a. Given a program with a dynamic instruction count of 1.0E8 instructions di ed into classes as follows: 25% class A, 15% class B, 25% class C, class D, and 10% class E which implementation is slower? P1 or P2? o. What is the global CPI for each implementation? c. Find the clock cycles required in both cases.

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2-Consider two different implementations of the same instruction set architecture.
The instructions can be divided into five classes according to their CPI (class A,
B, C, D and E). P1 with a clock rate of 2.5 GHz and CPIS of 5, 3, 4, 2, and 3 and
P2 with a clock rate of 2 GHz and CPIS of 3, 4, 2, 6, and 3.
a. Given a program with a dynamic instruction count of 1.0E8 instructions divid-
ed into classes as follows: 25% class A, 15% class B, 25% class C , 25%
class D, and 10% class E which implementation is slower? P1 or P2?
b. What is the global CPI for each implementation?
c. Find the clock cycles required in both cases.
Transcribed Image Text:2-Consider two different implementations of the same instruction set architecture. The instructions can be divided into five classes according to their CPI (class A, B, C, D and E). P1 with a clock rate of 2.5 GHz and CPIS of 5, 3, 4, 2, and 3 and P2 with a clock rate of 2 GHz and CPIS of 3, 4, 2, 6, and 3. a. Given a program with a dynamic instruction count of 1.0E8 instructions divid- ed into classes as follows: 25% class A, 15% class B, 25% class C , 25% class D, and 10% class E which implementation is slower? P1 or P2? b. What is the global CPI for each implementation? c. Find the clock cycles required in both cases.
Expert Solution
Step 1

B) 

Global CPI = Sum(CPIi*Fi) 

For P1, 

global CPI = 0.25*5+0.15*3+0.25*4+0.25*2+0.1*3

= 1.25+0.45+1+0.5+0.3

= 3.5

Global CPI for P2 = 

0.25*3+0.15*4+0.25*2+0.25*6+0.1*3

= 0.75+0.6+0.5+1.5+0.3

= 3.65

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