Question 33

Calculate currently allocated resources and available resources:

• Total tape drives $= 9$.
• Current allocation: $P_1 = 3, P_2 = 1, P_3 = 3$.
• Total allocated $= 3 + 1 + 3 = 7$.
• Available drives $= 9 - 7 = 2$.

Calculate the Need vector for each process ($\text{Need} = \text{Max} - \text{Allocation}$):

• $\text{Need}(P_1) = 8 - 3 = 5$
• $\text{Need}(P_2) = 6 - 1 = 5$
• $\text{Need}(P_3) = 5 - 3 = 2$

Run the safety algorithm with $\text{Available} = 2$:

1. Check $P_3$: $\text{Need}(P_3) = 2 \le 2$. $P_3$ can finish and release its allocated 3 tape drives.
   New available $= 2 + 3 = 5$.
2. Check $P_1$: $\text{Need}(P_1) = 5 \le 5$. $P_1$ can finish and release its allocated 3 tape drives.
   New available $= 5 + 3 = 8$.
3. Check $P_2$: $\text{Need}(P_2) = 5 \le 8$. $P_2$ can finish and release its allocated 1 tape drive.
   New available $= 8 + 1 = 9$.

A valid safe sequence $\langle P_3, P_1, P_2 \rangle$ exists. Therefore, the system is in a Safe state. A safe state is by definition Not Deadlocked.
This matches option B.