Real-Time Systems,
COSC-4301-01,
Lecture 2
Stefan Andrei
7/29/2017
COSC-4301-01, Lecture 2
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Reminder of last lecture
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Introduction in Real-Time Systems:
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Chapter 1 of [Cheng; 2002];
Chapter 1 of [Kopetz; 1997];
Chapter 1 of [Stankovic, Spuri, Ramamritham,
Buttazzo; 1998])
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Overview of this lecture
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Real-Time Scheduling and Schedulability
Analysis
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Schedulability test
Schedulability utilization
Optimal scheduler
Determining computation time
Uniprocessor scheduling
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Scheduling preemptable and independent tasks
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Fixed-priority schedulers: RM, DM
Dynamic-priority schedulers: EDF, LL
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Scheduling
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General definition (uniprocessor, multiprocessor,
distributed system):
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Given a set of (computer) tasks (a.k.a., processes),
scheduling is to determine when to execute which task, thus
determining the execution order of these tasks.
Example: T1=‘fetch children to school’, T2=‘do classes’,
T3=‘do shopping’ by the same person (uniprocessor).
For the multiprocessors and distributed systems, scheduling
means also to determine an assignment of these tasks to a
specific person in a team of people.
Example: T1 and T3 are done by processor P1 (e.g.,
spouse), and T2 is done by processor P2 (multiprocessor).
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Scheduling (cont)
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Is a central activity of a computer system, usually
performed by the operating system.
Is also necessary in many non-computer systems (e.g.,
assembly lines).
In non-real-time systems, the typical goal of scheduling
is to maximize average throughput (number of tasks
completed per unit time) and/or to minimize average
waiting time of the tasks.
In the case of real-time scheduling, the goal is to meet
the deadline of every task by ensuring each task can
complete execution by its specified deadline.
The deadline is obtained from the environmental
constraints imposed by the application.
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Scheduling analysis
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Is to determine whether a specific task of a
set of tasks satisfying certain constraints can
be successfully scheduled (completing
execution of every task by its specified
deadline) using a specific scheduler.
Schedulability test: validate whether a given
application can satisfy its specified deadlines
when scheduled according to a specific
scheduling algorithm.
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Schedulability test
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Is often done at compile time, before the computer
system and its tasks start their execution.
If the test can be performed efficiently, then it can be
done at run-time as an on-line test.
Types of schedulers:
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Compile-time (a.k.a., static) schedulers
Run-time (a.k.a., on-line or dynamic) schedulers.
Schedulable utilization: is the maximum utilization
allowed for a set of tasks that will guarantee a feasible
scheduling of this task set.
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Hard and soft real-time systems
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Hard real-time system:
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Requires that every task or task instance
completes its execution by its specified deadline;
Failure to do so even for a single task or task
instance may lead to catastrophic consequences.
Soft real-time system:
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Allows some tasks or task instances to miss their
deadlines, but a task or task instance that misses
its deadline may be less useful or valuable to the
system.
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Optimal scheduler
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Is one which may fail to meet the deadline of
a task only if no other scheduler can meet its
deadline.
Examples include uniprocessor earliestdeadline-first (EDF) and least-laxity-first (LLF)
schedulers for independent tasks with no
synchronization constraints and no resource
requirements.
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Real-time scheduling. Task’s
characterization
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Characterization of a task:
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c: computation time (a.k.a., WCET – Worst Case Execution
Time)
S: start time (a.k.a., release or ready time)
d: deadline (i.e., relative to the start time)
p: period (a.k.a., minimum separation)
Hence, a task T can be denoted as (S, c, d, p)
Some characterizations include also:
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D: absolute deadline (a.k.a., wall clock time deadline).
Usually D = S + d.
Thus, a task T can be denoted as (S, c, d, D, p)
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Types of tasks
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Single-instance task: executes only once.
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Periodic task: has many instances or iterations, and
there is a fixed period between two consecutive
releases of the same task.
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Example: a periodic task may perform signal processing of a
radar scan once every 2 seconds, so the period of this task is
2 seconds.
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Types of tasks (cont)
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Sporadic task: has zero or more instances, and
there is a minimum separation between two
consecutive releases of the same task.
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Example: a sporadic task may perform emergency
maneuvers of an airplane when the emergency button is
pressed, but there is a minimum separation of 20 seconds
between two emergency requests.
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An aperiodic task is a sporadic task with either a soft
deadline or no deadline.
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Determining S, d, p, and c
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The application and the environment in which the
application is embedded are main factors determining
S, d, and p.
c, the computation time of a task, is dependent on its
source code, object code, execution architecture,
memory management policies, and actual number of
page faults and I/O.
c is not just an upper bound on the execution time of
task code without interruption, but it has to include
the time of the central processing unit (CPU) for
handling page faults, I/O requests, etc.
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Determining c
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Is crucial to successfully scheduling it in a real-time
system.
An overly pessimistic estimation of c will result in
wasted CPU time, whereas an under-approximation
would result in missed deadlines.
First solution: testing the tasks and use the largest
value of c during these tests.
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Disadvantage: the largest value seen during testing may not
be the largest observed in the working system.
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Determining c
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Second solution: analyzing the source code (an
advantage is that these methods are safe).
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Disadvantage: using an overly simplified model of the CPU
may result in over-approximating c.
Third solution: running the tasks in systems with
several levels of memory components (e.g., cache,
main memory).
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Disadvantage: there are restrictions in their models and thus
the proposed analysis techniques cannot be applied in
systems not satisfying their constraints.
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Determining c
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Fourth solution: use a probability approach to model
WCET of a process.
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Main idea: model the distribution of c and use it to compute
a confidence level for any given c.
Advantage: in soft real-time systems, if the designer wants a
confidence of 99% on the estimation for WCET, he/she can
determine which WCET to use from the probability approach.
Disadvantage: is not recommended in hard real-time
systems as WCET is just an approximation.
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Scheduling
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Uniprocessor scheduling
Multiprocessor scheduling
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Uniprocessor scheduling. Basic
Assumptions
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Compile-time/static (i.e., priority of tasks do
not change during run-time);
Preemptive tasks (i.e., tasks can be
interrupted and resumed later);
No precedence constraints (i.e., all are
independent);
Negligible context-switching time;
Periodic tasks.
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Rate-Monotonic (RM) scheduling
strategy
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Is a fixed and static priority scheduling
strategy (Rate-Monotonic Scheduler - RMS).
The task’s priority is task’s fixed period.
RMS executes at any time instant the
instance of the ready task with the shortest
period first.
More formal, task Ji has a higher priority than
task Jk if and only if pi < pk.
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RM-scheduling strategy. Example
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J1: S1 = 0, c1 = 2, p1 = d1 = 5
J2: S2 = 1, c2 = 1, p2 = d2 = 4
J3: S3 = 2, c3 = 2, p3 = d3 = 20
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Figure 3.1 from [Cheng; 2005], page 45
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RM-scheduling strategy. Considerations
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U = c1/p1+ … + cn/pn
If a task set is RM-schedulable, then U ≤ 1.
So, if U > 1, then the task set is not RMschedulable.
In general, the RM-scheduling strategy is not
optimal because there exist schedulable task
sets that are not RM-schedulable.
There is a special class of periodic tasks sets
for which the RM-scheduling is optimal: their
periods pi are multiples of each other.
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RM-schedulability test 1
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Given a set of n independent, preemptable,
and periodic tasks on a uniprocessor such
that di ≥ pi and pi are multiples of each other,
let U be the total utilization of this tasks set
(U = c1/p1+ … + cn/pn).
The task set is RM-schedulable if and only if
U ≤ 1.
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RM-schedulability test 1. Example
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J1: S1 = 0, c1 = 1, d1 = p1 = 4
J2: S2 = 0, c2 = 1, d2 = p2 = 2
J3: S3 = 0, c3 = 2, d3 = p3 = 8
pi are exact multiples of each other (p2 < p1 <
p3, p1 = 2p2, p3 = 4p2).
U = c1/p1+ c2/p2 + c3/p3 = 1/4 +1/2 + 2/8 = 1.
By applying RM-Schedulability Test 1, then
this tasks set is RM-schedulable.
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RM-schedulability test 2
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Liu and Layland’s sufficient condition (1973):
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Given a set of n independent, preemptable, and
periodic tasks on a uniprocessor, let U be the total
utilization of this tasks set.
The task set is schedulable if U ≤ n(21/n – 1).
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Remarks:
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For n=2, n(21/n – 1) equals to 0.8284…
For n=3, n(21/n – 1) equals to 0.7797…
limnn(21/n – 1) = ln 2 ≈ 0.6931…
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RM-schedulability test 2. Example
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J1: S1 = 0, c1 = 1, p1 = 4
J2: S2 = 0, c2 = 1, p2 = 5
J3: S3 = 0, c3 = 3, p3 = 10
U = c1/p1+ c2/p2 + c3/p3 = 1/4 +1/5 + 3/10 = 0.75
By applying RM-Schedulability Test 2, this task
set is RM-schedulable.
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RM-schedulability test 2. Example
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J1: S1 = 0, c1 = 1, p1 = 4
J2: S2 = 0, c2 = 1, p2 = 3
J3: S3 = 0, c3 = 2, p3 = 6
U = c1/p1+ c2/p2 + c3/p3 = 1/4 +1/3 + 2/6 = 0.91666…
The RM-Schedulability Test 1 cannot be applied since
periods are not exact multiples of each other.
The RM-Schedulability Test 2 cannot be applied since
U > 3(20.33-1) ≈ 0.7797.
However, this tasks set is RM-schedulable…
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RM-schedulability test 2. Example (cont)
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J1: S1 = 0, c1 = 1, p1 = 4
J2: S2 = 0, c2 = 1, p2 = 3
J3: S3 = 0, c3 = 2, p3 = 6
J1
J2
J3
0 1 2 3 4 5 6 7 8 9 10 11 12
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RM-schedulability test 3
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Hence, it makes sense to look for stronger analytical
conditions to ensure the schedulability without running
the task set.
Necessary and sufficient condition requires checking
inequalities that depend on time t.
Let J1, …, Ji sorted increasingly according to their
period.
Let wi(t)=Σk=1i ck t / pk, 0 < t ≤ pi
Task Ji is RM-schedulable if and only if wi(t) ≤ t,
for a time instant t, t = k pj, j = 1, …, i, k = 1, …, pi/pj
If di ≠ pi, we replace pi by min(di, pi) above.
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RM-schedulability test 3. First Example
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We sort the jobs according to their period:
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J1: S1 = 0, c1 = 1, p1 = 3
J2: S2 = 0, c2 = 1, p2 = 3
J3: S3 = 0, c3 = 1, p3 = 3
U = 1/3 + 1/3 + 1/3 = 1, so Test 2 cannot be
applied.
We need to check the following inequalities:
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w1(3) ≤ 3,
w2(3) ≤ 3,
w3(3) ≤ 3.
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RM-schedulability test 3. First Example
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w1(3) = 1
w2(3) = 2
w3(3) = 3
Since i t such that wi(t) ≤ t, it means this
tasks set is RM-schedulable according to
Schedulability Test 3.
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RM-schedulability. Second Example
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1.
2.
Let us consider the following preemptable
task set T={J1, J2, J3}:
J1: S1 = 0, c1 = 1, p1 = d1 = 4
J2: S2 = 1, c2 = 2, p2 = d2 = 6
J3: S3 = 3, c3 = 4, p3 = d3 = 10
compute the utilization rate.
analyze whether T is feasible using RMscheduling method. In the affirmative case,
provide a schedule.
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RM-schedulability. Second Example
The utilization rate:
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U = C1/P1 + C2/P2 + C3/P3 = 1/4 + 2/6 + 4/10 =
59/60 = 0.98
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Schedulability Test 1
For this test U ≤ 1.
But the task periods are not exact multiples
of each other, so Schedulability Test 1
cannot be applied.
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RM-schedulability. Second Example
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Schedulability test 2
For this test, we check whether the utilization
rate (U) is less or equal than n(21/n -1).
Here n = 3, so n(21/n -1) = 3(21/3-1) =
0.7797…
Since U > n(21/n -1), the Schedulability Test 2
cannot be applied (as 0.98 … > 0.7797…).
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RM-schedulability. Second Example
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For Schedulability test 3, we start with the job J1 that
has the smallest deadline.
For J1:
i = 1, j = 1, k = 1
Thus t = k * pj = 1 * 4 = 4
w1(t) = c1 [t / p1] = 1
Now, J1 is RM-schedulable if and only if w1(t) ≤ 4.
Therefore, J1 is RM-schedulable.
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RM-schedulability. Second Example
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For J2:
i = 2; j  {1, 2}, k = 1. Thus t = k * pj where k = 1.
For j = 1: t = 1 * p1 = 4.
For j = 2: t = 1 * p2 = 6.
Hence, we get the values of t  {4, 6}.
Substituting the respective values of t and j, we get
the following conditions (at least one should hold):
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c1 + c2 ≤ kp1, that is, 3 ≤ 4 or
2c1 + c2 ≤ kp2, that is, 4 ≤ 6.
(In fact, both of these conditions hold.)
So, J2 is RM-schedulable together with J1.
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RM-schedulability. Second Example
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For J3:
i = 3; j  {1, 2, 3}, k  {1, 2}, and t = k * pj.
For k = 1, j  {1, 2, 3}, we have t  {4, 6, 10}.
For k = 2, j  {1, 2, 3}, we have t  {8, 12, 20}.
Hence t  {4, 6, 10, 8, 12, 20}.
Substituting the values of t and j, we get:
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c1 + c2 + c3 ≤ 4, that is, 7 ≤ 4, or
2c1 + c2 + c3 ≤ 6, that is, 8 ≤ 6, or
2c1 + 2c2 + c3 ≤ 8, that is, 10 ≤ 8, or
3c1 + 2c2 + c3 ≤ 10, that is, 11 ≤ 10, or
3c1 + 2c2 + 2c3 ≤ 12, that is, 15 ≤ 12, or
5c1 + 4c2 + 2c3 ≤ 20, that is, 21 ≤ 20.
Since none of the above condition holds, J3 is not RMschedulable together with J1, and J2.
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The RM scheduling attempt
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It can be seen that J3 misses its deadline at t=53.
J1
J2
J3 misses its
deadline
J3
0
3
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13
19
23
28
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43
53
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Dynamic-priority schedulers
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Earliest Deadline First (EDF)
Least Laxity First (LLF)
Reminder: An optimal scheduler is one which
may fail to meet a deadline of a task only if
no other schedule can meet its deadline.
Both above strategies are optimal.
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Earliest Deadline First (EDF or ED)
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Executes at every instant the ready task with
the earliest absolute deadline first (i.e., D).
Reminder: D = S + d.
If two tasks or more have the same deadlines,
EDF randomly selects one for execution next.
EDF is a dynamic-priority scheduler since task
priorities may change at run-time depending
on the nearness of their absolute deadline.
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EDF terminology
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[Krishna & Shin, 1997] call EDF a deadlinemonotonic (DM) scheduling algorithm;
[Liu, 2000] defines DM algorithm as a fixedpriority scheduler that assigns higher
priorities to tasks with shorter relative
deadlines.
[Cheng, 2002] and we use EDF or DM to
refer to dynamic-priority scheduling algorithm.
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Example. FIFO scheduler fails
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Four single-instance tasks (for simplicity):
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J1: S1 = 0, c1 = 4, D1 = 15
J2: S2 = 0, c2 = 3, D2 = 12
J3: S3 = 2, c3 = 5, D3 = 9
J4: S4 = 5, c4 = 2, D4 = 8
A FIFO (First-In-First-Out) scheduler (often
used in non-real-time operating systems)
executes the tasks in the order of their arrivals
and their deadlines are not considered.
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Example. FIFO scheduler fails
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The FIFO (First-In-First-Out) scheduler gives
an infeasible schedule (J3 misses its deadline
– D3 = 9; also, J4 misses its deadline – D4 =
8).
J1
J2
J3
J4
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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Example. EDF scheduler works
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At time 0, since D1 > D2, then J2 has a higher
priority than J1;
At time 2, since D3 < D2, then J2 is preempted
and J3 begins execution;
At time 5, since D4 < D3, then J3 is preempted
and J4 begins execution, and so on, so forth.
J2
J3
J4
J3
J2
J1
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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EDF algorithm is optimal
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Theorem [Dertouzos; 1974]. Given a set S of
independent (no resource contention or
precedence constraints) and preemptable
tasks with arbitrary start times and deadlines
on a uniprocessor, the EDF algorithm yields a
feasible schedule for S if and only if S has
feasible schedules.
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[Dertouzos; 1974] M.L. Dertouzos, Control Robotics: the
Procedural Control of Physical Processes, Information
Processing 74, North-Holland Publishing Company, 1974
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Least-Laxity-First (LL or LLF)
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Like EDF, the LL strategy is another run-time optimal
scheduler.
It is also known as Minimum-Laxity-First (MLF) or LeastSlack-Time-First (LST) algorithm.
Let c(i) denote the remaining computation time of a task
at time i.
Let d(i) denote the deadline of a task relative to the
current time i.
The laxity (or slack) of a task at time i is d(i)-c(i).
The laxity is the maximum time a task can delay
execution without missing its deadline in the future.
The LL scheduler executes at every instant the ready
task with the smallest laxity.
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Example: LL scheduling
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J1: S1 = 0, c1 = 2, d1 = 6 = p1
J2: S2 = 0, c2 = 3, d2 = 7 = p2
J3: S3 = 0, c3 = 2, d3 = 11 = p3
At i=0:
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c1(0)=2, d1(0)=6-i=6, l1(0)=6-2=4
c2(0)=3, d2(0)=7-i=7, l2(0)=7-3=4
c3(0)=2, d3(0)=11-i=11, l3(0)=11-2=9
The LL scheduler chooses the minimum laxity task (say J1).
At i=1:
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c1(1)=1, d1(1)=6-i=5, l1(1)=5-1=4
c2(1)=3, d2(1)=7-i=6, l2(1)=6-3=3
c3(1)=2, d3(1)=11-i=10, l3(1)=10-2=8
The LL scheduler chooses J2 since l2(1) is less l1(1) and l3(1).
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LL is optimal
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[Mok; 1983] The Least Laxity First Algorithm is
optimal for preemptable and independent tasks.
However, the LL algorithm has the disadvantage
of a potentially very large number of
preemptions, and it is no longer optimal if
preemption is not allowed [George et al.; 1996].
[Mok; 1983] A.K. Mok, Fundamental Design Problems of
Distributed Systems for the Hard Real-Time Environment, Ph.D.
Dissertation, MIT, 1983.
[George et al.; 1996] L. George, N. Rivierre, M. Spuri, Preemptive
and Non-Preemptive Real-Time Uni-Processor Scheduling,
Rapport de Recherche RR-2966, INRIA, France, 1996.
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Schedulability Test 4
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Let ci denote the computation time of task Ji.
For a set of n periodic preemptable tasks such that
the relative deadline di of each task is equal to or
greater than its respective period pi (di ≥ pi), a
necessary and sufficient condition for feasible EDF
and LL scheduling of this task set on a uniprocessor
is: U = c1/p1+ … + cn/pn ≤ 1.
Remark: for a task set containing some tasks whose
relative deadlines di are less than their periods, no
easy schedulability test exists with a necessary and
sufficient condition.
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Schedulability test 5

A sufficient condition for feasible EDF and LL
scheduling of a set of independent,
preemptable, and periodic tasks on a
uniprocessor is
c1/min(d1,p1)+ … + cn/min(dn,pn) ≤ 1.

Remarks:
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The term ci/min(di,pi) is the density of task Ji.
If the deadline and the period of each task are
equal (di = pi), then Schedulability Test 5 is the
same as Schedulability Test 4.
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Example 1
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J1: S1 = 0, c1 = 2, d1 = 6 = p1
J2: S2 = 0, c2 = 3, d2 = 7 = p2
J3: S3 = 0, c3 = 2, d3 = 11 = p3
U = 2/6 + 3/7 + 2/11 = 218/232 < 1 and di = pi
for any i.
Hence, this task set is feasible (according to
Schedulability Test 4).
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Example 2
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J1: S1 = 0, c1 = 1, d1 = 4, p1 = 5
J2: S2 = 0, c2 = 2, d2 = 5, p2 = 4
J3: S3 = 0, c3 = 1, d3 = 5, p3 = 5
Since d1 < p1, Schedulability Test 4 is not
applicable.
c1/min(d1,p1)+ c2/min(d2,p2) + c3/min(d3,p3) = 1/4
+ 2/4 + 1/5 = 19/20 < 1.
Since the above number is smaller than 1,
according to Schedulability Test 5, the above
task set is EDF and LL schedulable.
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Example 3
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J1: S1 = 0, c1 = 2, d1 = 6, p1 = 7
J2: S2 = 0, c2 = 3, d2 = 7, p2 = 6
J3: S3 = 0, c3 = 2, d3 = 11, p3 = 11
Since d1 < p1, Schedulability Test 4 is not
applicable.
c1/min(d1,p1)+ c2/min(d2,p2) + c3/min(d3,p3) =
2/6 + 3/6 + 2/11 = 67/66 > 1.
Schedulability Test 5 cannot be applied, as the
above number is greater than 1.
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Schedulability test 6
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[Krishna & Shin, 1997] Given a set of n independent,
preemptable, and periodic tasks on a uniprocessor, let:

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U be the total utilization of this tasks set (U = c1/p1+ … + cn/pn),
dmax be the maximum relative deadline among these tasks’
deadlines,
P is the least common multiple (LCM) of these tasks’ periods,
s(t) be the sum of the computation tasks with absolute
deadlines less than t.
This task set is not EDF-schedulable if either of the
following conditions holds:
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U>1
 t < min(P+dmax, (U/(1-U)) max1≤i≤n (pi-di)) such that s(t)>t.
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Revisiting Example 3
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Let us consider the following preemptable
task set T={J1, J2, J3}:
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J1: S1 = 0, c1 = 2, d1 = 6, p1 = 7
J2: S2 = 0, c2 = 3, d2 = 7, p2 = 6
J3: S3 = 0, c3 = 2, d3 = 11, p3 = 11
We study the applicability of Schedulability
Test 6:
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U = 2/7 + 3/6 + 2/11 = 0.9675…
The minimum t such that the inequality s(t) > t is
29 (the LCM is 462, etc).
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Revisiting Example 3
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We check whether  t such that s(t) > t, for
some t  {1, …, 29}.
5
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10
15
20
25
29
Since s(t) t, for any t  {1, …, 29}, then Test 6 checks
that the above task set could be EDF-schedulable.
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Comparing fixed and dynamic-priority
schedulers
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RM and DM algorithms are fixed-priority schedulers.
EDF and LL algorithms are dynamic-priority
schedulers.
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Reminder:
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A fixed-priority scheduler assigns the same priority to all
instances of the same task.
A dynamic-priority scheduler may assign different priorities
to different instances of the same task.
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Comparing fixed and dynamic-priority
schedulers
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No optimal fixed-priority scheduling algorithm
exists since we can find a schedulable task
set that cannot be scheduled by a fixedpriority algorithm.
Both EDF and LL algorithms are optimal
dynamic-priority schedulers for preemptable
tasks.
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Example 1
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J1: S1 = 0, c1 = 5, p1 = 10 = d1
J2: S2 = 0, c2 = 12, p2 = 25 = d2
Since U = 5/10 + 12/25 = 0.98 ≤ 1,
Schedulability Test 4 is satisfied, this task set is
feasible (EDF-schedule is next slide).
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Example 1. EDF schedule
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LCM(10,25)=50
At time 0, D1 = d1 + S1 = 10 and D2 = d2 + S2 = 25, so J1
has a higher priority than J2.
At time 5, J2 is the only task ready to begin.
At time 10, D1 = d1 + S1 = 20 and D2 = d2 + S2 = 25, so
J1 has a higher priority than J2.
At time 15, J1 is the only task ready to begin.
At time 20, D1 = d1 + S1 = 30 and D2 = d2 + S2 = 25, so
J2 has a higher priority than J1…
 Figure 3.5 from [Cheng; 2005], page 53
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Example 1 (cont)
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The RM scheduler leads to an infeasible schedule
(figure below).
J1 has a shorter period and thus a higher priority, so it
is always scheduled first (before J2).
The first instance of J2 is allocated only 10 time units
before its deadline of 25, so it finishes at time 27,
causing a miss of its deadline at 25.
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Figure 3.6 from [Cheng; 2005], page 54.
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Example 2
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J1: S1 = 0, c1 = 4, p1 = 10 = d1
J2: S2 = 0, c2 = 13, p2 = 25 = d2
Since U = 4/10 + 13/25 = 0.92 ≤ 1, Schedulability
Test 4 is satisfied, this task set is feasible (EDFschedule is below).
LCM(10,25)=50.

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Figure 3.8 from [Cheng; 2005], page 55.
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Example 2 (cont)
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The RM scheduler leads now to a feasible schedule.
Previously, J2 missed its deadline at time 25, so after
decreasing c1 from 5 to 4, J2 can be executed by time
25.
In fact, since 1 time unit remains available (between
time 24 and 25), c2 can be increased from 12 to 13.

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Figure 3.7 from [Cheng; 2005], page 55.
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Additional constraints
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May complicate the scheduling of tasks with
deadlines:
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Frequency of tasks requesting service periodically;
Precedence relations among tasks and subtasks;
Resources shared by tasks;
Whether task preemption is allowed or not (if tasks
are preemptable, we assume a task can be
interrupted only at discrete (integer) time instants).
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Summary
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Real-Time Scheduling and schedulability
analysis
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Schedulability test
Schedulability utilization
Optimal scheduler
Determining computation time
Uniprocessor scheduling
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Scheduling preemptable and independent tasks
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Fixed-priority schedulers: RM, DM
Dynamic-priority schedulers: EDF, LL
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Reading suggestions
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From [Cheng; 2002]
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Chapter 3 (Introduction, Sections 3.1, 3.2.1)
Chapters 3, 10 and 11 of [Kopetz; 1997]
Chapter 2 of [Stankovic, Spuri, Ramamritham,
Buttazzo; 1998]
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Coming up next
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Sporadic tasks
Scheduling nonpreemptable tasks
Scheduling nonpreemptable sporadic tasks
Scheduling nonpreemptable tasks with
precedence constraints
Communicating periodic tasks: deterministic
rendezvous model
Chapter 3 of [Cheng; 2002], Sections 3.2.1,
3.2.2, 3.2.3, 3.2.4, 3.2.5
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Thank you for your attention!
Questions?
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