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 TimeP1 0 12P2 2 4P3 3 6P4 8 5Q.The average waiting time (in milliseconds) of the processes is _________.a)4.5b)5.0c)5.5d)6.5Correct answer is option 'C'. Can you explain this answer? - EduRev Computer Science Engineering (CSE) Question

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Process Arrival Time Burst Time
P1 0 12
P2 2 4
P3 3 6
P4 8 5

Q. The average waiting time (in milliseconds) of the processes is _________.

a)
4.5
b)
5.0
c)
5.5
d)
6.5

Correct answer is option 'C'. Can you explain this answer?

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An operating system uses shortest remaining time first scheduling algo...

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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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 TimeP1 0 12P2 2 4P3 3 6P4 8 5Q.The average waiting time (in milliseconds) of the processes is _________.a)4.5b)5.0c)5.5d)6.5Correct answer is option 'C'. Can you explain this answer?

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