Consider processes A (arrival 0, burst 6), B (3, 2), C (5, 4), D (7, 6), E (10, 3), times in ms. Using shortest remaining time first scheduling, the average turnaround time is

1. A7.2
2. B8.2
3. C6
4. D9

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Correct answer: A. 7.2

Explanation

A runs 0–3; B (2) preempts and runs 3–5; A (3 left) runs 5–8; C runs 8–12 (E arriving at 10 needs 3 > C's 2 left); E runs 12–15; D runs 15–21. Turnaround: A 8, B 2, C 7, D 14, E 5. Average 36/5 = 7.2 ms.