A five-stage pipeline has stage delays of 50, 60, 80, 75, 80 and 70…
A five-stage pipeline has stage delays of 50, 60, 80, 75, 80 and 70 nanoseconds. The registers that are used between the pipeline stages have a delay of 10 nanoseconds each.The total time to execute 500 independent instructions on this pipeline, assuming there are no pipeline stalls, is _______ nanoseconds.
Answer: A. 45360 — Answer: 45360 ns Given: number of pipeline stages k = 5, number of instructions n = 500. Stage delays provided: 50, 60, 80, 75, 80, 70 ns. The cycle time is…
- A.
45360
- B.
40400
- C.
45270
- D.
45460
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Correct answer: A
Answer: 45360 ns
Given: number of pipeline stages k = 5, number of instructions n = 500.
Stage delays provided: 50, 60, 80, 75, 80, 70 ns. The cycle time is set by the slowest stage plus the register delay.
Compute cycle time tp = max(stage delays) + register delay = 80 ns + 10 ns = 90 ns.
Total time = (number of stages + number of instructions - 1) × tp = (k + n - 1) × tp = (5 + 500 - 1) × 90 = 504 × 90.
Compute: 504 × 90 = 45360 ns. (Or equivalently pipeline fill = 5 × 90 = 450 ns, plus remaining 499 × 90 = 44910 ns; total = 450 + 44910 = 45360 ns.)