Shortest Remaining Time First (SRTF) Scheduling with I/O Bound Processes: Optimizing CPU Utilization

Extend your understanding of Shortest Remaining Time First (SRTF) scheduling to include processes with I/O operations. This guide provides a step-by-step example illustrating how SRTF handles I/O bursts, impacting process scheduling and overall average waiting time.

Shortest Remaining Time First (SRTF) Scheduling with I/O Bound Processes

Understanding I/O Bound Processes in SRTF

Previously, we focused on CPU-bound processes in SRTF scheduling. Now, let's consider processes that also require I/O operations. These I/O operations cause the process to temporarily stop using the CPU while waiting for the I/O to complete. This example illustrates how SRTF handles such scenarios.

Example Scenario: Four Processes

We have four processes (P1, P2, P3, and P4) with the following arrival times and CPU burst times (including I/O bursts):

Process Id Arrival Time (Burst Time, IO Burst Time, Burst Time)
P1 0 (3, 2, 2)
P2 0 (1, 3, 1)
P3 3 (3, 1, 2)
P4 6 (5, 4, 5)

Gantt Chart Explanation

Let's step through the scheduling process:

  1. Time 0: P1 and P2 arrive. P2 has the shortest initial burst time (1), so it runs.
  2. Time 1: P2 completes its first burst and enters the waiting state for its I/O operation. P1 begins execution.
  3. Time 1-3: P1 runs as no other processes are ready.
  4. Time 3: P3 arrives. P1's remaining burst time is shorter than P3's, so P1 continues.
  5. Time 4: P1 completes its CPU burst and enters the waiting state. P2's I/O is complete, and its remaining burst time (1) is shorter than P3's, so P2 runs.
  6. Time 5: P2 completes. P3 now has the shortest remaining burst time and runs.
  7. Time 5-6: P3 runs.
  8. Time 6: P4 arrives. P1's I/O is complete. P3's remaining burst time is shorter than P1 and P4's, so P3 continues.
  9. Time 6-8: P3 runs.
  10. Time 8: P3 completes its CPU burst and enters the waiting state. P1 has the shortest remaining burst time and runs.
  11. Time 8-9: P1 runs.
  12. Time 9: P3's I/O is complete. P1 continues running (shortest remaining time).
  13. Time 9-10: P1 completes.
  14. Time 10-12: P3 runs to completion (no preemption).
  15. Time 12-17: P4 runs its first burst.
  16. Time 17-21: P4 waits for I/O.
  17. Time 21-26: P4 completes its remaining burst.

Results Summary

Process Id Arrival Time Total CPU Burst Time Completion Time Turnaround Time Waiting Time
P1 0 5 10 10 5
P2 0 2 5 5 3
P3 3 5 12 9 4
P4 6 10 26 20 10

Average waiting time: (5 + 3 + 4 + 10) / 4 = 5.5 units