GATE CS 2015 Set 3 — Question 57
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Operating System → Process Scheduling → Scheduling Numericals
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Question
For the processes listed in the following table, which of the following scheduling schemes will give the lowest average turnaround time?
| Process | Arrival Time | Processing Time |
|---|---|---|
| A | 0 | 3 |
| B | 1 | 6 |
| C | 4 | 4 |
| D | 6 | 2 |
Correct answer
(C) Shortest Remaining Time
Solution
We calculate the average turnaround time (TAT = Completion Time - Arrival Time) for each algorithm.1. Shortest Remaining Time (SRT) - Preemptive SJF:
TAT: A=3, B=14, C=4, D=4. Average = .2. FCFS:
Order: A, B, C, D.
Completion: A=3, B=9, C=13, D=15.
TAT: A=3, B=8, C=9, D=9. Average = .3. Non-preemptive SJF:
Completion: A=5, B=15, C=13, D=11.
TAT: A=5, B=14, C=9, D=5. Average = .Lowest is SRT (6.25).
- : P_A arrives (burst 3). Runs A.
- : P_B arrives (burst 6). A has 2 left. , continue A.
- : A finishes. Ready: {B(6)}. Run B.
- : P_C arrives (burst 4). B has 5 left. , preempt B. Run C.
- : P_D arrives (burst 2). C has 2 left. . Continue C (tie-breaking usually favors current or FCFS).
- : C finishes. Ready: {B(5), D(2)}. Run D.
- : D finishes. Ready: {B(5)}. Run B.
- : B finishes.
TAT: A=3, B=14, C=4, D=4. Average = .2. FCFS:
Order: A, B, C, D.
Completion: A=3, B=9, C=13, D=15.
TAT: A=3, B=8, C=9, D=9. Average = .3. Non-preemptive SJF:
- : Run A (3). Finishes at 3.
- : Ready {B(6)}. Run B. Finishes at 9.
- : Ready {C(4), D(2)}. Run D. Finishes at 11.
- : Ready {C(4)}. Run C. Finishes at 15.
Completion: A=5, B=15, C=13, D=11.
TAT: A=5, B=14, C=9, D=5. Average = .Lowest is SRT (6.25).
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