Four processes arrive at time 0 with burst times P1 = 6, P2 = 8, P3 = 7, P4 = 3. What is the average waiting time under non-preemptive SJF?
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Correct answer: (C) 7
Explanation
Order P4, P1, P3, P2 → waits 0, 3, 9, 16 → 28/4 = 7.
From DSSSB TGT Computer Science Full Mock Test 1
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