MadeEasy Test Series 2018: Operating System - Process Syncronization

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?

Comments

if second question is about starvation then it is possible na process gets starve ?

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progress always implies starvation?

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how is S1 true?

process P1 can initially execute P(Sx); and P(Sy); and can enter its CS. Now suppose P2 also wanted to enter the CS and so it will get blocked on semaphore Sx. Now when P1 is about to leave its CS and it executes Signal(Sx) and got preempted, then the process that is blocked on this semaphore, ie. P2 will get unblocked and execute wait(Sy). Now P2 again gets blocked on semaphore Sy and gets preempted. Now P1 is again gets scheduled and calls signal(Sy), so it will unblock P2 now which can now enter into CS. So even if P1 expresses its desire to enter into CS, but P2 will get the chance if it is willing to enter into CS too since blocked processes are maintained in a queue and so they will be unblocked also in the same order.

where am I going wrong?

Answer 1

S1 looks correct...bounded waiting may not be satisfied in a solution with semaphore but since we have progress so it's true

but S2 is false