semiannual expense of the debt?
5. A debt of $100,000 is contracted under the agreement that interest
at 6% shall be paid semiannually 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 semiannual
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 semiannually 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 90day, noninterestbearing 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 semiannually, 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 semiannual 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 semiannually. 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 semiannually. Then, commencing with a first payment at the
end of 10J years, semiannual 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 semiannually, 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 90day 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 semiannually 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 semiannual 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 semiannual 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 semiannually. (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 90day 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 semiannually.
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%, 20year 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. 9Qday n.gt;Q for $1000, bearing 6%
interest, is discounted at 8%?
142 MATHEMATICS OF INVESTMENT
66. X requests a 60day 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 semiannually, 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 semiannual deposits, the first due im
mediately and the last at the end of 7 years. Find the semiannual
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 semiannual
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% semiannually 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 semiannually, 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
semiannual 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 exsoldiers, 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 today (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 semiannual 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 semiannual
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 Bemiannual 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 semiannually. 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 semiannual deposits at (.04, m = 2). The corporation also
can raise money by issuing, at par, 15year, 7% annuity bonds redeemable
in semiannual installments, (a) Which method would entail the least
semiannual 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 semiannually 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 semiannual payments, the first due immediately, (a) What will
be the semiannual payment if the creditor, in computing it, uses the rate
146 MATHEMATICS OF INVESTMENT
5%, compounded semiannually? (6) What will be the semiannual
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 tomorrow. Hence, there are only two
possibilities to consider, and the probability of dying before tomorrow 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+ii. (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 $4 of the expectation is given by
A = PS(1 + if". (10)
NOTE. In the future, the arithmetical work in all examples will be per
formed by 5place logarithms.
Example 1. If money is worth 3$%, 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 nyear 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 ?iyear 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 4ryear 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
953! 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 Tiyear 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
Le
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)
na*  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
dwlS
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.*
"^
96s
,;.
^.,
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 nyear 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 5year term insurance gives no benefit unless
(jc) dies' within 5 years. Let A 1 ^ represent the present value of
an Tiyear term insurance for $1 on the life of (x}. To obtain the
value of A, assume that the company issues nyear 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;+nl
$dx+nl
f^s+nl
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 1year 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 3year 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 10year 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 10year 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 nyear 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 20year 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 nyear en
dowment insurance of $1 on the life of (x}. The present valuo
Axft is the sum of the present values of (a) the nyear term in
surance for $1 on the life of (x}, and of (6' the nyoar 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, 20year 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 20year term insurance for $1000
on the life of the man aged 23, by using (a) and (&).
'2. Find the net single premium for a 10year endowment insurance for
$5000 on the life of a inun aged 30.
3. (a) Find the net single premium for a 10year endowment insurance
for $3000 on the life of a man aged 26. (6) Find the net single premium
for a 10year term insurance for $3000 on his life, (c) From the results
of (a) and (&), find the present value of a 10year pure endowment of $3000
for the man.
f /4. How much 20year endowment insurance can a man aged 33 pur
chase for $3000 cash?
6. How much 10year 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 10year 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 10year 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
wpayment lif e
Whole life insurance
Annually for n years
III
?iyear term
7iyear term insurance
Annually for n years
rv
Tfcyear endowment
(a) Tiyear pure endowment
(6) 7iyear 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 npayment 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 nyear term insur
ance policy for $1 for (x) is given by
M x+n _

,..
(60)
Let P.^ be the net annual premium for an nyear endowment
policy for $1 for (x). The premiums paid by the man aged x form
an wyear 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 5payment 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 nyear 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 20payment life policy for $2000.
6. Find the net annual premium for a 10year term policy for $1000
for a man (a) aged 32 ; (&) aged 42 ; (c) aged 62.
7. Find the net annual premium for a 20year endowment policy for
$1000 for a person of your own age.
8. (a) Find the net annual premium for a 5year 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
1year 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 1year 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 1year term plan.
1 11. How much insurance on the 20payment 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 20year 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, wellmanaged 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 10year 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(jffaa).
LIFE INSURANCE
177
The annual premiums form a 28year 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) 10year 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 10year
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 1year 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 1year 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 5payment 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 JH 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 14year endowment insurance for $1000 for a man aged 30.
The remaining premiums form a 14year temporary annuity due. Let V be the
reserve at the end of 6 years. From equation 62,
1000(Jfao M + Z>) _ 39.086(^0  AT). (Formulas 46, 35)
Dao D$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, 10payment 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, 5payment 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, 20year 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 1year 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 wpayment 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 10year 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 10year 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 nyear 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 20year 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) 20payment life ; (b) 10year endow
ment; (c) ordinary whole life ; (d) 10year 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 8year 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 10payment life policy. At the end of
10 years he desires to convert it into a 20year 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 10year 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 logoW = 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:
lOgaJV* = 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 (15561630), 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 5place 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 5place 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 whichwe 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 6figure 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 5place 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,5place 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 (107) 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 882  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 8place 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(nl)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 TS = 5(1 r) = a ar n .
Hence, s = a f^ = 5^1 ( 18 )
Since I ar n1 , 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) " + (105) + (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  jrk
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 10year 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;
wpaymcnt life, 172; nyoar 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
npayment endowment policy, 172
Tirpaymont life policy, 172
7iyear 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 20.4
31.0 30.0 20.7
110
613
644
676
706
737
768
799
820
860
801
41
42
43
44
45
46
47
48
40
922
15229
534
836
16137
435
732
17026
319
953
259
664
866
167
465
761
056
348
983
290
694
897
197
495
791
085
377
+014
320
025
927
227
524
820
114
406
+045
351
655
967
256
554
850
143
435
+076
381
685
987
286
584
870
173
464
+106
412
715
+017
316
613
909
202
493
+137
442
746
+047
346
643
938
231
522
*168
473
770
+077
376
673
967
260
851
*198
503
806
+107
406
702
997
289
580
2
I
5
a
7
!
32 31 30
8.2 3Tl S70~
0.4 0.2 0.0
0.0 0.3 0.0
12.8 12,4 12,0
10.0 16.0 10,0
19.2 18.0 18.0
22.4 21.7 21.0
20.6 24.8 24.0
28.8 27.0 27.0
150
609
638
667
696
725
754
782
811
840
sto
IT '
1
2
8
4
, 5
6
7
8
ft
PP
N
1
2
3
4
5
6
1
8
9
]
pp
150
17609
638
667
696
725
754
782
811
840
800
51
52
53
898
18184
469
926
213
498
955
241
526
084
270
554
*013
298
583
*041
327
611
*070
355
639
+099
384
667
+127
412
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+1BO
441
724
29 28
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55
56
67
58
59
752
19033
312
590
800
20140
780
001
340
618
893
167
808
080
368
045
921
194
837
117
396
073
048
222
865
145
424
700
970
240
803
173
461
728
+008
270
021
201
479
756
+030
303
949
220
607
783
+058
330
077
257
535
811
+085
358
+006
285
562
838
+112
385
1
2
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17
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160
412
430
406
493
520
548
675
602
029
656
61
02
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710
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7 26
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564
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115
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107
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194
453
712
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220
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737
094
722
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248
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2
3
5
7
8
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IE
21
24
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096
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147
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72
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164
400
204
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724
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212
455
254
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237
479
270
527
773
+018
261
503
2
3
6
7
8
2.5
5.0
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10.0
12.5
15.0
17.5
20.0
22.5
180
C27
551
675
BOO
624
048
072
696
720
744
81
82
83
768
26007
245
702
031
200
810
055
203
840
070
310
804
102
340
888
120
304
912
160
387
036
174
411
950
198
435
083
221
458
Z
4 23
84
85
86
87
88
80
482
717
061
27184
416
646
50G
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207
430
660
520
704
908
231
402
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563
788
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254
485
716
670
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277
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738
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531
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023
858
+001
323
554
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340
577
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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
12273
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
217
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
4.7001 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
07063 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
0.4801 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
0.27312347
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.OlfiO0132
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 8081
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
0.8Ul.l flflao
0.7S211 (IS10
0.7l!1.1 7128
(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 4971
0.r>S20 0010
0.5430 337t
0.5083 4020
0.017!) (1162
0.0027 5400
0.5007 0223
0.5215 8347
0.4851 0303
0. llUDt (111(13
0.5S31 iioio
0./540S! 0888
0.5002 IS'.)?
o,i<m '.wit)
0.15] 21) .15011
o./iniii 2ii3. r >
o.fwoii on ir.
0.179H 70(18
0.M22 85ia
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 5113
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
0.3102 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 3100
0.3106 7130
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
0.211IM 58110
0.2302 8160
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 5118
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(121 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
0.0188 7Mfi
0.0460 4742
0.011 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 51K2
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
181.2780 6184
100
107
108
100
110
132.0280 7000
131.4828 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
140.0102 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
200.0444 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
201.0887 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
2w
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
218.3827 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
' (rfoCi + yi
"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
11.8360 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
3569406 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 2180
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
40.1817 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
44.4015 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
43.8410 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 7817
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 3011
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%
207
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
10.1010 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
00150 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
00148 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
00116 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
00099 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
0.61310295
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
00783 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
207
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%
QQ7
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
00302 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 0313
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 6113
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%
2V
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
1.0010 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. 0010 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 ', tfo
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
1.0021 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
0007 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
706691
.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
2.58157
8.17313
.140701
38.3101
2.48707
7.80700
.101551
0.00
41.7601
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
7.04081
,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 jO'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
70.7410
70.0230
2.08320
2.08100
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