两阶段串联可修系统的统计过程控制与视情维修整合研究.pdf

?34?9?Vol.34,No.92014?9?Systems Engineering Theory&PracticeSept.,2014?:1000-6788(2014)09-2339-11?:O213.1?:A?1,?1,?2(1.?,?210094;2.?,?210094)?,?,?(SPC)?.?,?-?,?,?,?SPC?,?;?,?,?SPC?;?,?,?,?;?,?.?;?-?;?;?;?Integration SPC and condition-based maintenance for two-stageseries repairable systemsZHONG Jian-lan1,MA Yi-zhong1,LIU Li-ping2(1.School of Economics and Management,Nanjing University of Science and Technology,Nanjing 210094,China;2.School ofBusiness,Nanjing Normal University,Nanjing 210094,China)AbstractFor two-stage series repairable systems,considering random shift,a problem of integratingstatistical process control(SPC)and condition-based maintenance is discussed.Firstly,the mean-residualjoint control chart is used to monitor the equipment and provides signals indicating equipment deterioration,and appropriate maintenance is performed.This leads to different scenarios.According to the law of totalprobability decomposition,the probability of occurrence associated with each scenario can be calculated.Secondly,taking into account manpower,cost of loss production and production rate,a procedure forcalculating average cost per time is presented according to renew theory.Then,a numerical example isgiven to illustrate the application of the proposed integrated model.The results show that the integratedmodel outperforms the maintenance model and the model using rules of thumb.Finally,a sensitivityanalysis is conducted to develop insights into process parameter,time parameters and cost parametersthat influence the integration efforts.Keywords two-stage series repairable systems;mean-residual joint control chart;condition-based main-tenance;manpower cost;random shift1?1.?(SPC)?,?,?;?SPC?,?.?,?.?,?,?.SPC?2,?,?SPC?,?:2013-01-10?:?(71471088,71211140350)?:?(1984),?,?,?,?,?:?;?(1964),?,?,?,?,?,?,?:?,?,E-mail:yzma-.2340?34?.Linderman?1?SPC?,?,?,?,?.?,?Linderman?35.?,?,?,?.?,?,?,?,?.Panagiotidou?6?(?)?,?.Yeung?7?X?.?,Wu?8?,Yin?9?.?,Wang10?.?,?.?11?,?,?,?.Liu?12?,?X?.?,?,?,?SPC?.?,?.?.?13?X?,?VPX?,?B&L?,?,?.?,?.?14.?,?SPC?,?.Wu?15?,?,?,?.Pandey?16?,?.?,?,?SPC?,?-?.?.2 SPC?,?,?X?,?Y?,?X?Y?.X?Y?17:Y=b0+b1x+(1)?,b0?,b1?,?,?E()=0,V ar()=2 0,?.?:?.?,?X N(X,20,1),?Y N(Y,20,2).?,?.?,?.?,?.?.?,?Ti?i?,?fi(t)=iexp(it),t 0,i=1,2(2)?Fi(t)=1 exp(it),t 0,i=1,2(3)?f(t1,t2)=f1(t1)f2(t2)(4)?F(t1,t2)=2?i=1(1 exp(iti)(5)?9?,?:?2341?1?,?X?X?X+1(1?=0),?Y?Y?Y+b11;?2?,?Y?Y?Y+2(2?=0);?,?X?X?X+1,?Y?Y?Y+b11+2.?,?,?.?,?“?”,?.2.1?-?SPC?18.?.?1?X?;?2?(?e)?,?e=Y?Y(6)?,e N(0,2).?21?,?-?(ZX Ze)?,?h?n?,?ZX,Ze:ZX=X XX/n,Z e=e/n(7)?ZXN(0,1),ZeN(0,1),?ZX?Ze?k1?k2.?=1+2 12(8)?i?i(i=1,2)?i=2(1 (ki)(9)?,?1?2?10=1(1 2)(10)?1?2?01=(1 1)2(11)?11=12(12)?i?i=(ki in)(ki in)(13)?i?i?Rayleigh?15,?f(i)=i22,iexp?i42,i?(14)?i?12?E(i)=?0if(i)di(15)?(8)(13),?i?p00i=(1 )i1(16)?1?2?i?p10i=i110(1 10)(17)?1?2?i?p01i=i101(1 01)(18)?i?p11i=i111(1 11)(19)2.2?.?k?,?,?(k+1)?(?TPM=(k+1)h)?;?,?;?,?(?1).2342?34?4?S4?3?S3?2?S2?1?S1?1?1 SPC?1?,?,?:?1?,?TPM?.?k?,?.?1?:P(S1)=k?i=1P(?)=k?i=1P(?|?)P(?)=1 F1(kh)1 F2(kh)k?i=11 F1(ih)1 F2(ih)p00i(20)?2?,?,?2?:P(S2)=k?i=11 F1(ih)1 F2(ih)p00i(21)?3?4?,?,?i(i=1,2)?,?.?1?,?2?ZX?(P(S31)?P(S31)=F1(kh)1 F2(kh)k?i=1F1(ih)F1(i 1)h)1 F2(ih)?1 i1?j=1p0i?ki+1?l=1p10l(22)?,?1?,?2?Ze?(P(S32)?P(S32)=1 F1(kh)F2(kh)k?i=11 F1(ih)F2(ih)F2(i 1)h)?1 i1?j=1p0i?ki+1?l=1p01l(23)?ZX Ze?(P(S33)?P(S33)=F1(kh)F2(kh)k?i=1F1(ih)F1(i 1)h)F2(ih)F2(i 1)h)?1 i1?j=1p0i?ki+1?l=1p11l(24)?,?3?P(S3)=3?j=1P(S3j)(25)?,?4?P(S4)=3?j=1P(S4j)(26)?9?,?:?2343?P(S41)=k?i=1F1(ih)F1(i 1)h)1 F2(ih)?1 i1?j=1p0i?ki+1?l=1p10lP(S42)=k?i=11 F1(ih)F2(ih)F2(i 1)h)?1 i1?j=1p0i?ki+1?l=1p01lP(S43)=k?i=1F1(ih)F1(i 1)h)F2(ih)F2(i 1)h)?1 i1?j=1p0i?ki+1?l=1p11l(27)3?,?,SPC?.?,?(?)?,?.?,?,?.?19,?(E(C)?(E(T)?ECT=E(C)E(T)(28)3.1?1?,?TPM?,?,?1?(TP):E(T|S1)=(k+1)h+TP(29)?2?,?,?2?(TF)?(TCM):E(T|S2)=hk?i=1ip00i1 F1(ih)1 F2(ih)+TF+TCM(30)?3?,?,?,?TPM?,?3?TPM?i(i=1,2)?(TPreM(j)?:E(T|S3j)=(k+1)h+TPreM(j),j=1,2,3(31)?,TPreM(3)=?2i=1TPreM(i)?.?4?TPM?.?i(i=1,2)?1,?fi(t|(k+1)h)=fi(t)Fi(k+1)h)=ieitFi(k+1)h),0 t (k+1)h,i=1,2(32)?f(t1,t2|(k+1)h)=f(t1,t2)F(k+1)h,(k+1)h),0 t1,t2(k+1)h(33)?,?i(i=1,2)?E(?|S4i)=?kh0tfi(t|(k+1)h)dt,i=1,2(34)?i(i=1,2)?E(?|S4i)=h ARLi i+TA+TPreM(i),i=1,2(35)?,ARLi?i?,i?ith?(i+1)th?i?.?Tin=1?(1+2)(kh+1/(1+2)e(1+2)kh1 e(1+2)(k+1)h(36)2344?34?E(?|S43)=h ARL3(1)+TA+TPreM(3)(37)?,(1)?ith?(i+1)th?(?1);ARL3?.?,?4?E(T|S4)=3?j=1E(T|S4j)(38)?,E(T|S41)=?kh0tf1(t|(k+1)h)dt+h ARL10 10+TA+TPreM(1)E(T|S42)=?kh0tf2(t|(k+1)h)dt+h ARL01 01+TA+TPreM(2)E(T|S43)=Tin+h ARL3(1)+TA+TPreM(3)(39)?,?(20)(27)?(29)(39),?E(T)=4?i=1E(T|Si)P(Si)(40)3.2?,?,?,?.?1(C11?,?:(2)(1)?,?1?2).?:1)?.?(C0)?;?,?(C1,C2,C3)?.Ci(i=0,1,2,3)?20?(?3.3?).2)?:?SPC?(Cman).?(a)?(b),?:Cman?(Cm,?)?SPC?(MSPC)?,?Cman=CmMSPC(41)?E(?|Si)=(a+bn)E(?|Si)+Cman,i=1,2,4(42)3)?CF,CA,?.4)?7,?,?(Cm)?(Clp)?(PR)?.?,?CPM=TP(PR Clp+Cm MM)+CFPM(43)?CCM=TC(PR Clp+Cm MM)+CFCM(44)?CPreM(j)=TPreM(j)(PR Clp+Cm MM)+CFPreM(j),j=1,2CPreM(3)=2?j=1CPreM(j)(45)MSPC(MM)?SPC(?)?(?)?,?(?)?SPC(?)?,?MSPC(MM)?;?(?)?SPC(?)?,?MSPC(MM)?1.?SPC?,?SPC?.?.?9?,?:?2345?1?1E(C|S1)C0(k+1)h+k(a+bn)+Cman+CPM2E(C|S2)C0h?ki=1ip00i1F1(ih)1F2(ih)+(a+bn)?ki=1ip00i1F1(ih)1F2(ih)+Cman+CF+CCME(C|S31)C0?kh0tf1(t|(k+1)h)dt+C1(k+1)h?kh0tf1(t|(k+1)h)dt+k(a+bn)+Cman+CPreM(1)3E(C|S32)C0?kh0tf2(t|(k+1)h)dt+C2(k+1)h?kh0tf2(t|(k+1)h)dt+k(a+bn)+Cman+CPreM(2)E(C|S33)C0Tin+C(11)(2)(1)+C3(k+1)h (2)(1)Tin+k(a+bn)+Cman+CPreM(3)E(C|S41)C0?kh0tf1(t|(k+1)h)dt+C1(h ARL1 1)+E(T|S41)h(a+bn)+Cman+CA+CPreM(1)4E(C|S42)C0?kh0tf2(t|(k+1)h)dt+C2(h ARL2 2)+E(T|S42)h(a+bn)+Cman+CA+CPreM(2)E(C|S43)C0Tin+C(11)(2)(1)+C3(h ARL3(2)+E(T|S43)h(a+bn)+Cman+CA+CPreM(3)?,?E(C)=4?i=1E(C|Si)P(Si)(46)3.3?,?.?,?,?.?21?C0=PR A0+A1220,2(47)A0?,A1?.?i(i=1,2)?,?Ci=PR?0Li(i)f(i)di,i=1,2(48)Ci(i=1,2)?i?i(i=1,2)?.?(C3)?C3=2?i=1Ci(49)?,Li(i)(i=1,2)?i?i?,?15,?Li(i)=2?d=1li,dli,d=?Ad(20,d+2d),d?=iAi(20,i+2i),d=i(50)?,li,d?i?i?d(d i)?.?,?i?i,?d(d i)?i?,?d(d i)?.4?21?.?1?(X),?2?(Y).?40?.?(xi,yi).?,?X?N(77.05,4.82),Y?N(95.755,82),?Y=11+1.1X+e=Y?Y(51)?,?e N(0,36.1216).?1=0.002?2=0.001?.?,?X?91.45,Y|X?29.03+1.1X.?,?,?,?,?2346?34?:TA=0.04,CS=1000;TCM=1,CFCM=1000;TPreM(1)=TPreM(2)=1,CPreM(1)=CPreM(2)=1000;?,?:TF=0.1,CF=250;?k?,?,?,?TP=0.8,CFPM=2000.?400,?20,?1.?,?A0=1,A1=1.2.?,?SPC?,?MATLAB?,?,?,?ECT?.?,?:MinimizeECT(n,h,k1,k2,k)Subject toARL0=370ARLuv 10E(1),E(2)40k1,k2,MSPC,MM 0n,k N+(52)?,?n=5,h=2.23,k1=3.48,k2=3.06,k=18,MSPC=0.1,MM=4,?ECT=5636?.?,?,?,?TPM?.?,?SPC?;?,?TPM,?,?,?1?1?3.?,?,?2.ECTmt=E(Cmt)E(Tmt)(53)?,E(Tmt)=E(Tmt|S1)P(S1)+3?j=1E(Tmt|S3j)P(S3j),E(Cmt)=E(Cmt|S1)P(S1)+3?j=1E(Cmt|S3j)P(S3j).?2?1E(Cmt|S1)(k+1)h+TPC0(k+1)h+CPME(Cmt|S31)(k+1)h+TPreM(1)C0?kh0tf1(t|(k+1)h)dt+C1(k+1)h?kh0tf1(t|(k+1)h)dt+CPreM(1)3E(Cmt|S32)(k+1)h+TPreM(2)C0?kh0tf2(t|(k+1)h)dt+C2(k+1)h?kh0tf2(t|(k+1)h)dt+CPreM(2)E(Cmt|S33)(k+1)h+TPreM(3)C0Tin+C(11)(2)(1)+C3(k+1)h (2)(1)Tin+CPreM(3)?,?6146?.?:SPC?,?.?,?,?,?,?;?,?,?.?,?SPC?,?,?SPC?.?(n=5,h=1,k1=3,k2=3)?.?,?k,MSPC,MM?,?,?k=41,MSPC=6,MM=4,?ECT=5928?.?:?,?.?9?,?:?23475?,?(1,2,1,2,PR),?(TF,TA,TP,TPreM,TCM)?(CF,CA,CP,CFPreM,CCM)?(n,h,k1,k2)?(k)?.?,?,?,?,?21?.?3?.?3?ABCDEFGHJK?1212PRTFTATPTCMTPreM?10.0020.0010.50.51011444?20.020.01225044888?LMNOPQRSTUV?CFCACFPMCFCMCFPreMabA0A1ClpCm?120010002400100050001011110010?2500300048003000100004052250040?,?PB?32?,?,?MATLAB?,?.?2?(Clp)?(Cm)?(PR)?(ETC)?.?4,?“+”?,?:“?”?,?.?,?n?1?2,?b?Cm?;?PR?A1?.(响应为）标准化效应的正态图标准化效应百分比mlp?2?(ETC)?4?nhk1k2kETC1?+?2?+?1?+?1?+?PR+?+TFTA+TPTCMTPreMCFCA+CFPM+CFCMCFPreMab?A0?A1+?Clp+Cm?+6?SPC?.ZX Ze?,?,?.?,?;?,?,?,?ZX Ze?.?,?,?:?,?;?,?,?.2348?34?1 Linderman K,McKone-Sweet K E,Anderson J C.An integrated systems approach to process control andmaintenanceJ.European Journal of Operation Research,2005,164(2):324340.2 Pandey D,Kulkarni M S,Vrat P.Joint consideration of production scheduling,maintenance and quality policies:A review and 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(1 F1(t)(1 F2(t)=1 e(1+2)t(A.3)?f(1)(t)=(1+2)e(1+2)t(A.4)?(1)=E(T(1)|T(2)(k+1)h)=k?i=0?(i+1)hih?(i+1)ht(1)(t(1)ih)g(t(1),t(2)dt(2)dt(1)=k?i=0e(1+2)(i+1)hh+1/1+1?2 1?(1+2)e(1+2)ihe1h?2+e2h?11/(1+2)(1 e1(k+1)h)(1 e2(k+1)h),(2)=E(T(2)|T(2)(k+1)h)=k?i=0?(i+1)hih?t(2)ih(t(2)ih)g(t(1),t(2)dt(1)dt(2)=k?i=0e(1+2)(i+1)h(h+11+2)e(1+2)ihe1h(h+11)+e2h(h+12)1112+11+2(1 e1(k+1)h)(1 e2(k+1)h)(A.5)?g(t(1),t(2)?T(1),T(2)?22.?,?Tin=?kh0tf(1)(t)dtF(1)(k+1)h)=1?(1+2)(kh+1/(1+2)e(1+2)kh1 e(1+2)(k+1)h(A.6)?2?C(11):C(11)=C1p(TAC1 TAC2|0 TAC1,TAC2 h)+C2p(TAC2 TAC1|0 TAC1,TAC2 h)=C1?h0?t20f1(t1)?h0f1(t1)dt1f2(t2)?h0f2(t2)dt2dt1dt2+C2?h0?t10f1(t1)?h0f1(t1)dt1f2(t2)?h0f2(t2)dt2dt2dt1=C1(1 e2h)+C2(1 e1h)C12+C211+21 e(1+2)h(1 e1h)(1 e2h)(A.7)