Subnetting Part II

The host portion of the IP address cannot be expressed as all 0s or all 1s. All 0s in the host portion of the address indicates the Network ID. All 1s in the host portion of the address indicates
the address for the broadcast on the subnet. Therefore, using our
example above, here are the valid host IDs for our two valid subnets172.16.64.0/18 and 172.16.128.0/18:172.16.64.1/18 - 172.16.127.254/18
172.16.128.1/18 - 172.16.191.254/18To understand the rule more clearly, convert the host addresses to
binary:64 1 127 254
x.y. 01000000.00000001 -x.y. 01111111.11111110
128 1 191 254
x.y. 10000000.00000001 -x.y. 10111111.11111110
 
To calculate the number of hosts or the number of networks for a given subnet mask, use the following formula: 2^(number of bits used) - 2 = number of networks/hosts (The ‘^’ means ‘raised to the power of’.)
So, given the fact that we have borrowed 2 bits to extend our network, we have 2^2 - 2 = 2 networks (remember, we are excluding the all 1 and all 0 networks).
We have 14 bits left to describe the hosts on each network; therefore we have 2^14 - 2 = 16,382 hosts per network (remember we are excluding host addresses with all 0s or all 1s).

The value of the lowest order bit in the subnet mask tells you two things: a) it tells you the first network ID for the subnets created by the extended subnet mask, b) it tells you the value
that you can add to one network ID to arrive at the next possible
network ID.
Let’s take our example extended subnet mask of 255.255.192.0.The 3rd octet expressed in binary with the decimal equivalents is:128 64 32 16 8 4 2 1
1 1 0 0 0 0 0 0The value of the lowest order bit is 64. Therefore, our first
network ID is 172.16.64.0/18. The next and last network ID is172.16.128.0/18 (64 + 64 = 128)To help you calculate subnet masks, you might want to memorize the
table on the right side of this page.That’s it. The first thing I do when I go into an exam situation where
I know I am going to be asked about subnetting is to draw out the
subnetting table on a blank piece of paper.
For the Microsoft 70-642: MCTS: Windows Server Network
Infrastructure, Configuring, you will find that you need to be able
to calculate subnets quickly and accurately. I have found this quick
method to be helpful, but particularly helpful on this exam.

To use the table, find the value for the subnet mask.
Let’s say you are subnetting a Class B address
with the subnet mask 255.255.248.0. Find 248 on
the chart and follow the lines to determine the
low order bit for the subnet. The value for the low
order bit for 248 is 8. That tells you two things: the
starting network ID and the difference between
one network ID and the next: 8. Therefore, given
a class B address of 172.16.0.0/21 (subnet mask of
255.255.248.0), the possible network IDs and host
ranges areNetwork IDs Host Ranges172.16.8.0/21 172.16.8.1 -
172.16.15.254
172.16.16.0/21 172.16.16.1 -
172.16.23.254
172.16.24.0/21 172.16.24.1 -
172.16.31.254
. . . .
172.16.240.0/21 172.16.240.1 -
172.16.247.254To determine the number of networks created by the
subnet mask, count the number of bits used for the
subnet mask. For a subnet mask of 248, the number
is 5 bits. Therefore, the number of networks is 2^5
- 2 = 30. To determine the number of hosts, count
the remaining bits. For a subnetted Class B address
with a subnet mask of 255.255.248.0, we have 3
+ 8 = 11 bits remaining. Therefore, we can have 2^11
- 2 = 2046 hosts per network.
So, to sum up, memorize the Subnetting Table above
and the following rules:1. You can’t have all 0s or all 1s in the host or
network portion of the address (unless the
hardware and software support it).
2. To determine the number of networks or hosts,
use the formula 2^(no. of bits used) - 2.
3. The low order bit in the subnet mask determines
the starting network ID and the difference between
possible network ID