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The semaphore variables full, empty and mutex are initialized to $0$, $n$ and $1$, respectively. Process P1 repeatedly adds one item at a time to a buffer of size $n$, and process P2 repeatedly removes one item at a time from the same buffer using the programs given below. In the programs, $K$, $L$, $M$ and $N$ are unspecified statements.

 $P_1$ while (1) { K; P(mutex); Add an item to the buffer; V(mutex); L; } $P_2$ while (1) { M; P(mutex); Remove an item from the buffer; V(mutex); N; }

The statements $K$, $L$, $M$ and $N$ are respectively

1. P(full), V(empty), P(full), V(empty)
2. P(full), V(empty), P(empty), V(full)
3. P(empty), V(full), P(empty), V(full)
4. P(empty), V(full), P(full), V(empty)

@ Sukhbir Singh, according to you, initially buffer is full. So, FULL=n and EMPTY=0. Now, if process P1 want to execute then due to P(EMPTY), semaphore EMPTY value becomes "-1" so, process P1 got blocked until process P2 perform V(EMPTY). So, in this scenario only P2 process can execute first.
we can’t swap K,p(mutex) and M,p(mutex) otherwise their will be deadlock
This question plays with words. Else it’s an easy level question.

$P_1$ is the producer. So, it must wait for full condition. But semaphore $\text{ full }$ is initialized to 0 and semaphore $\text{ empty }$ is initialized to $n$, meaning $full = 0$ implies no item and $empty=n$ implies space for $n$ items is available. So, $P_1$ must wait for semaphore $\text{ empty }$ - $K - P( \text{ empty })$ and similarly $P_2$ must wait for semaphore $\text{ full }$- $M - P( \text{ full })$. After accessing the critical section (producing/consuming item) they do their respective $V$ operation. Thus option D.
by

Incase if the question would have been full, empty and mutex are initialized to n, 0 and 1, then Option B) is the answer ...

whether it is fixed that full is for producer and empty is for consumer ;

Since P(Empty) at M then P2 remove an item from an Empty buffer, which fail the synchronisation problem that is P2 wants to remove the item which is not present inside the buffer. So option B and C are not the answer.

If P(Full) then P1 cant enter an item into the buffer, so option A is not the answer.

Hence option D ensures that P2 never remove an item from an empty buffer, while enter the item easily.

by

It ensures that P2 never removes an item from an empty buffer.

In any PRODUCER-CONSUMER problem this 4 criteria must satisfy

(3)When a writer is writing, no reader must be allowed.
(4)Multiple writers(more than 1) should not be allowed.

here 2 counting semaphores are used to keep track how much the buffer if empty or full they are b_full=0 ( means initially nothing in the buffer) & b_empty=100 (means initially empty cells in buffer is 100 or buffer is all empty)

now for K - Producer produced an item & he wants to insert into buffer but reader should not read meanwhile producer adding item to buffer for that (P(mutex)) now it downs the semaphore P(b_empty) means now buffer_empty=99 or there is 1 item inserted.

L - after inserting, it ups the semaphore V(b_full) means Buffer_full=1 or there is 1 item present in buffer.

now for M - it downs the "P(b_full)" semaphore means it has taken out 1 item from buffer and consumes item and

N - up the "V(b_empty)" to state that one more cell of buffer is empty now.

Nice Explanation

Thanks.