ee5900 advanced embedded system for smart infrastructure
DESCRIPTION
EE5900 Advanced Embedded System For Smart Infrastructure. RMS and EDF Scheduling. C’ i. t. d i. Laxity. Priority-driven Preemptive Scheduling. Assumptions & Definitions Tasks are periodic No aperiodic or sporadic tasks Job (instance) deadline = end of period No resource constraints - PowerPoint PPT PresentationTRANSCRIPT
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EE5900 Advanced Embedded System For Smart Infrastructure
RMS and EDF Scheduling
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Priority-driven Preemptive Scheduling
Assumptions & Definitions• Tasks are periodic • No aperiodic or sporadic tasks• Job (instance) deadline = end of period• No resource constraints• Tasks are preemptable
• Laxity (Slack) of a Task • Ti = di – (t + ci’) where di: deadline;
t : current time;ci’ : remaining computation time.
t
C’i
Laxity
di
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Rate Monotonic Scheduling (RMS)
• Schedulability check (off-line) - A set of n tasks is schedulable on a uniprocessor by the RMS algorithm if the processor utilization (utilization test)
ci is the execution time and pi is the period,The term n(21/n -1) approaches ln 2, (0.69 as n ).
- This condition is sufficient, but not necessary.
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RMS (cont.)
• Schedule construction (online)
- Task with the smallest period is assigned the highest priority (static priority).
- At any time, the highest priority task is
executed.
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RMS Scheduler - Example 1
Task set: Ti = (ci, pi) [computation time, period]
T1 = (2,4) and T2 = (1,8)
Schedulability check:
2/4 + 1/8 = 0.5 + 0.125 = 0.625 ≤ 2(√2 -1) = 0. 82
T11 T2
1 T12
0 2 3 4 6 8
Active Tasks :
{T1, T2}
Active Tasks :{T2}
Active Tasks :{T1}
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RMS scheduler – Example 2
Task set: Ti = (ci, pi)
T1 = (2,4) and T2 = (4,8)
Schedulability check:
2/4 + 4/8 = 0.5 + 0.5 = 1.0 > 2(√2 -1) = 0. 82
T11 T2
1 T12
0 2 3 4 6 8
Active Tasks :
{T1, T2}
Active Tasks :{T2}
Active Tasks :
{T2, T1}
T21
Active Tasks :{T2}
Some task sets that FAIL the utilization-based schedulability test are also schedulable under RMS
RMS is not optimal
• T1=(1,2) and T2=(2.5,5)
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Earliest Deadline First (EDF)
• Schedulability check (off-line)
- A set of n tasks is schedulable on a uniprocessor by the EDF algorithm if the processor utilization.
• This condition is both necessary and sufficient.
- Least Laxity First (LLF) algorithm has the same schedulability check.
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EDF/LLF (cont.)
• Schedule construction (online) – EDF/LLF: Task with the smallest deadline/laxity is
assigned the highest priority (dynamic priority).– At any time, the highest priority task is executed.
• It is optimal (i.e., whenever there is a feasible schedule, EDF can always compute it) when preemption is allowed and no resource constraint is considered. – Given any two tasks in a feasible schedule, if
they are not scheduled in the order of earliest deadline, you can always swap them and still generate a feasible schedule.
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EDF scheduler - Example
Task set: Ti = (ci, pi, di)
T1 = (1,3,3) and T2 = (4,6,6)
Schedulability check:
1/3 + 4/6 = 0.33 + 0.67 = 1.0
T11 T2
1 T12
0 1 5 6
Active Tasks :
{T1, T2}
Active Tasks :{T2}
Active Tasks :
{T2, T1}
Active Tasks :{T1}
Unlike RMS, Only those task sets which pass the schedulability test are schedulable under EDF
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T21
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Comparison of RMS and EDF
0 5 10 15 20 25 30 35
0 7 14 21 28 35
T1
T2
RMS schedule
0 5 10 15 20 25 30 35
0 7 14 21 28 35
T1
T2
EDF schedule
Deadline miss
Process Period, T WCET, C T1 5 2 T2 7 4
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Resource sharing
• Periodic tasks
• Task can have resource access
• Semaphore is used for mutual exclusion
• RMS scheduling
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Background – Task State diagram
• Ready State: waiting in ready queue
• Running State: CPU executing the task
• Blocked: waiting in the semaphore queue until the shared resource is free
• Semaphore types – mutex (binary semaphore), counting semaphore
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Task State Diagram
READY RUN
WAITING
Activate
scheduling
Preemption
Termination
Wait on busy
resource
Signal free
resource
Process/Task state diagram with resource constraints
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Priority Inversion Problem
Priority inversion is an undesirable situation in which a higher priority task gets blocked (waits for CPU) for more time than that it is supposed to, by lower priority tasks.
Example:
• Let T1 , T2 , and T3 be the three periodic tasks with decreasing order of priorities.
• Let T1 and T3 share a resource S.
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Priority Inversion - Example
• T3 obtains a lock on the semaphore S and enters its critical section to use a shared resource.
• T1 becomes ready to run and preempts T3. Then, T1 tries to enter its critical section by first trying to lock S. But, S is already locked by T3 and hence T1 is blocked.
• T2 becomes ready to run. Since only T2 and T3 are ready to run, T2 preempts T3 while T3 is in its critical section.
Ideally, one would prefer that the highest priority task (T1) be blocked no longer than the time for T3 to complete its critical section. However, the duration of blocking is, in fact, unpredictable because task T2 got executed in between.
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Priority Inversion example
T1
T2
T3T3
0
T3 is the only
active task
Preempted by higher priority
task T1
T1
Makes a request for resource S and gets blocked
T3
Preempted by higher priority
task T2
T2
T3
T3 completes
T1
Resource S is available and
T1 is scheduled
here
K1 K2K3
T2 completes
L1
Total blocking time for task T1 = (K2+K3) + (L1)
Highest priority
Least priority
Medium priority
T1 and T3 share resource
S
A higher priority task waits for a
lower priority task
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Priority Inheritance ProtocolPriority inheritance protocol solves the problem of priority inversion.
Under this protocol, if a higher priority task TH is blocked by a lower priority task TL, because TL is currently executing critical section needed by TH, TL temporarily inherits the priority of TH.
When blocking ceases (i.e., TL exits the critical section), TL resumes its original priority.
Unfortunately, priority inheritance may lead to deadlock.
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Resource access control - example
Task Ti ci pi cix ci
y ciz
T1 2 8 2 0 0
T2 4 12 0 4 0
T3 2 6 1 1 0
T2 and T3 have access to a shared resource R
cix : Task duration before entering the critical section
ciy : Critical section duration
ciz : Task duration after the critical section
ci = cix + ci
y + ciz
By RMS, T3 > T1 > T2
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Schedules
Locks R
Preempted by T3
T3 T1 T2 T3 T2 T1 T2 T3 T3 T2
T3 T1 T2 T3 T2 T3 T1 T3 T2
Direct blocking of T3 by T2
Priority inversion of T3 by T1
0 2 4 6 7 8 10 11 12 14 16
RMS Schedule
RMS Schedule with Priority Inheritance Protocol
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Direct blocking of T3 by T2
Inheritance blocking of T1 by T2
Release R
Task Ti ci pi cix ci
y ciz
T1 2 8 2 0 0
T2 4 12 0 4 0
T3 2 6 1 1 0
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Priority Inheritance Protocol – Deadlock Assume T2 has higher priority than T1
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Priority Ceiling Protocol• Priority ceiling protocol solves the priority inversion
problem without getting into deadlock.• For each semaphore, a priority ceiling is defined,
whose value is the highest priority of all the tasks that may lock it.
• When a task Ti attempts to execute one of its critical sections, it will be suspended unless its priority is higher than the priority ceiling of all semaphores currently locked by tasks other than Ti.
• If task Ti is unable to enter its critical section for this reason, the task that holds the lock on the semaphore with the highest priority ceiling is said to be blocking Ti and hence inherits the priority of Ti.
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Priority Ceiling Protocol - properties
• This protocol is the same as the priority
inheritance protocol, except that a task Ti can also be blocked from entering a critical section if any other task is currently holding a semaphore whose priority ceiling is greater than or equal to the priority of task Ti.
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Priority Celiling Protocol - Example
• For the previous example, the priority ceiling for both CS1 and CS2 is the priority of T2.
• From time t0 to t2, the operations are the same as before.
• At time t3, T2 attempts to lock CS1, but is blocked since CS2 (which has been locked by T1) has a priority ceiling equal to the priority of T2.
• Thus T1 inherits the priority of T2 and proceeds to completion, thereby preventing deadlock situation.
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Scheduling tasks with precedence relations
Scheduler{T1, T2}
Conventional task set
T1 T2
Modify task parameters in order to respect
precedence constraintsScheduler
task set with precedence constraints
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Modifying the task parameters for RMS
• While using the RMS scheduler the task parameters (ready time) need to be modified in order to respect the precedence constraints
• Rj* ≥ Max (Rj, Ri*) where Ri* is the modified ready time of the task Ti
• Priority Prioi ≥ Prioj
Ti Tj
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Modifying ready times for RMS: example
T11
T22
T32
T41
T53
Task Ri Ci Di
T1 0 1 5
T2 5 2 7
T3 0 2 5
T4 0 1 10
T5 0 3 12
Initial Task Parameters
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Modifying the Ready times for RMS
T11
T22
T32
T41
T53
Task Ri Ci Di
T1 0 1 5
T2 5 2 7
T3 0 2 5
T4 0 1 10
T5 0 3 12
Initial Task Parameters
R1 = 0 R2 = 5
R3 = 0R4 = 0
R5 = 0
R3’ = max(R1,
R3) R3’ = 0
R4’ = max(R1, R2,R4)
R4’ = 5
R5’ = max(R3’, R4’,R5)
R5’ = 5
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Modified Ready times for RMS
T11
T22
T32
T41
T53
Task Ri Ci Di
T1 0 1 5
T2 5 2 7
T3 0 2 5
T4 5 1 10
T5 5 3 12
Modified Task Parameters
R1 = 0 R2 = 5
R3’ = 0 R4’ =
5
R5’ = 5
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Assigning task priorities for RMS
T11
T22
T32
T41
T53
Task Ri Ci Di
T1 0 1 5
T2 5 2 7
T3 0 2 5
T4 5 1 10
T5 5 3 12
Modified Task Parameters
R1 = 0 R2 = 5
R3’ = 0 R4’ =
5
R5’ = 5
Assume all tasks of a connected component have the same period. Therefore, as per RMS all tasks will have a tie. We assign priorities to
break the ties.
Priority
3
4
2
1
0
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Modifying task parameters for EDF
• While using the EDF scheduler the task parameters need to be modified in order to respect the precedence constraints
• Rj* ≥ Max (Rj, (Ri* + Ci))
• Di* ≥ Min (Di, (Dj* – Cj))
Ti Tj
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Modifying the Ready times for EDF
T11
T22
T32
T41
T53
Task Ri Ci Di
T1 0 1 5
T2 5 2 7
T3 0 2 5
T4 0 1 10
T5 0 3 12
Initial Task Parameters
R1 = 0 R2 = 5
R3 = 0R4 = 0
R5 = 0
R3’ = max(R1 + C1, R3)
R3’ = 1
R4’ = max(R1+C1, R2+C2,R4)
R4’ = 7
R5’ = max(R3’+C3, R4’+C4,R5)
R5’ = 8
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Modifying the Ready times for EDF
T11
T22
T32
T41
T53
Task Ri Ci Di
T1 0 1 5
T2 5 2 7
T3 1 2 5
T4 7 1 10
T5 8 3 12
Modified Task Parameters
R1 = 0 R2 = 5
R3’ = 1 R4’ =
7
R5’ = 8
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Modifying the Deadlines for EDF
T11
T22
T32
T41
T53
Task Ri Ci Di
T1 0 1 5
T2 5 2 7
T3 1 2 5
T4 7 1 10
T5 8 3 12
Modified Task Parameters
D1 = 5 D2 = 7
D3 = 5D4 =
10
D5 = 12
D3’ = Min( (D5 – C5), D3)
D4’ = Min( (D5 – C5), D4)
D3’ = 5 D4’ =
9
D2’ = Min( (D4’ – C4), D2)
D2’ = Min( (D4’ – C4), (D3’ – C3), D1)
D2’ = 7
D1’ = 3
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Modifying the Deadlines for EDF
T11
T22
T32
T41
T53
Task Ri Ci Di
T1 0 1 3
T2 5 2 7
T3 1 2 5
T4 7 1 9
T5 8 3 12
Modified Task Parameters
D5 = 12
D3’ = 5 D4’ =
9
D2’ = 7
D1’ = 3