Real-Time Systems,
COSC-4301-01,
Lecture 3
Stefan Andrei
7/31/2017
COSC-4301-01, Lecture 3
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Reminder of Last 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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Overview of this lecture
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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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Sporadic tasks
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Sporadic tasks may be released at any time
instant, but a minimum separation exists
between releases of consecutive instances of
the same sporadic task.
Can be modeled by periodic tasks with
periods equal to the minimum separation.
In this way, we can re-use the previous
scheduling algorithms.
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Sporadic tasks (comparison with
periodic tasks)
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Unlike periodic tasks, sporadic tasks are released
irregularly or may not be released at all.
So, even though the scheduler allocates a time slice
to the periodic equivalent of a sporadic task, this
sporadic task may not actually be released.
When this sporadic task does request service, it
immediately runs if its release time is within its
corresponding scheduled time slice.
Otherwise, it waits for the next scheduled time slice
for running its periodic equivalent.
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Example 1
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Two periodic tasks J1 and J2 arriving at time 0, and
one sporadic task J3 with minimum separation time
of 60.
J1: c1 = 10, p1 = 50
J2: c2 = 15, p2 = 80
J3: c3 = 20, p3 = 60
A RM schedule is given below:
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Figure 3.9 from [Cheng; 2005], page 56
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The second approach
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Treat sporadic tasks as one periodic task JS with the
highest priority and a period chosen to accommodate
the minimum separations and computation
requirements of this collection of sporadic tasks.
Any sporadic task may run within the time slices
assigned to JS while the other periodic tasks run
outside of these time slices.
Example: JS: cS = 20, pS = 60
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Figure 3.10 from [Cheng; 2005], page 57
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The third approach
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Deferred server (DS) [Lehosczky, Sha, Strosnider;
1987]
DS is similar to second approach, with the
modification:
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The periodic task corresponding to the collection of sporadic
tasks is the deferred server.
When no sporadic task waits for service during a time
slice assigned to sporadic tasks, the processor runs
the other (periodic) tasks.
If a sporadic task is released, then the processor
preempts the currently running task and runs the
sporadic task for a time interval up to the total time
slice assigned to sporadic tasks.
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Deferred server technique. Example
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JS: cS = 20, pS = 60
Deferred server scheduler allocates 20 time units to
sporadic tasks every 60 time units.
Suppose a sporadic task J1 arrives with c1 = 30 at
time 20.
Since only 20 time units are available in each period
of 60 time units, it is scheduled to run from time 20
and preempted at time 40 (periodic tasks may run for
the rest of 20 units).
Then at time 60 (the start of second period of 60 time
units), J1 is scheduled to run for the other remaining
10 time units.
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Deferred server technique. Example (cont)
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Suppose a sporadic task J2 arrives with c2 = 50 at time 100
(because J1 is also a sporadic task that uses 10 time units
from the time of 20 units allocated to 60-120).
Since only 10 time units are still available in the second
period of 60 time units, it is immediately scheduled to run
(for period 60-120, only 20 units are deferred to sporadic
tasks).
At time 110, it is preempted (other periodic tasks may run).
At time 120, J2 runs for 20 time units, and at time 180, J2
runs for the remaining 20 time units.
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Figure 3.11 from [Cheng; 2005], page 57
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Analytical approach of sporadic tasks
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For a deferred server with an arbitrary priority
in a system of tasks scheduled using RM
algorithm, no schedulable utilization is known
that guarantees a feasible scheduling of the
system.
However, for the special case in which DS has
the shortest period among all tasks (that is, DS
has the highest priority), a schedulable
utilization exists: Schedulability Test 7
[Lehosczky, Sha, Strosnider; 1987],
[Strosnider, Lehosczky; 1995]
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Schedulability test 7
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Let pS and cS be the period and allocated time
for the DS.
Let US=cS/pS be the utilization of the server.
A set of n independent, preemptable, and
sporadic tasks with relative deadlines the same
as the corresponding periods on a
uniprocessor such that pS < p1 < …< pn < 2pS
and pn > pS + cS, is RM-schedulable if the total
utilization of this task set is at most U(n) = (n-1)
[((US+2)/(US+1))(1/(n-1))-1].
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Schedulability test 7. Example 1
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J1: c1 = 10, p1 = 50
J2: c2 = 15, p2 = 70
JS: cS = 10, pS = 40
US=cS/pS = 0.25
pS < p1 < …< pn < 2pS holds (n=2).
pn > pS + cS holds.
U(n) = (n-1) [((US+2)/(US+1))(1/(n-1))-1]=0.80
The total utilization of this task set is U=0.664, that is
less than U(n).
According to Schedulability test 7, the above task set
is RM-schedulable.
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Schedulability test 7. Example 2
Let us consider the following preemptable task set
(JS is a sporadic task):
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J1: S1 = 0, c1 = 1, p1 = d1 = 5
J2: S2 = 0, c2 = 2, p2 = d2 = 7
JS: SS = 0, cS = 1, pS = dS = 4
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1.
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compute the utilization rate.
check the applicability of RM-scheduling method
based on schedulability test 7.
run the RM-scheduler for the above task set.
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Schedulability test 7. Example 2
U = 0.735…
According to schedulability test 7, the following
conditions should be hold:
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ps<p1<p2<…< pn< 2ps translates to 4 < 5 < 7 < … < 8 (TRUE)
pn> ps+ cs translates to 7 > 4 + 1 (TRUE)
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Also, U(n) = (n-1) [((US+2)/(US+1))(1/(n-1))-1], where
US=1/4=0.25.
Therefore U(n) = 0.80, so U < U(n) (TRUE)
Hence, schedulability test 7 can be applied to the
above task set.
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Schedulability test 7. Example 2
The RM schedule is:
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J1
J2
Js
0
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Scheduling non-preemptable tasks
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So far we have assumed that tasks can be
preempted at any integer time instants.
In practice, tasks may contain critical sections
that cannot be interrupted.
Examples of critical sections:
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Access and modify shared variables;
Use shared resources (e.g., disks, printer).
An important goal is to reduce task waiting
time and context-switching time – [Lee,
Cheng; 1994].
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Scheduling non-preemptable tasks
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Using fixed-priority schedulers for non-real-time tasks
may potentially lead to the priority inversion problem
[Sha, Rajkumar, Lehoczky; 1990] which occurs when
a low-priority task with a critical section blocks a
higher-priority task for an unbounded or long period of
time.
The EDF and LL algorithms are no longer optimal if
the tasks are not preemptable.
However, if all the tasks start at time 0, then EDF
technique is still optimal.
So, an EDF algorithm may fail to meet a deadline of a
task set S even if another scheduler can produce a
feasible schedule for S (if starting time is not 0).
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Optimality for non-preemptable tasks
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Without preemption, we cannot transform a
non-EDF schedule into an EDF schedule by
interchanging computation blocks of different
tasks.
No priority-based scheduling algorithm is
optimal for non-preemptable tasks with
arbitrary start times, computation times, and
deadlines, even on a uniprocessor [Mok;
1984].
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Scheduling non-preemptive task sets
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When preemption is not allowed, the
scheduling problem is known to be NPcomplete.
However, if only non-idling schedulers are
considered (all tasks have their start times =
0), the problem is again tractable:
(Theorem 3.4, page 30 from [Stankovic et.
Al., 1997])
Non-preemptive non-idling EDF is optimal.
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Scheduling non-preemptable sporadic
tasks
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We transform the sporadic tasks into
equivalent periodic tasks.
A sporadic task (c, d, p) can be transformed
into and scheduled as a periodic task (c’, d’,
p’) if the following conditions hold:
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d ≥ d’ ≥ c
c’ = c
p’ ≤ d – d’ + 1
A proof can be found in [Mok; 1984].
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Schedulability test 8 [Mok; 1984]
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Let M = MP  MS be the set of all non-preemptable
tasks, where MP is the set of periodic tasks, and MS is
the set of sporadic tasks.
Let the nominal (initial) laxity li of task Ti be di -ci.
Each sporadic task Ti be (ci, di, pi) is replaced by an
equivalent periodic task T’i = (c’i, d’i, p’i) as follows:
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c’i = ci, d’i = ci, p’i = min(pi, li+1).
If we can find a feasible schedule for the resulting set
M’ of periodic tasks, we can schedule M without
knowing in advance the start (release or request) time
of the sporadic tasks in MS.
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Schedulability test 8. Example
Let us consider the following nonpreemptable task set (JS is a sporadic task):
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J1: S1 = 0, c1 = 2, p1 = d1 = 5
J2: S2 = 0, c2 = 2, p2 = d2 = 8
JS: SS = 0, cS = 2, pS = dS = 8
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compute the utilization rate.
check the applicability of RM-scheduling
method based on schedulability test 8.
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Schedulability test 8. Example
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U = 0.9
The laxity: l = ds - cs = 8 – 2 = 6.
Now, replacing the sporadic task
Js = (cs, ds, ps)
with equivalent periodic task
Js’ = (cs’, ds’, ps’),
where cs’= cs=2, ds’= cs=2, ps’=min(ps, ls + 1) =
min(8, 6+1) =7.
So, the new sporadic task as an equivalent
periodic task is JS’: SS’ = 0, cS’ = 2, pS’=7, dS’= 2.
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Schedulability test 8. Example
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Let us check the condition for schedulability
test 8.
d>=d’ means 8>=2 (TRUE)
c’=c means 2=2 (TRUE)
p’ <= d-d’+1 means 7 <= 8-2+1 (TRUE)
Since all the conditions hold true,
schedulability test 8 can be applied to the
above task set.
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Scheduling non-preemptable tasks
with precedence constraints
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We introduce precedence and mutual exclusion (nonpreemption)
constraints to the scheduling problem for single-instance tasks (tasks
that are neither periodic nor sporadic) on a uniprocessor.
A task precedence graph shows the required order of execution of a
set of tasks.
 A node represents a task (e.g., Ti) and directed edges (e.g., ) indicate
the precedence relationships between tasks.
 An edge Ti Tj means that Ti must complete before Tj can start.
 For a task Ti,
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incoming edges from predecessor tasks indicate all these predecessor
tasks have to complete execution before Ti can start execution;
outgoing edges to successor tasks indicate that Ti must finish execution
before the successor tasks can start execution.
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Scheduling algorithm A for tasks with
precedence constraints
A topological ordering of the tasks in a precedence
graph shows one allowable execution order of these
tasks.
Suppose we have a set of n one-instance tasks with
deadlines, all ready at time 0 and with precedence
constraints described by a precedence graph.
Algorithm A:
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1.
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Sort the tasks in the precedence graph in topological order (so
tasks with no in-edges come first). If two or more tasks can be
listed next, select the one with the earliest deadline, ties are
broken arbitrarily;
Execute tasks one at a time following this topological order.
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Algorithm A. Example 1
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Consider the set of tasks {T1, T2, T3, T4, T5,
T6} with the following:
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precedence constraints: T1  T2, T1  T3, T2 
T4, T2  T6, T3  T5, T3  T4, T4  T6
Computation times and deadlines:
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T1: c1 = 2, d1 = 4,
T2: c2 = 3, d2 = 7,
T3: c3 = 2, d3 = 10,
T4: c4 = 8, d4 = 18,
T5: c5 = 6, d5 = 24,
T6: c6 = 4, d6 = 28.
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Algorithm A. Example 1 (cont)
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The precedence graph is:
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Figure 3.14 from [Cheng; 2005],
page 60
The EDF-schedule for tasks with precedence
constraints is:
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Figure 3.16 from [Cheng; 2005], page 61
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Algorithm B
Is a variation of Algorithm A by:
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Considering the task with the latest deadline first
Shifting the entire schedule toward 0.
Algorithm B:
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Sort the tasks according to their deadlines in non-decreasing
order and label the tasks such that d1 ≤ d2 ≤ … ≤ dn.
Schedule task Tn in the time interval [dn – cn, dn]
While there is a task to be scheduled do
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4.
Suppose S is the set of all unscheduled tasks whose successors
have been scheduled. Schedule as late as possible the task with
the latest deadline in S.
Shift the tasks toward time 0 while maintaining the execution
order indicated at step 3.
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Refinement of Algorithm B
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Given a task set T={T1, …, Tn}, let us denote by seti the “starting
execution time” for Ti: task Ti executes in time interval [seti, seti
+ ci], where ci is its computation time;
Sort the tasks according to their deadlines in non-decreasing
order and label the tasks such that d1 ≤ d2 ≤ … ≤ dn.
setn= dn – cn; // Schedule Tn in the time interval [dn – cn, dn]
for (i = n – 1; i > 0; i--) seti = min{di – ci, seti+1 – ci};
shift = set1;
for (i = 1; i <= n; i++) {
1. old_seti = seti;
2. seti = seti – shift; // Schedule Tn in the interval [seti, seti+ci]
3. if (i < n) shift = shift + seti+1 – (old_seti + ci);
}
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Algorithm B. Example 1
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We re-consider the same example from slide 19, so
the first three steps of Algorithm B gives:
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Figure 3.15 from [Cheng; 2005], page 61
The schedule for tasks with precedence constraints
after shifting tasks is:
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Figure 3.16 from [Cheng; 2005], page 61
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Algorithms A, B. Example 2
Let us consider the set of non-preemptable tasks T
= {J1, J2, J3, J4} with the following:
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precedence constraints: J1  J2, J1  J3, J2  J4, J3  J4
computation times and deadlines:
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J1: c1 = 2, d1 = 5,
J2: c2 = 3, d2 = 7,
J3: c3 = 2, d3 = 10,
J4: c4 = 8, d4 = 18,
compute the task precedence graph.
investigate the applicability of Algorithms A and B to T.
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Algorithms A, B. Example 2
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The task precedence graph is:
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Algorithm A. Example 2
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The EDF schedule for this task set with the
precedence constraints:
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Algorithm B. Example 2
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According to Algorithm B, the tasks are sorted in
non decreasing order and tasks are labeled such
that d1 < d2 < … < dn and tasks are scheduled in
interval [dn-cn].
Running the refinement of Algorithm B, we get:
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set4 = 10, set3 = 8, set2 = 4, set1 = 2
shift = 2, set1 = 0,
shift = 2, set2 = 2,
shift = 3, set3 = 5,
shift = 3, set4 = 7.
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Communicating periodic tasks:
deterministic rendezvous model
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Allowing tasks to communicate each other complicates
the scheduling problem.
Interprocess communication leads to precedence
constraints not only between tasks but also between
blocks within these tasks.
Example:
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Ada programming language provides the primitive
rendezvous() to allow one task to communicate with another
at a specific point during the task execution.
Ada is used in the implementation of a variety of embedded
and real-time systems, including airplane avionics.
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Deterministic rendezvous model
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If a task A wants to communicate with task B, then task
A executes rendezvous(B). Task A then waits until
task B executes a corresponding rendezvous(A).
As a result, this pair of rendezvous() imposes a
precedence constraint between the computations of
tasks A and B by requiring all the computations prior to
the rendezvous() primitive in each task be completed
before the computations prior to the rendezvous()
primitive in the other task can start.
To simplify the scheduling strategy, we consider the
execution time of a rendezvous() primitive is zero or
that its execution time is included in the preceding
computation block.
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Deterministic rendezvous model (cont)
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A one-instance task can rendezvous with another
one-instance task.
However, it is semantically incorrect to allow a
periodic task and a sporadic task to rendezvous with
each other since the sporadic task may not run at all,
causing the periodic task to wait forever for the
matching rendezvous.
Two periodic tasks may rendezvous with each other,
but there are constraints on the lengths of their
periods to ensure correctness.
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Deterministic rendezvous model (cont)
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Two tasks are compatible if their periods are
multiples of each other.
To allow two (periodic) tasks to communicate in any
form, they must be compatible.
One attempt to schedule compatible and
communicating tasks is to use the EDF scheduler to
execute the ready task with the nearest deadline that
is not blocked due to a rendezvous.
The solution [Mok; 1983] to this scheduling problem
starts by building a database for the runtime scheduler
so that EDF algorithm can be used with dynamically
assigned task deadline.
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Deterministic rendezvous model (cont)
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Let L be the longest period.
Since the communicating tasks are
compatible, L is the same as the LCM (Lowest
Common Multiple) of these tasks’ periods.
We denote a chain of scheduling blocks
generated in chronological order for task Ti in
interval (0, L) by Ti (1), Ti (2), …, Ti (mi).
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Deterministic rendezvous model (cont)
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If there is a rendezvous constraint with Ti targeting Tj
between
Ti (k) and Ti (k + 1),
Tj (l) and Tj (l + 1),
then the following precedence relations are specified:
Ti (k)  Tj (l + 1),
Tj (l)  Ti (k + 1),
Within each task, the precedence constraints between
blocks are:
Ti (1)  Ti (2)  …  Ti (mi).
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Deterministic rendezvous model.
Example
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T1: c1 = 1, d1 = p1 = 12.
T2: c2, 1 = 1, c2, 2 = 2, d2 = 5, p2 = 6.
T3: c3, 1 = 2, c3, 2 = 3, d3 = p3 = 12.
T2 must rendezvouz with T3 after the first
scheduling block.
T3 must rendezvouz with T2 after the first and
second scheduling blocks.
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Deterministic rendezvous model.
Example’s solution
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L = 12
The precedence constraints between blocks
are:
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T2(1)  T2(2)  T2(3)  T2(4)
T3(1)  T3(2)
Rendezvous constraints with T2(1) targeting
T3(1): T2(1)  T3(2), T3(1)  T2(2)
Rendezvous constraints with T2(2) targeting
T3(2): T2(2)  T3(3), T3(2)  T2(3)
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Deterministic rendezvous model.
Example’s solution
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The EDF scheduler will provide the schedule (the first and
second occurrences of T2,1 stands for T2(1) and T2(3),
respectively, etc):
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Figure 3.18 from [Cheng; 2005], page 63
The solution is infeasible because T2(2) should be before 5
time units, and T2(4) should be before 5+6=11 time units.
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Deterministic rendezvous model (cont)
Algorithm for revising the deadlines to fix the
previous infeasibility decision:
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3.
4.
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Sort the scheduling blocks in [0, L] in reverse
topological order, so the block with the latest
deadline appears first.
Initialize the deadline of the k-th instance of Ti,j
with (k-1)pi+di.
Let S and S’ be scheduling tasks. Then revise dS =
min(dS , {dS’ - cS’ : S S’}).
Use the EDF scheduler to schedule the blocks
with the revised deadlines.
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Deterministic rendezvous model.
Example’s second solution
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We consider the previous example.
The initial deadlines for T1, T3,1, T3,2, T2,3, T2,4, T2,1,
T2,2 are 12, 12, 12, 11, 11, 5, and 5, respectively (in
topological order).
Since T3(1)  T2(2) is a rendezvous constraint, then
deadline of T3,1 is revised to 5-2=3.
Similarly, since T3(2)  T2(3) is a rendezvous
constraint, then deadline of T3,2 is revised to 111=10.
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Deterministic rendezvous model.
Example’s second solution
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The revised deadlines for T1, T3,1, T3,2, T2,3, T2,4, T2,1,
T2,2 are now 12, 3, 10, 9, 11, 3, and 5, respectively.
Using again the EDF scheduler, we get the following
feasible schedule:

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Figure 3.19 from [Cheng; 2005], page 64
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Summary
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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
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Reading suggestions
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Chapter 3 of [Cheng; 2002], Sections 3.2.1,
3.2.2, 3.2.3, 3.2.4, 3.2.5
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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Multiprocessor scheduling
Available scheduling tools
Available real-time operating systems
Chapter 3 of [Cheng; 2002], Sections 3.3,
3.4, 3.5
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Thank you for your attention!
Questions?
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