Consider processes P1 (arrival 0, burst 8), P2 (1, 4), P3 (2, 9), times in ms. What is the average waiting time with preemptive shortest job first scheduling?
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Correct answer: C. 4.66
Explanation
P1 0–1; P2 (4) preempts, 1–5; P1 5–12; P3 12–21. Waiting: P1 = 12 − 8 = 4, P2 = 0, P3 = 21 − 2 − 9 = 10. Average 14/3 ≈ 4.66.
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