Transcript Chapter 1 -- Introduction - Arizona State University

CSE 522
Real-Time Scheduling (2)
Computer Science & Engineering Department
Arizona State University
Tempe, AZ 85287
Dr. Yann-Hang Lee
[email protected]
(480) 727-7507
1
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
2
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
3
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
n
ei
U   1
i 1 pi
 Proof
 the only-if part is obvious
 the if part --- show if there is a job misses its deadline, then U > 1
4
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
 (t  k ) 
(t  i )ei
ei
ek
t

e

t
(

)  tU

k
pi
pk 
pi k i pk
k i 
 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]
B
A
0
r
Ji,c is released
C
D
t
Ji,c misses its deadline
5
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
6
General EDF Feasibility Analysis
 Processor demand: all processing request during a
period [ti, t2), i.e.,
 Loading factor:
h[t1 ,t2 ) 
u[t1 ,t2 ) 
C
k
t1  rk , d k t 2
h[t1 ,t2 )
t 2  t1
 Each set of real-time jobs is feasibly scheduled by
EDF iff
u  sup u[t1 ,t2 )  1
0t1 t 2
 How to compute loading factor?
7
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
u  u  1
 Need to show that
h[t1 ,t2 )
 t 2 t 1 Di 
  (1  
 )Ci  h[0,t2 t1 )
Di t 2  t 1
 Ti

 Hybrid task sets (with both periodic and sporadic
task) are schedulable under EDF if
n
Ci
1

i 1 minDi , Ti 
8
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
n
L( 0 )   Ci
i 1
( m 1)
L
 L( m ) 
 
Ci
i 1  Ti 
n
 Test loading factor at deadline instants that are less than L
9
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
1
2
3
6
2
4
8
2
5
9
3
7
t
h
u
4
2
0.5
5
4
0.8
7
7
1
10
9
0.9
13
11
0.85
16
16
1
WCET Deadline
10
Example of EDF Schedule
 A digital robot with EDF schedule
 control loop: ec  8ms at 100Hz
 BIST: eb  50ms
8 50
 given
u c  ub    1
10 pb
 BIST can be done every 250ms
 Add a telemetry task to send and receive messages with et 
15ms
 if BIST is done every 1000ms
 given
u c  ub 
8
50 15

 1
10 1000 Dt
 the telemetry task can have a relative deadline of 100ms
 sending or receiving must be separated at least 100ms
11
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
n
U 
i 1
ei
 n(21/ n  1)
pi
 a complete test  what is the worst response time given all possible
arrivals and preemptions
12
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
T1
T2
T1, T2
T2
T1
13
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)
t 
w i (t )  ei     ek
k 1  pk 
i 1
t
w(t)
 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 
t  ei     ek
k 1  pk 
i 1
time
Di
14
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:
 t 
w i , j (t )  j ei     ek
k 1  pk 
i 1
for ( j  1)pi  t  w i , j (t )
t  w i , j (t  ( j  1)pi )  ( j  1)pi
15
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-e1
 Else, max(e2)=p2-2e1
 U=e1/p1+(p1-e1)/p2
 U=e1/p1+(p2-2e1)/p2
p2
maximal e2
p2
T1
T1
0
p1
T1
2p1
16
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)
U
1
p2=p1
2p1
 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…
17
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
18
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 + e2 + . . . . + en-1 ) =2p1 - pn

U (
p
p2
p
2p
 1)  ( 3  1)  ....  ( n  1)  ( 1  1)
p1
p2
pn 1
pn
 q2,1  q3,2  .....  qn,n 1 
2
n
q2,1q3,2 ..... qn,n 1
 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
19
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
lpk
Tn
Tk
Tn
pn
20
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:
 an 
 ei    e j
j 1  p j 
i 1
an 1
i
where a0   e j
j 1
 Test Terminates when an+1 = an
 Task i is schedulable if its response time is before its
deadline: an ≤ pi
21
Sample Problem: UB Test
e
p
U
Task 1
20
100
0.200
Task 2
40
150
0.267
Task 3
100
350
0.286
 Total utilization is
.200 + .267 + .286 = .753 < U(3) =.779
 The periodic tasks in the sample problem are
schedulable according to the UB test.
22
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: 0.667 < U(2) = 0.828
 First two tasks are schedulable by UB test
 Utilization of all three tasks: 0.953 > U(3) = 0.779
 UB test is inconclusive
 Need to apply RT test
23
Example: Applying RT Test (2)
 Use RT test to determine if ‫ז‬3 meets its first
deadline:
i=3
3
a0   e j  e1  e2  e3  40  40  100  180
j 1
2 
 a0 
a0 
a1  ei    e j  e3    e j
j 1  p j 
j 1  p j 
180 
180 
 100  
( 40 )  
( 40 )  100  80  80  260


100 
150 
i 1
24
Example: Applying RT Test (3)
 a1 
 260 
 260 
a2  ei    e j  100  
( 40 )  
( 40 )  300


100 
150 
j 1  p j 
i 1
 a2 
 300 
 300 
a3  ei    e j  100  
( 40 )  
( 40 )  300


100 
150 
j 1  p j 
i 1
a3 = a2 = 300
Done!
 Task is schedulable using RT test.
 a3 = 300 < p3 = 350
25
Modeling Task Switching
Ui 
ei  2S
pi
C1
0
1
100
C1
2
C1
C1
C1 C1
200
C1
C1
C2
C1
C2
C1
S
e2
S
C2
3
S
40
e1
S
S
40
time
p1
p2
2p1
Two scheduling actions per task
(start of period and end of period)
26
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
1
2
4
200
60
HW
200
100
20
High
100
150
40
Medium
150
350
40
Low
350
27
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 =
Preemption form the
tasks that can hit more
than once (with period
less that pi)
e j ei
1



pi pi
jHn p j
Execution of a
task under test
e
k
kH1
Preemption from tasks
That can hit only once
(with period greater than
pi )
 Compare effective utilization against bound, U(n).
n = num(Hn) + 1
num(Hn) = the number of tasks in the set Hn
28
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 =
Preemption form the
tasks that can hit more
than once (with period
less that di)

jH n
ej
ei
1
 
p j di di
Execution of a
task under test
e
kH1
k
Preemption from tasks
That can hit only once
(with period greater than
di )
 Compare effective utilization against bound, U(n).
n = num(Hn) + 1
num(Hn) = the number of tasks in the set Hn
29
UB Test with Interrupt Priority: 3
 For 3, no tasks have a higher priority:
H = Hn =H1 = { }.
f3 =
0
e3
 0  U(1 )
p3
 Note that utilization bound is U(1): num(Hn) = 0.
e3
60

f3 = p 200  0.3  1.0
3
30
UB Test with Interrupt Priority: 1
 For 1, 3 has a higher priority: H = {3}; Hn = {};
H1 = {3}.
f1 = 0 
e1 1
  ek  U(1 )
p1 p1 k 3
 Note that utilization bound is U(1): num(Hn) = 0.
e1 e3
20
60

 0.800  1.0
f1 =  
p1 p1 100 100
31
UB Test with Interrupt Priority: 2
 For 2
: H={1, 3}; Hn={1 }; H1={3 };
ej
e
1
f2  
 2 
p2 p2
j 1 p j
e
k
 U( 2 )
k 3
 Note that utilization bound is U(2): num(Hn) = 1.
f2 
e1 e2 e3
20
40
60





 0.867  0.828
p1 p2 p3 100 150 150
32
UB Test with Interrupt Priority:
 For 4
: H={1, 2, 3}; Hn={1 , 2, 3 }; H1={};
f4 
ej
e4

 0  U( 4 )

p4
j 1 ,2 ,3 p j
 Note that utilization bound is U(4): num(Hn) = 3.
e1 e2
e3
e4
f4 



p1 p2
p3
p4

20
40
60
60



 0.882  0.756
100 150 200 350
33
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
34
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
 Li (a ) 
 a  Di  D j 
Wi (a, Li (a ))   min( 
,1  
 )C j
Tj
j  i , D j  a  Di
 Tj 


 Then, an iterative formula for Li(a)
L(i 0) (a )
0
L(i m 1) (a )
 Wi (a, L(i m ) (a ))  (1 
 The response time is:
rti (a )  max{Ci , Li (a )  a}
a
T  )Ci
 i
rti  max a 0 {rti (a )}
35
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
n
a  A  (  {kT j  D j  Di : k  0})  {0, L}
j 1, j  i
36
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
37
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
38
Project outline submissions
 5 late submissions; one on March 31
 One submitted as assignment 3
RTOS for
Real-time
AFDX
RTSJ Scheduling
Toyota
Real-time
automotive Communication Avionics
and
for
Others
UA
Hypervisor
Systems
on CAN bus Network
SCJ
multicore
8
5
8
2
16
9
12
3