The shortest job first algorithm is shown in the following image:
If it is shortest job first/shortest process next, shouldn't the order be:
P1 → P5 → P3 → P4 → P2 ? Since that's the order of lowest to highest service times.
Why does process 2 come second?
I know if we use burst times instead, that would be the order, but I have no idea what the differences between service time and burst times are.
Any help would be much appreciated explaining that graphic.
The image in the question follows the correct order which is:
P1 → P2 → P5 → P3 → P4
Explanation:
P1 is arrived at time = 0 , so it will be executed first. Service Time of this process is 3. So this process is completed at time=3.
At time=3, there is only one process that is arrived which is P2. All other processes arrive later. So this process is now executed. Service time of this process is 6, so this process is completed at time=3+6=9.
Now at time=9, there are three processes which are P3, P4 and P5 (which arrived at time= 4, 6 and 8 respectively). Since the service time of P5 is 2 which is minimum as compared to that of P3 and P4, so P5 is now executed and it gets completed at time=9+2=11.
At time=11, we have two processes which are P3 and P4 (which are arrived at time= 4 and 6 respectively). Since the service time of P3 is 4 which is less as compared to that of P4, so P4 is executed now and it gets completed at time= 11+4=15
At time=15, we have only one process which is P4. So it is executed now. Since service time of this process is 5, so it gets completed at time = 15+5 = 20
