Consider three concurrent processes P1, P2 and P3 as shown below, which access a shared variable D that has been initialized to 100. Refer to the table below. The processes are executed on a uniprocessor system running a time-shared operating system. If the minimum and maximum possible values of D after the three processes have completed execution are X and Y respectively, then the value of Y − X is _______.

Each statement $D = D + k$ consists of three atomic operations:

1. READ(D) into a local register

2. ADD constant $k$ to the local register

3. WRITE the local register value back to $D$

Initially, $D = 100$.

• $P_1$: $D = D + 20$

• $P_2$: $D = D - 50$

• $P_3$: $D = D + 10$

To minimize $D$ ($X$):

• $P_2$ reads $D = 100$ and is preempted before writing back ($D - 50 = 50$).

• $P_1$ and $P_3$ run to completion: $D$ becomes $100 + 20 + 10 = 130$.

• $P_2$ resumes and writes its local value $50$ to $D$.

• Therefore, minimum value $X = 50$.

To maximize $D$ ($Y$):

• $P_1$ reads $D = 100$ and is preempted before writing back ($D + 20 = 120$).

• $P_2$ and $P_3$ run to completion: $D$ becomes $100 - 50 + 10 = 60$.

• $P_1$ resumes and writes its local value $120$ to $D$.

• Alternatively, if $P_3$ reads $D=100$, $P_1$ finishes ($120$), $P_2$ finishes ($70$), and $P_3$ writes $110$, the maximum achieved is $120$ using $P_1$ or $130$ if $P_2$ is overwritten.

• Let's check: $P_2$ reads $D = 100$, preempted. Then $P_1$ and $P_3$ execute fully, setting $D = 100 + 20 + 10 = 130$. $P_2$ writes $50$—this was for minimum.

• If $P_2$ reads $D=100$ and is preempted, then $P_1$ and $P_3$ run sequentially: $D = 100 + 20 + 10 = 130$. If $P_2$ writes $50$ before $P_1$ or $P_3$ completes, the maximum value remaining at the end without interruption would simply be if $P_2$ runs first ($D=50$), and then $P_1$ and $P_3$ add $30$, resulting in $80$. But if $P_2$'s decrement is completely lost (overwritten), say $P_2$ reads $100$, $P_1$ reads $100$, $P_2$ writes $50$, then $P_1$ writes $120$, and then $P_3$ runs: $D = 120 + 10 = 130$.

• Thus, maximum value $Y = 130$.

Calculating $Y - X$: