計算機體系結構第一章1.11Availabilityisthemostimportantconsiderationfordesigningservers,followedcloselybyscalabilityandthroughput.Wehaveasingleprocessorwithafailuresintime(FIT)of100.Whatisthemeantimetofailure(MTTF)forthissystem?Ifittakes1daytogetthesystemrunningagain,whatistheavailabilityofthesystem?Imaginethatthegovernment,tocutcosts,isgoingtobuildasupercomputeroutofinexpensivecomputersratherthanexpensive,reliablecomputers.WhatistheMTTFforasystemwith1000processors?Assumethatifonefails,theyallfail.答：平均故障時間(MTTF)是一個可靠性度量方法，MTTF的倒數是故障率，一般以每10億小時運行中的故障時間計算(FIT)。因此由該定義可知1/MTTF=FIT/10A9,所以MTTF=10A9/100=10A7O系統可用性=MTTF/(MTTF+MTTR),其中MTTR為平均修復時間，在該題目中表示為系統重啟時間。計算10A7/(10^7+24)約等于1.由于一個處理器發(fā)生故障，其他處理器也不能使用，所以故障率為原來的1000倍，所以MTTF值為單個處理器MTTF的1/1000即10a7/1000=10a4o1.14Inthisexercise,assumethatweareconsideringenhancingamachinebyaddingvectorhardwaretoit.Whenacomputationisruninvectormodeonthevectorhardware,itis10timesfasterthanthenormalmodeofexecution.Wecallthepercentageoftimethatcouldbespentusingvectormodethepercentageofvectorization.Drawagraphthatplotsthespeedupasapercentageofthecomputationperformedinvectormode.Labelthey-axis“Netspeedup”andlabelthex-axis“Percentvectorization”.Whatpercentageofvectorizationisneededtoachieveaspeedupof2?Whatpercentageofthecomputationruntimeisspentinvectormodeifaspeedupof2isachieved?Whatpercentageofvectorizationisneededtoachievedone-halfthemaximumspeedupattainablefromusingvectormode?Supposeyouhavemeasuredthepercentageofvectorizationoftheprogramtobe70%.Thehardwaredesigngroupestimatesitcanspeedupthevectorhardwareevenmorewithsignificantadditionalinvestment.Youwonderwhetherthecompilerofvectorizationwouldthecompilerteamneedtoachieveinordertoequalanaddition2*speedupinthevectorunit(beyondtheinitial10*)?答：根據加速比定義可知，增強加速比=10，如果令增強比例為x,總加速比為y,則有y=l/(l-x+x/l0)。x的取值范圍為［0,1］；y的取值范圍為［0,10］。如下圖示：y=1/(1-x+x/10);當y=2時，x=5/9=55.6%;(5/9)/10/(1/2)=1/9=11.1%最大加速比理論上為10;最大加速比的一半就是5;y=1/(1-x+x/10);當y=5時，x=8/9=88.9%當前x=70%;y=1/(1-x+x/10);可知y=2.7;如果y=2x2.7=5.4;y=1/(1-x+x/10);可知x=0.91;第二章2.8ThefollowingquestionsinvestigatetheimpactofsmallandsimplecachesusingCACTIandassumea65nm(0.065^m)technology.Comparetheaccesstimesof64KBcacheswith64byteblocksandasinglebank.Whataretherelativeaccesstimesoftwo-wayandfour-waysetassociativecachesincomparisontoadirectmappedorganization?Comparetheaccesstimesoffour-waysetassociativecacheswith64byteblocksandasinglebank.Whataretherelativeaccesstimesof32KBand64KBcachesincomparisontoa16KBcache?Fora64KBcache,findthecacheassociativitybetween1and8withthelowestaveragememoryaccesstimegiventhatmissesperinstructionforacertainworkloadsuiteis0.00664fordirectmapped,0.00366fortwo-waysetassociative,0.00987forfour-waysetassociativeand0.foreight-waysetassociativecache.Overall,thereare0.3datareferencesperinstruction.Assumecachemissestake10nsinallmodels.Tocalculatethehittimeincycles,assumethecycletimeoutputusingCACTI,whichcorrespondstothemaximumfrequencyachchecanoperatewithoutanybubblesinthepipeline.答：直接映射：0.86ns；兩路組相聯：1.12ns；四路組相聯：1.37ns。兩路組相聯訪存時間是直接映射的1.12/0.86=1.30倍；四路組相聯訪存時間是直接映射的1.37/0.86=1.59倍。16KBcache的訪存時間為1.27ns,32KBcache為1.35ns,64KBcache為1.37ns。32KBcache的訪存時間是16KBcache訪存時間的1.35/1.27=1.06倍；64KBcache的訪存時間是16KBcache訪存時間的1.37/1.27=1.08倍；平均訪存時間=命中率X命中時間+缺失率X缺失代價；DM缺失率=0.00664/0.3=2.2%;2-way缺失率=0.00366/0.3=1.2%;4-way缺失率=0.00987/0.3=0.33%;8-way缺失率=0./0.3=0.09%;DM訪存所用時鐘周期=0.86ns/0.5ns向上取整=2;2-way訪存所用時鐘周期=1.12ns/0.5ns向上取整=3;4-way訪存所用時鐘周期=1.37ns/0.83ns向上取整=2;8-way訪存所用時鐘周期=2.03ns/0.79ns向上取整=3;DM缺失代價=10ns/0.5ns=20時鐘周期；2-way缺失代價=10ns/0.5ns=20時鐘周期；4-way缺失代價=10ns/0.83ns=13時鐘周期；8-way缺失代價=10ns/0.79ns=13時鐘周期；DM平均訪存時間=(1-0.22)X2+0.022X20X0.5=2.396X0.5=1.2ns；2-way平均訪存時間=(1-0.012)X3+0.012X20X0.5=3.2X0.5=1.6ns；4-way平均訪存時間=(1-0.0033)X2+0.0033X13X0.83=2.036X0.83=1.69ns8-way平均訪存時間=(1-0.0009)x3+0.0009x13x0.79=3x0.79=2.37ns。2.11ConsidertheusageofcriticalwordfirstandearlyrestartonL2cachemisses.Assumea1MBL2cachewith64byteblocksandarefillpaththatis16byteswide.AssumethattheL2canbewrittenwith16bytesevery4processorcycles,thetimetoreceivethefirst16byteblockfromthememorycontrolleris120cycles,eachadditional16byteblockfrommainmemoryrequires16cycles,anddatacanbebypasseddirectlyintothereadportoftheL2cache.IgnoreanycyclestotransferthemissrequesttotheL2cacheandtherequesteddatatotheL1cache.HowmanycycleswouldittaketoserviceanL2cachemisswithandwithoutcriticalwordfirstandearlyrestart?DoyouthinkcriticalwordfirstandearlyrestartwouldbemoreimportantforL1cachesorL2caches,andwhatfactorswouldcontributetotheirrelativeimportance?答：使用關鍵字優(yōu)先和提前啟動策略：第一次獲取16bytes所需時鐘周期120.不使用關鍵字優(yōu)先和提前啟動策略：第一次獲取16bytes所需時鐘周期120個，獲取其余字節(jié)所需的時鐘周期為16x(64B-16B)/16B=48個，所以一共所需的時鐘周期數為168個。有兩個關鍵因素：L1和L2缺失時，對平均訪存時間的影響；使用關鍵字優(yōu)先和提前啟動策略時減少的缺失比率。2.21Virtualmachinescanloseperformancefromanumberofevents,suchastheexecutionofprivilegedinstructions,TLBmisses,traps,andI/O.Theseeventsareusuallyhandledinsystemcode.Thus,onewayofestimatingtheslowdownwhenrunningunderaVMisthepercentageofapplicationexecutiontimeinsystemversususermode.Forexample,anapplicationspending10%ofitsexecutioninsystemmodemightslowdownby60%whenrunningonaVM.Fugure2.32liststheearlyperformanceofvarioussystemcallsundernativeexecution,purevirtualizationandparavirtualizationforLMbenchusingXenonanItanuumsystemwithtimesmeasuredinmicroseconds.WhattypesofprogramswouldbeexpectedtohavesmallerslowdownswhenrunningVMs?Ifslowdownswerelinearasafunctionofsystemtime,giventheslowdownabove,howmuchslowerwouldaprogramspending20%ofitsexecutioninsystemtimebeexpectedtorun?Whatisthemedianslowdownofthesystemcallsinthetableaboveunderpurevirtualizationandparavirtualization?Whichfunctionsinthetable