Directions (Q96–100): Consider the following set of processes with arrival time and CPU burst time given in milliseconds (ms):
| Process | Arrival Time | Burst Time |
|---|---|---|
| P1 | 0 | 5 |
| P2 | 2 | 3 |
| P3 | 2 | 2 |
| P4 | 5 | 3 |
| P5 | 6 | 1 |
What is the average waiting time for these processes with non-preemptive Shortest Job First (SJF) scheduling?
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Correct answer: D. 3.2 ms
Explanation
P1 0–5. At 5: P3 (2) is shortest → 5–7. At 7: P5 (1) → 7–8. Then P2 (3, arrived earlier than P4) 8–11, P4 11–14. Waiting: P1 0, P2 6, P3 3, P4 6, P5 1. Average = 16/5 = 3.2 ms.More System Software and Operating System questions
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