Last Updated :
26 Oct, 2018
Calculate the average waiting time for the following processes:
Process ID |
Arrival Time |
Bust Time |
P1 |
5 |
7 |
P2 |
2 |
9 |
P3 |
0 |
5 |
P4 |
0 |
3 |
P5 |
1 |
8 |
P6 |
15 |
25 |
Consider the preemptive shortest job first scheduling algorithm and all the time in nanoseconds.
(A) 7.10 ns
(B) 7.12 ns
(C) 7.16 ns
(D) None of these.
Answer: (C)
Explanation: Using preemptive shortest job first scheduling algorithm, Gantt chart will be
Since, Turn Round Time = Completion time – arrival time, so
P1 = 12 - 5 = 7
P2 = 29 - 2 = 27
P3 = 5 - 0 = 5
P4 = 3 - 0 = 3
P5 = 20 - 1 = 19
P6 = 54 - 15 = 39
And, Waiting time = Turn around time – burst time, so
P1 = 7 - 7 = 0
P2 = 27 - 9 = 18
P3 = 5 - 5 = 0
P4 = 3 -3 = 0
P5 = 19 - 8 = 11
P6 = 39 - 25 = 14
Process ID |
Arrival Time |
Bust Time |
Turn Arround Time |
Waiting Time |
P1 |
5 |
7 |
7 |
0 |
P2 |
2 |
9 |
27 |
18 |
P3 |
0 |
5 |
5 |
0 |
P4 |
0 |
3 |
3 |
0 |
P5 |
1 |
8 |
19 |
11 |
P6 |
15 |
25 |
39 |
14 |
Hence,
Average waiting time
= (18 + 11 +14 ) / 6 = 7.16 ns
So, option (C) is correct.
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