Computer Organization and Design: The Hardware/Software Interface

Info icon This preview shows pages 1–3. Sign up to view the full content.

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.
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

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
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern