Chapter 1 Solutions2 Chapter 2 Solutions6 Chapter 3 Solutions13 Chapter 4 Solutions33 Chapter 5 Solutions44 Chapter 6 Solutions50 Appendix A Solutions63 Appendix B Solutions83 Appendix C Solutions92 Copyright 2012 Elsevier, Inc. All rights reserved. Solutions to Case Studies and Exercises1 Copyright 2012 Elsevier, Inc. All rights reserved. 2Solutions to Case Studies and Exercises Case Study 1: Chip Fabrication Cost1 1.1a. 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. 1.2a. Profit = 416 0.65 $20 = $5408 b. Profit = 240 0.50 $25 = $3000 c.The Woods chip d. Woods chips: 50,000/416 = 120.2 wafers needed Markon chips: 25,000/240 = 104.2 wafers needed Therefore, the most lucrative split is 120 Woods wafers, 30 Markon wafers. 1.3a. No defects = 0.282 = 0.08 One defect = 0.28 0.72 2 = 0.40 No more than one defect = 0.08 + 0.40 = 0.48 b. $20 0.28 = Wafer size/old dpw = $23.33 Chapter 1 Solutions Yield1 0.303.89 4.0 - -+ 4 0.36= Dies per wafer 30 2() 2 1.5 - -= 30 sqrt 21.5() - -47154.4416= Yield1 0.301.5 4.0 - -+ 4 0.65= Dies per wafer 30 2() 2 2.5 - -= 30 sqrt 22.5() - -28342.1240= Yield1 0.302.5 4.0 - -+ 4 0.50= Defect Free single core1 0.751.99 2 4.0 - -+ 4 0.28= $20 Wafer size old dpw0.28 - -= x Wafer size 1/2old dpw0.48 - - $200.28 1/20.48 -= Copyright 2012 Elsevier, Inc. All rights reserved. Chapter 1 Solutions3 Case Study 2: Power Consumption in Computer Systems 1.4a80x = 66 + 2 2.3 + 7.9; x = 99 b. .6 4 W + .4 7.9 = 5.56 c.Solve the following four equations: seek7200 = .75 seek5400 seek7200 + idle7200 = 100 seek5400 + idle5400 = 100 seek7200 7.9 + idle7200 4 = seek5400 7 + idle5400 2.9 idle7200 = 29.8% 1.5a. b. c.200 W 11 = 2200 W 2200/(76.2) = 28 racks Only 1 cooling door is required. 1.6a.The IBM x346 could take less space, which would save money in real estate. The racks might be better laid out. It could also be much cheaper. In addition, if we were running applications that did not match the characteristics of these benchmarks, the IBM x346 might be faster. Finally, there are no reliability numbers shown. Although we do not know that the IBM x346 is better in any of these areas, we do not know it is worse, either. 1.7a.(1 8) + .8/2 = .2 + .4 = .6 b. c. ; x = 50% d. Exercises 1.8a.(1.35)10 = approximately 20 b. 3200 (1.4)12 = approximately 181,420 c.3200 (1.01)12 = approximately 3605 d. Power density, which is the power consumed over the increasingly small area, has created too much heat for heat sinks to dissipate. This has limited the activity of the transistors on the chip. Instead of increasing the clock rate, manufacturers are placing multiple cores on the chip. 14 KW 66 W2.3 W7.9 W+() - -183= 14 KW 66 W2.3 W2+7.9 W() - -166= Power new Power old - V0.60()2F0.60() V2F - -0.630.216= 1 .75 1x()x 2+ - -= Power new Power old - V0.75()2F0.60() V2F - -0.7520.60.338= Copyright 2012 Elsevier, Inc. All rights reserved. 4Solutions to Case Studies and Exercises e.Anything in the 1525% range would be a reasonable conclusion based on the decline in the rate over history. As the sudden stop in clock rate shows, though, even the declines do not always follow predictions. 1.9a.50% b. Energy = load V2. Changing the frequency does not affect energyonly power. So the new energy is load ( V)2, reducing it to about the old energy. 1.10a.60% b. 0.4 + 0.6 0.2 = 0.58, which reduces the energy to 58% of the original energy. c.newPower/oldPower = Capacitance (Voltage .8)2 (Frequency .6)/ Capacitance Voltage Frequency = 0.82 0.6 = 0.256 of the original power. d. 0.4 + 0 .3 2 = 0.46, which reduce the energy to 46% of the original energy. 1.11a.109/100 = 107 b. 107/107 + 24 = 1 c.need solution 1.12a.35/10000 3333 = 11.67 days b. There are several correct answers. One would be that, with the current sys- tem, one computer fails approximately every 5 minutes. 5 minutes is unlikely to be enough time to isolate the computer, swap it out, and get the computer back on line again. 10 minutes, 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 dow