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+39 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 $B$?

- $245.248.136.0/21 \text{ and } 245.248.128.0/22$
- $245.248.128.0/21 \text{ and } 245.248.128.0/22$
- $245.248.132.0/22 \text{ and } 245.248.132.0/21$
- $245.248.136.0/24 \text{ and } 245.248.132.0/21$

+37 votes

0

I have doubt in option b where address are same which will lead to clash and hence we are not choosing option b?

But they do have different subnet won't it work?

But they do have different subnet won't it work?

0

Will the partition diagram in the solution will be opposite as per the answer. Please explain if not I'm confused.

0

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 !)

(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 !)

+33 votes

+2

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

(4)245.248.136.0/21 and 245.248.132.0/22

Option (A) is the correct answer.

(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

(4)245.248.136.0/21 and 245.248.132.0/22

Option (A) is the correct answer.

0

(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 !)

+18 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 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.

+1

**The ramaining half (other than B and retained by ISP) is the n/w 245.248.132.0/22. Am I right? **

If I am right then this also could be an answer because ISP and B both have equal quarters and so we can give them any of the two quarters. BTW there is no option including 245.248.136.0/21 and 245.248.132.0/22

+2

AS per given option range could be like this :-

B range could be **(128 to 131)** //means 128.0 to 131.255

A range could be **(132 to 138) ** // means 131.0 to 138.255

rest chuck range could be **(139 to 255)** // means 139.0 to 255.255

+1

To divide the n/w into two halves we choose 1 bit from host id part i.e the part A. and to divide the half into another half we choose 2 bit from host i/d part as subnet id and it coosen as 00 because of the option i think .if you choose 01 then 132,if 10 then 136 ,if 11 then 140.

0

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.

(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.

0

(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

(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

+14 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**

+13 votes

ans should be a) , the mask are /21 and/22 respectively, acc. to the half and quarter chunks requirement... and option b) can creat problem as address can get clashed between the two..

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