In Rate Monotonic scheduling, if a task has a period of 80 ms and an execution time of 55 ms, then its processor utilization and task rate will be
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Correct answer: A. 0.69, 12.5 Hz
Explanation
The processor utilization of a periodic task is its execution time divided by its period: U = 55 ms / 80 ms = 0.6875, which rounds to approximately 0.69. The task rate is the reciprocal of the period: rate = 1 / 80 ms = 1 / (80 × 10⁻³ s) = 12.5 Hz. So the utilization is about 0.69 and the task rate is 12.5 Hz.More System Software and Operating System questions
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