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

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

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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
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L Solutions to Case Study Exercises
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L-2 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. b. c. $9.38 × .4 = $3.75 d. Selling price = ($9.38 + $3.75) × 2 = $26.26 Profit = $26.26 – $4.72 = $21.54 e. Rate of sale = 3 × 500,000 = 1,500,000/month Profit = 1,500,000 × $21.54 = $32,310,000 $1,000,000,000/$32,310,000 = 31 months 1.3 a. b. 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 = =
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L.1 Chapter 1 Solutions 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. b. c. 1.6 a. See Figure L.1. b. 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% b. 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 ( ) ---------------------------------------------------------------------- 146 = 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% = =
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L-4 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.
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