The concept of pipelining is most effective in improving performance if the…

Concept

In a k-stage pipeline every stage is driven by one common clock, so a single clock period must be long enough for the slowest stage: Tclk = max(τ1, τ2, …, τk). Throughput is 1/Tclk, so it is fixed by the largest stage delay alone and not by the other stages.

The work per task is the sum Στ = τ1 + … + τk. Because Στ ≤ k · max(τi), the limiting speedup of a k-stage pipeline is capped at k, and that cap is approached only when the stage delays are all equal.

Application

1. Write the per-task work of the unpipelined machine: it must run every stage in turn, taking Στ = τ1 + … + τk for one task.

2. Clock the k stages together: each stage must finish inside one period, so Tclk = max(τi), the slowest stage delay.

3. Fill and drain the pipeline for n tasks: the first task needs k periods and each later task adds one, giving a total of (k + n − 1) · Tclk.

4. Form the speedup over the unpipelined machine: S = n · Στ / [(k + n − 1) · Tclk], which tends to Στ / max(τi) as n grows large.

5. Bound that limiting ratio: Στ ≤ k · max(τi), so the limiting speedup is at most k, with equality exactly when every τi equals max(τi) — that is, when the stage tasks take about the same amount of time. For any finite n the fill-and-drain factor n / (k + n − 1) keeps the actual S strictly below that limit, whatever the split.

Cross-check

Four-stage splits of the same total work Στ = 8 ns, compared by their limiting speedup Στ / Tclk:

Contrast

• "require about the same amount of time": Στ = k · τ and Tclk = τ, so Στ / Tclk = k, the largest limiting speedup a k-stage pipeline can reach.

• "require different amount of time": Στ < k · max(τi), so the ratio falls below k; the shortfall is exactly the idle slack left in the faster stages every cycle.

• "require different amount of time with time difference between any two tasks being same": an evenly spaced set such as 0.5, 1.5, 2.5, 3.5 ns is still unequal, giving 8 / 3.5 ≈ 2.29 < 4 — a regular gap does not remove the slack.

• "require different amount of time with time difference between any two tasks being different": irregular delays such as 5, 1, 1, 1 ns give 8 / 5 = 1.60 < 4; the largest stage still sets the clock.

Pipelining therefore approaches its full k-fold benefit only when the stage tasks require about the same amount of time.