计算机组成与体系结构答案

1、Page LastUpdated:October2003Page LastUpdated:October2003TheEssentialsofComputerOrganizationandArchitectureLindaNullandJuliaLoburJonesandBartlettPublishers,2003Chapter5InstructorfsManualChapterObjectivesChapter5.ACloserLookatInstructionSetArchitectures,providesacloserlookatinstructionsetarchitectures

2、,includinginstructionformats,instructiontypes,endianess.andaddressingmodesInstruclionlevelpipeliningisintroducedaswell.Real-worldISAs(includingIntel,MIPS,andJava)arediscussedtoreinforcetheconceptspresentedinthechapterThischaptershouIdbecoveredafterChapter4.LecturesshouIdfocusonthefollowingpoints:Ins

3、tructionformatsImportantissueshereincludeinstructionlengthlittleversusbigendian,registerusage(andtheuseofstacks),andexpandingopcodesAddressingAlthoughaddressingisaninstructiondesignissue,therearemanyaspectstoconsider,themostimportantofwhichisaddressingmodesInstruction-levelpipeliningThefetch-decode-

4、executecyclecanbeoverlappedresukinginfasterexecutiontimeResourceconflicts,conditionalbranching,anddatadependenciescanslowthisprocessdown.Real-worldexampiesofISAs.Intel.MIPS,andtheJavavirtualmachineareallpresentedtoreinforcetheconceptsofthechapterRequiredLectureTimeTheimportantconceptsinChapter5canty

5、picallybecoveredin3lecturehoursHowever,ifateacherwantsthestudentstohaveamasteryofalltopicsinChapter5.8lecturehoursaremorereasonableIflecturetimeislimited,wesuggestthatthefocusbeoninstructionformats(includingbigversuslittleendian,useofregisters,andinstructionlength)andaddressing(withconsiderableatten

6、tionpaidtothevariousmodes).LectureTipsItisimportanttomotivatethelittleversusbigendiandebateManystudentscomprehendtheconcept,butdon*tseethesignificanceThevarioussoftwareapplicationslistedinthischapterwillhelptheinstructormakethispointStudentsoftenhavedifficultyunderstandingstackmachinesItisimportantt

7、oemphasizethatnotALLinstructionsonstackmachinehavezerooperands,butrather,theinstructionsthatallowforoperandsarelimitedTheconceptofexpandingopcodesissometimesdifficuItforstudentsaswell.Wesuggestgoingoverasma11exampleindetailWhencoveringtheaddressingmodes,wesuggestthatinstructorsincludemanyexamples,as

8、thisisoneconceptthattendstobeeasiertounderstandthroughexamplesStudentsoftenconfuseinstruction1evelpipeliningwithothertypesofpipeliningItisimportanttostressthattherearemanytypesofpipelining,butinthischapteronlypipeliningthefetch-decode-executecycleisaddressedAlthoughthereal-lifeexamplescanbeleftforst

9、udentstoread,thesecasestudiesprovidetheinstructorwithameanstotietheconceptsfromthischaptertogether,aswellasamethodformotivatingstudyoftheconceptsfromthischapterAnswerstoExercisesAssumeyouhaveamachinethatuses32bitintegersandyouarestoringthehexvalue1234ataddress0.a.Showhowthisisstoredonabigendianmachi

10、neb.Showhowthisisstoredonalittleendianmachinec.IfyouwantedtoincreasethehexvaluetoI23456、whichbyteassignmentwouIdbemoreefficient,bigorlittleendian?ExplainyouranswerAns.aandbAddress00011011BigEndian00001234LittleEndian3-412()()()()c.Littleendianismoreefficientbecausetheadditionalinformationsimplyneeds

11、tobeappendedWithbigendian,the”12”and”34”wouldneedtoshifttomaintainthecorrectbyteordering.ShowhowthefollowingvalueswouIdbestoredbymachineswith32bilwords,usinglittleendianandthenbigendianformatAssumeeachvaluestartsataddress1016Drawadiagramofmemoryforeach,placingtheappropriatevaluesinthecorrect(andlabe

12、led)memorylocations456789A116bOOOOO58Ai6c14148888“Ans.a.Address10i61116121613“BigEndian4567S9AlLittleEndianAl1丿6745b.Address1“111612“13“BigEndian()()()()05SALittleEndian()5小）（）（）c.Address10“111612“13“BigEndian14148888LittleEndian888814143.Thefirsttwobytesofa2Mx16mainmemoryhavethefollowinghexvalues:B

13、yte0isFEByte1is01Ifthesebytesholda16-bitiwoscomplementinteger,whatisitsactualdecimalvalueif:a.memoryisbigendian?b.memoryislittleendian?Ans.FE01I6=11111110000000012=-51110OIFEm=00000001111111102=51Oio4.Whatkindsofproblemsdoyouthinkendiannesscancauseifyouwishedtotransferdatafromabigendianmachinetoalit

14、tleendianmachine?Explain.Ans.Ifthemachinesreceivingthedatausesdifferentendian-nessthanthemachinesendingthedata.thevaluescanbemisinterpretedForexample,thevaluefromProblem3,sentasthevalue511onabigendianmachinewouIdbereadasthevalue510onalittleendianmachine5.ThePopulationStudiesInstitutemonitorsthepopul

15、ationoftheUnitedStatesIn2000,thisinstitutewroteaprogramtocreatefilesofthenumbersrepresentingthevariousstates,aswellasthetotalpopulationoftheU.S.Thisprogram,whichrunsonaMotorolaprocessor,projectsthepopulationbasedonvariousrules,suchastheaveragenumberofbirthsanddeathsperyearTheInstituterunstheprograma

16、ndthenshipstheoutputfilestostateagenciessothedatavaluescanbeusedasinputintovariousapplicationsHowever,onePennsylvaniaagency,runningallIntelmachines,encountereddifficulties.asindicatedbythefollowingproblemWhenthe32bitunsignedinteger1D2F37E816(representingtheoverallU.S.populationpredicationfor2003)isu

17、sedasinput,andtheagencysprogramsimplyoutputsthisinputvalue,theU.S.populationforecastfor2003isfartoolargeCanyouhelpthisPennsylvaniaagencybyexplainingwhatmightbegoingwrong?Ans.Hint:TheyarerunondifferentprocessorsIntelandMotorolausedifferentendian-nessIftheprogramontheIntelmachinedoesnotadjustforthis,t

18、hentheintegerisinterpretedincorrectly.6.TherearereasonsformachinedesignerstowantallinstructionstobethesamelengthWhyisthisnotagoodideaonastackmachine?Ans.TheonlyinstructionsonastackmachinethatneedtoaddressmemoryarepushandpopSoanoperandfieldisrequired,whichimpliestheinstructionfieldmustbedividedintoan

19、opcodeandanoperandHowever,theotherinstructionsneednotaccessmemoryandcanthusconsistofonlytheopcodeTomakethemallthesamelength,theseinstructionswouIdneedtobe”artificiallylengthened.7.Acomputerhas32-bitinstructionsand12-bitaddressesSupposethereare2502-addressinstructionsHowmany1-addressinstructionscanbe

20、formulated?ExplainyouranswerAns.Thereare2502addressinstructionsThereareonlyatotalof2562-addressinstructionsallowedifwehave32-bitinstructions(twoaddressestakeup24bits,leavingonly8bitsfortheopcode).Lookingatthe8bitopcode,assumebitpatterns00000000(0)through11111001(249)areusedforthe250two-addressinstru

21、ctionsThenthereare6bitpatternsleftforoneaddressinstructionsHowever,eachoneofthesecanusetheremaining12bitsgainedfromhavingonlyoneoperand,sowehave6*2ConvertthefollowingexpressionsfrominfixtoreversePolish(postfix)notation.“X*Y+W*Z+V*UbW*X+W*(U*V+Z)c(W(X+Y(U六V)/(U*(X+Y)Ans.XY*WZ*VU*+00000000111110011111

22、101011111010 xxxxxxxxxxxxxxxxxxxxxxxx000000000000111111111111xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx000000000000111111111111xxxxxxxxxxxxxxxxxxxxxxxx2502-addressinstructions212,I6*21*1-addressinstructions2】2WX*WUVWXYUV*9ConvertthefollowingexpressionsfromreversePolishnotationtoinfixnotationa.

23、WXYZ+*b.uVWXYZ+*+*+CXYZ+Vw_*z+Ans.W*(X+Y-Z)U+(V*(W+(X*(Y+Z)X+(Y+Z)*(V-W)+Z)10.a.Writethefollowingexpressioninpostfix(ReversePolish)notation.Remembertherulesofprecedenceforarithmeticoperators!A_B+C*(D*E_F)_G+H*KbWriteaprogramtoevaluatetheabovearithmeticstatementusingastackorganizedcomputerwithzero-ad

24、dressinstructions(soonlypopandpushcanaccessmemory).Ans.nAB-CDE*F-*+GHK*+/b.Theprogramwouldbe:PushAPushBSubtractPushCPushDPushEMultPushFSubtractMultAddPushGPushHPushKMultAddDivPopX11a.Inacomputerinstructionformat,theinstructionlengthis11bitsandthesizeofanaddressfieldis4bitsIsitpossibletohave:2-addres

25、sinstructions451-addressinstructions320addressinstructionsusingtheformat?Justifyyouranswer.b.Assumethatacomputerarchitecthasalreadydesigned6addressand24zeroaddressinstructionsusingtheinstructionformataboveWhatistheinaximumnumberofone-addressinstructionsthatcanbeaddedtotheinstructionset?Ans.a.YesThe2

26、-addressinstructionscouIdberepresentedOOOxxxxxxxxthrough1OOxxxxxxxx(using000through100foropcodes).The1-addressinstructionscouIduse1010000through1011111(16).1100000through1101111(16),and1110000through1111100(13more,foratotalof45).The0-addressinstructionscouIduseHill100000through11111101111(16),and111

27、11110000through111I1111111(16).Sowehave:000 xxxx100 xxxx101000010111111100000110111111100001111100XXXXIxxxxJxxxxxxxxxxxx16xxxxXXXXxxxx52-addressinstructions16451-addressinstructionsPage #LastUpdated:October20031616a320-addressinslmclions11111100000i11111101111J11111110000i11111111111Jb.Assumethetwad

28、dressinstructionsusebitpatterns000 xxxxxxxxthrough101xxxxxxxxAssumealsothatthezero-addressinstructionsareoftheformat11111101000through11111101111(8111111110000through11111110111(8).and111111iToOOthrough111I1111111(8)(Theseconstitutethelast16binarynumberspossiblewith11bits).Thenallinstructionsbeginni

29、ngwith110(1100000 xxxxthrough1101111xxxx)couIdbeoneaddressinstructions(16).Inaddition,1110000 xxxxthrough1111101xxxx_couIdbeoneaddressinstructions,givingus14more.foratotalof301-addressinstructions12.Whatisthedifferencebetweenusingdirectandindirectaddressing?GiveanexampleAns.Directaddressingprovidest

30、heactualmemoryaddressoftheoperandintheinstruction,whereasindirectaddressingprovides,aspartoftheinstruction,apointertoamemorylocation.Forexample.theinstructionLoadXinterpretedusingdirectaddressingwouIdgotomemorylocationXandloadthevaluefoundthereUsingindirectaddressing,memorylocationXwouIdbeusedasthee

31、ffectiveaddressofwhatshouIdactuallybeloadedSoifavalueof200werefoundatlocationX.thevaluelocatedataddress200wouIdbeloaded.Page LastUpdated:October2003I3SupposewehavetheinstructionLoad1000.GivenmemoryandregisterR1containthevaluesbelow:Memory10001100120013001400R1200Page #LastUpdated:October2003Page #La

32、stUpdated:October2003Ans.AssumingR1isimpliedintheindexedaddressingmode,determinetheactualvalueloadedintotheaccumulatorandfillinthetablebelow:SupposewehavetheinstructionLoadvaluesbelow:500.GivenmemoryandregisterR1containtheModeValueLoadedintoACImmediateDirectIndirectIndexedModeValueImmediate1000Direc

33、t1400Indirect1300Indexed1000Page #LastUpdated:October2003Page LastUpdated:October2003R1200100400500600700AssumingR1isimpliedintheindexedaddressingmode,determinetheactualvalueloadedintotheaccumulatorandfillinthetablebelow:ModeValueLoadedintoACImmediateDirectIndirectIndexedAns.Page #LastUpdated:Octobe

34、r2003Page #LastUpdated:October2003ModeValueImmediate500Direct100Indirect600Indexed800Anonpipelinedsystemtakes200nstoprocessataskThesametaskcanbeprocessedina5-segmentpipelinewithaclockcycleof40nsDeterminethespeedupratioofthepipelinefor200tasksWhatisthemaximumspeedupthatcouIdbeachievedwiththepipelineu

35、nitoverthenonpipelinedunit?Ans.SpeedUp=(200nsx200)/(5+200-1)(40ns)=40000/8160=4.91MaxSpeedUp=5Anonpipelinedsystemtakes100nstoprocessataskThesametaskcanbeprocessedina5-segmentpipelinewithaclockcycleof20nsDeterminethespeedupratioofthepipelinefor100tasksWhatisthetheoreticalspeedupthatcouIdbeachievedwit

36、hthepipelinesystemoveranonpipelinedsystem?Ans.SpeedUp=(100nsx100)/(5+100-1)(20ns)=10000/2080=4.8MaxSpeedUp=5Writecodetoimplementtheexpression:A=(B+C)*(D+E)on3,2飞1and0-addressmachinesInaccordancewithprogramminglanguagepractice,computingtheexpressionshouIdnotchangethevaluesofitsoperandsAns.AddRLBCAddR

37、2.DEMullA.R1.R23-addressmachineLoadRl,BAddRl,CLoadR2.EAddR2.EMultR2.RStoreA.R22-addressmachineLoadBAddCStoreTempLoadDAddEMultTempStoreA1-addressmachinePage #LastUpdated:October2003Page #LastUpdated:October200318.Adigitalcomputerhasamemoryunitwith24bitsperwordTheinstructionsetconsistsof150differentop

38、erationsAllinstructionshaveanoperationcodepart(opcode)andanPage LastUpdated:October2003addresspart(allowingforonlyoneaddress)Eachinstructionisstoredinonewordofmemorya.Howmanybitsareneededfortheopcode?bHowmanybitsareleftfortheaddresspartoftheinstruction?c.Whatisthemaximumallowablesizeformemory?dWhati

39、sthelargestunsignedbinarynumberthatcanbeaccommodatedinonewordofmemory?Ans.a.150instructionsimplies28(27willonlygiveusI28instructions),or8bitsfortheopcode.b24-8=16c2,or32Md24ISor2Pickanarchitecture(otherthanthosecoveredinthischapter).Doresearchtofindouthowyourarchitectureapproachestheconceptsintroduc

40、edinthischapter,aswasdoneforIntel.MIPS,andJavaAns.Noanswergiven-119Thememoryunitofacomputerhas256Kwordsof32bitseachThecomputerhasaninstructionformatwith4fields:anopcodefield;amodefieldtospecify1of7addressinginodes;aregisteraddressfieldtospecifyoneof60registers:andamemoryaddressfieldAssumeaninstructi

41、onis32bitslongAnswerthefollowing:Howlargemustthemodefieldbe?Howlargemusttheregisterfieldbe?Howlargemusttheaddressfieldbe?dHowlargeistheopcodefield?Ans.Weneedtoidentify1of7items,sotheremustbe3bits(23=8)60registersimplies6bits(26=64)c256K=28210=2%or18bitsd32-(3+6+18)=5bits20.Supposeaninstructiontakesf

42、ourcyclestoexecuteinanonpipelinedCPU:onecycletofetchtheinstruction,onecycletodecodetheinstruction,onecycletoperforintheALUoperation,andonecycletostoretheresuItInaCPUwitha4-stagepipeline,thatinstructionstilltakesfourcyclestoexecute、sohowcanwesaythepipelinespeedsuptheexecutionoftheprogram?Ans.Foronein

43、struction,thereisnospeedupThespeedupcomeswiththeparallelexecutionofmultipleinstructionsWhilethefirstinstructionisdecoding,thesecondcanbefetched;whilethefirstinstructionisperformingtheALUinstruction.thesecondcanbedecoding,andthethirdcanbefetched,etcSampieExamQuestions1Supposewehavetheinstructionlda80

44、0.Givenmemoryasfollows:Memory100011001200800900a.immediateb.directc.indirectWhatwouldbeloadedintotheACiftheaddressingmodefortheoperandis:Ans.a.800b900c1000Writethefollowingexpressioninpostfix(reversePolishnotation).RemembertherulesofprecedenceforarithmeticoperatorsX=A-B+C*(D*E-F)/(G+H*K)Ans.AB-CDE*F

45、-*+GHK*+/Anonpipelinedsystemtakes300nstoprocessataskThesametaskcanbeprocessedina5-segmentpipelinewithaclockcycleof60nsDeterminethespeedupratioofthepipelinefor100tasksWhatisthemaximumspeedupthatcouIdbeachievedwiththepipelineunitoverthenonpipelinedunit?Ans.SpeedUp=(300nsx100)/(5+100-1)(60ns)=30000/6240=4.8MaxSpeedUp=5b.32-7=25.c225,or32G.Showhow