Chapter 1, Problem 1.14.3E

Explanation of Solution

Given,

The number of Floating Point (FP) instructions is $\text{50}\text{×}{\text{10}}^{\text{6}}$

The number of INT instructions is $110\text{×}{\text{10}}^{\text{6}}$

The number of L/S instructions is $80\text{×}{\text{10}}^{\text{6}}$

The number of branch instructions is $16\text{×}{\text{10}}^{\text{6}}$

The Cycle Per Instructions (CPI) of FP, INT, L/S and branch instructions is 1, 1, 4 and 2 respectively.

The number of clock cycles can be calculated using the following formula

$\begin{array}{c}\text{Clock}\text{cycles}\text{=}{\text{CPI}}_{\text{fp}}\text{×}\text{Number}\text{of}\text{FP}\text{instructions}\\ \text{+}{\text{CPI}}_{\text{int}}\text{×}\text{Number}\text{of}\text{INT}\text{instructions}\\ \text{+}{\text{CPI}}_{\text{l/s}}\text{×}\text{Number}\text{of}\text{L/S}\text{instructions}\\ \text{+}{\text{CPI}}_{\text{branch}}\text{×}\text{Number}\text{of}\text{branch}\text{instructions}\end{array}$

We have,

40% of reduction in CPI of FP and INT instructions and 30% of reduction in CPI of L/S and branch instructions

Substitute, “$\text{1}\text{×}\text{0}\text{.6}$” for “CPI_fp”, “$\text{50}\text{×}{\text{10}}^{\text{6}}$” for “number of FP instructions”, “$\text{1}\text{×}\text{0}\text{.6}$” for “CPI_int”,”$\text{110}\text{×}{\text{10}}^{\text{6}}$” for “number of INT instructions”, “$4\text{×}\text{0}\text{.7}$” for “CPI_l/s” “$\text{80}\text{×}{\text{10}}^{\text{6}}$” for “number of L/S instructions”, “$2\text{×}\text{0}\text{.7}$” for “CPI_branch” and “$\text{16}\text{×}{\text{10}}^{\text{6}}$” for “number of branch instructions” is given below:

$\begin{array}{c}\text{Clock}\text{cycles}\text{=}\text{(1}\times 0\end{array}$