Given CPU time slice of 2ms and following list of processes.
| Process | Burst Time(ms) | Arrival Time |
| P1 | 3 | 0 |
| P2 | 4 | 2 |
| P3 | 5 | 5 |
Find average turnaround time and average waiting time using round robin CPU scheduling?
(A)
4, 0
(B)
5.66, 1.66
(C)
5.66, 0
(D)
7, 2
Answer: (B)
Explanation:
=> Given data in Round Robin CPU scheduling CPU time slice = 2 ms
=> Gantt Chart for given processes: P1,P2,P3
| P1 | P2 | P1 | P2 | P3 | P3 | P3 | |
| 0 | 2 | 4 | 5 | 7 | 9 | 11 | 12 |
| Process | Burst time (ms) |
Arrival time |
Completion time |
Turn around time |
Waiting time |
| P1 | 3 | 0 | 5 | 5 | 2 |
| P2 | 4 | 2 | 7 | 5 | 1 |
| P3 | 5 | 5 | 12 | 7 | 2 |
=> Now Average Turnaround Time = ( 5 + 5 + 7 ) / 3
=> = 5.66
=> Average Waiting Time = ( 2 + 1 + 2 ) / 3
=> = 1.66
So, option 2 is correct answer.
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