Question

c. Performance: Company A's processor and Company B's processor are benchmarked on the same program. The CPI of the component instructions that are executed in the program are shown above for each processor. Additionally, the number of instructions of each type is provided. Company A 5 Branch CPI Load/Store CPI Arithmetic CPI 7 1 Company B 4 Branch CPI Load/Store CPI Arithmetic CPI 9 1 Program Instruction Count Branch Load/Store Arithmetic CPL 1.2x109 1.67x1010 1.5x1011 i. Assume Company A has a clock rate of 1.2GHz, and Company B has a clock rate of 1.3GHz. Find the execution time of the benchmark program on the two processors. Which processor is faster and by how much? (2 points) ii. The company with the largest execution time reports that a new feature will be released in a processor revision that will improve the branch CPI by 50%. Using Amdahl's Law, determine the change in execution time. (2 points)

Answer 1

Sol.

For company A.

Instruction Count CPI

Branch 1.2 * 10⁹ 5

Load / Store 1.67 * 10¹⁰    7

Arithmetic 1.5* 10¹¹   1

Execution time = Instruction_count * CPI * Clock Cycle time

Execution time E_A =  (1.2 * 10⁹ * 5 +  1.67 * 10¹⁰ * 7 + 1.5* 10¹¹ *1) * 1.2 * 10^-9 = 327.48 sec

For company B.

Instruction Count CPI

Branch 1.2 * 10⁹ 4

Load / Store 1.67 * 10¹⁰ 9

Arithmetic 1.5* 10¹¹   1

Execution time E_B = (1.2 * 10⁹ * 4 +  1.67 * 10¹⁰ * 9 + 1.5* 10¹¹ *1) * 1.3 * 10^-9 = 396.63 sec

Company A processor is faster then company B.

Improved CPI By Company B by 50 % SO excution time of company improoved

Execution time E_B = (1.2 * 10⁹ * (4/2) +  1.67 * 10¹⁰ * (9/2) + 1.5* 10¹¹ *(1/2)) * 1.3 * 10-9 = 198.31 sec

SO overall Speed Up = Old Exec time / New Exec TIme

= 396.63/198.31 = 2.00 sec.

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