Discussion
Three tasks have execution times 1, 2 and 1 ms and periods 4, 6 and 8 ms. The utilisation is about
Answer: C
Correct answer: 0.71
\[ \frac{1}{4} + \frac{2}{6} + \frac{1}{8} = 0.25 + 0.333 + 0.125 = 0.708. \] The set demands about 71 per cent of the processor, which must now be compared against the bound for three tasks rather than against unity.
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