Consider a network with five nodes, N1 to N5, as shown as below. The network…
2011
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.
he 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, ∞. 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.
∞
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Correct answer: C
After the N2 − N3 cost drops and one update round completes, N3 has (3,2,0,2,5). When the link N1 − N2 then goes down, N2 immediately sets its distance to N1 to ∞. In the next round N3 recomputes its distances using its neighbours N2 and N4; the best path to N1N1N1 becomes via N4N4N4 with cost 2+8=10.
opology & direct link costs (from figure):
N1 − N2=1, N2 − 5=3, N5 − 4=4, N4 − 3=2, N3 − 2=6(then N2 − N3 is reduced to 2 by the event).
Given initial DV (before any change):
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)
Event 1: cost N2 − N3 reduces to 2. Between rounds only N2 and N3 change that entry:
N2 → (1,0,2,7,3)
N3 → (7,2,0,2,6)
Next round (all nodes exchange & update): using neighbours’ vectors, N3 updates to
(3,2,0,2,5) (paths considered via N2 and N4). Simultaneously other nodes update too.Event 2 (after that round): link N1 − N2 goes down. N2 immediately sets distance to N1 = ∞. So at start of the next round:
N2 is (∞,0,2,4,3) (only N2’s entry for N1 changed to ∞)
N4 remains (8,4,2,0,4) (from previous round)
Next round (we are asked after this round): N3N3N3 recalculates distances using neighbours N2 and N4:
via N2: cost = 2+N2[N1]=2+∞=∞
via N4: cost = 2+N4[N1]=2+8=10
So best cost to N1 from N3 = 10.
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