Post on 14-Jan-2016
description
Examples of periodic tasks
Audio sampling in hardware Audio sample processing Video capture and processing Feedback control (sensing and
processing) Navigation Temperature and speed monitoring
Scheduling periodic tasks
• Preemptive scheduling is an effective approach for scheduling real-time DSP systems– modularity simplifies the overall design
• Application can be viewed as a collection of independent tasks or jobs– complexity is reduced as the functionality
becomes encapsulated into a set of well defined tasks
Scheduling periodic tasks
• Systems designed using preemptive scheduling are also more maintainable– issue of changes to one task in the
system affecting other jobs in the system is removed
– New functionality can easily be added by adding a new task
Scheduling periodic tasks
• Preemptive scheduling approach also makes the system more efficient– preemptive scheduling is more efficient
at utilizing time slots that may not be fully utilized
• Scheduling algorithms– rate monotonic scheduling– deadline monotonic scheduling
cost of handling event C = 4
periodic arrivals. period T = 10
---- 10 ----
4 4 4
System Utilization = C/T = .40
Periodic Arrivals with Fixed Cost of Processing
System will be able to meet all deadlines. It can finish processing arrivals before the next arrival occurs.
1. periodic arrival, period T = 10 and C=4
2. periodic arrival, T=10 and C=3 ??
---- 10 ----
4 4 4
Can a second periodic event be accommodated?
1. periodic arrival, period T = 10 and C=4
2. periodic arrival, T=10 and C=3 ??
---- 10 ----
4 4 4
System Utilization C/T = .70
Can a second periodic event be accommodated?
1. periodic arrival, period T = 10 and C=4
2. periodic arrival, T=6 and C=3 ??
---- 10 ----
4 4 4
How about 2nd periodic event with T=6 and C=3?
1. periodic arrival, period T = 10 and C=4
2. periodic arrival, T=6 and C=3 ??
---- 10 ----
4 4 4
System Utilization C/T = .90
How about 2nd periodic event with T=6 and C=3?
---- 10 ----
4 4 4
--6--
If we process Event #1 before Event #2 then,
2nd event processing will not complete before the next comparable event occurs
Can’t Meet Deadline!
Event #1
Event #2
Task #1
Task #2
---- 10 ----
4
--6--
Event #1
Event #2
Try Event #2 before Event #1-
We still cannot complete task 1 before the next task 2 event occurs at t=6
unless...
Task #1
Task #2
---- 10 ----
4
--6--
Event #1
Event #2
Try Event #2 before Event #1-
We still cannot complete task 1 before the next task 2 event occurs at t=6
unless…we Interrupt task 1
Task #1
Task #2
---- 10 ----
4
--6--
Event #1
Event #2
Try Event #2 before Event #1-
We still cannot complete task 1 before the next task 2 event occurs at t=6
unless…we Interrupt task 1
Giving event #2 priority means that we can meet our deadline IF we preempt the processing of event #1 when event #2 occurs
Task #1
Task #2
Rate Monotonic Analysis
Rate Monotonic Analysis
• Assume a set of “n” periodic tasks– period Ti– worst case execution time Ci
• Rate-monotonic priority assignment– task with a shorter period (higher rate)
assigned a fixed higher priority
Rate Monotonic Analysis
• Rate Monotonic scheduling addresses how to determine whether a group of tasks, whose individual CPU utilization is known, will meet their deadlines– assumes a priority preemption
scheduling algorithm– assumes independent tasks (no
communication or synchronization)
Rate Monotonic Analysis
– restriction of no communication or synchronization may appear to be unrealistic, but there are techniques for dealing with this
– Each task is a periodic task which has a period T, which is the frequency with which it executes
Rate Monotonic Analysis
An execution time C, which is the CPU time required during the period
A utilization U, which is the ratio C/T• A task is schedulable if all its deadlines
are met (i.e., the task completes its execution before its period elapses.)– A group of tasks is considered to be
schedulable if each task can meet its deadlines
Rate Monotonic Analysis
• RMA is a mathematical solution to the scheduling problem for periodic tasks with known cost– assumption is that the total utilization
must always be less than or equal to 100%
• Any more and you are exceeding the capacity of the CPU
• Are you asking for more computing power than you have? IF so, forget it!
Rate Monotonic Analysis
• For a set of independent periodic tasks, the rate monotonic algorithm assigns each task a fixed priority based on its period, such that the shorter the period of a task, the higher the priority
Rate Monotonic Analysis• For three tasks T1, T2, and T3 with periods
of 5, 15 and 40 msec respectively the highest priority is given to the task, T1, as it has the shortest period, the medium priority to task T2, and the lowest priority to task T3– priority assignment is independent of the
applications “priority” i.e. how important meeting this deadline is to the functioning of the system or user concerns
Rate Monotonic Analysis• A mathematical solution to the scheduling
problem for Periodic Tasks with known Cost
• Tasks will have:– Cost (Time to complete a task)– Period (Time between events)– Utilization ( Cost/Period)
• Assumption– Total Utilization must always be <= 100%
3 levels of analysis using RMA
• Utilization bound test
• Completion time test
• Response time test
Utilization bound test
• If this simple rule is followed, then all tasks are guaranteed to meet their requirements if the following holds true;
)()12(/..../ /111 nUnTCTC n
nn where and are the execution time andperiod of task , respectively.
iCiT
it
Utilization bound test
• In this rule, the bound is 1.0 for harmonic task sets
• A task set is said to be harmonic if the periods of all its tasks are either integral multiples or sub-multiples of one another– On the average, for random Cs and Ts,
this number will be about 0.88.
Utilization bound test
• Theory is a worst case approximation
• For a randomly chosen group of tasks, it has been shown that the likely upper bound is 88%– Harmonic periods can give even higher
upper bounds– The algorithm is stable in conditions
where there is a transient overload
Utilization bound test
• In this case, there is a subset of the total number of tasks, namely those with the highest priorities that will still meet their deadlines
Example of UB test
Task t1: C1=20; T1= 100; U1 = .2 Task t2: C2=30; T2= 150; U2 = .2 Task t3: C3=60; T3= 200; U3 = .3
– The total utilization for this task set is .2 + .2 + .3 = .7. Since this is less than the 0.779 utilization bound for this task set, all deadlines will be met.
ExampleCan these 4 tasks be
scheduled?
– Can the system run and meet all hard deadlines?
Task Ci Ti Ui
1 3 10 .302 3 12 .253 4 16 .254 7 20 .35
ExampleExampleCan these 4 tasks be Can these 4 tasks be
scheduled?scheduled?
– Can the system run and meet all hard deadlines?
– NO! The Total Utilization = 115%
Task Ci Ti Ui
1 3 10 .302 3 12 .253 4 16 .254 7 20 .35
Example
Can these tasks always meet their deadlines?Total Utilization = 80%It MAY be possible - Rate Monotonic Scheduling applies!
Task Ci Ti Ui
1 6 20 .302 4 16 .253 3 12 .25
Rate Monotonic Theorem• For PERIODIC Tasks• Most frequent task gets highest
priority• THEOREM (Simple Version)
– IF the utilization of all tasks is less than or equal to 69%, then all tasks will ALWAYS meet their deadlines
Are These Tasks Schedulable?
Task Ci Ti Ui
1 2 20 .102 4 16 .253 3 12 .254 1 20 .05
Are These Tasks Are These Tasks Schedulable?Schedulable?
Task Ci Ti Ui
1 2 20 .102 4 16 .253 3 12 .254 1 20 .05
Yes. Total CPU Utilization is 65% < 69%
Are These Tasks Are These Tasks Schedulable?Schedulable?
Task Ci Ti Ui
1 2 20 .102 4 16 .253 3 12 .254 3 20 .15
Are These Tasks Are These Tasks Schedulable?Schedulable?
Task Ci Ti Ui
1 2 20 .102 4 16 .253 3 12 .254 1 20 .05
Total CPU Utilization is 65%
???
Exercise
• Using Rate Monotonic Scheduling, determine if the following task set is schedulable
Task Execution time Period 1 1 10 2 4 16 3 3 12 4 1 20
More on Rate Monotonic Analysis
Rate Monotonic TheoremRate Monotonic Theorem For PERIODIC Tasks Most frequent task gets highest
priority RMS THEOREM (Mathematical
Version) n periodic tasks scheduled by the rate
monotonic algorithm will always meet their deadlines if the total utilization of all tasks is less than
n (21/n - 1) this converges to ln2 = 69% for large n
Tasks Utilizationn n(2^(1/n) -1)1 12 0.828427123 0.779763154 0.756828465 0.743491776 0.734772297 0.72862668 0.724061869 0.7205376510 0.7177346311 0.7154519812 0.71355713
converges toward 69%
In a Nutshell:
The more tasks you try and schedule, the more slack time you must be willing to tolerate to mathematically guarantee schedulability
Priority Inversion
Taskh
Taskmed
Tasklow
Normalexecution
Execution incritical section
Priorityinversion
Unbounded Priority InversionUnbounded Priority Inversion
Taskh
Taskmed
Tasklow
Normalexecution
Execution incritical section
Priorityinversion
Taskmed
Uncontrolled priority inversion
Priority Inheritance Protocol
Taskh
Taskmed
Tasklow
Normalexecution
Execution incritical section
Priorityinversion
Execution incritical section athigher priority
What Happened on Mars ?
What happened on Mars ?
• Mars pathfinder “flawless” in early days of mission– unconventional landing with airbags– deployment of Sojourner rover– gathering and transmitting data back to earth
• A few days into the mission the Pathfinder began experiencing total system resets, each including losses of data
What happened on Mars ?
• Press reported these as “software glitches”
• VxWorks RTOS provides preemptive priority scheduling of tasks– tasks executed as threads– priorities assigned reflecting relative
urgency of the tasks
What happened on Mars ?What happened on Mars ?
busmanagementtask
information bus
mutex
meteorologicaldatagatheringtask
communicationtask
high priority -frequent execution
low priority -infrequent execution
medium priority
• Combination worked fine most of the time• Possible for interrupt to occur that caused
the long running medium priority task to be scheduled during the short interval while the high priority task was blocked waiting on the semaphore that the low priority task had.
What happened on Mars ?What happened on Mars ?
• Watchdog timer would go off, notice data bus task not in use for some time, conclude that something bad went wrong, and initiate a total system reset
• Classic case of priority inversion Classic case of priority inversion
What happened on Mars ?What happened on Mars ?
How was this debugged ?
• VxWorks can run in trace mode, recording interesting events.
• JPL engineers spent hours in lab trying to reproduce the problem on the ground.
• When finally reproduced, the trace data indicated the priority inversion problem
How was this problem corrected?
• Mutex object accepts boolean parameter indicating whether priority inheritance should be used
• Initialized with parameter off– if on, the low-pri thread would have
inherited the pri of the high-pri thread– medium pri thread would never have
been executed
How was this problem corrected?
• VxWorks has a C language interpreter that allows C commands to be executed on the fly
• JPL engineers left this in the software• Changed global variables by uploading a
short program to the spacecraft• No more system resets occurred after re-
programming!
Analysis and Lessons
• Diagnosing this problem as a black box would have been impossible– trace data was required
• Leaving debugging facilities in the system saved the day
• Time critical situations requires additional correctness measures even at the expense of some performance
Human nature, Deadline Pressures
• One or two system resets had occurred on the ground prior to launch
• Never reproducable or explainable• “it was probably caused by a hardware
glitch”• Engineer focus caused part of the problem
– extremely focused on ensuring quality and flawless operation of landing software
– the occasional glitch was dismissed
Importance of good Theory/Algorithms
• Some of the heros were people from CMU who published a paper years ago on the priority inversion problem– “An Approach to Real-Time
Synchronization” IEEE Transaction on Computers, Vol39, pp1175-1185, September 1990
End of session