A guide

1
SUBNETTING

• A guide
• By
• Brian Proctor

2

• IP Address classes
• Class A 0.X.X.X - 126.X.X.X
• Class B 128.X.X.X - 191.X.X.X
• Class C 192.X.X.X - 223.X.X.X
• Class D 224.X.X.X - 239.X.X.X - Multicast
• Class E 240.X.X.X - 256.X.X.X - Experimental

3

• Private address
• These are not used on or connected to the open
  internet
• Used on private networks (usually behind NAT)
• 10.x.x.x
• 172.16.x.x
• 192.168.x.x
• 127.x.x.x is a loopback address

4

• Subnetting a class C IP address
• Example 212.168.20.0
• In this example we will use a subnet mask of 224
• Ie a subnet mask of 255.255.255.224
• To achieve this we need to borrow 3 bits from the
  host portion of the IP address

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• Default Subnet mask in decimal 255.255.255.0
  Default Subnet mask in binary Class C
• NETWORK HOST
• 11111111.11111111.11111111.00000000
• 255 255 255 0
• Subnet mask of 255.255.255.224
• NETWORK HOST
• 11111111.11111111.11111111.11100000
• 255 255 255 224
• 3 BITS BORROWED
• FROM HOST

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• Calculating the size of subnets
• Host Octet bit value
• 128 64 32 16 8 4 2 1
• Bits borrowed
• 128 1 0 0 0 0 0 0
  0
• 192 1 1 0 0 0 0 0
  0
• 224 1 1 1 0 0 0 0
  0
• 240 1 1 1 1 0 0 0
  0
• 248 1 1 1 1 1 0 0
  0
• 252 1 1 1 1 1 1 0
  0
• 254 1 1 1 1 1 1 1
  0
• 255 1 1 1 1 1 1 1
  1

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• Host Octet bit value
• 128 64 32 16 8 4 2 1
• Count from the left 3 bits at look up to find the
  number of hosts per subnet.
• Count from the right 3 bits then look diagonally
  to find number of subnets.
• In this example 8 subnetworks with 32 hosts per
  network
• Usable networks 8 2 6
• Convention dictates that first and last is not
  used, but are valid ( network address and
  broadcast).

Hosts Networks
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• Alternative method for calculation
• Number of bits borrowed 3
• Calculation of subnetworks
• 3 bits (23 -2) (2 x 2 x 2) - 2 6
• Calculation of Hosts
• 5 bits left in host portion of IP address (8
  3 5)
• 5 bits (25 2)
• (2 x 2 x 2 x 2 x 2) 2 30 hosts per
  subnetworks

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• Resulting subnetworks
• Subnet 0 - 212.168.20.0 through 212.168.20.31
• Subnet 1 - 212.168.20.32 through 212.168.20.63
• Subnet 2 - 212.168.20.64 through 212.168.20.95
• Subnet 3 - 212.168.20.96 through 212.168.20.127
• Subnet 4 - 212.168.20.128 through 212.168.20.159
• Subnet 5 - 212.168.20.160 through 212.168.20.191
• Subnet 6 - 212.168.20.192 through 212.168.20.223
• Subnet 7 - 212.168.20.224 through 212.168.20.255
• Convention dictates that first and last is not
  used, but are valid ( network address and
  broadcast).

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• Applying the boolean and to the IP address and
  subnet mask to determine which subnet this
  address belongs (This is what the router does to
  forward traffic to the correct subnet)
• IP address 212.168.20.127
• Subnet mask 255.255.255.224
• 11010100.10101000.00010100.01111111
• 11111111.11111111.11111111.11100000
• 11010100.10101000.00010100.01100000
• 212. 168. 20. 96
• Therefore this IP address belongs to the
• 212.168.20.96 subnet - subnet 3 (4th subnet)

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• Subnetting a class B IP address
• Example 150.168.0.0
• In this example we will use a subnet mask of 224
• Ie a subnet mask of 255.255.224.0
• To achieve this we need to borrow 3 bits from the
  host portion of the IP address

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• Default Subnet mask in decimal 255.255.0.0
• Default Subnet mask in binary Class B
• 11111111.11111111.00000000.00000000
• 255 255 0 0
• Subnet mask of 255.255.224.0
• NETWORK HOST
• 11111111.11111111.11100000.00000000
• 255 255 224 0
• NETWORK HOST
• 3 BITS BORROWED
• FROM HOST

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• Host Octet bit value (2 Octects class B)
• 32768 16384 8192 4096 2048 1024 512
  256 128 64 32 16 8 4 2
  1
• Count from the left 3 bits and look up to find
  the number of hosts per subnet.
• Count from the right 3 bits then look diagonally.
• In this example 8 subnetworks with 8192 hosts per
  network
• Convention dictates that first and last is not
  used, but are valid ( network address and
  broadcast).

Hosts Networks
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• Alternative method for calculation
• Number of bits borrowed 3
• Calculation of subnetworks
• 3 bits (23) (2 x 2 x 2) 8
• 13 bits left Calculation of Hosts in host portion
  of IP address (16 3 13)
• 13 bits (213 2)
• (2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x
  2 ) 2
• 8190 hosts per subnetworks

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• Resulting subnetworks
• Subnet 0 - 212.168.0.0 through 212.168.31.255
• Subnet 1 - 212.168.32.0 through 212.168.63.255
• Subnet 2 - 212.168.64.0 through 212.168.95.255
• Subnet 3 - 212.168.96.0 through 212.168.127.255
• Subnet 4 - 212.168.128.0 through 212.168.159.255
• Subnet 5 - 212.168.160.0 through 212.168.191.255
• Subnet 6 - 212.168.192.0 through 212.168.223.255
• Subnet 7 - 212.168.224.0 through 212.168.255.255
• Convention dictates that first and last is not
  used, but are valid ( network address and
  broadcast).
• First and last host in each subnet unavailable

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Finding the xxx subnet

• Example find the 201st subnet from the address
  10.0.0.0/20 201-1200 (subnet 0)
• Binary of 200 11001000
• Subnetmask
• 11111111.11111111.11110000.0000000
• Binary of 10.0.0.0
• 00001010.00000000.00000000.00000000
• 1100 1000 then add
• 00001010.00001100.0100000.000000000
• 10. 12. 64. 0