09 CPU Scheduling

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Amity School of Engineering

B.Tech., CSE(3thSem.)

Operating Systems (Module-II)Topic: CPU Scheduling

ANIL SAROLIYA


B.Tech., CSE-IT(3rdSem.)


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Maximum CPU utilization obtained with multiprogramming

CPUI/O Burst CycleProcess execution consists of a cycleof CPU execution and

I/O wait

Program executionbegins with CPU burst and that is followed by an I/O burst CPU burst distribution

Basic Concepts

Alternating Sequence of

CPU And I/O Bursts


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Histogram of CPU-burst Times

Burst

Frequency

Duration (millisecond)


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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 non-preemptive

All other scheduling ispreemptive

Figure: Process State

once CPU given to the process it cannot be

preempted until it completes its CPU burst.

if a new process arrives with CPU

burst length less than remainingtime of current executing process,

preempt. This scheme is known as

the Shortest-Remaining-Time-First

(SRTF)

CPU Scheduler


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Dispatchermodule gives control of the CPU to the process selected by the short-

term scheduler; this involves:

context switching

switching to user mode

jumping to the proper location in the user program to restart that program

Dispatch latencytime it takes for the dispatcher to stop one process and start

another running

CPU utilizationkeep the CPU as busy as possible

Throughputnumber of processes that complete their execution per time unit

Turnaround timeamount of time to execute a particular process. The interval

from the time of submission of a process to the time of completion

periods spent waiting to get into memory + waiting in the ready queue

+ executing on the CPU + doing I/O

Waiting timeamount of time a process has been waiting in the ready queue. It

is the sum of the periods spent waiting in the ready queue

Response timeamount of time it takes from when a request was submitted

until the first response is produced, notoutput (for time-sharing environment)

Dispatcher

Scheduling Criteria


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Optimization Criteria

CPU utilization

Throughput

Turnaround time

Waiting time

Response time

Max

Min


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Process Burst Time

P1 24

P2 3

P3 3 Suppose that the processes arrive in the order: P1, P2, P3

The Gantt Chartfor the schedule is:

Waiting time for P1 = 0; P2 = 24; P3 = 27

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

Turnaround Time = Burst time + Waiting time

Process Turnaround Time

P1 24 + 0 = 24

P2 3 + 24 = 27

P3 3 + 27 = 30 Average Turnaround Time: (24 + 27+30)/3 = 27

P1 P2 P3

24 27 300

First-Come, First-Served (FCFS) Scheduling

Drawback:

High waiting time


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Suppose that the processes arrive in the order

P2, P3, P1

The Gantt chart for the schedule is:

Waiting time for P1 =6;P2= 0; P3 = 3

Average waiting time: (6 + 0 + 3)/3 = 3

Much better than previous case FCFS group effectshort process behind long process

P1P3P2

63 300



FCFS Scheduling (continued)


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Shortest-Job-First (SJF) 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:

nonpreemptiveonce CPU given to the process it cannot bepreempted 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 theShortest-Remaining-Time-First (SRTF)

SJF is optimalgives minimum average waiting time for a given

set of processes


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Example of Non-Preemptive SJF

Process Arrival Time Burst Time

P1 0 7

P2 2 4

P3 4 1

P4 5 4 SJF (non-preemptive)

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

P1 P3 P2

73 160

P4

8 12P2P1 P3 P41 2 4 5 6


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Example of Preemptive SJF

Process Arrival Time Burst Time

P1 0.0 7

P2 2.0 4

P3 4.0 1

P4 5.0 4

SJF (preemptive)

Average waiting time = (9 + 1 + 0 +2)/4 = 3

P1 P3P2

42 110

P4

5 7

P2 P1

16


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Round Robin (RR) Scheduling

One of the oldest, simplest, fairest and most widely used algorithms in round robin.

In this approach, a time slice is defined, which is a particular small unit of time.

In each time slice, CPU runs the current process until the end of the time slice.

if (that process is done)

it is discarded and the next one in the queue is dealt with.

else

it is halted and put at the back of the queue, then the next process in line is

addressed during the next time slice.


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Example: Round Robin (RR) Scheduling

Letstake four processes that arrive at the same time in this order andthe time quantum is 20

Process Burst Time

P1 53

P2

17

P3 68

P4 24

The Gantt chart is:

Typically, higher average turnaround than SJF, but having betterresponse time.

P1 P2 P3 P4 P1 P3 P4 P1 P3 P3

0 20 37 57 77 97 117 121 134 154 162

Waiting time for:

P1 = 57+24=81; P2 = 20

P3 = 37+40+17=94; P4 = 57+40=97

Average Waiting Time = (81+20+94+97)/4 = 73

Turnaround time=burst time + Waiting time

P1 = 53+81=134; P2 = 17+20 =37

P3 = 68+94=162; P4 = 24 + 97=121

Avg. Turnaround Time = 113.5 = 114 (Approx)


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Priority Scheduling

A priority number (integer) is associated with each process

The CPU is allocated to the process with the highest priority

(smallest integer highest priority)

Preemptive

Non-preemptive

SJF is also a priority scheduling where priority is the predicted next CPU burst

time (really hard to predict the next CPU burst)

Problem Starvationlow priority processes may never execute

Solution Agingas time progresses increase the priority of the process


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CPU scheduling more complex when multiple CPUs are available

Load sharingcan be achieve by this scheduling

Here, we will only concentrate on the systems that contains

Homogeneous processorswithin a multiprocessor Asymmetric multiprocessing only one processor accesses the

system data structures, reducing the need for data sharing

Symmetric multiprocessing every processor is self-scheduled.

Each processor has its own private ready queue.



Multiple Processor Scheduling


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Calculate the Turnaround (Completion) time, wait time and average of each for

FCFS, Round Robin and SJF scheduling algorithms and fill the blanks below.



Question


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Solution:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3

3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4

4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5

FCFS schedulingRepresents time sliceof 1 second

Waiting time for:

P1 =0, P2 =50, P3 =90, P4 =120, P5 =140

Average Waiting Time = 80

Turnaround time (Completion time)

P1 =50, P2 =90, P3 =120, P4 =140, P5= 150

Average Turnaround Time = 110


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Solution (Continued.)

Round Robin schedulingRepresents time sliceof 1 second

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 251 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1

2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 1 2 3 1 2 3 1

2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 1 2 1

2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 1 1 1 1 1 1 1


P1 =150, P2 =140, P3 =120, P4 =90, P5 =50


Waiting timefor

P1 =100, P2 =100, P3 =90, P4 =70, P5= 40



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Solution (Continued.)

SJF SchedulingRepresents time sliceof 1 second

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

5 5 5 5 5 5 5 5 5 5 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4

4 4 4 4 4 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

3 3 3 3 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1


P1 =150, P2 =100, P3 =60, P4 =30, P5 =10


Waiting timefor

P1 =100, P2 =60, P3 =30, P4 =10, P5= 0



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Solution (Continued.)The answer is:




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Thanks

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09 CPU Scheduling

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