subnet for novice

7/28/2019 Subnet For Novice

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Subnet for Novice

Network

Class

1st Octet MSD # ofNetwork bit

# of Host

bit

Assignable

IP ranges

Assignable

IP ranges

Broadcast

Address

Subnet Mask

A 0 - 127 0 8 24 N.H.H.H 1 - 126 127 255.0.0.0

B 128 - 191 10 16 16 N.N.H.H 129 - 190 191 255.255.0.0

C 192 - 223 110 24 8 N.N.N.H 193 - 222 223 255.255.255.0D 224 239 Reserved for IP Multicasting 225 - 238 238 N/A

E 240 254 Reserved for R & D 241 - 253 254 N/A

N = network address bits; H = host/node address bits (s+h, where s= # of subnet bits and h= # of host bits)

Class A Addressing Guide

CIDR Subnet Mask # of bits

Borrowed

Total # of Subnets Total # of Hosts Total Assignable

Hosts

/32 255.255.255.255 This is a single host route

/31 255.255.255.254 2 0

/30 255.255.255.252 22 4,194,304 4 2

/29 255.255.255.248 21 2,097,152 8 6

/28 255.255.255.240 20 1,048,576 16 14

/27 255.255.255.224 19 524,288 32 30

/26 255.255.255.192 18 262,144 64 62

/25 255.255.255.128 17 131,072 128 126

/24 255.255.255.0 16 65,536 256 254

/23 255.255.254.0 15 32,768 512 510

/22 255.255.252.0 14 16,384 1024 1022

/21 255.255.248.0 13 8192 2048 2046

/20 255.255.240.0 12 4096 4096 4094

/19 255.255.224.0 11 2048 8192 8190

/18 255.255.192.0 10 1024 16,384 16,382

/17 255.255.128.0 9 512 32,768 32,766

/16 255.255.0.0 8 256 65,536 65,534

/15 255.254.0.0 7 128 131,072 131,070/14 255.252.0.0 6 64 262,144 262,142

/13 255.248.0.0 5 32 524,288 524,286

/12 255.240.0.0 4 16 1,048,576 1,048,574

/11 255.224.0.0 3 8 2,097,152 2,097,150

/10 255.192.0.0 2 4 4,194,304 4,194,302

/9 255.128.0.0 1 2 8,388,608 8,388,606

/8 255.0.0.0 0 1 16,777,216 16,777,214

Class B Addressing Guide


Borrowed


Hosts

/30 255.255.255.252 14 16,384 4 2

/29 255.255.255.248 13 8192 8 6

/28 255.255.255.240 12 4096 16 14

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/27 255.255.255.224 11 2048 32 30

/26 255.255.255.192 10 1024 64 62

/25 255.255.255.128 9 512 128 126

/24 255.255.255.0 8 256 256 254

/23 255.255.254.0 7 128 512 510

/22 255.255.252.0 6 64 1024 1022

/21 255.255.248.0 5 32 2048 2046

/20 255.255.240.0 4 16 4096 4094/19 255.255.224.0 3 8 8192 8190

/18 255.255.192.0 2 4 16,384 16,382

/17 255.255.128.0 1 2 32,768 32,766

/16 255.255.0.0 0 1 65,536 65,534

Class C Addressing Guide


Borrowed


Hosts

/30 255.255.255.252 6 64 4 2

/29 255.255.255.248 5 32 8 6

/28 255.255.255.240 4 16 16 14

/27 255.255.255.224 3 8 32 30

/26 255.255.255.192 2 4 64 62

/25 255.255.255.128 1 2 128 126

/24 255.255.255.0 0 1 256 254

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IP Addressing Guide Class A, Class B, Class C

CIDR Subnet Mask # of bit borrowed

S = CIDR - N

Total Subnets = 2**S

or 2**(24-H/16-H/8-H)

Total Hosts Total Hosts

Assignable

/32 255.255.255.255 A B C A B C

/31 255.255.255.254

/30 255.255.255.252 22 14 6 4194304 16384 64 4 2

/29 255.255.255.248 21 13 5 2097152 8192 32 8 6/28 255.255.255.240 20 12 4 1048576 4096 16 16 14

/27 255.255.255.224 19 11 3 524288 2048 8 32 30

/26 255.255.255.192 18 10 2 262144 1024 4 64 62

/25 255.255.255.128 17 9 1 131072 510 2 128 126

/24 255.255.255.0 16 8 0 65536 256 256 254

/23 255.255.254.0 15 7 32768 128 512 510

/22 255.255.252.0 14 6 16384 64 1024 1022

/21 255.255.248.0 13 5 8192 32 2048 2046

/20 255.255.240.0 12 4 4096 16 4096 4094

/19 255.255.224.0 11 3 2048 8 8192 8190

/18 255.255.192.0 10 2 1024 4 16,384 16,382

/17 255.255.128.0 9 1 512 2 32,768 32,766

/16 255.255.0.0 8 0 256 65,536 65,534/15 255.254.0.0 7 128 131,072 131,070

/14 255.252.0.0 6 64 262,144 262,142

/13 255.248.0.0 5 32 524,288 524,286

/12 255.240.0.0 4 16 1,048,576 1,048,574

/11 255.224.0.0 3 8 2,097,152 2,097,150

/10 255.192.0.0 2 4 4,194,304 4,194,302

/9 255.128.0.0 1 2 8,388,608 8,388,606

/8 255.0.0.0 0 16,777,216 16,777,214

Mask Subnet Boundary

/30 252 248 244 240 236 232 228 224 220 216 212 208 204 200 196 192

188 184 180 176 172 168 164 160 156 152 148 144 140 136 132 128

124 120 116 112 108 104 100 96 92 88 84 80 76 72 68 64

60 56 52 48 44 40 36 32 28 24 20 16 12 8 4 0

/29 248 240 232 224 216 208 200 192 184 176 168 160 152 144 136 128

120 112 104 96 88 80 72 64 56 48 40 32 24 16 8 0

/28 240 224 208 192 176 160 144 128 112 96 80 64 48 32 16 0

/27 224 192 160 128 96 64 32 0

/26 192 128 64 0

/25 128 0

Note: A subnet address of a network always starts at boundary. A subnet address cannot starts in between two

boundaries.

Network Addresses are always EVEN Broadcast addresses are always ODD

First usable addresses are always ODD Last usable addresses are always EVEN

No host/node address can have (octets 2,3,4 or octets 3,4 or octet 4 containing) all 0s or all 1s

Octets containing all 0s are called Gateway address

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Octets containing all 1s are called Broadcast address

Host/node address cannot be represented by octets containing all 0s or all 1s

Subnet X = Not necessarily 1 network rather consists of a range/group of networks with sequential IP address

Number of bits (N) for subnet X = 2 ** N < = X or Strip of all leading zeros and count # of remaining digits of X binary

CIDR = Mask Bits = # of leading 1s in 32 bits subnet mask

Total Number of Hosts = 2 ** (32 CIDR #) (including network and broadcast IP address)

Total number of assignable hosts = 2 ** (32 CIDR #) - 2

Number of Subnet = Class A: 2**(24 h) or 2**s || Class B: 2**(16 h) or 2**s || Class C: 2**(8 h) or 2**s

Calculating Subnets and its IP ranges in a Class A Network

Example 1

Given network IP address (25.0.0.0), get 14 subnets and its IP address ranges.

Network IP address (25.0.0.0) represents a class A network because 1st octet (e.g. 25) is within 0 127.

For class A network, 1st octet (e.g. 25) represents network address and octets 2, 3, 4 (e.g. 0.0.0) represent host/node address.

For subnetting a class A network address 25.0.0.0, we need to consider the 1st host/node octet (2nd octet of the network

address) and find out # of lower bits required to represent 14 (required # of subnet).

The binary representation of the 1st host/node octet or the 2nd octet of the network IP address (e.g. 0) is

0 0 0 0 0 0 0 0

128 64 32 16 8 4 2 1

Therefore, lower order 4 bits (8+4+2+1=15) are required to get 14 subnets.

To get the corresponding subnet mask, add 4 higher order bits (128+64+32+16) = 240

Therefore, the subnet mask of network address 25.0.0.0 for having 14 subnets is 25.240.0.0.

Please note that, in general, the subnet mask for the class A network is 255.0.0.0. But since the network IP address 25.0.0.0

has been subneted to 14 subnets the corresponding subnet mask is 25.240.0.0. The subnet mask 25.240.0.0 indicates that

the class A network IP address 25.0.0.0 contains max of 2**4 2 = 14 subnets.

The added high order bits used in determining the subnet mask are 128, 64, 32, 16. The lowest of the high order bits is used

in determining the subnet address. The first subnet address is determined by adding the lowest of the high order bits to the

2nd octet of the given network address, that is 25.16.0.0. The subsequent subnet address will be determined just by adding16 to the 2nd octet of the previous subnet address. Therefore, the 14 subnet addresses are:

25.16.0.0 25.32.0.0 25.48.0.0 25.64.0.0 25.80.0.0 25.96.0.0 25.112.0.0

25.128.0.0 25.144.0.0 25.160.0.0 25.176.0.0 25.192.0.0 25.208.0.0 25.224.0.0

The number of host/node IP addresses in each subnet = 2**(24 4) 2 = 1048576 2 = 1048574

The IP ranges of individual subnet are as follows.

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Subnet # Subnet Addr Start Addr End Addr Broadcast Addr

1 25.16.0.0 25.16.0.1 25.31.255.254 25.31.255.255

2 25.32.0.0 25.32.0.1 25.47.255.254 25.47.255.255

3 25.48.0.0 25.48.0.1 25.63.255.254 25.63.255.255

4 25.64.0.0 25.64.0.1 25.79.255.254 25.79.255.255

5 25.80.0.0 25.80.0.1 25.95.255.254 25.95.255.255

6 25.96.0.0 25.96.0.1 25.111.255.254 25.111.255.255

7 25.112.0.0 25.112.0.1 25.127.255.254 25.127.255.255

8 25.128.0.0 25.128.0.1 25.143.255.254 25.143.255.255

9 25.144.0.0 25.144.0.1 25.159.255.254 25.159.255.255

10 25.160.0.0 25.160.0.1 25.175.255.254 25.175.255.255

11 25.176.0.0 25.176.0.1 25.191.255.254 25.191.255.255

12 25.192.0.0 25.192.0.1 25.207.255.254 25.207.255.255

13 25.208.0.0 25.208.0.1 25.223.255.254 25.223.255.255

14 25.224.0.0 25.224.0.1 25.239.255.254 25.239.255.255

Example 2


For class A network, one octet (1st octet) (e.g. 100) represents the network address

For class A network, 3 octets (octets 2, 3, 4) (e.g. 20.0.0) represent host/node address.

For subnetting class A network of address 100.20.0.0, we need to consider the 1st host/node octet (2nd octet of the network

address) and find out # of lower bits are required for 25 (required # of subnet).

The binary representation of the 1st host/node octet or the 2nd octet of the network IP address (that is 20) is

0 0 0 1 0 1 0 0

128 64 32 16 8 4 2 1

Therefore, lower order 5 bits (16+8+4+2+1=31) are required to get 25 subnets.

To get the corresponding subnet mask, add 5 higher order bits (128+64+32+16+8) = 248

Therefore, the subnet mask of 100.24.0.0 for having 25 subnets is 100.248.0.0.

Please note that, in general, the subnet mask for the class A network is 255.0.0.0. But since the network IP address

100.20.0.0 has been subneted to 25 subnets the corresponding subnet mask is 100.248.0.0.. The subnet mask 100.248.0.0

indicates that the class A network IP address 100.24.0.0 contains max of 2**5 2 = 30 subnets.

The added high order bits used in determining the subnet mask are 128, 64, 32, 16, 8. The lowest of the high order bits is

used in determining the subnet address. The first subnet address is determined by adding the lowest of the high order bits to

the 2nd octet of the given network address, that is 100.20.0.0. The subsequent subnet address will be determined just by

adding 8 to the 2nd octet of the previous subnet address. Therefore, the 25 subnet addresses are

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100.32.0.0 100.40.0.0 100.48.0.0 100.56.0.0 100.64.0.0

100.72.0.0 100.80.0.0 100.88.0.0 100.96.0.0 100.104.0.0

100.112.0.0 100.120.0.0 100.128.0.0 100.136.0.0 100.144.0.0

100.152.0.0 100.160.0.0 100.168.0.0 100.176.0.0 100.184.0.0

100.192.0.0 100.200.0.0 100.208.0.0 100.216.0.0 100.224.0.0




1 100.32.0.0 100.32.0.1 100.35.255.254 100.35.255.255

2 100.40.0.0 100.40.0.1 100.43.255.254 100.43.255.255

3 100.48.0.0 100.48.0.1 100.44.255.254 100.51.255.255

4 100.56.0.0 100.56.0.1 100.59.255.254 100.59.255.255

5 100.64.0.0 100.64.0.1 100.75.255.254 100.75.255.255

.. . . .. ..

.. . . .. ..

24 100.216.0.0 100.216.0.1 100.229.255.254 100.229.255.255

25 100.224.0.0 100.224.0.1 100.237.255.254 100.237.255.255

Calculating Subnets and its IP ranges in a Class B Network

Example 3


For class B network, octet 1st, 2nd (e.g. 131.100) represent the network address.

For class C network, octet 3rd, 4th (e.g. 0.0) represents host/node address.

For subnetting class B network of address 131.100.0.0 we need to consider the 1st host/node octet (3rd octet of the network


The binary representation of the 1st host/node octet or the 4th octet of the network IP address (that is 15) is

0 0 0 0 1 1 1 1

128 64 32 16 8 4 2 1

Therefore, lower order 4 bits (8+4+2+1=15) are required to get 13 subnets.

To get the corresponding subnet mask, add 4 higher order bits (128+64+32+16) = 240


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Please note that, in general, the subnet mask for the class B network is 255.255.0.0. But since the network IP address

131.100.0.0 has been subneted to 13 subnets the corresponding subnet mask is 255.255.240.0. The subnet mask

255.255.240.0 indicates that the class C network IP address 131.100.0.0 contains max of 2**4 2 = 14 subnets.

The added high order bits used in determining the subnet mask are 128, 64, 32, 16. The lowest of the high order bits is used

in determining the subnet address. The first subnet address is determined by adding the lowest of the high order bits to the

3rd octet of the given network address (131.100.0.0). The subsequent subnet address will be determined just by adding 16 to

the 3rd

octet of the previous subnet address. Therefore, the 13 subnet addresses are:

131.100.16.0 131.100.32.0 131.100.48.0 131.100.64.0 131.100.80.0

131.100.96.0 131.100.112.0 131.100.128.0 131.100.144.0 131.100.160.0

131.100.176.0 131.100.192.0 131.100.208.0

The number of host/node IP addresses in each subnet = 2**(16 4) 2 = 4096 - 2 = 4094



1 131.100.16.0 131.100.16.1 131.100.31.254 131.100.31.255

2 131.100.32.0 131.100.32.1 131.100.47.254 131.100.47.255

3 131.100.48.0 131.100.48.1 131.100.63.254 131.100.63.255

----- ---- ---- ----

----- ---- ---- ----

12 131.100.192.0 131.100.192.1 131.100.207.254 131.100.207.255

13 131.100.208.0 131.100.208.1 131.100.223.254 131.100.223.255

The last subnet is

14 131.100.224.0 131.100.224.1 131.100.239.254 131.100.239.255

Calculating Subnets and its IP ranges in a Class C Network

Example 5


For class C network, octet 1st, 2nd, 3rd (e.g. 195.20.45) represent the network address.

For class C network, octet 4th (e.g. 15) represents host/node address.

For subnetting class C network of address 195.20.45.15 we need to consider the 1st host/node octet (4th octet of the network


The binary representation of the 1st host/node octet or the 4th octet of the network IP address (that is 15) is

0 0 0 0 1 1 1 1

128 64 32 16 8 4 2 1

Therefore, lower order 3 bits (4+2+1=7) are required to get 5 subnets..

To get the corresponding subnet mask, add 3 higher order bits (128+64+32) = 224

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Please note that, in general, the subnet mask for the class C network is 255.255.255.0. But since the network IP address



The added high order bits used in determining the subnet mask are 128, 64, 32. The lowest of the high order bits is used in

determining the subnet address. The first subnet address is determined by adding the lowest of the high order bits to the 4 th

octet of the given network address (195.20.45.15). The subsequent subnet address will be determined just by adding 32 to

the 4th octet of the previous subnet address. Therefore, the 5 subnet addresses are:

195.20.45.47 195.20.45.79 195.20.45.111 195.20.45.143 195.20.45.175




1 195.20.45.47 195.20.45.48 195.20.45.77 195.20.45.78

2 195.20.45.79 195.20.45.80 195.20.45.109 195.20.45.110

3 195.20.45.111 195.20.45.112 195.20.45.141 195.20.45.142

4 195.20.45.143 195.20.45.144 195.20.45.173 195.20.45.174

5 195.20.45.175 195.20.45.176 195.20.45.208 195.20.45.209

The last subnet is

6 195.20.45.210 195.20.45.211 195.20.45.240 195.20.45.241

Loss of Usable IP address

The loss of usable IP address in the class C network due to subneting is a lot. That loss depends on # of subnets are created.

The smaller number subnet creation means more loss of usable IP address and the bigger number subnet creation means

less loss of usable IP address.

For this particular example, you lost two usable IP address in each subnet (subnet address, broadcast address) plus all

address in between 195.20.45.15 and 195.20.45.47.

Example 6


For class C network, octet 1st, 2nd, 3rd (e.g. 220.10.55) represent the network address.

For class C network, octet 4th (e.g. 0) represents host/node address.

For subnetting class C network of address 220.10.55.0, we need to consider the higher order host/node octet (here is 4 th

octet of the network address) and find out # of lower bits required for 2 (required # of subnet).

The binary representation of the 1

st

host/node octet or the 4

th

octet of the network IP address (that is 0) is0 0 0 0 0 0 0 0

128 64 32 16 8 4 2 1

Therefore, lower order 2 bits (2+1=3) are required to get 2 subnets..

To get the corresponding subnet mask, add 2 higher order bits (128+64) = 192

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Please note that, in general, the subnet mask for the class C network is 255.255.255.0. But since the network IP address



The added high order bits used in determining the subnet mask are 128, 64. The lowest of the high order bits is used in

determining the subnet address. The first subnet address is determined by adding the lowest of the high order bits to the 4 th

octet of the given network address (220.10.55.0). The subsequent subnet address will be determined just by adding 64 to

the 4th octet of the previous subnet address. Therefore, the 2 subnet addresses are:

220.10.55.64 195.20.45.128




1 220.10.55.64 220.10.55.65 220.10.55.126 220.10.55.127

2 220.10.55.128 220.10.55.129 220.10.55.190 220.10.55.191

Loss of Usable IP address

The loss of usable IP address in the class C network due to subneting is a lot. That loss depends on # of subnets are created.

The smaller number subnet creation means more loss of usable IP address and the bigger number subnet creation means

less loss of usable IP address.

For this particular example, you lost two usable IP address in each subnet (subnet address, broadcast address) plus all

address in between 220.10.55.0 and 220.10.55.64.

Mask to CIDR Conversion

Problem 1: Find CIDR when Subnet Mask is 255.255.248.0

The binary representation of subnet mask 255.255.248.0 is

11111111 11111111 11111000 00000000

# of leading 1s 8 8 5

Total # of leading 1s = 8 + 8 + 5 = 21

Therefore, CIDR = /21

Problem 2: Find CIDR when Subnet Mask is 255.255.255.192

The binary representation of subnet mask 255.255.255.192 is

11111111 11111111 11111111 11000000

# of leading 1s 8 8 8 2

Total # of leading 1s = 8 + 8 + 8 + 2 = 26

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Therefore, CIDR = /26

CIDR to Mask Conversion

Problem 3: Find subnet mask when CIDR is /23

23/8 = 2 with remainder 7

Mask = 8 leading 1 in 1st octet, 8 leading 1 in 2nd octet, 7 leading 1 in 3rd octet, 0 leading 1 in 4th octet

The binary representation of Mask is 11111111 11111111 11111110 00000000

The decimal representation of Mask is 255.255.254.0

Problem 4: Find subnet mask when CIDR is /29

29/8 = 3 with remainder 5

Mask = 8 leading 1 in 1st octet, 8 leading 1 in 2nd octet, 8 leading 1 in 3rd octet, 5 leading 1 in 4th octet

The binary representation of Mask is 11111111 11111111 11111111 11111000

The decimal representation of Mask is 255.255.255.248

Problem 5: Given a network address and subnet mask Find # of subnets and # of hosts per subnet.

Given the address - 172.16.0.0 /22

First we list the binary representation of network IP address and then the subnet mask.

Network IP address 10101100 00010000 00000000 00000000 Class B

|--------network--------| |-----------host----------|

Subnet mask = /22 (8+8+6+0) 11111111 11111111 11111100 00000000

|--------|-----------------|

p=6 q=10

Number of subnets:For class B network, 1st host octet or 3rd network IP address octet is used for subnet calculation. We look for # of high

order non zero in sequence bits in the 3rd octet of the mask or # of high order non zero in sequence bits in the host.

That is 6 ( /22 = 8 + 8 + 6).

Formula: = 2**p 2, where p = 6 is # of high order non zero in sequence bits in the 3rd octet in the mask

Therefore, the # of subnets = 2**6 2 = 64-2 = 62.

Number of hosts per subnet:

Formula: 2**q 2, where q is # of bits available for hosts = 10 (/22: 32 - 22 = 10)

or # of low order zero in sequence bits in the mask = 10

Therefore, # of hosts per subnet = 2**10 - 2 = 1024 - 2 = 1022

Problem 6: Given a network address (172.16.0.0 ), # of subnets, and hosts per subnet, Find the customers subnets.

(Requirements: need 6 departments with 2000 hosts per department)

First we list the network address and then the mask in binary:

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Network address 10101100 00010000 00000000 00000000 172.16.0.0 Class B

Network mask 11111111 11111111 00000000 00000000 255.255.0.0 Default class B net mask

For class B network, octet 1st, 2nd (e.g. 172.16) represent the network address.

For class B network, octet 3rd, 4th (e.g. 0.0) represents host/node address.

For subnetting class B network of address 172.16.0.0, we need to consider the higher order host/node octet (here is 3 rd octetof the network address) and find out # of lower bits required for 6 (required # of department which is same as subnet).

The binary representation of the 1st host/node octet or the 3rd octet of the network IP address (that is 0) is

0 0 0 0 0 0 0 0

128 64 32 16 8 4 2 1

Therefore, lower order 3 bits (4+2+1=7) are required to get 6 subnets.

To get the corresponding subnet mask, take 3 higher order non-zero in-sequence bits and add them (128+64+32) = 224

Therefore, the subnet mask of network address 172.16.0.0 is 172.16.224.0 that can be represented by 172.16.0.0/19The # of host per subnet = 2**(total host bits in class B - # of higher bits used for subnet) - 2

= 2**(16 3) 2 = 8192 2 = 8190 ------- enough host per subnet as requirement.

Problem 7: Given a network address (172.16.0.0) and subnet mask (255.255.224.0), list the valid subnets

First we list the mask then the address in binary:

First we list the network address and then the mask in binary:

Network address 10101100 00010000 00000000 00000000 172.16.0.0 Class B

Network mask 11111111 11111111 11100000 00000000 255.255.224.0

|---|-------|

3 5

As we know, for class B network the corresponding to the 1st host octet or corresponding to the 3 rd network octet of the

mask play an important role for calculating subnet ranges (# of subnets). Here, mask used 3 extra bits (3 non-zero

high order in-sequence bits) and 5 zero in-sequence bits. This 5 non-zero in-sequence bits signifies the subnet

ranges. Therefore, 2**5 = 32 is the range of subnet address or the subnet address interval. Therefore, we need to add

32 to the 1st host octet in order to subnet ranges. Therefore, the valid subnets are:

172.16.32.0 172.16.64.0 172.16.96.0 172.16.128.0 172.16.160.0 172.16.192.0 172.16.224.0

An even simpler way to do this is list the mask, say 255.255.255.224

As the fourth octet has been manipulated we subtract 224 from 256 = 32 so our networks will increment in 32's, 32,

64, 96 etc

Problem 8: Given a network address (192.168.1.44 ) and subnet mask (255.255.255.192) identify which subnet the

address belongs to (here address means 192.168.1.44).

First we covert the network address and then the subnet mask to binary

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Network addr 11000000 10101000 00000001 00101100 192.168.1.44 class C network

Network mask 11111111 11111111 11111111 11000000 255.255.255.292

Next we do a logical AND, where we compare the subnet mask to the address and where there is a 1 and a 1 we list

1, where there is a 1 and a 0 we list as 0, and where is a 0 and a 0 we list 0

So adding our two values we get:

11000000 10101000 00000001 00000000Converting our address back to binary we get: 192.168.1.0 - which is the subnet our address belongs to

Class C Subnet at a Glance

Example: /25 Network

Network/Subnet mask = /25; Total Subnet = 2; IP Addr per subnet = 128; Hosts per subnet = 126

Subnet Network Addr, Subnet GW Addr, and Broadcast Addr are generally reserved

Subnet # Subnet Network Addr Subnet GW Addr Subnet Start Addr Subnet End Addr Broadcast Addr

1 195.20.45.0 195.20.45.1 195.20.45.2 195.20.45.126 195.20.45.127

2 195.20.45.128 195.20.45.129 195.20.45.130 195.20.45.254 195.20.45.255





1 195.20.45.0 195.20.45.1 195.20.45.2 195.20.45.62 195.20.45.63

2 195.20.45.64 195.20.45.65 195.20.45.66 195.20.45.126 195.20.45.127

3 195.20.45.128 195.20.45.129 195.20.45.130 195.20.45.190 195.20.45.191

4 195.20.45.192 195.20.45.193 195.20.45.194 195.20.45.254 195.20.45.255





1 195.20.45.0 195.20.45.1 195.20.45.2 195.20.45.30 195.20.45.31

2 195.20.45.32 195.20.45.33 195.20.45.34 195.20.45.62 195.20.45.63

3 195.20.45.64 195.20.45.65 195.20.45.66 195.20.45.94 195.20.45.95

4 195.20.45.96 195.20.45.96 195.20.45.98 195.20.45.126 195.20.45.127

5 195.20.45.128 195.20.45.129 195.20.45.130 195.20.45.158 195.20.45.159

6 195.20.45.160 195.20.45.161 195.20.45.162 195.20.45.190 195.20.45.191

7 195.20.45.192 195.20.45.193 195.20.45.194 195.20.45.222 195.20.45.223

12


13/14

8 195.20.45.224 195.20.45.225 195.20.45.226 195.20.45.254 195.20.45.255





1 195.20.45.0 195.20.45.1 195.20.45.2 195.20.45.14 195.20.45.15

2 195.20.45.16 195.20.45.17 195.20.45.18 195.20.45.30 195.20.45.31

3 195.20.45.32 195.20.45.33 195.20.45.34 195.20.45.46 195.20.45.47

4 195.20.45.48 195.20.45.49 195.20.45.50 195.20.45.62 195.20.45.63

5 195.20.45.64 195.20.45.65 195.20.45.66 195.20.45.78 195.20.45.79

6 195.20.45.80 195.20.45.81 195.20.45.82 195.20.45.94 195.20.45.95

7 195.20.45.96 195.20.45.97 195.20.45.98 195.20.45.110 195.20.45.111

8 195.20.45.112 195.20.45.113 195.20.45.114 195.20.45.126 195.20.45.127

9 195.20.45.128 195.20.45.129 195.20.45.130 195.20.45.142 195.20.45.143

10 195.20.45.144 195.20.45.145 195.20.45.146 195.20.45.158 195.20.45.159

11 195.20.45.160 195.20.45.161 195.20.45.162 195.20.45.174 195.20.45.175

12 195.20.45.176 195.20.45.177 195.20.45.178 195.20.45.190 195.20.45.191

13 195.20.45.192 195.20.45.193 195.20.45.194 195.20.45.206 195.20.45.207

14 195.20.45.208 195.20.45.209 195.20.45.210 195.20.45.222 195.20.45.223

15 195.20.45.224 195.20.45.225 195.20.45.226 195.20.45.238 195.20.45.239

16 195.20.45.240 195.20.45.241 195.20.45.242 195.20.45.254 195.20.45.255





1 195.20.45.0 195.20.45.1 195.20.45.2 195.20.45.6 195.20.45.7

2 195.20.45.8 195.20.45.9 195.20.45.10 195.20.45.14 195.20.45.15

3 195.20.45.16 195.20.45.17 195.20.45.18 195.20.45.22 195.20.45.23

4 195.20.45.24 195.20.45.25 195.20.45.26 195.20.45.30 195.20.45.31

.. .. .. .. .. ..

30 195.20.45.232 195.20.45.233 195.20.45.234 195.20.45.238 195.20.45.239

31 195.20.45.240 195.20.45.241 195.20.45.242 195.20.45.246 195.20.45.247

32 195.20.45.248 195.20.45.249 195.20.45.250 195.20.45.254 195.20.45.255

13


14/14





1 195.20.45.0 195.20.45.1 195.20.45.2 195.20.45.2 195.20.45.3

2 195.20.45.4 195.20.45.5 195.20.45.6 195.20.45.6 195.20.45.7

3 195.20.45.8 195.20.45.9 195.20.45.10 195.20.45.10 195.20.45.11

4 195.20.45.12 195.20.45.13 195.20.45.14 195.20.45.14 195.20.45.15

.. .. .. .. .. ..

62 195.20.45.244 195.20.45.245 195.20.45.246 195.20.45.246 195.20.45.247

63 195.20.45.248 195.20.45.249 195.20.45.250 195.20.45.250 195.20.45.251

64 195.20.45.252 195.20.45.253 195.20.45.254 195.20.45.254 195.20.45.255

14

subnet for novice

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