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Consider the following table of arrival time and burst time for three processes P0, P1, P2:

Process arrival time Burst time
P0 0 ms 7
P1 1 ms 3
P2 2 ms 7


The pre-emptive shortest job first scheduling algorithm is used. Scheduling is carried out only at arrival or completion of a process. What is the average waiting time for the three processes? 

This question was previously asked in
UGC NET Computer Science (Paper 2) 17 June 2023 Official Paper
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  1. 3 ms 
  2. 3.67 ms
  3. 4.47 ms 
  4. 4 ms

Answer (Detailed Solution Below)

Option 2 : 3.67 ms
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Detailed Solution

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The correct answer is 3.67 ms

Key PointsIn Shortest Job First (SJF) scheduling, the process with the smallest burst time is processed next. In Preemptive SJF, if a new process arrives with CPU burst length less than remaining time of current executing process, CPU is preempted, and that new process starts execution.

Gantt chart

F1 SSC Priya 6-2-24 D16

Process arrival time Burst time Waiting time
P0 0 ms 7 3
P1 1 ms 3 0
P2 2 ms 7 8

So, the average waiting time is:

(3ms + 0ms + 8ms) / 3 = 3.67ms (rounded to nearest hundredth)

So, the average waiting time for the processes under a preemptive SJF scheduling algorithm is approximately 3.67ms.

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