MajorBacon’s 6 Steps to Easy Subnetting1.                  Read the question. Know what network ID you are starting with and what you want

to end up with in terms of hosts per network or number of desired networks.

2.                  Write your binary table on your paper. If you can multiply by two, you can do

this table

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 1

3.                  Use the magic formula A (2x >= your desired number of networks, where x is the

number of new ones in the new subnet mask) or magic formula B (2x-2 >= your

desired number of hosts, where x is the number of zeros left in the new subnet

mask)*

4.                  If your goal is a certain number of subnets, x is the number of new ones to add to

your subnet mask. The rest will be zeros.

If your goal is a certain number of hosts, x is the number of zeros remaining in the

subnet mask. The rest will be ones.

Based on this, write out your new subnet mask (in binary, counting ones or zeros as

necessary).

5.                  In order to figure out the number of hosts you have in each subnet, use 2x-2, where

x is the number of zeros in your subnet mask.

In order to figure out the total number of subnets you have, use 2x, where x is the

number of new ones (not total ones) in your subnet mask.

6.                  In order to determine your network IDs, start with your original network ID, then your

next network will be one step value away. How much is your step value? It’s

determined by the “least significant bit”, the last one in the subnet mask. The column

value of this bit will be the increment from one network to the next, in that same

octet! (Also, 256 minus the last positive octet will also yield the increment or step

value of the networks)

1.                  You have a Class C network, 202.5.4.0 that you want to subnet into 6 subnets.

2.                  We write down the all-important table:

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 1

3.                  2x >= your Goal, therefore 23 = 8 >= 6 desired subnets.

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 1

4.                  A Class C subnet mask is 255.255.255.0, so if we add 3 ones, it will be

255.255.255.11100000 or 255.255.255.224

5.                  We have five zeros in the subnet mask, so 25-2=30 hosts per subnet, and we

have added 3 ones to the subnet mask, so 23=8 new subnets.

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 1

6.                  Our increment is based on the least significant bit in the subnet mask, which in

binary was 255.255.255.11100000. If we examine the last octet compared to our

table we see that the last one is in the thirty-two column. (Also, 256-224=32)

27 26 25 24 23 22 21 20

128 64 32 16 8 4 2 1

1 1 1 0 0 0 0 0

So, our network started with 202.5.4.0/24 (the /24 being CIDR notation for a 24-

bit subnet mask). We added three network bits, so our new CIDR notation for

the first network would be

202.5.4.0/27. Our increment is 32 in the fourth octet, so our next network would

be

202.5.4.32/27, and the third

202.5.4.64/27, then

202.5.4.92/27, then

202.5.4.128/27, then

202.5.4.160/27, then

202.5.4.192/27, and finally

202.5.4.224/27.

Please note that if you count all of those subnets up, you have 8 of them. The

amount we predicted back in step 3. Great job!

Now it’s time to practice!!!

Then practice some more!!!

* Note regarding magic formula B: The original RFC standard for subnetting

(RFC 950) stated that the first and last subnets created were not permissible in

public use. This would make the formula for subnets 2x-2. The later superseding

RFC standard (RFC 1812) stated that these subnets were permissible for use.

You may find subnetting calculators that allow you to enable or disable this

standard, changing the subnet quantity and subnet ID list. As for testing

situations, Microsoft and Comptia will not test on this point (they won’t give you a

test question where it will matter), and CISCO will assume the later standard

unless otherwise specified.