GATE CS 2015 Set 1 Q38 — Instruction Pipelining

To find the speedup achieved by the pipelined processor, we compare the average execution time per instruction for both the non-pipelined and pipelined processors.

1. Non-Pipelined Processor

• Clock rate ($f_{\text{non-pipe}}$): $2.5\text{ GHz}$
• Clock cycle time ($T_{\text{non-pipe}}$): $\frac{1}{2.5\text{ GHz}} = 0.4\text{ ns}$
• Cycles Per Instruction ($\text{CPI}_{\text{non-pipe}}$): $4$

Average execution time per instruction ($\text{ET}_{\text{non-pipe}}$):

2. Pipelined Processor

• Clock rate ($f_{\text{pipe}}$): $2\text{ GHz}$
• Clock cycle time ($T_{\text{pipe}}$): $\frac{1}{2\text{ GHz}} = 0.5\text{ ns}$
• Cycles Per Instruction ($\text{CPI}_{\text{pipe}}$): Since there are no stalls in the pipeline, the ideal CPI in the steady state is $1$.

Average execution time per instruction in steady state ($\text{ET}_{\text{pipe}}$):

3. Speedup

The speedup ($S$) is given by the ratio of the execution times:

Thus, the speedup achieved is 3.2.