Current total allocation:
$R=8$
$S=4+8=12$.
$\therefore Available=(12-8,24-12)=(4,12)$.
Initial remaining needs:
$P_1:(0,8)$
$P_2:(0,15)$
$P_3:(8,12)$.
Part $1:$ Maximum request by $P_1$
Suppose $P_1$ requests $x$ units of $S$.
Its remaining need becomes: $(0,8-x)$.
Available becomes: $(4,12-x)$.
For every $x\leq8$:
$8-x\leq12-x$.
So $P_1$ itself can always be guaranteed to finish.
Its maximum remaining claim is only $8$ units.
$\boxed{\therefore x_1=8}$.
Part $2:$ Maximum request by $P_2$
Suppose $P_2$ receives $x$ units of $S$.
Available becomes: $(4,12-x)$.
Its remaining need becomes: $(0,15-x)$.
$P_2$ cannot be the first process to finish because $15-x>12-x$ for every $x$.
$P_3$ cannot finish first either because it still requires $8$ units of $R$, while only $4$ are available.
Therefore $P_1$ must finish first.
For $P_1$ to finish:
$8\leq12-x$.
$\Rightarrow x\leq4$.
At $x=4$:
Available becomes $(4,8)$.
$P_1$ can finish and release its current 4 units of $S$:
$Work=(4,12)$
Now $P_2$ has remaining need $(0,11)$, which can be met.
So, $\boxed{x_2=4}$.
Part $3:$ Maximum request by $P_3$
Suppose $P_3$ receives $x$ additional units of $S$.
Available becomes $(4,12-x)$.
$P_3$ still cannot finish first because it needs 8 additional units of $R$.
Again, $P_1$ must finish first.
This requires:
$8\leq12-x$
$\Rightarrow x\leq4$
After $P_1$ finishes,
$Work=(4,16-x)$.
Before $P_3$ can complete, we need $P_2$ to finish and release its 8 units of $R$.
For $P_2$ to finish:
$15\leq16-x$.
$\Rightarrow x\leq1$.
For $x=1$, a safe order is:
$P_1\rightarrow P_2\rightarrow P_3$.
$\boxed{\therefore x_3=1}$.
Thus, $(x_1,x_2,x_3)=\boxed{(8,4,1)}$
Answer : A
