Computer Networks MCQ - Computer Networks Section 3

36 In a packet switching network, packets are routed from source to destination along a single path having two intermediate nodes. If the message size is 24 bytes and each packet contains a header of 3 bytes, then the optimum packet size is

Consider optimum packet size which required minimum number of packet and does not waste any space.
Here consider each packet header because they are all work on packet switching network. 3 bytes is header size;
9 – 3 = 6 is data size
24/6 = 4 are the required packet
For case 1 : 24 packets are required
For case 2 : 8 packets are required
For case 3 : 6 packets are required

37 Suppose that it takes 1 unit of time to transmit a packet (of fixed size) on a communication link. The link layer uses a window flow control protocol with a window size of N packets. Each packet causes an ack or a nak to be generated by the receiver, and ack/nak transmission times are negligible. Further, the round trip time on the link is equal to N units. Consider time i > N. If only acks have been received till time i(no naks), then the goodput evaluated at the transmitter at time i(in packets per unit time) is

Goodput (is the application level throughput) = Successfully delivered data/ Time = (i-N)/i = 1- N/i.

38 Count to infinity is a problem associated with

This Problems can happen in distance vector routing protocol when a link or a router fails.

39 For the network given in the figure below, the routing tables of the four nodes A, E, D and G are shown. Suppose that F has estimated its delay to its neighbors, A, E, D and G as 8, 10, 12 and 6 msecs respectively and updates its routing table using distance vector routing technique.

Distance from F to F is 0 So option C is false. Using Distance Vector routing protocol, F → D → B distance is 20
So option B and D is also false.

40 A group of 15 routers are interconnected in a centralized complete binary tree with a router at each tree node. Router i communicates with router j by sending a message to the root of the tree. The root then sends the message back down to router j. The mean number of hops per message, assuming all possible router pairs are equally likely is

As per question, we have to count average hops per message.
Steps used:
1) Message goes up from sender to root.
2) Message comes down from root to destination
1) Average hops message goes to root – ((3 * 8) + (2 * 4) + (1 * 2) + (0 * 1))/15 = 2.267 // 3 * 8 means represents 3 hops & 8 routers for Bottom most level & So on.
2) Similarly average hops when message comes down -
((3 * 8) + (2 * 4) + (1 * 2) + (0 * 1))/15 = 2.267
So, Total Hops = 2.267 + 2.267 = 4.53.