Gate2017 cs Q19
| 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.
Note: Numerical Type question.
- Option : C
- Explanation :
Turn Around Time
P1 = 12-0 = 12
P2 = 6-3 = 3
P3 = 17-5 = 12
P4 = 8 - 6 = 2
Waiting Time
P1 = 12-7 = 5
P2 = 3-3 = 0
P3 = 12-5 = 7
P4 = 2 - 2 = 0
Average Waiting time = (7+0+5+0)/4 = 3.0
Therefore, option C is correct
