GO Classes DPP | OS | PROCESS SYNCHRONIZATION

Answer 1

1. Initial State:

• $S=3 ~(3$ resources available$)$.
   
• $Q=\emptyset$ (Empty).
   

2. $P_1, P_2, P_3$ arrive and execute $\verb|Wait|$ :

• All three find $S>0$.
   
• $S$ decrements three times: $3 \rightarrow 2 \rightarrow 1 \rightarrow 0$.
   
• Current State: $S=0 . P_1, P_2, P_3$ are holding resources. $Q=\emptyset$.
   

3. $P_4(L=40)$ and $P_5(L=50)$ arrive and execute $\verb|Wait|$ :

• $S=0$, so they cannot acquire the resource.
   
• They enter the Priority Queue $Q$.
   
• Since $Q$ is sorted by $L$ : $Q=\left[P_5(50), P_4(40)\right]$.
   
• Current State: $S=0 . Q=\left\{P_5, P_4\right\}$.

 

4. $P_1$ executes $\verb|Signal|$ :

• According to the logic: "If $Q$ is not empty, the process with the highest priority $L$ in $Q$ is woken up... $S$ remains $0$."
   
• $P_5$ has the highest priority ( $L=50$ ) in $Q$.
   
• Result: $P_5$ is woken up and takes $P_1$ 's resource.
   
• Current State: $S=0 . Q=\left\{P_4\right\}$.
   

5. $P_6(L=100)$ arrives and executes $\verb|Wait|$ :

• $S=0$, so $P_6$ must enter the queue $Q$.
   
• Since $Q$ is sorted by $L$ and $P_6$ has priority $100$, it moves to the front of the queue.
   
• Current State: $S=0 . Q=\left[P_6(100), P_4(40)\right]$.
   

A is False: $P_5$ (Priority 50) acquired the resource after $P_1$ signaled, not $P_4$.

B is True: $P_5$ acquired the resource because it was at the head of the queue when $P_1$ signaled. However, because $P_6$ arrived with a higher priority than $P_4, P_6$ will be the next to wake up when $P_2$ or $P_3$ signals. If more high-priority processes $(P_7, P_8 \ldots)$ continue to arrive, $P_4$ will stay in the queue indefinitely.

C is False: $P_6$ enters the queue. It must wait for the next $\verb|Signal|$ event $($from $P_2, P_3$, or $P_5)$ to acquire a resource. It cannot "snatch" the resource from $P_5$ during $P_1$ 's signal.

D is False: Strict priority queues are a classic cause of Starvation for low-priority processes.

Answer 2

B