August 2000
AN215 : ADVANCED NETWORKING

QUESTION 2

Total Marks: 15 Marks

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Question 2

The network layer may provide services using packet switching,message switching or circuit switching.

(a)What is the purpose of these switching techniques?[2 marks ]
(a) Multiplexing is a technique where several low speed lines are combined into a high-speed line [1] for long distance transmission[1].

(b)Describe the operation of message switching.[6 marks ]
(b) Time Division Multiplexing divides the bandwidth amongst the users
based on time slots, where each user is given a time slot to send his
frame after which the control is passed to the next user[1].
synchronous multiplexing[1]- all stations attached to the multiplexor are
given time slots on a rotational basis regardless of whether the station
has data to send or not.


diagram [1]
statistical multiplexing- only active stations attached to the statmux are
given time slots on a rotational basis [1].
Idle time slots


diagram [1]

(c)Is message switching suitable for interactive traffic?Explain your answer [3 marks ]
(c) CDM is a digital multiplexing technique based on the spread spectrum
theory of communications[1]. A signal is assigned a unique id code that
only the user recognizes. The receiver will then obtain the signal and
continue to track it[1]
CDM offers better privacy in transmission[1]
Used in wireless LANs [1] Also used in digital audio broadcasting.

(d)A 4096 byte message,which has to be sent.You have a choice between using a virtual circuit (VC) or a datagram network.The addressing verheads involved in the VC is a 2 byte VC identifier per packet that it takes 2 seconds to establish a logical path.On the other hand,the overheads in the datagram packet is 16 bytes.You may ignore any other overheads.Assuming it takes one second to transfer one packet across the network.Which type of network is more efficient Justify your choice.State any assumptions you wish to make.(Show all workings)[4 marks ]
(d) Synchronous TDM => 9600bps X 10 = 96kbps [2]
Statistical TDM => 9600 X 10 X (0.5/0.8) = 60kbps [2]
In each case, award [1] for setting up the calculation correctly, and [1] for
getting the right answer.