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# GATE | GATE-CS-2017 (Set 1) | Question 16

• Last Updated : 22 Jul, 2021

Consider the following CPU processes with arrival times (in milliseconds) and length of CPU bursts (in milliseconds) as given below:

If the pre-emptive shortest remaining time first scheduling algorithm is used to schedule the processes, then the average waiting time across all processes is _______ milliseconds.

Note: This questions appeared as Numerical Answer Type.
(A) 1
(B) 2
(C) 3
(D) 4

Explanation:

Turn Around Time

```P1 Â = 12-0 = 12
P2 = 6-3 = 3
P3 = 17-5 = 12
P4 = 8 - 6 = 2
```

Waiting Time

```P1 Â = 12-7 =Â 5
P2 = 3-3 =Â 0
P3 = 12-5 =Â 7
P4 = 2 - 2Â =Â 0
```

Average Waiting time = (7+0+5+0)/4 = 3.0

Therefore, option C is correct

Alternate Solution

Given, with arrival time and burst time:

Using (preemptive) shortest remaining time first algorithm, gantt chart is:

Therefore,
Average waiting time = ( 5 + 0 + 7 + 0 ) / 4 = 12 / 4 = 3

This explanation has been contributed by Mithlesh Upadhyay.

Watch GeeksforGeeks Video Explanation :

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