Discussion
For that same three-task set, the rate monotonic bound to compare against is
Answer: D
Correct answer: 0.780
\[ n\left(2^{1/n}-1\right) = 3\left(2^{1/3}-1\right) = 3 \times 0.2599 = 0.780. \] The bound falls with the number of tasks, towards \( \ln 2 = 0.693 \) as n grows without limit. Since 0.708 is below 0.780, this set is guaranteed schedulable under rate monotonic.
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