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SRTF GATE 2011 Example

If we talk about scheduling algorithm from the GATE point of view, they generally ask simple numerical questions about finding the average waiting time and Turnaround Time. Let's discuss the question asked in GATE 2011 on SRTF.

Q. Given the arrival time and burst time of 3 jobs in the table below. Calculate the Average waiting time of the system.

Process ID Arrival Time Burst Time Completion Time Turn Around Time Waiting Time
1 0 9 13 13 4
2 1 4 5 4 0
3 2 9 22 20 11

There are three jobs P1, P2 and P3. P1 arrives at time unit 0; it will be scheduled first for the time until the next process arrives. P2 arrives at 1 unit of time. Its burst time is 4 units which is least among the jobs in the queue. Hence it will be scheduled next.

At time 2, P3 will arrive with burst time 9. Since remaining burst time of P2 is 3 units which are least among the available jobs. Hence the processor will continue its execution till its completion. Because all the jobs have been arrived so no preemption will be done now and all the jobs will be executed till the completion according to SJF.

os SRTF GATE 2011 Example

                    Avg Waiting Time = (4+0+11)/3 = 5 units






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