Process Synchronization

Question

If there are n processes executing concurrently using binary semaphore S , (n-1) processes have the code

do{                                   

wait(S);

<c.s>

signal(s);

<r.s>}while(1);

the code for the n-th process i.e Pn is given by

do{                                   

signal(S);

<c.s>

wait(S);

<r.s>}while(1);

a) What is the max no. of processes that can be there in the critical section simultaneosly ?
A. 2

B . 3

C. n-1

D. n

Comments

The answer given is 3

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Try this question too: https://gateoverflow.in/96839/operating-systems-2-5

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If s=0 then there's only two process in critical section at a time

Answer 1

total three process can enter in critical section .consider p1 enter in CS and get preempted after nth process came and signal it now both of them in critical section now s=1 again p2 will came enter in CS in which P1 is already there so total  number of process $P_{1}$+$P_{2}$+$P_{N}$=3 so answer will be 3.

Answer 2

P1 executes its code and due to wait() monitor it got suspended and then Pn execute its code and signal() P1 and enter in critical section and now P1 also enter in critical section hence 2 processes are there in critical section at any point of time.

Comments

If s=1 then????