Chapter 1, Problem 1.7E

a.

Explanation of Solution

Given,

The instruction count f and execution time ${\text{T}}_{\text{exec}}$ of the compiler A is 1.0E9 and 1.1s respectively.

The instruction count f and execution time ${\text{T}}_{\text{exec}}$ of the compiler B is 1.2E9 and 1.5s respectively.

The time taken by the clock cycle is 1ns.

The average Cycles Per Instruction (CPI) for each program in a compiler can be calculated using the following formula is given below:

$\text{CPI}\text{=}\frac{{\text{T}}_{\text{exec}}\text{×}f}{\text{Number}\text{of}\text{instructions}}$

Substitute, “$\text{1}\text{×}{\text{10}}^{\text{9}}$” for “f”, “1.1” for “${\text{T}}_{\text{exec}}$” and “$1.0\times {10}^9$” for “Number of instructions” is given below:

$\begin{array}{c}\text{CPI}\text{=}\frac{\text{1}\text{.1}\text{×}\text{1}\text{×}{\text{10}}^9}{\text{1}}\end{array}$

b.

Explanation of Solution

Given,

The instruction count f and execution time ${\text{T}}_{\text{exec}}$ of the compiler A is 1.0E9 and 1.1s respectively.

The instruction count f and execution time ${\text{T}}_{\text{exec}}$ of the compiler B is 1.2E9 and 1.5s respectively.

The time taken by the clock cycle is 1ns.

The average Cycles Per Instruction (CPI) for each program in a compiler can be calculated using the following formula.

$\text{CPI}\text{=}\frac{{\text{T}}_{\text{exec}}\text{×}f}{\text{Number}\text{of}\text{instructions}}$

Substitute, “$\text{1}\text{×}{\text{10}}^{\text{9}}$” for “f”, “1.1” for “${\text{T}}_{\text{exec}}$” and “$1.0\times {10}^9$” for “Number of instructions” is given below:

$\begin{array}{c}\text{CPI}\text{=}\frac{\text{1}\text{.1}\text{×}\text{1}\text{×}{\text{10}}^9}{\text{1}\text{.0}\times {\text{10}}^9}\\ =1.1\end{array}$

Thus, the average CPI for each program of the compiler A is 1.1.

Substitute, “$\text{1}\text{×}{\text{10}}^{\text{9}}$” for “f”, “1.5” for “${\text{T}}_{\text{exec}}$” and “$1.2\times {10}^9$” for “Number of instructions” is given below:

$\begin{array}{c}\text{CPI}\text{=}\frac{\text{1}\text{.5}\text{×}\text{1}\text{×}{\text{10}}^9}{\text{1}}\end{array}$

c.

Explanation of Solution

The time required by the compiler can be calculated using the formula

$\text{Time}\text{=}\frac{\text{CPI}\text{×}\text{Number}\text{of}\text{instructions}}{f}$

Substitute, “1.1” for “CPI”, “$\text{6}\text{×}{\text{10}}^{\text{8}}$” for “number of instructions” and “$\text{1}\text{×}{\text{10}}^{\text{9}}$” for “f” in the above formula is given below:

$\begin{array}{c}{\text{T}}_{\text{new}}\text{=}\frac{\text{1}\text{.1}\text{×}\text{6}\text{×}{\text{10}}^{\text{8}}}{\text{1}\text{×}{\text{10}}^{\text{9}}}\\ \text{=}0.66\end{array}$

Thus, the time required by the compiler with 1.1 CPI and $\text{6}\text{×}{\text{10}}^{\text{8}}$ number of instructions is 0