Chapter 1, Problem 1.6E

a.

Explanation of Solution

Given,

The clock rate of the processor P1 is 2.5GHz

The clock rate of the processor P2 is 3GHz

The Cycles Per Instruction (CPI) of the classes A, B, C and D related to the processor P1 is 1, 2, 3 and 3 respectively.

The CPI of the classes A, B, C and D related to the processor P2 is 2, 2, 2 and 2 respectively.

The time required to complete the implementation can be calculated using the following formula is given below:

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

Substitute, “$\text{1}\times {\text{10}}^{\text{5}}\text{,}\text{2}\text{×}{\text{10}}^{\text{5}}\text{,}\text{5}\text{×}{\text{10}}^{\text{5}}\text{and}\text{2}{\text{×10}}^{\text{5}}$” for “Number of instructions of A, B, C and D” “1, 2, 3 and 3” for “CPI of A, B, C and D” and “2.5” for “Clock rate” is given below:

$\begin{array}{c}\text{Time}\text{=}\frac{{\text{(10}}^{\text{5}}\times \text{1)}\text{+}\text{(2}\text{×}{\text{10}}^{\text{5}}\text{×}\text{2)}\text{+}\text{(5}\text{×}{\text{10}}^{\text{5}}\text{×}\text{3)}\text{+}\text{(2}\text{×}{\text{10}}^{\text{5}}\text{×}\text{3)}}{\text{2}\text{.5}\text{×}{\text{10}}^{\text{9}}}\\ \text{=}\text{10}\text{.4}\text{×}{\text{10}}^{\text{-4}}\end{array}$

Thus, the time required by the processor P1 to complete the implementation is $\text{10}\text{.4}\text{×}{\text{10}}^{\text{-4}}\text{s}$

Substitute, “$\text{1}\times {\text{10}}^{\text{5}}\text{,}\text{2}\text{×}{\text{10}}^{\text{5}}\text{,}\text{5}\text{×}{\text{10}}^{\text{5}}\text{and}\text{2}{\text{×10}}^{\text{5}}$” for “Number of instructions of A, B, C and D” “2, 2, 2 and 2” for “CPI of A, B, C and D” and “2

b.

Explanation of Solution

Given,

The clock rate of the processor P1 is 2.5GHz

The clock rate of the processor P2 is 3GHz

The Cycles Per Instruction (CPI) of the classes A, B, C and D related to the processor P1 is 1, 2, 3 and 3 respectively.

The CPI of the classes A, B, C and D related to the processor P2 is 2, 2, 2 and 2 respectively.

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

$\text{Number}\text{of}\text{clock}\text{cycles}\text{=}\text{Number}\text{of}\text{instructions}\text{×}\text{CPI}$

Substitute, “$\text{1}\times {\text{10}}^{\text{5}}\text{,}\text{2}\text{×}{\text{10}}^{\text{5}}\text{,}\text{5}\text{×}{\text{10}}^{\text{5}}\text{and}\text{2}{\text{×10}}^{\text{5}}$” for “Number of instructions of A, B, C and D” and “1, 2, 3 and 3” for “CPI of A, B, C and D” is given below:

$\begin{array}{c}\text{Clock}\text{cycles}\text{=}{\text{(10}}^{\text{5}}\text{×}\text{1)}\text{+}\text{(2}\text{×}{\text{10}}^{\text{5}}\text{×}\text{2)}\text{+}\text{(5}\text{×}{\text{10}}^{\text{5}}\text{×}\text{3)}\text{+}\text{(2}\text{×}{\text{10}}^{\text{5}}\text{×}\text{3)}\\ \text{=}\end{array}$