Chapter 1, Problem 1.5E

a.

Explanation of Solution

Given,

The clock rate and the Cycles Per Instruction (CPI) of the processor P1 is 3 GHz and 1.5 respectively.

The clock rate and the CPI of the processor P2 is 2.5 GHz and 1.0 respectively.

The clock rate and the CPI of the processor P3 is 4 GHz and 2.2 respectively.

The performance of the processor in terms of instructions per second can be calculated using the following formula is given below:

$\text{Performance}\text{=}\frac{\text{Clock}\text{}\text{rate}}{\text{CPI}}$

Substitute, “$3\times {10}^9$” for “Clock rate” and “1.5” for “CPI” is given below:

Thus, the performance of the processor P1 is $\frac{\text{3}\text{×}{\text{10}}^{\text{9}}}{\text{1}\text{.5}}\text{=}\text{2}\text{×}{\text{10}}^{\text{9}}$ instructions/second.

Substitute, “$\text{2}\text{.5}\text{×}{\text{10}}^{\text{9}}$” for “Clock rate” and “1.0” for “CPI” is given below:

Thus, the performance of the processor P1 is $\frac{\text{2}\text{.5}\text{×}{\text{10}}^{\text{9}}}{\text{1}\text{.0}}\text{=}\text{2}\text{.5}\text{×}{\text{10}}^{\text{9}}$ instructions/second.

Substitute, “$\text{4}\text{×}{\text{10}}^{\text{9}}$” for “Clock rate” and “2.2” for “CPI” is given below:

Thus, the performance of the processor P1 is $\frac{\text{4}\text{×}{\text{10}}^{\text{9}}}{\text{2}\text{.2}}\text{=}\text{1}\text{.8}\text{×}{\text{10}}^{\text{9}}$ instructions/second.

By comparing the above performances of the processors, it is concluded that the processor P1 is having highest performance.

b.

Explanation of Solution

Given,

The clock rate and the Cycles Per Instruction (CPI) of the processor P1 is 3 GHz and 1.5 respectively.

The clock rate and the CPI of the processor P2 is 2.5 GHz and 1.0 respectively.

The clock rate and the CPI of the processor P3 is 4 GHz and 2.2 respectively.

Time taken by each processor to execute the required program is 10 seconds.

The number of cycles of the processor per 10 seconds can be calculated using the following formula is given below:

$\text{Number}\text{of}\text{cycles}\text{=}\text{Number}\text{of}\text{seconds}\text{×}\text{Clock}\text{rate}\text{of}\text{the}\text{processor}$

Substitute, “$3\times {10}^9$” for “Clock rate” and “10” for “Number of seconds” is given below:

Thus, the number of cycles of the processor P1 per 10 seconds is $\text{10}\text{×}\text{3}\text{×}{\text{10}}^{\text{9}}\text{=}\text{30}\text{×}{\text{10}}^{\text{9}}$.

Substitute, “$\text{2}\text{.5}\text{×}{\text{10}}^{\text{9}}$” for “Clock rate” and “10” for “Number of seconds” is given below:

Thus, the number of cycles of the processor P2 per 10 seconds is $\text{10}\text{×}\text{2}\text{.5}\text{×}{\text{10}}^{\text{9}}\text{=}\text{25}\text{×}{\text{10}}^{\text{9}}$.

Substitute, “$\text{4}\text{×}{\text{10}}^{\text{9}}$” for “Clock rate” and “10” for “Number of seconds” is given below:

Thus, the number of cycles of the processor P3 per 10 seconds is $\text{10}\text{×}\text{4}\text{×}{\text{10}}^{\text{9}}\text{=}\text{40}\text{×}{\text{10}}^{\text{9}}$.

The number of instructions of the processor per 10 seconds can be calculated using the following formula is given below:

$\text{Number}\text{of}\text{instructions}\text{=}\frac{\text{Number}\text{of}\text{cycles}}{\text{CPI}}$

Substitute, “$\text{30}\text{×}{\text{10}}^{\text{9}}$” for “Number of cycles” and “1.5” for “CPI” is given below:

Thus, the number of instructions of the processor P1 per 10 seconds is $\frac{\text{30}\text{×}{\text{10}}^{\text{9}}}{\text{1}\text{.5}}\text{=}\text{20}\text{×}{\text{10}}^{\text{9}}$.

Substitute, “$\text{25}\text{×}{\text{10}}^{\text{9}}$” for “Number of cycles” and “1.0” for “CPI” is given below:

Thus, the number of instructions of the processor P2 per 10 seconds is $\frac{\text{25}\text{×}{\text{10}}^{\text{9}}}{\text{1}\text{.0}}\text{=}\text{25}\text{×}{\text{10}}^{\text{9}}$.

Substitute, “$\text{40}\text{×}{\text{10}}^{\text{9}}$” for “Number of cycles” and “2.2” for “CPI” is given below:

Thus, the number of instructions of the processor P3 per 10 seconds is $\frac{\text{40}\text{×}{\text{10}}^{\text{9}}}{\text{2}\text{.2}}\text{=}\text{40}\text{×}{\text{10}}^{\text{9}}$.

c.

Explanation of Solution

Given,

The clock rate and the Cycles Per Instruction (CPI) of the processor P1 is 3 GHz and 1.5 respectively.

The clock rate and the CPI of the processor P2 is 2.5 GHz and 1.0 respectively.

The clock rate and the CPI of the processor P3 is 4 GHz and 2.2 respectively.

The number of instructions of the processor per 10 seconds can be calculated using the following formula is given below:

$\text{Number}\text{of}\text{instructions}\text{=}\frac{\text{Number}\text{of}\text{cycles}}{\text{CPI}}$

Substitute, “$\text{30}\text{×}{\text{10}}^{\text{9}}$” for “Number of cycles” and “1.5” for “CPI” is given below:

Thus, the number of instructions of the processor P1 per 10 seconds is $\frac{\text{30}\text{×}{\text{10}}^{\text{9}}}{\text{1}\text{.5}}\text{=}\text{20}\text{×}{\text{10}}^{\text{9}}$.

Substitute, “$\text{25}\text{×}{\text{10}}^{\text{9}}$” for “Number of cycles” and “1.0” for “CPI” is given below:

Thus, the number of instructions of the processor P2 per 10 seconds is $\frac{\text{25}\text{×}{\text{10}}^{\text{9}}}{\text{1}\text{.0}}\text{=}\text{25}\text{×}{\text{10}}^{\text{9}}$.

Substitute, “$\text{40}\text{×}{\text{10}}^{\text{9}}$” for “Number of cycles” and “2.2” for “CPI” is given below:

Thus, the number of instructions of the processor P3 per 10 seconds is $\frac{\text{40}\text{×}{\text{10}}^{\text{9}}}{\text{2}\text{.2}}\text{=}\text{18}\text{.18}\text{×}{\text{10}}^{\text{9}}$.

The CPI_new is calculated using the formula ${\text{CPI}}_{\text{new}}\text{=}{\text{CPI}}_{\text{old}}\text{×}\text{1}\text{.2}$

Substitute, “1.5” for “CPI_old” is given below:

Thus, the CPI_new of the processor P1 is $\text{1}\text{.5}\text{×}\text{1}\text{.2}\text{=}\text{1}\text{.8}$.

Substitute, “1.0” for “CPI_old” is given below:

Thus, the CPI_new of the processor P2 is $\text{1}\text{.0}\text{×}\text{1}\text{.2}\text{=}\text{1}\text{.2}$.

Substitute, “4.0” for “CPI_old” is given below:

Thus, the CPI_new of the processor P3 is $\text{2}\text{.2}\text{×}\text{1}\text{.2}\text{=}\text{2}\text{.6}$.

The required clock rate cam be calculated using a function f and the required formula is as follows:

$f\text{=}\text{Number}\text{of}\text{instructions}\text{×}\frac{{\text{CPI}}_{\text{new}}}{\text{time}}$

Substitute, “$\text{20}\text{×}{\text{10}}^{\text{9}}$” for “Number of instructions”, “1.8” for “CPI_new” and “7” for “time” is given below:

Thus, the required function of the processor P1 is $\begin{array}{c}f\text{(P1)}\text{=}\text{20}\text{×}{\text{10}}^{\text{9}}\text{×}\frac{\text{1}\text{.8}}{\text{7}}\\ \text{=}\text{5}\text{.14}\text{GHz}\end{array}$.

Substitute, “$\text{25}\text{×}{\text{10}}^{\text{9}}$” for “Number of instructions”, “1.2” for “CPI_new” and “7” for “time” is given below:

Thus, the required function of the processor P2 is $\begin{array}{c}f\text{(P2)}\text{=}\text{25}\text{×}{\text{10}}^{\text{9}}\text{×}\frac{\text{1}\text{.2}}{\text{7}}\\ \text{=}\text{4}\text{.28}\text{GHz}\end{array}$.

Substitute, “$\text{40}\text{×}{\text{10}}^{\text{9}}$” for “Number of instructions”, “2.6” for “CPI_new” and “7” for “time” is given below:

Thus, the required function of the processor P3 is $\begin{array}{c}f\text{(P3)}\text{=}\text{40}\text{×}{\text{10}}^{\text{9}}\text{×}\frac{\text{2}\text{.6}}{\text{7}}\\ \text{=}\text{6}\text{.75}\text{GHz}\end{array}$.