Double system parts optimization: static and dynamic model

Transcription

1 Double sysem pars opmizaon: sac and dynamic model 1 Inroducon Jan Pelikán 1, Jiří Henzler 2 Absrac. A proposed opmizaon model deals wih he problem of reserves for he funconal componens-pars of mechanism in order o increase is reliabiliy. he following facors are aken ino consideraon: he probabiliy of he failure-free run of a par wihou a reserve, he probabiliy of he failure-free run of a par wih a reserve, he mean value of losses caused by he par s malfuncon wihou a reserve, he mean value of losses caused by he par s malfuncon wih a reserve, coss of he purchase and mainenance of he reserve for he given pars. he values of hese pars failure probabilies are supposed o be known in advance, he losses caused by his failure are esmaed. Sascal independence of he failures of hose pars is supposed. In he model, he coss of he pars doubling are supposed o be limied o a fixed value. As a resul of he problem soluon, he pars of he model are sored ino wo groups: he pars which are o be doubled and he pars which are no. he sac model as well as he dynamic one, in which he failures are considered as Poisson evens are described, including numerical examples. Keywords: doubling of sysem pars, opmizaon model, probabiliy of failure, mean value of losses, sysem reliabiliy, sac and dynamic model. JEL classificaon: C44 AMS classificaon: 90C15 Complex mechanisms are composed of a grea number of componens. Each of he pars is responsible for he righ funconing of he whole sysem and vice versa, each par s failure can disurb he sysem or compleely pu i ou of operaon and cause damages in is effec. One of he ways of eliminang or a leas diminishing hese damages is doubling of some imporan pars. Having hese pars doubled, here is a possibiliy o exchange immediaely a non-funconal par by a funconal one (or in oher words, he failure is reduced only o a necessary me of a swich-over). On he oher side here are coss of doubling, primarily he price of a doubled par. For ha reason no all of he pars can be doubled, especially he expensive ones and also hose whose failure is no bringing so expensive damages. 2 Double sysem pars opmizaon model If we wan o know which pars are o be doubled, he following opmizaon model can be used. Firs we inroduce he presumpon of he model. Le us suppose n pars of he sysem (aggregaes, componens) Z 1,Z 2,...,Z n. Each of hese pars is characerized by: p i and p i probabilies of he failure-free run of Z i wihou and a wih reserve, q i and q i he mean value of losses caused by Z i s disorders wihou and wih a reserve, 1 Universiy of Economics Prague, Deparemen of Economerics, nám W. Churchilla 4, , Praha 3, Czech Republic, pelikan@vse.cz 2 Universiy of Economics Prague, Deparemen of Mahemacs, Ekonomická 957, , Praha 4, Czech Republic, henzler@vse.cz

2 c i coss of he purchase and mainenance of he reserve for Z i. Nex we suppose: a sascal independence of he failures of hose pars, he coss of he pars doubling are limied o he amoun K. Le us inroduce 0-1 variables x 1,x 2,...,x n, he variable x i involves he decision beween he doubling of Z i (x i = 1) or no (x i = 0), see [3]. oal coss of he reserves for he pars are n c i x i and since he resources for reserves are limied by K, so i has o be valid n c i x i K. (1) On ha condions we can: a) maximize he reliabiliy of he sysem, ha is failure-free run, b) minimize he mean value of he sum of losses caused by he par s disorders. In he case a) he probabiliy of he failure-free sae of he sysem is he produc of he probabilies of he failure-free saes of all he pars. he par Z i will be failure-free wih he probabiliy p i, if i has a reserve (x i = 1). If he par Z i is wihou reserve (x i = 0) hen he failure-free probabiliy is p i. Alogeher he probabiliy of he par Z i s failure-free sae can be pu in he form p i + (p i p i )x i (see [1]). Hence he oal probabiliy of he failure-free sae expresses he whole sysem s reliabiliy: n r(x) = [p i + (p i p i )x i ]. (2) Afer logarihming he funcon r(x) in order o make he objecve funcon linear we ge he objecve funcon in he form n log r(x) = log[p i + (p i p i )x i ]. (3) Since he expression log[p i + (p i p i )x i ] for x i = 0 equals log(p i ) and for x i = 1 equals log(p i ), we can wrie he expression log[p i + (p i p i )x i ] in he form (1 x i )log(p i ) + x i log(p i ) : n n n log r(x) = [(1 x i )log(p i ) + x i log(p i )] = log(p i ) + x i log(p i /p i ) (4) he maximum reliabiliy model is as follows: n log r(x) = [log(p i ) + x i log(p i /p i )], n c i x i K, x i {0,1}, i = 1,2,...,n. (5) In he case b) he mean value of he losses caused by he par Z i s failure is wihou he reserve q i and wih he reserve q i. he mean value of he oal losses is as follows: n n z(x) = [(1 x i )q i + x i q i ] = [q i x i q i ], where q i = q i q i. (6) he minimal losses model is as follows: n z(x) = [q i x i q i ] min, n c i x i K, x i {0,1}, i = 1,2,...,n. (7)

3 3 wo models Now, wo models can be disnguished. he firs one is a one-case siuaon, when he funcon operaes for a shor period - sac case. he second one is a long-me model for a longer me period - dynamic model. 3.1 Sac model he probabiliy of he componens failure Z i will be denoed as π i. Consequenly he probabiliy of he failure-free run of he par wihou a reserve is p i = 1 π i and he probabiliy of he failure-free run of he par wih a reserve is p i = 1 π 2 i. Le he loss caused by one failure of he par Z i be denoed by Q i ; hen he mean value of he loss is q i = π i Q i in case when here is no reserve, q i = π 2 i Q i in case when here is a reserve for he par Z i. Example 1. Le us have pars Z 1,Z 2,Z 3,Z 4,Z 5. Main characeriscs of hose pars are conained in able 1. he coss of he pars doubling are limied by he amoun K = 100. Z 1 Z 2 Z 3 Z 4 Z 5 p i π i = 1 p i πi p i = 1 πi c i Q i q i = Q i π i q i = Q i πi q i = q i q i able 1 Characeriscs of he sac model Reliabiliy model which maximizes he failure-free probabiliy is: log(r(x)) = log( ) + x 1 log(0.99/0.9) + x 2 log(0.96/0.8) + (8) +x 3 log(0.99/0.9) + x 4 log(0.9951/0.93) + x 5 log(0.91/0.9919) max, 80x x x x x 5 100, x 1,x 2,x 3,x 4,x 5 {0,1}. (9) By using sandard sofware LINGO we ge he opmal soluon x = (0,1,1,0,1) wih he failure-free probabiliy equal , which is maximal. From he resul follows ha i has o double Z 2,Z 3,Z 5. Model which minimizes he mean value of he oal losses is: z(x) = x 1 40x 2 30x 3 105x 4 81x 5 min, (10) 80x x x x x 5 100, x 1,x 2,x 3,x 4,x 5 {0,1}. (11) When we use again LINGO sysem, we ge he opmal soluon x = (1,0,0,0,1) wih he minimal value of losses in he mean value. According o his soluon only he pars Z 1 and Z 5 will be doubled. he differences in he soluons obained above we can explain by grea influence of he amoun of he loss in he opmal soluon in he second model. Firs soluon x = (0,1,1,0,1) means he mos reliable sysem, bu he loss is no minimal. Second soluon x = (1,0,0,0,1) gives us less reliable sysem, bu he loss is minimal. We can observe he values of reliabiliy, mean loss and coss of reserves for differen soluons in he able 2. he values wih a bulle are opmal for K =

4 soluon x reliabiliy losses doubling cos (0,0,0,0,0) (0,1,1,0,1) (1,0,0,0,1) (1,1,1,1,1) able 2 Sac model - differen soluons 3.2 Dynamic model Le us suppose ha he sysem s reliabiliy should be opmized wihin a period < 0, > and he periods beween he failures of he pars are exponenally disribued. Le he mean value of he period beween wo failures of he componen Z i be 1/λ i. he period of replacing he failed componen Z i by a new one has he fixed lengh i (see Figure 1). If he Z i is wihou a reserve, hen he probabiliy of he failure-free run due he par Z i is p i = exp( λ i ). (12) As he number of he failures wihin he period < 0, > is Poisson disribued wih he mean value λ i, he mean value of he loss caused by he par Z i in he case x i = 0 is equal o q i = λ i Q i, where he Q i is he loss caused by one failure of he par Z i. (13) Z i failure-free run 0 u 1 u 2 Figure 1 wo failures, u 1, u 2, in he sysem wihou a reserve par In case ha componen Z i has a reserve (x i = 1), he i is period necessary for he reserve componens exchange, where i << (see Figure 2). During he period i he failure-free probabiliy of he sysem (i.e. no failure of he reserve par) is exp( λ i i ), see [2]. Z i reserve 0 failure-free run u u u Figure 2 Failures in sysem wih a reserve par As he failures of a par are reaed as Poisson evens and herefore independen, he probabiliy of failure-free run of he sysem in case of k failures of he par Z i or is reserve is (exp( λ i i )) k = exp( kλ i i ). (14) herefore, he overall probabiliy of failure-free run of he sysem consisng of he par Z i wih a reserve in he period < 0, > is as follows

5 p i = exp( λ i ) i k=0 (λ i ) k exp( kλ i i ) = exp( λ i ) i k=0 (λ i exp( λ i i )) k where [ i ], i << is he op number of failures in he period < 0, >, exp( λ i ) (λi)k is he probabiliy of k failures of he par Z i or is reserve, exp(λ i [exp( λ i i ) 1]), (15) Le us find he logarihm of he reliabiliy funcon r(x); using he general formula (5) and supposing i << we have + n x i log i k=0 log r(x) = n log(p i ) + n x i log(p i /p i ) = (λ i ) k exp( kλ i i ) n λ i + n ( λ i ) + (16) n x i λ i exp( λ i i ). he mean value of he number of sysem failures due o he componen Z i s, i.e. q i can be esablished in he following way. he number of failures boh in he par Z i and is reserve is Poisson disribued. he me beween he wo failures is exponenally disribued wih parameer λ i. he sysem fails if he me inerval beween he failures of he par Z i and is reserve (or vice versa) is shorer hen he repair inerval i. he probabiliy of his even is equal 1 exp( λ i i ). Le he number of failures of he par Z i in he inerval < 0, > be k. hese failures can be aken as binomial evens, wih one oucome failure of he reserve of Z i wihin inerval i, i.e. sysem failure and oher oucome no such failure wihin i. hen probabiliy P(X i = r), where X i is he number of sysem failures due o Z i and is reserve, is as follows: i k=r [ P(X i = r) = i ] k=r ( k )(1 exp( λ i i )) r (exp( λ i i )) k r (λ i) k exp( λ i ) = r = (1 exp( λ i i )) r (λ i ) r exp( λ i ) r! i i k=r (λ i ) k r (k r)! exp( (k r)λ i i ), (17) E(X i ) = r P(X i = r) = (1 exp( λ i i ))λ i exp( λ i ) (1 exp( λ i i )) r 1 (λ i ) r 1 (r 1)! (λ i ) k r (k r)! r=0 exp( (k r)λ i i ) = (1 exp( λ i i ))λ i exp( λ i ) i r=1 [ ] 1 i s=0 (λ i ) s (1 exp( λ i i ))λ i, s! q i = E(X i ) Q i = (1 exp( λ i i ))λ i Q i. (18) Example 2. Le us have pars Z 1,Z 2,Z 3,Z 4,Z 5 (see able 3 for characeriscs; values of c i and Q i are he same as in he sac example). Le he overall me be = 5 [in hours]. he coss of he pars doubling are limied by he amoun K = 100. Reliabiliy model which maximizes he failure-free probabiliy is: n n n log(r(x)) = log(p i ) + x i log(p i /p i )] λ i + n x i λ i exp( λ i i ) = (19) = x x x x x max,

6 Z 1 Z 2 Z 3 Z 4 Z 5 λ i i [in hours] p i p i log(p i /p i ) c i Q i q i q i q i = q i q i able 3 Characeriscs of he dynamic model 80x x x x x 5 100, x 1,x 2,x 3,x 4,x 5 {0,1}. (20) By using LINGO sofware we ge he opmal soluon x = (0, 1, 0, 1, 1) wih he failure-free probabiliy equal 0,019958, which is maximal. From he resul follows ha i has o double Z 2,Z 4,Z 5. Model which minimizes he mean value of he oal losses is: z(x) = x x x x x 5 min, (21) 80x x x x x 5 100, x 1,x 2,x 3,x 4,x 5 {0,1}. (22) By LINGO we ge we ge he opmal soluon x = (0,1,0,1,1) wih he minimum loss equal In conrary o he sac model, in he dynamic model he opmal soluon on respec o boh maximum reliabiliy and minimum loss is he same (see able 4 for differen soluons). soluon x reliabiliy losses doubling cos (0,0,0,0,0) (0,1,0,1,1) (1,1,1,1,1) able 4 Dynamic model - differen soluons Acknowledgemens his research was suppored by GAČR gran No. GACR P403/12/1947 and he projec F4/18/2011 founded by he Inernal Gran Agency of he Universiy of Economics, Prague. References [1] Černý, M.: Binary Segmenon and Bonferroni-ype Bounds, Kyberneka 47 (2011), No. 1, ISSN [2] Dupač, V.: Privae communicaon. [3] Nemhauser, G. L., Wolsey, L. A.: Ineger and Combinaorical Opmizaon. John Wiley & Sons Inc., New York,