Now let us add another task of middle priority say M. Now the task priorities are in the order of L < M < H. In our example, M doesn’t share the same Critical Section (CS). In this case, the following sequence of task running would result in a ‘Priority Inversion’ problem.
L is running in CS; H also needs to run in CS; H waits for L to come out of CS; M interrupts L and starts running; M runs till completion and relinquishes control; L resumes and starts running till the end of CS; H enters CS and starts running.
