Computer Architecture Please show work and explain your answer. Thank you. Consi

Question

Computer Architecture

Please show work and explain your answer. Thank you.

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 2.5 GHz and CPIs of 1, 2, 3, and 3, and P2 with a clock rate of 3 GHz and CPIs of 2,2, 2, and 2. Given a program with a dynamic instruction count of 1.0E6 instructions divided into classes as follows: 10% class A, 20% class B, 50% class C, and 20% class D, which implementation is faster? What is the global CPI for each implementation? Find the clock cycles required in both cases.

Explanation / Answer

Given data

P1 : 2.5Ghz clock cycle CPIs 1 2 3 3
P2 : 3 Ghz clock cycle CPIs 2 2 2 2

A -> 10 %
B -> 20 %
C -> 50 %
D -> 20 %

a) Global cpi for each implementation is as follows:

For P1:
total clock cycles = 10^6 * (CPI's * (A,B,C,D))
==> 10^6 x ((1x10) + (2x20) + (3x50) + (3x20))
Total number of instructions = 10^6

So total clock cycles / Total number of instructions
===> 10^6 x ((1x10%) + (2x20%) + (3x50%) + (3x20%)) / 10^6 ===> 0.1 + 0.4 + 1.5 + 0.6 ==> 2.6
Global CPI = 2.6

For P2:
total clock cycles = 10^6 * (CPI's * (A,B,C,D))
==> 10^6 x ((2x10) + (2x20) + (2x50) + (2x20))
Total number of instructions = 10^6

So total clock cycles / Total number of instructions
===> 10^6 x ((2x10%) + (2x20%) + (2x50%) + (2x20%)) / 10^6 ===> 0.2 + 0.4 + 1 + 0.4 ==> 2.0
Global CPI = 2.0

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b)
Total clock cycles required for
P1:
10^6 x ((1x10%) + (2x20%) + (3x50%) + (3x20%)) ====> 2.6 x 10^6 clock cycles

P2:
10^6 x ((2x10%) + (2x20%) + (2x50%) + (2x20%)) ===> 2.0 x 10^6 clock cycles

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