计算机组成与设计硬件软件接口课后习题答案资料下载.pdf

计算机组成与设计硬件软件接口课后习题答案资料下载.pdf

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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-60ondssecminute-5.56ms=12-revolution12-=rev17200-minutesrev-60ondssecminute-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（200106）/10=20106forM1and（160106）/5=32106forM2.2.3WeknowthatCyclesperinstruction=Cyclespersecond/Instructionspersec-ond.ForM1wethushaveaCPIof200106cyclespersecond/20106instructionspersecond=10cyclesperinstruction.ForM2wehave300/32=9.4cyclesperinstruc-tion.2.4Wearegiventhenumberofcyclespersecondandthenumberofseconds,sowecancalculatethenumberofrequiredcyclesforeachmachine.IfwedividethisbytheCPIwellgetthenumberofinstructions.ForM1,wehave3seconds200106cy-cles/second=600106cyclesperprogram/10cyclesperinstruction=60106in-structionsperprogram.ForM2,wehave4seconds300106cycles/second=1200106cyclesperprogram/9.4cyclesperinstruction=127.7106instructionsperpro-gram.2.5M2istwiceasfastasM1,butitdoesnotcosttwiceasmuch.M2isclearlythema-chinetopurchase.2.6Ifwemultiplythecostbytheexecutiontime,wearemultiplyingtwoquantities,foreachofwhichsmallernumbersarepreferred.Forthisreason,costtimesexecutiontimeisagoodmetric,andwewouldchoosethemachinewithasmallervalue.Intheexample,weget$10,00010seconds=100,000forM1vs.$15,0005seconds=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=1.2timesfasterthan