Consider a set of 5 processes whose arrival time, CPU time needed and the priority are given below:
Process Arrival Time CPU Time Needed Priority
(in ms)
P1 0 10 5
P2 0 5 2
P3 2 3 1
P4 5 20 4
P5 10 2 3(smaller the number, higher the priority)
If the CPU scheduling policy is priority scheduling without preemption, the average waiting time will be
(A) 12.8 ms
(B) 11.8 ms
(C) 10.8 ms
(D) 9.8 ms
Answer: (C)
Explanation: Following is the Gantt diagram:
Waiting time = turn around time – burst time
Turn around time = completion time – arrival time
Arrival Time CPU Time Priority turnaround time Waiting Time
(in ms) Needed
P1 0 10 5 40 - 0 = 40 40 - 10 = 30
P2 0 5 2 5 - 0 = 5 5 - 5 = 0
P3 2 3 1 8 - 2 = 6 6 - 3 = 3
P4 5 20 4 28 - 5 = 23 23 -20 = 3
P5 10 2 3 30 - 10 = 20 20 - 2 = 18Average Waiting Time = (30 + 3 + 3 + 18)/ 5 = 10.8
So, option (C) is correct.
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