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Module 2: Processes & Scheduling

20247m

Consider the following set of processes with the length of the CPU Burst given in milliseconds:

Process Burst Time Priority
A 10 3
B 1 1
C 2 3
D 1 4
E 5 2

The processes are assumed to have arrived in order A, B, C, D, E all at time 0.
Draw the Gantt charts that illustrate the execution of FCFS and SJF scheduling algorithm. Show which algorithm results in the minimum average waiting time.

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FCFS vs SJF for A(10),B(1),C(2),D(1),E(5) arrival 0

FCFS order A,B,C,D,E: Gantt A0-10,B10-11,C11-13,D13-14,E14-19. WT: A0,B10,C11,D13,E14 => Avg WT=9.6, Avg TAT=13.4
SJF (non-preemptive, sorted by burst): order B(1),D(1),C(2),E(5),A(10): Gantt B0-1,D1-2,C2-4,E4-9,A9-19. WT: B0,D1,C2,E4,A9 => Avg WT=3.2, Avg TAT=7.0
Conclusion: SJF gives minimum avg waiting (3.2 vs 9.6). SJF is optimal for avg WT.

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