## An easier way of subnetting

Here’s a way of learning subnetting that I’ve come up with:

You should know this:

Class A Address: 8 network bits - 24 host bits; range is 0-127 in the first octet

Class B Address: 16 net bits - 16 host bits; range is 128-191 in the first oct

Class C Address: 24 net bits - 8 host bits; range is 192-223 in the first oct

Now, remember this

1 - 128

2 - 64

3 - 32

4 - 16

5 - 8

6 - 4

7 - 2

8 - 1

As long as you remember this (very easy) rule, you should be able to answer most subnetting questions in flash.

Explanation:

First part is the number of Host bits and the second part is the ‘increment’.

Example:

Let’s take the third line, 3 - 32.

This means, if the number of host bits is 3, you should increment the subnet by 32.

Let me elaborate:

192.168.23.46/27 - what subnet should this be in?

This is all you need to do:

27-24 = 3 (since this is a class C address, you subtract 24 network bits from 27 and that gives you 3 host bits)

3 should correspond to 32 (according to the

that means increment is 32

so the subnet range should be:

192.168.23.0

192.168.23.32

192.168.23.64

and so on…

the address given to us is 192.168.23.46, so that’ll fall in 192.168.23.32 subnet (as 46 lies between 32 and 64)

Easy.

Here’s another example.

192.168.1.50/28

Again, class C address. So this is what you do: 28-24=4

Golden rule: 4 corresponds to 16, so 16 should be the ‘increment’.

So here we go:

192.168.1.0

192.168.1.16

192.168.1.32

192.168.1.48

192.168.1.64

and so on…

our address has the last octet value of 50, which lies in the range 48 - 64, so the address is in the 192.168.1.48 subnet.

Let’s try a Class B example now:

What subnet does host 172.28.12.167/20 belong to?

Its a class B address so this is what we do: 20 - 16 = 4

4 should correspond to 32 going by the golden rule, so increment value is 32 here.

Since it’s a class B address, here’s how we can break the bits into Network and Host bits:

Typical Class B address: 8 Net bits - 8 Net bits (16) - 8 Host bits - 8 Host bits (16)

This example: 8 Net bits - 8 Net bits - 4 Host bits - 0 Host bits (we only have 4 Host bits, remember? 20-16 = 4)

So the increment would be in 3rd octet.

Here we go:

172.28.0.0

172.28.16.0

172.28.32.0 and so on..

Third octet in this example is 12, so it should lie in the first address (between 0 and 16): 172.28.0.0

If you’re preparing for a CCNA exam, this rule should help a lot, as you can easily remember it and after some practice, these values do get embedded in your brain. Saves you a lot of time.

If the Subnet mask has been provided in dotted format, just subtract the first non-zero octet value from right by 256 and that’ll give you the ‘increment’ value.

Example, 192.168.23.46 255.255.255.240

All you do is: 256-240 = 16. 16 is your increment value!

Another example, 172.15.23.187 255.255.248.0

All you do is: 256-248=8. 8 is your increment value and here’s what the subnets should look like:

172.15.0.0

172.15.8.0

172.15.16.0 and so on..

Hope this helps. Cheers!

You should know this:

Class A Address: 8 network bits - 24 host bits; range is 0-127 in the first octet

Class B Address: 16 net bits - 16 host bits; range is 128-191 in the first oct

Class C Address: 24 net bits - 8 host bits; range is 192-223 in the first oct

Now, remember this

**golden rule**:1 - 128

2 - 64

3 - 32

4 - 16

5 - 8

6 - 4

7 - 2

8 - 1

As long as you remember this (very easy) rule, you should be able to answer most subnetting questions in flash.

Explanation:

First part is the number of Host bits and the second part is the ‘increment’.

Example:

Let’s take the third line, 3 - 32.

This means, if the number of host bits is 3, you should increment the subnet by 32.

Let me elaborate:

192.168.23.46/27 - what subnet should this be in?

This is all you need to do:

27-24 = 3 (since this is a class C address, you subtract 24 network bits from 27 and that gives you 3 host bits)

3 should correspond to 32 (according to the

**golden rule**I gave you above)that means increment is 32

so the subnet range should be:

192.168.23.0

192.168.23.32

192.168.23.64

and so on…

the address given to us is 192.168.23.46, so that’ll fall in 192.168.23.32 subnet (as 46 lies between 32 and 64)

Easy.

Here’s another example.

192.168.1.50/28

Again, class C address. So this is what you do: 28-24=4

Golden rule: 4 corresponds to 16, so 16 should be the ‘increment’.

So here we go:

192.168.1.0

192.168.1.16

192.168.1.32

192.168.1.48

192.168.1.64

and so on…

our address has the last octet value of 50, which lies in the range 48 - 64, so the address is in the 192.168.1.48 subnet.

Let’s try a Class B example now:

What subnet does host 172.28.12.167/20 belong to?

Its a class B address so this is what we do: 20 - 16 = 4

4 should correspond to 32 going by the golden rule, so increment value is 32 here.

Since it’s a class B address, here’s how we can break the bits into Network and Host bits:

Typical Class B address: 8 Net bits - 8 Net bits (16) - 8 Host bits - 8 Host bits (16)

This example: 8 Net bits - 8 Net bits - 4 Host bits - 0 Host bits (we only have 4 Host bits, remember? 20-16 = 4)

So the increment would be in 3rd octet.

Here we go:

172.28.0.0

172.28.16.0

172.28.32.0 and so on..

Third octet in this example is 12, so it should lie in the first address (between 0 and 16): 172.28.0.0

If you’re preparing for a CCNA exam, this rule should help a lot, as you can easily remember it and after some practice, these values do get embedded in your brain. Saves you a lot of time.

If the Subnet mask has been provided in dotted format, just subtract the first non-zero octet value from right by 256 and that’ll give you the ‘increment’ value.

Example, 192.168.23.46 255.255.255.240

All you do is: 256-240 = 16. 16 is your increment value!

Another example, 172.15.23.187 255.255.248.0

All you do is: 256-248=8. 8 is your increment value and here’s what the subnets should look like:

172.15.0.0

172.15.8.0

172.15.16.0 and so on..

Hope this helps. Cheers!

An easier way of subnetting
Reviewed by VT
on
December 11, 2012
Rating: 5