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PartII:SolutionsGuide52InstructorsManualforComputerOrganizationandDesign1.1q1.2u1.3f1.4a1.5c1.6d1.7i1.8k1.9j1.10o1.11w1.12p1.13n1.14r1.15y1.16s1.17l1.18g1.19x1.20z1.21t1.22b1.23h1.24m1.25e1.26v1.27j1.28b1.29f1.30j1.31i1.32e1SolutionsPartII:SolutionsGuide531.33d1.34g1.35c1.36g1.37d1.38c1.39j1.40b1.41f1.42h1.43a1.44a1.45TimeforTimefor1.46Asdiscussedinsection1.4,diecostsriseveryfastwithincreasingdiearea.Con-siderawaferwithalargenumberofdefects.Itisquitelikelythatifthedieareaisverysmall,somedieswillescapewithnodefects.Ontheotherhand,ifthedieareaisverylarge,itmightbelikelythateverydiehasoneormoredefects.Ingeneral,then,dieareagreatlyaffectsyield(astheequationsonpage48indicate),andsowewouldexpectthatdiesfromwaferBwouldcostmuchmorethandiesfromwaferA.1.47ThedieareaofthePentiumprocessorinFigure1.16is91mm2anditcontainsabout3.3milliontransistors,orroughly36,000persquaremillimeter.Ifweassumetheperiodhasanareaofroughly.1mm2,itwouldcontain3500transistors(thisiscertainlyaveryroughestimate).SimilarcalculationswithregardtoFigure1.26andtheIntel4004resultin191transistorspersquaremillimeterorroughly19transistors.1.48WecanwriteDiesperwafer=f((Diearea)–1)andYield=f((Diearea)–2)andthusCostperdie=f((Diearea)3).Moreformally,wecanwrite:1.49Nosolutionprovided.1.50FromthecaptioninFigure1.16wehave198diesat100%yield.Ifthedefectdensityis1persquarecentimeter,thentheyieldisapproximatedby1/((1+1´.91/2)2)=.47.Thus198´.47=93dieswithacostof$1000/93=$10.75perdie.1.51Defectsperarea.12---revolution12---=rev15400------------´minutesrev-----------------------60´ondssecminute-----------------------5.56ms=12---revolution12---=rev17200------------´minutesrev-----------------------60´ondssecminute-----------------------4.17ms=CostperdieCostperwaferDiesperwaferyield´----------------------------------------------------------=DiesperwaferWaferareaDiearea-----------------------------=Yield11DefectperareaDiearea2¤´+()2--------------------------------------------------------------------------------------------=54InstructorsManualforComputerOrganizationandDesign1.521.531.54Nosolutionprovided.1.55Nosolutionprovided.1.56Nosolutionprovided.1980Diearea0.16Yield0.48Defectdensity17.041992Diearea0.97Yield0.48Defectdensity1.981992+1980Improvement8.62Yield11DefectsperareaDiearea2¤´+()2----------------------------------------------------------------------------------------------=PartII:SolutionsGuide552.1Forprogram1,M2is2.0(10/5)timesasfastasM1.Forprogram2,M1is1.33(4/3)timesasfastasM2.2.2Sinceweknowthenumberofinstructionsexecutedandthetimeittooktoexecutetheinstructions,wecaneasilycalculatethenumberofinstructionspersecondwhilerunningprogram1as(200´106)/10=20´106forM1and(160´106)/5=32´106forM2.2.3WeknowthatCyclesperinstruction=Cyclespersecond/Instructionspersec-ond.ForM1wethushaveaCPIof200´106cyclespersecond/20´106instructionspersecond=10cyclesperinstruction.ForM2wehave300/32=9.4cyclesperinstruc-tion.2.4Wearegiventhenumberofcyclespersecondandthenumberofseconds,sowecancalculatethenumberofrequiredcyclesforeachmachine.IfwedividethisbytheCPIwe’llgetthenumberofinstructions.ForM1,wehave3seconds´200´106cy-cles/second=600´106cyclesperprogram/10cyclesperinstruction=60´106in-structionsperprogram.ForM2,wehave4seconds´300´106cycles/second=1200´106cyclesperprogram/9.4cyclesperinstruction=127.7´106instructionsperpro-gram.2.5M2istwiceasfastasM1,butitdoesnotcosttwiceasmuch.M2isclearlythema-chinetopurchase.2.6Ifwemultiplythecostbytheexecutiontime,wearemultiplyingtwoquantities,foreachofwhichsmallernumbersarepreferred.Forthisreason,costtimesexecutiontimeisagoodmetric,andwewouldchoosethemachinewithasmallervalue.Intheexample,weget$10,000´10seconds=100,000forM1vs.$15,000´5seconds=75,000forM2,andthusM2isthebetterchoice.Ifweusedcostdividedbyexecutiontimeandassumewechoosethemachinewiththelargervalue,thenamachinewitharidiculous-lyhighcostwouldbechosen.Thismakesnosense.Ifwechoosethemachinewiththesmallervalue,thenamachinewitharidiculouslyhighexecutiontimewouldbecho-sen.Thistoomakesnosense.2.7Wewoulddefinecost-effectivenessasperformancedividedbycost.Thisisessen-tially(1/Executiontime)´(1/Cost),andinbothcaseslargernumbersaremorecost-effectivewhenwemultiply.2.8WecanusethemethodinExercise2.7,buttheexecutiontimeisthesumofthetwoexecutiontimes.SoM1isslightlymorecost-effective,specifically1.04timesmore.2SolutionsExecutionspersecondperdollarforM111310,000´----------------------------1130,000-------------------==ExecutionspersecondperdollarforM21915,000´-------------------------1135,000-------------------==56InstructorsManualforComputerOrganizationandDesign2.9Wedothisproblembyfindingtheamountoftimethatprogram2canberuninanhourandusingthatforexecutionspersecond,thethroughputmeasure.Withperformancemeasuredbythroughputforprogram2,machineM2is=1.2timesfasterthanM1.Thecost-effectivenessofthemachinesistobemeasuredinunitsofthroughputonprogram2perdollar,soCost-effectivenessofM1==0.053Cost-effectivenessofM2==0.043Thus,M1ismorecost-effectivethanM2.(MachinecostsarefromExercise2.5.)2.10ForM1thepeakperformancewillbeachievedw