Consider two different implementations of the same instruction set architecture. The instructions can be divided into four classes according to their CPI (class A, B, C and D). P1 with a clock rate of 3GHz and CPIs of 3, 2, 1, 4, and P2 with a clock rate of 2.5GHz and CPIs of 2, 2, 2, 2.Processor Class A Class B Class C Class DP1 3 2 1 4P2 2 2 2 2Given a program with a dynamic instruction count of 1.0E5 instructions divided into classes as follows: 10% class A, 30% class B, 40% class C, and 20% class D.1. Which implementation is faster?2. What is the global CPI for each implementation?3. Find the clock cycles required in both cases?

[tex]\text{P1 clock rate} = 3 \ GHz \\ \\\text{P2 clock rate} = 2.5 \ GHz \\\\\text{Class Division: 10 \% of class A,} \text{30 \% of class B, 40 \% of class C, 20 \% of class D} \\\\[/tex]

In point a:

Calculating P1 device mean CPI:

[tex]Average CPI = \frac{summation \times instruction \ count \times CPI}{total \ instruction \ count}\\[/tex]

[tex]= (10^5 \times 10 \% \times 3 + 10^5 \times 30 \% \times 2 + 10^5 \times 40 \% \times 2 + 10^5 \times 20%[/tex] [tex]\times 4) \times \frac{1} {1 \times 10^5}\\\\[/tex]

[tex]= \frac{(3 \times 10^4 + 6 \times 10^4 + 8 \times 10^4 + 8 \times 10^4)}{1 \times 10^5}\\\\ = \frac{25\times 10^4}{1 \times 10^5}\\\\= 2.5[/tex]

Estimating P2 device Average CPI:

[tex]= (1 \times 10^5 \times 10 \% \times2 + 1 \times 10^5 \times 30 \% \times 2 + 1 \times 10^5 \times 40 \% \times 2 + 1 \times 10^5 \times 20 \%[/tex][tex]\times 2) \times \frac{1}{ 1 \times 10^5}[/tex]

[tex]= \frac{(2 \times 10^4 + 6 \times 10^4 + 8 \times 10^4 + 4 \times 10^4)}{1 \times 10^5} \\\\= \frac{20 \times 10^4}{1 \times 10^5} \\\\= 2 \\[/tex]

[tex]\text{Execution time} = \frac{\text{instruction count} \times \text{average CPI}}{clock rate}}[/tex]

[tex]= \frac{1 \times 10^5 \times 2.5}{3} \ GHz \\\\ = \frac{1 \times 10^5 \times 2.5}{3 \times 10^9} \ sec \ as \ 1 \ GHz \\\\ = 10^9 \ Hz \\\\= 0.083 \ msec \ as \ 1 \ sec \\\\ = 10^3 \ msec[/tex]

Calculating processor execution time P2:

[tex]= \frac{1 \times 10^5 \times 2}{2.5} \ GHz \\\\= \frac{1 \times 10^5 \times 2}{2.5 \times 10^9} \ sec \ as\ 1 \ GHz \ = 10^9 \ Hz \\\\ = 0.080 \ msec \ as \ 1 \ sec \\\\ = 10^3 \ msec[/tex]

Since P2 is less than P1 for processor execution time, P2 is therefore faster than P1.

In point b:

Global processor P1 CPI = 2.5

Global processor P2 CPI = 2

In point c:

Finding Processor P1 clock cycles:

[tex]\text{Clock cycle = instruction count} \times \text{average CPI}[/tex]

[tex]= 1 \times 10^5 \times 2.5 \\\\ = 2.5 \times 10^5 \ cycles[/tex]

Finding Processor P2 clock cycles:

[tex]\text{Clock cycle = instruction count} \times \text{average CPI} \\\\[/tex]

[tex]= 1 \times 10^{5} \times 2\\\\ = 2 \times 10^{5} \ cycles\\\\[/tex]