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OS Process Scheduling Q & A #6
Question: Shown below is the workload for 5 jobs arriving at time zero in the order given below −
| Job | Burst Time |
|---|---|
| 1 | 10 |
| 2 | 29 |
| 3 | 3 |
| 4 | 7 |
| 4 | 12 |
Now find out which algorithm among FCFS, SJF And Round Robin with quantum 10, would give the minimum average time.
Answer: For FCFS, the jobs will be executed as:
| Job | Waiting Time |
|---|---|
| 1 | 0 |
| 2 | 10 |
| 3 | 39 |
| 4 | 42 |
| 5 | 49 |
| 140 |
The average waiting time is 140/5=28.
For SJF (non-preemptive), the jobs will be executed as:
| Job | Waiting Time |
|---|---|
| 1 | 10 |
| 2 | 32 |
| 3 | 0 |
| 4 | 3 |
| 5 | 20 |
| 65 |
The average waiting time is 65/5=13.
For Round Robin, the jobs will be executed as:
| Job | Waiting Time |
|---|---|
| 1 | 0 |
| 2 | 32 |
| 3 | 20 |
| 4 | 23 |
| 5 | 40 |
| 115 |
The average waiting time is 115/5=23.
Thus SJF gives the minimum average waiting time.
os_exams_questions_answers.htm
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