# 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

### 2 thoughts on “Numerical Solved by Shortest Job First Scheduling Algorithm”

• April 18, 2018 at 1:20 am
• 