Numerical Solved by Shortest Job First Scheduling Algorithm
Q8. Consider the following:
Arrive Time Process Burst Time
0 P1 10
1 P2 4
2 P3 3
3 P4 1
Draw Gantt Chart and calculate average turnaround and waiting time using Shortest Job First Scheduling Algorithm.
Grant Chart:
P1 | P2 | P4 | P2 | P3 | P1 |
0 1 3 4 6 9 18
Turnaround Time
P1=18-0=18
P2=6-1=5
P3=9-2=7
P4=4-3=1
Average Turnaround Time
=(18+5+7+1)/4
=31/4
=7.77 ms
Waiting Time
P1’s waiting time=9-1=8
P2’s waiting time=4-3=1
P3’s waiting time=6-2=4
P4’s waiting time=3-3=0
Average Waiting Time
=(8+1+4+0)/4
=13/4
=3.25 ms
Final answer for waiting time is wrong. It should be 3.25
Thanks. Now i have updated the results. Please do check it out.