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GATE | Gate IT 2007 | Question 61

  • Last Updated : 28 Jun, 2021

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) 3
(B) 4.26
(C) 4.53
(D) 5.26


Answer: (C)

Explanation: Average hops for a message to reach to root – (3∗8)+(2∗4)+(1∗2)+(0∗1)15 = 2.267
Where, 3 hops & 8 routers for Bottom-most level and similarly the others
Similarly average hops for a message to comes down to leaf- (3∗8)+(2∗4)+(1∗2)+(0∗1)15
{Same as above}

So, Total Hops = 2 * 2.267 = 4.53


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