Let us consider a statistical time division multiplexing of packets. The…

Solution overview: compute backlog by tracking how packets accumulate when arrivals exceed service capacity.

Step 1: Convert capacity to packets: 5000 bits per time unit = 5 packets per time unit (1000 bits per packet).

Step 2: Use the recurrence for backlog:

backlog_t = max(0, backlog_{t-1} + arrivals_t - 5), with backlog_0 = 0.

• Time 1: arrivals = 6 → backlog = max(0, 0 + 6 - 5) = 1

• Time 2: arrivals = 9 → backlog = max(0, 1 + 9 - 5) = 5

• Time 3: arrivals = 3 → backlog = max(0, 5 + 3 - 5) = 3

• Time 4: arrivals = 7 → backlog = max(0, 3 + 7 - 5) = 5

• Time 5: arrivals = 2 → backlog = max(0, 5 + 2 - 5) = 2

• Time 6: arrivals = 2 → backlog = max(0, 2 + 2 - 5) = 0

• Time 7: arrivals = 2 → backlog = max(0, 0 + 2 - 5) = 0

• Time 8: arrivals = 3 → backlog = max(0, 0 + 3 - 5) = 0

• Time 9: arrivals = 4 → backlog = max(0, 0 + 4 - 5) = 0

• Time 10: arrivals = 6 → backlog = max(0, 0 + 6 - 5) = 1

• Time 11: arrivals = 1 → backlog = max(0, 1 + 1 - 5) = 0

• Time 12: arrivals = 10 → backlog = max(0, 0 + 10 - 5) = 5

• Time 13: arrivals = 7 → backlog = max(0, 5 + 7 - 5) = 7

• Time 14: arrivals = 5 → backlog = max(0, 7 + 5 - 5) = 7

• Time 15: arrivals = 8 → backlog = max(0, 7 + 8 - 5) = 10

• Time 16: arrivals = 3 → backlog = max(0, 10 + 3 - 5) = 8

• Time 17: arrivals = 6 → backlog = max(0, 8 + 6 - 5) = 9

• Time 18: arrivals = 2 → backlog = max(0, 9 + 2 - 5) = 6

• Time 19: arrivals = 9 → backlog = max(0, 6 + 9 - 5) = 10

• Time 20: arrivals = 5 → backlog = max(0, 10 + 5 - 5) = 10

Sum of backlog values over 20 time units = 89 packets.

Average backlog = 89 / 20 = 4.45 packets per time unit.

Therefore the average number of backlogged packets per time unit during the given period is 4.45.