GATE | GATE-CS-2014-(Set-3) | Question 65
An operating system uses shortest remaining time first scheduling algorithm for pre-emptive scheduling of processes. Consider the following set of processes with their arrival times and CPU burst times (in milliseconds):
Process Arrival Time Burst Time
P1 0 12
P2 2 4
P3 3 6
P4 8 5
The average waiting time (in milliseconds) of the processes is _________.
(A) 4.5
(B) 5.0
(C) 5.5
(D) 6.5
Answer: (C)
Explanation:
Process Arrival Time Burst Time
P1 0 12
P2 2 4
P3 3 6
P4 8 5
Burst Time – The total time needed by a process from the CPU for its complete execution.
Waiting Time – How much time processes spend in the ready queue waiting their turn to get on the CPU
Now, The Gantt chart for the above processes is :
P1 - 0 to 2 milliseconds
P2 - 2 to 6 milliseconds
P3 - 6 to 12 milliseconds
P4 - 12 to 17 milliseconds
P1 - 17 to 27 milliseconds
Process p1 arrived at time 0, hence cpu started executing it.
After 2 units of time P2 arrives and burst time of P2 was 4 units, and the remaining time of the process p1 was 10 units,hence cpu started executing P2, putting P1 in waiting state(Pre-emptive and Shortest remaining time first scheduling).
Due to P1’s highest remaining time it was executed by the cpu in the end.
Now calculating the waiting time of each process:
P1 -> 17 -2 = 15
P2 -> 0
P3 -> 6 - 3 = 3
P4 -> 12 - 8 = 4
Hence total waiting time of all the processes is
= 15+0+3+4=22
Total no of processes = 4
Average waiting time = 22 / 4 = 5.5
Hence C is the answer.
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Last Updated :
22 Jul, 2021
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