Solving Fuzzy Linear Programming Problem With Fuzzy Relational Equation Constraint

Transcription

1 Intern. J. Fuzz Maeatcal Archve Vol., 0, -0 ISSN: 0 (P, 0 0 (onlne Publshed on 0 Septeber 0 Internatonal Journal of Solvng Fuzz Lnear Prograng Proble W Fuzz Relatonal Equaton Constrant Thangara Beaula and Saraswa.K Departent of Maeatcs, TBML College, Poraar, Tal Nadu Eal: edwnbeaula@ahoo.co.n Receved Jul 0; accepted August 0 Abstract. A lnear prograng proble w nu lnear obectve functon subect to a sste of fuzz relaton equatons usng a-n product s consdered. Soe procedures to reduce e proble are establshed and llustrated b a nuercal eaple. Kewords: Fuzz relaton equatons, Ma-n product,0- prograng AMS Maeatcs Subect Classfcaton (00: 0E, 0B0. Introducton Fuzz Relatons Equatons (FRE were ntroduced and appled to dagnoss probles n [,], fuzz relaton equatons on a fnte set were later consdered n [] and structure of e set of solutons of such equatons was studed. The structure of soluton sets of FRE w dfferent coposton operators was gven n 0s, see for nstance [,,]. It s well known now at e soluton set of fnte FRE w contnuous a t-nor s deterned b one au soluton and fnte nuber of nal solutons. However, t s not eas to obtan all nal solutons for a large scale proble because nuber of nal solutons a ncrease ver sharpl as of e proble sze ncreases. In recent several ears, ere are stll an results on developng a ore effectve algor for obtanng all nal solutons of FRE and ere has been a growng nterest n a class of nzng probles w fuzz relatonal equaton constrants. We attept to fnd an optal soluton to e fuzz lnear prograng proble w fuzz relatonal constrants whch s llustrated b nuercal eaple.. A necessar and suffcent condton for estence of soluton defnton. Consder ree fuzz bnar relatons P(X,Y, Q (Y,Z and R(X,Z, defned on e sets, X { / I} ; Y { / J} ; Z{z k / k K} w I N n, J N, and KN s. Let e

2 Solvng Fuzz Lnear Prograng Proble W Fuzz Relatonal Equaton Constrant Let p P (,, q k Q(, z k, r k R (,z k Ths eans at all entres n e atrces P,Q and R are real nubers n e unt nterval [0,]. Assue now at e ree relatons constraned w each oer n such a wa at P o QR ( where 'o' denotes e a-n coposton. Ths eans at a J n (p, q k r k ( for all and I and k k. That s, e atr equaton ( encopasses n s sultaneous equatons of e for (. ese equatons are referred to as fuzz relaton equatons.the set of all partcular atrces of e for P at satsf ( s called ts soluton and denote e set of all solutons as S (Q, R {P / P o Q R} ( Necessar condton for estence of solutons, consder equaton ( e., a n (p, q k r k and f a q k < r k J J. Soluton Meod Consder atr equatons ( of e spler for p o Q r where p [p / J], Q[q k / J, k k], r [r k / k k] e., p,q and r represent a fuzz set on, q fuzz relaton on Y Z, and a fuzz set on Z respectvel.in our dscusson e constrant equaton s ο A b where { / I}, A {a / I, J}, b {b / J} Denote e soluton set of p o Q o A b r as X (A, B { / o A b} When X (A,b φ, e au soluton [ / I] of ( ( s deterned as follows: σ (a,b ( n J b f a > b where σ ( a, b oerwse When deterned n s wa does not satsf oa b, en X (A,b φ. Assue now at A and b of oa b are gven, and at we have to deterne e set X (A,b of all nal solutons of e equaton. Assue furer at has been deterned b ( and has been verfed as e au soluton. When 0 for soe I, we a elnate s coponent for p as well as row fro atr A, clearl, 0 ples 0 for each X (A,b. Furerore, when b 0 for soe J, We a elnate s coponent fro b and e colun fro atr A, Snce each ( ρ ust satsf, n s case, e a n equaton represented b, e colun of Q, and b 0.. A necessar condton for optal soluton Ths secton recalls soe prelnares of t-nors. Soe propertes of a-t- nor fuzz relatonal equatons are presented.

3 Thangara Beaula and Saraswa.K T: [0,] [0,] such at for all α, β, δ [0,,] (a T ( α, β T ( β, α (b T ( α, T ( β, δ T (T ( α, β, δ (c T (α, β T ( α, δ whenever β δ, and (d T ( α, α. Defnton.. The fuzz lnear prograng proble w fuzz relatonal equaton constrant s defned as Mnze Z( c,,... I ( subect to X (A,b { [0,] / oa b} ( where c R s e coeffcent assocated w varable ; A[a ] s an n nonnegatve atr w a < ; b(b, b,.. b n s an n-densonal vector w 0< b <,,, n J and e operaton "o" represents e a- n coposton (or e a T coposton operator. Lea.. If X ( A, b, en for each J ere ests o I such at n ( o, a o b and n (, a b for ever I Proof. Snce o Ab, en {n(, a } b for J ( a I That eans for each J, n (, a b. And, n order to satsf e equalt constrant, ere ust est at least one o I such at n ( o, a o b Defnton.. For a soluton X ( A, b, we call o a bndng varable f n ( o, a o b for o I and n (,a b for all I. Lea.. Let ( I be e au soluton, and ( I be a soluton of (. If s bndng n e equaton, en s also bndng. However, f s not a bndng varable, s also non bndng for an soluton. Proof. For an soluton ( X ( A, b we have a I I {n(, a b, J Ths ples n (, a b, J. If s now bndng n e equaton, en n(,a b. also b n(,a n (, a b Ths ples at n. (, a b. Hence s also bndng n e equaton. On e oer hand, f s not bndng n an equaton en n (, a b holds for an soluton, we have n

4 Solvng Fuzz Lnear Prograng Proble W Fuzz Relatonal Equaton Constrant Lea.. Let ( I be e au soluton. If e cost co-effcent c 0, I, en s an optal soluton of proble. Proof. For an soluton n X (A,b we have 0 Snce c 0, I we have c c ( Therefore, s an optal soluton.. Separaton of fuzz LPP In s secton, we stud how to separate proble( -(nto two sub probles; and how to eld an optal soluton fro e au soluton and one of e nal solutons... Two sub probles of odel ( ( Fang et al [] showed at an optal soluton for odel ( ( w a n or a product coposton can be obtaned fro two sub- probles, whch are fored b separatng e negatve and non negatve coeffcent n e obectve functon. Consder e followng two probles. Mnze Z ( c subect to X A, b { [0,] / oa b} ( ( and Mnze Z ( c ( (0 where c f c < 0 o f c < 0 c and c I 0 f c 0 c f c 0 probles ( and (0are subected to e orgnal constrant. Furerore, c c + c, I. The au soluton ( I s an optal soluton for proble ( w optal value Z (. Addtonall, one nal soluton, sa subect to X A, b { [0,] / oa b} ( s an optal soluton for proble (0 w optal value Z (. A new vector s now defned b. I ( I f f It follows at < <. Hence value Z ( z ( + Z (. c < 0 c 0 I ( s a soluton of equaton ( w obectve

5 Thangara Beaula and Saraswa.K The reanng task s to show at as defned n ( s an optal soluton of e orgnal proble (- ( w optal value Z ( Ths can be seen fro e followng nequaltes: z( c ( c + c c + c c + c c + c c + c z ( + Z ( Z(.. An equvalent 0- nteger prograng proble. The followng nde sets are defned to fnd a nal soluton fro X(A,b to optze proble (0. I ( I / n (, a b, J J { J / n(, a b } I The nde set I ndcates e possble varables of at a be selected as a bndng varable n e equaton. The nde set J ndcates ose equatons at are satsfed b. The followng varables are defned to fnd nal soluton fro X(A,b to solve proble (0. f I I, J 0 oerwse Notabl, e varable corresponds to a possble selecton of e coponent of soe nal solutons at are bndng n e equaton. Snce each soluton ust satsf all equatons, a nal soluton can be transfored nto e selecton of one varable w value n each equaton. The 0- nteger prograng proble, whch s equvalent to proble (0 Z a Mnze Z ( ( c { b } subect to J 0 or, I, J 0, I Therefore, e obectve functon becoes c {, } J J, J ( ( a Moreover, onl ose ndces n J need to be consdered.

6 Solvng Fuzz Lnear Prograng Proble W Fuzz Relatonal Equaton Constrant I, J. The 0- nteger prograng odel for proble (0 s presented as follows: Mnze Z ( c a{ } subect to, I An optal soluton J J 0 or, I, 0, w I, J I ( corresponds to e stuaton where e varable J ( contanng e varable n proble s bndng n e equaton.. Rules for reducng proble (0 Consder e gven atr A(a w I and J..To develop a procedure of fndng an optal soluton, e followng nde set are gven for e value atr I { I / n(, a b }.Ths nde set contans I such at can be satsfed e equaton. Rule. If a sngleton I { } ests fro soe J, en s assgned to e coponent of an optal soluton Proof. The nde set I { } ples at e equaton onl can be satsfed b varable. Ths ples at e coponent of an soluton (hence, e varable ust be bndng n e equaton, elds ( snce b > 0 Based on Rule., e colun of M can be deleted fro furer consderaton. The correspondng row of n A can also be deleted. Rule. If Ip I q for soe p, q J n e value atr A, en e q colun of M can be deleted. Proof. Rule reveals at f a sngleton I { } ests for soe J n e atr A, and I I q for soe q J en e q equaton can be deleted. Furerore, e deleton can be perfored when I (A s not a sngleton. Lea.. If X ( A, b φ, J en I φ, J Proof. Fro lea., we know at ere ests at least one 0 I at can satsf constrant, erefore I ust contan at least one eleent.

7 Thangara Beaula and Saraswa.K Rule. If p,q,r I, J and does not belongs to an oer I t, t J and t such at c p p > cq q > cr r en an optal soluton ( has 0 I p q Proof. Gven at p,q,r I Ths ples. p, q, r satsfed e equaton, e does not satsf an oer equaton.also snce c p p > cq q > cr r In order to satsf e. equaton, we need onl one varable w nal cost coeffcent. We set p q 0. If Rule s appled n e process of fndng an optal soluton en e rows of atr A at are assocated w p, q can be deleted. Rule. Durng e process of fndng an optal soluton. If s s an undecded decson varable such at s I, J en an optal soluton ( has 0 I s Proof. Snce s s an undedcated decson varable and s I, J.Ths ples s does not satsf an equaton. an optal soluton ( has 0 I s.if rule s appled en e correspondng rows of A can be deleted... An algor Based on e concepts dscussed before, we present an algor for fndng an optal soluton. Step. Check e necessar condton for estence of solutons. If a > b J, contnue, oerwse stop, (A,bφ and proble ( ( has a J no soluton. Step. Copute e vector ( I b (. Step. Check e consstenc b verfng wheer oa b stop n case of nconsstenc (If consstent, en ( I s e au soluton Step. For two sub probles as probles ( and (0 Step. Fnd optal soluton for proble (0 Step.. Copute nde set I (A for all J for e gven value atr A. Step.. Appl Rules - to deterne e values of decson varables as an as possble. Delete e correspondng rows and or coluns n A (Thus reducng e sze of e proble Denote e reduced sub atr b A agan. If all decson varables have been set, en go to step. Step.. Take e (reanng value atr A. Eplo e branch and bound eod to solve for e reanng undecded decson varables. Step. Generate optal solutons for e orgnal proble fro optal solutons of probles ( and (0 b (... Nuercal eaple

8 Solvng Fuzz Lnear Prograng Proble W Fuzz Relatonal Equaton Constrant Mnze z ( subect to oa b 0,,... ( where,,,,,,,, A b (0., 0.0, 0., 0., 0., 0., 0., 0., 0. Step. Snce a > b J, e necessar condton s satsfed. a J Step. Copute e vector, 0., 0., 0., 0., 0., 0., 0., 0., 0. Step. Snce oa b.e., Ma ( a, b, J { } (n I The proble s solvable and s e au soluton. Step. For two sub probles as probles ( and (0 e followng sub proble P, s gven as proble ( w negatve co-effcent n e obectve functon. P: Mnze Z ( 0. subect to oa b The oer sub proble, P, s gven as proble (0 w non negatve coeffcents n e obectve functon. P: Mnze z ( subect to oa b Step. Fnd optal soluton for proble (0

9 Consder e gven atr. Thangara Beaula and Saraswa.K A (Snce denote n (, a b We know at I { I / n(, a b }. Therefore I( A I Step.. Appl Rules - to deterne e values of as an decson varables as possble. Delete e correspondng rows and / or coluns n A. For e gven atr A, e nde set I I ( A { } ndcate at e varable s e onl bndng varable n e nd and e equaton let ( I be an optal soluton of sub proble P. Then can be assgned b rule. s also bndng n equatons, and (or coluns,, of A Hence, ese coluns and e correspondng row, can be deleted fro atr A. After deleton e reduced atr A becoes. A {, }, I ( A { }, I( A {, }, I ( A {,,, }, I( A { }, I( A {,,,, } {, }, I {, }, I {, }

10 Solvng Fuzz Lnear Prograng Proble W Fuzz Relatonal Equaton Constrant Snce I {,}, I {,}, I {,,,,}, I {,}. The reduced atr A s equvalent to equatons w eght varables. The nde set to e reduced atr A s such at I I ( A. So colun or of A can be deleted b Rule. Also snce,, I and,, I,,, w c 0. > c 0. > c 0.. we set 0 b rule. Also, e reduced value atr A has w I, J. We set 0 b rule. After deletng colun or and e correspondng rows of e atr A at are assocated w, and e reduced atr A becoes, A I {,}, I {,, }, I {,}. Snce Rules - cannot be appled to e current atr A and fve reanng,,, are undecded goto e net step. varables { }, Step.. Take e (reanng value atr A. Eplo e branch and bound eod to solve for e reanng undecded decson varables. Snce, Mn Z oa B, o b,,,, where,,,, ( A

11 Thangara Beaula and Saraswa.K {,}, I {,, }, I ( {,}. I A J A {,}, J {}, J {}, J {}, J ( {,}. Its correspondng 0- nteger progra s Mnze z.( (, +.( ( Subect to a a ( J J + a ( ( a ( J J J 0 a + + ( ( St op 0 a 0

12 Solvng Fuzz Lnear Prograng Proble W Fuzz Relatonal Equaton Constrant Now, consder e frst constrant equaton. Eer, or has to be. Ths elds nodes and. If en erefore e lower bound of node s 0... Also, f en. And e lower bound of node s 0... Fro node we can branch furer to eer node or node or node w or or respectvel f en erefore e lower bound of node s calculated b. + ( If en and e lower bound of node s. (0.. If en and e lower bound of node s. + ( Fro node we can branch furer to eer node or node w or respectvel. Ths s equvalent to addng anoer constrant. Snce s added constrant s e last one, we obtan e eact obectve values nstead of e lower bounds. If en and e obectve value of node s. + ( If en and e obectve value of node s. + (. 0.. Snce, Z ( at node and s greater an e lower bound of node, we can branch node to eer node or node w or If en and e obectve value of node s. + (. 0.. If en and e obectve value of node s. + (. 0.. Snce Z ( at node,, and s greater an e lower bound of node we can branch node to eer node 0 or node w or. If en and e obectve value of node 0 s.0 + ( If en and e obectve value of node s.0 + (. 0.. Snce Z ( at node s equal to e lower bound of node, we can stop branchng to node. Moreover Z ( at node and node elds e optal value. Fgure shows e B & B of e gven proble. Fro e above dscusson, we get e two optal solutons.

13 Thangara Beaula and Saraswa.K (0, 0, 0., 0., 0, 0., 0, 0. and (0., 0, 0,0, 0, 0., 0, 0. For e sub proble p w obectve value z ( z (.. Now, at all e decson varables have been deterned go to e net step. Step. Generate optal solutons for e orgnal proble fro optal solutons of proble ( and (0 b (. Notabl, onl varable of e sub proble P has a negatve co-effcent n e obectve functon. Hence e au soluton ( I s an optal soluton w optal value Z ( c 0. for sub proble p. on e oer hand two optal solutons and are gven for sub proble P w optal value Z ( Z (.. Cobnng ese optal solutons derved fro sub probles P and P b ( elds two optal solutons and as follows. (0, 0, 0., 0., 0., 0, 0., 0, 0, 0., 0, 0. and (0., 0, 0,0., 0, 0, 0., 0, 0. REFERENCES. E. Czogala, J. Drcwnak and W. Pedrcz, Fuzz relaton equatons on a fnte set, Fuzz Sets and Sstes, (, -0.. E.Czogale and W.Predrucz, On dentfcaton n fuzz sstes and ts applcatons n control probles, Fuzz Sets and Sstes, (,.. S.-C. Fang and G. L, Solvng fuzz relaton equatons w a lnear obectve functon, Fuzz Sets and Sstes, 0 (, 0-.. H.C. Lee and S.M. Gue, On e optal ree ter ulteda streang servces, Fuzz Optzaton and Decson Makng, (00, -.. E. Sanchez, Resoluton of coposte fuzz relaton equatons, Infor and Control, 0 (, -.. E. Sanchez, Inverses of fuzz relatons: Applcatons to possblt dstrbuton and edcal dagnoss, Fuzz Sets and Sstes, (,.. L.A. Zadeh, Fuzz Sets, Inforaton and Control, ( (, -.