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Official PYQ · 2014

GATE CS 2014 Set 1 Q55 — Instruction Pipelining

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Consider two processors P1P_1 and P2P_2 executing the same instruction set. Assume that under identical conditions, for the same input, a program running on P2P_2 takes 25%\text{25\%} less time but incurs 20%\text{20\%} more CPI (clock cycles per instruction) as compared to the program running on P1P_1. If the clock frequency of P1P_1 is 1GHZ\text{1GHZ}, then the clock frequency of P2P_2 (in GHz) is ______.
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Answer: 1.6

The execution time TT of a program on a processor is given by the formula:

T=I×CPIfT = \frac{I \times \text{CPI}}{f}

where II is the instruction count, CPI\text{CPI} is the clock cycles per instruction, and ff is the clock frequency.

Let the parameters for processor P1P_1 be T1T_1, CPI1\text{CPI}_1, and f1=1 GHzf_1 = 1\text{ GHz}. Since both processors execute the same program, the instruction count II remains the same.

For processor P2P_2, we are given:

  • Execution time: T2=0.75×T1T_2 = 0.75 \times T_1
  • CPI: CPI2=1.20×CPI1\text{CPI}_2 = 1.20 \times \text{CPI}_1

Using the execution time formula for both processors:

T2T1=I×CPI2/f2I×CPI1/f1\frac{T_2}{T_1} = \frac{I \times \text{CPI}_2 / f_2}{I \times \text{CPI}_1 / f_1} 0.75=1.20×CPI1f2×f1CPI10.75 = \frac{1.20 \times \text{CPI}_1}{f_2} \times \frac{f_1}{\text{CPI}_1} 0.75=1.20×f1f20.75 = 1.20 \times \frac{f_1}{f_2}

Substituting f1=1 GHzf_1 = 1\text{ GHz}:

f2=1.200.75×1 GHz=1.6 GHzf_2 = \frac{1.20}{0.75} \times 1\text{ GHz} = 1.6\text{ GHz}

Thus, the clock frequency of P2P_2 is 1.6 GHz.

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