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The clock interrupt handler on a certain computer requires $2\: msec$ (including process switching overhead) per clock tick. The clock runs at $60\: Hz.$ What fraction of the CPU is devoted to the clock?
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Answer:

$2\;\text{msec}$ $\;60 $ times a second $=120 \;\frac{\text{msec}}{\text{sec}}=12\%\;\text{of the CPU time}$

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