IP address summarization

Unanswered Question
Dec 18th, 2008

Hello all,

just need an explanation why C & D are the correct answer for the following 2 questions please:

10. Given the addresses 10.1.8.0/24 and 10.1.9.0/24, which of the following is the best summary?

a. 10.0.0.0/8

b. 10.1.0.0/16

c. 10.1.8.0/23-correct answer

d. 10.1.10.0/24

11. Given the addresses 10.1.138.0/27, 10.1.138.64/26, and 10.1.138.32/27, which of the

following is the best summary?

a. 10.0.0.0/8

b. 10.1.0.0/16

c. 10.1.138.0/24

d. 10.1.138.0/25-correct answer

I have this problem too.
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Jon Marshall Fri, 12/19/2008 - 03:11

Mohammed

No 10

=====

/23 = 255.255.254.0

256 - 254 = 2

So with a /23 you have 2 x /24 networks.

10.1.8.0/24

10.1.9.0/24

No 11

=====

/27 = 255.255.255.224

/26 = 255.255.255.192

256 - 224 = 32

256 - 192 = 64

So

10.1.138.0/27 = 32

10.1.138.64/26 = 64

10.1.138.32/27 = 32

32 + 64 + 32 = 128

/25 = 255.255.255.128

256 - 128 = 128

If that is unclear then we can do it in binary if you want.

Jon

mhasabal Sat, 12/20/2008 - 13:54

John,

thank you so much for the reply.

i am trying to avoid doing it binary cause it takes a long time.

i do understand #11

but #10 i don't understand the part "So with a /23 you have 2 x /24 networks."

thanks again for your help

Jon Marshall Sun, 12/21/2008 - 05:16

Mohammed

10.1.8.0 255.255.255.0

10.1.9.0 255.255.255.0

Now you need to summarise these with just one entry. So a class C subnet mask of 255.255.255.0 will not work.

Which octet changes in the 2 subnets ie.

10 is the same for both so 10.

1 is the same for both so 10.1

the 3rd octet is the one that changes so we can derive a subnet mask so far of

255.255.

so at the moment we have

10.1 255.255

Now we know we cannot use 255 in the 3rd octet for the subnet mask because this would only allow 1 of the above subnets ie.

either 10.1.8 OR 10.1.9 but not both

If you wrote this out in binary it would become obvious but we need a value in the 3rd octet of the subnet mask that covers 2 class C networks

256 - 254 = 2

so we know that we can get 2 networks by making the 3rd octet 254. Again if we did this in binary it would be obvious but we are trying to give you an easier/quicker way to do it.

So the subnet mask now is 255.255.254. and this subnet mask covers 10.1.8 AND 10.1.9

Add a 0 at the end of the subnet mask and we end up with

10.1.8.0 255.255.254.0

A very quick bit of binary - you don't need to remember this it's just so you can understand how we get /23

128 64 32 16 8 4 2 1

if each of the 8 values had the binary bit on then they would all add up to 255. To get 254 we simply turn the last bit off ie. the value of 1 in the above is turned off. So the first 7 values above are on the last value is off.

Finally :-), using the subnet mask 255.255.254

255 = 8 bits on

255 = 8 bits on

254 = 7 bits on (last bit off)

8 + 8 + 7 = 23

Hope this makes sense.

Jon

subnetip1 Thu, 12/25/2008 - 14:31

Hi,

Let's say that for Example 11, /27 and /26 are 32 IPs and 64 IPs, and when you look at their network address, these 3 blocks belong to the same first 128 IPs, which means a /25. C is too big and others are much too big. Same story for Example 10.

You can test yourself at subnetip.com; it's free and there is no registration.

Good luck!

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