IP Addressing
and
Subnetting
Workbook
Version 2.0
Instructor's Edition
1111111U
10010101
ooo:
11010011
IP Address Classes
Class A 1 - 127 (Network 127 is reserved for loopback and internal testing)
Leading bit pattern oooooooo.oooooooo.oooooooo.oooooooo
Network . Host Host Host
Class B 128 - 191 Leading bit pattern 10 10000000.00000000.00000000.00000000
Network . Network . Host Host
Class C 192-223 Leading bit pattern 110 11000000.00000000.00000000.00000000
Network . Network . Network . Host
Class D 224-239 (Reserved for multicast)
Class E 240 - 255 (Reserved for experimental, used for research)
Private Address Space
Class A 10.0.0.0 to 10.255.255.255
Class B 172.16.0.0 to 172.31.255.255
Class C 192.168.0.0 to 192.168.255.255
Default Subnet Masks
Class A 255.0.0.0
Class B 255.255.0.0
Class C 255.255.255.0
Produced by: Robb Jones
Special Thanks to Melvin Baker and Jim Dorsch
for taking the time to check this workbook for errors,
and to everyone who has sent in suggestions to improve the series.
Workbooks included in the series:
IP Addressing and Subnetting Workbooks
ACLs - Access Lists Workbooks
VLSM Variable-Length Subnet Mask Workbooks
Instructors (and anyone else for that matter) please do not post the Instructors version on public websites.
When you do this you are giving everyone else worldwide the answers. Yes, students look for answers this way.
It also discourages others; myself included, from posting high quality materials.
Inside Cover
Binary To Decimal Conversion
128 64 32 16 8 4 2 1 Answers Scratch Area
10 10
1110 1
146 4
11111111 255 2
/
110 1
11110 1
10
10
110
11110
11110
1110
1
1
146
1 1
1 79
1 1
255
1
197
1
246
1 1
19
1
129
1
^9
120
24 O
1 1
59
1 1
7
00011011
27
10101010
17 O
01101111
III
11111000
248
00100000
32
01010101
85
00111110
62
00000011
3
11101101
237
11000000
192
128 a
16 32
2_ 16
1 19
/
1
1
o
1
1
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238
-128
1 IO
34
-32
2
-64
-2
46
-32
O
14
-8
6
-4
2
-2
o
Decimal To Binary Conversion
Use all 8 bits for each problem
128 64 32 16 8 4 2 1 = 255^ Scratch Area
238
34
123
50
255
200
10
138
1
13
250
107
224
114
192
172
100
119
57
98
179
2
Address Class Identification
Address
Class
10.250.1.1
A
150.10.15.0
B
192.14.2.0
C
148.17.9.1
B
193.42.1.1
C
126.8.156.0
A
220.200.23.1
C
230.230.45.58
P
177.100.18.4
B
119.18.45.0
A
249.240.80.78
B
199.155.77.56
C
117.89.56.45
A
215.45.45.0
C
199.200.15.0
C
95.0.21.90
A
33.0.0.0
A
158.98.80.0
B
219.21.56.0
C
Network & Host Identification
Circle the network portion
of these addresses:
177^L0Oi8.4
^119^.8. 45.0
(209.240.80)78
(199.155.77)56
(117)89.56.45
(215.45.45)0
(192.200.15)0
(95)0.21.90
(33)0.0.0
(^8^98)80.0
(217.21.56)0
(5)250.1.1
(15010)15.0
(192.14.2)0
(148^17)9.1
(193 .42 j)l
^26;8.156.0
(220.200.23)1
4
Circle the host portion of
these addresses:
1Q.(L5.123.5
171.2.(199.31
198.125.87(177)
223.250.200(222,
17(45.222.45)
126(201.54.23j)
191.41(35^112)
155.25(1 69^22 7)
192.15.155ff
123(f02.45.254)
148.17(^15^
100(25X1)
195.0.21(98
25(250.135.46)
171.102Q77/77
55.(g 50.5.5 )
218.155.230(14)
10(250.1.1)
Network Addresses
Using the IP address and subnet mask shown write out the network address:
188.10.18.2 /3B • IO • ° ' °
255.255.0.0
10.10.48.80
255.255.255.0
192.149.24.191
255.255.255.0
150.203.23.19
255.255.0.0
10.10.10.10
255.0.0.0
186.13.23.110
255.255.255.0
223.69.230.250
255.255.0.0
200.120.135.15
255.255.255.0
27.125.200.151
255.0.0.0
199.20.150.35
255.255.255.0
191.55.165.135
255.255.255.0
28.212.250.254
255.255.0.0
IO .
IO .43 .O
/92
. 149
. 24 .O
150
. 203
.O .O
IO .
O . O
. O
I8G
. 13 .
23 . C?
223
. 69 .
O . <?
200
. 120
' . 135 . tf
27 .
O . C?
. C?
/99
. 20 .
150 . C?
/<?/
. 5~5~ .
/65~ . O
28 .
212 .
O . O
Host Addresses
Using the IP address and subnet mask shown write out the host address:
188.10.18.2
255.255.0.0
10.10.48.80
255.255.255.0
222.49.49.11
255.255.255.0
128.23.230.19
255.255.0.0
10.10.10.10
255.0.0.0
200.113.123.11
255.255.255.0
223.169.23.20
255.255.0.0
203.20.35.215
255.255.255.0
117.15.2.51
255.0.0.0
199.120.15.135
255.255.255.0
191.55.165.135
255.255.255.0
48.21.25.54
255.255.0.0
o
.O.I8.2
o
.0.0.80
o
.O .O . //
o
. O . 230 . /9
o
. IO . IO . IO
o
.O .O . //
o
.0.23.20
o
.O.O. 215
o
. /5 . 2 .9/
o
.O.O. 135
o
.O.O. 135
o
.O .25 .54
Default Subnet Masks
Write the correct default subnet mask for each of the following addresses:
177.100.18.4
119.18.45.0
191.249.234.191
223.23.223.109
10.10.250.1
126.123.23.1
223.69.230.250
192.12.35.105
77.251.200.51
189.210.50.1
88.45.65.35
128.212.250.254
193.100.77.83
125.125.250.1
1.1.10.50
220.90.130.45
134.125.34.9
95.250.91.99
255 .
255
' .o .o
255
. O .
o . o
255 .
255
.o .o
255 . 255 .
255 . O
255
. O .
O . O
255
. O .
O . O
255 . 255 .
255 . O
255 . 255 .
255 . O
255
. o .
O . O
255 .
255
.O .O
255
. O .
O . O
255 .
255
.O .O
255 . 255 .
255 . O
255
. O .
O . O
255
. O .
O . O
255 . 255 .
255 . O
255 .
255
.O .O
255
. O .
O . O
ANDING With
Default subnet masks
Every IP address must be accompanied by a subnet mask. By now you should be able to look
at an IP address and tell what class it is. Unfortunately your computer doesn't think that way.
For your computer to determine the network and subnet portion of an IP address it must
"AND" the IP address with the subnet mask.
Default Subnet Masks:
Class A 255.0.0.0
Class B 255.255.0.0
Class C 255.255.255.0
ANDING Equations:
1 AND 1 = 1
1 AND =
0AND1=0
AND =
Sample:
What you see...
IP Address:
192 . 100 . 10 . 33
What you can figure out in your head.
Address Class:
Network Portion:
Host Portion:
192 . 100 . 10 . 33
192 . 100 . 10 . 33
In order for you computer to get the same information it must AND the IP address with
the subnet mask in binary.
IP Address:
Default Subnet Mask:
AND:
Network
Host
11000000.01100100.00001010
11111111.01111111.11111111
00100001
00000000
11000000.01100100.00001010
00000000
(192 . 100 . 10 . 33)
(255 . 255 . 255 . 0)
(192 . 100 . 10 . 0)
ANDING with the default subnet mask allows your computer to figure out the network
portion of the address.
8
ANDING With
Custom subnet masks
When you take a single network such as 192.100.10.0 and divide it into five smaller networks
(192.100.10.16, 192.100.10.32, 192.100.10.48, 192.100.10.64, 192.100.10.80) the outside
world still sees the network as 192.100.10.0, but the internal computers and routers see five
smaller subnetworks. Each independent of the other. This can only be accomplished by using
a custom subnet mask. A custom subnet mask borrows bits from the host portion of the
address to create a subnetwork address between the network and host portions of an IP
address. In this example each range has 14 usable addresses in it. The computer must still
AND the I P address against the custom subnet mask to see what the network portion is and
which subnetwork it belongs to.
IP Address:
Custom Subnet Mask:
192 . 100 . 10 .
255.255.255.240
Address Ranges:
192.10.10.0 to 192.100.10.15
192.100.10.16 to 192.100.10.31
192.100.10.32 to 192.100.10.47 (Range in the sample below)
192.100.10.48 to 192.100.10.63
192.100.10.64 to 192.100.10.79
192.100.10.80 to 192.100.10.95
192.100.10.96 to 192.100.10.111
192.100.10.112 to 192.100.10.127
192.100.10.128 to 192.100.10.143
192.100.10.144 to 192.100.10.159
192.100.10.160 to 192.100.10.175
192.100.10.176 to 192.100.10.191
192.100.10.192 to 192.100.10.207
192.100.10.208 to 192.100.10.223
192.100.10.224 to 192.100.10.239
192.100.10.240 to 192.100.10.255
Network
Sub
Network Host
IP Address:
Custom Subnet Mask:
AND:
11000000.01100100.00001010
11111111.01111111.11111111
11000000.01100100.00001 0^6
Four bits borrowed from the host
portion of the address for the
custom subnet mask.
(192 . 100 . 10 . 33)
The ANDING process of the four borrowed bits
shows which range of IP addresses this
particular address will fall into.
In the next set of problems you will determine the necessary information to determine the
correct subnet mask for a variety of I P addresses.
How to determine the number of subnets and the
number of hosts per subnet
Two formulas can provide this basic information:
Number of subnets = 2 s (Second subnet formula: Number of subnets = 2 s - 2)
Number of hosts per subnet = 2 h - 2
Both formulas calculate the number of hosts or subnets based on the number of binary bits
used. For example if you borrow three bits from the host portion of the address use the
number of subnets formula to determine the total number of subnets gained by borrowing the
three bits. This would be 2 3 or 2 x 2 x 2 = 8 subnets
To determine the number of hosts per subnet you would take the number of binary bits used in
the host portion and apply this to the number of hosts per subnet formula If five bits are in the
host portion of the address this would be 2 5 or 2x2x2x2x2 = 32 hosts.
When dealing with the number of hosts per subnet you have to subtract two addresses from
the range. The first address in every range is the subnet number. The last address in every
range is the broadcast address. These two addresses cannot be assigned to any device in
the network which is why you have to subtract two addresses to find the number of usable
addresses in each range.
For example if two bits are borrowed for the network portion of the address you can easily
determine the number of subnets and hosts per subnets using the two formulas.
/9f. 223 . 50
O O
o o
The number of subnets
created by borrowing 2
bitsis2 2 or2x2 = 4
subnets.
O O
o o
The number of hosts created by
leaving 6 bits is 2 6 - 2 or
2x2x2x 2x2x2 = 64 -2 = 62
usable hosts per subnet.
What about that second subnet formula:
Number of subnets = 2 s - 2
In some instances the first and last subnet range of addresses are reserved. This is similar to
the first and last host addresses in each range of addreses.
The first range of addresses is the zero subnet. The subnet number for the zero subnet is
also the subnet number for the classful subnet address.
The last range of addresses is the broadcast subnet. The broadcast address for the last
subnet in the broadcast subnet is the same as the classful broadcast address.
10
Class C Address unsubnetted:
115. 223 .50.0
195.223.500 to 195. 223. 50. 25 5
Class C Address subnetted (2 bits borrowed):
Notice that the subnet and
broadcast addresses match.
195. 223 . 50 .O O O O O O
(iwAid r^ r ) (0) 195 .223.50. "
( 1) I95.223.50.G4
(2) 195.223.50.123
(l*v*lU r^ r ) (3) 195.223.50.192
O
*t t
t<
t<
o 195 .223.50. G3
o 195 .223.50. 127
o 195 .223.50. 19 1
o 195 .223.50.255
The primary reason the the zero and broadcast subnets were not used had to do pirmarily with
the broadcast addresses. If you send a broadcast to 195.223.255 are you sending it to all 255
addresses in the classful C address or just the 62 usable addresses in the broadcast range?
The CCNA and CCENT certification exams may have questions which will require you to
determine which formula to use, and whehter or not you can use the first and last subnets. Use
the chart below to help decide.
When to use which formula to determine the number of subnets
Use the 2 s - 2 formula and don't use the
zero and broadcast ranges if...
Use the 2 s formula and use the zero and
broadcast ranges if...
Classful routing is used
Classless routing or VLSM is used
RIP version lis used
RIP version 2, EIGRP, or OSPF is used
The no ip subnet zero command is
configured on your router
The ip subnet zero command is
configured on your router (default setting)
No other clues are given
Bottom line for the CCNA exams; if a question does not give you any clues as to whether or not
to allow these two subnets, assume you can use them.
This workbook has you use the number of subnets = 2 s formula.
11
Custom Subnet Masks
Problem 1
Number of needed subnets 14
Number of needed usable hosts 14
Network Address 192.10.10.0
Address class
Default subnet mask
255 . 255 . 255 . O
Custom subnet mask
255 . 255 . 255 . 240
Total number of subnets
1G
Total number of host addresses
l<o
Number of usable addresses
14
Number of bits borrowed
Show your work for Problem 1 in the space below.
25~6 128 G4 32
Number of
Sublets ~24 8 IG
128 G4 32 IG
112 .10.10,0000
Number of
IG
8 4
2 ~
Hosts
32
G4 128
25G
8
4 2
1 -
&fh\A.ru \/fi.lu\es
o o o o
J28
Add the binary value Cy4
numbers to the left of the line to
create the custom subnet mask. 3 2
+ T6
24 O
I S Observe the total number of
' ° hosts.
— . Subtract 2 for the number of
/ hT usable hosts.
12
Custom Subnet Masks
Problem 2
Number of needed subnets 1000
Number of needed usable hosts 60
Network Address 165.100.0.0
Address class
B
Default subnet mask
255 . 255 .O.O
Custom subnet mask
255 . 255 . 255 . /92
Total number of subnets
IP 24
Total number of host addresses
G4
Number of usable addresses
G2
Number of bits borrowed
IO
Show your work for Problem 2 in the space below.
B
^
^
^
so
-fe.
\o
.
^
Vi
^j
.
"Q
"Q
Q
<S\
Number of
so
so
-fe.
so
^
Hosts
1
|
1
1
1
1
1
|
1
|
1
I
1
1
1
1
Number of
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Sublets
2
4
8
16
32
64
128
256.
t'^Art,
< i/a.lu\es
- 128 64
32
16
8
4
2
1 .
Add the binary value
numbers to the left of the line to
create the custom subnet mask
128
128
G4
+G4
32
16
192
8
4
2
+ 1
255
256 128
— V^
^ 4s.
128 64
IG5 . IOO .0 0000000.00
64 32 16
8
^i -fe. so (^
Q "Q "— \>i
4s. vS »s so
!S ^ M *
so Sj^
32 16 8 4 2 1
o o o o o o
£.£/. Observe the total number of
*> ' hosts.
-2
Subtract 2 for the number of
& 2 usable hosts.
13
Custom Subnet Masks
Problem 3
Network Address 148.75.0.0 /26
Address class H_
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
126 indicates the total number of
bits used for the network and
subnetwork portion of the
address. All bits remaining belong
to the host portion of the address.
299 . 299 .O.O
IP 2 4
G4
G2
IO
299 . 299 . 299 . /92
Number of
Hosts
Number of
Sublets
Show your work for Problem 3 in the space below.
I
I
I
I
I
I
-fe.
I
I
I
-fe.
I
I
-fe.
SO
I
I
-fe.
I
I
\o
I
I
I
I
2 4 8 IG 32 (>4 128 25G.
25G 128
— to
Bi**.ry values ~ 128 G4 32
16 8 4 2 I . 128 G4
148 .75.00000000.00
Add the binary value
numbers to the left of the line to
create the custom subnet mask.
128
128
G4
+G4
32
l<o
192
8
4
2
+ /
G4 32
IG
8
4
2
* *
<»
«">
6^
Q <3
. —
VjJ
vj
^
-b. -s>
-£>
OS
(5 s
vjJ
cs <S^
Vsi
-fe.
OS
<S^
32 IG 8 4 2 1
o o o o o o
£.£/. Observe the total number of
^ ' hosts.
-2
Subtract 2 for the number of
yy £. usable hosts.
299
1024
' Subtract 2 for the total number of
~ 2 subnets to get the usable number of
subnets.
1,022
14
Custom Subnet Masks
Problem 4
Number of needed subnets 6
Number of needed usable hosts 30
Network Address 195.85.8.0
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
C
255
. 255
. 255
. O
255
. 255
255
. 224
8
32
30
Show your work for Problem 5 in the space below.
25G 128 G4
Number of
Sublets ~243
128 G4 32
115 . 85 . 8 . O O O
32 IG 8 4 2
/G 32 G4 128 25 G
IG 8 4 2
Nv\r*ber of
Hosts
I ~ BtKA.ru \/a.tu\es
o o o o o
123
G4
+ 32
224
32
-2
30
8
-2
15
Custom Subnet Masks
Problem 5
Number of needed subnets 6
Number of needed usable hosts 30
Network Address 210.100.56.0
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
C
255 . 255 . 255 . O
255 . 255 . 255 . 224
8
32
30
Show your work for Problem 4 in the space below.
25G 128 G4
Number of
Sublets -248
128 G4 32
2IO . IOO . 56 . O O O
224
32 16 8 4 2
IG 32 G4 128 25 6
IG 8 4 2
Number of
Hosts
I ~ BtKA.ru i/*.lv\es
o o o o o
123
G4
8
32
+ 32
-2
-2
30
16
Custom Subnet Masks
Problem 6
Number of needed subnets 126
Number of needed usable hosts 131,070
Network Address 118.0.0.0
Address class 2
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
255 .O.O.O
255 . 254. O . O
128
131,072
13 1,070
7
Show your work for Problem 6 in the space below.
Number of
Hosts
NtA^ber of
StA^^ets
"■3 ^
I
-*-> M\ V3
*. V» "^
tS *. V5
■* Vi "—
-J 03 ■*■
C^ CS -fc.
I I I
I
I I I I I
I I I I I
8 lb 32 b4 128
Bi^ru ve,Ues - 1 28 b4 32 lb
8
1 18. O O O O O O O
128
G4
32
IG
8
4
+ 2
yj
254
i
29b
S3
I I
I
I I
— Kl
«^ -fe.
-fc. 03
- "<3
I I
I I
128 b4 32 lb
-fe. V3 --
® A M
I I
& -
I
29b 128 b4 32 lb 8
_ V3 ^\
U3 5^ VJ
— V3 4i
VJ -fc. 03
128 b4 32 lb 8
—
vo
*.
u
■Q
"
*.
03
-s
*.
<S\
.
yj
-J
^
(S
5-
^
*.
o.oooooooo.oooooooo
128 131,072
-2 -2
l2<o 131,070
17
Custom Subnet Masks
Problem 7
Number of needed subnets 2000
Number of needed usable hosts 15
Network Address 178.100.0.0
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
B
255 . 255 .O.O
255 . 255 . 255 . 224
2,048
32
30
II
Show your work for Problem 7 in the space below.
Number of
Hosts
Number of
Sw.bi^e'ts
I
Vi
I
I
4
I
I
i
I
8
I
I
I
4s.
I
I
4x
I
I
^
Wi
I
256 128 64
16 32 64 128 256.
— ^
^ 4s.
4s.
OS
Bi^ry values ~ 128 64 32
16 8 4 2 I . 128 64 32
178 OOOOO.OOO
128
64
32
16
8
4
2
+ 1
2,048
-2
2P4G
32
-2
30
32 16 8
4*.
so
OS
4s.
Vi
SO
16 8 4 2 1
o o o o o
18
25 5
Custom Subnet Masks
Problem 8
Number of needed subnets 3
Number of needed usable hosts 45
Network Address 200.175.14.0
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
255 . 255 . 255 . O
255 . 255 . 255 . /92
G4
G2
Show your work for Problem 8 in the space below.
25G 128
Number of
Sublets -24
128 G4
200 . 175 . 14 . O O
G4 32 lb 8 4 2
8 lt> 32 G4 128 25 6
32 16 8 <f 2
o o o o o
NiA^ber of
Hosts
I ~ BtK^ru values
123
+G4
24 O
4
-2
G4
-2
<o2
19
Custom Subnet Masks
Problem 9
Number of needed subnets 60
Number of needed usable hosts 1,000
Network Address 128.77.0.0
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
B
255 . 255 .O.O
255 . 255 . 252 . O
64
1,024
1,022
6
Show your work for Problem 9 in the space below.
Number of
Hosts
Number of
Su\b^e~ts
I
2
I
I
I
I
4
CO
4s.
I
I
I
I
8
CO 4s.
B
dc
128 G4 32 /6 8
4s.
CO
I I I
I I I
I I I
I I I
IG 32 G4
123 .77.000000
I
I
I
5G
128
G4 32
IG
8
4
2
to ■*
CO
5^
\0
<?\
Q
Q "Q
.
V)J
vj
OA
• —
\o
4s. -S>
^s>
CO
<^
yJ
^
4s.
co cT*
*0
4s.
CO
<S^
I I
128 25G.
2 I . 128 G4 32 IG 8 4 2 1
00.00000000
20
J28
64
32
16
8
+4
252
64
-2
1,024
-2
62 1,022
Custom Subnet Masks
Problem 10
Number of needed usable hosts 60
Network Address 198.100.10.0
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
C
255 . 255 . 255 . O
255 . 255 . 255 . 192
G4
G2
Show your work for Problem 10 in the space below.
25G 128
Number of
Sublets ~ 2 4
128 G4
198 . IOO . IO . O O
64 32 16 8 4 2
8 lt> 32 G4 128 25 6
32 16 8 <f 2
o o o o o
Number of
Hosts
I ~ BtK^ru values
128
192
G4
-2
G2
4
-2
21
Custom Subnet Masks
Problem 11
Number of needed subnets 250
Network Address 101.0.0.0
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
255 .O.O.O
255 . 255 .O.O
256
65,536
65,534
8
Show your work for Problem 11 in the space below.
Number of
Hosts
Number of
_ SiAb^erts
4i
-r> ^\ «o
*. ^> «\
*.
^
CO ^ VJ
^
. —
""J\ Vi —
VI
>J\
-i CO ■*>.
1
1
C^ CO -fc^
1 1 1
-s
I I
I I
I I I
I I I
2 4 8
-128 b4 32
lb 32 b4 128 25b
lb 8 4 2 I
IOI. oooooooo
M S~ CO *
"-~i \>i "— "<3
voJ qn CO *S> ,3
C^ CO -fc. ^ Q^
I I
I
I
I I
— Vi
— M -fe.
Vi *. CO
I I
I I
-te. CO
"<3
S*
*S> CO
M 4i
-fe. ^ —
CO -te. VJ
I I
25b 128 b4 32 lb 8
I
v>) CT-
-1 ■*■ CO
M -te.
-fe.
— M *.
4s. 3> -»
co -^ -»*.
CO
128 b4 32 lb 8 4
128 b4 32 lb 8
.OOOOOOOO. OOOOOOOO
123
64
32
16
8
4
2
256
65,536
+ 1
-2
-2
255 254 65,534
22
Custom Subnet Masks
Problem 12
Number of needed subnets 5
Network Address 218.35.50.0
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
C
255 . 255 . 255 . O
255 . 255 . 255 . 224
8
32
30
Show your work for Problem 12 in the space below.
Number of
Sw.bi^e'ts
2/8 . 35 . 50
25G
128 G4
2
4
8
128
G4
32
o
o
o
32 16 8 4 2
l(> 32 G4 128 25G
IG 8 4 2
o o o o
Number of
Hosts
I ~ BtK^ru values
128
G4
G4
4
+ 32
-2
-2
224
G2
23
Custom Subnet Masks
Problem 13
Number of needed usable hosts 25
Network Address 218.35.50.0
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
255
. 255
. 255
. O
255
. 255
255
. 224
8
32
30
Show your work for Problem 13 in the space below.
Number of
2/8 . 35 . 50
25G
128 G4
2
*
8
128
G4
32
o
o
o
32 16 8 4 2
l(> 32 (>4 128 25G
l(> 8 4 2
o o o o
Number of
Hosts
I ~ Bi^^ru values
24
128
G4
8
32
+ 32
-2
-2
224
30
Custom Subnet Masks
Problem 14
Number of needed subnets 10
Network Address 172.59.0.0
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
B
255 . 255 .O.O
255 . 255 . 240 . O
16
4,096
4,094
Show your work for Problem 14 in the space below.
Number of
Hosts
Number of
Sublets
I
I
I
I
I
4
so
4s.
I
I
I
I
8
Q3
16
B/tsAry values ~ 128 (,4 32 16
172 . 59 .O O O O
4s.
I
I
4s.
I
I
4s.
I
I
\^>
I
I
I
I
256 128 64 32 16 8
32 64 128 256.
— ^
Vi 4s.
^
4s.
so
<S^
Q
Q
. —
MJ
VJ
vSN
4s.
^s>
-s>
CO
•5 s
MJ
W
<S^
v^
4s.
c»
<S^
8 4 2 I . 128 64 32 16 8 4 2 1
0000.00000000
128
64
32
+ 16
24 O
16
-2
14
4,096
-2
4,094
25
Custom Subnet Masks
Problem 15
Number of needed usable hosts 50
Network Address 172.59.0.0
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
B
255 . 255 .O.O
255 . 255 . 255 . 192
1,024
G4
G2
JO
Show your work for Problem 15 in the space below.
%, ^
^ ^
<s^
CO
4s.
\si
• —
*J\ ~-^i
vjj
.
"Q
'Q
Q
<S\
Number of
v>3 c7>
CO
4s.
4s.
CO
4s.
>z>
Hosts
1 1
I |
1
|
1
|
1
1
1
1
1
|
1
1
Number of
1 1
1 1
1
1
1
1
1
1
1
1
1
1
1
1
Sublets
~ 2 4
8
/6
32
G4
128
25G.
25<o 128
— ^
Vi 4s.
Bi^ry v^Ues ~ 128 G4 32 Ho
8 4 2 I . 128 G4
172 O O O O . O O
128
G4
32
16
8
2
G4 32
IG
8
4
2
K> *
CO
5*
\6
Q Q
. —
VjJ
Vi
vy»
4s. -S>
~s>
CO
5 s
vjj
co c7*
Wi
4s.
CO
<S^
32 IG 8 4 2 1
o o o o o o
128
+G4
G4
-2
1,024
-2
255 192 G2 IP22
26
Custom Subnet Masks
Problem 16
Number of needed usable hosts 29
Network Address 23.0.0.0
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
255 .O.O.O
255 . 255 . 255 . 224
524,233
32
30
19
Show your work for Problem 16 in the space below.
-fc.
4i
^3
03
V3 _
^ —
<$•
03
-t
KJ
NiA^ber of
*
M\
K>
"* ^
"-1
5 s
Ji
.9
4t
>J\
Hosts —
-fe.
1
1
1
1
1
1
1
1
4i VJ
1 1
1 1
5 s
1
1
58
1
-fe.
1
1
1
1
5"
1
1
1
1
1
NiAr*ber of
1
1
1
1
1 1
>J\
to
5 s
Vi
-fe.
<s
5 s
Ks
4s.
oa
«*
Sublets ~
2
4
a
lb
32
64
128 256
&it\e.ru va.tiAes "
128
64
32
16
8
4
2 /
. 128
64
32
16
8
4
2
/
23.00000000.00000000
25i.
• 128 64
-fe.
128
64
32
o
o o
32 16 8
4 2
— M -*>■
Q "O —
* ^> •»
<» -1 -fe.
••J\ ■— V
-1 ^ <3
6 s VJ -«i
16 8 4
2 1
o o o
o o
123
G4
+ 32
224
32 524,233
-2 -2
30 524,23G
27
Subnetting
Problem 1
Number of needed subnets 14
Number of needed usable hosts 14
Network Address 192.10.10.0
Address class _ C-
Default subnet mask 25"5" . 255 . 255 . O
Custom subnet mask 25"5" . 255 . 255 . 24Q
Total number of subnets (&_
Total number of host addresses L&_
Number of usable addresses UL
Number of bits borrowed JJ_
What is the 4th
subnet range? l92.IO.IO.4S to I92.IO.IO.G3
What is the subnet number
for the 8th subnet? 192 . IO . IO . 1 12
What is the subnet
broadcast address for
the 13th subnet? 192 . IO . IO . 207
What are the assignable
addresses for the 9th
subnet? 192. IO JO .129 to 192. IO JO .142
28
Show your work for Problem 1 in the space below.
Number of
SiAbi^e'ts
256
2
128 64
* 8
32
lb
N
16 8 4 2 - H
32 64 128 256
timber
osts
of
128
64
32
16
8 4 2 1 ~ Si^^ru values
192. IO . IO .
o
o
o
o
o o o o
C'->,
o
o
o
o
n2.lo.loo
to
192.10.10.15
(A
o
o
o
1
l92.IO.IO.Ko
to
l92.IO.IO.3l
(A
o
o
I
o
l92.IO.IO.32
to
192. IO. I0.47
(f\
o
o
1
1
n2.IO.IO.4S
to
I92.IO.IO.G3
(?\
o
1
o
o
I92.IO.IO.G4
to
192. IO. I079
?M
o
1
o
1
192. IO. IO.SO
to
192. IO. I0.95
(, 7 l
o
1
I
o
192.10.10.9(0
to
I92.IO.IO.1 1 1
(, 3 \
o
1
1
1
192. IO. IO. 1 12
to
192. IO. IO. 127
(?\
I
o
o
o
192. IO. IO. 123
to
192. IO. IO. 143
(P\
1
o
o
1
192. IO. IO. 144
to
192. IO. IO. 159
(J'l
1
o
I
o
I92.IO.IO.I<oO
to
192. IO. IO. 175
(12)
1
o
1
1
I92.IO.IO.I7G
to
192. IO. IO. 19 1
(P\
1
1
o
o
192. IO. IO. 192
to
192. IO. I0.207
(J4)
1
1
o
1
192. IO. I0.203
to
l92.IO.IO.223
(A
1
1
I
o
192. IO. 10.224
to
l92.IO.IO.239
(li>)
1
1
1
1
192. IO. I0.240
to
192. IO. 10.255
128
<o4
32
Ko
Ko
Custom subnet <-> w*-,
mask £. i C/
-2
Usable subnets . .
-2
Usable hosts . .
The binary value of the last bit borrowed is the range. In this
problem the range is 16.
The first address in each subnet range is the subnet number.
The last address in each subnet range is the subnet broadcast
address.
29
Subnetting
Problem 2
Number of needed subnets 1000
Number of needed usable hosts 60
Network Address 165.100.0.0
Address class &
Default subnet mask 25"5" . 255 . O . O
Custom subnet mask 25"5" . 255 . 255 . 192
Total number of subnets 1 ,024
Total number of host addresses £t
Number of usable addresses ^^
Number of bits borrowed (&
What is the 15th
subnet range? IG5 .IOO .3.128 -to 14,5.100.3.191
What is the subnet number
for the 6th subnet? /££" • IOO . I . G4
What is the subnet
broadcast address for
the 6th subnet? /££" • IOO . I . 127
What are the assignable
addresses for the 9th
subnet? I G5. IOO. 2. 1 -to IG5.IOO.O.G2
30
Show your work for Problem 2 in the space below.
n^> k-> ^^> K^f*
f^CMCs^ (Y>c\|<5sUs ^MCn^ (Vtf\j<5sUs
v£>^^M ^^^M ^"^.M vS^^M
£>£>£>£> <<« CM* MM M rr>* rW^'rW
&S&& ^'Cbc>cb &'&£>£> o>'o>'c>c>'
C^SS C^OCbCb OCbCbCb C^CbCbCb
^^^^ ^^^^ ^^^^ ^^^^
lxslxslxslx\ Un Irs Itn trs UsUsUsUs Un Un Ixs U\
v£>vS>v$vS> v$vS>vS>vS> vS>vS>v£>v£> v£>vS>v£>vS>
($3<°
6^'
32,
^7&8
Oj(o
CM
Sfc
o q q o
vvw
00 CM
>-CM<5^
• * • •
• * • •
Q
vvw
00 CM
"^CM<5n
• * • •
• * • •
00 CM
^hCM<5^
£>^.^.
CM CM CM CM
• * • •
Q
vvw
00 CM
^hCM<5^
n\ rc\ n\ n\
• * • •
UslxsUslxs
lf\lxsUsUs
Us Us Us Us
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V
-^ CM
CM CM
OS
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OS
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CM
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Ci — Cb
8,
tft
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3 2,^ 8
CM CM fc
CM 0o f\
fc&fcfc
CM «v V lx\ V? IV 00 <5n $ $ Jj 5? < U? ^
00 ^h CM vS 00 ^h- CM
CM v£> (vs ^
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00 >-
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CM
$3<°
s
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g m
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o
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co -a
cu -a
c *
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CD CJ
"a co
CO ir
^ CD
CD
CD
cy
31
Subnetting
Problem 3
Number of needed subnets 2
Network Address 195.223.50.0
Hint: It is possible to borrow
one bit to create two subnets.
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
C
255 . 255 . 255 . O
255 . 255 . 255 . 123
128
I2G
What is the 2nd
subnet range? 115 -223 .50 J 28 - 115 .223 .50 .25 5
What is the subnet number
for the 2nd subnet?
What is the subnet
broadcast address for
the 1st subnet?
195 .223.50. 1 28
195 .223.50. 127
What are the assignable
addresses for the 1st
subnet? 195 .223 .50 . 1 - 195 .223 .50 .I2G
32
Show your
NiAv*ber of
Si^bi^e'ts ~ 2
128
115. 223 .50.0
work for Problem 3 in the space below.
Number of
128 G4 32 IG 8 4 2 - Hosts
4 8 IG 32 G4 128 25G
G4 32 IG 8 4 2 1 - B^ry v*.Ues
o o o o o o o
(,) o
(2) 1
195 .223.50.0 to 195 .223.50. 127
I95.223.50.I2S to 195 .223 .50 .255
33
Subnetting
Problem 4
Number of needed subnets 750
Network Address 190.35.0.0
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
B
255 . 255 .O.O
255 . 255 . 255 . /92
1,024
G4
G2
JO
What is the 15th
subnet range? _ U0.3*.3.I28 -to /9Q.35.3J9/
What is the subnet number
for the 13th subnet?
What is the subnet
broadcast address for
the 10th subnet?
What are the assignable
addresses for the 6th
subnet?
WO. 35. 3 O
WO.35.2.127
WO.35.LG5 -to I90.35.I.I2G
34
Show your work for Problem 4 in the space below.
n^[* k^ i^^ k^^
v£> ^ "*>; <N [ vS> ^ ^ CM ^ "^ ^ fM v£> "•*. *>. C\|
C> d' Ci Oi < "^ S "^ M* ^i M| N h\ fW «*[ n\
6 6 fc 6 6 6 6 6 6 6 6 6 6 fc 6 fc
ooqqooqqqo
Q Q Q Q Q
Q
<
tft*
b
vvvvvvvvvvvvvvvv
>
3 27^
CM
fc
°& <N CXiCN °& <N CXiCM
Xh CNI <5n ^ c\ <5n . "^ M <5n ^ CNJ <5n
00
b^ 2
>
M vO ^. ^. f^ vO ^ ^ vO ^. ^. Q vO -*: "^
c%
0> Ci Ci Ci ^' < ^' "■>•" N* C\| CNI CN| (t\ pr>' rr{ rt\
IxslfslfslfslfslxslfslfslfslfslfslfslxslxslXNlXN
CM
4p* b
fS^fS^fS^fS^fS^fS^fS^fS^fS^fS^fS^fS^fS^fS^fS^fS^
fY\
OCbOOOCiOOCbOCNOC^CbOCi
>
2<* 9
M
^^^^^^^^^^^^^^^^
v&
CY\
^^^^^^^^^^^^^^^^
CO
CM
10^
fc
O^C^^O^Cb^Cb^C^^Cb^Ci^
v£>
a \1
00
CM
Us
fc
^^CbCb^^O&^^CiCb^^
5 iz-
^^^"-CiCN&C^^^^
,p*< -
oo
e\i
CN
fc
*-^^^^^^^^
ip4* -
V
>
fc
^1\ n T^^^^*k*oo*^"$^*^*$ x ^*S^*^
4 p c ' b -
esi
00
e," 2 -
vS
v£>
fc
,<,?« ~
00
CNI
^ ^ ^. CXi>-CN|
3i^ e '
V
00
fc
CXi^hCN|vS>CXi^hCN|^
CN|v£>fs^^ +■
M CNlvS
CM
Irs
CM
btf 3 '*'
^ N
CM
fc
l
(
(
>CN|
CM
V ^
^ ^
^
^
Un
;^
^5
^
•
5
5
5 2
s (A
$
^
^
o&
35
Subnetting
Problem 5
Number of needed usable hosts 6
Network Address 126.0.0.0
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
255 .O.O.O
255 . 255 . 255 . 248
2,097,152
8
21
What is the 2nd
subnet range?
What is the subnet number
for the 5th subnet?
What is the subnet
broadcast address for
the 7th subnet?
What are the assignable
addresses for the 10th
subnet?
I2GPP.8 -to I2GPP.I5
I2GPP.32
I2GPP.55
I2GPP73 to I2GPP78
36
Show your work for Problem 5 in the space below.
e
<5n r^
^ cm
**********
**********
e> e> e> e> e> e> e> e> e> e>
C\C\C\C\C\C\C\C\CNjC\C\C\CMC\CNjCN)
e
O Ci o
6 e e
CM
l¥
-e
^- CM Cb
v£> ^hCNC3Cftv£>^hCNCbCXivS>Cs^.fN
el e e e e e e e e e e
***********
. e> e> e> e> e> e> e> e> e> e> e>
C\C\C\C\C\C\C\C\C\C\C\C\C\C\C\jC\i
e
e e e e e
e e e e e
00
CN
^
1 ^'°
\p*
,24
13'
20»
IP* 4 -
2P 4 °-
4 p°l i >-
3 2.7 &8 -
(c^
I?'
2 fc2
5
,072-
,44.
2 4.2^"
lP*
,/52-
?p4-
5
£772
^
$3b
3 27^
-4p1 h
0>
-e
N e
*e
*e
^?e
- 2 p4» se
e
e
. lP *4 ^e
1-7 *
$ I* CM
^ e
s N e
J*e
cm ? rv
5 vD
v
\
V
^s ^s s^
s
5
CMf^>-l^v£>ts.cxi^
CvCj^ljV^
C30 >- CM \£> CX}
CM vS> (^ ^ ^
CO
CM
CXi ^h CM v£> CXi ^h CM
CM v£> (^ ^
i-
CM
CXi CM
I
vS>
37
c&
Subnetting
Problem 6
Number of needed subnets 10
Network Address 192.70.10.0
Address class.
Default subnet mask.
Custom subnet mask.
Total number of subnets.
Total number of host addresses.
Number of usable addresses.
Number of bits borrowed
C
255 . 255 . 255 . O
255 . 255 . 255 . 240
IG
l<o
14
4
What is the 9th
subnet range? 1^270.10.128 to 19270.10.143
What is the subnet number
for the 4th subnet? I9270.I0.43
What is the subnet
broadcast address for
the 12th subnet? 19270.10.191
What are the assignable
addresses for the 10th
subnet? 19270.10.145 to I9270.IO.I53
38
Show your work for Problem 6 in the space below.
N<AmL
■>er of
Number of
256
128 64
32
16 8 4 2 -
Hosts
Subi^e'ts
2
4
8
16
32 64 128 256
128
64
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16
8 4 2 1 - B
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192 . 70 . 10 .
o
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to
19270.10.15
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to
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1
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I9270.I0.32
to
I9270.I0.47
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I9270.I0.43
to
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1
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I9270.IO.G4
to
19270.1079
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19270.10.95
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192.70. I0.9G
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I9270.IO.I 1 1
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19270.10.127
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I9270.IO.I23
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1
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19270.10.159
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19270.10.175
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I9270.IO.I92
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123
240
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14
39
Subnetting
Problem 7
Network Address 10.0.0.0 /16
Address class
Default subnet mask
255 .O.O.O
Custom subnet mask
255 . 255 .O.O
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
256
65,536
65,534
8
What is the 11th
subnet range?
IO.IO.O.O to IO.IO.255 .255
What is the subnet number
for the 6th subnet?
What is the subnet
broadcast address for
the 2nd subnet?
/0.5.0.0
IO. 1.255 .255
What are the assignable
addresses for the 9th
subnet?
IOB.O.I -to IOB.255 .254
40
Show your work for Problem 7 in the space below.
CO
IN
5/2.
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41
Subnetting
Problem 8
Number of needed subnets 5
Network Address 172.50.0.0
B
255 . 255 .O.O
Address class __
Default subnet mask _
Custom subnet mask _
Total number of subnets
# 19 7
Total number of host addresses '
Number of usable addresses ' '
Number of bits borrowed
255 . 255 . 224 . O
8
What is the 4th
subnet range? /72.5Q.96P -to I72.5QJ27 .255
What is the subnet number
for the 5th subnet? I72.50J23P
What is the subnet
broadcast address for
the 6th subnet? 172 .50 J4 1 .25 5
What are the assignable
addresses for the 3rd
subnet? I72.50.G4.I -to 17 2. 50 35 .254
42
Show your work for Problem 8 in the space below.
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43
Subnetting
Problem 9
Number of needed usable hosts 28
Network Address 172.50.0.0
B
255 . 255 .O.O
Address class __
Default subnet mask _
Custom subnet mask _
Total number of subnets ' ^
~Z7
Total number of host addresses
Number of usable addresses _
Number of bits borrowed
255 . 255 . 255 . 224
30
II
What is the 2nd
subnet range? I725QP.32 to I72.500.G3
What is the subnet number
for the 10th subnet? I72.5Q.1 .32
What is the subnet broadcast
address for
the 4th subnet? I72.50O.I27
What are the assignable
addresses for the 6th
subnet? I72.50P.IGI -to I72.500.no
44
Show your work for Problem 9 in the space below.
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45
Subnetting
Problem 10
Number of needed subnets 45
Network Address 220.100.100.0
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
255 . 255 . 25 5 . O
255 . 255 . 255 . 252
G4
4
What is the 5th
subnet range? 220JOO.IOO.I6 to 220.100.100.1^
What is the subnet number
for the 4th subnet? 220.IOO.IOO.I2
What is the subnet
broadcast address for
the 13th subnet? 220.IOO.I00.5 1
What are the assignable
addresses for the 12th
subnet? 220.100.100.^5 to 220.IOO.IOO.4G>
46
Show your work for Problem 10 in the space below.
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47
Subnetting
Problem 11
Number of needed usable hosts 8,000
Network Address 135.70.0.0
B
255 . 255 .O.O
Address class __
Default subnet mask _
Custom subnet mask _
Total number of subnets _
q I a y
Total number of host addresses '
Number of usable addresses ' ''
Number of bits borrowed
255 . 255 . 224 . O
8
What is the 6th
subnet range? 13^70. IGOP -to 13570.1^1.255
What is the subnet number
for the 7th subnet? 135 70.192.0
What is the subnet
broadcast address for
the 3rd subnet?
What are the assignable
addresses for the 5th
subnet?
1357035.255
13570.123.1 to 13570. 159.254
48
Show your work for Problem 11 in the space below.
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49
Subnetting
Problem 12
Number of needed usable hosts 45
Network Address 198.125.50.0
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
255 . 255 . 25 5 . O
255 . 255 . 255 . 192
4
G4
G2
What is the 2nd
subnet range? 198 J 25 .50.64 to 98 J 25 .50. 1 27
What is the subnet number
for the 2nd subnet? 198 .1 25 .50 .G4
What is the subnet
broadcast address for
the 4th subnet? 198 . 1 25 .50 .25 5
What are the assignable
addresses for the 3rd
subnet? 198.1 25 .50. 1 29 to 198.125 .50 J90
50
Show your work for Problem 12 in the space below.
Number of
25G 128
G4 32 16 8 4
2
~ Hosts
Number of
Sublets -24
8 lt> 32 G4 128
25G
128 G4
32 16 8 <f 2
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193 .
125 .50.00
o o o o o
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198.125.50.0
to
198.125 .50.G3
(2) 1
198.1 25. 50.G4
to
198.125.50.127
(3) 1 O
198.12550.128
to
198. 125. 50.191
(4) 1 1
198.125 50.1 9 2
to
198.12550255
128
192
G4
-2
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51
Subnetting
Problem 13
Network Address 165.200.0.0 /26
12
Address class
Default subnet mask _
Custom subnet mask
255 . 255 .O.O
255 . 255 . 255 . 192
I 024
Total number of subnets '_
64
Total number of host addresses
Number of usable addresses
Number of bits borrowed
62
JO
What is the 10th
subnet range? I $5.200.2 64 to 165 .2QQ.2.I27
What is the subnet number
for the 11th subnet? 165 .200.2.128
What is the subnet
broadcast address for
the 1023rd subnet? 165 .200.25 5 .19 1
What are the assignable
addresses for the 1022nd
subnet? 165 .200.25 5 .65 to 165 .200.25 5 .126
52
rork tor Problem 13 in the space below.
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53
Subnetting
Problem 14
Number of needed usable hosts 16
Network Address 200.10.10.0
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
C
255 . 255 . 255 . O
255 . 255 . 255 . 224
8
32
30
What is the 7th
subnet range? 200.10.10.112 -to 200JOJ0.223
What is the subnet number
for the 5th subnet? 200 . IO. JO. J 28
What is the subnet
broadcast address for
the 4th subnet? 200.IO.IO.1 27
What are the assignable
addresses for the 6th
subnet? 200.IO.IO.lbl to 200./0./O.I90
54
Show your work for Problem 14 in the space below.
Number of
25i
128
32 lb 8 4
2 ~
Hosts
Number of
5tAbr*efs ~ 2
4
8
16 32 64 128
256
128
64
32
16 8 4 2
1 ~
BtKa.ru values
200 . IO . IO . O
o
o
o o o o
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200. 10. 10 D
to
200.10.10.31
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200.10.10.32
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200.10.10.64
to
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200.10.10.128
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200.10.10.160
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200. 10.10. 19 2
to
200.10.10.223
(8) 1
1
1
200.10.10.224
to
200.10.10.255
128
+ 32
224
32
-2
30
55
Problem 15
Subnetting
Network Address 93.0.0.0 \19
A
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
255 .O.O.O
255 . 255 . 224 . O
2,048
8, 192
3,190
II
What is the 15th
subnet range? 13-/W to 93.1.223.255
What is the subnet number
for the 9th subnet?
What is the subnet
broadcast address for
the 7th subnet?
What are the assignable
addresses for the 12th
subnet?
93.I.O.O
93P.223.255
93.I.9G.I -to 93.1.127.254
56
Show your work for Problem 15 in the space below.
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57
Practical Subnetting 1
Based on the information in the graphic shown, design a network addressing scheme that will
supply the minimum number of subnets , and allow enough extra subnets and hosts for
100% growth in both areas. Circle each subnet on the graphic and answer the questions
below.
IP Address 172.16.0.0
Address class
Custom subnet mask
Minimum number of subnets needed
B
255.255.2240
Extra subnets required for 100% growth jH j_
(Round up to the next whole number)
Total number of subnets needed -
8
Number of host addresses
in the largest subnet group
Number of addresses needed for
100% growth in the largest subnet
(Round up to the next whole number)
Total number of address
GO
+
GO
needed for the largest subnet _=
_ 120
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Research
IP address range for Marketing
IP address range for Management
IP address range for Router A
to Router B serial connection
17 2.1 GPP to
172.31.255
I72.IG.32P to
I72.G3.255
I72./G.G4P to
172.95.255
l72.IG.9G.Oto
172.127.255
58
Show your work for Practical Subnetting 1 in the space below.
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Practical Subnetting 2
Based on the information in the graphic shown, design a network addressing scheme that will
supply the minimum number of hosts per subnet , and allow enough extra subnets and
hosts for 30% growth in all areas. Circle each subnet on the graphic and answer the questions
below.
IP Address 135.126.0.0
TTsh Department
15 Hosts
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 30% growth
(Round up to the next whole number)
Total number of subnets needed
Number of host addresses
in the largest subnet group
B
255.255.255.224
+
= 7
20
Number of addresses needed for r
30% growth in the largest subnet _+ g
(Round up to the next whole number)
Total number of address _ r
needed for the largest subnet = ^^
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Tech Ed 135 .I2G.O.O to I35.I26.Q.3I
IP address range for English I35.I2G.Q.32 to I35.I2G.O.G3
IP address range for Science 135 .I2G.O&4 to I35.I26P.95
IP address range for Router A __ ,„,^„, , ,^r~ ,-,,* ,^-,
to Router B serial connection I35.J26.Q.96 to 135 .I26.Q.I27
IP address range for Router A ,.,_ ,„, _ ,_ , f-i _ ,„, ^ ,^ n
to Router B serial connection 135 .12^.0 .123 to 135 .I26P.I59
60
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Practical Subnetting 3
Based on the information in the graphic shown, design a classfull network addressing scheme
that will supply the minimum number of hosts per subnet , and allow enough extra subnets
and hosts for 25% growth in all areas. Circle each subnet on the graphic and answer the
questions below.
Marketing
50 Hosts
ft | Jt
Address class
Custom subnet mask
Minimum number of subnets needed
255255.2550
Extra subnets required for 25% growth J^ j_
(Round up to the next whole number)
Total number of subnets needed =
Number of host addresses
in the largest subnet group
185
Number of addresses needed for w 7
25% growth in the largest subnet + Y '
(Round up to the next whole number)
Total number of address 9:?9
needed for the largest subnet — ***
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Sales
IP address range for Marketing
IP address range for Administrative
IP address range for Router A
to Router B serial connection
l72.l6P.Oto I72.I6P.255
172. 16. IP
to 172. 16. 1.255
I72.I6.2P
to 172.16.2.255
I72.I6.3P
to 172.16.3.255
62
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Practical Subnetting 4
Based on the information in the graphic shown, design a network addressing scheme that will
supply the minimum number of subnets , and allow enough extra subnets and hosts for 70%
growth in all areas. Circle each subnet on the graphic and answer the questions below.
F0/0
IP Address 135.126.0.0
SO/0/0.
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 70% growth
(Round up to the next whole number)
Total number of subnets needed
Number of host addresses
in the largest subnet group
255.255.2400
+ 4
- 9
325
Number of addresses needed for n „ Q
70% growth in the largest subnet + ^<£°
(Round up to the next whole number)
Total number of address ca
needed for the largest subnet =
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for New York 135 .I26P.O to 135.126.15.255
IP address range for Washington D. C. 135 . 1 26.16. to 135.126.31.25 5
IP address range for Dallas I35.J26.32.Q to 135.126.47.25 5
IP address range for Router A ._,,_ In , sio^j. i->cr ms s -> ncc
to Router B serial connection 1 35 .U6.48P to 1 3t> .116.63.1'? ?
IP address range for Router A ._,,- m, ,,1*^. iic? ms -jv nccr
tn RnntPr C. QPrial rnnnprtinn I 3? .1 ^b.b't .U to 13? . I £k> ./ 1 .£? ?
64
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Practical Subnetting 5
Based on the information in the graphic shown, design a network addressing scheme that will
supply the minimum number of hosts per subnet , and allow enough extra subnets and
hosts for 100% growth in all areas. Circle each subnet on the graphic and answer the
questions below.
IP Address 210.15.10.0
Address class
Custom subnet mask
Minimum number of subnets needed
255.255.255.192
Extra subnets required for 100% growth J^
(Round up to the next whole number)
Total number of subnets needed
= 4
Number of host addresses ? n
in the largest subnet group
Number of addresses needed for ? n
100% growth in the largest subnet _+
(Round up to the next whole number)
Total number of address , n
needed for the largest subnet =
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Router F0/0 Port 2 IP .15 JO O to 2IP.I5.IP.63
IP address range for Router F0/1 Port 2IP.I5.IP.64 to 2 IP. 15 JO. 127
66
Show your work for Problem 5 in the space below.
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256
128
64 32 16 8 4
2 ~
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Sis.bi^e'ts ~ 2
4
8 16 32 64 128
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67
Practical Subnetting 6
Based on the information in the graphic shown, design a network addressing scheme that will
supply the minimum number of subnets , and allow enough extra subnets and hosts for 20%
growth in all areas. Circle each subnet on the graphic and answer the questions below.
IP Address 10.0.0.0
ET ^ Techn ology
Buildirft
320 Hosts'
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 20% growth _+ _
(Round up to the next whole number)
_ Q
Total number of subnets needed - '
255.240.0.0
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Technology
IP address range for Science
IP address range for Arts & Drama
IP Address range Administration
IP address range for Router A
to Router B serial connection
IP address range for Router A
to Router C serial connection
IP address range for Router B
to Router C serial connection
lOPPPto IP. 15. 255. 255
IO.KoP.Oto IQ.3 1.255.255
IO.32.OP to IP. 47. 255. 255
IO.4S.OP to IO.(o3.255.255
IO.G4PP to IO. 79. 25 5. 25 5
I0.80PP to IO. 95, 255. 255
I03<o.0.0 to IO. 1 11.255.255
68
Show your work for Problem 6 in the space below.
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69
Practical Subnetting 7
Based on the information in the graphic shown, design a network addressing scheme that will
supply the minimum number of hosts per subnet , and allow enough extra subnets and
hosts for 125% growth in all areas. Circle each subnet on the graphic and answer the
questions below.
b-^ IP Address 177.135.0.0
L0/0/0
Marketi
75 Hosts
Address class
Custom subnet mask
Minimum number of subnets needed
B
255.255.252P
4
Extra subnets required for 125% growth +
(Round up to the next whole number)
Total number of subnets needed
Number of host addresses
in the largest subnet group
= 9
363
Number of addresses needed for v_,
125% growth in the largest subnet + "? t
(Round up to the next whole number)
Total number of address _ ,_
needed for the largest subnet = ° > '
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Router A Port F0/0 177.135 OP to 177.135.3.255
IP address range for Research 177 .135 .4 P to 177.135.7.255
IP address range for Deployment 177. 135. 8P to 177 .135 J 1 .255
IP address range for Router A .-,-. ._,,- .,,_, , .__ ,- c lc ozrzr
to Router B serial connection 177. lit? .UP to 177.13'? .15 25 5
70
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Practical Subnetting 8
Based on the information in the graphic shown, design a network addressing scheme that will
supply the minimum number subnets , and allow enough extra subnets and hosts for 85%
growth in all areas. Circle each subnet on the graphic and answer the questions below.
IP Address 192.168.1.0
Address class
Custom subnet mask
Minimum number of subnets needed
25 5. 25 5. 25 5. 224
Extra subnets required for 85% growth _j^
(Round up to the next whole number)
Total number of subnets needed
Number of host addresses
in the largest subnet group
= 6
13
Number of addresses needed for ln
85% growth in the largest subnet + >*
(Round up to the next whole number)
Total number of address « £ _
needed for the largest subnet _=__ -?
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Router A F0/0 {$_ l68.I.Ott 192.168.1.31
IP address range for New York 192. 168. 1 .32 to 192. 168. 1. 63
IP address range for Router A IQn ,, , , y/ , , an iso i ac
to Router B serial connection 19 2. 1 68. 1. 61 to 19 2. 1 68. 1. 95
72
Show your work for Problem 8 in the space below
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25G 128 64
32 IG 8 4 2
- Ho
sis
Number of
Sublets ~248
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73
Practical Subnetting 9
Based on the information in the graphic shown, design a network addressing scheme that will
supply the minimum number of hosts per subnet , and allow enough extra subnets and
hosts for 15% growth in all areas. Circle each subnet on the graphic and answer the questions
below.
IP Address 148.55.0.0
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 15% growth
(Round up to the next whole number)
Total number of subnets needed
Number of host addresses
in the largest subnet group
255.255.2400
+
= 6
2300
Number of addresses needed for 3 ^ _
15% growth in the largest subnet + ^t?
(Round up to the next whole number)
Total number of address n , ^^
needed for the largest subnet = ^^-
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Ft. Worth 148.55 PP. to 148.55.15.255
IP address range for Dallas 148.55 .16.0. to 148.55.31.255
IP address range for Router A .32.0. to 148.5 '.47.255
to Router B serial connection
IP address range for Router A 148.55. 48P. to 148.55.63.255
to Router C serial connection
IP address range for Router C 148.55.64 P. to 148.55.79.255
74 to Router D serial connection
Show your work for Problem 9 in the space below.
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75
Practical Subnetting 10
Based on the information in the graphic shown, design a network addressing scheme that will
supply the minimum number of subnets , and allow enough extra subnets and hosts for
110% growth in all areas. Circle each subnet on the graphic and answer the questions below.
Address class
Custom subnet mask
Minimum number of subnets needed
255.255.255.240
Extra subnets required for 110% growth J^
(Round up to the next whole number)
Total number of subnets needed
Number of host addresses
in the largest subnet group
= 9
MO
Number of addresses needed for /cr A
110% growth in the largest subnet + '?"
(Round up to the next whole number)
Total number of address „ QA
needed for the largest subnet = *
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Sales/Managemnt
IP address range for Marketing
IP address range for Research
IP address range for Router A
to Router B serial connection
l72.l6P.Oto 172. 16. 15. 255
l72.l6.Ko.Oto
172.16.31.255
I72.I6.32P to
172.1647.255
172.1648.0 to
172.16.63.255
76
Show your work for Problem 10 in the space below.
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77
Valid and Non-Valid IP Addresses
Using the material in this workbook identify which of the addresses below are correct and
usable. If they are not usable addresses explain why.
IP Address: 0.230.190.192
Subnet Mask: 255.0.0.0
Reference Page Inside Front Cover
IP Address: 192.10.10.1
Subnet Mask: 255.255.255.0
Reference Pages 28-29
IP Address: 245.150.190.10
Subnet Mask: 255.255.255.0
Reference Page Inside Front Cover
IP Address: 135.70.191.255
Subnet Mask: 255.255.254.0
Reference Pages 48-49
The network f C? cfi.r\r\of be O ,
OK
2.45 is reserved for
experir*erctfi.t u\Se.
This is the brofi.dcfi.st address
for "this rfi.r\qe.
IP Address: 127.100.100.10 127 is reserved for (oopbfi.ck
Subnet Mask: 255.0.0.0
Reference Pages Inside Front Cover
IP Address: 93.0.128.1
Subnet Mask: 255.255.224.0
Reference Pages 56-57
IP Address: 200.10.10.128
Subnet Mask: 255.255.255.224
Reference Pages 54-55
IP Address: 165.100.255.189
Subnet Mask: 255.255.255.192
Reference Pages 30-31
IP Address: 190.35.0.10
Subnet Mask: 255.255.255.192
Reference Pages 34-35
IP Address: 218.35.50.195
Subnet Mask: 255.255.0.0
Reference Page Inside Front Cover
IP Address: 200.10.10.175 /22
Reference Pages 54-55 and/or Inside Front Cover
testis
OK
This is the Sublet address for the
3rd v\S^ble '«.*qe of 200 IO.IO.O
nj^
This address is te.ke* from the / ' irst
rg.*qe for this Sublet which is iweiid .
This hfi.s a c(fi.ss B sublet
mfi.
s£,
A ctfi.ss C- address must tASe fi.
rv\ir\iw\tAry\ of 24 bets.
IP Address: 135.70.255.255
Subnet Mask: 255.255.224.0
Reference Pages 48-49
78
This is a broadcast fi.ddress.
IP Address Breakdown
256 Hosts
0-255
0-127
128-255
0-63
64-127
128-191
192-255
0-15
16-31
32-47
48-63
64-79
80-95
96-111
112-127
128-143
144-159
160-175
176-191
192-207
208-223
224-239
240-255
0-7
8-15
16-23
24-31
32-39
40-47
48-55
56-63
64-71
72-79
80-87
88-95
96-103
104-111
112-119
120-127
128-135
136-143
144-151
152-159
16-167
168-175
176-183
184-191
192-199
200-207
208-215
216-223
224-231
232-239
240-247
248-255
0-3
4-7
8-11
12-15
16-19
20-23
24-27
28-31
32-35
36-39
40-43
44-47
48-51
52-55
56-59
60-63
64-67
68-71
72-75
76-79
80-83
84-87
88-91
92-95
96-99
100-103
104-107
108-111
112-115
116-119
120-123
124-127
128-131
132-135
136-139
140-143
144-147
148-151
152-155
156-159
160-163
164-167
168-171
172-175
176-179
180-183
184-187
188-191
192-195
196-199
200-203
204-207
208-211
212-215
216-219
220-223
224-227
228-231
232-235
236-239
240-243
244-247
248-251
252-255
79
Visualizing Subnets Using
The Box Method
The box method is the simplest way to visualize the breakdown of
subnets and addresses into smaller sizes.
Start with a square. The whole square
is a single subnet comprised of 256
addresses.
/24
255.255.255.0
256 Hosts
1 Subnet
Split the box in half and you get two
subnets with 128 addresses,
/25
255.255.255.128
128 Hosts
2 Subnets
Divide the box into quarters and you
get four subnets with 64 addresses,
/26
255.255.255.192
64 Hosts
4 Subnets
80
128
63
191
64
192
127
255
Split each individual square and you
get eight subnets with 32 addresses,
121
255.255.255.224
32 Hosts
8 Subnets
Split the boxes in half again and you
get sixteen subnets with sixteen
addresses,
/28
255.255.255.240
16 Hosts
16 Subnets
The next split gives you thirty two
subnets with eight addresses,
/29
255.255.255.248
8 Hosts
32 Subnets
The last split gives sixty four subnets
with four addresses each,
/30
255.255.255.252
4 Hosts
64 Subnets
31
32
63
128
159
160
191
64
95
96
127
192
223
224
255
32
128
160
15
47
143
175
16
48
144
176
31
63
159
191
64
96
192
224
79
111
207
239
80
112
208
240
95
127
223
255
8
32
40
128
136
160
168
7
15
39
47
135
143
167
175
16
24
48
56
144
152
176
184
23
31
55
63
151
159
183
191
64
72
96
104
192
200
224
232
71
79
103
111
199
207
321
239
80
38
112
120
208
216
240
248
87
95
119
127
215
223
247
255
8
32
40
128
136
160
168
3
11
35
43
131
139
163
171
4
12
36
14
132
140
164
172
7
15
39
47
135
143
167
175
16
24
48
56
144
152
176
184
19
27
51
59
147
155
179
187
20
28
52
30
148
156
180
188
23
31
55
63
151
159
183
191
64
72
96
104
192
200
224
232
67
75
99
107
195
203
227
235
68
76
100
108
196
204
228
236
71
79
103
111
199
207
321
239
80
88
112
120
208
216
240
248
83
91
115
123
211
219
243
251
84
92
116
124
212
220
244
252
87
95
119
127
215
223
247
255
81
Class A Addressing Guide
CIDR
# of Bits
Borrowed
Subnet
Mask
Total # of
Subnets
Total # of
Hosts
Usable # of
Hosts
/8
255.0.0.0
l
16,777,216
16,777,214
/9
l
255.128.0.0
2
8,388,608
8,388,606
/10
2
255.192.0.0
4
4,194,304
4,194,302
111
3
255.224.0.0
8
2,097,152
2,097,150
112
4
255.240.0.0
16
1,048,576
1,048,574
113
5
255.248.0.0
32
524,288
524,286
114
6
255.252.0.0
64
262,144
262,142
us
7
255.254.0.0
128
131,072
131,070
116
8
255.255.0.0
256
65,536
65,534
117
9
255.255.128.0
512
32,768
32,766
118
10
255.255.192.0
1,024
16,384
16,382
119
11
255.255.224.0
2,048
8,192
8,190
/20
12
255.255.240.0
4,096
4,096
4,094
121
13
255.255.248.0
8,192
2,048
2,046
122
14
255.255.252.0
16,384
1,024
1,022
123
15
255.255.254.0
32,768
512
510
124
16
255.255.255.0
65,536
256
254
125
17
255.255.255.128
131,072
128
126
126
18
255.255.255.192
262,144
64
62
127
19
255.255.255.224
524,288
32
30
128
20
255.255.255.240
1,048,576
16
14
129
21
255.255.255.248
2,097,152
8
6
130
22
255.255.255.252
4,194,304
4
2
Class B Addressing Guide
CIDR
# of Bits
Borrowed
Subnet
Mask
Total # of
Subnets
Total # of
Hosts
Usable # of
Hosts
116
255.255.0.0
l
65,536
65,534
117
l
255.255.128.0
2
32,768
32,766
118
2
255.255.192.0
4
16,384
16,382
119
3
255.255.224.0
8
8,192
8,190
120
4
255.255.240.0
16
4,096
4,094
121
5
255.255.248.0
32
2,048
2,046
122
6
255.255.252.0
64
1,024
1,022
123
7
255.255.254.0
128
512
510
124
8
255.255.255.0
256
256
254
125
9
255.255.255.128
512
128
126
126
10
255.255.255.192
1,024
64
62
127
11
255.255.255.224
2,048
32
30
128
12
255.255.255.240
4,096
16
14
129
13
255.255.255.248
8,192
8
6
130
14
255.255.255.252
16,384
4
2
Class C Addressing Guide
CIDR
# of Bits
Borrowed
Subnet
Mask
Total # of
Subnets
Total # of
Hosts
Usable # of
Hosts
124
255.255.255.0
l
256
254
125
l
255.255.255.128
2
128
126
126
2
255.255.255.192
4
64
62
127
3
255.255.255.224
8
32
30
128
4
255.255.255.240
16
16
14
129
5
255.255.255.248
32
8
6
130
6
255.255.255.252
64
4
2
82
Inside Cover
Live Music Archive
Librivox Free Audio
Metropolitan Museum
Cleveland Museum of Art
Internet Arcade
Console Living Room
Books to Borrow
Open Library
TV News
Understanding 9/11