Real Time Operating Systems Schedulability - Part 2 Course originally developed by Maj Ron Smith 12/20/2015Dr Alain Beaulieu1

Real Time Operating Systems

Schedulability - Part 2

Course originally developed by Maj Ron Smith

04/21/23 Dr Alain Beaulieu 1

Outline

Review of RMA and Utilization Bound Testing

Revisiting Critical Instances

Time Demand Analysis

Schedulability & Aperiodic Tasks

Deadline Monotonic Priority Algorithm

Schedulability & Bandwidth Preserving Servers

Deferrable Server Schedulability

04/21/23 Dr Alain Beaulieu

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 underlying

assumptions of the test are consistent can lead to false confidence


Alternatives

When utilization based tests are inconclusive, we can always 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

impractical for most large systems and is not easily automated As experienced in the first homework

alternative: time-demand analysis


Revisiting Critical Instants

Recall that a critical instant is defined as the instant in time that a job is released such that it has the maximum 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 in a 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


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


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


Time-Demand Analysis (Di ≤ pi)

Under time-demand analysis we: predict the worst case response time for each task

going from highest priority to lowest then compare each task’s response time to its deadline if all worst case response times are less than their

respective deadlines, the system is schedulable

Recall that a tasks response time is its completion time minus its release time

This technique is sometimes referred to as response time analysis


Worst-Case Response Time

Alternatively, the response time of a task is defined to be the sum of its own worst case computation time and its maximum interference ->

wi = ei + Ii (1)

where Ii is the maximum interference* that task i can experience in any time interval [t, wi )

*The condition for maximum interference obviously occurs when all higher priority tasks are released at the same time as task i (in other words, at a critical instant)


Counting Releases

Consider task i and a higher priority task j now, the number of releases of task j in time

interval 0 to wi can be derived as follows:

# of releases j = wi/pj (2)

ceiling function[ 0, wi )

0 100 200

i

j 40

40

40

5

pipj

# of releases = 2

wi


Determining Interference

From this, the maximum interference of task j on task i in interval 0 to wi is given by:

max interference (task j) = wi/pj ej (3)

but there may be other higher priority tasks,

therefore:

Ii = wi/pj ej (4)

[ 0, wi )

jhp(i)


Calculating Response Time

Substituting equation (4) into (1) gives the general expression for response time:

wi = ei + wi/pj ej (5)

Note the following issue (problem):

wi is on both sides of this “strange” equation

Before we attempt to solve this equation, let’s

first see if we can understand it

jhp(i)


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


Solving the Response Time Equation

Essentially, form a recursive relationship with equation (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 wi

n+1

if win+1 = wi

n , then solution found -> wi = win

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

jhp(i)


Response Time Analysis Example

Given T = { (7,3), (12, 3), (20,5) }

Use RMA scheduling Apply the utilization based schedulabilty test If necessary, use time-demand analysis to

determine whether the system is schedulable

exercise to be completed in classexercise to be completed in class


Schedulability & Asynchronous Tasks

The simple task model that we have been able to deal with thus far is restrictive in several ways. Not being able to handle asynchronous tasks is a major restriction.

One approach is to make asynchronous tasks resemble periodic tasks consider that an asynchronous task’s minimum inter-

arrival time can be treated like a period, T with just this assumption one can use RMA to extend to

both types of tasks, BUT


Deadline Monotonic Priority Ordering

The simple task model assumption that D=p is also unrealistic for most asynchronous tasks often, an asynchronous task will occur infrequently (large

inter-arrival time) but must be serviced quickly -> D < p therefore priority assignment based upon p will usually

not satisfy the requirement to meet D

The deadline monotonic priority ordering (DMPO) scheme is introduced as follows: the shorter the task deadline, the higher the priority


DMPO Example

Task ei pi Di i wi *

1 3 20 5 1 3 2 3 15 7 2 6

3 4 10 10 3 10 4 3 20 20 4 20

* Response times are calculated using the methods presented earlier using equation (6)

Note that a rate monotonic ordering would not schedule these tasks


Schedulability & BP Servers

Recall that a bandwidth preserving server is a type of polling server that permits the budget of the server to be preserved until later in the period should an asynchronous job arrive

We described two general classes of bandwidth preserving server: deferrable server sporadic server

How do we determine the schedulability of the system in the presence of bandwidth preserving servers?


Recall: Deferrable Servers

A deferrable server is a bandwidth preserving server with period ps and execution budget es which preserves its budget if idle, and replenishes its budget fully every period

Budgets are not allowed to accumulate from period to period

Since the deferrable server is purely periodic, it can be assigned priority along with all other periodic tasks … and, for reasons that will now be explained, we must

ensure that the priority of the deferrable server is higher than all other periodic tasks


Recall the Deferrable Server Example

The schedule for this system is given below:T= {(2, 3.5, 1.5, 3.5), (6, 0.5) }TDS = (3,1)

A1 = (2.5, 2)

0 2 4

T1

T2

TDS

6 8

1/2

1/2

1 1

1/2

1/2 1 1/2

1/2

2.5 5.5

Note: = {TDS, T1, T2}


Critical Instants & Systems with Deferrable Servers

In order to use time demand analysis on a system with a deferrable server, a critical instant, t0, must be found as follows: at t0 a job of task i is released

a job in every higher priority task is also released at t0

a (higher priority) deferrable server has: a full budget es at t0

an asynchronous job is released at t0,

a full asynchronous backlog exists for an extended time (beyond wi), and

the next replenishment is at t0 + es


Critical Instants in Systems with Deferrable Servers

To understand the critical instant in the presence of a deferrable server, study the graph below note how the worst case release time for any periodic

task occurs when TDS has (and uses) its full budget just prior to getting a full replenishment

t0

T1

T2

TDS es es es es es


Time-Demand Analysis with Deferrable Servers (Di ≤ pi)

Understanding this formula:

wi = ei + es + (wi-es) / ps es + wi/pk ekkhp(i), excluding TDS

execution of deferrable server immediately following critical instant





interference of the deferrable server, TDS, in the interval (t0+es , wi]





interference of all k higher priority tasks, excluding the deferrable server, TDS


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 all aperiodic 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)}


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 A1A2


Exercise Solution

determine the response time of A1

wA1 = 8.5 - 0.5 = 8

wA2 = 8.75 - 2.5 = 6.25

determine the worst case response time of all periodic tasks

w11 = e1 + es + (w1

0-es) / ps es where w10 = 1.5

= 1.5 + 1 + (1.5 -1) / 4 (1) = 3.5w1

2 = 2.5 + (3.5 -1) / 4 (1) = 3.5

therefore w1 = 3.5 D1 (worst-case)


Exercise Solution

determine the worst case response time of all periodic tasks (continued)

w21 = e2 + es + (w2

0-es) / ps es + (w20) / p1 e1 w2

0 = 2

= 2 + 1 + (2 -1) / 4 (1) + 2 / 5 (1.5) = 5.5w2

2 = 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-case conditions, we can not guarantee the system schedulable


References

[1] Liu, J.W.S., “Real-Time Systems”, Prentice-Hall, 2000.

[2] Burns, A. and Wellings, A., “Real-Time Systems and Programming Languages”, Chapter 13, Addison Wesley, 1997

[3] TimeSys Corp, “The Concise Handbook of Real-Time Systems”, Version 1.0, 1999


Documents

Real Time Operating Systems Schedulability - Part 2 Course originally developed by Maj Ron Smith 12/20/2015Dr Alain Beaulieu1