Solution - Speedup Ratio of Pipeline (Q6a, 2024)

Given:

• Non-pipeline time per task $t_n = 50$ sec
• Number of pipeline segments $k = 6$
• Pipeline clock cycle time $t_p = 10$ sec
• Number of tasks $n = 100$

1. Time without pipeline (Sequential):
$$T_{non-pipe} = n \times t_n = 100 \times 50 = 5000 \text{ sec}$$

2. Time with pipeline:
The first task takes $k \times t_p$ to fill the pipeline, then one task completes every $t_p$.

$$T_{pipe} = k \cdot t_p + (n-1) \cdot t_p = (k + n -1) \cdot t_p$$
$$T_{pipe} = (6 + 100 - 1) \times 10 = 105 \times 10 = 1050 \text{ sec}$$

3. Speedup Ratio ($S$):
$$S = \frac{T_{non-pipe}}{T_{pipe}} = \frac{5000}{1050} \approx 4.76$$

Interpretation:

• Theoretical maximum speedup = $t_n / t_p = 50/10 = 5$. As $n \to \infty$, $S \to 5$.
• For $n=100$, speedup is 4.76, i.e., pipeline is 4.76 times faster than non-pipeline for 100 tasks.
• Efficiency $\eta = S/k = 4.76/6 \approx 0.793$ or $79.3\%$.

Answer: Speedup $\approx$ 4.76 (1050 sec vs 5000 sec). If $n$ is very large, speedup approaches 5.

Note: If clock cycle were $t_n/k \approx 8.33$ sec ideally, speedup would approach 6. Here $t_p=10$ sec > ideal, so overhead reduces speedup.