Configure TCP/IP - MCSA Windows Server 2012 R2: Installation and Configuration Study Guide - page 414

There are four subnets, because 2 2 = 4. The valid third-octet values for the subnets are
0, 64, 128, and 192 (256 - 192 = 64, so the incremental value of the third octet is 64).
However, there are 14 bits (0s) left over for host addressing. This gives you 16,382 hosts
per subnet (2 14 - 2 = 16,382).
The valid subnets and hosts are as follows:

Subnet Hosts Broadcast

x.y.0.0 x.y.0.1 through x.y. 63.254 x.y.63. 255

x.y.64.0 x.y.64.1 through x.y.127. 254 x.y.127.255

x.y.128.0 x.y.128.1 through x.y.191.254 x.y.191.255

x.y.192 .0 x.y.192.1 through x.y.255.254 x.y.255.255

You can add another bit to the subnet mask, making it 11111111.11111111.11100000.
00000000, or 255.255.224.0. This gives you eight subnets (2 3 = 8) and 8,190 hosts. The
valid subnets are 0, 32, 64, 96, 128, 160, 192, and 224 (256 - 224 = 32). The subnets, valid
hosts, and broadcasts are listed here:

Subnet Hosts Broadcast

x.y.0.0 x.y.0.1 through x.y.31. 254 x.y.31. 255

x.y.32 .0 x.y.32.1 through x.y.63.254 x.y.63. 255

x.y.64.0 x.y.64.1 through x.y.95. 254 x.y.95. 255

x.y.96.0 x.y.96.1 through x.y.127.254 x.y.127.255

x.y.128.0 x.y.128.1 through x.y.159.254 x.y.159.255

x.y.160.0 x.y.160.1 through x.y.191.254 x.y.191.255

x.y.192 .0 x.y.192.1 through x.y.223.254 x.y.223.255

x.y. 224.0 x.y.224.1 through x.y.255.254 x.y.255.255

The following are the breakdowns for a 9-bit mask and a 14-bit mask:

If you use 9 bits for the mask, it gives you 512 subnets (2 9 ). With only 7 bits for hosts,
you still have 126 hosts per subnet (2 7 - 2 = 126). The mask looks like this:

■

11111111.11111111.11111111.10000000, or 255.255.255.128

■

If you use 14 bits for the subnet mask, you get 16,384 subnets (2 14 ) but only two hosts
per subnet (2 2 - 2 = 2). The subnet mask would look like this:

11111111.11111111.11111111.11111100, or 255.255.255.252