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  • homework3


    Suppose the burst time is as follows:
    P1:4  P2:5 P3:9 P4:6 P5:2
    for FCFS: the average waiting time is:(4+9+18+24)/5=10
    for Shortest-job-First:  the average waiting time is: 7.2

    then add arrival time for each process like:
    P1:1 P2:2 P3:9 P4:0 P5:13
    then for Shortest-job-First algorithm the average waiting time is:(1+5+10+0+0)/5=3.2

    and for Priority Scheduling we add priorities for each process:
    P1:1 P2:2 P3:4 P4:3 P5:5
    the average


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