Homework Answers
Utilization = (20/100)+(30/150)+(90/200) = 0.85
The sufficient condition for 3 processors under which we can conclude that system is schedulable is
U = 3(2^(1/3)-1) = 0.77976
0.85> 0.77976 (utilization should be less than U is necessary condition)
System with higher utilization is not schedulable with rate monotonic algorithm.
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