EE357-HW5_solutions-Nazarian-Fall09

EE357-HW5_solutions-Nazarian-Fall09 - EE 357 Homework 5...

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1 EE 357 Homework 5 Fall 2009 Nazarian Name: ___________________________________________ Assigned Tuesday, Nov. 17 Due: Monday, Nov. 30 (EEB340 at 5pm) Score: ________ Performance and Amdahl’s Law 1) a) 1.3.1: Exec. Time = 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 / 1E9 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. 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. 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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2 b) 1.5.1 Clock rate Peak Performance a 1.0 GHz 1000 MIPS 1.5 GHz 750 MIPS b 1.0 GHz 1000 MIPS 1.5 GHz 1500 MIPS 1.5.2 Clock rate CPI IPS = F / CPI a P1 1.0 GHz 14/6=2.33 1E9/2.33= 0.43 GIPS P2 1.5 GHz 16/6=2.67 1.5E9/2.5 = 0.56 GIPS b P1 1.0 GHz 10/6=1. 67 1.0E9/1.67 = 0.60 GIPS P2 1.5 GHz 14/6=2.33 1.5E9/2 = 0.64 GIPS For A P2 is = 21/16 = 1.31 or (.56/.43=1.30) times faster
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EE357-HW5_solutions-Nazarian-Fall09 - EE 357 Homework 5...

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