Numerical Solved by SJF
Q5. Consider the set of process P1,P2,P3,P4 and P5 having burst time as 10,1,2,1 and 5 ms and priority 5,1,3,4 and 2. The processes are assumed to have arrived at time 0, then draw Gantt Chart and calculate average turnaround and waiting time using Shortest Job First Scheduling Algorithm.
Ans. Solving by SJF
Process Burst Time Priority Time
P1 10 5
P2 1 1
P3 2 3
P4 1 4
P5 5 2
Gantt Chart
P2 | P4 | P3 | P5 | P1 |
0 1 2 4 9 19
Turnaround Time
P1=19
P2=1
P3=4
P4=2
P5=9
Average Turnaround Time
=(19+1+4+2+9)/5
=35/5
=7 milliseconds
Waiting Time
P1’s waiting time=9
P2’ waiting time=0
P3’s waiting time=2
P4’swaiting time=1
P5’s waiting time=4
Average Waiting Time
=(9+0+2+1+4)/5
=16/5
=3.2 ms