December 1999
AN215 : ADVANCED NETWORKING

QUESTION 3

Total Marks: 15 Marks

• Question 1
• Question 2
• Question 4
• Question 5
• Cover Page

SUGGESTED SOLUTIONS
Solutions and allocated marks are indicated in green.
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Switching is needed in data communications because it is not possible to have a point-to-point connection between all computers on most networks.

List the three steps in circuit switching.
call set-up (1 mark),data transfer (1 mark),call termination (1 mark).

Both message and packet switching techniques employ the ‘store and forward ’ technique.Describe the differences between these two techniques.

Award marks for any of the following points,or other relevant issues.To gain the mark,a genuine comparison must be made;if comparisons are very brief,only up to two marks should be awarded.For a very detailed comparison on any point,award an extra mark, subject to the maximum for this question part.

Overhead Low for message switching,just one header and one trailer;high for packet switching as each packet must contain full addressing information (1 mark).

Security Low for message switching,because entire message is stored at intermediate nodes,making it vulnerable to snooping; higher for packet switching as each packet may take a different route,so any particular node will receive only partial information (1 mark).

Buffers Large buffer is needed for message switching,as the unit of transfer is the message,so nodes need to hold whole messages;in packet switching,nodes need to buffer only small packets (1 mark).

Delays High delays in message switching,subject to propagation and queueing delays; low delays occur in packet switching,making it suitable for interactive traffic (1 mark).

A public switched telephone network is circuit-switched.This technique dedicates a physical path between the two parties.Does circuit switching provide good bandwidth utilization?Explain your answer.
No (1 mark) because during idle periods when no party is talking (sending data) the bandwidth utilization is 0%(1 mark).Since this is a dedicated path, no other stations (callers)can utilize the path (1 mark).

Suppose you have to send a 4 096 byte message.You may use a virtual circuit which has a set-up time of three seconds and a transmission rate of 256 bytes per second. Each packet has a 2-byte VC number as an overhead.

Alternatively,you could use a datagram network where each packet may contain up to 240 bytes of data and 16 bytes of overheads and addressing information.Assume that one packet takes one second to pass through the network.

Which approach will be more efficient?Justify your answer by showing your working. State any additional assumptions you make.
Packet switching: 4096 /240 =17.06 ,so 18 packets will be required,taking 18 seconds (2 marks).

Virtual circuit: 4096 /(256 -2 )=16.1 ,so 17 seconds,plus three second set-up,making 20 seconds (2 marks).

Datagram/packet switching will be more efficient (1 mark)[in fact,probably much more efficient,because the packets will travel in pipelined fashion,and maybe in parallel.]