Case Study

An organization has been assigned the network number 193.1.1.0/24 and it needs to define six subnets. The largest subnet is required to support 25 hosts.

The first step is to determine the number of bits required to define the six subnets. Because a network address can be subnetted only along binary boundaries, subnets must be created in blocks of powers of two 2 (2¹), 4 (2²), 8 (2³), 16 (2⁴), and so on. It is impossible to define an IP address block such that it contains exactly six subnets. For this example, the network administrator must define a block of 8 (2³) and have two unused subnets that can be reserved for future growth.

Because 8 = 2³, three bits are required to enumerate the eight subnets in the block. In this example, the organization is subnetting a /24 so that it will need three more bits, or a /27, as the extended-network-prefix. A 27-bit extended-network-prefix can be expressed in dotted-decimal notation as 255.255.255.224. This is illustrated in Figure 3-4.

Figure 3-4. Defining the Subnet Mask/Extended-Prefix Length

A 27-bit extended-network-prefix leaves 5 bits to define host addresses on each subnet. This means that each subnetwork with a 27-bit prefix represents a contiguous block of 2⁵ (32) individual IP addresses. However, because the all-0s and all-1s host addresses cannot be allocated, there are 30 (2⁵-2) assignable host addresses on each subnet.

The eight subnet numbers for this example are given in the following list. The italicized portion of each address identifies the extended-network-prefix, while the bold digits identify the 3 bits representing the subnet-number field:

 Base Net: 11000001.00000001.00000001.00000000 = 193.1.1.0/24 Subnet #0: 11000001.00000001.00000001.000 00000 = 193.1.1.0/27 Subnet #1: 11000001.00000001.00000001.001 00000 = 193.1.1.32/27 Subnet #2: 11000001.00000001.00000001.010 00000 = 193.1.1.64/27 Subnet #3: 11000001.00000001.00000001.011 00000 = 193.1.1.96/27 Subnet #4: 11000001.00000001.00000001.100 00000 = 193.1.1.128/27 Subnet #5: 11000001.00000001.00000001.101 00000 = 193.1.1.160/27 Subnet #6: 11000001.00000001.00000001.110 00000 = 193.1.1.192/27 Subnet #7: 11000001.00000001.00000001.111 00000 = 193.1.1.224/27 

An easy way to check whether the subnets are correct is to ensure that they are all multiples of the Subnet #1 address. In this case, all subnets are multiples of 32: 0, 32, 64, 96, etc.

The broadcast address for Subnet #2 is the all-1s host address:

 11000001.00000001.00000001.010 11111 = 193.1.1.95 

Note that the broadcast address for Subnet #2 is exactly one less than the base address for Subnet #3 (193.1.1.96). This is always the case the broadcast address for Subnet #n is one less than the base address for Subnet #(n+1).

The broadcast address for Subnet #6 is the all-1s host address: 11000001.00000001.00000001.110 11111 = 193.1.1.223. Again, the broadcast address for Subnet #6 is exactly one less than the base address for Subnet #7 (193.1.1.224).