六西格玛绿带培训教材(PPT 217页)1.ppt

1、1,版本:1.00日期:May2003,6西格玛绿带培训教材ONE,2,DAY1第一天(定义阶段):-6西格玛及精简优化与COQ(質量成本）的关系COQ的脑力风暴-FirstPassYieldExerciseII初始直通率的練習IPO（輸入輸出流程）andflowdiagram（IPO和流程圖）FlowanalysisofdroppingcardsontotargetRepeatingtheexercise重復練習Resultsanddiscussions結論和檢討-西格玛培训中的某些质量改进工具脑力风暴技术第一天結束wrapup通過以上的教學引導學員對品質成本的認識運用六西格瑪就是有效的降低

2、品質成本,課程安排,3,DAY2第二天(VarianceReduction降低变差的理解):-ThepowerofPlatochartandthe80/20rules柏拉圖表的功能和80/20的規則ConstructionofaPlatoChartusingcomputerflowdiagramanditsassociatedsymbols流程圖和其制作符號含義Two实例offlowdiagram(usingacommonscenario)兩個流程圖的實例（使用通用的情節）Barriersthathinder6西格玛implementation阻礙開展執行六西格瑪的因素-Break-Whati

3、sFMEA什么是FMEAExampleofFMEA關于FMEA的實例GroupexerciseonFMEAofbarriersto6西格玛implementationFMEApresentations關于FMEA的介紹-Lunch-Conceptofprecisionandaccuracy對准確和准確的理解（Cp和Ca）Howdoesitlinktothemeanand标准偏差（如何將平均值和標准偏差聯系起來）Precisionandaccuracyexample(i.e.Selectionoffundmanager准確和精確的實際例子CatapultexerciseI彈弓拋物發射器的思維練習

4、一-Break-Introducingconceptofvariancereduction(i.e.PF/CE/CNX/FMEA/SOP)介紹降低變差的觀念Variancereduction脑力风暴exerciseforCatapult用彈弓發射器進行降低變差的腦力風暴練習CatapultexerciseII彈弓拋物發射器思維練習二ComputationofCatapultexerciseresultaftervariancereduction評估計算彈弓發射器游戲中的數據來了解降低變差的含義Discussionofvariancecontributors討論降低變差的意義第二天wrapup在

5、以上學習中通過彈弓發射器游戲的了解在游戲中掌握了解在六西格瑪中(VarianceReduction）降低变差重要性,4,DAY3(MeasurePhase測量階段):Recapofstatisticalterminology全新的統計學朮語Histogramandanormaldata對直方圖和常態數據的理解Constructionofhistogram對直方圖的解釋Transformationofdata數據的轉換CalculateCp,Cpkfromnon-normaldata計算非正態數據的Cp,Cpk-Break-Theimportanceofgoodmeasurement正確的測量方

6、法的重要性Direct和indirectmeasurement(i.e.Introductiontoscatterdiagram)Riskofwronginterpretation錯誤解釋的風險-Underandvarianceconceptinmeasurementsystem在變異范圍內的測量系統的觀念-IntroductiontoGaugerepeatabilityandreproducibility(GRProject/CandidateSelection,CandidateSelection,Tofillin;Characteristicsof6西格玛candidateScore-s

7、heet和SummaryLeadershipValuesScore-sheet6西格玛CandidatesLeadershipValuesSummary6西格玛Candidates,ProjectSelection,Tofillin;6西格玛ProjectSelectionSummaryEaseofImplementationAssessmentROIImpactAssessment,Candidatesmustscoregreaterthan2.5ptsforeachtoqualify,Candidatesmustscoreatleast0.9ptsintheinthe6西格玛Project

8、SelectionSummarytoqualify,6西格玛ProjectSummary,Matchingrightprojecttotherightpeople,31,Phase2;6西格玛培训和applicationof6西格玛工具s,CollectBaselineDataonProject,yieldCOQCostCycleTimeInventorylevel,Tofillin;6西格玛R0ProjectReviewSheet6西格玛COQTemplate6西格玛ProjectProgressReportTogetallrelevantpartiesapprovalsignatories

9、,Attend6西格玛GB/BBAcademic培训,Applicationof6西格玛工具toProject(s),PF/CE/CNX/SOPsFMEAMSAPARETOPROBABILITYDISTRIBUTIONANOVADOESTATISTICALINTERVALSPC,CreateCertificationTemplate,FollowDMAICPresentationSummaryRevA,tocompleteeachphaseoftheproject,流程to6西格玛GB/BBcertification,32,Phase3;CertificationofCandidate,Com

10、pleteClosureTechnicalReport,Presentationof6西格玛CertificationProject(s),SiteAssessmentofProject(s),IssuePlaque/CertificateAchievementtocandidate,UpdateCandidateLORCareerProfileon培训和Recognition,FinancetoverifyprojectsavingsChampions和MBBtoassesscandidateunderstanding和applicationof6西格玛工具sCandidateCertifi

11、cationEvaluationForm,流程to6西格玛GB/BBcertification,FinancetoverifyprojectsavingsChampions和MBBtoassesscandidateunderstanding和applicationof6西格玛工具sCandidateCertificationEvaluationForm,33,RequirementsforCertification,Completionof6西格玛培训courseSuccessfulprojectcompletion(goalachievement和documentation)Demonstr

12、ationontheunderstandingof6西格玛工具sEffective和successfulcompletionofstepsto“holdthegain”Completionofeachprojectwithin1yr绿带Certification;completionof2projectswithminimumsavingofUS$25,000perproject黑带Certification;completionof2projectswithminimumsavingofUS$100,000perproject,交付ablesforCertification,Demonstr

13、ationofsix-sigmathough流程Completionof6西格玛R0ProjectReviewSheetCompletionof6西格玛ProjectProgressReportCompletionof6西格玛COQTemplateCompletionofDMAICPresentationSummaryCompletionofClosureTechnicalReport,34,Certification结构,CandidateCertificationBoardSiteLeaderChampionsMaster黑带CertificationBoardReview流程Candid

14、atetodistributetheClosureReportSummarytotheboardatleastoneweekbeforethereviewdate(maybewaivedatthediscretionofthesiteleader)MBBtoactasthechairoftheboardCandidatetopresentprojectdetailsusingusingthought流程mapwithemphasisonhoweach工具wasappliedQuestions,clarifications和reviewbytheboard,35,Certification结构(

15、cont),BoardmemberstoratecandidateusingCertificationEvaluationFormPassif;CandidateTechnicalAssessmentisgreaterthan20ptsforGreenbelt和greaterthan40ptsforBlackbelt流程Variation和MeasurableResultsEvaluationisatleast4ptsforeachcategoryInformcandidateoftheoutcome,36,SampleForms,Characteristicsof6西格玛Candidates

16、,LeadershipValuesScore-sheet,37,SampleForms;LeadershipValuesSummary,38,SampleForm;ClosureTechnicalReport,39,SampleForm;CandidateCertificationEvaluationForm,40,Definitionofa流程A流程isanactivitywhichutilizeinputsfromexternalsource和transformthemintodesiredoutput(s).,Example:Manufacturing流程es(i.e.wirebondi

17、ng,injectionmolding,glasssawing)Financial流程es(i.e.doublebookkeeping,产品costing)HR流程es(i.e.recruiting,ranking和appraisal,培训)Dailyactivity(i.e.parkingacar,buyinglunch,brushingyourteeth),41,IPO(Input-流程-Output)Diagram,PeopleMaterialEquipmentPoliciesProceduresMethodsEnvironment,PerformaserviceProducea产品Co

18、mpleteatask,utilizingexternalINPUTStoachievethedesiredOUTPUT(S),Avisualrepresentationofa流程whichlistsinputvariables和outputcharacteristics,42,WhatisthepurposeofIPO,Ahighlevelinterpretationofa流程,whichenableeaseofunderstanding,throughoutliningthe关系hipsbetweeninputvariables和outputresponse(s).,43,Whatisad

19、istribution?,Itisapatternformbythecollectionofdata,groupingtheoutcomehorizontally(x-axis),和indicatestheobservedfrequencyoftheoutcomevertically(y-axis).Instatistic,thisgenerateatheoreticalpatternwherebyinformationofentirepopulationcanbeobtainedfromobservinglimitedsamples.,44,Characteristicofanormaldi

20、stributionIthasasinglepeak和abellshapecurveItisadistributionforcontinuousdata-Themean(average)isisatthecenterofthecurveThevariancedescribethespreadingofdata,45,为什么isdistributionimportanttoa流程Theoutputofa流程canmostofthetimeassociateswithastatisticaldistribution,givingopportunityforengineertoanalyzethed

21、atastatistically,hencearrivingconclusionwithastatisticalconfidence和atalowercost.,46,流程CapabilityStudy,流程capabilitypotential,CpBasedontheassumptionsthat:,Cp=流程capabilitypotentialCpk=流程capabilityindexItisameasurementofthecapabilityofa流程,byindexingthe流程naturaltolerancewithrespecttothedevicespecificatio

22、n(i.e.customertolerance),流程isnormal,Itisa2-sidedspecification,流程meaniscenteredtothedevicespecification,Spreadinspecification,Naturaltolerance,47,流程CapabilityIndex,Cpk,1.Basedontheassumptionthatthe流程isnormal2.Anindexthatcomparethe流程centerwithspecificationcenter,Thereforewhen,Cpk0.30Checkifthespecific

23、ationlimitsisreasonableorattainableIfP/TOT0.30Checktheitemsthatwerepartofthemeasurementsystemstudy和seeiftheyarerepresentingatleast80%oftheactualtotal流程variability.CheckIfthemeasurementsystemequipmentisthebestcondition和isperforminguptospecifications.(ordowehavenochoicebuttouseit)IfP/TOLorP/TOTcloseto

24、0.3butifthe流程isoperatingathighcapabilityCpk2,thenthemeasurementsystemismostprobablynottheproblem,ImprovingPoorMeasurementSystem,136,Discussingrouponhowtomeasurethediameterofthecatapultball和determinethe流程steps.InyourgroupmeasurethecircumferenceofthegroupofCatapultBall提供dtoyou.Eachgroupistomeasurethe1

25、0Ballsthreetimes,n=3.Collectthedatafromtheothertwogroups.Oncethedataisavailable,combinetheresult和conducttheGRforstatisticaldependenceevent,=0.3/0.4=0.75,165,SolvingtheproblemwithatableWhatistheprobabilitythatadrawnboxisstriped,giventhattheboxisred?,Symbolically=P(Striped|Red),Whatistheprobabilitytha

26、tthedrawnboxisred?,Whatistheprobabilityofdrawingaredstripedbox?,Whatistheconditionalprobabilityofthedrawnboxtobestripedwhendrawnred?,=P(Striped和Red)/P(Red)=0.4/0.6=0.667,166,“Jointprobabilitiesunderstatisticaldependence”,Weknowthattheformulaforconditionalprobabilityunderstatisticaldependenceis:P(A|B

27、)=P(A和B)/P(B),IfweshiftP(A和B)totheleft和P(A|B)totheright,wewillhave:P(A和B)=P(A|B)xP(B),167,Withreferencetothescenarioof10boxesinabag,whatistheprobabilitytodrawagreendottedbox?P(Dotted和Green)=P(Dotted|Green)xP(Green)=0.75x0.4=0.3,168,PosteriorProbability(BayesTheorem),Posteriorprobabilitydefinesthatce

28、rtainprobabilitieswerealteredafterthepeopleinvolvedobtainedadditionalinformation.Thisnewprobabilitiesareknownasrevised,orposteriorprobabilities.,AChefhasformulatedanewchickenrecipe和basedonhisexperiencewithformulatingasteakrecipe,theprobabilityofsellingarecipewellistoincreasetheamountofgarlicpowderin

29、theingredients.However,afterintroducingthenewrecipeforoneweek,hefoundthatthenewchickenrecipedoesnotsellaswellastheoldone,hemustthereforerevisehispriorprobabilities和useotheringredientsintherecipe.,CalculatingposteriorprobabilitiesAssumingthatwehaveequalnumbersof2typesof“biased”diceinabowl.Onetypeofth

30、ebiaseddiewillrolloutavalueofFive40%ofthetime,和theothertypewillrolloutthevalueFive,70%ofthetime.Ifonediceisdrawn和rolltothevalueFive,whatistheprobabilitythatitisatype1dice?,169,Type1Type2,0.40.7,Calculatingposteriorprobabilities(Cont),Tofindtheprobabilitythatthedicewehavedrawnistype1,weshouldadoptthe

31、formulaforconditionalprobabilityunderstatisticaldependence:P(B|A)=P(BA)/P(A),P(Type1|Five)=P(Type1,Five)/P(Five)=0.20/0.55=0.364,Thereforetheprobabilityofdrawingatype1diceis0.364,170,Calculatingposteriorprobabilities(Cont),Assuch,whatistheprobabilityofdrawingatype1dicebeforethedicewasrolled?0.5,What

32、istherevisedprobabilityofdrawingatype1diceafterthedicewasrolled和havingavalueofFive?0.364,Ifwearetorollthesamediceagain和achieveavalueofFive,whatistheprobabilitythatitisatype1dice?,P(Type1|2Fives)=P(Type1,2Fives)/P(2Fives),Type1Type2,0.4x0.4=0.160.7x0.7=0.49,=(0.08/0.325),=0.246,171,Furtherexampleofpo

33、steriorprobabilities更多的實際例子,Kennethisamanagerofaautomation设计company和haspreviously8engineeringstaffsreportingtohim.Basedonhisassessmentsonthe8engineers,heisonlysatisfiedwiththeperformanceof6ofhisengineers.Hefurtherconcludedfrompast5yearsobservationthatifaprojectwasassignedtothoseengineerswhohadmethis

34、workexpectation,85%ofthetimetheprojectwillbecompletedbeforedateline.Ontheotherhand,the2engineerswhoseperformanceisunsatisfactory,hasonly35%ofthetimecompletedtheirprojectsbeforedateline.Oneofthe2“unsatisfactory”engineerhadresignedfromthecompany8monthsago,和hisreplacement,Joehadsincethenjointhecompanyf

35、oraperiodof6months,和itisnowtimeforKennethtoreviewJoesprobationstatus,whichhewouldliketodothisthroughprobabilitystudy.Duringthepass6monthsinthecompany,Joehadsuccessfullycompleted3projectsbeforethegivendateline.Basedonthisinformation,whatisthechancethatJoesactualabilityisuptoKennethexpectation?,172,Fu

36、rtherexampleofposteriorprobabilities(cont),Event=EmployingthecorrectorincorrectpersonStrike=Successinclosingtheprojectbeforedateline,P(Correct|3Strikes)=P(Correct,3Strikes)/P(3Strikes),=(0.4606/0.4713),=0.9773,IfJoehascompleted3successiveprojectsbeforedateline,theposteriorprobabilitythatheisthecorre

37、ctcandidateforthejobis0.9773(or97.73%).,CorrectIncorrect,0.853=0.61410.353=0.0429,173,Furtherexampleofposteriorprobabilities(cont),IfJoedoesnotmeethisdatelinetargetforthe4thproject,doesitmeanthatKennethhasmakethewrongjudgement,和whatistheprobabilityassociated?,CorrectIncorrect,0.853x0.15=0.09210.353x

38、0.65=0.0279,P(Correct|SSSN)=P(Correct,SSSN)/P(SSSN),=(0.0691/0.0761),=0.9080,IfJoedoesnotmeetthedatelineforhis4thprojects,thenKennethwillbe90.8%surethatJoeisthecorrectcandidateforthejob.,174,175,IntroductiontoProbabilityDistribution,Whatisprobabilitydistribution?Itisatheoreticalfrequencydistribution

39、thatdescribehowoutcomeareexpectedtovary,和areusefulmodelsinmakinginferencesunderconditionsofuncertainty.,176,ExampleofstatisticalinferenceusingprobabilitydistributionMartinisoneoftheauthorizedmagazinedistributorsforthemonthlybusinessjournal“FinancialTimes”.Thedistributioncontractwiththemagazinepublis

40、herhasstatedclearlythatnobackissuesareallowedinthestoreinordertopreventreadersfromobtaining/misledbyback-datedmarketinformation,causingcreditability和liabilityimpacttothepublisher.Duetothepopularityofthemagazine,alldistributorareallowedtoplaceonlyoneordereverymonth,和thepublisherwillnotrefundanyexcess

41、orderplacedbythedistributorifthemagazinewasnotsoldoutforanyparticularmonth.EachmagazinecostUS$2,和thedistributorearnUS$3fromeachcopysold.Inordertomaximizehisprofitbyreducingloss,Martindecidetouseprobabilitystudytohelphimmakehisdecisiononthenumberofmagazinetoordereachmonth.Hecollectedhissaledataoverap

42、eriodof1yearasbelow:,177,Exampleofstatisticalinferenceusingprobabilitydistribution(cont)Thereare2typesofpossiblelosses;costofobsolescence(i.e.excessorderthathastobethrowaway),和costoflossopportunity(i.e.costoflosingsales).,Costofobsolescence=(MonthlyOrder-MonthlySale)xUS$2xP(M_Sale)wherenegativevalue

43、equaltonoloss,Costoflossopportunity=(MonthlySale-MonthlyOrder)xUS$3xP(M_Sale)wherenegativevalueequaltonoloss,TotalLosses=ObsolescenceCost+OpportunityCost,178,Exampleofstatisticalinferenceusingprobabilitydistribution(cont),Thereforeinordertomaximizehisprofit,Martinwillneedtoorder12000magazinepermonth

44、.,179,TypesofProbabilityDistributions,Whatisthedifferencebetweenfrequency和probabilitydistributions?Frequencydistributionisalistingofobservedfrequenciesforalltheoutcomesthatoccurredwhenanexperimentwasdone,whileprobabilitydistributionisalistingoftheprobabilitiesforallthepossibleoutcomesifthesameexperi

45、mentistobecarriedout.,DiscreteprobabilitydistributionsAdistributionofprobabilitieswhichallowedtotakeonalimitednumberofvaluesforitsoutcomes.(i.e.therollingofadice)BinomialdistributionPoissondistribution,ContinuousprobabilitydistributionsAdistributionofprobabilitieswhichallowedtotakeonaanynumberofvalu

46、eswithinagivenrangeofoutcomes.(i.e.therollingofadice)NormaldistributionStudentTdistribution,180,BinomialDistributions,Whatisthebinomialdistributions?Aprobabilitydistributionofdiscreterandomvariable,resultingfromanexperimentknownasBernoulli流程.,Conditionsfortheuseofbinomialdistributions(i.e.Bernoullit

47、rial)?Eachexperimentaltrialhasonlytwopossibleoutcomes:successorfailure,headortail.Probabilityoftheoutcomeofanytrialremainsfixedovertime(i.e.statisticalindependence).和theoutcomeofoneeventdoesnotaffecttheoutcomeofthesubsequentevent.,Characteristicofbinomialdistributionp=Probabilityofsuccessq=Probabili

48、tyoffailurer=Numberofsuccessdesiredn=Numberofundertakentrials,181,Probabilitymassfunctionforbinomialdistributionn!Probabilityofrsuccessesinntrials=-xprxqn-rr!(n-r)!,ExampleofusingbinomialdistributionPaulisamanufacturingmanagerofadiskdrivecompanywhohasfive流程engineersreportingtohim.Heisfacingasituatio

49、nwherehisengineersareoftenlateforwork,和after某些observation,hehasdeterminedthatthereisa0.4chancethatanyoneengineerwillbelate.Ifatanyonetime,the产品ionwillrequired3流程engineerstorunsmoothly,whatistheprobabilitythathis产品ionwillrunsmoothlyatpresentsituation?,Probabilityofsuccess(late),p=0.4Probabilityoffail

50、ure(early),q=0.6Numberoftrial(engineers),n=5Numberofsuccess(late),r=2orless,182,n!Probabilityofrlatearrivaloutofnstudent=-xprxqn-rr!(n-r)!,Exampleofusingbinomialdistribution(cont),Forr=0,5!P(0)=-(0.40)(0.65)0!(5-4)!=0.07776,Forr=1,5!P(1)=-(0.41)(0.65-1)1!(5-1)!=0.2592,Forr=2,5!P(2)=-(0.42)(0.65-2)2!

51、(5-2)!=0.3456,Forr=3,5!P(0)=-(0.43)(0.65-3)3!(5-3)!=0.2304,Forr=4,5!P(1)=-(0.44)(0.65-4)4!(5-4)!=0.0768,Forr=5,5!P(2)=-(0.45)(0.65-5)5!(5-5)!=0.01024,Totalprobability=0.0776+0.2592+0.3456+0.2304+0.0768+0.01024=1,183,Exampleofusingbinomialdistribution(cont),P(r650=10.99911=0.00089,Assuchthereisonly0.

52、089%chancethatarandomcandidatewilltakemorethan650hourstocompletethe培训.,198,Whatistheprobabilitythatacandidateselectedatrandomwilltakesbetween360hours和510hourstocompletethe培训?,Exampleofusingnormaldistribution(cont),Z1=(Xm)/=(360400)/80=-0.5,Z2=(Xm)/=(510400)/80=1.375,Assuch60.69%chancethatarandomcand

53、idatewilltakebetween360hours和510hourstocompletethe培训.,199,NormaldistributionasanbinomialdistributionapproximationNormaldistributioncansometimesbeusedasanapproximationforbinomialdistribution,givingtheconditions:a)np5b)nq5,ExampleWhatistheprobabilityofgetting5,6,7,or8tailsin10tossesoffaircoin?,p=0.5,q

54、=0.5,n=10,r=5,6,7,or8Frombinomialprobabilitydistributionstable,P(5,6,7,8)=P(5)+P(6)+P(7)+P(8)=0.2461+0.2051+0.1172+0.0439=0.6123,200,Toimprovetheaccuracyoftheapproximation,subtract和addacontinuitycorrectionfactorof0.5totherange5to8respectively.Theareaunderthisrangeiswhatweareinterestedin.,201,Hence,P

55、(4.5x8.5)=0.9864-0.3745=0.6119(approxequalto0.6123bybinomialdistribution),202,IntroductiontoSampling取樣的介紹,Whatispopulationinstatistic?Apopulationinstatisticreferstoallitemsthathavebeenchosenforstudy.,Whatisasampleinstatistic?Asampleinstatisticreferstoaportionchosenfromapopulation,bywhichthedataobtai

56、ncanbeusedtoinferontheactualperformanceofthepopulation,Population,Sample2,Sample6,Sample8,Sample1,Sample3,Sample7,Sample4,Sample5,203,Samplingdistribution-adistributionofsamplemeansIfyoutake10samplesoutofthesamepopulations,youwillmostlikelyendupwith10differentsamplemeans和sample标准偏差s.ASamplingdistrib

57、utiondescribestheprobabilityofallpossiblemeansofthesamplestakenfromthesamepopulation.,204,Whensamplesizeincreases,thestandarderror(orthestddeviationofsamplingdistribution)willgetsmaller.,Samplingdistribution(cont)取樣分配,205,CentralLimitTheorem中心線規則,206,Exampleofsamplingdistribution,Thepopulationdistri

58、butionofannualincomeofengineersisskewednegatively.Thisdistributionhasameanof$48,000,和a标准偏差of$5000.Ifwedrawasampleof100engineers,whatistheprobabilitythattheiraverageannualincomeis$48700和more.,207,Exampleofsamplingdistribution(cont),Therefore,mean=48000sigma=500X=50000Z=(48700-48000)/500=700/500=1.4,Fromthestandardizednormaldistributiontable,P(X$48700)=0.9192Therefore;P(X$48700)=1-0.9192=0.0808Thus,wehavedeterminedthatithasonly8.08%chancefortheaverageannualincome