Suppose that the following processes arrive for execution at the times…
2026
Suppose that the following processes arrive for execution at the times indicated. Each process will run for the amount of time listed in the given table. Use non-preemptive scheduling and answer the following:
(a) What is the average turnaround time for these processes with the FCFS scheduling algorithm?
(b) What is the average turnaround time for these processes with the SJF scheduling algorithm?
Process | Arrival Time | Burst Time |
|---|---|---|
P1 | 0.0 | 8 |
P2 | 0.4 | 4 |
P3 | 1.0 | 1 |
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(a) FCFS Scheduling
Execution Order:
P1 → P2 → P3
Completion Times:
P1: 0 to 8
P2: 8 to 12
P3: 12 to 13
Turnaround Time = Completion Time − Arrival Time
Process | Completion Time | Arrival Time | Turnaround Time |
|---|---|---|---|
P1 | 8 | 0.0 | 8 |
P2 | 12 | 0.4 | 11.6 |
P3 | 13 | 1.0 | 12 |
Average Turnaround Time:
(8 + 11.6 + 12) / 3
= 31.6 / 3
= 10.53
FCFS Average Turnaround Time = 10.53
(b) Non-Preemptive SJF Scheduling
At time 0, only P1 is available, so it executes first.
After P1 completes, both P2 and P3 are ready. Since P3 has the shortest burst time, it executes before P2.
Execution Order:
P1 → P3 → P2
Completion Times:
P1: 0 to 8
P3: 8 to 9
P2: 9 to 13
Process | Completion Time | Arrival Time | Turnaround Time |
|---|---|---|---|
P1 | 8 | 0.0 | 8 |
P2 | 13 | 0.4 | 12.6 |
P3 | 9 | 1.0 | 8 |
Average Turnaround Time:
(8 + 12.6 + 8) / 3
= 28.6 / 3
= 9.53
SJF Average Turnaround Time = 9.53
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