3. Consider the following set of processes, with the arrival times and the lengt

Question

3. Consider the following set of processes, with the arrival times and the length of the CPU-burst times given in milliseconds and priority Arrival Process Time Burst Time Priority 32 P2 P3 P4 P5 12 28 36 18 4 25 a) Draw a Gantt chart that shows the execution of these processes using the following scheduling algorithms Nonpreemptive Priority Preemptive Priority Preemptive SJF RR with q=8 and q=16 b) Calculate the average waiting time and the average turnaround time for each scheduling above c) Calculate the number of context switches for the processes under RR CPU scheduling

Explanation / Answer

Nonpreemptive Priority

Gantt Chart

0 P1 32 P2 40 P5 65 P3 83 P4 100

Average waiting time = (0 + 32 + 40 + 65 + 83)/5 = 44

Average Turnaround time (32 + 40 + 65 + 83 +100)/5 = 64

Preemptive Priority

Gantt Chart

0 P1 32 P3 50 P4 67 P5 92 P2 100

Average waiting time = ((0-0) + (32-28) + (50-36) + (67-46) + (92-12))/5 = 23.8

Average Turnaround time ((32-0) + (50-28) + (67-36) + (92-46) +(100-12))/5 = 43.8

Preemptive SJF

Gantt Chart

0 P1 12 P2 20 P1 40 P4 57 P3 75 P5 100

Average Waiting Time: ((20-0-12) + (12-12) + (57-28) + (40-36) + (75-46))/5 = 14

Average Turnaround Time: ((40-0) + (20-12) + (75-28) + (57-36) + (100-46))/5 = 34

RR with q=8

Gantt Chart

0 P1 8 P1 16 P2 24 P1 32 P3 40 P4 48 P5 56 P1 64 P3 72 P4 80 P5 88 P3 90 P4 91 P5 99 P5 100

Average Waiting Time = 29.40

Average Turnaround Time = 49.40

RR with q=16

Gantt Chart

0 P1 16 P2 24 P1 40 P3 56 P4 72 P5 88 P3 90 P4 91 P5 100

Average Waiting Time = 24.60

Average Turnaround Time = 44.60

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