MadeEasy Test Series: Operating System - Process Synchronization

Question

Consider two processes P1 and P2 accessing shared variable ‘x’ and ‘y’ protected by two binary semaphores Sx and Sy respectively both initialized to 1. ‘wait’ and ‘signal’ denote the usual semaphore operators, where wait decreaments the semaphore value, and signal increments the semaphore value.

S1: P = Wait (Sx), Q = Wait (Sy), R = Signal (Sx) and S = Signal (Sx) may lead to starvation.
S2: P = Wait (Sy ), Q = Wait (Sy), R = Signal (Sx) and S = Signal (Sx) may lead to race condition.

Which of the following is true?

How S1 lead to starvation? There is no While loop after performing the last signal operation, the process is not going to do these operations again. Also both these process executing two simple statement which will take constant time.

Comments

For S1 there is no deadlock, as for both processes, the first statement is Wait($S_x$) and the second statement is wait($S_y$). but for S2 we may have a deadlock.

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yes , agree.

Answer 1

S1 not providing wound and wait

S2  provide deadlock because p1 will be  blocked there are two wait (Sy)  Sy value becomes 0 after p2 also get blocked if start from P1

But if start from P2 only it perform it's operation