[J.L.Hennessy & D.A.Patterson] - Computer Architecture_A Quantitative Approach (4th Edition) (So

[J.L.Hennessy& - L.1 L.2 L.3 L.4 L.5 L.6 Chapter 1 Solutions Chapter 2 Solutions Chapter 3 Solutions Chapter 4 Solutions Chapter 5 Solutions

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
L.1 Chapter 1 Solutions L-2 L.2 Chapter 2 Solutions L-7 L.3 Chapter 3 Solutions L-20 L.4 Chapter 4 Solutions L-30 L.5 Chapter 5 Solutions L-46 L.6 Chapter 6 Solutions L-52
Background image of page 1

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

View Full DocumentRight Arrow Icon
L Solutions to Case Study Exercises
Background image of page 2
L-2 n Appendix L Solutions to Case Study Exercises Case Study 1: Chip Fabrication Cost 1.1 a. b. c. The Sun Niagara is substantially larger, since it places 8 cores on a chip rather than 1. 1.2 a. c. $9.38 × .4 = $3.75 d. Selling price = ($9.38 + $3.75) × 2 = $26.26 Proft = $26.26 – $4.72 = $21.54 e. Rate oF sale = 3 × 500,000 = 1,500,000/month Proft = 1,500,000 × $21.54 = $32,310,000 $1,000,000,000/$32,310,000 = 31 months 1.3 a. Prob oF one deFect = 0.29 × 0.71 7 × 8 = 0.21 Prob oF two deFects = 0.29 2 × 0.71 6 × 28 = 0.30 Prob oF one or two = 0.21 × 0.30 = 0.51 c. 0.71 8 = .06 (now we see why this method is inaccurate!) L.1 Chapter 1 Solutions Yield 1 0.7 × 1.99 4.0 ------------------------ +   4 0.28 == Yield 1 0.75 × 3.80 4.0 --------------------------- + 4 0.12 Yield 1 0.30 × 3.89 4.0 + 4 0.36 Dies per waFer π × 30 2 () 2 3.89 ----------------------------- = π × 30 sqrt 2 × 3.89 ---------------------------------- 182 33.8 148 = = Cost per die $500 148 × 0.36 ------------------------- $9.38 Yield 1 .7 × 1.86 4.0 --------------------- + 4 0.32 Dies per waFer π × 30 2 2 1.86 = π × 30 sqrt 2 × 1.86 380 48.9 331 = = Cost per die $500 331 × .32 ---------------------- $4.72 Yield 1 .75 × 3.80 8 4.0 ------------------------------- + 4 0.71 Prob oF error 1 0.71 0.29
Background image of page 3

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

View Full DocumentRight Arrow Icon
L.1 Chapter 1 Solutions n L - 3 d. 0.51 / 0.06 = 8.5 e. x × $150 + 8.5 x × $100 – (9.5 x × $80) – 9.5 x × $1.50 = $200,000,000 x = 885,938 8-core chips, 8,416,390 chips total Case Study 2: Power Consumption in Computer Systems 1.4 a. .70 x = 79 + 2 × 3.7 + 2 × 7.9 x = 146 b. 4.0 W × .4 + 7.9 W × .6 = 6.34 W c. The 7200 rpm drive takes 60 s to read±seek and 40 s idle for a particular job. The 5400 rpm disk requires 4±3 × 60 s, or 80 s to do the same thing. There- fore, it is idle 20% of the time. 1.5 a. c. 1.6 a. See Figure L.1. Sun Fire T2000 c. More expensive servers can be more compact, allowing more computers to be stored in the same amount of space. Because real estate is so expensive, this is a huge concern. Also, power may not be the same for both systems. It can cost more to purchase a chip that is optimized for lower power consumption. 1.7 a. 50% c. Sun Fire T2000 IBM x346 SPECjbb 213 91.2 SPECweb 42.4 9.93 Figure L.1 Power/performance ratios. 14 KW 79 W 2.3 W 7.0 W ++ () ----------------------------------------------------------- 158 = 14 KW 79 W 2.3 W 2 × 7.0 W ---------------------------------------------------------------------- 1 4 6 = MTTF 1 9 × 10 6 ----------------- 8 × + 1 4500 ----------- 1 3 × 10 4 + 8 × 2000 300 + 9 × 10 6 ------------------------------------ 16301 9 × 10 6 == 1 Failure rate -------------------------- 9 × 10 6 16301 522 hours = Power new Power old V × 0.50 2 × F × 0.50 V 2 × F ------------------------------------------------------------- 0.5 3 0.125 = .70 1 x x 2 + -------------------------------- ; x 60%
Background image of page 4
L-4 n Appendix L Solutions to Case Study Exercises d. Case Study 3: The Cost of Reliability (and Failure) in Web Servers 1.8 a. 14 days × $1.4 million/day = $19.6 million $4 billion – $19.6 million = $3.98 billion b. Increase in total revenue: 4.8±3.9 = 1.23 In the fourth quarter, the rough estimate would be a loss of 1.23 × $19.6 mil- lion = $24.1 million. c. Losing $1.4 million × .50 = $700,000 per day. This pays for $700,000±$7,500 = 93 computers per day. d. It depends on how the 2.6 million visitors are counted. If the 2.6 million visitors are not unique, but are actually visitors each day summed across a month: 2.6 million × 8.4 = 21.84 million transactions per month. $5.38 × 21.84 million = $117 million per month. If the 2.6 million visitors are assumed to visit every day: 2.6 million × 8.4 × 31 = 677 million transactions per month. $5.38 × 677 million = $3.6 billion per month, which is clearly not the case, or else their online service would not make money.
Background image of page 5

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

View Full DocumentRight Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/18/2010 for the course EECS 453 taught by Professor Proflee during the Spring '10 term at Seoul National.

Page1 / 65

[J.L.Hennessy& - L.1 L.2 L.3 L.4 L.5 L.6 Chapter 1 Solutions Chapter 2 Solutions Chapter 3 Solutions Chapter 4 Solutions Chapter 5 Solutions

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online