1. CSE 522 Real-Time Scheduling (2)
Computer Science & Engineering Department Arizona State University Tempe, AZ 85287
Dr. Yann-Hang Lee (480)

2. Priority-Driven Scheduling of Periodic Tasks
Why priority-driven scheduling
use priority to represent urgency
easy implementation of scheduler (compare priorities and dispatch tasks)
tasks can be added or removed easily
no direct control of execution instant
How can we analyze the schedulability if we don’t know when a task is to be executed
Let’s begin a deterministic case in one processor
independent periodic tasks
deadline = period
preemptable
no overhead for context switch

3. Priority-Driven Schedules
Assign priority when jobs arrive
static -- all jobs of a task have a fixed priority
dynamic -- different priorities to individual jobs of a task
relative priorities don’t change while jobs are waiting
Static priority schedules
Rate-monotonic -- the smaller a task priority, the higher its priority
Deadline-monotonic
Dynamic priority schedules
EDF -- earliest deadline first
LSTF -- least slack time first
Schedulable utilization:
a scheduling algorithm can feasibly schedule any sets of priority tasks if the total utilization is equal to or less than the schedulable utilization of the algorithm

4. EDF Schedule
A optimal algorithm under single processor and preemptable tasks
How do we know a set of periodic tasks are schedulable under EDF ?
If we know the schedulable utilization SU of EDF, then any sets of tasks are schedulable when U  SU
Theorem: A set of n periodic tasks can be scheduled by EDF iff
Proof
the only-if part is obvious
the if part --- show if there is a job misses its deadline, then U > 1

5. EDF Schedule
Let the 1st unschedulable job be Ji,c that misses its deadline at t and the processor is never idle during [0,t]
how much execution we have done before t
Case 1: no type D jobs and let’s ignore type C jobs
Case 2: if type D jobs exist and are executed before r
let t’ be the latest instant that type D jobs execute
how much execution we have done in [t’, t]
Ji,c is released
Ji,c misses its deadline t r A B C D

6. Extension of EDF Schedulable Utilization
If Di  pi, EDF is schedulable iff U  1
What can we do if Di < pi
density of task k : k = ek / min(pk,Dk)
EDF is schedulable if the total density is equal to or less than 1
proof: if there is a job missing its deadline, then the total density > 1
there is no “only-if” part ---- if the total density > 1, EDF may or may not schedulable
If Di  pi, LSTF is schedulable iff U  1
Predicable for single-processor preemptive schedule of independent tasks
Robust
independent of phases
periods are lower bound  applicable to sporadic tasks with minimum separations

7. General EDF Feasibility Analysis
Processor demand: all processing request during a period [ti, t2), i.e.,
Loading factor:
Each set of real-time jobs is feasibly scheduled by EDF iff
How to compute loading factor?

8. Sporadic Tasks under EDF
minimum separation Ti ≠ Di
Worst possible scenario: arrives synchronously at the maximum rate
A sporadic task set is feasibly scheduled by EDF iff
Need to show that
Hybrid task sets (with both periodic and sporadic task) are schedulable under EDF if

9. EDF Schedulability Test
If Di ≤ Ti and synchronous task sets
loading factor u[0,t)≤1 for all t
L: synchronous busy period, can be compute iteratively by
Test loading factor at deadline instants that are less than L

10. EDF Scheduling Example
L0=7, L1=9, L2=11, L3=14, L4=16
the set of points to be tested: D={4,5,7,10,13,16}
Task(i)
Period
WCET
Deadline 1 6 2 4 2 8 5 3 9 3 7 t h u 4 2
0.5 5
0.8 7 1 10 9
0.9 13 11
0.85 16

11. Example of EDF Schedule
A digital robot with EDF schedule
control loop: ec  8ms at 100Hz
BIST: eb  50ms
given
BIST can be done every 250ms
Add a telemetry task to send and receive messages with et  15ms
if BIST is done every 1000ms
the telemetry task can have a relative deadline of 100ms
 sending or receiving must be separated at least 100ms

12. Rate-Monotonic Algorithm
A base case: no additional overhead, simple periodic tasks with pi =Di
Assign priorities according their periods
Ti has a higher priority that Tk if i < k ( pi < pk )
Is RM optimal?  if there is a feasible fixed-priority schedule, then RM is feasible
How do we know RM is feasible  schedulability test
Results:
RM is optimal if pi  Di
sufficient condition  utilization test
a complete test  what is the worst response time given all possible arrivals and preemptions

13. Critical Instant
Critical instant of Ti: a job of Ti arriving at the instant has a maximum response time
If we can find the critical instant of Ti, then
check whether all jobs of Ti meet their deadlines
let’s increase ei until the maximum response time = Di
 schedulable utilization
In-phase instant is critical: all higher priority tasks are released at the same instant of Ji,c (assume all jobs are completed before the next job of the same task is released.)
which T2 has the maximum response time
T T2
T1, T2
T T1

14. Schedulability Test: Time-Demand Analysis
Consider in-phase instant only
If Ji is done at t, then the total work must be done in [0,t] is (from Ji and all higher priority tasks)
Can we find a t  Di such that
wi(t)  t
cannot check all t  [0, Di]
check all arrival instants and Di
The completion time of Ji satisfies t
w(t)
time Di

15. Schedulability Test
EDF has a schedulable utilization of 1, how about RMS
If Di=pi, the schedulable utilization exists
if U  n ( 21/n - 1 ), done
else do time-demand analysis
if Di < pi, do time-demand analysis
if Di > pi, there may be more than one jobs of task i in the system
examine all jobs of task i in a level-i busy interval (in-phase)
the following equations represent:

16. Schedulable Utilization of RMS
Must be less than 1
Let’s consider two tasks and deadline=period
T2 can only be executed when T1 is not in the system
Let p2 < 2p1. What is the maximum schedulable e2
If p2 < p1+e1, max(e2)=p1-e  U=e1/p1+(p1-e1)/p2
Else, max(e2)=p2-2e1  U=e1/p1+(p2-2e1)/p2 T1
p p1 p2
maximal e2

17. Schedulable Utilization of RMS
Given e1, p1, and p2, plot U
The minimum U occurs when
p2=p1+e1
where
U= e1/p1+(p1-e1)/(p1+e1)
What is the minimum U
take the derivative wrt to p1 and set dU/dp1=0
we will get e1=(21/2-1)p1 and U=0.828…
p2=p p1 U 1

18. Schedulable Utilization of RMA
U  n ( 21/n - 1 )
Is there a case that is feasible and gives the minimal schedulable utilization
When pn  2 p1
processor must be busy in [0,pn]
become unscheduable if increase ei
processor will be idle if increase pi T1 T2 T3 T4 T5 p5

19. Schedulable Utilization of RMA
What do we have from the timeline diagram
ek = pk+1 - pk for k=1,2,..,n-1
en = pn - 2(e1 + e en-1 ) =2p1 - pn
Can we increase e1 and decrease ek by the same amount
still schedulable for the 1st arrival of all tasks
utilization is higher
Can we decrease e1 and increase ek
When will U be minimum when q2,1=q3,2= = 2/n

20. Schedulable Utilization of RMA
When pn > 2pk and schedulable
construct a new task set that is schedulable and pn > 2pk
the original set has a higher utilization Tk Tn pn
lpk

21. Schedulability: RT Test
Theorem: for a set of independent, periodic tasks, if each task meets its first deadline, with worst-case task phasing, the deadline will always be met.
Response Time (RT) test: let an = response time of task i. an may be computed by the following iterative formula:
Test Terminates when an+1 = an
Task i is schedulable if its response time is before its deadline: an ≤ pi

22. Sample Problem: UB Test
Task 1 20
100
0.200
Task 2 40
150
0.267
Task 3
350
0.286
Total utilization is
= .753 < U(3) =.779
The periodic tasks in the sample problem are schedulable according to the UB test.

23. Example: Applying RT Test (1)
If we increase the compute time of ז1 from 20 to 40; is the task set still schedulable?
Utilization for the first task : 40/100=0.4 < U(1)
Utilization of first two tasks: < U(2) = 0.828
First two tasks are schedulable by UB test
Utilization of all three tasks: > U(3) = 0.779
UB test is inconclusive
Need to apply RT test

24. Example: Applying RT Test (2)
Use RT test to determine if ז3 meets its first deadline:
i = 3

25. Example: Applying RT Test (3)
a3 = a2 = Done!
Task is schedulable using RT test.
a3 = 300 < p3 = 350

26. Modeling Task Switching

27. Schedulability with Interrupts
Interrupt processing can be inconsistent with rate monotonic priority assignment.
interrupt handler executes with high priority despite its period
interrupt processing may delay execution of tasks with shorter periods
Effects of interrupt processing must be taken into account in schedulability model.
Task(i)
Period(p)
WCET(e)
Priority
Deadline(D) 3
200 60 HW 1
100 20
High 2
150 40
Medium 4
350
Low

28. UB Test with ANY FIXED PRIORITY
Test is applied to each task.
Determine effective utilization (fi) of each task i using when di=pi
fi =
Compare effective utilization against bound, U(n).
n = num(Hn) + 1
num(Hn) = the number of tasks in the set Hn
Preemption form the tasks that can hit more than once (with period less that pi)
Execution of a task under test
Preemption from tasks
That can hit only once
(with period greater than pi )

29. UB Test with Interrupt Priority
Test is applied to each task.
Determine effective utilization (fi) of each task i using when di<pi (ei must be done within di, not pi)
fi =
Compare effective utilization against bound, U(n).
n = num(Hn) + 1
num(Hn) = the number of tasks in the set Hn
Preemption form the tasks that can hit more than once (with period less that di)
Execution of a task under test
Preemption from tasks
That can hit only once
(with period greater than di )

30. UB Test with Interrupt Priority: 3
For 3, no tasks have a higher priority:
H = Hn =H1 = { }.
f3 =
Note that utilization bound is U(1): num(Hn) = 0.
f3 =

31. UB Test with Interrupt Priority: 1
For 1, 3 has a higher priority: H = {3}; Hn = {};
H1 = {3}.
f1 =
Note that utilization bound is U(1): num(Hn) = 0.
f1 =

32. UB Test with Interrupt Priority: 2
For 2 : H={1, 3}; Hn={1 }; H1={3 };
Note that utilization bound is U(2): num(Hn) = 1.

33. UB Test with Interrupt Priority:
For 4 : H={1, 2, 3}; Hn={1 , 2, 3 }; H1={};
Note that utilization bound is U(4): num(Hn) = 3.

34. Is There a RT Analysis for EDF
Does a critical instant exist?
releasing task instance i at the start of the busy period may not lead to its worst-case response time.
i arrives at a with deadline d, is finished in t2
t1 the last time no pending instances with arrival time earlier than ti and deadline ≤ d
[t1, t2) –the busy period of instances with deadlines ≤ d
“shift left” all instances of other tasks to start at t1  increase busy period

35. RT Analysis for EDF (1) Deadline busy period where WCET happens:
only tasks with deadlines less than and equal to d get executed
started at 0 for the instance of i arrived at a: Li(a)
higher priority workload
Then, an iterative formula for Li(a)
The response time is:

36. RT Analysis for EDT (2) Choice of “a”:
Li(a) > a (arrives before deadline busy period ends)
a+Di coincides with the absolute deadlines of other tasks
in the interval [0, L-Ci), L is the busy period of synchronous arrivals

37. Summary of Feasibility Analysis (1)
A set of tasks {Ci, Di, Ti,, i}
Dynamic priority -- EDF is optimal
Di ≥ Ti -- utilization bound on slide #4 (iff)
Di < Ti -- all loading factor less than 1 on slide 7 (iff)
Synchronous or Sporadic – busy period analysis and testing algorithm on slide #9 (iff)
Asynchronous – if the cooresponding synchronous task set is schedulable.
worst case execution time – slide #34
necessary condition: U ≤ 1
sufficient condition: ∑(Ci / min{Di , Ti }) ≤ 1

38. Summary of Feasibility Analysis (2)
Static priority
Rate monotonic – optimal if Di ≥ Ti
Deadline monotonic – optimal if Di < Ti
Necessary condition: U ≤ 1
Synchronous
Sufficient condition
RM and DM -- ∑(Ci / min{Di , Ti }) ≤ n(21/n-1 ) (slide #12)
fixed priorities – effective utilization for each task ≤ n(21/n-1 ) (slide #28)
worst case response time (i.e. 1st arrival) ≤ Di (slide #21) (iff)
Asynchronous
the conditions for synchronous case are sufficient

39. Project outline submissions
5 late submissions; one on March 31
One submitted as assignment 3
Toyota UA
RTOS for automotive Systems
Real-time Communication on CAN bus
AFDX Avionics Network
Real-time Hypervisor
RTSJ and SCJ
Scheduling for multicore
Others 8 5 2 16 9 12 3