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GATE | GATE CS Mock 2018 | Question 16
  • Last Updated : 10 Jan, 2018

Consider the following CPU processes with arrival times (in milliseconds) and length of CPU bursts (in milliseconds) except for process P4 as given below:

Process Arrival Time   Burst Time
P1       0              5
P2       1              1
P3       3              3
P4       4              x

If the average waiting time across all processes is 2 milliseconds and pre-emptive shortest remaining time first scheduling algorithm is used to schedule the processes, then find the value of x ?
(A) 1
(B) 2
(C) 4
(D) None of these


Answer: (B)

Explanation: If we take value of x is 2, then we have gantt chart as
Capture11111

So, completion time of P1, P2, P3, and P4 are 6, 2, 11, and 8 respectively.
Turn around time of P1, P2, P3, and P4 are 6, 1, 8, and 4 respectively.
Waiting time of P1, P2, P3, and P4 are 1, 0, 5, and 2 resectively.
Therefore, average waiting time = (1+0+5+2) / 4 = 8/2 = 2

Option (B) is correct.

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