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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CPI MFP = .1 * 6 + .15 * 4 + .05 * 20 + .7 * 2 = 3.6
CPI MNFP = .1 * 60 + .15 * 40 + .05 *100 + .7 *2 = 18.4
Note that for CPI MNFP, we used 60, 40, and 100 rather than 30, 20, and 50. This is
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 Spring '04
 Kubiatowicz
 Computer Architecture, Clock signal, clock cycle, die area, RightState

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