Consider a network with five nodes, N1 to N5, as shown below.
The network used 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.
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 no N! in the distance vector of N3?
Consider a network with five nodes, N1 to N5, as shown below.
The network used 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.
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 no N! in the distance vector of N3?
The correct option is B 10
In the next round, N3 will receive distance form N2 to N1 as infinite. It will receive distance form N4 to N1 as 8.
So it will update distance to N1 as 8+2=10.
In the next round, N3 will receive distance form N2 to N1 as infinite. It will receive distance form N4 to N1 as 8.
So it will update distance to N1 as 8+2=10.
