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20 votes
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Consider routing table of an organization’s router shown below:

$$\begin{array} {|l|l|l|} \hline \text{Subnet Number} & \text{Subnet Mask} & \text{Next Hop} \\\hline 12.20.164.0 & 255.255.252.0 & \text{R1} \\\hline  12.20.170.0 & 255.255.254.0 & \text{R2} \\\hline 12.20.168.0 & 255.255.254.0 & \text{Interface 0} \\\hline 12.20.166.0 & 255.255.254.0 & \text{Interface 1} \\\hline \text{default} & & \text{R3} \\\hline \end{array}$$

Which of the following prefixes in $\text{CIDR}$ notation can be collectively used to correctly aggregate all of the subnets in the routing table?

  1. $12.20.164.0/20$
  2. $12.20.164.0/22$
  3. $12.20.164.0/21$
  4. $12.20.168.0/22$
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@Arjun Sir

please correct the answer “A” to “B, D” on this page. 

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Can anyone say Why C And D can't collectively used here!? I think B and C are generating same sort of lowest and highest address.
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5 Answers

24 votes
24 votes

 GATE official initial key has given B,D as answers. 

I strongly feel this question should be challenged to give marks for all because this question requires certain assumptions.

In this answer, I have explained the reason why options B and D are correct and the required assumptions for their correctness.

I have prepared a draft to submit. Please find it here – here

 

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Please watch this short video if you want to understand suppernetting. 
Video explanation for route aggregation – Route aggregation or Supernetting | Easiest explanation

Four networks are given in question as follows –

$\textbf{Network1: } 12.20.164.0/22  $

$\textbf{Network2: } 12.20.170.0/23  $

$\textbf{Network3: } 12.20.168.0/23 $

$\textbf{Network4: } 12.20.166.0/23 $

 

 

Four options are also given, following is range of options along with range of networks –

Option A covers everything (all networks) but it also covers lot more extra IPs.
Option C and D  collectively aggregate everything (all networks) but  also covers few more extra IPs.

Option B and D  collectively aggregate everything (all networks) without covering extra IPs.

Hence B and D should be answer.

Still, there is a possibility of marks to all
Explaining in short (Please read the above draft for full detail)-

Two reasons – 

  1. Whats wrong with option A or (C,D) ? – they are just covering extra IPs but it is given in kurose that we can cover extra IPs if these IPs are unallocated. (Please see above draft where kurose screenshots are attached.)
  2. Even If we agree that B and D are more precise in terms of not covering extra IPs still option B and D requires an assumption about topology of network (given below)

Now we explain the assumption about topology of network to make  option B, D  correct. – 

 

Option D combines network 2 and 3. But you can combine only when Next Hop is same for both networks.

$$\begin{array} {|l|l|l|} \hline
\text{Subnet Number} & \text{Subnet Mask} & \text{Next Hop} \\\hline
12.20.164.0 & 255.255.252.0 & \text{R1} \\\hline 
12.20.170.0 & 255.255.254.0 & \color{green}{\text{R2}} \\\hline
12.20.168.0 & 255.255.254.0 & \color{green}{\text{Interface 0}} \\\hline
12.20.166.0 & 255.255.254.0 & \text{Interface 1} \\\hline
\text{default} & & \text{R3} \\\hline \end{array}$$

THE VERY FIRST CONDITION IN AGREGGRATION IS TO HAVE SAME NEXT HOP. OTHERWISE AGREGGRATION IS NOT POSSIBLE.

$\color{green}{\text{R2}}$ and $\color{green}{\text{Interface 0}} $ should be same. Which means following topology – 

If network 2 and 3 are on same side of router R then only we can combine them in routing table of R.

What if networks 2 and 3 are connected to different interfaces?

In this case we can not even combine network 2 and 3 hence none of the option should be correct.

First Figure : R2 = Interface 0

Second Figure : R2 ≠ Interface 0

Both Fiigures are very much possible.

Option B and D are correct subject to two assumptions - 

  1. We should have topology given in first diagram
  2. We are not allowed to cover extra IPs. 

 

edited by

4 Comments

So you think b,d has to be correct right ?, I strongly feel that it has to be they won't change the keys and especially gives MTA, I dont know why there is so much buzz about this, Also in the document they donot followed the rules of aggregation, I mean why they are being given as the rules if we donot have to follow them. Also option a cannot be correct ever because it is one of the host id's it can never be the subnet id if it will be .160 at last instead of 164, then It will be correct, because then we can have .160 as one of the subnet id, we cannot make .164 in last part by only /20. But idk why they complicated it so much..It is a simple aggregation based question just given in table format and some mixed language, according to the rules b and d will be correct and also I dont see them changing the keys, Also your link says the same about it.

Also the mst question hate to say it, I am gonna lose marks but they won't change it.. 

@aspirant22

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@N.S. Same here.

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edited by

@Sachin Mittal 1 Sir, I understood all the explanation and ambiguities, but I have one question, In Option (A) 12.20.164.0 /20   How could they write such combination? Because if we apply mask on NID itself, then it will give 12.20.160.0, Same happening with option C. Please explain this & correct me if I’m wrong.

 

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19 votes
19 votes

12.20.164.0/22  → 12.20.10100100.00000000

12.20.170.0/23  → 12.20.10101010.00000000

12.20.168.0/23  → 12.20.10101000.00000000

12.20.166.0/23  → 12.20.10100110.00000000

 

12.20.170.0 and 12.20.168.0 can be combined to 12.20.10101000.0 (12.20.168.0/22).

12.20.166.0/23 comes inside 12.20.164.0/22. So, 12.20.164.0/22 can be used for route aggregation of both.

12.20.168.0/22 and 12.20.164.0/22 can’t be combined further.

So, answer is option B and D.


Question is asking which prefixes can be collectively used to correctly aggregate all of the subnets in the routing table. Not which options independently does it.

Also, Option A contains IP address 12.20.175.1 which doesn’t belong to the specified organization.

edited by

1 comment

Can you please explain, why C and D cannot be right. C contains 1st and last subnets, just like B. So answer can be C and D.

I know I am wrong. Please telll me where am I making a mistake.
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5 votes
5 votes
A is the answer by combining the last common bits by the principle of supernetting.

1 comment

It is not, a can never be correct, How can you make .164 in 3 octet by just /20. It is one of the host id's it can never be a subnet id, Also the routing table stores the subnet or network id's.

You are blindly applying the rules, but it is not the case you have to think carefully in this question..
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0 votes
0 votes

i think this is right .

Answer:

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