cpu sheduling

Operating System Concepts Prasad Sawant

CPU Scheduling

Prof.Prasad Sawant Lecturer MCA

MACS College Pune


Alternating Sequence of CPU And I/O Bursts


CPU Scheduler

• Selects from among the processes in memory that are ready to execute, and allocates the CPU to one of them.

• CPU scheduling decisions may take place when a process:1.Switches from running to waiting state.2.Switches from running to ready state.3.Switches from waiting to ready.4.Terminates.

• Scheduling under 1 and 4 is nonpreemptive.• All other scheduling is preemptive.


Dispatcher

• Dispatcher module gives control of the CPU to the process selected by the short-term scheduler.


Scheduling Criteria

• CPU utilization – keep the CPU as busy as possible

• Throughput – # of processes that complete their execution per time unit

• Turnaround time – amount of time to execute a particular process

• Waiting time – amount of time a process has been waiting in the ready queue


Keep in mind

• Waiting time (WT)=start time(ST)-Arrival time (AT)• Finish time (FT)=start time(ST)+Burst Time(BT)• Total turn around time(TOT)=Finish time(FT)-Arrival time (AT )


Optimization Criteria

• Max CPU utilization• Max throughput• Min turnaround time • Min waiting time • Min response time


First-Come, First-Served (FCFS) SchedulingSuppose that the processes arrive in the order: P1 , P2 , P3

0 10 20 30

P1 24

P2 27

P3 30 Waiting time for P1 = 0;

P2 = 24; P3 = 27Average waiting time: (0 + 24 + 27)/3 = 17

schedule

schedule

schedule

Process Bust Time

P1 24

P2 3

P3 3


Process Bust TimeP1 24P2 3P3 3

First-Come, First-Served (FCFS) Scheduling Gantt Chart

Suppose that the processes arrive in the order: P1 , P2 , P3

Waiting time for P1 = 0; P2 = 24; P3 = 27Average waiting time: (0 + 24 + 27)/3 = 17

schedule

schedule

schedule

0 P1 24 P2 27 P3 30


Exercise

• Find Average waiting time for • Suppose that the processes arrive in the

order: P2 , P3 , P1

Process Burst Time

P1 24

P2 3

P3 3


Shortest-Job-First (SJR) Scheduling

• Associate with each process the length of its next CPU burst. Use these lengths to schedule the process with the shortest time.

• Two schemes: – nonpreemptive – once CPU given to the process it cannot be

preempted until completes its CPU burst.– preemptive – if a new process arrives with CPU burst length less

than remaining time of current executing process, preempt. This scheme is know as the Shortest-Remaining-Time-First (SRTF).

• SJF is optimal – gives minimum average waiting time for a given set of processes.


Example of Non-Preemptive SJFProcess A T Burst Time

P1 0.0 7

P2 2.0 4

P3 4.0 1

P4 5.0 4

0 P1 7 P3 8 P2 12 P4 16

schedule

schedule

schedule

schedule

Average waiting time = (0+6+3+7)/4=4


Example of Non-Preemptive SJFProcess AT BT ST WT(ST-AT) FT(ST+BT) TOT(FT-AT)

P1 0 7 0 0 7 7P2 2 4 8 6 12 10P3 4 1 7 3 8 4P4 5 4 12 7 16 11

P1 7 P3 8 P2 12 P4 16

Average waiting time = (0+6+3+7)/4=4

Average TOT =(7+10+4+11)/4=8


Preemptive SJFProcess A.T. B.T S.T W.T

P1 0 8

P2 1 4

P3 2 9

P4 3 5

0 P1 1 P2 5

Pschedule

P4 10

schedule

7

P1 17

schedule

P3 26

schedule

10

1

17

5

10-1=9

1-1=0

17-2=15

5-3=2


Priority SchedulingProcess BT Priority ST WT

P1 10 3

P2 1 1 High

P3 2 4

P4 1 5

P5 5 2

Avg waiting time

0 P2 1

0

P5 6 P1 16 P3 18 P4 19

Schedule

Schedule

Schedule

Schedule

Schedule

6

0

16

18

1

6

0

16

18

1

41/5=8.2 ms


SJF Preemptive Priority Scheduling Process BT AT Priority ST WT

P1 8 0 4

P2 6 1 6

P3 7 3 3

P4 9 3 1(high)

Avg Waiting Time

0 P1 3

P 5

P4 12

scheduled

P3 19

scheduled

P1 24

scheduled

P2 30

scheduled

19

24

12

3

19-3=16

24-1=23

12-3=9

3-3=0

48/4=12 ms


Round Robin Scheduling

• Each process gets a small unit of CPU time (time quantum), usually 10-100 milliseconds. After this time has elapsed, the process is preempted and added to the end of the ready queue.

• If there are n processes in the ready queue and the time quantum is q, then each process gets 1/n of the CPU time in chunks of at most q time units at once. No process waits more than (n-1)q time units.


Time Quantum and Context Switch Time

Quantum Context Switch Process Time 10

12

6

1

0

1

9

0 10

6 10

2 3 4 5 6 7 8 9 101


Round Robin Time Quantum 4Process Burst Time TOT

P1 24

P2 3P3 3

0 P1 4

24-4=20

P2 7

scheduled

P3 10

scheduled

P1 14

20-4=16

P1 18

16-4=12

P1 22

12-4=8

P1 26

8-4=4

P1 30

4-4=0scheduled 30

7

10


Questions

1. Define the difference between pre-emptive and non-pre-emptive scheduling.

2. Explain the concept of a priority used in scheduling. Why is priority working usually chosen for real time processes?

3. Define by difference between preemptive and non-emptive scheduling.

4. Comment on the principle disadvantage of each of these scheduling methods: FCFS, SJF, RR


Bibliography

• Operating System Principle-Peter Galvin Galvin

• Prof.S.G.Lakhdive (Dept .Computer Sci )Prof.Ramkirshna More A.C.S College Akurdi

• Mr. Abhishek Nagar Web Administrator at Symbiois


Thanks you

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