31,309 views
67 votes
67 votes

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 $A$ and $B$?

  1. $245.248.136.0/21 \text{ and } 245.248.128.0/22$
  2. $245.248.128.0/21 \text{ and }  245.248.128.0/22$
  3. $245.248.132.0/22 \text{ and }  245.248.132.0/21$
  4. $245.248.136.0/24 \text{ and }  245.248.132.0/21$

9 Answers

Best answer
58 votes
58 votes

 

Correct option will be A

edited by
71 votes
71 votes

The different subnet splittings possible are shown above. So, the answer would be $(A)$

edited by
22 votes
22 votes

The prefix is given is 20. So, the first 20 bits denote the network id bits, and the next 12 bits indicate host bits. To divide the network into two halves, we borrow bits from the host bits. Hence, for half (2 partitions), the prefix will be 20 + 1 and for 4 partitions, the prefix will be 20 + 2 = 22. So prefix for A will be 21 and prefix for B will be 22. This eliminates options C and D.

For A, we need to consider the 21st bit. It can have values 0 or 1. Let's take its value as 1. So, the network id becomes 245.248.136.0/21.

For B, we need to consider the 21st and 22nd bits. We can have four possibilities: 00, 01, 10, 11. Out of these, 10, and 11 belong to network A. So, we take 00 ie 245.248.128.0/22 (Network B) and 01 as 245.248.132.0/22.(Rest)

Option A

20 votes
20 votes

20 bit net id  so 12 bit host id  

Address allocation to A is $2^{12}/2 = 2^{11}$

Address allocated to B is $2^{12}/4 = 2^{10}$

set  12th bit to 1 so address for A is 245.248.136.0/21

now for B set 12th bit and 13th bit to 0 so address for B is 245.248.128.0/22 as in option A. 

Option B is wrong as host addresses are overlapping. 

Answer:

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