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    Question

    A system is using the Round Robin (RR) scheduling

    algorithm with a time quantum of 4 milliseconds. The processes P1, P2, and P3 arrive at times 0ms, 2ms, and 4ms respectively, with burst times of 5ms, 7ms, and 10ms. Assuming the context switch overhead is negligible, at what time does P2 complete its execution?
    A 16 ms Correct Answer Incorrect Answer
    B 18 ms Correct Answer Incorrect Answer
    C 20 ms Correct Answer Incorrect Answer
    D 22 ms Correct Answer Incorrect Answer
    E 24 ms Correct Answer Incorrect Answer

    Solution

    In the Round Robin scheduling algorithm, each process gets an equal share of the CPU for a fixed time quantum. If a process doesn't complete within its quantum, it is added back to the ready queue. Step-by-step timeline: 1. At time 0ms, P1 starts executing and uses the CPU for 4ms (quantum time). Remaining burst for P1: 1ms. 2. At time 4ms, P2 starts and executes for 4ms. Remaining burst for P2: 3ms. 3. At time 8ms, P3 executes for 4ms. Remaining burst for P3: 6ms. 4. At time 12ms, P1 resumes and completes its remaining 1ms. Now, P2 is next. 5. At time 13ms, P2 executes its remaining 3ms and completes at 20ms. Hence, P2 finishes execution at 20ms. Why Other Options Are Incorrect: • A) 16 ms: Incorrect because P2 has a remaining burst time after its first execution. It cannot complete before P1 and P3 are processed. • B) 18 ms: Incorrect as this doesn't consider the remaining execution time of P3 before P2 resumes. • D) 22 ms & E) 24 ms: These are overestimated values that do not align with the timeline for quantum-based processing.

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