Why do RTOSes not have any implementation to prevent priority inversion for semaphore even though it exists for mutex.
Semaphores do not need to prevent priority inversion?
The same situation happens both on uC/OS and GreenHills RTOS.
Thanks in advance.
Priority inversion occurs when a low-priority task owns a semaphore,
and a high-priority task is forced to wait on the semaphore until the
low-priority task releases it. If, prior to releasing the semaphore,
the low priority task is preempted by one or more mid-priority tasks,
then unbounded priority inversion has occurred because the delay of
the high-priority task is no longer predictable. This defeats Deadline
Monotonic Analysis (DMA) because it is not possible to predict if the
high-priority task will meet its deadline.
Sharing a critical resource between high and low priority tasks is not
a desirable design practice. It is better to share a resource only
among equal priority tasks or to limit resource accesses to a single
resource server task. Examples are a print server task and a file
server task. We have long advocated this practice. However, with the
layering of increasingly diverse and complicated middleware onto
RTOSs, it is becoming impractical to enforce such simple strategies.
Hence, in the interest of safety, it is best to implement some method
of preventing unbounded priority inversion.
Check full link at http://www.smxrtos.com/articles/techppr/mutex.htm
Regards,
Otacon
Related
Is it guaranteed that by using FCFS scheduling, the 'system' will not be in deadlock?
Thanks in advance!
The four conditions for a deadlock are:
Mutual exclusion: Irrespective of the scheduling algorithm, resources can be possessed by one process without sharing.
Hold and wait: In this condition, processes can wait for other resources while holding onto one resource. This is possible with any scheduling algorithm.
No preemption: FCFS is non-preemptive. That is, processes executing a critical section of their code cannot be forced to stop.
Circular wait: Processes are waiting for another process to release a resource in a circular fashion. This, again, is irrespective of the scheduling algorithm
Hence, FCFS does not guarantee that the system will not be in deadlock. If the four conditions are met, a deadlock will occur.
Deadlocks are caused by resource locking, not scheduling order. FCFS doesn’t guarantee that your threads will always grab resources in sequence, so the answer to your question is no.
I know what a binary semaphore is: it is a flag when is set to 1 by an ISR of an interrupt.
But what is a semaphore when we are using a pre-emptive kernel, say FreeRTOS? Is it the same as binary semaphore?
it is a flag when is set to 1 by an ISR of an interrupt.
That is neither a complete nor accurate description of a semaphore. What you have described is merely a flag. A semaphore is a synchronisation object; there are three forms provided by a typical RTOS:
Binary Semaphore
Counting Sempahore
Mutual Exclusion Semaphore (Mutex)
In the case of a binary semaphore, there are two operations give and take. A task taking a semaphore will block (i.e. suspend execution and allow other lower or equal priority threads to run threads to run) until some other thread or interrupt handler gives the semaphore. Binary semaphores are used to signal between threads and from ISRs to threads. They are often used to implement deferred interrupt handlers, so that an ISR can ve bery short, and the handler benefit from RTOS mechanisms that are not allowed in an ISR (anything that blocks or suspends execution).
Multiple threads may block on a single semaphore, but only one of those tasks will respond take the semaphore. Some RTOS have a flush operation (VxWorks for example) that puts all threads waiting on a semaphore in the ready state simultaneously - in which case they will run according to the priority scheduling scheme.
A Counting Semaphore is similar to a Binary Semaphore, except that it can be given multiple times, and tasks may take the semaphore without blocking until the count is zero.
A Mutex is used for resource locking. It is possible to use a binary semaphore for this, but a mutex provides features that make this safer. The operations on a mutex are lock and unlock. When a thread locks a mutex, and another task attempts to lock the same mutex, the second (and any subsequent) task blocks until the first task unlocks it. This can be used to prevent more than one thread accessing a resource (memory or I/O) simultaneously. A thread may lock a mutex multiple times; a count is maintained, so that it must be unlocked an equal number of times before the lock is released. This allows a thread to nest locks.
A special feature of a mutex is that if a thread with the lock is a lower priority that a task requesting the lock, then the lower priority task is boosted to the priority of the higher in order to prevent a priority inversion where a middle priority task may preempt the low priority task with the lock increasing the length of time the higher priority task must wait this rendering the scheduling non-deterministic.
The above descriptions are typical; specific RTOS implementations may differ. For example FreeRTOS distinguishes between a mutex and a recursive mutex, the latter supporting the nestability feature; while the first is marginally more efficient where nesting is not needed.
Semaphores are not just flags, or counts. They support send and wait operations. A user-space thread can wait on a semaphore without unnecessary and unwanted polling and be made ready/running 'immediately' when another thread, or an appropriately-designed driver/ISR, sends a unit.
By 'appropriately-designed driver/ISR', I mean one that can perform a send() operation and then exit via the OS scheduler whenever it needs to set a waiting thread ready/running.
Such a mechanism is vitally important on preemptive kernels because it allows them to achieve very good I/O performance without wasting time, CPU cycles and memory-bandwidth on polling. Non-preemptive systems are hopelessly slow, latency-ridden and wasteful at I/O and this is why they are essentially no longer used and why we put up with all the synchro/locking/queueing etc issues.
How many tasks are needed for priority inversion to happen??..As per as my understanding we need atleast 3.....or can we have it only with 2 task??
I actually went through a book : modern operating system by Andrew Tanenbaum . I knew only when 3 task are there in some patern as u all know ..a priority inversion can happen..however ..I found the book says only 2 two task 1-low and the 1-high can also cause the same..so I am confused...
You need one high-priority task which waits for a resource held by a low-priority task, while a mid-priority task is running.
So yes, you need three.
priority inversion can occur with 2 thread also,
Ex:-
Higer priority task waiting on a low priority task which holds a spinlock effectively disabling preemption of the task by higher priority one.
In a multitasking operating system context, sometimes you hear the term round-robin scheduling. What does it refer to?
What other kind of scheduling is there?
Round Robin Scheduling
If you are a host in a party of 100 guests, round-robin scheduling would mean that you spend 1 minute (a fixed amount) per guest. You go through each guest one-by-one, and after 100 minutes, you would have spent 1 minute with each guest. More on Wikipedia.
There are many other types of scheduling, such as priority-based (i.e. most important people first), first-come-first-serve, earliest-deadline-first (i.e. person leaving earliest first), etc. You can start off by googling for scheduling algorithms or check out scheduling at Wikipedia
Timeslicing is inherent to any round-robin scheduling system in practice, AFAIK.
I disagree with InSciTek Jeff's implication that the following is round-robin scheduling:
That is, each task at the same priority in the round-robin rotation can be allowed to run until they reach a resource blocking condition before yeilding to the next task in the rotation.
I do not see how this could be considered round-robin. This is actually preemptive scheduling. However, it is possible to have a scheduling algorithm which has elements of both round-robin and preemptive scheduling, which VxWorks does if round-robin scheduling and preemption are both enabled (round-robin is disabled by default). The way to enable round-robin scheduling is to provide a non-zero value in kernelTimeSlice.
I do agree with this statement:
Therefore, while timeslicing based scheduling implies round-robin scheduling, round-robin scheduling does not require equal time based timeslicing.
You are right that it doesn't require equal time. Preemption can muck with that. And actually in VxWorks, if a task is preempted during round-robin scheduling, when the task gets control again it will execute for the rest of the time it was allocated.
Edit directed at InSciTek Jeff (I don't have comment privileges)
Yes, I was referring to task locking/interrupt disabling, although I obviously didn't express that very well. You preempted me (ha!) with your second comment. I hope to debate the more salient point, that you believe round-robin scheduling can exist without time slicing. Or did you just mean equal time based time slicing? I disagree with the former, but agree with the latter. I am eager to learn. Thanks.
Edit2 directed at Jeff:
Round-robin can exist without timeslicing. That is exactly what happens in VxWorks when kernelTimeSlice is disabled (zero).
I disagree with this statement. See this document section 2.2.3 with the heading Round-Robin Scheduling.
Round-robin scheduling uses time
slicing to achieve fair allocation of
the CPU to all tasks with the same
priority. Each task, in a group of
tasks with the same priority, executes
for a defined interval or time slice.
Round-robin scheduling is enabled by
calling kernelTimeSlice( ), which
takes a parameter for a time slice, or
interval. [...] If round-robin
scheduling is enabled, and preemption
is enabled for the executing task, the
system tick handler increments the
task's time-slice count.
Timeslicing is inherent in round-robin scheduling. Otherwise you are relying on a task to give up CPU control, which round-robin scheduling is intended to solve.
The answers here and even the Wikipedia article describe round-robin scheduling to inherently include periodic timeslicing. While this is very common, I believe that Round-Robin scheduling and timeslicing are not exactly the same thing. Certainly, for timeslicing to make sense, round-robin schedling is implied when rotating to each task, however you can do round-robin scheduling without having timeslicing. That is, each task at the same priority in the round-robin rotation can be allowed to run until they reach a resource block condition and only then having the next task in the rotation run. In other words, when equal priority tasks exist, the reschedling points are not time pre-emptive.
The above idea is actually realized specifically in the case of Wind River's VxWorks kernel. Within their priority scheme, tasks of each priority run round robin but do not timeslice without specifically enabling that feature in the kernel. The reason for this flexibility is to avoid the overhead of timeslicing tasks that are already known to run into a block within a well bounded time.
Therefore, while timeslicing based scheduling implies round-robin scheduling, round-robin scheduling does not require equal time based timeslicing.
An opinion. It seems that we are intertwining two mechanisms into one. Assuming only the OP's original assertion "In a multitasking operating system context" then
1 - A round robin scheduler always schedules the next item in a circular queue.
2 - How the scheduler regains control to perform the scheduling is separate and unrelated.
I don't disagree that the most prevalent method for 2 is time-slicing / yield waiting for resource, but as has been noted there are others. If I am not mistaken the first Mac's didn't utilize time-slicing, they used voluntary yield / yield waiting for resource (20+ year old brain cells can be wrong sometimes;).
Round robin is a simple scheduling algorithm where time is divided evenly among jobs without priority.
For example - if you have 5 processes running - each process will be allowed to run for 1/5 a unit of time before another process is allowed to run. Round robin is typically easy to implement in an OS.
Actaully, you are getting confused with Preemptive scheduling and Round robin. Infact RR is part of Preemptive scheduling.
Round Robin scheduling is based on time sharing also known as quantum (max time given by CPU to any process in one go). There are multiple processes(which require different time to complete aka burst time) in a queue and CPU has to process them all so it keeps switching between processes to give every process equal time based on the quantum value. This type of scheduling is known as Round Robin scheduling.
Checkout this simple video to understand round robin scheduling easily: https://www.youtube.com/watch?v=9hw-_qJ55K4
Had an interesting discussion with some colleagues about the best scheduling strategies for realtime tasks, but not everyone had a good understanding of the common or useful scheduling strategies.
For your answer, please choose one strategy and go over it in some detail, rather than giving a little info on several strategies. If you have something to add to someone else's description and it's short, add a comment rather than a new answer (if it's long or useful, or simply a much better description, then please use an answer)
What is the strategy - describe the general case (assume people know what a task queue is, semaphores, locks, and other OS fundamentals outside the scheduler itself)
What is this strategy optimized for (task latency, efficiency, realtime, jitter, resource sharing, etc)
Is it realtime, or can it be made realtime
Current strategies:
Priority Based Preemptive
Lowest power slowest clock
-Adam
As described in a paper titled Real-Time Task Scheduling for Energy-Aware Embedded Systems, Swaminathan and Chakrabarty describe the challenges of real-time task scheduling in low-power (embedded) devices with multiple processor speeds and power consumption profiles available. The scheduling algorithm they outline (and is shown to be only about 1% worse than an optimal solution in tests) has an interesting way of scheduling tasks they call the LEDF Heuristic.
From the paper:
The low-energy earliest deadline first
heuristic, or simply LEDF, is an
extension of the well-known earliest
deadline first (EDF) algorithm. The
operation of LEDF is as follows: LEDF
maintains a list of all released
tasks, called the “ready list”. When
tasks are released, the task with the
nearest deadline is chosen to be
executed. A check is performed to see
if the task deadline can be met by
executing it at the lower voltage
(speed). If the deadline can be met,
LEDF assigns the lower voltage to the
task and the task begins execution.
During the task’s execution, other
tasks may enter the system. These
tasks are assumed to be placed
automatically on the “ready list”.
LEDF again selects the task with the
nearest deadline to be executed. As
long as there are tasks waiting to be
executed, LEDF does not keep the pro-
cessor idle. This process is repeated
until all the tasks have been
scheduled.
And in pseudo-code:
Repeat forever {
if tasks are waiting to be scheduled {
Sort deadlines in ascending order
Schedule task with earliest deadline
Check if deadline can be met at lower speed (voltage)
If deadline can be met,
schedule task to execute at lower voltage (speed)
If deadline cannot be met,
check if deadline can be met at higher speed (voltage)
If deadline can be met,
schedule task to execute at higher voltage (speed)
If deadline cannot be met,
task cannot be scheduled: run the exception handler!
}
}
It seems that real-time scheduling is an interesting and evolving problem as small, low-power devices become more ubiquitous. I think this is an area in which we'll see plenty of further research and I look forward to keeping abreast!
One common real-time scheduling scheme is to use priority-based preemptive multitasking.
Each tasks is assigned a different priority level.
The highest priority task on the ready queue will be the task that runs. It will run until it either gives up the CPU (i.e. delays, waits on a semaphore, etc...) or a higher priority task becomes ready to run.
The advantage of this scheme is that the system designer has full control over what tasks will run at what priority. The scheduling algorithm is also simple and should be deterministic.
On the other hand, low priority tasks might be starved for CPU. This would indicate a design problem.