Question
Download Solution PDFConsider 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?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFThe 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
| 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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