After looking around the internet over the past few months I found the following method works best:
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
NNNNNNNN . NNNNNNNN . HHHHHHHH . HHHHHHHH Class B
NNNNNNNN . NNNNNNNN . NNNNNNNN . HHHHHHHH Class C
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 2^3 = 8 which gives us our block size i.e. 2 to the power of 3 equals 8.
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 2^5 = 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. 2^4 = 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 that subnetting 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 2^4 = 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) = 2^7 = 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 remembering 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.
Source (http://www.techexams.net/forums/ccna-ccent/38772-subnetting-made-easy.html)
Spending time using this method you can quickly determine the Network ID and Broadcast Address.
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 2^5 = 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.
172.16.128.0 would be the next subnet so subtract 1 from third octet, 128- 1 = 127. The Broadcast Address, for subnet 172.16.96.0 is 172.16.127.255
What subnet does 172.16.116.4/ 255.255.255.224 sit on?
Sweet we almost have to do no math to solve this one. First identify which number in the subnet mask is not 255. That's right it's 224.
Subtract 256- 224= 32 is our block size.
(Additional fun! 2^5 = 32. Looking at the 8 16 24 32
We would use 32-5= 17. Now we could rewrite the problem 172.16.116.4/17
Hopefully this looks familiar. This is unessary in solving the question, but it help if develop the thinking method)
Count up block size. 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.
172.16.128.0 would be the next subnet so subtract 1 from third octet, 128- 1 = 127. The Broadcast Address, for subnet 172.16.96.0 is 172.16.127.255
