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Consider Dijkstra's algorithm in the link state routing protocol at node u,professor Ram first sets the route for each directly connected node v to be the link connecting u to v.Ram then implements the rest of the algorithm correctly,aiming to produce a minimum-cost routes,but does not change the routes to the directly connected nodes.In this network, u has atleast two directly connected nodes and there is more than one path between any two nodes.Assume that all link costs are non-negative.Which of the following statements is false of u's routing table?

A)There are topologies and link costs where the majority of the routes to other nodes will be incorrect.

B)There are topologies and link costs where no routing table entry (other than from u to itself) will be correct.

C)There are topologies and link costs where all routing table entry (other than from u to itself) will be correct.

D)Both (A) and (B)
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professor Ram first sets the route for each directly connected node

This doesn't change any thing i guess because we are getting the distance directly without any effort (like sending hello packets) . So option c??

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ans given option --B

+1 vote

@srestha MA'AM,

A: True

route nodes are incorrect for the directly connected nodes, bz even if other path of less weight  exist we are not considering, this makes the wrong entry.

B: False

some entries and topology are correct for the vertices which is not directly connected from u and conected to each other with less weight, for neighbour vertices it will have correct routing entry

C: I know leaving the path (U-> directly conneted )will be correct but unable to understand this phrase in this sentence

"other than from u to itself", please guide

this is really good question

A)There are topologies and link costs where the majority of the routes to other nodes will be incorrect. True

Here professor connected u to v1 & v2 & v3  as they are directly connected to u.

but what if there is edge between v1 and v2 having weight 1

also what if there is edge between v1 and v3 having weight 1

then above topology will be wrong.

B)There are topologies and link costs where no routing table entry (other than from u to itself) will be correct. False

professor Ram connecting u to all its connected vertices(suppose there are n such vertices ). And at later stage even if we discovered that out of n connected vertices of u , n-1 have better path. but still there will be 1 vertices which is  correctly connected to u. Otherwise u will not be connected at all

Lets take above figure only

in that we discovered better path from u to v2 & from u to v2. But still entry u to v1 has not change.

Hence there will be atleast one correct entry.

C)There are topologies and link costs where all routing table entry (other than from u to itself) will be correct. True

Consider above graph only and assume that apart from edges shown in figure no other edge exist.