A computer has two processors, $M_1$ and $M_2$. Four processes $P_1, P_2, P_3, P_4$ with CPU bursts of 20, 16, 25, and 10 milliseconds, respectively, arrive at the same time and these are the only processes in the system. The scheduler uses non-preemptive priority scheduling, with priorities decided as follows: • $M_1$ uses priority of execution for the processes as, $P_1 > P_3 > P_2 > P_4$, i.e., $P_1$ and $P_4$ have highest and lowest priorities, respectively. • $M_2$ uses priority of execution for the processes as, $P_2 > P_3 > P_4 > P_1$, i.e., $P_2$ and $P_1$ have highest and lowest priorities, respectively. A process $P_i$ is scheduled to a processor $M_k$, if the processor is free and no other process $P_j$ is waiting with higher priority. At any given point of time, a process can be allocated to any one of the free processors without violating the execution priority rules. Ignore the context switch time. What will be the average waiting time of the processes in milliseconds?

Let all processes arrive at $t = 0$ ms.

### Process Details:

• $P_1$: Burst time = $20\text{ ms}$

• $P_2$: Burst time = $16\text{ ms}$

• $P_3$: Burst time = $25\text{ ms}$

• $P_4$: Burst time = $10\text{ ms}$

### Priority Orders:

• For $M_1$: $P_1 > P_3 > P_2 > P_4$

• For $M_2$: $P_2 > P_3 > P_4 > P_1$

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### Step-by-step Scheduling:

1. At $t = 0\text{ ms}$:

   • Both $M_1$ and $M_2$ are free.

   • Among the available processes $\{P_1, P_2, P_3, P_4\}$:

     • The highest priority process for $M_1$ is $P_1$. Thus, $P_1$ is allocated to $M_1$ for duration $20\text{ ms}$ (executes from $t = 0$ to $t = 20$).

     • Among the remaining processes $\{P_2, P_3, P_4\}$, the highest priority process for $M_2$ is $P_2$. Thus, $P_2$ is allocated to $M_2$ for duration $16\text{ ms}$ (executes from $t = 0$ to $t = 16$).

2. At $t = 16\text{ ms}$:

   • Processor $M_2$ finishes $P_2$ and becomes free.

   • Waiting processes are $\{P_3, P_4\}$.

   • For $M_2$, priority order is $P_3 > P_4$. Therefore, $P_3$ is scheduled on $M_2$ for duration $25\text{ ms}$ (executes from $t = 16$ to $t = 41$).

3. At $t = 20\text{ ms}$:

   • Processor $M_1$ finishes $P_1$ and becomes free.

   • The only remaining process waiting is $P_4$.

   • Therefore, $P_4$ is scheduled on $M_1$ for duration $10\text{ ms}$ (executes from $t = 20$ to $t = 30$).

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### Waiting Times: $$\begin{aligned} \text{Waiting Time for } P_1 &= 0 - 0 = 0\text{ ms} \\ \text{Waiting Time for } P_2 &= 0 - 0 = 0\text{ ms} \\ \text{Waiting Time for } P_3 &= 16 - 0 = 16\text{ ms} \\ \text{Waiting Time for } P_4 &= 20 - 0 = 20\text{ ms} \end{aligned}$$

### Average Waiting Time:

Hence, the correct option is (A).