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CHINESE JOURNAL OF MECHANICAL ENGINEERING 594 Vol. 22,aNo. 4,a2009 DOI: 10.3901/CJME.2009.04.594, available online at ; Reliability Simulation and Design Optimization for Mechanical Maintenance LIU Deshun*, HUANG Liangpei, YUE Wenhui, and XU Xiaoyan Hunan Provincial Key Laboratory of Health Maintenance for Mechanical Equipment Hunan University of Science and Technology, Xiangtan 411201, China Received September 8, 2008; revised April 16, 2009; accepted April 30, 2009; published electronically May 5, 2009 Abstract: Reliability model of a mechanical product system will be newly reconstructed and maintenance cost will increase because failed parts can be replaced with new components during service, which should be accounted for in system design. In this paper, a reliability model and reliability-based design optimization methodology for maintenance are presented. First, based on the time-to-failure density function of the part of the system, the age distributions of all parts of the system during service are investigated, a reliability model of the mechanical system for maintenance is developed. Then, reliability simulations of the systems with Weibull probability density functions are performed, the system minimum reliability and steady reliability for maintenance are defined based on reliability simulation during the life cycle of the system. Thirdly, a maintenance cost model is developed based on replacement rates of the parts, a reliability-based design optimization model for maintenance is presented, in which total life cycle cost is considered as design objective and system reliability as design constrain. Finally, the reliability-based design optimization methodology for maintenance is used to design of a link ring for the chain conveyor, which shows that optimal design with the lowest maintenance cost can be obtained, and minimum reliability and steady reliability of the system can satisfy requirement of system reliability during service of the chain conveyor. Key words: maintenance, reliability, simulation, design optimization is improved1014. 1 Introduction Maintenance starts at design. Obviously, design methodology for maintenance, which is one of best effective maintenance means in the life-cycle of a product, During the life cycle of a mechanical product, attracts many researchers interests. However, research on maintenance, which is implemented on the judgment of design for maintenance is mainly centralized on two fields. practical states, preservation and reconstruction of some One is maintainability evaluation on product design certain states for the product, is very important to keep the alternatives, the other is some peculiar structures of parts product available and prolong its life. Studies on designed for convenient maintenance. For example, maintenance for mechanical products are roughly classified computer-aided maintainability evaluation tools for product into the following three catalogs. 11 design , product assembly and disassembly simulation (1) How to formulate maintenance policy or (and) how 12 programs for maintenance ,airplane design for to optimize maintenance periods considering system maintenance13, and so on. But studies on design reliability and maintenance cost, e.g., when system methodologies considering product reliability, maintenance reliability is subjected to some certain conditions, cost and maintenance policy are seldom reported. SHU and maintenance policy and optimal maintenance interval are determined to make maintenance cost lowest14 . FLOWER once pointed out that reckoning in labor cost and production interval cost, design decision of alternatives of (2) To develop maintenance methods and tools to ensure the part would be influenced. However, subsequent system maintenance to both low cost and short repair time, 15 59 research reports have not been presented . In this paper, such as special maintenance toolboxes developed . based on the time-to-failure density function of the part, (3) To design for maintenance(DFM), namely during distributions of service age of parts for a mechanical design procedure, system maintainability is evaluated and system that undergoes maintenance are investigated. Then the reliability model of the mechanical system is * Corresponding author. E-mail: liudeshun This project is supported by National Basic Research Program of China reconstructed and simulated. Finally, a novel design (973 Program, Grant No. 2003CB317001), Scientific Research Fund of optimization methodology for maintenance is developed Hunan Provincial Education Department of China (Grant No. 07A018), Hunan Provincial Natural Science Foundation of China (Grant No. and illustrated by means of design of a link ring for the 07JJ5074), and National Natural Science Foundation of China (Grant No. chain conveyor. 50875082) CHINESE JOURNAL OF MECHANICAL ENGINEERING 595 age to reenter the first box. 2 Reconstruction of Reliability Model of Initially, all parts are new and zero age in the first box. Mechanical System for Maintenance That is, at t0 0 , the portion in the first box is 2.1 Model assumptions p 0 (t 0 ) 1 . (1) After a mechanical system runs some time, due to replacement of fail parts, primary reliability model is At t , age fractions of the first box and the second 1 inapplicable to changed system, thus the reliability model box are represented as should be reconstructed. The mechanical system discussed in this paper has following characteristics. p (t ) p (t ) 1= f (x )dx , (1) System consists of a large number of same type parts, 1 1 0 0 0 in which the number of parts is constant during the whole (2) p (t ) p (t ) f (x )dx . life cycle of the system. 0 1 0 0 0 (2) The time-to-failure density distribution functions of all parts are the same, also, replacement parts have the Portions of both age boxes survive and advance to the same failure distribution functions as the original parts next age box, and portions of failed parts from both boxes (3) Failure of each part is a random independent event, replaced by new parts appear in the first box. i.e., failure of one part does not affect failure of other parts At t 2 , the proportions of the first three boxes are 2 in the system. calculated as follows: For example, a chain conveyor widely used in many industries consists of a large number of same round rings, 2 p (t ) p (t ) 1= f (x )dx , same link sheets and same scrape boards. Their respective 2 2 1 1 0 numbers are constant after the chain conveyor is put into p (t ) p (t ) 1= f (x )dx , the service. Also, each part, being subjected to similar work 1 2 0 1 0 (3) conditions and similar failure states, has the same or 2 p (t ) p (t ) f (x )dx =+p (t ) f (x )dx , identical density distribution of time to failure. Moreover, 0 2 1 1 0 0 1 0 replacement parts have failure time density function same or identical to the original parts during the service of the chain conveyor. So, at tn n , portions of parts in each box are calculated by using the following equations: 2.2 Reliability modeling for maintenance Reliability of a mechanical system depends on its parts, n p n (t n ) p n1(t n1) 1= f (x )dx , yet reliability and failure probability of which rest on their 0 service ages. Herein, according to the density distribution ( n1) p n1(t n ) p n2 (t n1) 1= f (x )dx , function of time to failure of the part, part service age 0 distribution of the mechanical system is calculated, then (n2) p n2 (t n ) p n3 (t n1 ) 1= f (x )dx , reliability model of the mechanical system for maintenance 0 is developed. During the service of a mechanical system, some parts that fail require to be replaced in time, hence 3 (4) age distribution of parts of the mechanical system p 3 (t n ) p 2 (t n1 ) 1= f (x )dx , 0 undergoing maintenance has been changed. Supposed that 2 after the mechanical system runs some time t n , where p (t n ) p (t n ) 1= f (x )dx , n 2 1 1 0 is time between maintenance activities, i.e., = maintenance interval, the unit of can be hours, days, p (t ) p (t ) 1 f (x)dx , 1 n 0 n1 0 months, or years. If p (t ) represents age proportion of i n n1 parts at tn with age i , thus age distribution of parts at p (t ) p (t ) (i+1) f (x )dx . time t denotes matrix p (t ),p (t ), p (t ), , 0 n i n1 0 n 0 n 1 n i n i 0 p n (t n ) .The failure density function of parts and current age distribution of parts in the system determine age Where p 0 (t n ) is the fraction of population of parts with distribution at next time, or the portion of the contents of age 0 at tn , representing parts that have just been put into each bin that survive to the next time step. An age service. It means that p 0 (t n ) is failure rate of parts, or distribution obtained at each time step for each part replacement rate of failed parts. In other word ，the fractions population determines failure rate for the following time of parts in the first box at t ,t , ,t are new parts that 0 1 n step. To find failure probability of parts the failure density replace these failed parts. function is integrated from zero to tn .The portion of the A series system consists of N parts that have the same population that survives advances to the next age box, and failure density distribution, each part is just a series unit, the portion that fail is replaced by new parts to become zero and each unit is relatively independent. In series system the 596 YLIU Deshun, et al: Reliability Simulation and Design Optimization for Mechanical MaintenanceY failure of any one unit results in system failure, in 3 plots the influence of the scale parameter of Weibull according to the principle of probability multiplication, the distribution on system reliability, and four curves represent reliability of series systems becomes four different type parts corresponding to a constant value of equal to 4 paired with value of 8,10,12,15 n i p i (t n )N respectively. Fig. 4 reveals how the shape parameter of R (tn ) 1= 0 f (x )dx . (5) Weibull distribution affects system reliability, and Weibull i 0 distribution parameters of five curves are 10, 1,2,3,4,5 . Correspondingly, their replacement rate Since the number of parts that comprise the system is curves of systems parts for these time-to-failure density constant, here, the system reliability of the mechanical distribution functions are plotted in Fig. 5. Additionally, in system for maintenance is defined as Fig. 3Fig. 5, maintenance interval is 1. N R (tn ) R (tn ) n i p i (t n )N N 1 0 f (x )dx i 0 n i p i (t n ) 1 0 f (x )dx . (6) i 0 From above to see, as long as the time-to-failure density function and maintenance interval are given, service age distributions of parts and system reliability could be obtained by simulation. Fig. 1. Weibull probability distributions 3 Replacement Rate and Reliability Simulation for Maintenance 3.1 Weibull distribution of time to failure The Weibull probability density function is widely used in failure modeling in mechanical parts and electronic components. Here the Weibull distribution with two parameters is used to simulate reliability of the system that is undergoing maintenance, that is, the time-to-failure density function of systems constituted parts is 1 f (x ) x exp x ,0 x . (7) = Fig. 2. System reliability R(t) with In Eq. (7), is the shape parameter, is the scale parameter. x is time, whose unite can be hours, days, or years. Five failure density functions with their Weibull parameters 10, 1,2,3,4,5are described in Fig. 1. It is shown that is large, before service age of parts arrives at the expected value, failure probability of parts is extremely low. Whereas, is small, many parts fails in short time of service. 3.2 Reliability simulation Different maintenance interval of the mechanical system and different time-to-failure density function of its parts are selected to simulate reliability of the system shown as Fig. Fig. 3. System reliability R(t) with 2Fig. 4. Fig. 2 shows how simulation time step (maintenance interval) affects system reliability, the plots Several characteristics of these figures are of interest. shown correspond to maintenance interval 0.5,1,2 , First, the reliability and replacement rate eventually reaches and with Weibull distribution parameters 4, 10 . Fig. steady state. This agrees with Drenicks Theorem, which CHINESE JOURNAL OF MECHANICAL ENGINEERING 597 states the superposition of an infinite number of independent dependence on is not surprising, higher values of for a given set of yield higher values for expect time to failure and thus lower replacement rate and higher reliability. More interestingly, with the increase of the value of , the steady values of replacement rate decrease and the steady values of reliability increase. Fourth, the degree of oscillation of system reliability depends on the parameters of Weibull distribution. Although the influence of on oscillations can be neglected, the influence of on oscillations should be paid special attention to. Bigger value of denotes that failure rate of parts is lower before service time of parts reaches expected life time, and the majority of parts Fig. 4. System reliability R(t) with prolong use time, thus, the steady value of system reliability becomes higher. However, in this case, the majority of parts fail at quite centralized time, so minimum value of system reliability is lower. It is suggested that , denoting concentrative degree of failure time distribution, is a sensitive parameter. The influence of on steady value of reliability is different from and contrary to that of on minimum value of reliability. Therefore, selection of appropriate should be paid special attention to in design, because both steady value and minimum reliability coincidentally meet design requirements. 3.3 Definitions Simulation results show that system reliability varies Fig. 5. Part replacement rate p 0(t) during service. The reliability of a system experiences several oscillations, sometimes is maximum value and then equilibrium renewal process is homogeneous Poisson minimum value, finally reaches steady value. Oscillations process. During the initial stage of system service, parts of of system reliability periodically decay, and the period is the system are “new”, then, become “old”. The portion of about the expected life time of parts (for Weibull parts that fail gradually increases, thus the part replacement rate increases and system reliability will drop distribution, the parameter approximates expected life monotonically. With the replacement of a significant at big ). For design and maintenance of mechanical portion of the population, portion of parts that fail will systems, minimum value and steady value of system decrease, thus the part replacement rate will drop and the reliability are of importance. Minimum reliability of the system reliability will rise until this oscillation is over and system appears at beginning stage, but steady reliability next oscillation begins. After some oscillations, the value of the system appears after running a long time. Here, population becomes more age-diversified with each to conveniently discuss later, minimum reliability and oscillation, and the age distribution approaches steady. At steady reliability of the system for maintenance are defined that time, the oscillations in replacement rate and system based on simulation results of system reliability shown as reliability diminish. Compared Fig. 4 with Fig. 5, it is in Fig. 6. shown that the trend of replacement rate is contrary to the change of system reliability. When system reliability increases, part replacement rate reduces. Otherwise, as system reliability reduces, part replacement rate increases. Secondly, the steady state value and the degree of oscillation of the system reliability depend on maintenance interval. As Fig. 2 shows, the reliability rises as maintenance interval decreases since parts that fail are being replaced more quickly. The shorter the maintenance interval is, the higher reliability is, and the smaller oscillations are. However, frequent repairs will result in higher maintenance cost. Thirdly, the steady state value of the system reliability depends on the parameters of Weibull distribution. The Fig. 6. System reliability parameters definition 598 YLIU Deshun, et al: Reliability Simulation and Design Optimization for Mechanical MaintenanceY As it appears at initial phase, minimum reliability of the including production cost and maintenance cost, is system can be found in discrete reliability values of represent as simulation results from t 0 to t 2 . At this time, m C c =+ c p (t ) +c . (10) minimum reliability Rm is defined as 0 1 0 i 2 i 1 R min R(t ) ,i 0,1, , n. (8) m ( i ) In Eq. (10), C is total cost of life cycle of the system for per part in the system. c , c ,c denote coefficient of 0 1 2 Supposed that some simulation time is T , and 0 part production cost, coefficient of replacement cost and Rmax ,Rmin represent maximum value and minimum value coefficient of preparation cost respectively, and these of t T ,T +2 respectively. Once when ratio of coefficients can be confirmed by statistical analysis of field 0 0 maximum reliability value and minimum reliability value datum. m M / , where M represents life of the system. Rmin / Rmax is satisfied, system reliability is regarded The first term of right-hand side of Eq. (10) represents as arriving at steady value at time T . Thus system production cost of the system, the second term of 0 reliability, or called as steady reliability, is defined as right-hand side of Eq. (10) represents maintenance cost of the system. In Eq. (9), c1 c0 , because part replacement Rs (Rmax =+Rmin ) / 2 , (9) cost includes not only production cost of the part that replaces the failed part, but also costs that are spent for 1 is the stabilization requirement, which could resources, and indirect cost caused by replacement. usually be 98%. If T does not exist, system reliability Obviously, the cost that Eq. (10) denotes is not absolute 0 will be unsteady. cost, but relative cost. Eq. (10) is also represented as 4 Reliability-based Design and Optimization m C c =+mc +c p (t ) . (11) Modeling for Maintenance 0 2 1 0 i i 1 A reliability-based design optimization model for 4.2 Model of reliability-based design and optimization maintenance is presented to make a trade-off between the Supposed that a type part of the system has n design system reliability and life-cycle cost of parts that includes alternatives, X (x , x , , x ) , their failure density maintenance cost, in which the above models are helpful to 1 2 n calculate part replacement rate of the system, minimum functions are expressed as F (f 1(t ), f 2 (t ), , f n (t ) corresponding to each alternative. reliability and system reliability. In the model, the cost of 0 For a fixed maintenance interval , its reliability-based life cycle is considered as a design objective, and the design optimization model for maintenance is represented reliability of the system is considered as design constraint. as follows: The task is to find a design having the minimum cost and satisfying the constraints. min C ( x), x X , 0 R R (12) 4.1 Model of life cycle cost s.t. , m m Life cycle costs of mechanical systems include 0 . R R s s production costs and maintenance costs. System maintenance costs are from items listed as follows: (1) cost Apparently, the minimum life cycle cost and reliability of parts replacement, (2) operation cost including cost of obtained from the above model is responding to the fixed resources spent (i.e. labor, equipment) for replacing parts, period. For any one of n design alternatives, its cost and (3) indirect cost resulting from production interrupt caused reliability depend on the maintenance interval . The by replacing parts, and (4) preparation work cost for achievable minimum cost could be obtained from the replacing parts16.The foregoing three items are concerned optimization of the maintenance interval. For the optimal with the number of replacing parts every time of maintenance interval, namely, maintenance interval is maintenance. The more parts replaced will consume more optimized to minimized the life cycle cost, thus resource, occupy more production time, thus bring reliability-based design and optimization model for tremendous loss and increase maintenance cost. The last maintenance is expressed as follows: item is not concerned with the number of replacing parts min C( x,), x X , but times of maintenance or replacement. As a result, 0 R R maintenance costs of mechanical systems are classified as s.t. , (13) m m cost considering part replacement number and cost 0 . R R s s considering maintenance times. In this way, for a mechanical system with a constant number of parts N , In Eq. (12) and Eq. (13), C is obtained from Eq. (10) or after it runs for time M , its life cycle cost model, Eq. (11). Rm ,Rs denote minimum reliability and steady CHINESE JOURNAL OF MECHANICAL ENGINEERING 599 reliability of the system respectively. R 0 ,R 0 are allowable m s reliability values of the system. In general, R 0 (0.75 m 0.95)R 0 , which implies system reliability allows to vary in s some certain degree during whole life cycle, but variation scope is not over 5%25% of steady reliability. 4.3 Design optimization based on system reliability simulation Obviously, system steady reliability, minimum reliability and part replacement rate in design models can be derived from reliability simulation. Therefore, design optimization for maintenance is a design methodology based on simulation. In design models, input conditions of reliability simulation are the time to failure density distribution functions of the system part F , system service life M and coefficients of life cycle cost are c , c ,c . For 0 1 2 maintenance of fixed interval, input conditions add in fixed maintenance interval 0 . Times of maintenance are clearly equal to 0 M / during whole life cycle. As to the situation that maintenance interval needs to be optimized, times of maintenance are rounded M / to obtain at different maintenance interval. In addition, Design alternatives for the system must satisfy requirement of system reliability, Fig. 7. Chart of design optimization for maintenance thus R 0 ,R 0 are given. Finally, an optimal design m s alternative and its minimum reliability, steady reliability Table 1. Design alternatives and their parameters and life cycle cost are outputted. Scheme x c /CNY c /CNY c /CNY The flow chart of design optimization for maintenance is 0 1 2 shown as Fig. 7, in which two models of design 1 5 10 3 7 0.5 optimization for maintenance are integrated. Most possibly, 2 4 12 5 10 0.5 the solutions from the above model and model is 3 4 20 10 18 0.5 usually different. Suppose that the requirement of minimum reliability and 5 Design Demonstration steady reliability is R 0 0.85, R 0 0.75 . Considering that s m system maintenance interval is selected from a series of There are three design alternatives for link rings of chain equivalent difference values, discrete optimization method conveyors, the service life M of which is equal to 100 is adopted. Simulation results of two design models for months. The density distribution function of the time to maintenance are listed in Table 2. Fig. 8 to Fig. 11 failure of rings is the Weibull distribution, and their illustrates that system reliability and total life cycle costs distribution parameters and cost coefficients of life cycle vary with service time of the system. are listed in Table 1 as below. Table 2. Simulation results of design alternatives Identical cycle Optimization cycle Scheme x 0 0 /month Rm Rs C/CNY /month Rm Rs C /CNY 1 1.0 0.726 6 0.853 1 131.138 4 0.8 0.773 4 0.880 7 146.887 9 2 1.0 0.791 7 0.867 7 157.628 8 1.0 0.791 7 0.867 7 157.628 8 3 1.0 0.865 4 0.925 2 180.037 0 1.8 0.782 0 0.858 8 141.505 8 0 Notes: is interval of design model for fixed cycle understood that there could not be a design alternative that * maintenance Eq. (12), and is optimum interval of design would meet system reliability constrains for an model for optimizing cycle maintenance Eq. (13). inappropriate fixed maintenance interval. When system Seeing from simulation results listed in Table 2, when maintenance interval is optimized, optimum design system maintenance interval is fixed 0 1 , optimum alternative derived from Eq. (13) is alternative x3 . In the design alternative derived from Eq. (12) is alternative x2 . case, all design alternatives meet the requirements of Alternative x1 does not satisfy system reliability system reliability, and total life costs of alternative x3 is constrains, and total life costs of alternative x2 is lower the lowest, correspondingly system maintenance interval * than that of alternative x3 . From the example, it could be is 1.8 in the alternative x3 . It is shown that variable 600 YLIU Deshun, et al: Reliability Simulation and Design Optimization for Mechanical MaintenanceY maintenance cycle police leads to different choice of design highest total life cycle costs. alternatives, and total life costs can be reduce by optimizing maintenance interval. Several interesting results could be found from Fig. 8Fig. 11. Fig. 11. Life cycle costs simulation of design alternatives for an optimum maintenance interval (2) When maintenance interval is optimized, selection of optimum interval is based on the premise of satisfying Fig. 8. Reliability simulation of design alternatives for a fixed maintenance interval requirements of system reliability. As to alternative x1 , in order to meet requirements of system reliability, * maintenance interval decreases, 0.8 , but its total life cost increases somewhat. For alternative x2 , maintenance interval keeps constant after optimization, also, which means that interval 1 is an optimum interval for this alternative. For alternative x3 , due to optimization, maintenance interval increases, 1.8 , and the difference between system reliability and design requirements reduces, thus it has lower total life cycle costs. Besides, three design alternatives are optimized, their curves of system reliability and total life cycle costs trend to centralization and consistence, and difference of costs among three alternatives reduces. Fig. 9. Life cycle costs simulation of design alternatives (3) When the system requires high reliability, correspo- for a fixed maintenance interval ndingly, maintenance interval will reduce and maintenance costs will rise. On the contrary, when system requires low reliability, correspondingly, maintenance interval will delay, so maintenance costs will reduce, as the decrease of system maintenance costs is subject to system reliability requirement. The steady value and minimum value of system reliability monotonously reduce with the increase of maintenance interval, also, total life cycle costs reduce with the increase of maintenance interval. As a result, minimum interval that steady value and minimum value of system reliability satisfy design requirements will obtain minimum total life cycle costs for the design alternative. It must be pointed out that system reliability of the design alternative Fig. 10. Reliability simulation of design alternatives is not equal to but little more than the requirement values for an optimum maintenance interval due to adoption of discrete optimization. 0 (4) When design alternatives of the system are decided, (1) When a fixed interval ( 1) is determined, system reliability of alternative x2 not only satisfies all design the optimum choice of design alternatives depends on not requirements but also approaches to requirement value. only maintenance interval but also requirement of system Reliability of alternative x1 satisfies the requirement of reliability and system service life. For example, when interval is fixed ( 0 steady reliability, but does not satisfy the requirement of 1), and system reliability required minimum reliability in spite of its lowest cost. Though reduces from R 0 0.75 to R 0 0.70 , the optimum m m alternative x3 satisfies the requirement of system reliability, design alternative derived from Eq. (12) is alternative x1 either steady reliability or minimum reliability, it has instead of alternative x2 . When system service life switches CHINESE JOURNAL OF MECHANICAL ENGINEERING 601 from M 100 to 50, the optimum design alternative Engineering, 1999, 6(2): 155171. obtained from Eq. (13) is alternative x replacing alternative 10 GAN Maozhi. Maintainability design and validation M. Beijing: 1 x3 shown as Fig. 11. It means that, since parts made by Defence Industry Publishing House, 1995. (in Chinese) high quality materials have long service life, the design 11 WELP E G, LINDEMANN U, LU Xinsheng. Support instructions alternative obtains lower total life cycle costs in spite of for a recyclable and maintainable designJ. Engineering Design, their higher production costs. 2002, 9(2): 7788. (in Chinese) 12 ZHOU Hong, GAN Maozhi, LIU Anqing, et al. Maintainability 6 Conclusions design of product based on concurrent engineeringJ. Journal of Machine Design , 2003, 20(9): 35. (in Chinese) Maintenance is one of critical jobs during life cycle of 13 CHEN Wen. Design principles of aircraft for maintenanceJ. the product. Replacement of parts will cause the change of Aviation Engineerging & Mainienance , 2001, 204(6): 25. (in system reliability and life cycle costs. Based on the Chinese) time-to-failure density function of parts, steady reliability, 14 SHI Quan. Report of cases about maintainability designM. Beijing: minimum reliability and life cycle costs can be obtained by Defence Industry Publishing House, 2001. (in Chinese) means of reconstruction of reliability model and simulation 15 SHU L H, FLOWR W C. Reliability modeling in design for of system reliability. This paper develops reliability-based remanufactureJ. ASME Journal of Mechanical Design , 1998, design optimization methodology for maintenance, in 120(12): 620627. which total life cycle costs are regarded as the design object 16 SUN Youzhao, ZUO Hongfu, WANG Wei. A quantitative prediction and system reliability as design constrains. It provides a method for controlling the maintenance cost of civil aviation aircraft new approach to make a trade-off between the reliability enginesJ. Journal of Applied Sciences , 2003, 21(4): 401405. (in and total life cycle costs of the mechanical system in design Chinese) optimization for maintenance. Biographical notes Re