Consider three processes: P0 (arrival 0 ms, burst 7 ms), P1 (arrival 1 ms, burst 3 ms), P2 (arrival 2 ms, burst 7 ms). Pre-emptive shortest job first scheduling is used, and scheduling happens only at arrival or completion of a process. What is the average waiting time?

1. A3 ms
2. B3.67 ms
3. C4.47 ms
4. D4 ms

Pick an option to check yourself, or open the answer below.

Show answer & explanation

Correct answer: B. 3.67 ms

Explanation

P0 runs 0–1; P1 (3) preempts and runs 1–4; then P0 (6 left) runs 4–10; P2 runs 10–17. Waiting: P0 = 10 − 7 = 3, P1 = 0, P2 = 17 − 2 − 7 = 8. Average 11/3 ≈ 3.67 ms.