Operating Systems · Question 22 · Medium
Operating Systems question 22
Process (arrival time, burst time) in ms: P1 (0, 7), P2 (2, 4), P3 (4, 1), P4 (5, 3). Using preemptive Shortest Remaining Time First, what is the average waiting time?
- 3.75 ms
- 4.50 ms
- 5.25 ms
- 2.75 ms
Answer and explanation
Answer: D. 2.75 ms
Explanation: Gantt chart: P1 0-2, P2 2-4, P3 4-5, P2 5-7, P4 7-10, P1 10-15. Waiting = completion - arrival - burst: P1 8, P2 1, P3 0, P4 2. Average = 11 / 4 = 2.75 ms.
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