Skip to main content

Pipeline speedup ratio derivation

Short answer

For a k-segment pipeline, n tasks take (k+n-1) clock cycles while non-pipelined takes n*k. Speedup S = n*k/(k+n-1). As n becomes large, S ≈ k, so the maximum speedup equals the number of pipeline segments.

Derive the expression showing that the speed up ratio equals the number of segments in a pipeline.

Let a pipeline have k segments, each taking one clock time tp, and let n tasks be executed.

  • Time for one task in a non-pipelined unit: tn = k · tp (all segments done in series).
  • Time for n tasks without pipeline: T1 = n · tn = n · k · tp.
  • Time for n tasks with pipeline: the first task takes k·tp, each of the remaining n - 1 comes out one clock later: T2 = (k + n - 1) · tp.

Speedup ratio:

 S = T1 / T2 = n·k·tp / ((k + n - 1)·tp) = n·k / (k + n - 1)

When the number of tasks is very large (n >> k − 1), k + n − 1 ≈ n, so

 S = n·k / n = k

So the maximum speedup equals the number of segments k. Example: k = 4, n = 100: S = 400/103 = 3.88 ≈ 4. (Assumes tn = k·tp, equal segment times and no stalls.)

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Computer Organization and Architecture chapter 4 questions →