Let us consider a statistical time division multiplexing of packets. The…
2007
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
Answer: C. 4.53 — To find the average number of backlogged packets per time unit, first calculate the total data transmitted and backlog for each of the 20 time units. Each…
- A.
3
- B.
4.26
- C.
4.53
- D.
5.26
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Correct answer: C
To find the average number of backlogged packets per time unit, first calculate the total data transmitted and backlog for each of the 20 time units. Each active source sends a 1000-bit packet, and the multiplexer capacity is 5000 bits per time unit. Backlog occurs when total data exceeds 5000 bits. The number of backlogged packets per time unit is the excess data (in bits) divided by 1000. Sum up backlogged packets over all time units and divide by 20 to get the average.
For each time unit, compute: (number of active sources × 1000) - min((number of active sources × 1000), 5000). Then divide the excess by 1000 to get backlogged packets. Sum these values and divide by 20.
For example, in the first time unit: 6 sources × 1000 = 6000 bits. Excess = 6000 - 5000 = 1000 bits. Backlogged packets = 1000 / 1000 = 1 packet.
Repeat for all 20 time units, sum the backlogged packets, and divide by 20 to get the average.
The final average is 4.53 packets per time unit.