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+27 votes
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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 $A$ 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. $245.248.136.0/24 \text{ and }  245.248.132.0/21$
asked in Computer Networks by Boss (18.3k points) | 5.2k views
0
ip is starts from 245.248.128.0/20 not 254.248.128.0

8 Answers

+30 votes
Best answer

Correct option will be A

answered by (255 points)
edited by
0
Yes its a correct division . Even i am geeting the same :D
+5

Range of A will be till 245.248.143.255/21. Seems like typing error.

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?
+19 votes

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

answered by Active (4.8k points)
edited by
+1
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.
0
Thanks for pointing out. I mistyped. I'll correct it ;)
0

(2)245.248.136.0/21 and 245.248.140.0/22  

Shouldn't this be 245.248.136.0/21 and  245.248.132.0/22

0
Yup...u r right.

Corrected now.
+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..
answered by Active (1.1k points)
+13 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. 

answered by Boss (31.9k points)
+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 

0

@pooja mam 
why we  set 12th bit and 13th bit to  0  for B  ?

+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.
0
@Warrior please explain it more .
+10 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

answered by Active (1.5k points)
+3 votes

If take ip :245.248.128.0/21 then B have choice : 245.248.136.0/22 or 245.248.140.0/22

if A ip : 245.248.136.0/21 then B have choice : 245.248.128.0/22 or 245.248.132.0/22

 

So, correct answer A

answered by (367 points)
+2 votes

Answer is A

Duplicate of GATE 2012 -34

https://gateoverflow.in/1752/gate2012_34

answered by Boss (35k points)
0 votes

Ans is A

answered by Junior (507 points)
edited by
Answer:

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