University of Phoenix ECON 240 EarlHedgehog195

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Chapter 1 Solutions Case Study 1: Chip Fabrication Cost 1.1 a. Yield¼1/(1 + (0.042))14¼0.34 b. It is fabricated in a larger technology, which is an older plant. As plants age, their process gets... tuned, and the defect rate decreases. 1.2 a. Phoenix: Dies per wafer ¼ πð Þ 45=2 2
=2ð Þ π45 =sqrt 2ð Þ¼ 2 79570:7 ¼ 724:5 ¼ 724 Yield ¼ 1=ð Þ 1+ 0ð Þ :04 2 14 ¼ 0:340 Profit ¼ 724 0:34 30 ¼ $7384:80 b. Red Dragon: Dies per wafer ¼ πð Þ 45=2 2
=2ð Þ π45 =sqrt 2ð Þ¼ 1:2 132591:25 ¼ 1234 Yield ¼ 1=ð Þ 1+ 0ð Þ :04 1:2 14 ¼ 0:519 Profit ¼ 1234 0:519 15 ¼ $9601:71 c. Phoenix chips: 25,000/724¼34.5 wafers needed Red Dragon chips: 50,000/1234¼40.5 wafers needed Therefore, the most lucrative split is 40 Red Dragon wafers, 30 Phoenix wafers. 1.3 a. Defect-free single core¼Yield¼1/(1 + (0.040.25))14¼0.87 Equation for the probability that N are defect free on a chip: #combinations (0.87)N (10.87)8N Yield for Phoenix4 : (0.39 + 0.21 + 0.06 + 0.01)¼0.57 Yield for Phoenix2 : (0.001 + 0.0001)¼0.0011 Yield for Phoenix1 : 0.000004 b. It would be worthwhile to sell Phoenix4 . However, the other two have such a low probability of occurring that it is not worth selling them. # defect-free # combinations Probability 8 7 6 5 4 3 2 1 0 1 0.32821167 0.39234499 0.20519192 0.06132172 0.01145377 0.00136919 0.0001023 4.3673E-06 8.1573E-08 8 28 56 70 56 28 8 1 c. $20 ¼ Wafer size odd dpw0:28 Step 1: Determine how many Phoenix4 chips are produced for every Phoenix8 chip. There are 57/33 Phoenix4 chips for every Phoenix8 chip¼1.73 $30 + 1:73 $25 ¼ $73:25 Case Study 2: Power Consumption in Computer Systems 1.4 a. Energy: 1/8. Power: Unchanged. b. Energy: Energynew/Energyold¼(Voltage 1/8)2 /Voltage2 ¼0.156 Power: Powernew/Powerold¼0.156 (Frequency 1/8)/Frequency¼0.00195 c. Energy: Energynew/Energyold¼(Voltage 0.5)2 /Voltage2 ¼0.25 Power: Powernew/Powerold¼0.25 (Frequency 1/8)/Frequency¼0.0313 d. 1 core¼25% of the original power, running for 25% of the time. 0:25 0:25 + 0ð Þ :25 0:2 0:75 ¼ 0:0625 + 0:0375 ¼ 0:1 1.5 a. Amdahl’s law: 1/(0.8/4 + 0.2)¼1/(0.2 + 0.2)¼1/0.4¼2.5 b. 4 cores, each at 1/(2.5) the frequency and voltage Energy: Energyquad/Energysingle¼4 (Voltage 1/(2.5))2 /Voltage2 ¼0.64 Power: Powernew/Powerold¼0.64 (Frequency 1/(2.5))/Frequency¼0.256 c. 2 cores + 2 ASICs vs. 4 cores ð Þ 2+ 0ð Þ :2 2 =4 ¼ ð Þ 2:4 =4 ¼ 0:6 1.6 a. Workload A speedup: 225,000/13,461¼16.7 Workload B speedup: 280,000/36,465¼7.7 1/(0.7/16.7 + 0.3/7.7) b. General-purpose: 0.70 0.42 + 0.30¼0.594 GPU: 0.70 0.37 + 0.30¼0.559 TPU: 0.70 0.80 + 0.30¼0.886 c. General-purpose: 159 W + (455 W159 W) 0.594¼335 W GPU: 357 W + (991 W357 W) 0.559¼711 W TPU: 290 W + (384 W290 W) 0.86¼371 W d. Speedup A B C GPU 2.46 2.76 1.25 TPU 41.0 21.2 0.167 % Time 0.4 0.1 0.5 2 ■ Solutions to Case Studies and Exercises GPU: 1/(0.4/2.46 + 0.1/2.76 + 0.5/1.25)¼1.67 TPU: 1/(0.4/41 + 0.1/21.2 + 0.5/0.17)¼0.33 e. General-purpose: 14,000/504 ¼ 27.8  28 GPU: 14,000/1838¼7.628 TPU: 14,000/861¼16.317 d. General-purpose: 2200/504¼4.374, 14,000/(4 504)¼6.747 GPU: 2200/1838¼1.21, 14,000/(1 1838)¼7.628 TPU: 2200/861¼2.562, 14,000/(2 861)¼8.139 Exercises 1.7 a. Somewhere between 1.410 and 1.5510, or 28.980x b. 6043 in 2003, 52% growth rate per year for 12 years is 60,500,000 (rounded) c. 24,129 in 2010, 22% growth rate per year for 15 years is 1,920,000 (rounded) d. Multiple cores on a chip rather than faster single-core performance e. 2¼x 4 , x¼1.032, 3.2% growth 1.8 a. 50% b. Energy: Energynew/Energyold¼(Voltage 1/2)2 /Voltage2 ¼0.25 1.9 a. 60% b. 0.4 + 0.60.2¼0.58, which reduces the energy to 58% of the original energy c. newPower/oldPower¼½Capacitance(Voltage0.8)2 (Frequency0.6)/½ CapacitanceVoltageFrequency¼0.82 0.6¼0.256 of the original power. d. 0.4 + 0.32¼0.46, which reduces the energy to 46% of the original energy 1.10 a. 109 /100¼107 b. 107 /107 + 24¼1 c. [need solution] 1.11 a. 35/10,0003333¼11.67 days b. There are several correct answers. One would be that, with the current system, one computer fails approximately every 5 min. 5 min is unlikely to be enough time to isolate the computer, swap it out, and get the computer back on line again. 10 min, however, is much more likely. In any case, it would greatly extend the amount of time before 1/3 of the computers have failed at once. Because the cost of downtime is so huge, being able to extend this is very valuable. c. $90,000¼(x+x+x+ 2x)/4 $360,000¼5x $72,000¼x 4th quarter¼$144,000/h Chapter 1 Solutions ■ 3 1.12 a. See Figure S.1. b. 2¼1/((1x) +x/20) 10/19¼x¼52.6% c. (0.526/20)/(0.474 + 0.526/20)¼5.3% d. Extra speedup with 2 units: 1/(0.1 + 0.9/2)¼1.82. 1.82 2036.4. Total speedup: 1.95. Extra speedup with 4 units: 1/(0.1 + 0.9/4)¼3.08. 3.08 2061.5. Total speedup: 1.97 1.13 a. old execution time¼0.5 new + 0.510 new¼5.5 new b. In the original code, the unenhanced part is equal in time to the enhanced part (sped up by 10), therefore: (1x)¼x/10 1010x¼x 10¼11x 10/11¼x¼0.91 1.14 a. 1/(0.8 + 0.20/2)¼1.11 b. 1/(0.7 + 0.20/2 + 0.103/2)¼1.05 c. fp ops: 0.1/0.95¼10.5%, cache: 0.15/0.95¼15.8% 1.15 a. 1/(0.5 + 0.5/22)¼1.91 [Show More]

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