Chapter 1, Problem 1.14.1E

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 is given below:

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

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

$\begin{array}{c}\text{Clock}\text{cycles}\text{=}\text{(1}\text{×}\text{50}\text{×}{\text{10}}^{\text{6}}\text{)}\text{+}\text{(1}\text{×}\text{110}\text{×}{\text{10}}^{\text{6}}\text{)}\text{+}\text{(4}\text{×}\text{80}\text{×}{\text{10}}^{\text{6}}\text{)}\text{+}\text{(2}\text{×}\text{16}\text{×}{\text{10}}^{\text{6}}\text{)}\\ \text{=}\text{512}\text{×}{\text{10}}^{\text{6}}\end{array}$

Therefore, the number of clock cycles is $\text{512}\text{×}{\text{10}}^{\text{6}}$.

The time taken to execute the clock cycles can be calculated using the following formula is given below:

${\text{T}}_{\text{CPU}}\text{=}\frac{\text{Clock}\text{cycles}}{\text{Clock}\text{rate}}$

Substitute, “$\text{512}\text{×}{\text{10}}^{\text{6}}$” for “clock cycles” and “$\text{2}\text{×}{\text{10}}^9$” for “clock rate” is given below:

$\begin{array}{c}{\text{T}}_{\text{CPU}}\text{=}\frac{\text{512}\text{×}{\text{10}}^{\text{6}}}{\text{2}\text{×}{\text{10}}^9}\\ \text{=}\text{0}\text{.256}\text{seconds}\end{array}$

The time required to execute the clock cycles is 0