GATE CS 2016 Set 2 Q.43 (CS Q.33): Instruction Pipelining and Hazards

Consider a 3 GHz (gigahertz) processor with a three-stage pipeline and stage latencies $\tau_1, \tau_2,$ and $\tau_3$ such that $\tau_1 = 3\tau_2 / 4 = 2\tau_3$ . 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.

Your answer

Official answer and solution

Accepted answers: 3.9 to 4.1
From: Official GATE 2016 answer key

Solution

Determine the original pipeline bottleneck stage latency from the processor frequency, then recalculate after splitting.

Given a 3-stage pipeline operating at clock frequency $f = 3\text{ GHz}$ :

\text{Clock cycle time } T = \frac{1}{f} = \frac{1}{3\text{ GHz}} = \frac{1}{3}\text{ ns}

The cycle time of an unbuffered pipeline is bounded by the longest stage:

T = \max(\tau_1, \tau_2, \tau_3)

We are given the stage latency relations:

\tau_1 = \frac{3}{4}\tau_2 = 2\tau_3

Express all stage latencies in terms of $\tau_2$ :

$\tau_1 = 0.75 \tau_2$
$\tau_3 = \frac{\tau_1}{2} = \frac{3}{8}\tau_2 = 0.375 \tau_2$

Clearly, $\tau_2$ is the longest stage. Therefore, the original cycle time is:

T = \tau_2 = \frac{1}{3}\text{ ns}

This gives:

$\tau_1 = \frac{3}{4} \times \frac{1}{3} = \frac{1}{4}\text{ ns} = 0.25\text{ ns}$
$\tau_3 = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}\text{ ns} = 0.125\text{ ns}$

The longest stage $\tau_2$ is split into two equal stages:

\tau_{2a} = \tau_{2b} = \frac{\tau_2}{2} = \frac{1/3}{2} = \frac{1}{6}\text{ ns} \approx 0.167\text{ ns}

The new 4-stage pipeline has latencies $\{\tau_1 = 1/4, \tau_{2a} = 1/6, \tau_{2b} = 1/6, \tau_3 = 1/8\}$ .
The new cycle time is the maximum of the new stage latencies:

T_{\text{new}} = \max\left(\frac{1}{4}, \frac{1}{6}, \frac{1}{6}, \frac{1}{8}\right) = \frac{1}{4}\text{ ns}

The new clock frequency is:

f_{\text{new}} = \frac{1}{T_{\text{new}}} = \frac{1}{1/4\text{ ns}} = 4\text{ GHz}

Final answer

Common trap

Assuming the new cycle time is halved ( $1/6$ ns, yielding $6$ GHz), forgetting that $\tau_1 = 1/4$ ns now becomes the new bottleneck.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2016 Computer Science and Information Technology (CS) Session 2
Question: 43 of 65, printed as CS Q.33
Section: Computer Science and Information Technology

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