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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 addresses 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 to B?
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- 245.248. 136.0/21 and 245.248. 128.0/022
- 245.248. 128.0/21 and 245.248. 128.0/22
- 245.248. 132.0/22 and 245.248. 132.0/21
- 245.248. 136.0/24 and 245.248. 132.0/21
- 245.248. 136.0/21 and 245.248. 128.0/022
Correct Option: A
Given, IP addresses 245.248.128.0/20 = 11111111.11111111.11110000.00000000
Now, total number of hosts we have = 212
Half of the host we have to give to organization A that means 212 hosts we have to give.
So, 12th bit from right or 1st bit of host ID of given network have to be done either 0 or 1.
NID = 11110101.11111000.10000000.00000000
So, for B network ID will be
So, bit number 22 could be 0 or 1.
⇒ 245.248.136.0/ 22 or ⇒ 245.248.140.0/ 22
So, none of them are in the option for organization B.
Case (A) : If we fix the 21st bit as 0.
245.248.128.0/ 21
For remaining address we have to put 21st bit as 1 because 0 is used in organization A. So, remaining address.
For organization B, we need half addresses so but of 11 bit, we need 10 bit for organization B.
From case A Answers would be
A : 245-248.128.0/2
B : 245.248.136.0/ 22
or Not present in options
B : 245.248.140.0/ 22
Case B : For organization A in bit 21st, now take
⇒ 245.248.136.0/ 22
So, 1 in 21st position is used now for remaining we need to take 0 in 21st position.
Out of this add half of them we need to give to organization B. So, in position 22nd, it could be 0 or 1.
⇒ 245.248.128.0/ 22
or 
⇒ 245.248.132.0/ 22
In Case B we get
A : 245-248.136.0/21
B : 245.248.128.0/ 22
or Anyone is correct
B : 245.248.132.0/ 22
