Consider following process details scheduled for execution. Consider pre-emptive shortest job first scheduling strategy. Find average waiting time.

Process	Arrival Time	Burst time
P0	0	8
P1	1	4
P2	2	9
P3	3	5

The correct answer is: option 1: 6.5 ms

Concept:

We are using the Preemptive Shortest Job First (SJF) scheduling algorithm, also known as Shortest Remaining Time First (SRTF). In this approach, the process with the shortest remaining burst time is scheduled next, even if it means preempting the currently running process.

Given process data:

Step-by-step execution using Gantt Chart (Preemptive SJF):

Timeline:

• 0–1: P0 runs (remaining 7)
• 1–5: P1 arrives (shortest), so preempts P0 and finishes
• 5–10: P3 (shorter than P0 & P2)
• 10–17: P0 resumes (remaining 7) and finishes
• 17–26: P2 runs and finishes

Completion Times:

• P0: CT = 17
• P1: CT = 5
• P2: CT = 26
• P3: CT = 10

Turnaround Time (TAT) = Completion Time – Arrival Time

• P0: 17 – 0 = 17
• P1: 5 – 1 = 4
• P2: 26 – 2 = 24
• P3: 10 – 3 = 7

Waiting Time (WT) = TAT – Burst Time

• P0: 17 – 8 = 9
• P1: 4 – 4 = 0
• P2: 24 – 9 = 15
• P3: 7 – 5 = 2

Total Waiting Time = 9 + 0 + 15 + 2 = 26

Average Waiting Time = 26 / 4 = 6.5

Hence, the correct answer should be: option 2: 6.5 ms