session 4 - cpu scheduling - finals

8/3/2019 Session 4 - CPU Scheduling - Finals

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What is in this topic?

To find out how to get a process attached to a

processor.

It centers around efficient algorithms that performwell.

The design of a scheduler is concerned with makingsure all users get their fair share of the resources.

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CPU Scheduling


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5: CPU-Scheduling

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CPU Scheduling

ConceptsMultiprogramming A number of programs can be inmemory at the same time. Allows

overlap of CPU and I/O.

Jobs are programs that run without

user interaction.

User are programs that may have user

interaction.

Process is the common name for both.

CPU - I/O burst cycle Characterizes process execution,

which alternates, between CPU

and I/O activity. CPU times are

generally much shorter than I/O

times.

Preemptive Scheduling An interrupt causes currently

running process to give up the

CPU and be replaced by anotherprocess.


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CPU SCHEDULING

5: CPU-Scheduling

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The 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


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SCHEDULERS

Process migrates among several queues

Device queue, job queue, ready queue

Scheduler selects a process to run from these queues

Long-term scheduler: load a job in memory

Runs infrequently

Short-term scheduler:

Select ready process to run on CPU

Should be fast


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WHEN DOES SCHEDULER RUN?

Non-preemptive minimum

Process runs until voluntarily relinquish CPU

process blocks on an event

process terminates

process yields

Preemptive minimum

All of the above, plus:

Event completes: process moves from blocked to ready

Timer interrupts Implementation: process can be interrupted in favor of another

New Ready RunningExit

Waiting


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Note usage of the words DEVICE, SYSTEM, REQUEST, JOB.

UTILIZATION The fraction of time a device is in use. ( ratio of in-use time / totalobservation time )

THROUGHPUT The number of job completions in a period of time. (jobs / second )

SERVICE TIME The time required by a device to handle a request. (seconds)

QUEUEING TIME Time on a queue waiting for service from the device. (seconds)

RESIDENCE TIME The time spent by a request at a device.RESIDENCE TIME = SERVICE TIME + QUEUEING TIME.

RESPONSE TIME Time used by a system to respond to a User Job. ( seconds )

THINK TIME The time spent by the user of an interactive system to figure out thenext request. (seconds)

The goal is to optimize both the average and the amount of variation.

CPU SCHEDULINGCriteria For

Performance

Evaluation


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THE PERFECT CPU SCHEDULER

Minimize latency: response or job completion time

Maximize throughput: Maximize jobs / time.

Maximize utilization: keep I/O devices busy.

Fairness: everyone makes progress, no one starves


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SCHEDULING ALGORITHMS FCFS

First-come First-served (FCFS) (FIFO)

Jobs are scheduled in order of arrival

Non-preemptive

Problem: Average waiting time depends on arrival order

Advantage: really simple!

time

P1 P2 P3

0 16 20 24

Process Service Time

P1 16

P2 4

P3 4


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EXAMPLE DATA:

Process Arrival Service

Time Time

1 0 8

2 1 4

3 2 9

4 3 5

0 8 12 21 26

P1 P2 P3 P4

FCFS

Average Turnaround Time= ( (8-0) + (12-1) + (21-2) + (26-3) )/4 = 61/4 = 15.25 s

CPU SCHEDULING Scheduling

Algorithms

Average Waiting Time= ( (0-0) + (8-1) + (12-2) + (21-3) )/4 = 36/4 = 9 s


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SHORTEST JOB FIRST SCHEDULING - NON-

PREEMPTIVE

Example: Process Arrival Time Burst Time

P1

0 7

P2

2 4

P3

4 1

P4 5 4 Non preemptive SJF

P1 P3 P2

7

P1(7)

160

P4

8 122 4 5

P2(4)

P3(1)

P4(4)

P1s wating time = 0

P2s wating time = 6

P3s wating time = 3

P4s wating time = 7

Ave. Turnaround Time= ( (7 - 0) + (12 - 2) + (8 - 4) + (16 - 5) )/4 = 32/4 = 8.0 s

Ave. Waiting Time= ( (0 - 0) + (8 - 2) + (7 - 4) + (12 - 5) )/4 = 16/4 = 4.0 s


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SHORTEST JOB FIRST SCHEDULING -

PREEMPTIVE

Example: Process Arrival Time Burst Time

P1

0 7

P2

2 4

P3

4 1

P4

5 4

Preemptive SJF

P1 P3P2

42 110

P4

5 7

P2 P1

16

P1(7)

P2(4)

P3(1)

P4(4)

P1s wating time = 9

P2s wating time = 1

P3s wating time = 0

P4s wating time = 2

P1(5)

P2(2)

Ave. Waiting Time= ( (11 - 2 - 0) + (5 - 4 - 2) + (4 - 4) + (7 - 5) )/4 = 12/4 = 3.0 s

Ave. Turnaround Time= ( (16 - 0) + (7 - 2) + (5 - 4) + (11 - 5) )/4 = 28/4 = 7.0 s


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CPU SCHEDULING

5: CPU-Scheduling

13

EXAMPLE DATA:


Time Time (s)

1 0 8

2 1 4

3 2 9

4 3 5

0 5 10 17 26

P2 P4 P1 P3

Preemptive Shortest Job First

Ave. Turnaround Time= ( (17-0) + (5-1) + (26-2) + (10-3) )/4 =52/4 = 13.0 s

P1

1

Scheduling

Algorithms

Ave. Waiting Time= ( (10-1-0) + (1-1) + (17-2) + (5-3) )/4 = 26/4 = 6.5 s


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CPU SCHEDULING

5: CPU-Scheduling

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PRIORITY BASED SCHEDULING:

y Assign each process a priority. Schedule highest priority first. All processes

within same priority are FCFS.

y Priority may be determined by user or by some default mechanism. The

system may determine the priority based on memory requirements, time

limits, or other resource usage.

y Suitable if some jobs are absolutely more important than the other, and

must be serviced rst.

y Starvation occurs if a low priority process never runs.

Scheduling

Algorithms


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PRIORITY SCHEDULING

Priority Scheduling

Choose next job based on priority

For SJF, priority = expected CPU burst

Can be either preemptive or non-preemptive

Problem:

Starvation: jobs can wait indefinitely

Solution to starvation

Age processes: increase priority as a function of waiting time


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PRIORITY SCHEDULING

The example assumed that all processes arrived

at 0 second.

Ave. Turnaround Time= ( (16 - 0) + (1 - 0) + (18 - 0) + (19 - 0) + (6 - 0) )/5=52/4 = 12.0 s

Ave. Waiting Time= ( (6 - 0) + (0 - 0) + (16 - 0) + (18 - 0) + (1 - 0) )/5= 41/5= 8.2 s


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PRIORITY SCHEDULING (NON-

PREEMPTIVE)

Ave. Waiting Time NP= ( (1 - 0) + (0 - 0) + (16 - 0) + (18 - 0) + (11 - 3) )/5= 43/5= 8.6 s

Ave. Turnaround Time NP= ( (11 - 0) + (1 - 0) + (18 - 0) + (19 - 0) + (16 - 3) )/5= 62/5= 12.4 s

0 11 16 18

P2 P5 P3P1

1

P4

19


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PRIORITY SCHEDULING

(PREEMPTIVE)

0 3 8 16 18

P1 P5 P1 P3P2

1 19

P4

Ave. Waiting Time NP= ( (8 3 - 0) + (0 - 0) + (16 - 0) + (18 - 0) + (3 - 3) )/5= 39/5= 7.8 s

Ave. Turnaround Time NP= ( (16 - 0) + (1 - 0) + (18 - 0) + (19 - 0) + (8 - 3) )/5=59/5= 11.8 s


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ROUND ROBIN:

y Use a timer to cause an interrupt after a predeterminedtime. Preempts if task exceeds its quantum.

y Train of eventsDispatch

Time slice occurs OR process suspends on event

Put process on some queue and dispatch next

y Use numbers in last example to find queueing andresidence times. (Use quantum = 4 sec.)

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CPU SCHEDULING SchedulingAlgorithms


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CPU SCHEDULING

5: CPU-Scheduling

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EXAMPLE DATA:


Time Time

1 0 8

2 1 4

3 2 9

4 3 5

0 8 12 16 26

P2 P3 P4 P1

Round Robin, quantum = 4, no priority-based preemption

Ave. Turnaround Time= ( (20-0) + (8-1) + (26-2) + (25-3) )/4 = 74/4 = 18.5

P1

4

P3 P4

20 24 25

P3

Scheduling

Algorithms

Note:

Example violates rules for

quantum size since most

processes dont finish in

one quantum.


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ROUND ROBIN

Round Robin (RR)

Often used for timesharing

Ready queue is treated as a circular queue (FIFO)

Each process is given a time slice called a quantum It is run for the quantum or until it blocks

RR allocates the CPU uniformly (fairly) across participants.

If average queue length is n, each participant gets 1/n


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CSS430 CPU Scheduling

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Each process is given CPU time in turn, (i.e. time quantum: usually 10-100milliseconds), and thus waits no longer than ( n 1 ) * time quantum

time quantum = 20

Process Burst Time Wait Time

P1

53 57 +24 = 81

P2

17 20

P3

68 37 + 40 + 17= 94

P4

24 57 + 40 = 97

ROUND ROBIN SCHEDULING

P1 P2 P3 P4 P1 P3 P4 P1 P3 P30 20 37 57 77 97 117 121 134 154 162

Average wait time = (81+20+94+97)/4 = 73

57

20

37

57

24

40

40

17

P1(53)

P2(17)

P3(68)

P4(24)

P1(33) P1(13)

P3(48) P3(28) P3(8)

P4(4)


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RR WITH TIME QUANTUM = 20

Process Burst Time

P1 53

P2

17

P3 68P4 24

The Gantt chart is:

Higher average turnaround than SJF,

But better response time

P1 P2 P3 P4 P1 P3 P4 P1 P3 P3

0 20 37 57 77 97 117 121 134 154 162


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RR: CHOICE OF TIME QUANTUM

Performance depends on length of the timeslice If timeslice time is set too high

attempting to amortize context switch cost, you get FCFS.

i.e. processes will finish or block before their slice is up anyway

If its set too low youre spending all of your time context switching between threads.

Timeslice frequently set to ~100 milliseconds

Moral:

Context switch is usually negligible (< 1% per timeslice)otherwise you context switch too frequently and lose allproductivity

session 4 - cpu scheduling - finals

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