Consider the following CPU processes with arrival times (in milliseconds) and length of CPU bursts (in milliseconds) except for process P4 as given below:
Process Arrival Time Burst Time P1 0 5 P2 1 1 P3 3 3 P4 4 x
If the average waiting time across all processes is 2 milliseconds and pre-emptive shortest remaining time first scheduling algorithm is used to schedule the processes, then find the value of x ?
(D) None of these
Explanation: If we take value of x is 2, then we have gantt chart as
So, completion time of P1, P2, P3, and P4 are 6, 2, 11, and 8 respectively.
Turn around time of P1, P2, P3, and P4 are 6, 1, 8, and 4 respectively.
Waiting time of P1, P2, P3, and P4 are 1, 0, 5, and 2 resectively.
Therefore, average waiting time = (1+0+5+2) / 4 = 8/2 = 2
Option (B) is correct.
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