執(zhí)行時(shí)間(latency等待時(shí)間)課件

PerformanceMeasurement1PerformanceExecutiontime執(zhí)行時(shí)間（latency等待時(shí)間）:Timebetweenthestartandthecompletionofanevent

一個(gè)事件從開(kāi)始到結(jié)束所經(jīng)過(guò)的時(shí)間Performance1/(Executiontime)

性能與執(zhí)行時(shí)間成反比Throughput吞吐量(bandwidth帶寬)：Totalamountofworkdoneinagiventime

給定時(shí)間內(nèi)完成的全部工作1PerformanceMeasurement1PerfoPerformanceMeasurement1MachineXisn%fasterthanMachineY:

機(jī)器X比機(jī)器Y快n%2PerformanceMeasurement1MachiPerformanceMeasurement2Example: MachineArunsaprogramin10seconds, MachineBrunsthesameprogramin15seconds, Ais__%fasterthanB.3PerformanceMeasurement2ExampMaketheCommonCaseFastPerhapsthemostimportantandpervasiveprincipleofcomputerdesignistomakethecommoncasefast:Inmakingadesigntrade-off,favorthefrequentcaseovertheinfrequentcase.計(jì)算機(jī)設(shè)計(jì)的最重要的原則就是：加快經(jīng)常性發(fā)生事件的執(zhí)行速度。4MaketheCommonCaseFastPerhaMaketheCommonCaseFastImprovingthefrequentevent,ratherthantherareevent,willobviouslyhelpperformance.Overflowcaseandnooverflowcaseinaddition提高頻繁事件的執(zhí)行速度，而不是提高罕見(jiàn)事件的執(zhí)行速度，將帶來(lái)明顯的性能上的提高例如加法運(yùn)算中的溢出和非溢出情況5MaketheCommonCaseFastImproAmdahl’sLaw1Amdahl’sLawstatesthattheperformanceimprovementtobegainedfromusingsomefastermodeofexecutionislimitedbythefractionofthetimethefastermodecanbeused.阿姆達(dá)定律表明：通過(guò)改進(jìn)某模式得到的整體性能提高，受限于該改進(jìn)模式所占的運(yùn)行時(shí)間比例。6Amdahl’sLaw1Amdahl’sLawstaAmdahl’sLaw2Speedup（加速比） = Performanceforentiretaskusingtheenhancementwhenpossible（改進(jìn)后完成整個(gè)任務(wù)的性能） Performanceforentiretaskw/ousingtheenhancement（改進(jìn)前完成整個(gè)任務(wù)的性能） = Executiontimeforentiretaskw/ousingtheenhancement（改進(jìn)前完成整個(gè)任務(wù)的時(shí)間） Executiontimeforentiretaskusingtheenhancementwhenpossible（改進(jìn)前完成整個(gè)任務(wù)的時(shí)間）7Amdahl’sLaw2Speedup（加速比）7Amdahl’sLaw3Executiontimenew =Executiontimeoldx

wherefE:fractionofenhancement

sE:improvementgainedbythe enhancementmode

即：新的執(zhí)行時(shí)間=

原來(lái)執(zhí)行時(shí)間x

8Amdahl’sLaw3ExecutiontimeneAmdahl’sLaw3

Speedup=

即：加速比＝原來(lái)的執(zhí)行時(shí)間/新的執(zhí)行時(shí)間1

＝

9Amdahl’sLaw3 9Amdahl’sLaw4Example:Anenhancementrun10timesfasterthantheoriginalmachine,butitisusable40%ofthetime,thenthespeedup=__. Sol:fE=0.4

sE=10 Speedup =1/((1-0.4)+0.4/10) =1.5610Amdahl’sLaw4Example:AnenhaAmdahl’sLawcanalsobeappliedtocomparetwoCPUdesignalternatives,forexample:

Implementationsoffloating-point(FP)squarerootvarysignificantlyinperformance,especiallyamongprocessorsdesignedforgraphics.SupposeFPsquareroot(FPSQR)isresponsiblefor20%oftheexecutiontimeofacriticalgraphicsbenchmark.OneproposalistoenhancetheFPSQRhardwareandspeedupthisoperationbyafactorof10.TheotheralternativeisjusttotrytomakeallFPinstructionsinthegraphicsprocessorrunfasterbyafactorof1.6;FPinstructionsareresponseibleforatotalof50%oftheexecutiontimefortheapplication.Comparethesetwodesignalternatives.Amdahl’sLawcanalsobeappliedtocomparetwoCPUdesignalternatives,forexample:

Implementationsoffloating-point(FP)squarerootvarysignificantlyinperformance,especiallyamongprocessorsdesignedforgraphics.Amdahl’sLaw也可以用于比較兩種設(shè)計(jì)不同的CPU，特別是對(duì)于處理圖形的處理器來(lái)說(shuō)，求浮點(diǎn)數(shù)平方根的不同實(shí)現(xiàn)方法在性能上有很大差異。11Amdahl’sLawcanalsobeappliAmdahl’sLawcanalsobeappliedtocomparetwoCPUdesignalternatives,forexample:

Implementationsoffloating-point(FP)squarerootvarysignificantlyinperformance,especiallyamongprocessorsdesignedforgraphics.SupposeFPsquareroot(FPSQR)isresponsiblefor20%oftheexecutiontimeofacriticalgraphicsbenchmark.OneproposalistoenhancetheFPSQRhardwareandspeedupthisoperationbyafactorof10.TheotheralternativeisjusttotrytomakeallFPinstructionsinthegraphicsprocessorrunfasterbyafactorof1.6;FPinstructionsareresponseibleforatotalof50%oftheexecutiontimefortheapplication.Comparethesetwodesignalternatives.SupposeFPsquareroot(FPSQR)isresponsiblefor20%oftheexecutiontimeofacriticalgraphicsbenchmark.OneproposalistoenhancetheFPSQRhardwareandspeedupthisoperationbyafactorof10.TheotheralternativeisjusttotrytomakeallFPinstructionsinthegraphicsprocessorrunfasterbyafactorof1.6;FPinstructionsareresponseibleforatotalof50%oftheexecutiontimefortheapplication.Comparethesetwodesignalternatives.

例如，求浮點(diǎn)數(shù)平方根的操作，在一個(gè)標(biāo)準(zhǔn)測(cè)試程序中占總執(zhí)行時(shí)間的20%。一種方法是改進(jìn)FPSQR硬件，將它的操作速度提10倍。另一種方法是將所有圖形處理器中的FP指令的執(zhí)行速度都提高1.6倍，這些FP指令在總的執(zhí)行時(shí)間中占50%比較這兩種設(shè)計(jì)方法。12Amdahl’sLawcanalsobeappliAnswer:wecancomparethesetwoalternativesbycomparingthespeedups:ImprovingtheperformanceoftheFPoperationsoverallisslightlybetterbecauseofthehigherfrequency.Answer:wecancomparethesetwoalternativesbycomparingthespeedups:

（可以通過(guò)計(jì)算加速比來(lái)進(jìn)行比較）

ImprovingtheperformanceoftheFPoperationsoverallisslightlybetterbecauseofthehigherfrequency.（可見(jiàn)提高所有FP操作的性能的方案要好，這是由于它們的執(zhí)行頻率較高）13Answer:wecancomparethesetAmdahl’sLaw6ExtremeCases極限情況fE=0Speedup=1fE=1Speedup=sE

fE增強(qiáng)比例

sE增強(qiáng)加速比14Amdahl’sLaw6ExtremeCases極限CPUPerformance1Mostcomputersareconstructedusingaclockrunningataconstantrate多數(shù)計(jì)算機(jī)的運(yùn)行都基于一個(gè)固定頻率的時(shí)鐘信號(hào)Referredtobylength/time,e.g.,10ns,

orrate,e.g.,100MHzms=10–3sec,s=10–6sec,ns=10–9secHz=1/sec,KHz=103Hz,MHz=106Hz,

GHz=109Hz

Clockcycletime=1/clockrate15CPUPerformance1MostcomputerCPUPerformance2CPI(clockcycleperinstruction每條指令時(shí)鐘周期數(shù))(程序CPU時(shí)鐘周期數(shù))(程序指令數(shù))CPUtimeforaprogram=CPUclockcyclesforaprogramxclockcycletime(執(zhí)行程序花費(fèi)的CPU時(shí)鐘周期數(shù))(時(shí)鐘周期時(shí)間)16CPUPerformance2CPI(clockcCPUPerformance3CPIxInstructionCountx1/(clockrate)=CPUtimeBUT,noteveryinstructiontakesthesamenumberofclockcyclestoexecute.Taketheaverage.執(zhí)行指令花費(fèi)的時(shí)鐘周期數(shù)并不相同，這里取平均值17CPUPerformance3CPIxInstrucCPUPerformance4CPI

n:numberofdifferentinstructionsinaprogram CPIi:CPIofinstructioni fi:frequencyofinstructioniinaprogram

n

即∑(第i條指令的CPI×該指令在全部指令中占的比例)

i=118CPUPerformance4CPI18CPUPerformance5Example:Operations frequencyclockcycleADD 60% 1LOAD 40% 2CPIoverall=_____ 1.419CPUPerformance5Example:1.419CPUPerformance6Example:Agivenprogramconsistsofa100-instructionloopthatisexecuted42times.Ifittakes16000cyclestoexecutetheprogramonagivensystem,whatarethatsystem’sCPIfortheprogram?一個(gè)程序由一個(gè)循環(huán)組成，循環(huán)內(nèi)100條指令，循環(huán)執(zhí)行42次，在某個(gè)特定的系統(tǒng)執(zhí)行這個(gè)程序花費(fèi)16000周期，則這個(gè)系統(tǒng)執(zhí)行這個(gè)程序的CPI是多少？Thetotalnumberofinstructionsexecutedis:

100×42=4200.SotheCPIis:

16000／4200=3.81.20CPUPerformance6Example:ThetImproveCPUPerformance1HowdoweimproveCPUperformance那么我們?cè)鯓犹岣逤PU性能呢？i.e.,reduceCPUtime?Again,CPUtime=CPIxInstructionCountx1/(clockrate)So,wewantto _____CPI _____InstructionCount _____clockrate _____clockcycletime我們可以減少CPI、IC、clockcycletime或增加clockrate21ImproveCPUPerformance1HowdImproveCPUPerformance2Clockrate增加時(shí)鐘頻率的方法HardWaretechnology硬件技術(shù)Organization組織結(jié)構(gòu)CPI減少CPI的方法OrganizationInstructionsetarchitecture指令集InstructionCount減少IC的方法InstructionsetarchitectureCompilertechnology編譯技術(shù)22ImproveCPUPerformance2ClockMIPS1MIPS:MillionInstructionPerSecond每秒百萬(wàn)指令MIPS指令數(shù)執(zhí)行時(shí)間23MIPS1MIPS:MillionInstructioMIPS2GivenMIPS, MIPS Executiontime Performance已知MIPS:則:執(zhí)行時(shí)間＝指令數(shù)/(MIPS×106)因此，如果MIPS增加，則執(zhí)行時(shí)間減少，性能增強(qiáng)24MIPS2GivenMIPS,24MIPS3Advantage:Easytounderstand(especiallybycustomers)容易理解DisadvantagesDifficulttocompareMIPSofcomputerswithdifferentinstructionsetsMIPS依賴于指令集，不同指令集的計(jì)算機(jī)不能比較MIPSMIPSvariesbetweenprogramsonthesamecomputer同一計(jì)算機(jī)上的MIPS可能因程序而異MIPScanvaryinverselytoperformance

(e.g.floating-pointinstructionexecutedbyhardwareorsoftware)MIPS可能與性能相反25MIPS3Advantage:25MIPS4Whenrunningaparticularprogram,computerAachieves100MIPSandcomputerBachieves75MIPS.However,computerAtakes60stoexecutetheprogram,whilecomputerBtakesonly45s.Howisthispossible?執(zhí)行一個(gè)具體的程序時(shí)，計(jì)算機(jī)A的MIPS為100而計(jì)算機(jī)B的MIPS為75。然而執(zhí)行這個(gè)程序計(jì)算機(jī)A花費(fèi)60s，而計(jì)算機(jī)B花費(fèi)45s，為什么？26MIPS4WhenrunningaparticulaMIPS5Solution:MIPSmeasurestherateatwhichaprocessorexecutesinstructions,butdifferentprocessorarchitecturesrequiredifferentnumbersofinstructionstoperformagivencomputation.IfcomputerAhadtoexecutesignificantlymoreinstructionsthancomputerBtocompletetheprogram,itwouldbepossibleforcomputerAtotakelongertoruntheprogramthanprocessorBdespitethefactthatcomputerAexecutesmoreinstructionspersecond.27MIPS5Solution:MIPSmeasuresMIPS5解答:MIPS是評(píng)價(jià)處理器執(zhí)行指令速度的一個(gè)標(biāo)準(zhǔn),但是對(duì)于一個(gè)給定的計(jì)算，不同體系結(jié)構(gòu)的處理器需要不同數(shù)量的指令來(lái)進(jìn)行計(jì)算。在執(zhí)行本程序時(shí)，如果計(jì)算機(jī)A所需要執(zhí)行的指令數(shù)比計(jì)算機(jī)B多，那么盡管計(jì)算機(jī)A的MIPS比計(jì)算機(jī)B大,它仍然可能需要比計(jì)算機(jī)B更長(zhǎng)的執(zhí)行時(shí)間28MIPS5解答:MIPS是評(píng)價(jià)處理器執(zhí)行指令速度的一個(gè)OtherMeasurementsMFLOPS: Millionsoffloatingpointoperationspersecond

每秒百萬(wàn)條浮點(diǎn)指令29OtherMeasurementsMFLOPS:29InternetResources30InternetResources30PerformanceMeasurement1PerformanceExecutiontime執(zhí)行時(shí)間（latency等待時(shí)間）:Timebetweenthestartandthecompletionofanevent

一個(gè)事件從開(kāi)始到結(jié)束所經(jīng)過(guò)的時(shí)間Performance1/(Executiontime)

性能與執(zhí)行時(shí)間成反比Throughput吞吐量(bandwidth帶寬)：Totalamountofworkdoneinagiventime

給定時(shí)間內(nèi)完成的全部工作31PerformanceMeasurement1PerfoPerformanceMeasurement1MachineXisn%fasterthanMachineY:

機(jī)器X比機(jī)器Y快n%32PerformanceMeasurement1MachiPerformanceMeasurement2Example: MachineArunsaprogramin10seconds, MachineBrunsthesameprogramin15seconds, Ais__%fasterthanB.33PerformanceMeasurement2ExampMaketheCommonCaseFastPerhapsthemostimportantandpervasiveprincipleofcomputerdesignistomakethecommoncasefast:Inmakingadesigntrade-off,favorthefrequentcaseovertheinfrequentcase.計(jì)算機(jī)設(shè)計(jì)的最重要的原則就是：加快經(jīng)常性發(fā)生事件的執(zhí)行速度。34MaketheCommonCaseFastPerhaMaketheCommonCaseFastImprovingthefrequentevent,ratherthantherareevent,willobviouslyhelpperformance.Overflowcaseandnooverflowcaseinaddition提高頻繁事件的執(zhí)行速度，而不是提高罕見(jiàn)事件的執(zhí)行速度，將帶來(lái)明顯的性能上的提高例如加法運(yùn)算中的溢出和非溢出情況35MaketheCommonCaseFastImproAmdahl’sLaw1Amdahl’sLawstatesthattheperformanceimprovementtobegainedfromusingsomefastermodeofexecutionislimitedbythefractionofthetimethefastermodecanbeused.阿姆達(dá)定律表明：通過(guò)改進(jìn)某模式得到的整體性能提高，受限于該改進(jìn)模式所占的運(yùn)行時(shí)間比例。36Amdahl’sLaw1Amdahl’sLawstaAmdahl’sLaw2Speedup（加速比） = Performanceforentiretaskusingtheenhancementwhenpossible（改進(jìn)后完成整個(gè)任務(wù)的性能） Performanceforentiretaskw/ousingtheenhancement（改進(jìn)前完成整個(gè)任務(wù)的性能） = Executiontimeforentiretaskw/ousingtheenhancement（改進(jìn)前完成整個(gè)任務(wù)的時(shí)間） Executiontimeforentiretaskusingtheenhancementwhenpossible（改進(jìn)前完成整個(gè)任務(wù)的時(shí)間）37Amdahl’sLaw2Speedup（加速比）7Amdahl’sLaw3Executiontimenew =Executiontimeoldx

wherefE:fractionofenhancement

sE:improvementgainedbythe enhancementmode

即：新的執(zhí)行時(shí)間=

原來(lái)執(zhí)行時(shí)間x

38Amdahl’sLaw3ExecutiontimeneAmdahl’sLaw3

Speedup=

即：加速比＝原來(lái)的執(zhí)行時(shí)間/新的執(zhí)行時(shí)間1

＝

39Amdahl’sLaw3 9Amdahl’sLaw4Example:Anenhancementrun10timesfasterthantheoriginalmachine,butitisusable40%ofthetime,thenthespeedup=__. Sol:fE=0.4

sE=10 Speedup =1/((1-0.4)+0.4/10) =1.5640Amdahl’sLaw4Example:AnenhaAmdahl’sLawcanalsobeappliedtocomparetwoCPUdesignalternatives,forexample:

Implementationsoffloating-point(FP)squarerootvarysignificantlyinperformance,especiallyamongprocessorsdesignedforgraphics.SupposeFPsquareroot(FPSQR)isresponsiblefor20%oftheexecutiontimeofacriticalgraphicsbenchmark.OneproposalistoenhancetheFPSQRhardwareandspeedupthisoperationbyafactorof10.TheotheralternativeisjusttotrytomakeallFPinstructionsinthegraphicsprocessorrunfasterbyafactorof1.6;FPinstructionsareresponseibleforatotalof50%oftheexecutiontimefortheapplication.Comparethesetwodesignalternatives.Amdahl’sLawcanalsobeappliedtocomparetwoCPUdesignalternatives,forexample:

Implementationsoffloating-point(FP)squarerootvarysignificantlyinperformance,especiallyamongprocessorsdesignedforgraphics.Amdahl’sLaw也可以用于比較兩種設(shè)計(jì)不同的CPU，特別是對(duì)于處理圖形的處理器來(lái)說(shuō)，求浮點(diǎn)數(shù)平方根的不同實(shí)現(xiàn)方法在性能上有很大差異。41Amdahl’sLawcanalsobeappliAmdahl’sLawcanalsobeappliedtocomparetwoCPUdesignalternatives,forexample:

Implementationsoffloating-point(FP)squarerootvarysignificantlyinperformance,especiallyamongprocessorsdesignedforgraphics.SupposeFPsquareroot(FPSQR)isresponsiblefor20%oftheexecutiontimeofacriticalgraphicsbenchmark.OneproposalistoenhancetheFPSQRhardwareandspeedupthisoperationbyafactorof10.TheotheralternativeisjusttotrytomakeallFPinstructionsinthegraphicsprocessorrunfasterbyafactorof1.6;FPinstructionsareresponseibleforatotalof50%oftheexecutiontimefortheapplication.Comparethesetwodesignalternatives.SupposeFPsquareroot(FPSQR)isresponsiblefor20%oftheexecutiontimeofacriticalgraphicsbenchmark.OneproposalistoenhancetheFPSQRhardwareandspeedupthisoperationbyafactorof10.TheotheralternativeisjusttotrytomakeallFPinstructionsinthegraphicsprocessorrunfasterbyafactorof1.6;FPinstructionsareresponseibleforatotalof50%oftheexecutiontimefortheapplication.Comparethesetwodesignalternatives.

例如，求浮點(diǎn)數(shù)平方根的操作，在一個(gè)標(biāo)準(zhǔn)測(cè)試程序中占總執(zhí)行時(shí)間的20%。一種方法是改進(jìn)FPSQR硬件，將它的操作速度提10倍。另一種方法是將所有圖形處理器中的FP指令的執(zhí)行速度都提高1.6倍，這些FP指令在總的執(zhí)行時(shí)間中占50%比較這兩種設(shè)計(jì)方法。42Amdahl’sLawcanalsobeappliAnswer:wecancomparethesetwoalternativesbycomparingthespeedups:ImprovingtheperformanceoftheFPoperationsoverallisslightlybetterbecauseofthehigherfrequency.Answer:wecancomparethesetwoalternativesbycomparingthespeedups:

（可以通過(guò)計(jì)算加速比來(lái)進(jìn)行比較）

ImprovingtheperformanceoftheFPoperationsoverallisslightlybetterbecauseofthehigherfrequency.（可見(jiàn)提高所有FP操作的性能的方案要好，這是由于它們的執(zhí)行頻率較高）43Answer:wecancomparethesetAmdahl’sLaw6ExtremeCases極限情況fE=0Speedup=1fE=1Speedup=sE

fE增強(qiáng)比例

sE增強(qiáng)加速比44Amdahl’sLaw6ExtremeCases極限CPUPerformance1Mostcomputersareconstructedusingaclockrunningataconstantrate多數(shù)計(jì)算機(jī)的運(yùn)行都基于一個(gè)固定頻率的時(shí)鐘信號(hào)Referredtobylength/time,e.g.,10ns,

orrate,e.g.,100MHzms=10–3sec,s=10–6sec,ns=10–9secHz=1/sec,KHz=103Hz,MHz=106Hz,

GHz=109Hz

Clockcycletime=1/clockrate45CPUPerformance1MostcomputerCPUPerformance2CPI(clockcycleperinstruction每條指令時(shí)鐘周期數(shù))(程序CPU時(shí)鐘周期數(shù))(程序指令數(shù))CPUtimeforaprogram=CPUclockcyclesforaprogramxclockcycletime(執(zhí)行程序花費(fèi)的CPU時(shí)鐘周期數(shù))(時(shí)鐘周期時(shí)間)46CPUPerformance2CPI(clockcCPUPerformance3CPIxInstructionCountx1/(clockrate)=CPUtimeBUT,noteveryinstructiontakesthesamenumberofclockcyclestoexecute.Taketheaverage.執(zhí)行指令花費(fèi)的時(shí)鐘周期數(shù)并不相同，這里取平均值47CPUPerformance3CPIxInstrucCPUPerformance4CPI

n:numberofdifferentinstructionsinaprogram CPIi:CPIofinstructioni fi:frequencyofinstructioniinaprogram

n

即∑(第i條指令的CPI×該指令在全部指令中占的比例)

i=148CPUPerformance4CPI18CPUPerformance5Example:Operations frequencyclockcycleADD 60% 1LOAD 40% 2CPIoverall=_____ 1.449CPUPerformance5Example:1.419CPUPerformance6Example:Agivenprogramconsistsofa100-instructionloopthatisexecuted42times.Ifittakes16000cyclestoexecutetheprogramonagivensystem,whatarethatsystem’sCPIfortheprogram?一個(gè)程序由一個(gè)循環(huán)組成，循環(huán)內(nèi)100條指令，循環(huán)執(zhí)行42次，在某個(gè)特定的系統(tǒng)執(zhí)行這個(gè)程序花費(fèi)16000周期，則這個(gè)系統(tǒng)執(zhí)行這個(gè)程序的CPI是多少？Thetotalnumberofinstructionsexecutedis:

100×42=4200.SotheCPIis:

16000／4200=3.81.50CPUPerformance6Example:ThetImproveCPUPerformance1HowdoweimproveCPUperformance那么我們?cè)鯓犹岣逤PU性能呢？i.e.,reduceCPUtime?Again,CPUtime=CPIxInstructionCountx1/(clockrate)So,wewantto _____CPI _____InstructionCount _____clockrate _____clockcycletime我們可以減少CPI、IC、clockcycletime或增加clockrate51ImproveCPUPerformance1HowdImproveCPUPerformance2Clockrate增加時(shí)鐘頻率的方法HardWaretechnology硬件技術(shù)Organization組織結(jié)構(gòu)CPI減少CPI的方法OrganizationInstructionsetarchitecture指令集InstructionCount減少IC的方法InstructionsetarchitectureCompilertechnology編譯技術(shù)52ImproveCPUPe