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Need help with Summarization

newnetman
Level 1
Level 1

Hello All,

There was a need to summarize the following addresses

3.3.8.3 - 3.3.11.3/24

and also

4.4.12.4 - 4.4.15.4/24.

My answers to the two summaries were 3.3.4.0/22 and 4.4.4.0/22

But unfortunately the book says the answer is 3.3.8.0 and 4.4.12.0/22

Will greatly appreciate thorogh explanation as to why this was so.

Thank you all

1 Accepted Solution

Accepted Solutions

Think of it this way:

3.3.8.0/24

3.3.9.0/24

3.3.10.0/24

3.3.11.0/24

How many subnets do you have? 4

How many times to the power of 2 can I use to get 4? 2

Subtract that 2 from 24 = 22

Take the first subnet and add the 22 as your new mask

3.3.8.0/22

Another example would be:

172.16.0.0/24

172.17.0.0/24

172.18.0.0/24

172.19.0.0/24

172.20.0.0/24

172.21.0.0/24

172.22.0.0/24

172.23.0.0/24

How many subnets? 8

How many powers of 2? 3

Subtract 3 from mask: 24-3 21

Answer: 172.16.0.0/21

It does get weirder when you cross boundaries. For instance:

Here's a good document for you to look at:

http://subnettingmadeeasy.blogspot.com/2007/11/router-summarization.html

HTH

--John

HTH, John *** Please rate all useful posts ***

View solution in original post

8 Replies 8

Jon Marshall
Hall of Fame
Hall of Fame

You are trying to summarise 4 networks in each case ie.

3.3.8

3.3.9

3.3.10

3.3.11

/22 = 255.255.252.0

252 gives you 4 networks ie.

256 - 252 = 4

so 3.3.4.0/22 =

3.3.4

3.3.5

3.3.6

3.3.7

All the above applies to 4.4.12 - 4.4.15

Does this makes sense ?

Jon

Hi Jon,

I'm grateful for the reply but just wished your response could address my query more precisely?. All i want to know is if i am correct or the book author is, and if so with explanation as to how

3.3.8.3/24

3.3.9.3/24

3.3.10.3/24

3.3.11.3/24

could become 3.3.8.0/22

And how

4.4.12.4/24

4.4.13.4/24

4.4.14.4/24

4.4.15.4/24

could be become 4.4.12.0/22?

Hence my answer was 3.3.4.0/22 and 4.4.4.0/22 respectfully.

Any further clarification shall be appreciated.

Regards

Oh sorry, i thought i had explained, my fault.

The book is correct.

Your answer is

3.3.4.0 255.255.252.0

255.255 match the 3.3 and they won't change.

4 matches the 252 and this will change.

252 means you can have 4 different values ie.

256 - 252 = 4

So starting with

3.3.4.0 that is 1

3.3.5.0 that is 2

3.3.6.0 that is 3

3.3.7.0 that is 4

so your answer 3.3.4.0/22 covers the networks

3.3.4.0/24

3.3.5.0/24

3.3.6.0/24

3.3.7.0/24

Now you know with 252 you can have 4 values so

3.3.8.0 is 1

3.3.9.0 is 2

3.3.10.0 is 3

3.3.11.0 is 4

which is the exact range the question asked for. So the book is correct.

Jon

Hello Jon,

Most grateful again for your time to explain. Nevertheless, if you understood my last query that means i am correct while the book is wrong?.

Here again is what i came up with

"Hence my answer was 3.3.4.0/22 and 4.4.4.0/22"

Which is exactly the same as yours.

Thanks

Sorry, i'm making a real mess of explaining this - perhaps someone else could step in ?

Note i am dealing with 3.3.x.x networks but it applies to 4.4.x.x networks as well.

Your original query was

There was a need to summarize the following addresses

3.3.8.3 - 3.3.11.3/24

Now i'm assuming you mean you need to summarize

3.3.8.0/24

3.3.9.0/24

3.3.10.0/24

3.3.11.0/24

Now my previous posts were trying to show that your answer

3.3.4.0/22 covers the following networks

3.3.4.0/24

3.3.5.0/24

3.3.6.0/24

3.3.7.0/24

which is not the networks you are asked to summarise. So the actual answer is 3.3.8.0/22 which covers

3.3.8.0/24

3.3.9.0/24

3.3.10.0/24

3.3.11.0/24

It could be me, am i missing something obvious ?

Jon

Compare your answer with a typical /30 we often use for a serial connection by dividing a class C network: for example 3.3.3.8 / 30.

When you use this network, you have a network address (3.3.3.8), two usable addresses (.9 and .10) and a broadcast address (.11). In this example you have divided the range of available addresses (256) into 64 smaller networks with each 4 addresses by using 6 bits of the fourth octet to define new networks (hence /30 (24+6).

Your case at hand does the same but for the third octet. You divide 3.3.0.0 /16 into 64 networks which include each 4 networks. You use 6 bits of the third octet (hence /22 (16+6) to create 64 networks with the following network addresses: 3.3.0.0, 3.3.4.0,3.3.8.0, 3.3.12.0 (...), 3.3.252.0

Your solution uses .4 as network address for the new subnet, which would mean (since you have a /22 mask) your network includes 3.3.4.0, 3.3.5.0, 3.3.6.0 and 3.3.7.0. Since you need to summarize 8 thru 11, your first network must be .8

The reference above is excellent, I also liked

http://www.coxpc.com/content/3com_Chuck_Semeria.htm, for the included tasks.

HTH, Thomas

Think of it this way:

3.3.8.0/24

3.3.9.0/24

3.3.10.0/24

3.3.11.0/24

How many subnets do you have? 4

How many times to the power of 2 can I use to get 4? 2

Subtract that 2 from 24 = 22

Take the first subnet and add the 22 as your new mask

3.3.8.0/22

Another example would be:

172.16.0.0/24

172.17.0.0/24

172.18.0.0/24

172.19.0.0/24

172.20.0.0/24

172.21.0.0/24

172.22.0.0/24

172.23.0.0/24

How many subnets? 8

How many powers of 2? 3

Subtract 3 from mask: 24-3 21

Answer: 172.16.0.0/21

It does get weirder when you cross boundaries. For instance:

Here's a good document for you to look at:

http://subnettingmadeeasy.blogspot.com/2007/11/router-summarization.html

HTH

--John

HTH, John *** Please rate all useful posts ***
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