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GATE | GATE CS 2019 | Question 50

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  • Last Updated : 22 Jul, 2021

Consider the following four processes with arrival times (in milliseconds) and their length of CPU burst (in milliseconds) as shown below:



These processes are run on a single processor using preemptive Shortest Remaining Time First scheduling algorithm. If the average waiting time of the processes is 1 millisecond, then the value of Z is __________.

Note: This was Numerical Type question.
(A) 2
(B) 3
(C) 1
(D) 4


Answer: (A)

Explanation: Using shortest remaining time (SRTF) first CPU scheduling algorithm,

Let Z = 1, then gantt chart will be,

Average waiting time,

= {(4-0-3) + (2-1-1) + (8-3-3) + (5-4-1)} / 4
= (1 + 0 + 2 + 0) / 4
= 3 / 4
= 0.75 

Now, let Z = 2, then gantt chart will be,

Average waiting time,

= {(4-0-3) + (2-1-1) + (9-3-3) + (6-4-2)} / 4
= (1 + 0 + 3 + 0) / 4
= 4 / 4
= 1 

So, answer is 2.

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