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58
Practical Subnetting 1
F0/1F0/0
S0/0/0 S0/0/1Router A
Router B
Based on the information in the graphic shown, design a network addressing scheme that willsupply the minimum number of subnets, and allow enough extra subnets and hosts for100% growth in both areas. Circle each subnet on the graphic and answer the questionsbelow.
Marketing24 Hosts
Management15 Hosts
F0/0
Reasearch60 Hosts
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 100% growth
Total number of subnets needed
Number of host addresses in the largest subnet group
Number of addresses needed for100% growth in the largest subnet
Total number of addressneeded for the largest subnet
IP address range for Research
IP address range for Marketing
IP address range for Management
IP address range for Router Ato Router B serial connection
IP Address 172.16.0.0
_____________________________
_____________________________
_________
_________
_________
_________
_________
_________
_____________________________
_____________________________
_____________________________
_____________________________
(Round up to the next whole number)
(Round up to the next whole number)
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
B255.255.224.0
44
8
60
60
120
+
+
=
=
172.16.0.0 to 172.31.255172.16.32.0 to 172.63.255172.16.64.0 to 172.95.255
172.16.96.0 to 172.127.255
59
Show your work for Practical Subnetting 1 in the space below.
172
. 16
. 0
0
0 0
0
0
0 0
. 0
0
0
0 0
0
0
017
2 .
16 .
0 0
0
0
0 0
0
0 .
0
0 0
0
0
0 0
0
172
. 16
. 0
0
0 0
0
0
0 0
. 0
0
0
0 0
0
0
017
2 .
16 .
0 0
0
0
0 0
0
0 .
0
0 0
0
0
0 0
0
172
. 16
. 0
0
0 0
0
0
0 0
. 0
0
0
0 0
0
0
012
8 6
4 3
2 16
8
4
2
1
Num
ber
of S
ubne
ts
-
2
4
8
16 3
2 6
4 1
28
25
6 .
. 25
6 12
8 6
4 32
16
8
4
2
128
64
32
16
8
4
2
1
.... .
512
Bin
ary
valu
es -
Num
ber
ofH
osts
-
102420484,0968,192
16,384
32,76865,536
512
1,024
2,048
4,096
8,192
16,384
32,768
65,536
0 1 0 1 0 1 0 1
1 1 0 0 1 1
1 1 1 1
172
.16.
0.0
172
.16.
32.0
172
.16.
64.0
172
.16.
96.0
172
.16.
128.
017
2.1
6.16
0.0
172
.16.
192
.017
2.1
6.2
24.
0
to to to to to to to to
172
.16.
31.2
55
172
.16.
63.2
55
172
.16.
95.2
55
172
.16.
127.
25
517
2.1
6.15
9.2
55
172
.16.
191.
25
517
2.1
6.2
23.
25
517
2.1
6.2
55
.25
5
(0)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
60 x1.0 604
x1.0 4
60
Practical Subnetting 2
F0/0S0/0/0
S0/0/1Router A
Based on the information in the graphic shown, design a network addressing scheme that willsupply the minimum number of hosts per subnet, and allow enough extra subnets andhosts for 30% growth in all areas. Circle each subnet on the graphic and answer the questionsbelow.
Science Lab10 Hosts
Tech Ed Lab20 Hosts
F0/0
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 30% growth
Total number of subnets needed
Number of host addresses in the largest subnet group
Number of addresses needed for30% growth in the largest subnet
Total number of addressneeded for the largest subnet
IP address range for Tech Ed
IP address range for English
IP address range for Science
IP address range for Router Ato Router B serial connection
IP address range for Router Ato Router B serial connection
IP Address 135.126.0.0
_____________________________
_____________________________
_________
_________
_________
_________
_________
_________
_____________________________
_____________________________
_____________________________
_____________________________
_____________________________
(Round up to the next whole number)
(Round up to the next whole number)
Router BS0/0/1
Router C
English Department15 Hosts
F0/1
F0/1
S0/0/0
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
+
+
=
=
52
7
20
6
26
B255.255.255.224
135.126.0.0 to 135.126.0.31135.126.0.32 to 135.126.0.63135.126.0.64 to 135.126.0.95
135.126.0.96 to 135.126.0.127
135.126.0.128 to 135.126.0.159
61
Show your work for Problem 2 in the space below.
135
. 12
6 . 0
0
0
0 0
0
0
0 .
0 0
0
0
0 0
0
013
5. 1
26
. 0
0 0
0
0
0 0
0
. 0
0
0 0
0
0
0 0
135
. 12
6 . 0
0
0
0 0
0
0
0 .
0 0
0
0
0 0
0
013
5. 1
26
. 0
0 0
0
0
0 0
0
. 0
0
0 0
0
0
0 0
135
. 12
6 . 0
0
0
0 0
0
0
0 .
0 0
0
0
0 0
0
012
8 6
4 3
2 16
8
4
2
1
Num
ber
of S
ubne
ts
-
2
4
8
16 3
2 6
4 1
28
25
6 .
. 25
6 12
8 6
4 32
16
8
4
2
128
64
32
16
8
4
2
1
.... .
512
Bin
ary
valu
es -
Num
ber
ofH
osts
-
102420484,0968,192
16,384
32,76865,536
512
1,024
2,048
4,096
8,192
16,384
32,768
65,536
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 1 0 0 1 1 0 0 1 1 0 0 1 1
1 1 1 1 0 0 0 0 1 1 1 1
1 1 1 1 1 1 1 1
. . . . . . . . .
135.
126.0
.013
5.12
6.0.32
135.
126.0
.6413
5.12
6.0.96
135.
126.0
.128
135.
126.0
.160
135.
126.0
.192
135.
126.0
.224
135.
126.1
.013
5.12
6.1.32
135.
126.1
.6413
5.12
6.1.96
135.
126.1
.128
135.
126.1
.160
135.
126.1
.192
135.
1261
.224
to to to to to to to to to to to to to to to to
135.
126.0
.3113
5.12
6.0.63
135.
126.0
.9513
5.12
6.0.12
713
5.12
6.0.15
913
5.12
6.0.19
113
5.12
6.0.2
2313
5.12
6.0.2
5513
5.12
6.1.31
135.
126.1
.6313
5.12
6.1.95
135.
126.1
.127
135.
126.1
.159
135.
126.1
.191
135.
126.1
.223
135.
126.1
.255
(0)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
5 x.3 1.5
(Rou
nd u
p to
2)
20 x.3 6
62
Practical Subnetting 3Based on the information in the graphic shown, design a classfull network addressing schemethat will supply the minimum number of hosts per subnet, and allow enough extra subnetsand hosts for 25% growth in all areas. Circle each subnet on the graphic and answer thequestions below.
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 25% growth
Total number of subnets needed
Number of host addresses in the largest subnet group
Number of addresses needed for25% growth in the largest subnet
Total number of addressneeded for the largest subnet
IP address range for Sales
IP address range for Marketing
IP address range for Administrative
IP address range for Router Ato Router B serial connection
_____________________________
_____________________________
_________
_________
_________
_________
_________
_________
_____________________________
_____________________________
_____________________________
_____________________________
(Round up to the next whole number)
(Round up to the next whole number)
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
+
+
=
=
F0/0
Administrative30 Hosts
Sales185 Hosts
F0/0
IP Address 172.16.0.0
S0/0/1
Marketing50 Hosts
F0/1 S0/0/0Router A
Router B
41
5
185
47
232
B255.255.255.0
172.16.0.0 to 172.16.0.255172.16.1.0 to 172.16.1.255172.16.2.0 to 172.16.2.255
172.16.3.0 to 172.16.3.255
63
Show your work for Problem 3 in the space below.
172
. 16
. 0
0 0
0
0
0 0
0
. 0
0
0 0
0
0
0 0
172
. 16
. 0
0 0
0
0
0 0
0
. 0
0
0 0
0
0
0 0
172
. 16
. 0
0 0
0
0
0 0
0
. 0
0
0 0
0
0
0 0
172
. 16
. 0
0 0
0
0
0 0
0
. 0
0
0 0
0
0
0 0
172
. 16
. 0
0 0
0
0
0 0
0
. 0
0
0 0
0
0
0 0
128
64
32 16
8
4
2
1
Num
ber
of S
ubne
ts
-
2
4
8
16 3
2 6
4 1
28
25
6 .
. 25
6 12
8 6
4 32
16
8
4
2
128
64
32
16
8
4
2
1
.... .
512
Bin
ary
valu
es -
Num
ber
ofH
osts
-
102420484,0968,192
16,384
32,76865,536
512
1,024
2,048
4,096
8,192
16,384
32,768
65,536
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 1 0 0 1 1 0 0 1 1 0 0 1 1
1 1 1 1 0 0 0 0 1 1 1 1
1 1 1 1 1 1 1 1
. . . . . . . . .
172.
16.0
.017
2.16
.1.0
172.
16.2
.017
2.16
.3.0
172.
16.4
.017
2.16
.5.0
172.
16.6.
017
2.16
.7.0
172.
16.8.
017
2.16
.9.0
172.
16.10
.017
2.16
.11.0
172.
16.12
.017
2.16
.13.0
172.
16.14
.017
2.16
.15.0
to to to to to to to to to to to to to to to to
1172
.16.0
.255
1172
.16.1.
255
1172
.16.2
.255
1172
.16.3.
255
1172
.16.4
.255
1172
.16.5
.255
1172
.16.6.
255
1172
.16.7.
255
1172
.16.8.
255
1172
.16.9.
255
1172
.16.10
.255
1172
.16.11
.255
1172
.16.12
.255
1172
.16.13
.255
1172
.16.14
.255
1172
.16.15
.255
(0)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
22
5x.2
55
6.2
5(R
ound
up
to 5
7)
4x..
25 1
64
Practical Subnetting 4
F0/0 S0/0/0S0/0/1Router A
Based on the information in the graphic shown, design a network addressing scheme that willsupply 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.
Dallas150 Hosts New York
325 Hosts
F0/0
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 70% growth
Total number of subnets needed
Number of host addresses in the largest subnet group
Number of addresses needed for70% growth in the largest subnet
Total number of addressneeded for the largest subnet
IP address range for New York
IP address range for Washington D. C.
IP address range for Dallas
IP address range for Router Ato Router B serial connection
IP address range for Router Ato Router C serial connection
IP Address 135.126.0.0
_____________________________
_____________________________
_________
_________
_________
_________
_________
_________
_____________________________
_____________________________
_____________________________
_____________________________
_____________________________
(Round up to the next whole number)
(Round up to the next whole number)
Router BS0/0/1
Router C F0/0F0/1
S0/0/0
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
+
+
=
=
Washington D.C.220 Hosts
54
9
325
228
553
B255.255.240.0
135.126.0.0 to 135.126.15.255135.126.16.0 to 135.126.31.255135.126.32.0 to 135.126.47.255
135.126.48.0 to 135.126.63.255
135.126.64.0 to 135.126.79.255
65
Show your work for Problem 4 in the space below.
135
. 12
6 . 0
0
0
0 0
0
0
0 .
0 0
0
0
0 0
0
013
5. 1
26
. 0
0 0
0
0
0 0
0
. 0
0
0 0
0
0
0 0
135
. 12
6 . 0
0
0
0 0
0
0
0 .
0 0
0
0
0 0
0
013
5. 1
26
. 0
0 0
0
0
0 0
0
. 0
0
0 0
0
0
0 0
135
. 12
6 . 0
0
0
0 0
0
0
0 .
0 0
0
0
0 0
0
012
8 6
4 3
2 16
8
4
2
1
Num
ber
of S
ubne
ts
-
2
4
8
16 3
2 6
4 1
28
25
6 .
. 25
6 12
8 6
4 32
16
8
4
2
128
64
32
16
8
4
2
1
.... .
512
Bin
ary
valu
es -
Num
ber
ofH
osts
-
102420484,0968,192
16,384
32,76865,536
512
1,024
2,048
4,096
8,192
16,384
32,768
65,536
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 1 0 0 1 1 0 0 1 1 0 0 1 1
1 1 1 1 0 0 0 0 1 1 1 1
1 1 1 1 1 1 1 1
. . . . . . . . .
135.
126.0
.013
5.12
6.16.0
135.
126.3
2.0
135.
126.4
8.013
5.12
6.64.
013
5.12
6.80.
013
5.12
6.96.0
135.
126.1
12.0
135.
126.1
28.0
135.
126.1
44.0
135.
126.1
60.0
135.
126.1
76.0
135.
126.1
92.0
135.
126.2
08.0
135.
126.2
24.0
135.
126.2
40.0
to to to to to to to to to to to to to to to to
135.
126.1
5.25
513
5.12
6.31.2
5513
5.12
6.47.2
5513
5.12
6.63.2
5513
5.12
6.79.2
5513
5.12
6.95.
255
135.
126.1
11.2
5513
5.12
6.127
.255
135.
126.1
43.2
5513
5.12
6.159
.255
135.
126.1
75.2
5513
5.12
6.191
.255
135.
126.2
07.2
5513
5.12
6.223
.255
135.
126.2
39.2
5513
5.12
6.125
5.25
5
(0)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
66
Practical Subnetting 5Based on the information in the graphic shown, design a network addressing scheme that willsupply the minimum number of hosts per subnet, and allow enough extra subnets andhosts for 100% growth in all areas. Circle each subnet on the graphic and answer thequestions below.
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 100% growth
Total number of subnets needed
Number of host addresses in the largest subnet group
Number of addresses needed for100% growth in the largest subnet
Total number of addressneeded for the largest subnet
IP address range for Router F0/0 Port
IP address range for Router F0/1 Port
_____________________________
_____________________________
_________
_________
_________
_________
_________
_________
_____________________________
_____________________________
(Round up to the next whole number)
(Round up to the next whole number)
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
+
+
=
=
F0/0
Science Room10 Hosts
Tech Ed Lab18 Hosts
English classroom15 Hosts
F0/1
Art Classroom12 Hosts
IP Address 210.15.10.0
22
4
30
30
60
C255.255.255.192
210.15.10.0 to 210.15.10.63
210.15.10.64 to 210.15.10.127
67
Show your work for Problem 5 in the space below.
128 64 32 16 8 4 2 1 - Binary values
Number of Subnets - 2 4 8 16 2 4 8 16 2 4 8 16 2 4 8 16 2 4 8 16 32 64 128 256
Number of256 128 64 32 16 8 4 2 - Hosts
210. 15 . 10 . 0 0 0 0 0 0 0 0210. 15 . 10 . 0 0 0 0 0 0 0 0210. 15 . 10 . 0 0 0 0 0 0 0 0210. 15 . 10 . 0 0 0 0 0 0 0 0210. 15 . 10 . 0 0 0 0 0 0 0 001
1 01 1
210.15.10.0210.15.10.64210.15.10.128210.15.10.192
210.15.10.63210.15.10.127210.15.10.191210.15.10.255
totototo
(0)(1)(2)(3)
68
Practical Subnetting 6Based on the information in the graphic shown, design a network addressing scheme that willsupply 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.
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 20% growth
Total number of subnets needed
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 Ato Router B serial connection
IP address range for Router Ato Router C serial connection
IP address range for Router Bto Router C serial connection
IP Address 10.0.0.0
_____________________________
_____________________________
_________
_________
_________
_____________________________
_____________________________
_____________________________
_____________________________
_____________________________
_____________________________
_____________________________
(Round up to the next whole number)
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
+
=
F0/0
S0/0/1Router A
Administration35 Hosts
TechnologyBuilding320 HostsF0/0 Router B
S0/0/1
Router C
F0/1
F0/1
S0/0/0
Science Building225 Hosts
S0/0/0
S0/0/1S0/0/0Art & Drama
75 Hosts
72
9
A255.240.0.0
10.0.0.0 to 10.15.255.25510.16.0.0 to 10.31.255.25510.32.0.0 to 10.47.255.25510.48.0.0 to 10.63.255.255
10.64.0.0 to 10.79.255.255
10.80.0.0 to 10.95.255.255
10.96.0.0 to 10.111.255.255
69
Show your work for Problem 6 in the space below.
Num
ber
of S
ubne
ts
-
2
4
8
16
32
64
12
8 2
56
.10
. 0 0
0 0
0 0
0 0
. 0
0 0
0 0
0 0
0 . 0
0 0
0 0
0 0
010
. 0 0
0 0
0 0
0 0
. 0
0 0
0 0
0 0
0 . 0
0 0
0 0
0 0
010
. 0 0
0 0
0 0
0 0
. 0
0 0
0 0
0 0
0 . 0
0 0
0 0
0 0
010
. 0 0
0 0
0 0
0 0
. 0
0 0
0 0
0 0
0 . 0
0 0
0 0
0 0
010
. 0 0
0 0
0 0
0 0
. 0
0 0
0 0
0 0
0 . 0
0 0
0 0
0 0
0
. 25
6 12
8 6
4 3
2
16
8
4
2
Bin
ary
valu
es -
128
64
32
16
8
4
2
1
. . . . . 1
28
64
32
16
8
4
2
1
. . . . . 1
28
64
32
16
8
4
2
1
-
5121,0242,0484,0968,192
16,38432,76865,536
131,072262,144524,2881,048,5762,097,1524,194,304
.
512
1,024
2,048
4,096
8,192
16,384
32,768
65,536
131,072
262,144
524,288
1,048,576
2,097,152
4,194,304
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 1 0 0 1 1 0 0 1 1 0 0 1 1
1 1 1 1 0 0 0 0 1 1 1 1
1 1 1 1 1 1 1 1
Num
ber
ofH
osts
10.0
.0.0
10.1
6.0.
010
.32.
0.0
10.4
8.0.
010
.64.
0.0
10.8
0.0.
010
.96.
0.0
10.11
2.0.
010
.128
.0.0
10.1
44.0
.010
.160
.0.0
10.1
76.0
.010
.192
.0.0
10.2
08.0
.010
.224
.0.0
10.2
40.0
.0
to to to to to to to to to to to to to to to to
10.15
.255
.255
10.3
2.25
5.25
510
.47.
255.
255
10.6
3.25
5.25
510
.79.
255.
255
10.9
5.25
5.25
510
.111.2
55.2
5510
.127.
255.
255
10.1
43.2
55.2
5510
.159.
255.
255
10.17
5.25
5.25
510
.191.2
55.2
5510
.207
.255
.255
10.2
23.2
55.2
5510
.239
.255
.255
10.2
55.2
55.2
55
(0)
(1)
(2)
(3)
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(7)
(8)
(9)
(10)
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70
Practical Subnetting 7Based on the information in the graphic shown, design a network addressing scheme that willsupply the minimum number of hosts per subnet, and allow enough extra subnets andhosts for 125% growth in all areas. Circle each subnet on the graphic and answer thequestions below.
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 125% growth
Total number of subnets needed
Number of host addresses in the largest subnet group
Number of addresses needed for125% growth in the largest subnet
Total number of addressneeded for the largest subnet
IP address range for Router A Port F0/0
IP address range for Research
IP address range for Deployment
IP address range for Router Ato Router B serial connection
_____________________________
_____________________________
_________
_________
_________
_________
_________
_________
_____________________________
_____________________________
_____________________________
_____________________________
(Round up to the next whole number)
(Round up to the next whole number)
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
+
+
=
=
Marketing75 Hosts
IP Address 177.135.0.0
Administration33 Hosts Sales
255 Hosts
Research135 Hosts
F0/0S0/0/0 F0/0
F0/1
S0/0/0Router A
Router B
Deployment63 Hosts
45
9
363
454
817
B255.255.252.0
177.135.0.0 to 177.135.3.255177.135.4.0 to 177.135.7.255177.135.8.0 to 177.135.11.255
177.135.12.0 to 177.135.15.255
71
Show your work for Problem 7 in the space below.
177.
135
. 0
0
0 0
0
0
0 0
. 0
0
0
0 0
0
0
017
7.13
5 .
0 0
0
0
0 0
0
0 .
0
0 0
0
0
0 0
0
177.
135
. 0
0
0 0
0
0
0 0
. 0
0
0
0 0
0
0
017
7.13
5 .
0 0
0
0
0 0
0
0 .
0
0 0
0
0
0 0
0
177.
135
. 0
0
0 0
0
0
0 0
. 0
0
0
0 0
0
0
012
8 6
4 3
2 16
8
4
2
1
Num
ber
of S
ubne
ts
-
2
4
8
16 3
2 6
4 1
28
25
6 .
. 25
6 12
8 6
4 32
16
8
4
2
128
64
32
16
8
4
2
1
.... .
512
Bin
ary
valu
es -
Num
ber
ofH
osts
-
102420484,0968,192
16,384
32,76865,536
512
1,024
2,048
4,096
8,192
16,384
32,768
65,536
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 1 0 0 1 1 0 0 1 1 0 0 1 1
1 1 1 1 0 0 0 0 1 1 1 1
1 1 1 1 1 1 1 1
. . . . . . . . .
177.1
35.0
.017
7.135
.4.0
177.1
35.8.
017
7.135
.12.0
177.1
35.16
.017
7.135
.20.
017
7.135
.24.
017
7.135
.28.0
177.1
35.32
.017
7.135
.36.0
177.1
35.4
0.0
177.1
35.4
4.0
177.1
35.4
8.017
7.135
.52.
017
7.135
.56.0
177.1
35.60
.0
to to to to to to to to to to to to to to to to
177.1
35.3.
255
177.1
35.7.
255
177.1
35.11
.255
177.1
35.15
.255
177.1
35.19
.255
177.1
35.2
3.255
177.1
35.2
7.255
177.1
35.31
.255
177.1
35.35
.255
177.1
35.39
.255
177.1
35.4
3.255
177.1
35.4
7.255
177.1
35.5
1.255
177.1
35.5
5.25
517
7.135
.59.2
5517
7.135
.63.2
55
(0)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
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72
Practical Subnetting 8
F0/0 S0/0/0S0/0/1Router A
Based on the information in the graphic shown, design a network addressing scheme that willsupply 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.
New York8 Hosts
F0/0
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 85% growth
Total number of subnets needed
Number of host addresses in the largest subnet group
Number of addresses needed for85% growth in the largest subnet
Total number of addressneeded for the largest subnet
IP address range for Router A F0/0
IP address range for New York
IP address range for Router Ato Router B serial connection
IP Address 192.168.1.0
_____________________________
_____________________________
_________
_________
_________
_________
_________
_________
_____________________________
_____________________________
_____________________________
(Round up to the next whole number)
(Round up to the next whole number)
Router B
F0/1
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
+
+
=
=
Boston5 Hosts
Research & Development8 Hosts
33
6
13
12
25
C255.255.255.224
192.168.1.0 to 192.168.1.31192.168.1.32 to 192.168.1.63
192.168.1.64 to 192.168.1.95
73
Show your work for Problem 8 in the space below.
128 64 32 16 8 4 2 1 - Binary values
Number of Subnets - 2 4 8 16 2 4 8 16 2 4 8 16 2 4 8 16 2 4 8 16 32 64 128 256
Number of256 128 64 32 16 8 4 2 - Hosts
192. 168 . 1 . 0 0 0 0 0 0 0 0192. 168 . 1 . 0 0 0 0 0 0 0 0192. 168 . 1 . 0 0 0 0 0 0 0 0192. 168 . 1 . 0 0 0 0 0 0 0 0192. 168 . 1 . 0 0 0 0 0 0 0 001
1 01 1
1 0 01 0 11 1 01 1 1
192.168.1.0192.168.1.32192.168.1.64192.168.1.96192.168.1.128192.168.1.160192.168.1.192192.168.1.224
192.168.1.31192.168.1.63192.168.1.95192.168.1.127192.168.1.159192.168.1.1191192.168.1.223192.168.1.255
totototototototo
(0)(1)(2)(3)(4)(5)(6)(7)
Practical Subnetting 9
F0/0
S0/0/0S0/0/1Router A
Based on the information in the graphic shown, design a network addressing scheme that willsupply the minimum number of hosts per subnet, and allow enough extra subnets andhosts for 15% growth in all areas. Circle each subnet on the graphic and answer the questionsbelow.
Dallas1500 Hosts
F0/0
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 15% growth
Total number of subnets needed
Number of host addresses in the largest subnet group
Number of addresses needed for15% growth in the largest subnet
Total number of addressneeded for the largest subnet
IP address range for Ft. Worth
IP address range for Dallas
IP address range for Router Ato Router B serial connection
IP address range for Router Ato Router C serial connection
IP address range for Router Cto Router D serial connection
IP Address 148.55.0.0
_____________________________
_____________________________
_________
_________
_________
_________
_________
_________
_____________________________
_____________________________
_____________________________
_____________________________
_____________________________
(Round up to the next whole number)
(Round up to the next whole number)
Router BS0/0/1
Router C
F0/1
S0/0/0
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
+
+
=
=
Router D S0/0/0S0/0/1
74
Ft. Worth2300 Hosts
51
6
2300
345
2645
B255.255.240.0
148.55.0.0. to 148.55.15.255148.55.16.0. to 148.55.31.255148.55.32.0. to 148.55.47.255
148.55.48.0. to 148.55.63.255
148.55.64.0. to 148.55.79.255
75
Show your work for Problem 9 in the space below.
148.
55
. 0
0
0 0
0
0
0 0
. 0
0
0
0 0
0
0
014
8. 5
5 .
0 0
0
0
0 0
0
0 .
0
0 0
0
0
0 0
0
148.
55
. 0
0
0 0
0
0
0 0
. 0
0
0
0 0
0
0
014
8. 5
5 .
0 0
0
0
0 0
0
0 .
0
0 0
0
0
0 0
0
148.
55
. 0
0
0 0
0
0
0 0
. 0
0
0
0 0
0
0
012
8 6
4 3
2 16
8
4
2
1
Num
ber
of S
ubne
ts
-
2
4
8
16 3
2 6
4 1
28
25
6 .
. 25
6 12
8 6
4 32
16
8
4
2
128
64
32
16
8
4
2
1
.... .
512
Bin
ary
valu
es -
Num
ber
ofH
osts
-
102420484,0968,192
16,384
32,76865,536
512
1,024
2,048
4,096
8,192
16,384
32,768
65,536
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 1 0 0 1 1 0 0 1 1 0 0 1 1
1 1 1 1 0 0 0 0 1 1 1 1
1 1 1 1 1 1 1 1
. . . . . . . . .
148.5
5.0.
014
8.55.
16.0
148.5
5.32
.014
8.55.
48.0
148.5
5.64
.014
8.55.
80.0
148.5
5.96
.014
8.55.
112.
014
8.55.
128.0
148.5
5.14
4.0
148.5
5.16
0.0
148.5
5.17
6.014
8.55.
192.
014
8.55.
208.0
148.5
5.22
4.0
148.5
5.24
0.0
to to to to to to to to to to to to to to to to
148.5
5.15
.255
148.5
5.31
.255
148.5
5.47
.255
148.5
5.63
.255
148.5
5.79
.255
148.5
5.95
.255
148.5
5.11
1.255
148.5
5.12
7.255
148.5
5.14
3.255
148.5
5.15
9.255
148.5
5.17
5.25
514
8.55.
191.2
5514
8.55.
207.2
5514
8.55.
223.2
5514
8.55.
239.2
5514
8.55.
255.
255
(0)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
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76
Practical Subnetting 10Based on the information in the graphic shown, design a network addressing scheme that willsupply the minimum number of subnets, and allow enough extra subnets and hosts for110% 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
Extra subnets required for 110% growth
Total number of subnets needed
Number of host addresses in the largest subnet group
Number of addresses needed for110% growth in the largest subnet
Total number of addressneeded for the largest subnet
IP address range for Sales/Managemnt
IP address range for Marketing
IP address range for Research
IP address range for Router Ato Router B serial connection
IP Address 172.16.0.0
_____________________________
_____________________________
_________
_________
_________
_________
_________
_________
_____________________________
_____________________________
_____________________________
_____________________________
(Round up to the next whole number)
(Round up to the next whole number)
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
+
+
=
=
F0/0S0/0/0
S0/0/1Router A
F0/0
Router B
F0/1
Sales115 Hosts
Management25 Hosts
Research35 Hosts
Marketing56 Hosts
45
9
140
154
294
B255.255.255.240
172.16.0.0 to 172.16.15.255172.16.16.0 to 172.16.31.255172.16.32.0 to 172.16.47.255
172.16.48.0 to 172.16.63.255
77
Show your work for Problem 10 in the space below.
172
.16
. 0
0 0
0
0
0 0
0
. 0
0
0 0
0
0
0 0
172
.16
. 0
0 0
0
0
0 0
0
. 0
0
0 0
0
0
0 0
172
.16
. 0
0 0
0
0
0 0
0
. 0
0
0 0
0
0
0 0
172
.16
. 0
0 0
0
0
0 0
0
. 0
0
0 0
0
0
0 0
172
.16
. 0
0 0
0
0
0 0
0
. 0
0
0 0
0
0
0 0
128
64
32 16
8
4
2
1
Num
ber
of S
ubne
ts
-
2
4
8
16 3
2 6
4 1
28
25
6 .
. 25
6 12
8 6
4 32
16
8
4
2
128
64
32
16
8
4
2
1
.... .
512
Bin
ary
valu
es -
Num
ber
ofH
osts
-
102420484,0968,192
16,384
32,76865,536
512
1,024
2,048
4,096
8,192
16,384
32,768
65,536
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 1 0 0 1 1 0 0 1 1 0 0 1 1
1 1 1 1 0 0 0 0 1 1 1 1
1 1 1 1 1 1 1 1
. . . . . . . . .
172.
16.0
.017
2.16
.16.0
172.
16.32
.017
2.16
.48.0
172.
16.64
.017
2.16
.80.0
172.
16.96
.017
2.16
.112.
017
2.16
.128.0
172.
16.14
4.0
172.
16.16
0.0
172.
16.17
6.017
2.16
.192.
017
2.16
.208
.017
2.16
.224
.017
2.16
.240
.0
to to to to to to to to to to to to to to to to
172.
16.15
.255
172.
16.31
.255
172.
16.4
7.255
172.
16.63
.255
172.
16.79
.255
172.
16.95
.255
172.
16.11
1.255
172.
16.12
7.255
172.
16.14
3.255
172.
16.15
9.255
172.
16.17
5.25
517
2.16
.191.2
5517
2.16
.207
.255
172.
16.2
23.2
5517
2.16
.239
.255
172.
16.2
55.2
55
(0)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
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Valid and Non-Valid IP Addresses
Using the material in this workbook identify which of the addresses below are correct andusable. If they are not usable addresses explain why.
IP Address: 0.230.190.192 ________________________________Subnet Mask: 255.0.0.0 ________________________________
IP Address: 192.10.10.1 ________________________________Subnet Mask: 255.255.255.0 ________________________________
IP Address: 245.150.190.10 ________________________________Subnet Mask: 255.255.255.0 ________________________________
IP Address: 135.70.191.255 ________________________________Subnet Mask: 255.255.254.0 ________________________________
IP Address: 127.100.100.10 ________________________________Subnet Mask: 255.0.0.0 ________________________________
IP Address: 93.0.128.1 ________________________________Subnet Mask: 255.255.224.0 ________________________________
IP Address: 200.10.10.128 ________________________________Subnet Mask: 255.255.255.224 ________________________________
IP Address: 165.100.255.189 ________________________________Subnet Mask: 255.255.255.192 ________________________________
IP Address: 190.35.0.10 ________________________________Subnet Mask: 255.255.255.192 ________________________________
IP Address: 218.35.50.195 ________________________________Subnet Mask: 255.255.0.0 ________________________________
IP Address: 200.10.10.175 /22 ________________________________________________________________
IP Address: 135.70.255.255 ________________________________Subnet Mask: 255.255.224.0 ________________________________
The network ID cannot be 0.
OK
245 is reserved forexperimental use.
This is the broadcast addressfor this range.
127 is reserved for loopbacktesting.
OK
This is the subnet address for the3rd usable range of 200.10.10.0
OK
This address is taken from the firstrange for this subnet which is invalid.
This has a class B subnetmask.
A class C address must use aminimum of 24 bits.
This is a broadcast address.
78
Reference Pages 28-29
Reference Page Inside Front Cover
Reference Pages 48-49
Reference Pages 54-55
Reference Page Inside Front Cover
Reference Page Inside Front Cover
Reference Pages 54-55 and/or Inside Front Cover
Reference Pages 56-57
Reference Pages Inside Front Cover
Reference Pages 34-35
Reference Pages 30-31
Reference Pages 48-49
0-127
128-255
0-34-78-11
12-1516-1920-2324-2728-3132-3536-3940-4344-4748-5152-5556-5960-6364-6768-7172-7576-7980-8384-8788-9192-9596-99
100-103104-107108-111112-115116-119120-123124-127128-131132-135136-139140-143144-147148-151152-155156-159160-163164-167168-171172-175176-179180-183184-187188-191192-195196-199200-203204-207208-211212-215216-219220-223224-227228-231232-235236-239240-243244-247248-251252-255
/308+8+8+6
255.255.255.2524 Hosts
/298+8+8+5
255.255.255.2488 Hosts
/288+8+8+4
255.255.255.24016 Hosts
/278+8+8+3
255.255.255.22432 Hosts
/268+8+8+2
255.255.255.19264 Hosts
/258+8+8+1
255.255.255.128128 Hosts
/248+8+8
255.255.255.0256 Hosts
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
0-15
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
0-63
64-127
128-191
192-255
0-255
IP Address Breakdown
79
/24255.255.255.0
256 Hosts1 Subnet
Start with a square. The whole squareis a single subnet comprised of 256addresses.
Visualizing Subnets UsingThe Box Method
The box method is the simplest way to visualize the breakdown ofsubnets and addresses into smaller sizes.
/25255.255.255.128
128 Hosts2 Subnets
/26255.255.255.192
64 Hosts4 Subnets
80
Split the box in half and you get twosubnets with 128 addresses,
Divide the box into quarters and youget four subnets with 64 addresses,
81
/27255.255.255.224
32 Hosts8 Subnets
Split each individual square and youget eight subnets with 32 addresses,
/30255.255.255.252
4 Hosts64 Subnets
/29255.255.255.248
8 Hosts32 Subnets
/28255.255.255.240
16 Hosts16 Subnets
Split the boxes in half again and youget sixteen subnets with sixteenaddresses,
The next split gives you thirty twosubnets with eight addresses,
The last split gives sixty four subnetswith four addresses each,
82
# of BitsBorrowed
012345678910111213141516171819202122
SubnetMask
255.0.0.0255.128.0.0255.192.0.0255.224.0.0255.240.0.0255.248.0.0255.252.0.0255.254.0.0255.255.0.0
255.255.128.0255.255.192.0255.255.224.0255.255.240.0255.255.248.0255.255.252.0255.255.254.0255.255.255.0
255.255.255.128255.255.255.192255.255.255.224255.255.255.240255.255.255.248255.255.255.252
Total # ofSubnets
1248163264128256512
1,0242,0484,0968,19216,38432,76865,536131,072262,144524,288
1,048,5762,097,1524,194,304
Total # ofHosts
16,777,2168,388,6084,194,3042,097,1521,048,576524,288262,144131,07265,53632,76816,3848,1924,0962,0481,02451225612864321684
Usable # ofHosts
16,777,2148,388,6064,194,3022,097,1501,048,574524,286262,142131,07065,53432,76616,3828,1904,0942,0461,02251025412662301462
# of BitsBorrowed
01234567891011121314
SubnetMask
255.255.0.0255.255.128.0255.255.192.0255.255.224.0255.255.240.0255.255.248.0255.255.252.0255.255.254.0255.255.255.0
255.255.255.128255.255.255.192255.255.255.224255.255.255.240255.255.255.248255.255.255.252
Total # ofSubnets
1248163264128256512
1,0242,0484,0968,19216,384
Total # ofHosts65,53632,76816,3848,1924,0962,0481,02451225612864321684
Usable # ofHosts65,53432,76616,3828,1904,0942,0461,02251025412662301462
Class C Addressing Guide# of Bits
Borrowed0123456
SubnetMask
255.255.255.0255.255.255.128255.255.255.192255.255.255.224255.255.255.240255.255.255.248255.255.255.252
Total # ofSubnets
1248163264
Total # ofHosts
25612864321684
Usable # ofHosts
25412662301462
CIDR/8/9/10/11/12/13/14/15/16/17/18/19/20/21/22/23/24/25/26/27/28/29/30
CIDR/16/17/18/19/20/21/22/23/24/25/26/27/28/29/30
CIDR/24/25/26/27/28/29/30
Class B Addressing Guide
Class A Addressing Guide
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Inside Cover
![IP Sub Sup Netting F5[1]Tarun](https://img.pdfslide.us/doc/110x75/5452ea1baf79590c308b5246/ip-sub-sup-netting-f51tarun.jpg)
