Computer Networks: CIDR based question
 

Hey,
I hope, I'm allowed to ask this question. First thing I want to tell you that it's not a homework question. Actually I appeared in an exam and I tried to solve this question but ...
Q-An ISP has the following chunk of CIDR-based IP address available with it: 245.248.128.0/20. The ISP wants to give half of this chunk of addresses to Organization A and a quarter to Organization B, while remaining with itself. Which of the following is a valid allocation of addresses to A and B?
A) 245.248.136.0/21 and 245.248.128.0/22
B) 245.248.128.0/21 and 245.248.128.0/22
A) 245.248.132.0/22 and 245.248.132.0/21
A) 245.248.136.0/24 and 245.248.132.0/21

My approach: Only in option B, bitwise logical AND operation on IP addresses with their subnet gives the same address. So I think it's right option but I'm not sure.I don't know what is the right approach to solve this question.Any suggestions will be appreciated.
TIA!

No, it's A. (fingers crossed!!) In B, the subnets overlap, so it's illegal. You clearly need a /21 for "half" and a /22 for "quarter". In A the TOP half of the /20 goes to Org A, and the very bottom quarter goes to Org B, with the middle bit kept behind.

I concur, half of a /20 is /21, a quarter is /22 (which I'm sure is just a typo in chris's post) so D is out.
The networks in B overlap. The networks in C overlap.
So A is the answer.

To double check, /20 gives us a block of 16 class C's = 245.248.128.0 > 245.248.143.0
245.248.128.0/21 is an 8 block which would give 245.248.128.0 > 245.248.135.0
The remainder is 245.248.136.0/21 and if we split this with a /22 we get two blocks of 4 class C's which are 245.248.136.0/22 and 245.248.140.0/22

So giving away 245.248.128.0/21 and 245.248.136.0/22 leaves 245.248.140.0/22.

Looks good.

Typo? What typo?!

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