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

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