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# GATE | Gate IT 2007 | Question 60

Let us consider a statistical time division multiplexing of packets. The number of sources is 10. In a time unit, a source transmits a packet of 1000 bits. The number of sources sending data for the first 20 time units is 6, 9, 3, 7, 2, 2, 2, 3, 4, 6, 1, 10, 7, 5, 8, 3, 6, 2, 9, 5 respectively. The output capacity of multiplexer is 5000 bits per time unit. Then the average number of backlogged of packets per time unit during the given period is
(A) 5
(B) 4.45
(C) 3.45
(D) 0

Explanation: Â

he capacity of multiplexer is 5000 bits per unit time. This means there are 5 packets per unit time since each source transmits a packet of 1000 bits in a unit time.

If the number of packets transmitted is larger than 5 then the extra packets are backlogged. This number gets added to the next number and further backlog is calculated.

 Number of packets (+backlog from previous) Backlog 6 1 9+1 5 3+5 3 7+3 5 2+5 2 2+2 0 2 0 3 0 4 0 6 1 1+1 0 10 5 7+5 7 5+7 7 8+7 10 3+10 8 6+8 9 2+9 6 9+6 10 5+10 10 Total Backlog 89

Â

Average number of backlogged packets = 89/20 = 4.45

This solution is contributed by Kriti Kushwaha .

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