GATE CS 2016 Set 2 — Question 57
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Operating System → Process Scheduling → Scheduling Numericals
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Question
Consider the following processes, with the arrival time and the length of the CPU burst given in milliseconds. The scheduling algorithm used is preemptive shortest remaining-time first.
The average turn around time of these processes is ________ milliseconds.
| Process | Arrival Time | Burst Time |
|---|---|---|
| 0 | 10 | |
| 3 | 6 | |
| 7 | 1 | |
| 8 | 3 |
Correct answer
8.2 to 8.3
Solution
The scheduling algorithm is Preemptive Shortest Remaining Time First (SRTF).Gantt Chart Execution:
1.Time 0: arrives (Burst=10). starts.
2.Time 3: arrives (Burst=6). has run for 3ms, remaining burst = . Comparing remaining times: . is preempted. starts.
3.Time 7: arrives (Burst=1). has run for 4ms (from 3 to 7), remaining burst = . Comparing remaining times: . is preempted. starts.
4.Time 8: arrives (Burst=3). has run for 1ms (from 7 to 8), remaining burst = . completes.
Ready queue: . Comparing remaining times: . resumes.5.Time 10: runs for 2ms (from 8 to 10), remaining burst = . completes.
Ready queue: . Comparing remaining times: . starts.6.Time 13: runs for 3ms (from 10 to 13), remaining burst = . completes.
Ready queue: . resumes.7.Time 20: runs for 7ms (from 13 to 20), remaining burst = . completes.
Completion Times (CT):- : 20 ms
- : 10 ms
- : 8 ms
- : 13 ms
- : ms
- : ms
- : ms
- : ms
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