SubNet Masking

This came from Poor Roger's Home Page at http://home.earthlink.net/~rogergoodman.

Poor Roger’s New & Improved Method for
SubNet Masking

Quickly scribble down the above table in this fashion:

Write the digits 7, 6, ..., 0.
Under the 0 at the end of the line write a "1." Move to the left and double it, "2." Etc.
Next line down. Write "128." Add to this the next above value (64) & write result (192). Etc.
Next line down write the digits 1, 2, ..., 8.

How to use:

1. Write down the default subnet mask as determined by class A/B/C leaving off the trailing zeros.

The mask must relate to the SubNet, not Host, portion. This requires an extra step.
Go to Step 2a.

2a. Case A: You know the number of Host ID’s required.

Since the mask must relate to the SubNet and not the Hosts portion, an extra step is needed.
Method: Find the number of bits needed for the Hosts portion & see what that leaves for SubNet.
Let n = the required number of Host ID's.

Do Step 3 which gives you the number of bits required in the Hosts portion of the octet.
This tells you what part of your 8-bit octet must be used for the Network portion of the mask.
So for the SubNet portion the remainder of those 8-bits are available:

Let N = 8 - (the number of bits required for the Hosts portion of the mask).
This N is the new number of bits available for the SubNet portion of the mask.

Now go to Step 4 with your new number of bits required and continue on from there.

2b. Case B: You know the number of SubNets required.

Method: Find out the number of bits this requires and convert that to a SubNet mask value.
Let n = the required number of SubNets.

3. Add 2 (=n+2) to allow for losing the disallowed all zero-bits and all one-bits values.

On the second line of your 1st table find the value equal to or next greater than n+2.
Look above that value to find the number of bits required.
If Case A: Continue where you left of in Step 2a Case A.
If Case B: Continue with Step 4.

4. Find that same number of bits required on the bottom line (NET High Order BITS).

5. Write that value as a continuation of the mask you started in Step 1. If not 4 octets add ". 0's" to finish.

To quickly produce the above table, do the following:

The exception is the 1st (A) range that ends at 126 because "127" is reserved for loopback.
This gives us the ranges A 1-126, B 128-191, C 192-223.

This came from Poor Roger's Home Page at http://home.earthlink.net/~rogergoodman.