Assume, we have the workload as shown below. All 5 processes arrive at time 0, in the order given below. The length of the CPU burst time is given in milliseconds
Process : P1 P2 P3 P4 P5
Burst time : 10 29 3 7 12
Considering the FCFS, SJF and RR (q=10 ms) scheduling algorithms, which algorithm would give the minimum average turnaround time.
Ans: The Gantt-chart for FCFS scheduling is
P1 P2 P3 P4 P5
0 10 39 42 49
61
Turnaround time = Finished Time – Arrival Time Turnaround time for process P1 = 10 – 0 = 10 Turnaround time for process P2 = 39 – 0 = 39 Turnaround time for process P3 = 42 – 0 = 42 Turnaround time for process P4 = 49 – 0 = 49 Turnaround time for process P5 = 61 – 0 = 61
Average Turnaround time = (10+39+42+49+61)/5 = 40.2 The Gantt-chart for SJF scheduling is
P3 P4 P1 P5 P2
0 3 10 20 32
61
Turnaround time for process P1 = 3 – 0 = 3 Turnaround time for process P2 = 10 – 0 = 10 Turnaround time for process P3 = 20 – 0 = 20 Turnaround time for process P4 = 32 – 0 = 32 Turnaround time for process P5 = 61 – 0 = 61
Average Turnaround time = (3+10+20+32+61)/5 = 25.2 The Gantt-chart for RR scheduling is
P1 P2 P3 P4 P5 P2 P 5 P2
0 10 20 23 30 40
50 52 61
Turnaround time for process P1 = 10 – 0 = 10
Turnaround time for process P2 = 61 – 0 = 61 Turnaround time for process P3 = 23 – 0 = 23 Turnaround time for process P4 = 30 – 0 = 30 Turnaround time for process P5 = 52 – 0 = 52
Average Turnaround time = (10+61+23+30+52)/5 = 44.2 So SJF gives minimum turnaround time.
