Following Questions have been asked in GATE 2010 CS exam.
1) One of the header fields in an IP datagram is the Time to Live (TTL) field. Which of the following statements best explains the need for this field?
(A) It can be used to priortize packets
(B) It can be used to reduce delays
(C) It can be used to optimize throughput
(D) It can be used to prevent packet looping
Time to Live can be thought as an upper bound on the time that an IP datagram can exist in the network. The purpose of the TTL field is to avoid a situation in which an undeliverable datagram keeps circulating.
2) Suppose computers A and B have IP addresses 10.105.1.113 and 10.105.1.91 respectively and they both use the same netmask N. Which of the values of N given below should not be used if A and B should belong to the same network?
The last octets of IP addresses of A and B are 113 (01110001) and 91 (01011011). The netmask in option (D) has first three bits set in last octet. If netmask has first 3 bits set, then these bits nmust be same in A and B, but that is not the case. In simple words, we can say option (D) is not a valid netmask because doing binary ‘&’ of it with addresses of A and B doesn’t give the same network address. It must be same address as A and B are on same network. See this for more details.
3) Consider a network with 6 routers R1 to R6 connected with links having weights as shown in the following diagram
All the routers use the distance vector based routing algorithm to update their routing tables. Each router starts with its routing table initialized to contain an entry for each neighbour with the weight of the respective connecting link. After all the routing tables stabilize, how many links in the network will never be used for carrying any data?
We can check one by one all shortest distances. When we check for all shortest distances for Ri we don’t need to check its distances to R0 to Ri-1 because the network graph is undirected.
Following will be distance vectors of all nodes.
Shortest Distances from R1 to R2, R3, R4, R5 and R6
R1 (5, 3, 12, 12, 16)
Links used: R1-R3, R3-R2, R2-R4, R3-R5, R5-R6
Shortest Distances from R2 to R3, R4, R5 and R6
R2 (2, 7, 8, 12)
Links used: R2-R3, R2-R4, R4-R5, R5-R6
Shortest Distances from R3 to R4, R5 and R6
R3 (9, 9, 13)
Links used: R3-R2, R2-R4, R3-R5, R5-R6
Shortest Distances from R4 to R5 and R6
R4 (1, 5)
Links used: R4-R5, R5-R6
Shortest Distance from R5 to R6
Links Used: R5-R6
If we mark, all the used links one by one, we can see that following links are never used.
4) Suppose the weights of all unused links in the previous question are changed to 2 and the distance vector algorithm is used again until all routing tables stabilize. How many links will now remain unused?
Following will be distance vectors of all nodes
R1 (2, 3, 9, 10, 11)
Links used: R1-R2, R1-R3, R2-R4, R4-R5, R4-R6
R2 (2, 7, 8, 9)
Links used: R2-R3, R2-R4, R4-R5, R4-R6
R3 (9, 9, 11)
Links used: R3-R2, R2-R4, R3-R5, R4-R6
R4 (1, 2)
Links used: R4-R5, R4-R6
Links Used: R5-R4, R4-R6
If we mark, all the used links one by one, we can see that all links are used except the following link.
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Please write comments if you find any of the answers/explanations incorrect, or you want to share more information about the topics discussed above.
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