Computer Organization and Design: The Hardware/Software Interface

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CS152 Homework 1 Solutions – Spring 2004 1.51 Yield = 1 / ((1 + (Defects per area * Die Area / 2))^2) Thus, if die area increases, defects per area must decrease. 1.52 Solving the yield equation for Defects per area, we get: Defects per area = (1 / (sqrt(yield) – 1) * (2 / Die Area) 1.53 In 1980, the yield = 48% and the Die Area = 0.16 from figure 1.31. In 1992, the yield = 48% and the Die Area = 0.97 from figure 1.31. Thus, Defects per area in 1980 = (1 / (sqrt(0.48) – 1) * (2 / .16) = 5.5 / unit area (in sq. cm) Defects per area in 1992 = (1 / (sqrt(0.48) – 1) * (2 / .97) = .914 / unit area (in sq. cm) Improvement = 5.5 / .914 = 601.7% improvement. 2.3 If you get confused in these questions, just remember to do unit analysis. CPI M1 = 10 seconds * (200e6 cycles / sec) * (1 / 200e6 instructions) = 10 cycles per instruction CPI M2 = 5 seconds * (300e6 cycles / sec) * (1 / 160e6 instructions) = 9.375 cycles per instruction 2.15 MIPS = Instruction Count / (Execution time * 10^6) Since Execution Time = Instruction Count * CPI * (1 / Clock Cycle) MIPS = Clock Cycle / (CPI * 10^6) First we need to find CPI of each machine.
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homework 1 S04 solutions - CS152 Homework 1 Solutions...

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