GATE | Gate IT 2005 | Question 91

Q85 Part_B Consider a simple graph with unit edge costs. Each node in the graph represents a router. Each node maintains a routing table indicating the next hop router to be used to relay a packet to its destination and the cost of the path to the destination through that router. Initially, the routing table is empty. The routing table is synchronously updated as follows. In each updation interval, three tasks are performed.

A node determines whether its neighbours in the graph are accessible. If so, it sets the tentative cost to each accessible neighbour as 1. Otherwise, the cost is set to ∞.
From each accessible neighbour, it gets the costs to relay to other nodes via that neighbour (as the next hop).
Each node updates its routing table based on the information received in the previous two steps by choosing the minimum cost.

For the graph given above, possible routing tables for various nodes after they have stabilized, are shown in the following options. Identify the correct table.

Table for node A

A	-	-
B	B	1
C	C	1
D	B	3
E	C	3
F	C	4

Table for node C

A	A	1
B	B	1
C	-	-
D	D	1
E	E	1
F	E	3

Table for node B

A	A	1
B	-	-
C	C	1
D	D	1
E	C	2
F	D	2

Table for node D

A	B	3
B	B	1
C	C	1
D	-	-
E	E	1
F	F	1

Continuing from the earlier problem, suppose at some time t, when the costs have stabilized, node A goes down. The cost from node F to node A at time (t + 100) is

(A)

>100 but finite

(B)

∞

(C)

(D)

>3 and ≤100

Answer

Please comment below if you find anything wrong in the above post
Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Last Updated :

Share your thoughts in the comments

GeeksforGeeks

Trending in News