Chapter 1, Problem 1.12.4E

Explanation of Solution

Given,

The processor P1 includes clock rate of 4 GHz with 0.9 Cycles Per Instructions (CPI) and requires $\text{5}\text{×}{\text{10}}^{\text{9}}$ number of instructions.

The processor P2 includes clock rate of 3 GHz with 0.75 CPI and requires $1\text{×}{\text{10}}^{\text{9}}$ number of instructions.

The time taken by the processor to complete the process can be calculated using the following formula is given below:

$\text{Time taken}\text{=}\frac{\text{Number}\text{of}\text{instructions}\text{×}\text{CPI}}{\text{Clock}\text{rate}}$

Substitute, “$\text{5}\text{×}{\text{10}}^{\text{9}}$” for “Number of instructions”, “0.9” for “CPI” and “$\text{4}\text{×}{\text{10}}^{\text{9}}$” for “clock rate” in the above formula is given below:

$\begin{array}{c}\text{T(P1)}\text{=}\frac{\text{5}\text{×}{\text{10}}^{\text{9}}\text{×}\text{0}\text{.9}}{\text{4}\text{×}{\text{10}}^{\text{9}}}\\ \text{=}\text{1}\text{.125}\end{array}$

Thus, the time required by the processor P1 is 1.125 seconds.

Substitute, “$1\text{×}{\text{10}}^{\text{9}}$” for “Number of instructions”, “0.75” for “CPI” and “$3\text{×}{\text{10}}^{\text{9}}$” for “clock rate” in the above formula is given below:

$\begin{array}{c}\text{T(P2)}\text{=}\frac{1\text{×}{\text{10}}^{\text{9}}\text{×}\text{0}\text{.75}}{3\text{×}{\text{10}}^{\text{9}}}\\ \text{=}0.25\end{array}$

Thus, the time required by the processor P2 is 0.25 seconds.

The Millions of Floating Points Per Second (MFLOPS) of the processor can be calculated using the following formula is given below:

$\text{MFLOPS}\text{=}\frac{\text{Number}\text{of}\text{FP}\text{operations}\text{×}{\text{10}}^{\text{-6}}}{\text{Time}\text{taken}}$

We have,

Only 40% of the instructions are included in the execution