solutions to alternate set of case study exercises 2006

solutions to alternate set of case study exercises 2006 -...

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Chapter 1 Solutions 2 Chapter 2 Solutions 6 Chapter 3 Solutions 19 Chapter 4 Solutions 29 Chapter 5 Solutions 45 Chapter 6 Solutions 50
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Solutions to Alternate Case Study Exercises
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2 n Solutions to Alternate Case Study Exercises Case Study 1: Chip Fabrication Cost 1.1 a. 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. Profit = 416 × 0.65 × $20 = $5408 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. 1.3 a. No defects = 0.28 2 = 0.08 One defect = 0.28 × 0.72 × 2 = 0.40 No more than one defect = 0.08 + 0.40 = 0.48 $20 × 0.28 = Wafer size/old dpw = $23.33 Chapter 1 Solutions Yield 1 0.30 × 3.89 4.0 --------------------------- +   4 0.36 == Dies per wafer π × 30 2 () 2 1.5 ----------------------------- = π × 30 sqrt 2 × 1.5 ------------------------------- 471 54.4 416 = = Yield 1 0.30 × 1.5 4.0 ------------------------ + 4 0.65 Dies per wafer π × 30 2 2 2.5 = π × 30 sqrt 2 × 2.5 283 42.1 240 = = Yield 1 0.30 × 2.5 4.0 + 4 0.50 Defect – Free single core 1 0.75 × 1.99 2 4.0 ---------------------------------- + 4 0.28 $20 Wafer size old dpw × 0.28 ------------------------------------ = x Wafer size 1/2 × old dpw × 0.48 --------------------------------------------------- $20 × 0.28 1/2 × 0.48 -------------------------
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Chapter 1 Solutions n 3 Case Study 2: Power Consumption in Computer Systems 1.4 a. .80 x = 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% 1.5 a. c. 200 W × 11 = 2200 W 2200 / (76.2) = 28 racks Only 1 cooling door is required. 1.6 a. The IBMx346 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. 1.7 a. (1 – 8) + .8/2 = .2 + .4 = .6 c. ; x = 50% d. Case Study 3: The Cost of Reliability (and Failure) in Web Servers 1.8 a. This date represents 3.6/108 = 3.3% of that quarter’s sales. Percentage of online sales is 106 million/3.9 billion = 2.7% Therefore, 3.3% of the 2.7% of the $4.8 billion dollars in sales for 4th quarter in 2005 is $4.3 million for that one day. $130 million × .20 × .01 = $260,000 14 KW 66 W 2.3 W 7.9 W ++ () ----------------------------------------------------------- 183 = 14 KW 66 W 2.3 W 2 × 7.9 W --------------------------------------------------------------------- 1 6 6 = Power new Power old -------------------------- V × 0.60 2 × F × 0.60 V 2 × F ------------------------------------------------------------- 0.6 3 0.216 == = 1 .75 1 x x 2 + -------------------------------- = Power new Power old V × 0.75 2 × F × 0.60 V 2 × F 0.75 2 × 0.6 0.338 =
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4 n Solutions to Alternate Case Study Exercises c. Assuming the 4.2 million visitors are not unique, but are actually the unique visitors each day summed across a month: 4.2 million × 8.9 = 37.4 million transactions per month $5.38 × 37.4 million = $201 million per month 1.9 a. FIT = 10 9 / MTTF MTTF = 10 9 / FIT = 10 9 = 150 = 6,666.667 b. 1.10 MTTF = 100 = 6 , 666 , 667 = 100 = 66 , 667 1.11 a. Assuming that we do not repair the computers, we wait for how long it takes for 20 , 000 × 2 / 5 = 8000 computers to fail.
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solutions to alternate set of case study exercises 2006 -...

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