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+24 votes
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Consider a network with $6$ routers $R1$ to $R6$ connected with links having weights as shown in the following diagram.


Suppose the weights of all unused links are changed to $2$ and the distance vector algorithm is used again until all routing tables stabilize. How many links will now remain unused?

  1. $0$
  2. $1$
  3. $2$
  4. $3$

 

asked in Computer Networks by Veteran (111k points)
edited by | 2.6k views

6 Answers

+28 votes
Best answer

First we need to find which are the unused links in the graph
For that we need not make distance vector tables,
We can do this by simply looking into the graph or else DVT can also give the answer.
So, $R1-R2$ and $R4-R6$ will remain unused.

Now If We changed the unused links to value $2$.
$R5-R6$ will Now remain unused.

So, the correct answer is option B).

answered by Active (4.5k points)
edited by
+3
I think no link will remain unused as routing tables of r6 and r5 will not be updated for 2+1=3 because their distance already 3.
0
Yes the answer should be (a)
0
@bad_engineer

when cost of two paths are same then we consider hop count which is less for R5-R6.So, there are no unused links right?
+18 votes

Only one link is not used

answered by Boss (22.4k points)
+6 votes

The links R1-R2  and R4-R6 will never be used for data transfer because there are shorter paths available in any case.
If those two link weights are changed to 2, now only one link ie R5-R6 will never be used.

answered by Junior (551 points)
+6 votes
Option-B

SImple reason for this.Use your intuition like below way

Find all the shortest path from each node to other and mark your visited edge .Then automatically in first searching u will find that only two edge are unused example edge R1---->R2 as well as R4---->R6.

When u change it into 2,Then again apply your intuition to find out shortest path from each node to other in this satuation u again find on link is unused.
answered by Loyal (9.2k points)
0 votes

Ans is B 

now

answered by Junior (507 points)
edited by
0 votes
As DVR is same as Dijkstra algorithm, so simply make Minimum cost spanning tree and then first confirm it by manually traversing for each source to Destination to confirm that it's same as that we'll get using SSSP, as Greedy need not be Optimal, so after it just the ones not in mst will be unused.
answered by (127 points)
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