GATE 2017 CS – Question 34
Consider the following CPU processes with arrival times (in milliseconds) and length of CPU bursts (in milliseconds) as given below:
| Process | Arrival time | Burst time |
|---|---|---|
| P1 | 0 | 7 |
| P2 | 3 | 3 |
| P3 | 5 | 5 |
| P4 | 6 | 2 |
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.
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Correct answer: 3
Explanation
P1 runs from 0. At time 3, P1 has 4 left and P2 needs 3, so P2 runs from 3 to 6. At 6, P4 (2) is shortest, so it runs 6 to 8. Then P1 (4 left) runs 8 to 12, and finally P3 runs 12 to 17. The waiting times are P1: $12 - 7 = 5$, P2: $6 - 3 - 3 = 0$, P3: $17 - 5 - 5 = 7$, P4: $8 - 6 - 2 = 0$. The average is $\frac{5 + 0 + 7 + 0}{4} = 3$.