AB 冗余系统内存使用详解

UnderstandControlLogixRedundancyMemoryUsageCreationDate:6-30-04DISCLAIMERBecauseofthevarietyofusesforthisinformation,theuserofandthoseresponsibleforapplyingthisinformationmustsatisfythemselvesastotheacceptabilityofeachapplicationanduseoftheprogram.InnoeventwillAllen-BradleyCompanyberesponsibleorliableforindirectorconsequentialdamagesresultingfromtheuseorapplicationofthisinformation.Theillustrations,chartsandexamplesshowninthisapplicationnoteareintendedsolelytoillustratetheprinciplesofprogrammablecontrollersandsomeofthemethodsusedtoapplythem.Particularlybecauseofthemanyvariablesandrequirementsassociatedwithanyparticularinstallation,Allen-BradleyCompanycannotassumeresponsibilityorliabilityforactualusebasedupontheillustrationsusedandapplications.Allen-BradleyCompanyassumesnopatentliabilitywithrespecttouseofinformation,circuits,equipment,orsoftwaredescribedinthistext.Reproductionofthecontentsoftheapplicationnote,inwholeorinpart,withouttheexpresswrittenconsentoftheAllen-BradleyCompanyisprohibited.DocumentPurposeThisdocumentisnotamanualortrainingmaterial,butanApplicationNote,whichcouldbeusefulinhelpingaRockwellAutomationcustomerwithunderstandingControlLogixRedundancymemoryusage.IntendedAudienceThisdocumentistobeusedbyRockwellAutomationemployees/customerssupportingandsellingControlLogixRedundancy.ConceptofApplicationNoteInaControlLogixcontroller(L55andL6X)therearetwoseparateareasofmemory,I/Omemoryanddataandlogicmemory.WhenRedundancyisenabled,certainareasofmemorywillapproximatelydoubleinusage;thoseareI/Omemoryanddatamemory.Wewilldiscusswhytheseareasofmemorydouble.FirstletuslookatanexampleusinganL55M13toillustrate:Non-RedundantSystem:I/Omemory:Total208KbytesUsed75KbytesDataandLogicmemory:Total1.5MegUsed250Kbytesdataand250KbyteslogicNowwhenRedundancyisenabledtheamountofI/Omemoryanddatamemorywilldouble.RedundantSystem:I/Omemory:Total208KbytesUsed125KbytesDataandLogicmemory:Total1.5MegUsed500Kbytesdataand250KbyteslogicAnotherexampleoftheI/OanddatamemorydoublingcanalsobeshownbyusingV13softwareandtheofflinememoryestimationtool.Firstwewilllookatthememoryestimationofablanknon-redundantcontroller.ThatisnoI/O,data,orlogic.Thisisonlyanestimation.Estimated120hlemmy_ Total: Estimated120hlemmy_ Total: 229,376 bytesI □Free; 211744 byte?-1 OUsed; 17,632 byte?-|*AManUsed: 17,632bytesLargedBlockFree; 211744byte?EstirriatedData占厂idLogicMernory--Total:1.605,632Lyles--Free;1,580,592切怕$-Used;24740Lvtes:U*MaxUsed:24,740bytesLargestBlockFree; 1^80,592bytesBaselineThefigureabovewillgiveusabaselinefromwhichtowork.Thereissomememoryusedevenifthecontrollerprogramisempty.NowwewilladdsomeI/Ototheprogram.ThefollowingI/Owasadded.一-住]I/lJinnfiriiirArinn-J口」i17E6-匚wmpjD匚卜旧1_El°]2[0]175GCNCR/DF：匚P1OT匚1S[1]1756-IBSE/BMl5l.y]iy-.p.-iH-izynmii[3]1756IB32/BM12巳r+l1756-IB3Z/BM155L"--J1 M14岁[&]1756IB32/BMIS°|r?l17I36-ID3Z/DMIG5 [k]I/Hh-IH.-iz/HMl/3旧」1756-IB32/BM1BnJriOl1756-ID3E/□M19-J.1[ii]I/hh-iPJHH；ljKI-MijIl-l-i巳口」17E6-OB32DInJ[2]17SGOD32Dll日r-311756-OB3ZDiZ\■5iL斗」1muinnJ[r；]17C5GOD32DU日r611756-OB3ZDi5\■5|y」ijruK「if.巳[S]17E6OB32D17nJr^l17S6-OD3ZDIG5[HQI/bh-LjH.-i^-U14Redundancynotenabled:EstimatedI/OMemoryEstimatedDataandLogicMemory--Total:229.376bytes-Total:1,605,632bytes--Free:209,400bytes--Free:1.56L164bytes-Used:19,803bytes-Used:44,463bytes-■MaxUsed:19,000bytes-•MaxUsed:44,460bytesLargestBlockFree:209,400bytesLargestBlockFree:1.56L164bytesFigure1Redundancyenabled:

EstimatedI/OMemoryEstimatedDataandLogicMemory-—-Total:229,376bytes—-Total:1,605,632bytes--Free:207,560bytes--Free:1,551,136bytes-Used:21,008bytes-Used:54,496bytes:-MaxUsed:2LS08bytes:■MaxUsed:54,496bytesLargestBlockFree:207,560bytesLargestBlockFree:1,551,136bytesFigure2BysubtractingtheMaxUsedbytesunderEstimatedI/OMemoryfromfiguresBaselineand1youcanseethat2256byteswereaddedtomemory.ThenbysubtractingtheMaxUsedbytesunderEstimatedI/OMemoryfromfiguresBaselineand2youcanseethat4176byteswereaddedtomemory,approximatelydoublethesize.Nowwewillnowadda10,000elementDINTarraytotheprogram.Redundancynotenabled:EstimatedI/OMemoryLargestBlockFree:229.376bytes209,400bytes19,888bytes19,000bytes209,400bytesEstimatedDataandLogicMemory1,605,632b^tesEstimatedI/OMemoryLargestBlockFree:229.376bytes209,400bytes19,888bytes19,000bytes209,400bytesEstimatedDataandLogicMemory1,605,632b^tesL521.084bvtes04,548bytes04,540b^tesLargestBlockFree:L521,084bvtesFigure3Redundancyenabled:EstimatedI/OMemoryEstimatedDataandLogicMemory-—-Total:229,376bvtes—-Total:1,605,632b^tes--Free:207,568bytes--Free:1,471,028bytes-Used:21,000b^tes-Used:134,604b^tes-■嘴MawUsed:21,003b^tes-■嘴MawUsed:134,604b^tesLargestBlockFree:207,568bytesLargestBlockFree:L471.028bytesFigure4BysubtractingtheMaxUsedbytesunderEstimatedDataandLogicMemoryfromfigures1and3youcanseethat39880byteswereaddedtomemory,approximatelythesizeofthe10,000elementDINTarray.ThenbysubtractingtheMaxUsedbytesunderEstimatedDataandLogicMemoryfromfigures2and4youcanseethat80108byteswereaddedtomemory,approximatelydoublethesizeofthe10,000elementdintarray.RockwellAutomationThetwoexamplesaboveconfirmthatI/Oanddatamemoryareapproximatelydoubledwhenredundancyisenabled.ThequestionthatcomesoutofallthisiswhyisI/OandDatamemoryusagedoublewhenredundancyisenabled?Eachsectionofmemorywillbeaddressedseparately.WhyisI/Omemorydoubled?Simplytohavebumplessoutputsforthehighestpriorityusertask.Anotherwaytosaythisisitpreventsamomentaryreversalofoutputsimmediatelyafteraswitchoverforoutputsofthehighpriorityusertask.ThebasicobjectiveoftheoutputhandlingapproachintheredundantLogixControlleristoensurethatnodataissenttotheoutputmodulesfromtheprimaryLogixcontrollerwithoutthesecondaryLogixcontrollerhavingthesamevaluesinitsoutputimage.Thisisaccomplishedviaanoutputbufferingmechanism.Therearetwocopiesoftheoutputimage:1.ProgramOutputImage2.OutputTransmitImageTheProgramOutputImage(POI)isthecopyoftheoutputsthataredirectlyaccessedbytheuserprogram.Allinstructions,bothoutput(write)instructionsandinput(read)instructionswhichreferenceoutputdatavalues,willusethePOI.Thesecondbuffer,theOutputTransmitImage(OTI),isthecopyoftheoutputsthatareactuallysenttotheoutputmodules.Ifthereisaqualifiedsecondarycontrollerinthesystem,thenthechangesintheprimarycontroller'sPOIarefirstsenttothesecondarycontroller.TheOTIintheprimaryisnotupdatedwiththePOI'sdatauntilthesecondarycontrolleracknowledgesthatithasreceivedthedataandmovedthedatatoitsownOTI.IfthereisnosecondarycontrollerpresentthedatawillbecopiedfromthePOItotheOTIattheendofeveryprogram.SobecauseofthecreationofthePOIwedoubletheamountofI/Omemory.Butyoumayaskyou'veonlytalkedabout“outputs”whatabout“inputs”.IfoutputsaretheonlyI/OthatneedtobebufferedthenmyI/Omemoryshouldonlyincreasebytheamountusedforoutputs.Yesthatistrue,butforincreasedperformanceitiseasierandfastertomakeanexactcopyofI/O