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1、?34?9?Vol.34, No.92014?9?Systems Engineering Theory 2.?,?210094)?,?,?(SPC)?.?,?-?,?,?,?SPC?,?;?,?,?SPC?;?,?,?,?;?,?.?;?-?;?;?;?Integration SPC and condition-based maintenance for two-stage series repairable systemsZHONG Jian-lan1, MA Yi-zhong1, LIU Li-ping2(1. School of Economics and Management, Nan
2、jing University of Science and Technology, Nanjing 210094, China; 2. School of Business, Nanjing Normal University, Nanjing 210094, China)AbstractFor two-stage series repairable systems, considering random shift, a problem of integrating statistical process control (SPC) and condition-based maintena
3、nce is discussed. Firstly, the mean-residual joint 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 total probability decomposition, the probabil
4、ity of occurrence associated with each scenario can be calculated. Secondly, taking into account manpower, cost of loss production and production rate, a procedure for calculating average cost per time is presented according to renew theory. Then, a numerical example is given to illustrate the appli
5、cation of the proposed integrated model. The results show that the integrated model outperforms the maintenance model and the model using rules of thumb. Finally, a sensitivity analysis is conducted to develop insights into process parameter, time parameters and cost parametersthat influence the int
6、egration 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:
7、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?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
8、)?, e N(0,2).?21?,?-?(ZX Ze)?,?h?n?,?ZX, Ze:ZX=X X X/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) =i 22,iexp? i 42,i? (14)?i?12?E(i) =?0if(i)di(15)?(8)(13),
9、?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 i
