I'll begin on very basic subnetting skills based on network scenarios. I'll start with the first one.

An organization has purchased a class C address block 215.21.4.0 and would like to use it to address a network of 5 subnets. The first thing we want to look at is how many host bits should we borrow. The number of host bits can be found by using 5=(2^x)-2. x would then be 3.

The subnet mask began with 255.255.255.0. We'll add 3 bits for subnetting, so we'll end up with 255.255.255.224. We'll now look at the increment. To find the increment, take the value of the least significant "1" in your subnet mask.

255.255.255.224 = 11111111.11111111.11111111.11100000.

The least significant "1" in this case would be 32. Therefore, you'll have subnets that look like this:

0)215.21.4.0 (.1 to .30)

1)215.21.4.32 (.33 to .62)

2)215.21.4.64 (.65 to .94)

3)215.21.4.96 (.97 to .126)

4)215.21.4.128 (.129 to .158)

5)215.21.4.160 (.161 to .190)

6)215.21.4.192 (.193 to .222)

7)215.21.4.224 (.225 to .254)

The first and last subnet cannot be used in legacy products because they would think 215.21.4.0 refers to the entire block, and 215.21.4.255 would broadcast to the entire block. However, currently if your router supports the subnet-zero command, then the first and last would be usable.

Let's do the next scenario. Convert 192.5.20.0/24 into 50 subnets. We'll need to find the subnet bits, which is 50=(2^x)-2. x would then be 6.

We started with mask 255.255.255.0. Now we'll have 255.255.255.252. Now let's find the increment. In 11111100, the increment is 4. We'll then have subnets that look like:

0)192.5.20.0 (.1 and .2)

1)192.5.20.4 (.5 and .6)

2)192.5.20.8 (.9 and .10)

...

10)192.5.20.40 (.41 and .42)

...

30)192.5.20.120 (.121 and .122)

...

50)192.5.20.200 (.201 and .202)

...

63)192.5.20.252 (.253 and .254)

Now the next example. We'll go through this quick. 150.3.0.0/16, and we'll need 100 networks. 100 networks will require 7 subnet bits, so we'll have 255.255.254.0. Increment is 2. We'll have:

0)150.3.0.0 (.0.1 to .1.254)

1)150.3.2.0 (.2.1 to .3.254)

2)150.3.4.0 (.4.1 to .5.254)

...

100)150.3.200.0 (.200.1 to .201.254)

127)150.3.254.0 (.254.1 to .255.254)

Here's another one. 10.0.0.0/8 to be separated into 500 networks. 500 networks will require 9 subnet bits. The mask would therefore be 255.255.128.0. The networks would then be:

0)10.0.0.0 (0.0.1 to 0.127.254)

1)10.0.128.0 (0.128.1 to 0.255.254)

2)10.1.0.0 (1.0.1 to 1.127.254)

3)10.1.128.0 (1.128.1 to 1.255.254)

4)10.2.0.0 (2.0.1 to 2.127.254)

I'll move through these quickly now.

Breaking 200.1.1.0/24 into 40 subnets would require 6 subnet bits which would give the mask of 255.255.255.252. With the increment of 4, the subnets would be:

200.1.1.0 (.1 and .2)

200.1.1.4 (.5 and .6)

200.1.1.8 (.9 and .10)

200.1.1.12 (.13 and .14)

Breaking 199.9.10.0/24 into 14 subnets would require 4 subnet bits which would give the mask of 255.255.255.240. With the increment of 16, the subnets would be:

199.9.10.16 (.17 to .30)

199.9.10.32 (.33 to .46)

199.9.10.48 (.49 to .62)

199.9.10.64 (.62 to .94)

Breaking 170.50.0.0/16 into 1000 subnets would require 10 subnet bits which would give the mask of 255.255.255.192. With the increment of 64, the subnets would be:

170.50.0.0 (0.1 to 0.62)

170.50.0.64 (0.65 to 0.126)

170.50.0.128 (0.129 to 0.190)

170.50.0.192 (0.193 to 0.254)

170.50.1.0 (1.1 to 1.62)

Breaking 12.0.0.0/8 into 25 subnets would require 5 subnet bits which would give the mask of 255.248.0.0. With the increment of 8, the subnets would be:

12.0.0.0 (0.0.1 to 7.255.254)

12.8.0.0 (8.0.1 to 15.255.254)

12.16.0.0 (16.0.1 to 23.255.254)

12.24.0.0 (24.0.1 to 31.255.254)

## Saturday, April 3, 2010

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