GATE CSE 2011 | Question: 53

Question

Consider a network with five nodes, $N1$ to $N5$, as shown as below.

The network uses a Distance Vector Routing protocol. Once the routes have been stabilized, the distance vectors at different nodes are as follows.

• N1: $(0, 1, 7, 8, 4)$
• N2: $(1, 0, 6, 7, 3)$
• N3: $(7, 6, 0, 2, 6)$
• N4: $(8, 7, 2, 0, 4)$
• N5: $(4, 3, 6, 4, 0)$

Each distance vector is the distance of the best known path at that instance to nodes, $N1$ to $N5$, where the distance to itself is $0$. Also, all links are symmetric and the cost is identical in both directions. In each round, all nodes exchange their distance vectors with their respective neighbors. Then all nodes update their distance vectors. In between two rounds, any change in cost of a link will cause the two incident nodes to change only that entry in their distance vectors.

The cost of link $N2-N3$ reduces to $2$ (in both directions). After the next round of updates, the link $N1-N2$ goes down. $N2$ will reflect this change immediately in its distance vector as cost, $\infty$. After the NEXT ROUND of update, what will be the cost to $N1$ in the distance vector of $N3$ ?

• A. $3$
• B. $9$
• C. $10$ 
• D. $\infty$

Comments

"In each round, all nodes exchange their distance vectors(OLD) with their respective neighbors. Then all nodes update their distance vectors."

---

for people who are getting confused by the wording of the question, refer the original question:

 

 

---

Problem with this question is that using words like "after next round of updates"

Which confuses that what round they are referring in both questions. As each round determines what table is shared between nodes.

Answer 1

Answer is (c) 10

Answer 2

one needs to carefully read the first line of the second question.

" After the update in the previous question.

and also below line of common data for both ques:

"any change in cost of a link will cause the two incident nodes to change in only that entry their distance vectors."

 

It is this update one must refer to

i.e after the update in prev question means only the update of changing N2 ->N3 cost to 2 in N2 and N3->N2 to 2 in N3.

and not consider the tables after the 'next round of updates'

Next round of updates is used only to compute the answer of first question and thats it.

Hence N4 will exchange its old table (8,7,2,0,4) with N3 and hence N3 will update N3->N1 to min(8+2=10,2+infinity)