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Posted on 2008-11-11

does network + exam require one to know how to subnet ? do they give you a network ID and ask you to subnet it into 6 networks, for instance, and then find network ID, first valid address, last valid address, and broadcast address.

thank you very much. and I am referencing current exam NOT The 2009 version.

thank you very much. and I am referencing current exam NOT The 2009 version.

8 Comments

I've never taken the Network+ exam so I couldn't say whether or not they'd ask you those types of subnetting questions. However, I do know a farily easy way to calculate those types of answers without using a subnet calculator.

Let me know if you're interested in seeing the method that I use and I'll post a couple of examples.

could you please?

I am using book's method but if you know of a quick method please let me know. the book goes on to say that 'don't worry about subnetting too much as net + does not concentrate on that" but you never know.

1 of book example is subnet network ID 150.87.0.0 into "6" subnets.

thank you very much

First thing I do is identify the starting network mask. In your example, 150.87.0.0 is a Class B so that would be 255.255.0.0. So written out in binaryl, we know that mask is 11111111.11111111.00000000

Lay out that host portion and start counting from left to right:

0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0

1 2 4 8 16 32 64 128. 256 512 1024...

Ok, now, for every bit we turn on, we multiply by 2. So:

Turning on the first bit we get: 1 x 2 = 2 subnets

Turning on the second we get: 2 x 2 = 4 subnets

Turning on the third bit we get: 4 x 2 = 8 -2 = 8 subnets AHA - that gives us enough to satisfy 6!

So now our network portion, in binary, looks like this:

1 1 1 1 1 1 1 1 . 1 1 1 1 1 1 1 1 . 1 1 1 0 0 0 0 0 . 0 0 0 0 0 0 0 0

So in dotted decimal, we have a new mask of: 255.255.224.0

To find the ranges, we count back right to left until we hit the first bit that turned on in the network portion. So;

1 1 1 0 0 0 0 0

128 64 32 16 8 4 2 1 So counting right to left, we turned on the 32 bit, so our ranges will be increments of 32's.

150.87.0.0 thru 150.87.31.255

150.87.32.0 thru 150.87.63.255

150.87.64.0 thru 150.87.95.255

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150.87.224.0 thru 150.87.255.255

So, there's the method I use. To summarize:

1. Find your default network mask

2. Write out the host portion in binary

3. Count left to right by powers of 2 until you have a number large enough for the number of subnets asked for

4. Determine how many bits you turned on and then convert back to decimal to get the new mask

5. Using the binary again, count right to left until you come to the first bit that you turned on. That bit value will correspond to your range increments.

I'll post a Class C example in the next message.

Class C example:

Given 192.168.23.0, break this into 30 subnets.

So, we have:

192.168.23.0

255.255.255.0

The host portion of the mask in binary is:

0 0 0 0 0 0 0 0

1 2 4 8 16 32 64 128

Count left to right and multiply by 2:

1 x 2 = 2 subnets

2 x 2 = 4 subnets

4 x 2 = 8 subnets

8 x 2 = 16 subnets

16 x 2 = 32 subnets - OK, we stop here!

Now, write out in binary so we can see how many bits we turned on:

1 1 1 1 1 0 0 0 = 248 in decimal

Our new mask is then:

255.255.255.248

To find our subnet ranges, count right to left until we come to the first bit that we turned on:

1 1 1 1 1 0 0 0

128 64 32 16 8 4 2 1

Looks like our increments will be in 8's!

192.168.23.0 thru 192.168.23.7

192.168.23.8 thru 192.168.23.15

192.168.23.16.0 thru 192.168.23.31

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192.168.23.248 thru 192.168.23.255

Now sometimes, they ask you to identify the first useable IP address in the third range (assuming ip subnet zero) or something similar. In this case, we would get:

192.168.23.17

I'll post a Class A example in the next message.

Sorry....I had the Class A example nearly finished and then something happened and I lost the message I was working on!

If you need the Class A example, let me know and I can re-type it. Essentially, with the Class A, they generally ask you for a large number of subnets like 2000. When you're counting from left to right and doubling, you'll end up turning on all of the bits in the second octet but you still won't have enough subnets (128 x 2 = 256). So, continue to the next octet and keep doubling. You'd get:

256 x 2 = 512

512 x 2 = 1024

1024 x 2 = 2048 There! Now we've got enough for 2000 subnets.

So our new Class A mask would be: 255.255.224.0 because we turned on all of the bits in the second octet and the first three bits in the third octet.

I hope these examples make sense. Please let me know if you need any clarification!

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