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Long Division Table

This is an example of a very LOOOOOONG and tedious way to find a subnet number. At the end of his example, I will show you the easy way to get the same answer.

This is a quote from his web page ---

Finding A Particular Subnet

By: Drew Campbell ­ Community College of Beaver County

IT Manager & CNAP Regional Academy Instructor

 We've learned how to subnet different classes of addresses, but how do we find a particular subnet? Sure, you can write out each of the network numbers, but what if you need to find a subnet in the hundreds? Or thousands? Or TEN THOUSANDS?!?

Let's take a look

 Given: 162.124.0.0 / 25

Find the following:

1. Class
2. Bits Borrowed
3. Subnet Mask
4. Number of subnets & useable subnets
5. Number of hosts & useable hosts
6. The 460^th subnet

  2. Bits borrowed: If the first 25 bits (from left to right) are 1's, we get:

11111111.11111111.11111111.10000000

A Class B address takes up the first 2 octets (16 bits), when subtracted from 25, leaves us with 9 bits (borrowed).

11111111.11111111.11111111.10000000

( 9 we borrowed)

This gives us 512 subnets, with 510 useable.

So we can find the 460^th.

So 2^7 (2*2*2*2*2*2*2) = 128 hosts per subnet, 126 useable per subnet.

Tough question? Sure. You could sit & write out ALL the network numbers, but there's an easier way. There are others ways, but this is the one I know:

1. First, let's find our increment. The last non-zero octet in the subnet mask is the fourth. (255.255.255.128)

We'll subtract it from 256 (which is the number of possibilities in an 8-bit number => range is 0 ­ 255, 256 possibilities)

256 ­ 128 = 128, our network increment.

460 * 128 = 58880

1. Now, we'll divide 2 into 58880 and each additional answer and keep track of the remainder [R], which will make up our binary subnetwork:

2 | 58880

2 | 29440 R 0 (Note: The colors I'm using represent

2 | 14720 R 0 the answer to each division

2 | 7360 R 0 and the corresponding remainder.

2 | 3680 R 0 So, the first step is "2 into 58880

2 | 1840 R 0 equals 29440, with a remainder

2 | 920 R 0 0", and so on )

2 | 460 R 0

2 | 230 R 0

2 | 115 R 0

2 | 57 R 1

2 | 28 R 1

2 | 14 R 0

2 | 7 R 0

2 | 3 R 1

2 | 1 R 1

0 R 1

Now, lets write out the remainders, reading from bottom to top:

1110011000000000

Since we work in 8-bit octets, let's group the remainders the same way, starting from RIGHT to LEFT:

ß11100110 00000000

*Note: You won't always have an even number of bits. If you see you only have 4, 5, 6, whatever left at the beginning, put trailing zero's in front.

From right to left grouping: 1101 00111100

Put zeros in front of the first group: 00001101 00111100

Now, let's look at the original problem again:

162.124.0.0 / 25 ; Subnet mask is 255.255.255.128

..and our remainders (grouped): 11100110 00000000

and in decimal: 230 . 0

 Since we borrow into the 3^rd & 4^th octet (and the 1^st & 2^nd are network-given Class B), our groups represent the 3^rd & 4^th octet addresses:

162.124.230.0 = our 460th subnet.

========================================================================================================

Wow, that was tiresome! Hope I did all my division and remainder carrying correct or I'll have to do it again.

Now let's do the same thing the easy way -

What is the 460th subnet of 162.124.0.0/25?

We know that 9 bits were borrowed leaving us with 7 bits for hosts:
162.124._ _ _ _ _ _ _ _ . _ | _ _ _ _ _ _ _

Starting with the last bit borrowed, plug in the 460 (IN BINARY! Right to left) and you have this:

162.124. 1 1 1 0 0 1 1 0 . 0 |_ _ _ _ _ _ _