Consider the following process:

Process Id Arrival Time Burst Time
P0 2 7
P1 5 10
P2 7 5
P3 8 13

What will be completion time for completion time for P1 when round-robin scheduling algorithm is used with a time quantum of 2 ns?
(A) 12.5 ns
(B) 13.25 ns
(C) 15.75 ns
(D) 21 ns


Answer: (B)

Explanation: Using Round-robin Algorithm

Gantt chart will be:

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Ready queue:

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Turn around time = Completion time – arrival time
P0 = 17 – 2 = 15
P1 = 27 – 5 = 22
P2 = 30 – 7 = 23
P3 = 37 – 9 = 28
Waiting time = Turn around time – burst time
P1 = 15 – 7 = 8
P2 = 22 – 10 = 12
P3 = 23 – 5 = 18
P4 = 28 – 13 = 15

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Process Id Arrival Time Burst Time Turn Around Time Waiting Time
P0 2 7 15 8
P1 5 10 22 12
P2 7 5 23 18
P3 8 13 28 15

Average waiting time = (8 + 12 + 18 + 15 ) / 4 = 53 / 4 = 13.25
So, option (B) is correct.


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  • Last Updated : 26 Oct, 2018

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