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Module 2: Processes & Scheduling

201914m

Consider the following process :

Process Arrival Time Burst Time
P1 0.0 7
P2 2.0 4
P3 4.0 1
P4 5.0 4

Considering non-preemptive and preemptive SJF algorithm, find out average waiting time and average turnaround time in both cases.

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SJF calculations (Arrival+ Burst)

P1 AT0 BT7, P2 AT2 BT4, P3 AT4 BT1, P4 AT5 BT4.
Non-preemptive SJF: At 0: P1 runs 0-7. At 7, ready P2(4),P3(1),P4(4) → P3 7-8, then P2 8-12, P4 12-16. WT: P1 0, P2 6, P3 3, P4 7 → Avg WT=4. Avg TAT = (7+10+4+11)/4=8.
Preemptive SJF (SRTF): 0-2 P1 (rem5), at2 P2 (4 vs5) → P2? Actually P1 rem5 vs P2 4 → preempt to P2 2-4 (rem2), at4 P3(1) arrives shortest → P3 4-5, back to P2 5-7, P4 arrives 5 (4) vs P1(5) → P4 7-11, P1 11-16. WT: P1 9, P2 1, P3 0, P4 2 → Avg WT=3. Avg TAT=7. Details with Gantt: Non-preemptive Gantt P1|P3|P2|P4, Preemptive P1(0-2)|P2(2-4)|P3(4-5)|P2(5-7)|P4(7-11)|P1(11-16).

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