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Tuesday, December 11, 2012

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 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!
5 RakshaTec: 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...

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