Address Masks That Don’t End on an Eight-Bit Boundary

If we recall that a mask of 255.255.255.0 actually represents a binary number then we can easily understand how the mask value need not end on an eight-bit boundary. For example, consider a Class B network 140.6.0.0. In this network I want to have up to 1024 hosts per subnetwork. This means I will need 10 bits to represent the host portion of the address (since 2 raised to the 10th power = 1024). I need 10 “0” bits in the mask to represent the host portion, the remainder will be split between the network and subnetwork identifiers. Since Class B uses the first 16 bits to identify the network this leaves 6 bits to represent the subnetwork. I can have up to 64 different values in the subnet field and up to 1024 values in the host field. The address mask looks like this:

11111111 11111111 11111100 00000000

There are three fields that are defined in this mask. Because we are applying this to a Class B network the first 16 bits represent the network. The next 6 bits represent the subnetwork, and the last 10 bits represent the host.

When each octet is converted to decimal the mask value becomes:

11111111 11111111 11111100 00000000
255. 255. 252. 0

Any number of bits can be used to identify the subnetwork. Some other examples (in Class B) include:

  • 255.255.255.128 – Nine bits for the subnet, seven bits for the host
  • 255.255.255.192 – Ten bits for the subnet, six bits for the host

The exact same scheme applies to Class A but only the first eight bits are included in the network part. This allows many more combinations for subnetting. Some examples in Class A include:

  • 255.255.0.0 – (Typical) Eight bits for the subnet, sixteen bits for the host.
  • 255.252.0.0 – Six bits for the subnet, eighteen bits for the host.
  • 255.255.252.0 – Fourteen bits for the subnet, ten bits for the host.

When considering the subnetting in a network it is necessary to convert the dotted-decimal octet strings for the addresses and the mask back into binary to compare the fields and evaluate the results.