GATE CS 2014 Set 1 Q.53 (CS Q.43): Instruction Pipelining and Hazards

Consider a 6-stage instruction pipeline, where all stages are perfectly balanced. Assume that there is no cycle-time overhead of pipelining. When an application is executing on this 6-stage pipeline, the speedup achieved with respect to non-pipelined execution if 25% of the instructions incur 2 pipeline stall cycles is ______________________.

Your answer

Official answer and solution

Official answer: 4
From: Official GATE 2014 answer key

Solution

Use the pipeline speedup formula comparing non-pipelined execution time to pipelined execution time with stalls.

Given parameters:

Number of stages $k = 6$ .
Non-pipelined execution time per instruction: $T_{\text{non-pipe}} = k = 6$ cycles.
Pipelined execution:
- In an ideal pipeline with no stalls, $\text{CPI}_{\text{ideal}} = 1$ .
- 25% (0.25) of instructions incur 2 stall cycles.
- Remaining 75% of instructions incur 0 stall cycles.
- Average stall cycles per instruction = $0.25 \times 2 = 0.5$ cycles.
- Effective $\text{CPI}_{\text{pipe}} = 1 + 0.5 = 1.5$ cycles per instruction.

Speedup achieved:

\text{Speedup} = \frac{T_{\text{non-pipe}}}{T_{\text{pipe}}} = \frac{k}{\text{CPI}_{\text{pipe}}} = \frac{6}{1.5} = 4

Thus, the speedup is 4.

Final answer

Common trap

Adding stalls directly to the stage count $k$ instead of using the formula $\text{Speedup} = k / \text{CPI}$ .

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

Source

Paper: GATE 2014 Computer Science and Information Technology (CS) Session 1
Question: 53 of 65, printed as CS Q.43
Section: Computer Science and Information Technology

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