Illustrated TCP/IP by Matthew G. Naugle Wiley Computer Publishing, John Wiley & Sons, Inc. ISBN: 0471196568   Pub Date: 11/01/98

Chapter 53
Let’s Try One

You have been assigned an address of 150.5.0.0. You need 75 subnets with at least 75 hosts per subnet. What is the acceptable subnet mask?

The first step is to find out how many bits are needed for 75 subnets. In binary, 6 bits represent 64 possible subnets (2 ⁿ6). Not enough. Seven bits is 128 (0 inclusive) and this is the number of bits that we will use. Plus it gives us room for expansion. This leaves 9 bits for hosts, which allows for 510 hosts per subnet.

We start to assign subnets from the left and work to the right. We assign the hosts from the right and work to the left.

You must also define the broadcast address for a subnet. If you wanted to send something to all hosts on a subnet then the hosts field must be set to all 1s. With this subnet mask, there are 9 bits of 1s for an all–hosts broadcast address.

Another example (not shown in the slide) would be to define the mask for a network to support 40 hosts per subnet using the class address 195.1.10.0. First, we determine that this address is a Class C address and that only the last octet can be used for subnetting. Forty hosts is represented by 2ⁿ6, which allows for 60 hosts. This may seem like a lot, but the nearest mask would be 2ⁿ 5, which would give us 30 possible host IDs, and this is not enough. Forty converted to binary is 101000. However, in the conversion we must remain contiguous and we cannot interleaf host and subnet bits. Therefore, we move the left 6 bits and then we can consume all 5 bits to the right. However, this only leaves 2 bits for a subnet. We can have 4, 2 bits left, subnets with 62 (2ⁿ6) –2 hosts per subnet. If the site needed more subnets, we would have to assign more Class C addresses to the site.

Let’s Try One