GATE CS 2014 Set 3 — Question 42
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Operating System → Process Scheduling → Scheduling Numericals
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Question
An operating system uses shortest remaining time first scheduling algorithm for pre-emptive scheduling of processes. Consider the following set of processes with their arrival times and CPU burst times (in milliseconds):
The average waiting time (in milliseconds) of the processes is _________.
| Process | Arrival Time | Burst Time |
|---|---|---|
| P1 | 0 | 12 |
| P2 | 2 | 4 |
| P3 | 3 | 6 |
| P4 | 8 | 5 |
Correct answer
5.5 to 5.5
Solution
Using Shortest Remaining Time First (SRTF) scheduling:Gantt Chart Analysis:
ms.
- Time 0: P1 arrives (Burst 12). P1 starts.
- Time 2: P2 arrives (Burst 4). P1 has run for 2ms, remaining = 10. P2(4) < P1(10), so preempt P1. P2 starts.
- Time 3: P3 arrives (Burst 6). P2 has run for 1ms, remaining = 3. P2(3) < P3(6). Continue P2.
- Time 6: P2 finishes. Ready queue: P1(10), P3(6). P3(6) < P1(10). P3 starts.
- Time 8: P4 arrives (Burst 5). P3 has run for 2ms, remaining = 4. P4(5) > P3(4). Continue P3.
- Time 12: P3 finishes. Ready queue: P1(10), P4(5). P4(5) < P1(10). P4 starts.
- Time 17: P4 finishes. Ready queue: P1(10). P1 starts.
- Time 27: P1 finishes.
- P1: 27
- P2: 6
- P3: 12
- P4: 17
- P1:
- P2:
- P3:
- P4:
- P1:
- P2:
- P3:
- P4:
ms.
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