Simple slide on cpu scheduling in operating system. not so descriptive, only key points and examples

1. CPU Scheduling

2. Contents
 CPU Scheduling
 Scheduling Criteria
 Scheduling Algorithms
 First Come First Served Scheduling
 Shortest Job First Scheduling
 Priority Scheduling
 Round Robin Scheduling

3. CPU Scheduling
 It is the change of ready state to running state of the process.
CPU scheduling decisions may take place when a process:
 Switches from running to waiting state.
 Switches from running to ready state.
 Switches from waiting to ready.
 Terminates

4. Scheduling Criteria
 CPU utilization : keep the CPU as busy as possible.
 Turn around Time: amount of time to execute a particular
process.
 Waiting time : amount of time a process has been waiting in
the queue.
 Response time : amount of time it takes from when a request
was submitted until the first response is produced not output.

5. Scheduling Algorithms
 First Come First Served Scheduling
 Shortest Job First Scheduling
 Priority Scheduling
 Round Robin Scheduling

6. First Come First Served
Suppose that the processes arrive in the order:
 P1 , P2 , P3
 The Gantt Chart for the schedule is:
 Waiting time for P1 = 0; P2 = 24; P3 = 27
 Average waiting time: (0 + 24 + 27)/3 = 17
Process Burst Time
P1 24
P2 3
P3 3
p1 p2 p3
0 24 27 30

7. First Come, First Served
 Suppose that the processes arrive in the order :
 P2 , P3 , P1 (P1:24,P2:3,P3:3)
 The Gantt chart for the schedule is:
0 3 6 30
 Waiting time for P1 = 6; P2 = 0; P3 = 3
 Average waiting time: (6 + 0 + 3)/3 = 3
 Much better than previous case
 Convoy effect - short process behind long process
p2 p3 p1

8. Shortest Job First
Normal SJF
The Gantt Chart for SJF (Normal) is:
0 3 7 14
Average waiting time = (0 + 3 + 7)/3 = 3.33
Process Burst Time
P1 7
P2 3
P3 4
p2 p3 p1

9. Shortest Job First
Non-Preemptive SJF
 The Gantt Chart for SJF (non-preemptive) is:
0 7 8 12 16
 Average waiting time = (0 + 6 + 3 + 7)/4 = 4
Process Arrival Time Burst
Time
P1 0.0 7
P2 2.0 4
P3 4.0 1
P4 5.0 4
p1 p3 p2 p4

10. Shortest Job First
 Preemptive SJF
 The Gantt Chart for SJF (preemptive) is:
0 2 4 5 7 11 16
 Average waiting time = (9 + 1 + 0 +2)/4 = 3
Process Arrival Time Burst
Time
P1 0.0 7
P2 2.0 4
P3 4.0 1
P4 5.0 4
p1 p2 p3 p2 p4 p1

11. Priority
 Gantt Chart
0 1 6 16 18 19
 Average waiting time = (6 + 0 + 16 + 18 + 1)/5 = 8.2
Process Burst Time Priority
P1 10 3
P2 1 1
P3 2 4
P4 1 5
P5 5 2
p2 p5 p1 p3 p4

12. Round Robin
 Quantum time = 4 milliseconds
 The Gantt chart is:
0 4 7 10 14 18 22 26 30
 Average waiting time = {[0+(10-4)]+4+7}/3 = 5.6
Process Burst Time
P1 24
P2 3
P3 3
p1 p2 p3 p1 p1 p1 p1 p1

13. Thank You