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07/10/2020 1

Lecture 11

Chapter 2 Processes and Threads

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Three- level Scheduling

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Robin (RR) Scheduling

Processes are dispatched in a FIFO manner

but are given a limited amount of CPU time

called a time-slice or a quantum.

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Robin (RR) Scheduling

1.

list of runable processes

2.

list of runable processes after B uses up its quantum

3.

Length of quantum too short means too much context switching.

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Example: The time slot could be 100 milliseconds. If a job1 takes a total time of 250ms to complete, the round-robin scheduler will suspend the job after 100ms and give other jobs their time on the CPU. Once the other jobs have had their equal share (100ms each), job1 will get another allocation of CPU time and the cycle will repeat. This process continues until the job finishes and needs no more time on the CPU.

Job1 = Total time to complete 250ms (quantum 100ms). 1.First allocation = 100ms.

2.Second allocation = 100ms.

3.Third allocation = 100ms but job1 self-terminates after 50ms. 4.Total CPU time of job1 = 250ms.

Round Robin (RR) Scheduling

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Round Robin (RR) Scheduling

APPLICATION

Using round-robin scheduling allots a slice of time to each

process that is running. In a computer for example, the user

starts three applications,

Email

, a web browser, and a

word processor

. These applications are loaded into system

memory as processes and each is allowed to run without the

user considering which applications are running in the

background.

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Burst Waiting Turnaround Process Time Time Time

P1 24 6 30

P2 3 4 7

P3 3 7 10

Average - 5.66 15.66

Time Quantum= 4ms (without priority)

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Time Quantum= 1ms (without priority)

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Process

Burst

P1

8 P2

6 P3

1 P4

9 P5

3 Q-

Assume that you have the following processes all arriving at

time 0:

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Process B.T Priority W.T T.T

P1 8 4 12 20

P2 6 1 16 22

P3 1 2 8 9

P4 9 2 18 27

P5 3 3 13 16

average 67/5=13.4 94/5=18.8

P1

P2

P3

P4

P5

P1

P2

P4

0 4 8 9 13 16 20 22 26 27

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Process

Burst

Priority

P1

8

4 P2

6

1 P3

1

2 P4

9

2 P5

3

3 Q-

Assume that you have the following processes all arriving at

time 0:

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P2

P3

P4

P5

P1

P2

P4

P1

P4

0 4 5 9 12 16 18 22 26 27

Process B.T Priority W.T T.T

P2 6 1 12 18

P3 1 2 4 5

P4 9 2 18 27

P5 3 3 9 12

P1 8 4 18 26

average 12.2 17.6

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Time Quantum= 2ms (without priority)

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Process

Arrival

Burst Time

P1

0

3 P2

1

2 P3

2

1

Total Waiting Time = (WT of Process 1)+ (WT of Process 2) + (WT of Process 3) = (3+1+2)s

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Processes Burst time Arrival time

P1 8 0.0

P2 4 1.0

P3 2 2.0

P4 5 3.0

Q-

Assume that you have following processes with arrival time and burst time given in milliseconds.

Draw a Gantt chart to determine average process waiting time of each process using round robin scheduling algorithm (Quantum Time = 2ms). Ignore process switching overhead.

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Priority Scheduling

1. Each process is assigned a priority, and priority is

allowed to run. Equal-Priority processes are

scheduled in FCFS order.

2. The shortest-Job-First (SJF) algorithm is a special

case of general priority scheduling algorithm.

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Priority Scheduling

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Burst Waiting Turnaround

Process Time Priority Time Time

P1 10 3 6 16

P2 1 1 0 1

P3 2 4 16 18

P4 1 5 18 19

P5 5 2 1 6

Average - - 8.2 12

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Process Priority AT BT

P1 2 0 4

P2 4 1 2

P3 6 2 3

P4 10 3 5

P5 8 4 1

P6 12 5 4

P7 9 6 6

Priority Scheduling-Non-Preemptive

EXAMPLE-3

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P1

P4

P6

P7

P5

P3

P2

Process W.T TAT

P1 0 4

P2 22 24

P3 18 21

P4 1 6

P5 15 16

Priority Scheduling-Non-Preemptive

SOLUTION

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Process Priority AT BT

P1 2 0 4

P2 4 1 2

P3 6 2 3

P4 10 3 5

P5 8 4 1

P6 12 5 4

P7 9 6 6

Priority Scheduling-Preemptive

EXAMPLE- 4

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P1

P2

P3

P4

P6

P4

P7

P5

P3

P2

P1

Process W.T TAT RT

P1 21 25 0

P2 19 21 0

P3 16 19 0

P4 4 9 0

P5 14 15 14

Priority Scheduling-Preemptive

SOLUTION

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Priority Scheduling

In this example, Tasks B and C are at the same priority, which is higher than that of Task A.

Task A is running when Task B becomes ready to run. Task A is preempted so that Task B can run. While Task B is running, Task C also becomes ready to run. The scheduler therefore allows Task B to run for a time slice period. After this period expires, the scheduler gives Task C an opportunity to run. Then Task C completes its work. Because Task B still has work to complete and has the highest priority of all ready tasks, the scheduler allows Task B to run. Once Task B completes its work, Task A can finish.