Preemptive SJF Process Arrival Time Burst Time P1 0 8 P2 1 5

Preemptive SJF

Process Arrival Time Burst Time

P₁ 0 8

P₂ 1 5

P₃ 2 3

P₄ 3 1

Preemptive SJF Gantt Chart ?

Average waiting time ?

Shortest Job First(SJF) Scheduling

Process

Arrival Time

Burst Time

Preemptive SJF Gantt Chart is-

0 1 2 3 4 6 10 17 (time)

Average waiting time

Here we have to found average waiting time for each process

waiting time for process P1=9

waiting time for process P2=4

waiting time for process P3=1

waiting time for process P4=0

sum of all process waiting time= 9+4+1+0

Number of Process=4

Average waiting time= (sum of all process waiting time)/ Number of Process

Average waiting time= (9+4+1+0)/4

Average waiting time= 3.75

Additional Info:-

(P1 stop just after 1 second and again start on 10 second so waiting time is 10-1=9)

(P2 arrive at 1 sec, stop just after 2 second and again start on 6 second so waiting time is 6-2=4)

(P3 arrive at 2 sec, stop just after 3 second and again start on 4 second so waiting time is 4-3=1)

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Thanks a lot