GATE CS 2025 Set 2 — Question 26
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Operating System → Process Scheduling → Scheduling Numericals
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Question
Processes arrive in that order at times 0, 1, 2, and 8 milliseconds respectively, and have execution times of 10, 13, 6, and 9 milliseconds respectively. Shortest Remaining Time First (SRTF) algorithm is used as the CPU scheduling policy. Ignore context switching times.Which ONE of the following correctly gives the average turnaround time of the four processes in milliseconds?
Correct answer
(D) 19
Solution
We are given the following processes with their Arrival Times (AT) and Burst Times (BT). The scheduling algorithm is Shortest Remaining Time First (SRTF), which is the preemptive version of SJF.
Gantt Chart Execution:
| Process | AT | BT |
|---|---|---|
| 0 | 10 | |
| 1 | 13 | |
| 2 | 6 | |
| 8 | 9 |
1.Time 0: arrives. Remaining time: . starts.
2.Time 1: arrives. Remaining times: . Since , continues.
3.Time 2: arrives. Remaining times: . Since has the shortest remaining time (), is preempted and starts.
4.Time 8: finishes execution (). arrives. Remaining times: . is done. Comparing remaining times, is the shortest (). resumes.
5.Time 16: finishes execution (). Remaining times: . is the shortest. starts.
6.Time 25: finishes execution (). Remaining time: . starts.
7.Time 38: finishes execution ().
Completion Times (CT):- : 8
- : 16
- : 25
- : 38
- :
- :
- :
- :
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