An Internet Service Provider(ISP) has the following chunk of CIDR-based IP addresses available with it:245.248.128.0/20. The ISP wants to give half of this chunk of addreses to Organization A, and a quarter to Organization B, while retaining the remaining

An Internet Service Provider(ISP) has the following chunk of CIDR-based IP addresses available with it:245.248.128.0/20. The ISP wants to give half of this chunk of addreses to Organization A, and a quarter to Organization B, while retaining the 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
(C) 245.248.132.0/22 and 245.248.132.0/21
(D) 245.248.136.0/22 and 245.248.132.0/21

 

5Comments
karmjit joshi @karmjitjoshi
12 Jan 2017 01:42 pm

?

 

Sumit Verma @sumitkgp
12 Jan 2017 04:27 pm

Correct option will be A. The below partition of addresses will lead to the correct option.
We have block of ip addresses as: 245.248.128.0/20
=11110101.11111000.10000000.00000000/20 
NID is 20 bits, so we can devide the network into two parts by taking 21st bit into NID.
So we have two block now:
P. 11110101.11111000.10001000.00000000/21 = 245.248.136.0/21
Q. 11110101.11111000.10000000.00000000/21 = 245.248.128.0/21
Take P for Organization A.
Now let's devide Q into two parts again. For that we have to inculde one more bit, i.e, 22nd bit into NID.
Q1: 11110101.11111000.10001000.00000000/22 = 245.248.132.0/22
Q2: 11110101.11111000.10000000.00000000/22 = 245.248.128.0/22
Q2 can be used for organization B.

karmjit joshi @karmjitjoshi
12 Jan 2017 02:28 pm

Thx sir

Vivek Justin Manukonda @vivekjustinmanu
17 Jan 2019 05:31 pm
correction---> Q1: 11110101.11111000.10000100.00000000/22 = 245.248.132.0/22
mmham @menakhaled
9 Jan 2018 08:30 pm

I guess there is something wrong in Q1, in the third octet 

it should be 10000100 to get the NID 245.248.132.0/22