10-15-2008 09:05 AM - edited 03-03-2019 11:56 PM
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
Solved! Go to Solution.
10-15-2008 01:27 PM
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
10-15-2008 09:25 AM
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
10-15-2008 10:18 AM
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
10-15-2008 10:32 AM
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
10-15-2008 12:28 PM
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
10-15-2008 12:41 PM
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
10-15-2008 12:59 PM
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
10-15-2008 09:40 AM
hi
check :http://www.cisco.com/web/about/ac123/ac147/archived_issues/ipj_9-1/ip_addresses.html
hope this will help
10-15-2008 01:27 PM
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
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