What is priority inversion?

Question

I\'ve heard the phrase \'priority inversion\' in reference to development of operating systems.

What exactly is priority inversion?

What is the problem it\'s

Answer 1

[ Assume, Low process = LP, Medium Process = MP, High process = HP ]

LP is executing a critical section. While entering the critical section, LP must have acquired a lock on some object, say OBJ. LP is now inside the critical section.

Meanwhile, HP is created. Because of higher priority, CPU does a context switch, and HP is now executing (not the same critical section, but some other code). At some point during HP's execution, it needs a lock on the same OBJ (may or may not be on the same critical section), but the lock on OBJ is still held by LP, since it was pre-empted while executing the critical section. LP cannot relinquish now because the process is in READY state, not RUNNING. Now HP is moved to BLOCKED / WAITING state.

Now, MP comes in, and executes its own code. MP does not need a lock on OBJ, so it keeps executing normally. HP waits for LP to release lock, and LP waits for MP to finish executing so that LP can come back to RUNNING state (.. and execute and release lock). Only after LP has released lock can HP come back to READY (and then go to RUNNING by pre-empting the low priority tasks.)

So, effectively it means that until MP finishes, LP cannot execute and hence HP cannot execute. So, it seems like HP is waiting for MP, even though they are not directly related through any OBJ locks. -> Priority Inversion.

A solution to Priority Inversion is Priority Inheritance -

increase the priority of a process (A) to the maximum priority of any other process waiting for any resource on which A has a resource lock.

Answer 2

Let me make it very simple and clear. (This answer is based on the answers above but presented in crisp way).

Say there is a resource R and 3 processes. L, M, H. where p(L) < p(M) < p(H) (where p(X) is priority of X).

Say

• L starts executing first and catch holds on R. (exclusive access to R)
• H comes later and also want exclusive access to R and since L is holding it, H has to wait.
• M comes after H and it doesn't need R. And since M has got everything it wants to execute it forces L to leave as it has high priority compared to L. But H cannot do this as it has a resource locked by L which it needs for execution.

Now making the problem more clear, actually the M should wait for H to complete as p(H) > p(M) which didn't happen and this itself is the problem. If many processes such as M come along and don't allow the L to execute and release the lock H will never execute. Which can be hazardous in time critical applications

And for solutions refer the above answers :)

Answer 3

A scheduling challenge arises when a higher-priority process needs to read or modify kernel data that are currently being accessed by a lower-priority process—or a chain of lower-priority processes. Since kernel data are typically protected with a lock, the higher-priority process will have to wait for a lower-priority one to finish with the resource. The situation becomes more complicated if the lower-priority process is preempted in favor of another process with a higher priority. As an example, assume we have three processes—L, M, and H—whose priorities follow the order L < M < H. Assume that process H requires resource R,which is currently being accessed by process L.Ordinarily,process H would wait for L to finish using resource R. However, now suppose that process M becomes runnable, thereby preempting process L. Indirectly, a process with a lower priority—process M—has affected how long process H must wait for L to relinquish resource R. This problem is known as priority inversion.It occurs only in systems with more than two priorities,so one solution is to have only two priorities.That is insufficient for most general-purpose operating systems, however. Typically these systems solve the problem by implementing a priority-inheritance protocol. According to this protocol, all processes that are accessing resources needed by a higher-priority process inherit the higher priority until they are finished with the resources in question.When they are finished,their priorities revert to their original values. In the example above, a priority-inheritance protocol would allow process L to temporarily inherit the priority of process H,thereby preventing process M from preempting its execution. When process L had finished using resource R,it would relinquish its inherited priority from H and assume its original priority.Because resource R would now be available, process H—not M—would run next. Reference :ABRAHAM SILBERSCHATZ

Answer 4

Suppose an application has three threads:

Thread 1 has high priority.
Thread 2 has medium priority.
Thread 3 has low priority.

Let's assume that Thread 1 and Thread 3 share the same critical section code

Thread 1 and thread 2 are sleeping or blocked at the beginning of the example. Thread 3 runs and enters a critical section.

At that moment, thread 2 starts running, preempting thread 3 because thread 2 has a higher priority. So, thread 3 continues to own a critical section.

Later, thread 1 starts running, preempting thread 2. Thread 1 tries to enter the critical section that thread 3 owns, but because it is owned by another thread, thread 1 blocks, waiting for the critical section.

At that point, thread 2 starts running because it has a higher priority than thread 3 and thread 1 is not running. Thread 3 never releases the critical section that thread 1 is waiting for because thread 2 continues to run.

Therefore, the highest-priority thread in the system, thread 1, becomes blocked waiting for lower-priority threads to run.

Answer 5

Imagine three (3) tasks of different priority: tLow, tMed and tHigh. tLow and tHigh access the same critical resource at different times; tMed does its own thing.

1. tLow is running, tMed and tHigh are presently blocked (but not in critical section).
2. tLow comes along and enters the critical section.
3. tHigh unblocks and since it is the highest priority task in the system, it runs.
4. tHigh then attempts to enter the critical resource but blocks as tLow is in there.
5. tMed unblocks and since it is now the highest priority task in the system, it runs.

tHigh can not run until tLow gives up the resource. tLow can not run until tMed blocks or ends. The priority of the tasks has been inverted; tHigh though it has the highest priority is at the bottom of the execution chain.

To "solve" priority inversion, the priority of tLow must be bumped up to be at least as high as tHigh. Some may bump its priority to the highest possible priority level. Just as important as bumping up the priority level of tLow, is dropping the priority level of tLow at the appropriate time(s). Different systems will take different approaches.

When to drop the priority of tLow ...

1. No other tasks are blocked on any of the resources that tLow has. This may be due to timeouts or the releasing of resources.
2. No other tasks contributing to the raising the priority level of tLow are blocked on the resources that tLow has. This may be due to timeouts or the releasing of resources.
3. When there is a change in which tasks are waiting for the resource(s), drop the priority of tLow to match the priority of the highest priority level task blocked on its resource(s).

Method #2 is an improvement over method #1 in that it shortens the length of time that tLow has had its priority level bumped. Note that its priority level stays bumped at tHigh's priority level during this period.

Method #3 allows the priority level of tLow to step down in increments if necessary instead of in one all-or-nothing step.

Different systems will implement different methods depending upon what factors they consider important.

• memory footprint
• complexity
• real time responsiveness
• developer knowledge

Hope this helps.