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Scheduling

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Consider a uniprocessor system which has to execute an infinite sequence of processes which arrive periodically at the rate of 6 processes per minute and each process requires an average of 9 seconds of service time. The fraction of time CPU busy is _________ (in sec).
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6 processes per minute => 6 processes for 60 seconds

1 process => 10 seconds

it means at time 0 assume P1 has arrived and at time 10 P2 arrives and so on...

Service time = 9 seconds.

Fraction of time cpu busy = 9 / 10 = 0.9

Fraction of time cpu idle = 0.1.
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