Question

Given Data:
- Two processors: $M_1$ and $M_2$
- Four processes: $P_1, P_2, P_3, P_4$
- CPU bursts: $P_1 = 20$, $P_2 = 16$, $P_3 = 25$, $P_4 = 10$
- Priority order for $M_1$: $P_1 > P_3 > P_2 > P_4$
- Priority order for $M_2$: $P_2 > P_3 > P_4 > P_1$
- Scheduling: Non-preemptive priority scheduling

Step 1: Assigning Processes to Processors
At $t = 0$, both processors are free. The highest-priority process for each processor is scheduled:
- $M_1$ gets $P_1$ (highest priority on $M_1$, burst = 20 ms).
- $M_2$ gets $P_2$ (highest priority on $M_2$, burst = 16 ms).

Step 2: Scheduling Remaining Processes
- $P_2$ finishes at $t = 16$, so $M_2$ picks the next highest-priority process, $P_3$ (burst = 25 ms).
- $P_1$ finishes at $t = 20$, so $M_1$ picks $P_4$ (burst = 10 ms).
- $P_4$ finishes at $t = 30$, and $P_3$ finishes at $t = 41$.

Step 3: Calculate Waiting Times
- $P_1$ starts at 0, so waiting time = $0$ ms.
- $P_2$ starts at 0, so waiting time = $0$ ms.
- $P_3$ starts at 16, so waiting time = $16$ ms.
- $P_4$ starts at 20, so waiting time = $20$ ms.

Step 4: Compute Average Waiting Time
$$\text{Average waiting time} = \frac{(0 + 0 + 16 + 20)}{4} = \frac{36}{4} = 9 \text{ ms}.$$

Final Answer: The average waiting time of the processes is $9$ milliseconds.

• A. 9.00
• B. 8.75
• C. 6.50
• D. 7.50

Hint:

In priority scheduling, the waiting time depends on the order in which processes are executed. The waiting time for each process is the total time it waits for other higher-priority processes to finish.

Answer 1

The Correct Option is A

Step 1: Initial Setup and Process Priorities
- Processor $M_1$ priority order: $P_1 > P_3 > P_2 > P_4$.
- Processor $M_2$ priority order: $P_2 > P_3 > P_4 > P_1$.
- CPU burst times:
$P_1 = 20$ ms, $P_2 = 16$ ms, $P_3 = 25$ ms, $P_4 = 10$ ms.

Step 2: Process Scheduling
- $M_1$ schedules $P_1$ first (highest priority), running for 20 ms.
- $M_2$ schedules $P_2$ first (highest priority), running for 16 ms.
- Remaining processes: $P_3$ (25 ms), $P_4$ (10 ms).
- $M_1$ next schedules $P_3$ (second priority), running for 25 ms.
- $M_2$ next schedules $P_4$ (second priority), running for 10 ms.

Step 3: Calculating Waiting Times
- $P_1$: Starts at $t=0$, waiting time = 0 ms.
- $P_2$: Starts at $t=0$, waiting time = 0 ms.
- $P_3$: Starts after $P_2$ finishes, at $t=16$, so waiting time = 16 ms.
- $P_4$: Starts after $P_1$ finishes, at $t=20$, so waiting time = 20 ms.

Step 4: Calculating Average Waiting Time
Total waiting time = $0 + 0 + 16 + 20 = 36$ ms.
Average waiting time = $\frac{36}{4} = 9$ ms.

Final Answer: The average waiting time is 9 milliseconds.