Consider a 3 GHz (gigahertz) processor with a three-stage pipeline and stage latencies $au_1$, $au_2$, and $au_3$ such that $au_1 = 3 au_2/4 = 2 au_3$. If the longest pipeline stage is split into two pipeline stages of equal latency, the new frequency is _________ GHz, ignoring delays in the pipeline registers.

The clock frequency of the processor is $f = 3\text{ GHz}$, so the initial clock cycle time is:

Given the relation:

$$au_1 = \frac{3}{4} au_2 = 2 au_3$$

Expressing each stage latency in terms of $au_2$:

$$au_1 = \frac{3}{4} au_2, \quad au_3 = \frac{3}{8} au_2$$

Thus, $au_2$ is the longest stage. The clock cycle time of a pipeline is determined by the maximum stage latency (ignoring register delays):

$$T = au_2 = \frac{1}{3}\text{ ns}$$

From this, we find:

• $

au_1 = $\frac{3}{4}$ \times $\frac{1}{3}$ = $\frac{1}{4}$$\text{ ns}$ = 0.25$\text{ ns}$$

• $

au_3 = $\frac{3}{8}$ \times $\frac{1}{3}$ = $\frac{1}{8}$$\text{ ns}$ = 0.125$\text{ ns}$$

When the longest stage $au_2$ is split into two equal stages, each new stage has latency:

$$au_2' = \frac{ au_2}{2} = \frac{1}{6}\text{ ns} \approx 0.167\text{ ns}$$

The new stage latencies are $au_1 = 0.25\text{ ns}$, $au_2' = 0.167\text{ ns}$, $au_2' = 0.167\text{ ns}$, and $au_3 = 0.125\text{ ns}$.

The new bottleneck stage is now $au_1 = \frac{1}{4}\text{ ns}$.

Therefore, the new clock frequency is:

$$f_{\text{new}} = \frac{1}{\max( au_1, au_2', au_3)} = \frac{1}{1/4\text{ ns}} = 4\text{ GHz}$$