ee357_hw5_sol

# ee357_hw5_sol - EE 357 Homework 5 Name_Solutions Due Note...

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EE 357 Homework 5 Name: ___Solutions_________________________________ Due: Score: ________ Note: Attach all work to receive full credit Performance 1) Exercise 1.3 (1.3.1-1.3.6) in CO&D, 4th Ed. page 59. a) 1.3.1: Exec. Tim = IC * CPI * Clock Period = IC * CPI * 1/ClockFreq. Perf. = 1 / Exec. Time. EXECUTION TIME PERF. P1 2 GHz 1.5 IC*1.5/2E9=0.75*IC / 1E9 1.33 E9 instruc. / sec. P2 1.5 GHz 1.0 IC*1/1.5E9=0.67*IC / 1E- 1.5 E9 instruc. / sec. (best performance) P3 3 GHz 2.5 IC*2.5/3E9=0.83*IC / 1E9 1.2 E9 instruc. / sec. b) 1.3.2: IC = Exec. Time * Freq / CPI 10 seconds of exec. time => Number of instructions P1 2 GHz 1.5 2E9 * 10 = 20E9 cycles 20E9 / 1.5 = 13.33E9 instructs. P2 1.5 GHz 1.0 1.5E9 * 10 = 15E9 cycles 15E9 / 1 = 15E9 instrucs . P3 3 GHz 2.5 3E9 * 10 = 30E9 cycles 30E9 / 2.5 = 12E9 instrucs. c) 1.3.3: Original Time = IC * CPI / Freq. New: 0.7 Time = IC * 1.2 CPI / x*Freq => x = 1.2/.7 = 1.71 New Clock Rate P1 2 GHz 1.5 1.71 * 2 GHz = 3.43 GHz P2 1.5 GHz 1.0 1.71 * 1.5GHz = 2.565 GHz P3 3 GHz 2.5 1.71 * 3GHz = 5.13 GHz

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ee357_hw5_sol - EE 357 Homework 5 Name_Solutions Due Note...

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