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UGC-NET | UGC NET CS 2018 July – II | Question 59

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  • Last Updated : 07 Sep, 2018
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Consider the following three processes with the arrival time and CPU burst time given in milliseconds:

Process       Arrival Time        Burst Time
P1     0                  7
P2     1                  4
P3     2                  8

The Gantt Chart for preemptive SJF scheduling algorithm is _________.

Answer: (B)

Explanation: P1 arrive at 0 it will be served by CPU for 1 unit. After 1 unit P2 arrived ans it is shortest So it will execute for 4 unit, after its execution There are 2 processes out of which P1 is shortest so it will execute for 6 unit after its execution P1 will execute for 8 unit.
Gantt chart for preemptive SJF scheduling algorithm is:

So, option (B) is correct.

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