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

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1、?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-

2、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 integratingst

3、atistical 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

4、 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 numeri

5、cal 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

6、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),?,?,?,?,?:?;?(196

7、4),?,?,?,?,?,?,?:?,?,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?

8、,?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,?

9、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?i4

10、2,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(i

11、h)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)=

12、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)

13、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

14、)?.?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)k

15、h1 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

16、)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(4

17、4)?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?kh0t

18、f1(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?kh0

19、tf2(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,

20、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;T

21、CM=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=

22、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+TP

23、reM(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,TPre

24、M,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,?“

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38、arts using quadratic loss functionJ.InternationalJournal of Quality and Reliability Management,1997,14(9):948966.?1?2122,?(1),(2)?Tin:TA1:?1?;TA2:?2?;?9?,?:?2349Ti:TAi(i=1,2)?;?i?,?fi(t|(k+1)h)=fi(t)Fi(k+1)h)=ieitFi(k+1)h),0 t (k+1)h,i=1,2(A.1)?1=k?i=0?(i+1)hih(t1 ih)f(t1|(k+1)h)dt1=?ki=0?(i+1)hih(t

39、1 ih)f1(t1)dt1F1(k+1)h)=?ki=0(1?1)1 (1h+1)e1he1ih1 e1(k+1)h,2=k?i=0?(i+1)hih(t2 ih)f(t2|(k+1)h)dt2=?ki=0?(i+1)hih(t2 ih)f2(t2)dt2F2(k+1)h)=?ki=0(1?2)1 (2h+1)e2he2ih1 e2(k+1)h(A.2)?T(1)=min(T1,T2),T(2)=max(T1,T2),T(1)?F(1)(t)=1 (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)

40、(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

41、(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)