Operating Systems · Question 7 · Medium
Operating Systems question 7
Process (arrival time, burst time) in ms: P1 (0, 8), P2 (1, 4), P3 (2, 2), P4 (3, 1). Using non-preemptive Shortest Job First, what is the average waiting time?
- 7.00 ms
- 5.50 ms
- 4.25 ms
- 6.75 ms
Answer and explanation
Answer: B. 5.50 ms
Explanation: P1 runs 0-8 because it is alone at time 0. Then the shortest job goes first: P4 8-9, P3 9-11, P2 11-15. The waiting times are P1 0, P4 5, P3 7 and P2 10, so the average is 22/4 = 5.5 ms. 7.00 is the FCFS answer.
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