GATE CSE 2012 | Question: 34, ISRO-DEC2017-32

Question

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 $B$?

• A. $245.248.136.0/21 \text{ and } 245.248.128.0/22$
• B. $245.248.128.0/21 \text{ and }  245.248.128.0/22$
• C. $245.248.132.0/22 \text{ and }  245.248.132.0/21$
• D. $245.248.136.0/24 \text{ and }  245.248.132.0/21$

Comments

ip is starts from 245.248.128.0/20 not 254.248.128.0

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Awesome Question :-)

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This is solved by eliminating options and I understood it, is it also possible and safe to assume that if we use another approach (using VLSM on the available address) It depends upon question setters how they approached it?

Either they went on subnetting from

A – B – ISP or B – A – ISP or B – ISP – A

So we have to figure out which approach they used? (Although this is not efficient)

Answer 1

 

Correct option will be A

Comments

But Since in Subnetting we require atleast two bits to divide it into 2 subnets so we can't do subnetting here at the first place ! As the bits we have is (12-11) = 1. And with 1 bit we cant do subnetting unlike most of us have done in their answers ! , so i think the answer should be subnetting not possible. Right ?

(The logic is we cant use 0,00,000 and 1,11,111 etc. for subnetting as all zeros will conflict with the original network address and all ones will conflict with the broadcast address of original network !)

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I totally get the answer but is there any kind of options elemenating method.. because its kind of lengthy method

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The ip address is 245.xxx.xxx.xxx  but you written 254.xxx.xxx.xxx.

Also in half partition the range must be from 245.248.136.0 -  245.248.143.255.

Seems like typing error ⚠️ .

Answer 2

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

Comments

But Since in Subnetting we require atleast two bits to divide it into 2 subnets so we can't do subnetting here at the first place ! As the bits we have is (12-11) = 1. And with 1 bit we cant do subnetting unlike most of us have done in their answers ! , so i think the answer should be subnetting not possible. Right ?

(The logic is we cant use 0,00,000 and 1,11,111 etc. for subnetting as all zeros will conflict with the original network address and all ones will conflict with the broadcast address of original network !)

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Thank you krish

Very nice explaination

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${\color{Red} Best\ Solution}$

Answer 3

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

Answer 4

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

now for B set 12^th bit and 13^th 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. 

Comments

There are 4 possible ways.

(1)245.248.128.0/21 and 245.248.136.0/22

(2)245.248.128.0/21 and 245.248.140.0/22

(3)245.248.136.0/21 and 245.248.128.0/22

(2)245.248.136.0/21 and 245.248.140.0/22

Option (A) is the correct answer.

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@Warrior please explain it more .

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(1)245.248.128.0/21 and 245.248.136.0/22    ------------(1)

(2)245.248.128.0/21 and 245.248.140.0/22  --------------(2)

(3)245.248.136.0/21 and 245.248.128.0/22   ---------------(3)

(2)245.248.136.0/21 and 245.248.140.0/22  --------------(4)

Option (A) is the correct answer because if you make see split (1) and (2) both B address i.e 245.248.136.0/22 & 245.248.140.0/22 are not matching any option as B should have 245.248.128.0/22. So only way to do this as per options given is 245.248.136.0/21 and 245.248.128.0/22