IP Addressing and Subnetting...0) Network Host IP Address: Default Subnet Mask: AND: 8 ANDING With Custom subnet masks When you take a single network such as 192.100.10.0 and divide
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Frederick County Career & Technology CenterCisco Networking Academy
Frederick County Public SchoolsFrederick, Maryland, USA
Special Thanks to Melvin Baker and Jim Dorschfor 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 WorkbooksACLs - Access Lists Workbooks
VLSM Variable-Length Subnet Mask IWorkbooks
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.
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 . 0 . 0
255 . 0 . 0 . 0
255 . 255 . 0 . 0
255 . 255 . 255 . 0
255 . 0 . 0 . 0
255 . 0 . 0 . 0
255 . 255 . 255 . 0
255 . 255 . 255 . 0
255 . 0 . 0 . 0
255 . 255 . 0 . 0
255 . 0 . 0 . 0
255 . 255 . 0 . 0
255 . 255 . 255 . 0
255 . 0 . 0 . 0
255 . 0 . 0 . 0
255 . 255 . 255 . 0
255 . 255 . 0 . 0
255 . 0 . 0 . 0
7
ANDING WithDefault subnet masks
Every IP address must be accompanied by a subnet mask. By now you should be able to lookat 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 255.0.0.0Class B 255.255.0.0Class C 255.255.255.0
ANDING Equations:1 AND 1 = 11 AND 0 = 00 AND 1 = 00 AND 0 = 0
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 outsideworld still sees the network as 192.100.10.0, but the internal computers and routers see fivesmaller subnetworks. Each independent of the other. This can only be accomplished by usinga custom subnet mask. A custom subnet mask borrows bits from the host portion of theaddress to create a subnetwork address between the network and host portions of an IPaddress. In this example each range has 14 usable addresses in it. The computer must stillAND the IP address against the custom subnet mask to see what the network portion is andwhich subnetwork it belongs to.
Address Ranges: 192.10.10.0 to 192.100.10.15192.100.10.16 to 192.100.10.31192.100.10.32 to 192.100.10.47 (Range in the sample below)192.100.10.48 to 192.100.10.63192.100.10.64 to 192.100.10.79192.100.10.80 to 192.100.10.95192.100.10.96 to 192.100.10.111192.100.10.112 to 192.100.10.127192.100.10.128 to 192.100.10.143192.100.10.144 to 192.100.10.159192.100.10.160 to 192.100.10.175192.100.10.176 to 192.100.10.191192.100.10.192 to 192.100.10.207192.100.10.208 to 192.100.10.223192.100.10.224 to 192.100.10.239192.100.10.240 to 192.100.10.255
In the next set of problems you will determine the necessary information to determine thecorrect subnet mask for a variety of IP addresses.
Four bits borrowed from the hostportion of the address for thecustom subnet mask.
The ANDING process of the four borrowed bitsshows which range of IP addresses thisparticular address will fall into.
SubNetwork HostNetwork
IP Address:Custom Subnet Mask:
AND:
9
How to determine the number of subnets and thenumber of hosts per subnet
Two formulas can provide this basic information:
Number of subnets = 2 (Second subnet formula: Number of subnets = 2 - 2)
Number of hosts per subnet = 2 - 2
Both formulas calculate the number of hosts or subnets based on the number of binary bitsused. For example if you borrow three bits from the host portion of the address use thenumber of subnets formula to determine the total number of subnets gained by borrowing thethree bits. This would be 2 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 inthe host portion and apply this to the number of hosts per subnet formula If five bits are in thehost portion of the address this would be 2 or 2 x 2 x 2 x 2 x 2 = 32 hosts.
When dealing with the number of hosts per subnet you have to subtract two addresses fromthe range. The first address in every range is the subnet number. The last address in everyrange is the broadcast address. These two addresses cannot be assigned to any device inthe network which is why you have to subtract two addresses to find the number of usableaddresses in each range.
For example if two bits are borrowed for the network portion of the address you can easilydetermine the number of subnets and hosts per subnets using the two formulas.
The number of hosts created byleaving 6 bits is 2 - 2 or2 x 2 x 2 x 2 x 2 x 2 = 64 - 2 = 62usable hosts per subnet.
6
What about that second subnet formula:
Number of subnets = 2 - 2
In some instances the first and last subnet range of addresses are reserved. This is similar tothe 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 isalso the subnet number for the classful subnet address.
The last range of addresses is the broadcast subnet. The broadcast address for the lastsubnet in the broadcast subnet is the same as the classful broadcast address.
Notice that the subnet andbroadcast addresses match.
Use the 2 - 2 formula and don’t use thezero and broadcast ranges if...
Classful routing is used
RIP version 1 is used
The no ip subnet zero command isconfigured on your router
Use the 2 formula and use the zero andbroadcast ranges if...
Classless routing or VLSM is used
RIP version 2, EIGRP, or OSPF is used
The ip subnet zero command isconfigured on your router (default setting)
No other clues are given
When to use which formula to determine the number of subnetss s
The primary reason the the zero and broadcast subnets were not used had to do pirmarily withthe broadcast addresses. If you send a broadcast to 195.223.255 are you sending it to all 255addresses 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 todetermine which formula to use, and whehter or not you can use the first and last subnets. Usethe chart below to help decide.
Bottom line for the CCNA exams; if a question does not give you any clues as to whether or notto allow these two subnets, assume you can use them.
This workbook has you use the number of subnets = 2 formula.s
Add the binary value numbers to the left of the line tocreate the custom subnet mask.
B
255 . 255 . 0 . 0
255 . 255 . 255 . 192
1,024
64
62
10
Observe the total number ofhosts.
Subtract 2 for the number ofusable hosts.
64-262
128643216842
+1255
128 64 32 16 8 4 2 1 .....
512
Binary values -
Number ofHosts -
128+64192
102420484,0968,192
16,384
32,76865
,536
512
1,024
2,048
4,096
8,192
16,384
32,768
65,5
36
Custom Subnet Masks
Problem 3Network Address
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
148.75.0.0 /26
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 3 in the space below.
B
255 . 255 . 0 . 0
255 . 255 . 255 . 192
1,024
64
62
10
/26 indicates the total number ofbits used for the network andsubnetwork portion of theaddress. All bits remaining belongto the host portion of the address.
192.10.10.0 to 192.10.10.15192.10.10.16 to 192.10.10.31192.10.10.32 to 192.10.10.47192.10.10.48 to 192.10.10.63192.10.10.64 to 192.10.10.79192.10.10.80 to 192.10.10.95192.10.10.96 to 192.10.10.111192.10.10.112 to 192.10.10.127192.10.10.128 to 192.10.10.143192.10.10.144 to 192.10.10.159192.10.10.160 to 192.10.10.175192.10.10.176 to 192.10.10.191192.10.10.192 to 192.10.10.207192.10.10.208 to 192.10.10.223192.10.10.224 to 192.10.10.239192.10.10.240 to 192.10.10.255
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
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Num
ber
of S
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2
4
8
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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
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32.0
172
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64.0
172
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96.0
172
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128.
017
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192
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to to to to to to to to
172
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55
172
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172
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172
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191.
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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
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5. 1
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135
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8 6
4 3
2 16
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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.
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.224
135.
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.013
5.12
6.1.32
135.
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.6413
5.12
6.1.96
135.
126.1
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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
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135.
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.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
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172
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. 16
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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
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017
2.16
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172.
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2.16
.9.0
172.
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2.16
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172.
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2.16
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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
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255
1172
.16.2
.255
1172
.16.3.
255
1172
.16.4
.255
1172
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1172
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255
1172
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255
1172
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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
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0
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5. 1
26
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135
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135
. 12
6 . 0
0
0
0 0
0
0
0 .
0 0
0
0
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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.
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)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
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)
(11)
(12)
(13)
(14)
(15)
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
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)
(14)
(15)
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)
(12)
(13)
(14)
(15)
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.