Consider the two functions incr and decr shown below.incr(){ wait(s); X = X+1; signal(s); } decr(){ wait(s); X = X-1; signal(s); }There are 5 threads each invoking incr once, and 3 threads each invoking decr once, on the same shared variable $X$. The initial value of $X$ is 10. Suppose there are two implementations of the semaphore $s$, as follows: **I-1:** $s$ is a binary semaphore initialized to 1. **I-2:** $s$ is a counting semaphore initialized to 2. Let $V_1, V_2$ be the values of $X$ at the end of execution of all the threads with implementations **I-1**, **I-2**, respectively. Which one of the following choices corresponds to the minimum possible values of $V_1, V_2$, respectively?

### Implementation I-1 (Binary Semaphore initialized to 1): A binary semaphore initialized to $1$ provides strict mutual exclusion for the critical sections. At any point in time, at most one thread can execute inside its critical section.

Since each thread executes atomically with respect to the others:

• $5$ increments add $+5$ to $X$.

• $3$ decrements subtract $-3$ from $X$.

Because mutual exclusion is guaranteed, no race condition can occur, so the minimum (and only) possible value of $V_1$ is 12.

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### Implementation I-2 (Counting Semaphore initialized to 2): A counting semaphore initialized to $2$ allows up to $2$ threads to enter their critical sections concurrently. This enables race conditions where thread updates can be overwritten.

Each operation $X = X \pm 1$ comprises three steps: Read(X), Modify, and Write(X).

To minimize the final value $V_2$, we want all $5$ increments to be overwritten while all $3$ decrements take effect:

1. Thread $D_1$ enters the critical section ($s$ becomes $1$), executes Read(X) and gets $10$, computes $10 - 1 = 9$, and is preempted before writing back.

2. Since $s = 1$, all $5$ incr threads ($I_1$ to $I_5$) execute one by one, updating $X$ sequentially from $10$ to $15$.

3. Thread $D_1$ resumes and writes its stored value $9$ to $X$, completely overwriting all $5$ increments.

4. Threads $D_2$ and $D_3$ execute normally one after another, decrementing $X$ from $9 \to 8 \to 7$.

Since $X$ starts at $10$ and there are at most $3$ decrement operations, the value of $X$ cannot decrease below $10 - 3 = 7$.

Thus, the minimum possible value of $V_2$ is 7.

Therefore, the minimum possible values of $(V_1, V_2)$ are (12, 7), which corresponds to option (C).