Exam (elaborations) TEST BANK FOR Computer Architecture A Quantitative Approach 4th Edition By John L. Hennessy & David Patterson (Solution manual)

Exam (elaborations) TEST BANK FOR Computer Architecture A Quantitative Approach 4th Edition By John L. Hennessy & David Patterson (Solution manual) 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 L Solutions to Case Study Exercises 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 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. Therefore, 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 × 106 ------------------ + 8 × 1 4500 ----------- 1 3 × 104 + ------------------ 8 × 2000 + 300 9 × 9 × 106 = = ------------------ 1 Failure rate --------------------------- 9 × 106 16301 = ------------------ = 522 hours = Power new Power old -------------------------- (V × 0.50)2 × (F × 0.50) V2 × F ------------------------------------------------------------- 0.53 = = = 0.125 .70 (1 – x) + x ⁄ 2 = -------------------------------- ; x = 60% 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 million = $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. 1.9 a. FIT = 109⁄ MTTF MTTF = 109⁄ FIT = 109⁄ 100 = 10,000,000 b. 1.10 Using the simplifying assumption that all failures are independent, we sum the probability of failure rate of all of the computers: Failure rate = 1000 × 10–7 = 10–4 = FIT = 105, therefore MTTF = = 104 1.11 a. Assuming that we do not repair the computers, we wait for how long it takes for 3,334 computers to fail. 3,334 × 10,000,000 = 33,340,000,000 hours b. Total cost of the decision: $1,000 × 10,000 computers = $10 million Expected benefit of the decision: Gain a day of downtime for every 33,340,000,000 hours of uptime. This would save us $1.4 million each 3,858,000 years. This would definitely not be worth it. Power new Power old -------------------------- (V × 0.70)2 × (F × 0.50) V2 × F ------------------------------------------------------------- 0.72 = = × 0.5 = 0.245 Availability MTTF MTTF + MTTR -------------------------------------- 107 107 + 24 = = -------------------- = about 100% 105 109 -------- 109 105 -------- L.1 Chapter 1