《计算机体系结构-量化研究方法》-第五版-习题答案

Chapter1Solutions2Chapter2Solutions6Chapter3Solutions13Chapter4Solutions33Chapter5Solutions44Chapter6Solutions50AppendixASolutions63AppendixBSolutions83AppendixCSolutions92Copyright©2012Elsevier,Inc.Allrightsreserved.SolutionstoCaseStudiesandExercises1Copyright©2012Elsevier,Inc.Allrightsreserved.2■SolutionstoCaseStudiesandExercisesCaseStudy1:ChipFabricationCost11.1a.b.Itisfabricatedinalargertechnology,whichisanolderplant.Asplantsage,theirprocessgetstuned,andthedefectratedecreases.1.2a.Profit=416×0.65×$20=$5408b.Profit=240×0.50×$25=$3000c.TheWoodschipd.Woodschips:50,000/416=120.2wafersneededMarkonchips:25,000/240=104.2wafersneededTherefore,themostlucrativesplitis120Woodswafers,30Markonwafers.1.3a.Nodefects=0.282=0.08Onedefect=0.28×0.72×2=0.40Nomorethanonedefect=0.08+0.40=0.48b.$20×0.28=Wafersize/olddpw=$23.33Chapter1SolutionsYield10.303.89×4.0---------------------------+⎝⎠⎛⎞4–0.36==Diesperwaferπ302⁄()×21.5-----------------------------=π30×sqrt21.5×()-------------------------------–47154.4416=–=Yield10.301.5×4.0------------------------+⎝⎠⎛⎞4–0.65==Diesperwaferπ302⁄()×22.5-----------------------------=π30×sqrt22.5×()-------------------------------–28342.1240=–=Yield10.302.5×4.0--------------------------+⎝⎠⎛⎞4–0.50==Defect–Freesinglecore10.751.992⁄×4.0----------------------------------+⎝⎠⎛⎞4–0.28==$20Wafersizeolddpw0.28×------------------------------------=xWafersize1/2olddpw×0.48×--------------------------------------------------$200.28×1/20.48×-------------------------==Copyright©2012Elsevier,Inc.Allrightsreserved.Chapter1Solutions■3CaseStudy2:PowerConsumptioninComputerSystems1.4a..80x=66+2×2.3+7.9;x=99b..6×4W+.4×7.9=5.56c.Solvethefollowingfourequations:seek7200=.75×seek5400seek7200+idle7200=100seek5400+idle5400=100seek7200×7.9+idle7200×4=seek5400×7+idle5400×2.9idle7200=29.8%1.5a.b.c.200W×11=2200W2200/(76.2)=28racksOnly1coolingdoorisrequired.1.6a.TheIBMx346couldtakelessspace,whichwouldsavemoneyinrealestate.Theracksmightbebetterlaidout.Itcouldalsobemuchcheaper.Inaddition,ifwewererunningapplicationsthatdidnotmatchthecharacteristicsofthesebenchmarks,theIBMx346mightbefaster.Finally,therearenoreliabilitynumbersshown.AlthoughwedonotknowthattheIBMx346isbetterinanyoftheseareas,wedonotknowitisworse,either.1.7a.(1–8)+.8/2=.2+.4=.6b.c.;x=50%d.Exercises1.8a.(1.35)10=approximately20b.3200×(1.4)12=approximately181,420c.3200×(1.01)12=approximately3605d.Powerdensity,whichisthepowerconsumedovertheincreasinglysmallarea,hascreatedtoomuchheatforheatsinkstodissipate.Thishaslimitedtheactivityofthetransistorsonthechip.Insteadofincreasingtheclockrate,manufacturersareplacingmultiplecoresonthechip.14KW66W2.3W7.9W++()------------------------------------------------------------183=14KW66W2.3W2++7.9W×()---------------------------------------------------------------------166=PowernewPowerold--------------------------V0.60×()2F0.60×()×V2F×-----------------------------------------------------------0.630.216===1.751x–()x2⁄+--------------------------------=PowernewPowerold--------------------------V0.75×()2F0.60×()×V2F×-----------------------------------------------------------0.7520.6×0.338===Copyright©2012Elsevier,Inc.Allrightsreserved.4■SolutionstoCaseStudiesandExercisese.Anythinginthe15–25%rangewouldbeareasonableconclusionbasedonthedeclineintherateoverhistory.Asthesuddenstopinclockrateshows,though,eventhedeclinesdonotalwaysfollowpredictions.1.9a.50%b.Energy=½load×V2.Changingthefrequencydoesnotaffectenergy–onlypower.Sothenewenergyis½load×(½V)2,reducingittoabout¼theoldenergy.1.10a.60%b.0.4+0.6×0.2=0.58,whichreducestheenergyto58%oftheoriginalenergy.c.newPower/oldPower=½Capacitance×(Voltage×.8)2×(Frequency×.6)/½Capacitance×Voltage×Frequency=0.82×0.6=0.256oftheoriginalpower.d.0.4+0.3×2=0.46,whichreducetheenergyto46%oftheoriginalenergy.1.11a.109/100=107b.107/107+24=1c.[needsolution]1.12a.35/10000×3333=11.67daysb.Thereareseveralcorrectanswers.Onewouldbethat,withthecurrentsys-tem,onecomputerfailsapproximatelyevery5minutes.5minutesisunlikelytobeenoughtimetoisolatethecomputer,swapitout,andgetthecomputerbackonlineagain.10minutes,however,ismuchmorelikely.Inanycase,itwouldgreatlyextendtheamountoftimebefore1/3ofthecomputershavefailedatonce.Becausethecostofdowntimeissohuge,beingabletoextendthisisveryvaluable.c.$90,000=(x+x+x+2x)/4$360,000=5x$72,000=x4thquarter=$144,000/hr1.13a.Itanium,becauseithasaloweroverallexecutiontime.b.Opteron:0.6×0.92+0.2×1.03+0.2×0.65=0.888c.1/0.888=1.1261.14a.SeeFigureS.1.b.2=1/((1–x)+x/10)5/9=x=0.56or56%c.0.056/0.5=0.11or11%d.Maximumspeedup=1/(1/10)=105=1/((1–x)+x/10)8/9=x=0.89or89%Copyright©2012Elsevier,Inc.Allrightsreserved.Chapter1Solutions■5e.Currentspeedup:1/(0.3+0.7/10)=1/0.37=2.7Speedupgoal:5.4=1/((1–x)+x/10)=x=0.91Thismeansthepercentageofvectorizationwouldneedtobe91%1.15a.oldexecutiontime=0.5new+0.5×10new=5.5newb.Intheoriginalcode,theunenhancedpartisequalintimetotheenhancedpartspedupby10,therefore:(1–x)=x/1010–10x=x10=11x10/11=x=0.911.16a.1/(0.8+0.20/2)=1.11b.1/(0.7+0.20/2+0.10×3/2)=1.05c.fpops:0.1/0.95=10.5%,cache:0.1