CPU Scheduling Exercises

Transcription

1 CPU Scheduling Exercises NOTE: All time in these exercises are in msec. Processes P 1, P 2, P 3 arrive at the same time, but enter the job queue in the order presented in the table. Time quantum = 3 msec Process Arrive Time Burst Time Response Time Waiting Time P = 6 P = 3 P = 6 Gantt chart: P 1 P 1 P 1 P 2 P 2 P 2 P 3 P 3 P 3 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 Average waiting time: ( ) / 3 = 5 msec We define context switch overhead as the ratio between the time taken by context switch and the time taken by all the processes. Assume that each context switch takes 1 msec, what is the context switch overhead here? 3 / 30 = 0.1 = 10% What if quantum = 30, what is the average waiting time? With quantum = 30, average waiting time = 17 msec, and the execution order is P 1, P 2, P 3. RR becomes FCFS, resulting in higher average waiting time. Page 1 of 6

2 Fair-Share: Users are guaranteed to get their fair-share of the CPU time. User 1 has four processes A, B, C and D, and user 2 has only 1 process E. All processes arrive at the same time, but enter the job queue in the order presented in the table. Assume that round-robin scheduling, with time quantum = 4 msec, is used, and user 1 is entitled 66.66% of the CPU time and user 2 is entitled 33.33% of the CPU time. (The context switch overhead is negligible.) Process Burst Time Response Time Waiting Time A = 26 B = 20 C = 24 D = 16 E = 24 Gantt chart: A A A A B B B B E E E E C C C C D D D A E E E E B B B B C C C A E E E E E E E E E E E E E E E E Average waiting time for user 1: Average response time for user 1: ( ) / 4 = 21.5 msec ( ) / 4 = 27.5 msec Average waiting time for user 2: 24 msec Average response time for user 2: 48 msec Try this example with 50% of allocation to each of the 2 users, and compare the results. User 2 receives better waiting time and response time. This example shows the benefit of proportional allocation of the CPU time. However, it is difficult to predict the proportion among users since processes are generated on the fly. Page 2 of 6

3 Multilevel Feedback Queue: Processes may be moved up and down in the priority hierarchy depends on the CPU time that they have consumed. Letʼs consider a multilevel feedback queue, as shown in the figure, with three queues. Processes in the high priority queue are scheduled using RR with time quantum = 8 msec. Processes in the medium priority queue are scheduled using RR with time quantum = 16 msec. Processes in the low priority queue are served in the FCFS order. Processes P 1, P 2, P 3 and P 4 arrive at the same time, but enter the job queue in the order presented in the table. Process Arrive Time Priority Burst Time Response Time Waiting Time P 1 0 High = 48 P 2 0 High = 54 P 3 0 Low = 64 P = 52 Gantt chart: P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 2 P 2 P 2 P 2 P 2 P 2 P 2 P 2 P 4 P 4 P 4 P 4 P 4 P 4 P 4 P 4 P 4 P 4 P 4 P 4 P 4 P 4 P 4 P 4 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 1 P 2 P 2 P 2 P 2 P 2 P 2 P 2 P 2 P 2 P 2 P 2 P 2 P 2 P 2 P 2 P 2 P 3 P 3 P 3 P 3 P 4 P 4 P 4 P 4 P 1 P 1 P P 1 P 1 P 1 P 2 P 2 P 2 P Average waiting time: Average response time: ( ) / 4 = 54.5 msec ( ) / 4 = 75 msec Page 3 of 6

4 Time Queues Processes 0 High P 1 = 30, P 2 = 28 P 4 = 20 Low P 3 = 4 8 High P 2 = 28 P 4 = 20, P 1 = 22 Low P 3 = 4 16 High P 4 = 20, P 1 = 22, P 2 = 20 Low P 3 = 4 32 High P 1 = 22, P 2 = 20 Low P 3 = 4, P 4 = 4 48 High P 2 = 20 Low P 3 = 4, P 4 = 4, P 1 = 6 64 High Low P 3 = 4, P 4 = 4, P 1 = 6, P 2 = 4 68 High Low P 4 = 4, P 1 = 6, P 2 = 4 72 High Low P 1 = 6, P 2 = 4 78 High Low P 2 = 4 Page 4 of 6

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