Operating Systems · Question 6 · Medium
Operating Systems question 6
Three processes arrive at time 0 in the order P1, P2, P3 with CPU burst times 6, 3 and 1 ms. With Round Robin scheduling and a time quantum of 2 ms, what is the average waiting time?
- 1.67 ms
- 5.00 ms
- 4.33 ms
- 3.67 ms
Answer and explanation
Answer: C. 4.33 ms
Explanation: The Gantt chart is P1 0-2, P2 2-4, P3 4-5, P1 5-7, P2 7-8, P1 8-10. Completion times are 10, 8 and 5, so the waiting times are 10-6 = 4, 8-3 = 5 and 5-1 = 4. The average is 13/3 ≈ 4.33 ms.
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