esd unit iii lecture 03 task scheduling rate monotonic.ppt

7/28/2019 ESD UNIT III Lecture 03 Task Scheduling Rate Monotonic.ppt

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SHRI SANT GAJANAN MAHARAJ COLLEGE OF ENGINEERINGDepartment of Electronics & Telecommunication Engineering

Real Time Task SchedulingUnit-III::Real Time Systems & RTOS


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Scheduling of Tasks

A real time system executes several tasks

concurrently

The sequence in which tasks can run

depends upon scheduling algorithms

Programmer needs to write code for defining

tasks and algorithm

The scheduler within RTOS runs the tasks

as per the algorithm


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Task Scheduling Algorithms


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RTOS Scheduler


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Schedulability Problem

Given: A set of tasks {Ti}

Periodic

Sporadic (with minimum inter arrival time)

Aperiodic Synchronization requirements

Precedence

Mutual exclusion

Wanted:

Test to determine if set schedulable

Scheduling algorithm


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Schedulability Test

Task set complexity

Sufficient NecessaryExact

If a sufficient schedulabilitytest is positive, these tasksare definitely schedulable

If a necessary schedulabilitytest is negative, these tasks

are definitely not schedulable


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Scheduling Algorithm

Is there an algorithm that will find afeasible schedule in bounded time ?

Is the algorithm optimal, i.e. will it find a

schedule whenever one exists ?


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Scheduling Algorithm

Classification

Best effort vs. Guaranteed Static vs. Dynamic

Pre-emptive vs. Non pre-emptive

Centralized vs. Distributed


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Scheduling AlgorithmBest Effort vs. Guaranteed

Best effort: Algorithm can fail

No feasible schedule

Not enough time to find one.

Guaranteed: All tasks will meet

deadlines

Task set is schedulable Algorithm will always find feasible schedule


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Scheduling AlgorithmStatic vs. Dynamic

Static Scheduling:

All scheduling decisions at compile time.

Temporal task structure fixed.

Precedence and mutual exclusion satisfied by the

schedule (implicit synchronization).

One solution is sufficient.

Any solution is a sufficient schedulability test.

Run time task dispatcher looks up a table.


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Scheduling AlgorithmStatic vs. Dynamic

Dynamic Scheduling:

All scheduling decisions at run time.

Based upon set of ready tasks.

Mutual exclusion and synchronization enforced by

explicit synchronization constructs. Benefits:

Flexibility.

Only actually used resources are claimed.

Disadvantages:

Guarantees difficult to support Replica determinism hard to enforce

Computational resources required for scheduling


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Scheduling AlgorithmPreemptive vs. Non pre-emptive

Preemptive Scheduling:

Event driven.

Each event causes interruption of running tasks.

Choice of running tasks reconsidered after eachinterruption.

Benefits:

Can minimize response time to events.

Disadvantages:

Requires considerable computational resources for

scheduling


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Scheduling AlgorithmPreemptive vs. Non pre-emptive

Non pre-emptive Scheduling:

Tasks remain active till completion

Scheduling decisions only made after task completion.

Benefits:

Reasonable when Less computational resources needed for scheduling

Disadvantages:

Shortest guaranteed response time =

longest task time + shortest task time + switching


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Round Robin Scheduling

Each process is assigned a quantum (time interval)it is allowed to run.

Process only runs for quantum or until blocked or

finished Analysis needs to be done to pick a good quantum

since a context switch is a good deal of overhead. Setting quantum can result in too many process

switches and lowers CPU efficiency Setting too long may cause poor response

Assumes all processes are equally important


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Rate Monotonic Algorithm

Assumptions

1. No non pre-emptible parts in a task, and

negligible preemption cost2. Resource constraint on CPU time only

3. No precedence constraints among tasks

4.All tasks periodic5. Relative deadline = period


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RM Parameters

Let Ti be period of task

Ci = Execution time

Di = Deadline

Utilization ratio Ui = Ci/Ti


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RM Algorithm

Order tasks according to their period

Task with shortest period get highest

priority

e.g. Period Priority

10 1 (highest)12 2

15 3

20 4 (lowest)

RM E l


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RM Example

Rate monotonic with two tasks

C1=2 , T1=D1=5 C2=4 , T2=D2=7


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RM Scheduling Example 1

Property indicating whether a real-time

system (a set of real-time tasks) can meet

their deadlines

(4,1)

(5,2)

(7,2)


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Real-Time Scheduling

Determines the order of real-time task

executions

Static-priority scheduling (RM)

(4,1)

(5,2)

(7,2)

5

5

10

10 15

15


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RM (Rate Monotonic)

Optimal static-priority scheduling

It assigns priority according to period A task with a shorter period has a higher priority

Executes a job with the shortest period

(4,1)

(5,2)

(7,2)

5

5

10

10 15

15

T1

T2

T3


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RM (Rate Monotonic)


(4,1)

(5,2)

(7,2)

5

5

10

10 15

15

T1

T2

T3


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RM (Rate Monotonic)


(4,1)

(5,2)

(7,2)

Deadline Miss !

5

5

10

10 15

15

T1

T2

T3


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Rate Monotonic Scheduling Example 2

Response time

Duration from released time to finish time

(4,1)

(5,2)

(10,2)

5

5

10

10 15

15

T1

T2

T3


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Response Time

Response time Duration from released time to finish time

(4,1)

(5,2)

(10,2)

Response Time

5

5

10

10 15

15

T1

T2

T3


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Response Time

Response Time (Ri)

(4,1)

(5,2)

(10,2)

( )k i

ii ki

T HP T k

RR C C

T

5

5

10

10

T1

T2

T3


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RM - Schedulability Analysis

Real-time system is schedulable under

RM

if and only ifRi Tifor all task(Ti,Ci)


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RM Utilization Bound

Real-time system is schedulable under

RM if

Ui n (21/n-1)


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RM Utilization Bound: An Example

Real-time system is schedulable under RM if

Ui n (21/n-1)

Example: T1(4,1), T2(5,1), T3(10,1),

Ui = 1/4 + 1/5 + 1/10

= 0.55

3 (21/3-1) 0.78

Thus, {T1, T2, T3} is schedulable under RM.


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RM Utilization Bounds

0.5

0.6

0.7

0.8

0.9

1

1.1

1 4 16 64 256 1024 4096

The Number of Tasks

Utilization

RM Utilization Bound

Real-time system is schedulable under RMif

Ui n (21/n-1)


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Rate Monotonic Algorithm Analysis

Amongst a set of tasks, the task that occurs most frequently is assigned

highest priority

Priority of each task is fixed, i.e., Static priority

Preemptive scheduling

Proposed by Liu & Leyland (1973)

RM is considered optimal in the sense that if a set of tasks cannot be

scheduled by this algorithm, it cannot be scheduled by any other

algorithm that assigns static priorities

Despite being optimal, RMS has a limitation: CPU utilization is bounded

and it is not always possible to fully maximize CPU resources.


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Properties of Utilization Bound U(n) = n(21/n-1)

n Bound for Utilization

1 100% One task

2 83% Two tasks case

3 78%

4 76%

69.3% = (ln 2) Infinite task case


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CPU Utilization and Utilization Bound

In RM, one tasks CPU Utilization Ui = Ci/Ti

n tasks CPU Utilization U = Ci/Ti = Ui + U2+ + Un Utilization Bound for n tasks U(n)= n (21/n-1)

Utilization Bound Theorem:

Consider a set of n independent periodic tasks. They areschedulable under RM ifU U(n)

Ci/Ti n (21/n-1)

Conclusion:01]IfU is less than 0.693, the tasks are definitely schedulable as per RM

02] If 0.693 < U < 1, then RM schedule is still valid but there is no

guarantee of success

03] IfU > 1, the tasks are definitely not schedulable as per RM


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Necessary Conditions for RM Schedulability

01] Individual UtilizationFor each task, Ci Ti. This means that for each task, the

execution time should be equal or less than its period.

02] Total Utilization U = [Ci / Ti] 1

This indicates that sum of all utilization ratios cannot

exceed 1. [Ci / Ti] = 1 means that CPU is 100% time

loaded. It cannot do any extra task.


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Worst Case Response Time Ri for RM

( )i

iii j

Tj HP T j

RR C CT

Here HP(Ti) is set of higher priority tasks than T

i

In this equation Ri appears both on left as well as right side

of equality sign

This implies that the equation should be solved by multipleIteration Method


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Worst Case Response Time for RM

( 1)

( )i

nin i i j

Tj HP T j

RR C C

T

The previous equation has been rewritten

R(n+1)i is the (n+1)th approximation to worst case response

time Rni and is found in the term of nth approximation

Multiple iterations will converge R(n+1)i = Rni

For iteration calculation we can begin with initial value R0i = 0


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Numerical on Worst Case Response Time of Tasks

Task C has highest priority, Task Bmedium and Task A has lowest priority

Lowest priority task A has tasks B and C inhp (A) set. This is because tasks B and Chave higher priorities than task A

Step-I Calculation of R


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Step-I Calculation of RA

Lowest priority task A

has tasks B and C in hp

(A) set. This is because

tasks B and C havehigher priorities than

task A

In order to start

iteration, we choose R0A= 0

Now subsequent

iterations are calculated

as shown:

( 1)

( )i

nin i i j

Tj HP Tj

RR C C

T


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Step-II Calculation of RB

Medium priority task B has only one task i.e. Task C in hp (B) set. This isbecause task C has higher priority than task B

Hence RB = 20

St III C l l ti f R d C l i


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Step-III Calculation of RC and Conclusion

There is no task having higher priority than Task C. Soset hp (C) is empty, Hence RC = CC = 10

So far we have calculated RA = 52, RB = 20 and RC = 10

Now the table of tasks is rewritten with worst-caseresponse times added in last column

Conclusion:

In each of the three tasks it is seen that Ri DiHence this task set is schedulable under RM


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How to solve RM Numerical

Three tasks T1, T2 and T3 have periods of 50, 100 and 200 and execution

times as 10, 20 and 50 respectively. Assume Period Ti = Deadline Di

(a)What is utilization of each task?(b)What is total CPU Utilization Ui?

(c) Calculate Utilization Bound U(n)

(d) Mention priority of each task

(e) Is the task set schedulable as Utilization Bound Test?

(f) Draw Time Line Diagram of the three tasks(g)Draw Time Line Diagram of RM Scheduling sequence of the three

tasks

(h)Find out worst case response times RT1, RT2 and RT3

(i) With reference to worst case response times, is the task set RM

schedulable?


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Step-I Calculate Ui and Ui

Ui = 0.20 + 0.20 + 0.25 = 0.65

U(n) = U(3) = 3(21/3 1) = 0.78

Since U < U(n), the task set is RM schedulable as perUtilization Bound Test

Task with lowest period has highest priority. Hence T1 has

highest priority, T2 medium and T3 lowest priority

Step-II Time Line Diagram of RM Schedule of Three Tasks


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Step-II Time Line Diagram of RM Schedule of Three Tasks


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Summary of RM TestsUtilization bound test is simple but conservative

Worst Case Response Timetest is more exactbut also more complicated

UB and Worst Case Response Time tests share

the same limitations:

all tasks run on a single processor all tasks periodic and non-interacting

deadlines always at the end of the period

no interrupts

rate monotonic priorities assigned

zero context switch overhead

tasks do not suspend themselves

esd unit iii lecture 03 task scheduling rate monotonic.ppt

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