System Software and Operating System - Process Management

6. Consider a set of 5 processes whose arrival time. CPU time needed and the priority are given below

Process PriorityArrival Time (in ms)CPU Time Needed (in ms)Priority
P10105
P2052
P3231
P45204
P51023
smaller the number, higher the priority.
If the CPU scheduling policy is SJF with pre-emption, the average waiting time will be

  • Option : C
  • Explanation : Scheduling order will be
    P2 , P3 , P1 , P5 , P1, P4
    Waiting time of processes will be
    P2 = 0
    P3= 5-2=3
    P1=10+2=12
    P5=0
    P4= 15
    Average waiting time will be = (0+3+12+0+15)/5= 30/5=6ms
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7. Consider a set of 5 processes whose arrival time. CPU time needed and the priority are given below

Process PriorityArrival Time (in ms)CPU Time Needed (in ms)Priority
P10105
P2052
P3231
P45204
P51023
smaller the number, higher the priority.
If the CPU scheduling policy is priority scheduling without pre-emption, the average waiting time will be

  • Option : C
  • Explanation : 30 + 0 + 3 + 3 + 18 divided by 5, i.e. 10.8 ms.
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8. Consider a set of 5 processes whose arrival time. CPU time needed and the priority are given below

Process PriorityArrival Time (in ms)CPU Time Needed (in ms)Priority
P10105
P2052
P3231
P45204
P51023
smaller the number, higher the priority.
If the CPU scheduling policy is priority sche duling with pre-emption, the average waiting time will be

  • Option : B
  • Explanation : Here the process which will start at the initial millisecond will be P2 as it has more priority that P1.

    ms  Process
    0 to 2      P2 (P2 completed 2 ms here)
    2 to 5        P3  (No wait for P3)
    5 to 8      P2  (P2 had to wait 3 ms to get executed )
    8 to 10       P4 (P4 had to wait 3 ms to get started)
    10 to 12     P5 (No wait for P5)
    12 to 30    P4 (P4 had to wait 2 ms to complete its remaining)
    30 to 40   P1 (Was waiting for 30 ms)
    So, waiting time---- P1 -30 P2 -3 P3 -0 P4 -5 P5 -0
    Average---- (30+3+0+5+0)/5= 7.6 ms
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