Exam Questions Submitted by Archana Gupta




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Exam Questions

Submitted by Archana Gupta
Question 1

The following Hamming coded (single bit correction) string was received: 010101010111. Is there an error? If yes, in which position?


Answer

0 1 0 1 0 1 0 1 0 1 1 1



positons 1 2 3 4 5 6 7 8 9 10 11 12

The string is interpreted as check bits at positions that are in bold.

3 = 1+2

5 = 1+4


6 = 2+4

7 = 1+2+4

9 = 1+8

10 = 2+8

11 = 1+2+8

12 = 4+8


A) Hence for the check bit 1 we look at bits 3,5,7,9,11 => according to our string received check bit 1 = 0 + 0 + 0 + 0 + 1 = 1 which is not true.
B) For check bit 2 we look at bits 3,6,7,10,11 =>according to our string received check bit 2 = 0 + 1+ 0 + 1 + 1 = 1 which is true.
C) For check bit at position 4 we look at bits 5,6,7,12 => according to our string received check bit 4 = 0 + 1 + 0 + 1 = 1 which is not true.
D) For check bit at position 8 we look at bits 9,10,11,12 => according to our string received check bit 4 = 0 + 1 + 1 + 1 = 1 which is true.
From the above it is observed that check bits at position 1 and 4 are incorrect implying that an error has occurred. The two common positions they share are bit positions 5 and 7. But bit position 7 is also used for calculation of check bit 2 which has been determined to be correct. As a result there is an error at bit position 5.

Question 2

Consider an error free 64 kbps satellite channel used to send 512 byte data frames in one direction, with very short acknowledgements coming back the other way. Given round trip propagation delay for satellite channels is 540msec, what id the maximum throughput for window sizes of 1,7,15 and 127?
Answer
Sliding window – send specified amount of frames, wait for one ack for that set of frames
512 bytes x 8 bits/B = 4096 bits per frame

4096/64000 bps = 64 msec to send one frame

Round trip delay = 540 msec
Window size 1: send 4096 bits per 540msec

4096 bits / 540 msec = 7.585 x 103 bps throughput


Window size 7: 7585 x 7 = 53096 bps
Window size 9 and greater: 7585 x 9 = 68265 bps but the maximum capacity is 64 kbps so for window sizes greater than 9 the maximum throughput is 64 kbps
Question 3

What is the difference between leaky bucket algorithm and token bucket algorithm?
Answer
The difference lies in the way they shape traffic. Leaky bucket algorithm does not allow idle hosts to save up permission to send large bursts later. Token bucket algorithm on the other hand does allow saving up to a maximum size of the bucket, n. This implies that bursts of up to n packets can be sent at once, allowing some burstiness in the output stream and giving faster response to sudden bursts of input.

Also when bucket fills up the token bucket algorithm discards tokens whereas leaky bucket algorithm discards packets.


Question 4

A university computer science department has three Ethernet segments, connected by two transparent bridges into a linear network. One day the network administrator quits and is hastily replaced by someone from the computer center, which is an IBM token ring shop. The new administrator, noticing that the ends of the network are not connected, quickly orders a new transparent bridge and connects both loose ends to it, making a closed ring. What happens next?
Answer

Bridges operate at data link layer and break apart networks into multiple LAN’s. One can program what data it forwards and what it doesn’t, looks up receiving address in a hash table. Hence when the transparent bridge is connected it would build its own hash table and one of the other bridges would go into standby mode for backup in case the other goes down.


Question 5

Apply the Bellman-Ford algorithm to the following network. Produce a table like the following, for each iteration of the algorithm showing the routing information held by each node.



Answer
First iteration
Destination

Source(below)ABCDEFGA-B,3UnknownD,3UnknownUnknownUnknownBA,3-C,2UnknownUnknownUnknownUnknownCUnknownB,2-D,2E,2UnknownUnknownDA,3UnknownC,2-E,6UnknownUnknownEUnknownUnknownC,2D,6-F,4G,3FUnknownUnknownUnknownUnknownE,4-UnknownGUnknownUnknownUnknownUnknownE,3Unknown-


Second iteration
Destination

Source(below)ABCDEFGA-B,3B,5D,3D,9UnknownUnknownBA,3-C,2C,4C,4UnknownUnknownCB,5B,2-D,2E,2E,6E,5DA,3C,4C,2-C,4E,10E,9ED,9C,4C,2C,4-F,4G,3FUnknownUnknownE,6E,10E,4-E,7GUnknownUnknownE,5E,9E,3E,7-



Third iteration
Destination

Source(below)ABCDEFGA-B,3B,5D,3B,7D,13D,12BA,3-C,2C,4C,4C,8C,7CB,5B,2-D,2E,2E,6E,5DA,3C,4C,2-C,4C,8C,7EC,7C,4C,2C,4-F,4G,3FE,13E,8E,6E,8E,4-E,7GE,12E,7E,5E,7E,3E,7-


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