Scheduling Algorithms

In the diagram above, process P₁, which arrives first, will start running on the CPU. At millisecond 1, when process P₂ with a shorter burst time arrives, the Scheduler will forcefully remove P₁ from the CPU and let P₂ run instead. Since the next arriving processes, P₃ and P₄, have longer burst times than this of P₂, the Scheduler won't remove P₂ from the CPU. After P₂ finishes running, the Scheduler will run the remaining processes as follows: P₄ (since its burst time is 5 ms,) then P₁ (since its remaining burst time is 8 - 1 = 7 ms,) and finally P₃ (since its burst time is the longest: 9 ms.)

Because the preemptive version of the SJF algorithm relies on the remaining burst time, we often call this algorithm the shortest remaining time first (SRTF) scheduling.

Finally, the average waiting time in this schedule is ((0 - 0) + (10 - 1) + (1 - 1) + (17 - 2) + (5 - 3)) / 4 = 26 / 4 = 6.5 ms. The reason we do subtraction inside the parentheses is that we subtract the arrival times from the wait times. If we use the cooperative version of SJF, the average waiting time would be 7.75 ms, and for FCFS it would be 8.75 ms.

The average resopnse time is ((0 - 0) + (1 - 1) + (5 - 3) + (17 - 2)) / 4 = 4.25 ms.