05s_schedulability_part2

7/31/2019 05s_schedulability_part2

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Utilization based schedulability tests

pros: usually involves a simple computational test

mathematically rigorous

Seminal paper

cons: not always conclusive

sufficient but not necessary

only valid if the workload model and the underlyingassumptions of the test are consistent can lead to false confidence

11/6/2012 Dr Alain Beaulieu


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Alternatives

When utilization based tests are inconclusive, we canalways construct a timing graph of the system construct graph over the interval [0, max {pi}]

if all tasks meet their deadlines within their first period,the system is schedulable

assumes that all tasks are in phase

the problem with this approach is that it will be impracticalfor most large systems and is not easily automated

As experienced in the first homework alternative: time-demand analysis



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Revisiting Critical Instants

Recall that a critical instant is defined as the instantin time that a job is released such that it has themaximum response time of all jobs in the task think of it as the worst possible time (busiest) to release on

the processor

Theorem: in a fixed priority system (where every job ina task completes before another job in the same task releases),the critical instant for Ti occurs when it releases at

the same time as a job in each higher priority task



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Critical Instants

Example 1: T= { (3,1), (5, 1), (10, 2) }

The critical instant for tasks T2 and T3 is 0

0 2 4 6 12

T1

T2

T3

8 10

1

1

1

1 1 1

1 1

1 1

In fact for in phase tasks, a critical instant always exists at t = 0



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Critical Instants

Example 2: T= { (1,3,1,3), (5, 1), (10, 2) }

The critical instant for tasks T2 and T3 is 10

0 2 4 6 12

T1

T2

T3

8 10

1

1

1

1 1 1

1 1

1 1



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Understanding Response Time

T = { (100, 20), (150, 30), (250, 60) }

w3 = e3 + w3/p1 e1 + w3/p2 e2

= 60 + 2(20) + 1(30) {by viewing the graph}

= 130

0 100 200

T1

T2

T3

w3 = 130



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Solving the Response Time Equation

Essentially, form a recursive relationship withequation (5) and solve iteratively:

win+1 = ei + wi

n / pj ej (6)

where initial (seed) value wi0 = ei

The algorithm then becomes:

solve for successive values of win+1

if win+1 = win , then solution found -> wi = win

if win+1 > Di , then task i cannot meet its deadline

jhp(i)



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Exercise

Draw the schedule for the system below (go out far enough in time to include as a minimum the

completion of the first job in each periodic task and allaperiodic jobs)

Determine the response time of A1 Determine the worst case response time of all

periodic tasks

Is the system schedulable? T={ (5, 1.5), (7, 2) }

TDS = ( 4, 1) , A = {(0.5, 2.5), (2.5, 0.25)}



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Exercise Solution

draw the schedule for the system belowT={ (5, 1.5), (7, 2) }

TDS = ( 4, 1) , A = {(0.5, 2.5), (2.5, 0.25)}

Tasks

T2

T1

TDS

6 8 t0 2 4

A1 A1 A1A

2



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Exercise Solution

determine the response time of A1wA1 = 8.5 - 0.5 = 8

wA2 = 8.75 - 2.5 = 6.25

determine the worst case response time of allperiodic tasks

w11 = e1 + es + (w1

0-es) / ps es where w10 = 1.5

= 1.5 + 1 + (1.5 -1) / 4 (1) = 3.5

w12 = 2.5 + (3.5 -1) / 4 (1) = 3.5

therefore w1 = 3.5 D1 (worst-case)



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Exercise Solution

determine the worst case response time of allperiodic tasks (continued)

w21 = e2 + es + (w20-es) / ps es + (w20) / p1 e1 w20 = 2

= 2 + 1 + (2 -1) / 4 (1) + 2 / 5 (1.5) = 5.5

w22 = 3 + (5.5 -1) / 4 (1) + 5.5 / 5 (1.5) = 8

w23 = 3 + (8 -1) / 4 (1) + 8/ 5 (1.5) = 8

therefore w2 = 8 > D2 (worst-case)

is the system schedulable?No since all tasks do not meet their deadlines under worst-caseconditions, we can not guarantee the system schedulable


05s_schedulability_part2

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