ITPEC FE Subject A October 2025 Question 13
Five processes A, B, C, D and E arrive at almost the same time in the order of A, B, C, D and E. The processes have estimated running times of 10, 6, 2, 8, and 4 milliseconds, respectively. Which of the following is the average process turnaround time with the First-Come-First-Served (FCFS) scheduling algorithm?
Explanation
20 ms — in FCFS each process finishes after everything that arrived before it, and with simultaneous arrival its turnaround is that finish time.
Average turnaround: (10 + 16 + 18 + 26 + 30) ÷ 5 = 100 ÷ 5 = 20 ms
Answer (c)
Why not others:
- •(a) 14 — 70 ÷ 5: the average waiting time (start times 0, 10, 16, 18, 26) instead of the turnaround; shortest-job-first turnaround also sums to 70
- •(d) 22 — 110 ÷ 5: runs the longest job first (A, D, B, E, C) instead of in arrival order
- •(b) 19.2 — needs a total of 96, but the FCFS completion times are fixed by the arrival order and sum to 100
Key rule: Turnaround = completion time − arrival time = waiting time + running time; with simultaneous arrival it is the running total of running times in execution order.
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