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

The net work uses a Distance Vector Routing protocol. Once the routes have stabilized, the distance vectors at different nodes are as following 

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.

1. The cost of link N2-N3 reduces to 2 in (both directions). After the next round of updates, what will be the new distance vector at node, N3

(A) (3. 2, 0, 2, 5) 

(B) (3, 2, 0, 2, 6) 

(C) (7, 2, 0, 2, 5) 

(D) (7, 2, 0, 2, 6)

2. After the update in the previous question, the link N1-N2 goes down. N2 will reflect this change immediately in its distance vector as cost, ∞ . 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) ∞

1 Answer

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Best answer

1. Correct option is (A) (3. 2, 0, 2, 5)

2. Correct option is (C) 10

N3 has neighbors N2 and N4 

N2 has made entry ∞ 

N4 has the distance of 8 to N1 

N3 has the distance of 2 to N4 

So 2 + 8 = 10

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