Discussion
For those two tasks, the response time of the lower priority one under rate monotonic is
Answer: A
Iterating \( R = C_B + \lceil R/T_A \rceil C_A \) from \( R = 4 \) gives \( 4 + 1 \times 2 = 6 \), then \( 4 + 2 \times 2 = 8 \), then 8 again, so the response time is 8 ms against a deadline of 20 ms. Stopping at the first iterate is the usual error, and it understates the answer.
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