GATE CS 2016 Set 2 Q33 — Instruction Pipelining

The clock cycle time of a pipeline is determined by the latency of its longest stage. Let the stage latencies be $\tau_1, \tau_2,$ and $\tau_3$. We are given the relationship:

Expressing all stage latencies in terms of $\tau_1$:

• $\tau_1 = \tau_1$
• $\tau_2 = \frac{4}{3}\tau_1 \approx 1.33\tau_1$
• $\tau_3 = \frac{1}{2}\tau_1 = 0.5\tau_1$

Comparing the values, the longest stage is $\tau_2$. Therefore, the clock cycle time of the original pipeline is: $$T = \tau_2$$

Given that the original frequency is $3\text{ GHz}$, the clock cycle time $T$ is: $$T = \tau_2 = \frac{1}{3}\text{ ns}$$

Using this, we can find the latency of the other stages: $$\tau_1 = \frac{3}{4}\tau_2 = \frac{3}{4} \times \frac{1}{3} = \frac{1}{4}\text{ ns}$$ $$\tau_3 = \frac{1}{2}\tau_1 = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}\text{ ns}$$

The longest stage ($\tau_2$) is split into two pipeline stages of equal latency. The latency of each of these new stages is: $$\tau_{2a} = \tau_{2b} = \frac{\tau_2}{2} = \frac{1/3}{2} = \frac{1}{6}\text{ ns}$$

The latencies of the four stages in the new pipeline are: $$\tau_1 = \frac{1}{4}\text{ ns}, \quad \tau_{2a} = \frac{1}{6}\text{ ns}, \quad \tau_{2b} = \frac{1}{6}\text{ ns}, \quad \tau_3 = \frac{1}{8}\text{ ns}$$

The new clock cycle time $T'$ is determined by the maximum latency among the new stages: $$T' = \max\left(\frac{1}{4}, \frac{1}{6}, \frac{1}{6}, \frac{1}{8}\right) = \frac{1}{4}\text{ ns}$$

Thus, the new frequency $f'$ is: $$f' = \frac{1}{T'} = \frac{1}{1/4\text{ ns}} = 4\text{ GHz}$$