Subnetting Made Easy - Critically Acclaimed!

We need to start with the fundamentals of IP addressing. An IP address is made up of 32 bits, split into 4 octets (oct = 8, yes?). Some bits are reserved for identifying the network and the other bits are left to identify the host.

There are 3 main classes of IP address that we are concerned with.

Class A	Range 0 - 127 in the first octet (0 and 127 are reserved)
Class B	Range 128 - 191 in the first octet
Class C	Range 192 - 223 in the first octet

Below shows you how, for each class, the address is split in terms of network (N) and host (H) portions.

NNNNNNNN	.	HHHHHHHH	.	HHHHHHHH	.	HHHHHHHH	Class A Address
NNNNNNNN	.	NNNNNNNN	.	HHHHHHHH	.	HHHHHHHH	Class B Address
NNNNNNNN	.	NNNNNNNN	.	NNNNNNNN	.	HHHHHHHH	Class C Address

At each dot I like to think that there is a boundary, therefore there are boundaries after bits 8, 16, 24, and 32. This is an important concept to remember.

We will now look at typical questions that you may see on subnetting. More often than not they ask what a host range is for a specific address or which subnet a certain address is located on. I shall run through examples of each, for each class of IP address.

What subnet does 192.168.12.78/29 belong to?

You may wonder where to begin. Well to start with let's find the next boundary of this address.

Our mask is a /29. The next boundary is 32. So 32 - 29 = 3. Now 2³ = 8 which gives us our block size.

We have borrowed from the last octet as the 29th bit is in the last octet. We start from zero and count up in our block size. Therefore it follows that the subnets are:-

192.168.12.0
192.168.12.8
192.168.12.16
192.168.12.24
192.168.12.32
192.168.12.40
192.168.12.48
192.168.12.56
192.168.12.64
192.168.12.72
192.168.12.80
.............etc

Our address is 192.168.12.78 so it must sit on the 192.168.12.72 subnet.

What subnet does 172.16.116.4/19 sit on?

Our mask is /19 and our next boundary is 24. Therefore 24 - 19 = 5. The block size is 2⁵ = 32.

We have borrowed into the third octet as bit 19 is in the third octet so we count up our block size in that octet. The subnets are:-

172.16.0.0
172.16.32.0
172.16.64.0
172.16.96.0
172.16.128.0
172.16.160.0
.............etc

Our address is 172.16.116.4 so it must sit on the 172.16.96.0 subnet. Easy eh?

What subnet does 10.34.67.234/12 sit on?

Our mask is 12. Our next boundary is 16. Therefore 16 - 12 = 4. 2⁴ = 16 which gives us our block size.

We have borrowed from the second octet as bit 12 sits in the second octet so we count up the block size in that octet. The subnets are:-

10.0.0.0
10.16.0.0
10.32.0.0
10.48.0.0
.............etc

Our address is 10.34.67.234 which must sit on the 10.32.0.0 subnet.

Hopefully the penny is starting to drop and you are slapping the side of your head realising that you were a fool to think it was hard. We will now change the type of question so that we have to give a particular host range of a subnet.

What is the valid host range of of the 4th subnet of 192.168.10.0/28?

Easy as pie! The block size is 16 since 32 - 28 = 4 and 2⁴ = 16. We need to count up in the block size in the last octet as bit 28 is in the last octet.

192.168.10.0
192.168.10.16
192.168.10.32
192.168.10.48
192.168.10.64
.................etc

Therefore the 4th subnet is 192.168.10.48 and the host range must be 192.168.10.49 to 192.168.10.62, remembering that the subnet and broadcast address cannot be used.

What is the valid host range of the 1st subnet of 172.16.0.0/17?

/17 tells us that the block size is 2^24-17 = 2⁷ = 128. We are borrowing in the 3rd octet as bit 17 is in the 3rd octet. Our subnets are:-

172.16.0.0
172.16.128.0

The first subnet is 172.16.0.0 and the valid host range is 172.16.0.1 to 172.16.127.254. You must remember not to include the subnet address (172.16.0.0) and the broadcast address (172.16.127.255).

What is the valid host range of the 7th subnet of address 10.0.0.0/14?

The block size is 4, from 16 - 14 = 2 then 2² = 4. We are borrowing in the second octet so count in the block size from 0 seven times to get the seventh subnet.

The seventh subnet is 10.24.0.0. Our valid host range must be 10.24.0.1 to 10.27.255.254 again remebering not to include our subnet (10.24.0.0) and the broadcast address (10.27.255.255).

What if they give me the subnet mask in dotted decimal?

If you're lucky and they give you a mask in dotted decimal format then you should have an even easier time. All you need again is your block size.

Let's say they have given a mask of 255.255.255.248 and you wish to know the block size. Here's the technique:

1. Starting from the left of the mask find which is the first octet to NOT have 255 in it.

2. Subtract the number in that octet from 256 to get your block size (e.g. above it is 256 - 248 = block size of 8).

3. Count up from zero in your block size in the octet identified in step 1 as you have learned above (the example above would be in the last octet).

Another example is a mask of 255.255.192.0 - you would simply count up in 256 - 192 = 64 in the third octet.

One more example is 255.224.0.0 - block size is 256 - 224 = 32 in the second octet.

What other questions may they ask?

You may find they ask for how many bits you need to borrow for a certain amount of subnets, the subnet mask needed for a certain number of hosts, or the number of hosts per subnet. THESE ARE ALL EASY TO CALCULATE! All you need to remember is that you borrow bits for subnets and reserve bits for hosts.

There are two simple formulas:

Number of subnets = 2ⁿ where n is the number of bits borrowed

Number of hosts = 2^{(32 - n)} - 2 where n is the number of bits in your subnet mask


Let's think of some questions. How many bits do you need to borrow to accommodate 6 subnets? No matter what address you are given the maths is still the same. The formula is 6 = 2ⁿ so you must find n which in this case is 3 as n = 2 gives only 4 subnets and n = 3 gives 8 subnets. Simply add n to your mask for your new subnet mask. For example, if you had a /24 address and you wanted 8 subnets then your mask will be 24 + 3 = /27.

What subnet mask should you use if you wanted 60 hosts per subnet? The formula is 60 = 2^{(32 - n)} - 2 so you must find n which is 26. This is easy to find as you know that 2⁶ - 2 = 62 so simply subtract 6 from 32 to get the 26. Therefore your mask is /26.

Lastly the number of hosts per subnet. How many hosts per subnet in the address 172.16.0.0/23? You have a /23 address therefore you formula is x = 2^{(32 - 23)} - 2 = 2⁹ - 2 = 510.

Another typical question they may ask will be giving you an IP address and mask and asking how many subnets and hosts there are from that address, for example:

Question: How many subnets and hosts per subnet can you get from the network 172.30.0.0/28?

From this you only need two pieces of information:

1. The default subnet mask of the address class.
2. The subnet mask in the question

Using the example above we know that:

1. The default subnet mask is /16 as the address given is a class B address
2. The subnet mask in the question is /28

The number of subnets = 2 ^ (subnet_mask_in_question - default_subnet_mask)
The number of hosts = (2 ^ (32 - subnet_mask_in_question)) - 2

For our example question the number of subnets = 2 ^ (28 - 16) = 2 ^ 12 = 4096 subnets.
The number of hosts = (2 ^ (32 - 28)) - 2 = (2 ^ 4) - 2 = 14 hosts per subnet

Let's use another address: 192.168.1.0/29

We know that:

1. The default subnet mask is /24 as the address given is a class C address
2. The subnet mask in the question is /29

The number of subnets = 2 ^ (29 - 24) = 2 ^ 5 = 32 subnets.
The number of hosts = (2 ^ (32 - 29)) - 2 = (2 ^ 3) - 2 = 6 hosts per subnet

Finally, let's use another address: 10.1.1.0/24

We know that:

1. The default subnet mask is /8 as the address given is a class A address
2. The subnet mask in the question is /24

The number of subnets = 2 ^ (24 - 8) = 2 ^ 16 = 65536 subnets.
The number of hosts = (2 ^ (32 - 24)) - 2 = (2 ^ 8) - 2 = 254 hosts per subnet

Easy isn't it?

What now?

Now it's time to go and pick up those books again and go straight to the practice questions, completely by-passing any of their techniques. Use my method and you will be laughing!

If you are unsure that you have the correct answers why not download a subnet calculator to double-check your answers? There is a great one by 3Com and can be downloaded from here.

Happy subnetting!