Consider a computer with two processes, H, with high priority, and L, with low priority. The scheduling rules are such that H is run whenever it is in ready state. At a certain moment, with L in its critical region, H becomes ready to run (e. g., an I/O operation completes). H now begins busy waiting, but since L is never scheduled with H is running, L never gets the chance to leave its critical region, so H loops forever. This situation is sometimes referred to as the priority inversion problem. If instead of priority scheduling, we use round-robin scheduling, will we encounter the same problem?

Question

Steve Ponce

Computers & Technology

22 September, 00:08

Consider a computer with two processes, H, with high priority, and L, with low priority. The scheduling rules are such that H is run whenever it is in ready state. At a certain moment, with L in its critical region, H becomes ready to run (e. g., an I/O operation completes). H now begins busy waiting, but since L is never scheduled with H is running, L never gets the chance to leave its critical region, so H loops forever. This situation is sometimes referred to as the priority inversion problem. If instead of priority scheduling, we use round-robin scheduling, will we encounter the same problem?

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Answers (1)

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Mason Kemp · Answer 1

No

Explanation:

We know that Priority inversion can be seen as a scheduling situation in which higher priority task gets preempted by lower priority task. And that this is one of the main problem in priority scheduling.

If we overlook Priority scheduling and instead we decided to use Round Robin scheduling we will never encounter the priority inversion problem. As, in Round Robin scheduling the process generally use a certain amount of time or time slices for each processes before it is force to wait for the next process.