Optimization of fuzzy relation equations with max-product composition

Fuzzy Sets and Systems 118 (2001) 509–517

www.elsevier.com/locate/fss

Optimization of fuzzy relation equations with max-product composition Jiranut Loetamonphong, Shu-Cherng Fang ∗ Department of Industrial Engineering and Graduate Program in Operations Research, North Carolina State University, Raleigh, NC 27695-7906, USA Received May 1998; received in revised form September 1998

Abstract An optimization problem with a linear objective function subject to a system of fuzzy relation equations using maxproduct composition is considered. Since the feasible domain is non-convex, traditional linear programming methods cannot be applied. We study this problem and capture some special characteristics of its feasible domain and the optimal solutions. Some procedures for reducing the original problem are presented. The problem is transformed into a 0 –1 integer program which is then solved by the branch-and-bound method. For illustration purpose, an example of the procedures is provided. c 2001 Elsevier Science B.V. All rights reserved..

Keywords: Fuzzy relation equations; Max-product composition; Branch-and-bound method; Integer programming

1. Introduction The notion of fuzzy relation equations based upon the max–min composition was ÿrst investigated by Sanchez [13]. He studied conditions and theoretical methods to resolve fuzzy relations on fuzzy sets deÿned as mappings from sets into complete Brouwerian lattices. Some theorems for existence and determination of solutions of certain basic fuzzy relation equations were presented in his work. However, the solution obtained in that work is only the greatest element (or the maximum solution) derived from the max–min (or min–max) composition of fuzzy relations. Sanchez’s work has shed some light on this important subject. Since then, many researchers have been trying to explore the problem and ∗

Corresponding author.

develop solution procedures [1,4,5,7,9 – 12]. The “max–min” composition [17] is commonly used when a system requires conservative solutions in the sense that the goodness of one value cannot compensate the badness of another value. In reality, there are situations that allow compensatability among the values of a solution vector. In this case, the min operator is not the best choice for the intersection of fuzzy sets. Instead, the “max-product” composition is preferred since it can yield better, or at least equivalent, results [6,14,18]. Note that when the intersection connector acts noninteractively, it can be uniquely deÿned by the min connector, but when the connector is interactive, it is application dependent and cannot be deÿned universally [2]. Some outlines for selecting an appropriate connector has been provided by Yager [16]. Recently, Bourke and Fisher [3] extended the study of an inverse solution of a system of fuzzy relation

c 2001 Elsevier Science B.V. All rights reserved. 0165-0114/01/$ - see front matter PII: S 0 1 6 5 - 0 1 1 4 ( 9 8 ) 0 0 4 1 7 - 5

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equations with max-product composition. They provided theoretical results for determining the complete solution sets as well as the conditions for the existence of resolutions. Their results showed that such complete solution sets can be characterized by one maximum solution and a number of minimal solutions. Since the total number of minimal solutions has a combinatorial nature in terms of the problem size, an ecient solution procedure is always in demand. Motivated by the work of [7], we are interested in studying the optimization problem with a linear objective function subject to a system of fuzzy relation equations with the “max-product” composition. Let A = [aij ]; 06aij 61, be an (m × n)-dimensional fuzzy matrix, b = (b1 ; : : : ; bn ); 06bj 61, be an n-dimensional vector, I = {1; 2; : : : ; m}, and J = {1; 2; : : : ; n}. In this paper, a system of fuzzy relation equations deÿned by A and b is denoted by x ◦ A = b;

(1)

where “◦” represents the max-product composition. The resolution of (1) is a set of solution vectors x = (x1 ; : : : ; xm ), 06xi 61, such that max{xi · aij } = bj i∈I

for j ∈ J:

(2)

Let c = (c1 ; : : : ; cm ) ∈ Rm be an m-dimensional vector where ci represents the weight (or cost) associated with variable xi , for i ∈ I . The optimization problem we are interested in has the following form: minimize

Z=

m X

ci xi

i=1

subject to

x ◦ A = b;

(3)

06xi 61: Note that the characteristics of the solution sets obtained by using the max–min operator and the maxproduct operator are similar, i.e., when the solution set is not empty, it can be completely determined by a unique maximum solution and a ÿnite number of minimal solutions [3,4,10]. Since the solution set can be non-convex, traditional linear programming methods, such as the simplex and interior-point algorithms [8], cannot be applied to this problem. In this paper we study the solution set of system (1) and solve problem (3). Even though the feasible domain of problem (3) requires the determination of the

set of minimal solutions, by using the special structure of the problem, we can ÿnd an optimal solution without explicitly generating every minimal solution. The rest of the paper is organized as follows. Section 2 explores the feasible domain of a system of fuzzy relation equations with max-product composition. Section 3 studies the eect of the cost vector and shows that problem (3) can be divided into two parts; one with non-negative cost coecients and the other with negative cost coecients. In Section 4, some procedures for reducing the problem are introduced. Some considerations for decomposing the problem into sub-problems are given in Section 5. Finally, problem (3) is transformed into a 0 –1 integer program and the branch-and-bound method is applied. The algorithm is outlined and illustrated by an example. 2. Characterization of the feasible domain We denote the solution set of problem (1) by X (A; b) = {(x1 ; : : : ; xm ) | xi ∈ [0; 1]; i ∈ I , and x ◦ A = b}. Deÿne X = [0; 1]m . For x1 ; x2 ∈ X , we say x1 6x2 if and only if xi1 6xi2 ; ∀i ∈ I . In this way, the operator “6” forms a partial order relation on X and (X; 6) becomes a lattice. xˆ ∈ X (A; b) is called a maximum solution if x6xˆ for all x ∈ X (A; b). Also, xÄ ∈ X (A; b) is called a minimal solution if x6x, Ä for any x ∈ X (A; b), implies x = x. Ä When X (A; b) is not empty, it can be completely determined by a unique maximum solution and a ÿnite number of minimal solutions [3,4,10]. The maximum solution can be obtained by applying the following operation [3].   n ^ (4) xˆ = A ˜ b =  (aij ˜ bj ) ; j=1

where aij ˜ bj =



i∈I

1 if aij 6bj ; bj =aij if aij ¿ bj ;

(5)

a ∧ b = min(a; b): Denote the set of all minimal solutions by XÄ (A; b). The complete set of solution, X (A; b), is obtained by [ {x ∈ X | x6x6 Ä x}: ˆ (6) X (A; b) = x∈ Ä XÄ (A; b)

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511

In analog to what is presented in [7], we have the following lemma.

For each solution f, Jfi corresponds to a set of constraint indices that can be satisÿed by xi .

Lemma 1. If x ∈ X (A; b), then for each j ∈ J there exists i0 ∈ I such that xi0 · ai0 j = bj and xi · aij 6bj ; ∀i ∈ I .

Lemma 4. For j; j 0 ∈ Jfi ; bj =aij = bj0 =aij0 .

Proof. Since x ◦ A = b, then max{xi · aij } = bj i∈I

for j ∈ J:

That means for each j ∈ J; xi · aij 6bj . And, in order to satisfy the equality constraint, there must exist at least one i0 ∈ I such that xi0 · ai0 j = bj . Deÿnition 1. For a solution x ∈ X (A; b), we call xi0 a binding variable if xi0 · ai0 j = bj for i0 ∈ I and xi · aij 6bj , for all i ∈ I . When the solution set of (1) is not empty, i.e., X (A; b) 6= ∅, we deÿne Ij = {i ∈ I | xˆi · aij = bj };

∀j ∈ J;

 = I1 × I2 × · · · × In :

(7) (8)

Here Ij corresponds to a set of xi ’s that can satisfy constraint j of the fuzzy relation equations. And, the set  represents all combinations of the binding variables such that every combination can satisfy every fuzzy relation constraint. Let each combination be represented by f = (f1 ; f2 ; : : : ; fn ) ∈ ; with fj ∈ Ij ; ∀j ∈ J . Lemma 2. If X (A; b) 6= ∅, then Ij 6= ∅; ∀j ∈ J . From Lemma 1, we know that there exists at least one i0 ∈ I that can satisfy constraint j. Therefore, Ij must contain at least one element. Lemma 3. If kIj k = 1, then xˆi = xÄi = bj =aij , for i ∈ Ij . Proof. In order to satisfy constraint j, xi has to be equal to bj =aij . Since xi is the only variable that can satisfy constraint j, it can take on only one value, i.e. bj =aij . Therefore, xˆi = xÄi = bj =aij .

From Lemma 4, we deÿne F :  → [0; 1]m such that, for i ∈ I ,  bj =aij ∀j ∈ Jfi 6= ∅; (10) Fi (f) = 0 if Jfi = ∅: Parallel to Theorem 1 of [7], we have the following result. Theorem 1. Given that X (A; b) 6= ∅, (1) If f ∈ , then F(f) ∈ X (A; b). (2) For any x ∈ X (A; b), there exists f ∈  such that F( f)6x. Consequently, we obtain the following corollary. Corollary 1. XÄ (A; b) ⊆ F() ⊆ X (A; b). 3. Eects of the cost vector The optimization problem (3) can be decomposed into two problems, namely m X

minimize

Z0 =

ci0 xi

subject to

x ◦ A = b;

i=1

(11)

06xi 61 and minimize

Z 00 =

m X

ci00 xi

i=1

Deÿne index sets Jfi = { j ∈ J | fj = i};

Proof. Let j and j 0 be indices in Jfi . From the definition of Jfi , since xi can satisfy all constraints in Jfi , that means xi · aij = bj and xi · aij0 = bj0 . Therefore, bj =aij = bj0 =aij0 .

subject to ∀i ∈ I:

(9)

x ◦ A = b; 06xi 61;

(12)

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where c0 = (c10 ; c20 ; : : : ; cm0 ) and c00 = (c100 ; c200 ; : : : ; cm00 ) are deÿned such that, ∀i ∈ I , 

ci 0  0 ci00 = ci

ci0 =

if ci ¿0; if ci ¡0;

(13)

if ci ¿0; if ci ¡0:

Apparently, the cost vector c = c0 + c00 and the objective value Z = Z 0 + Z 00 . Intuitively, when all the costs are non-positive, since xi ’s are non-negative and the problem is to minimize the objective value, we should make xi as large as possible. Therefore, we have the following result. Lemma 5. For problem (12) with ci00 60; ∀i ∈ I , xˆ is an optimal solution. Proof. For any x ∈PX (A; b) 6= ∅,P we know that m m 06x6x. ˆ Therefore, i=1 ci00 xi ¿ i=1 ci00 xˆi and xˆ is an optimal solution. Lemma 6. For problem (11) with ci0 ¿0; ∀i ∈ I , one of the minimal solutions is an optimal solution. Proof. For any x ∈ X (A; b) 6= ∅, we know from (6) that there exists a minimal solution xÄ0 ∈ XÄ (A; b) such 0 ˆ Since we have that xP Ä0 6x6x. Pmci ’s 0are non-negative, Pm 0 m 0 0 c x Ä 6 c x 6 c x ˆ . We then that i i i=1 i i i=1 s i i=1 i P Pm 0 m ∗ 0 ∗ choose xÄ such that i=1 ci xÄi = min{ i=1 ci xÄi | xÄ ∈ P P Ä b)}. Therefore, we have m ci0 xi¿ m ci0 xÄ0i ¿ X i=1 i=1 P(A; m 0 ∗ i=1 ci xÄi ; ∀x ∈ X (A; b). From Lemmas 5 and 6, the ÿnal solution x∗ = obtained by combining xˆ and xÄ∗ is

(x1∗ ; x2∗ ; : : : ; xm∗ ) xi∗ =



xÄ∗i if ci ¿0; xˆi if ci 60;

∀i ∈ I:

(14)

Parallel to Theorem 2 of [7], we have the following result. Theorem 2. If X (A; b) is not empty and x∗ is deÿned according to (14), then x∗ is an optimal solution (3) with an optimal value Pmof the problem Pm Z ∗ = i=1 ci xi∗ = i=1 (ci00 xˆi + ci0 xÄ∗i ).

4. Problem reduction Taking advantage of the special structure studied in the previous section, we now introduce some procedures to reduce the size of the original problem so that the eort to solve the problem is minimized. The key idea behind these reduction procedures is that some of the xi ’s can be determined immediately without solving the problem but just by identifying the special characteristic of the problem. Special cases which we can eliminate from consideration are as follows. Case I: ci 60. From Lemma 5 and Eq. (14), we know that xi∗ = xˆi , if ci 60. Hence, we can take any parts that are associated with these xˆi ’s out of consideration. Here we deÿne Iˆ = {i ∈ I | ci 60};

(15)

Jˆ = { j ∈ J | xˆi · aij = bj ; ∀i ∈ Iˆ }:

(16)

In other words, Jˆ is a set of indices of constraints which can be satisÿed by a set of xˆi ’s for i ∈ Iˆ. Having deÿned Jˆ and Iˆ, we now eliminate row i; i ∈ Iˆ, and column j; j ∈ Jˆ from matrix A as well as the jth element, j ∈ Jˆ, from vector b. Let A0 and b0 be the updated fuzzy matrix and fuzzy vector, respectively. Deÿne J 0 = J \Jˆ. J 0 represents a reduced set of constraints. Case II: Ij has only one element. Consider constraint j ∈ J 0 . If Ij contains only one element, it means that only one xi ; i ∈ Ij ; can satisfy the jth constraint. From Lemma 3, we set xi = bj =aij . Deÿne I = {i ∈ Ij | kIj k = 1; j ∈ J 0 };

(17)

J = { j ∈ J 0 | xi · aij = bj ; i ∈ I }:

(18)

Again, we can eliminate row i; i ∈ I, and column j; j ∈ J, from the updated fuzzy matrix A0 as well as the jth element, j ∈ J, from the updated vector b0 . Let A00 and b00 be the reduced fuzzy matrix and fuzzy vector corresponding to A0 and b0 , respectively. We also need to update . Deÿne J 00 = J 0 \J. J 00 is an index set of constraints which need to be solved

J. Loetamonphong, S.-C. Fang / Fuzzy Sets and Systems 118 (2001) 509–517

later by Q the branch-and-bound method. The updated 00 = j∈J 00 Ij . The branch-and-bound algorithm will be performed on these A00 and b00 . If b00 is empty, then all constraints have been taken care of. Therefore, in order to minimize the objective value, since we are now left with positive ci ’s, we can assign the minimum value, i.e. zero, to all xi ’s whose values have not been assigned yet. When b00 is not empty, we need to proceed further. Details will be discussed in the next section. 5. An algorithm for ÿnding an optimal solution Before we discuss the 0–1 integer program and the branch-and-bound (B&B) method, observe that sometimes the problem can be decomposed into several parts which can be solved separately. This leads to smaller problems which means fewer levels to be explored by the B&B method. 5.1. Decomposition of the problem In order to identify whether the problem is decomposable, consider a set of constraints, say B, which can be satisÿed by a certain set of variables, say XB . If the decision to choose which variable in the set XB to satisfy a constraint in B does not impact the decision on the rest of the problem, then we can extract this part from the whole problem. Let k be the number of sub-problems, 16k6kJ 00 k. Deÿne

= {Ij | j ∈ J 00 };    \  Ij 6= ∅ ;

l = Ij  00  j∈J

l ∩ l0 = ∅;

0

l 6= l ;

= 1 ∪ 2 ∪ · · · ∪ k ; l =

Y

Ij ;

(19) l = 1; : : : ; k;

(20) (21) (22) (23)

Fig. 1. The decomposed problem.

In this way, l contains sets of Ij ’s which have some element(s) in common and we can decompose the original problem into k sub-problems. I (l) and J (l) correspond to sets of indices of variables and constraints, respectively, on which the B&B method is performed for sub-problem l. Example 1. Suppose J 00 = {5; 6; 8; 9}; I5 = {2; 3; 4}; I6 = {1; 7}; I8 = {7; 8}; I9 = {3; 5; 6}. From Fig. 1, the original problem can be decomposed into two sub-problems, with 1 = {I5 ; I9 } and

2 = {I6 ; I8 }. 1 = I5 × I9 ; 2 = I6 × I8 . I (1) = {2; 3; 4; 5; 6}; J (1) = {5; 9} and I (2) = {1; 7; 8}; J (2) = {6; 8}. Note that the cardinality of the solution set is reduced from 3 × 2 × 2 × 3 = 36 to 3 × 3 + 2 × 2 = 13. 5.2. 0–1 Integer program and branch-and-bound method Without loss of generality, we only have to consider the sub-problems in the form of problem (11) with every ci ¿0. For simplicity, we refer I (l) and J (l) by I and J , respectively. Corollary 1 implies that XÄ (A; b) ⊆ F(). Therefore, solving problem (11) is equivalent to ÿnding an f∗ ∈  such that m X i=1

ci0 Fi (f∗ ) = min f∈

 (24)

J (l) = { j | Ij ∈ l }:

(25)

( m X

) ci0 Fi (f)

:

(26)

i=1

Recall the deÿnition of Ij in (7). We deÿne variables for a 0 –1 integer program

Ij ∈ l

I (l) = {i | i ∈ Ij ; Ij ∈ l };

513

xij =

1 if i is chosen from Ij ; 0 otherwise;

∀i ∈ I; j ∈ J: (27)

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Problem (11) can be transformed into the following 0 –1 integer programming problem.   m  X bj ci0 max · xij minimize Z0 = j∈J aij i=1 m X xij = 1; ∀j ∈ J; subject to (28) i=1

xij = 0 or 1; ∀i ∈ I; j ∈ J; xij = 0;

∀i; j with i 6∈ Ij :

The constraints of problem (28) require that for each j ∈ J , there exists exactly one i ∈ Ij , with xij = 1, to satisfy constraint j of the original problem. The condition of xij = 1; i ∈ Ij , implies fj = i. Therefore, f = ( f1 ; : : : ; fn ) is a feasible solution of problem (28). The non-binding variables are set to zero so that the problem is minimized. An optimal solution obtained from solving problem (28) yields an f∗ ∈  for problem (26). Consequently, this f∗ provides an optimal solution for problem (11). In order to solve problem (28), we apply the concept of branch-and-bound approach [15]. We can start from any constraint and then branch to the next level. Branching to the next level corresponds to adding another constraint which in turn tightens the feasible domain. The worst case scenario of the B&B method is that we enumerate all possible solutions. However, after evaluating some nodes, we, in general, can eliminate some branches. For example, if the solution of the partial branch we are evaluating is larger than the best solution obtained from a complete branch, then we can stop evaluating the current branch. 5.3. An algorithm Based on the concepts discussed before, we present an algorithm for ÿnding an optimal solution of problem (3). Algorithm 1 (Algorithm for ÿnding an optimal solution). Step 1: Find the maximum Vn solution of (1). Compute xˆ = A ˜ b =[ j=1 (aij ˜ bj )]i∈I according to (4) Step 2: Check feasibility. If xˆ ◦ A = b, continue. Otherwise, stop! X (A; b) = ∅ and problem (3) has no feasible solution.

Step 3: Compute index sets. Compute Ij = {i ∈ I | xˆi · aij = bj }; ∀j ∈ J , which represents a set of xi ’s that can satisfy constraint j of the fuzzy relation equations. Step 4: Arrange cost vector. Deÿne c0 and c00 according to (13). Step 5: Perform problem reduction. Compute Iˆ = {i ∈ I | ci 60} and Jˆ = { j ∈ J | xˆi · aij = bj ; i ∈ Iˆ}. Eliminate row i; i ∈ Iˆ and column j; j ∈ Jˆ from matrix A to obtain A0 . Also, eliminate the jth element, j ∈ Jˆ, from vector b to obtain b0 . Assign an optimal value xi∗ = xˆi , for i ∈ Iˆ. If b0 is empty, assign zero to unassigned xi∗ and go to Step 11. Otherwise, compute J 0 = J \Jˆ and proceed to the next step. Step 6: Find singleton Ij . Compute I = {i ∈ Ij | kIj k = 1; j ∈ J 0 } and J = { j ∈ 0 J | xi · aij = bj ; i ∈ I}. Eliminate row i; i ∈ I and column j; j ∈ J from matrix A0 to obtain A00 . Also, eliminate the jth element, j ∈ J, from vector b0 to obtain b00 . Assign xi∗ = bj =aij , for i ∈ I and i ∈ Ij . If b00 is empty, assign zero to unassigned xi∗ and go to Step 11. Otherwise, compute J 00 = J 0 \J and proceed to the next step. Step 7: Decompose the problem. Decompose the problem by computing equations (19) – (25). Step 8: Deÿne sub-problems. For each sub-problem l, deÿne problem (11) and its corresponding 0 –1 integer program using (28). Step 9: Solve the integer program(s). Solve each integer program by using the branchand-bound method. Step 10: Generate an optimal solution of subproblem. For each sub-problem l, deÿne fl = (fj ); j ∈ J (l) with fj = i if xij = 1. Generate F( f∗ ) via formula (10). Deÿne xÄ∗ = (xÄ∗1 ; : : : ; xÄ∗m ) with xÄ∗i = Fi (f∗ ). Step 11: Generate an optimal solution. Combine xÄ∗ with the solution obtained from (5) and (6) to yield an optimal solution of problem (3). 5.4. Example Consider the following optimization problem. minimize Z = − 4x1 + 3x2 + 2x3 + 3x4 + 5x5 + 2x6 + x7 + 2x8 + 5x9 + 6x10

J. Loetamonphong, S.-C. Fang / Fuzzy Sets and Systems 118 (2001) 509–517

subject to

from A and the jth element, j ∈ Jˆ, from b. Consequently, we obtain

x ◦ A = b; 06xi 61; i = 1; : : : ; 10;

where 

0:6  0:5   0:1   0:1   0:3 A =  0:8   0:4   0:6   0:2 0:1

0:2 0:6 0:9 0:6 0:8 0:4 0:5 0:3 0:5 0:3

0:5 0:9 0:4 0:2 0:8 0:1 0:4 0:4 0:7 0:6

0:3 0:5 0:7 0:5 0:8 0:1 0:8 0:3 0:4 0:6

0:7 0:8 0:5 0:4 0:8 0:2 0:4 0:1 0:9 0:6

0:5 0:9 0:7 0:1 0:5 0:8 0:7 0:2 0:9 0:4

515



0:2 0:3 0:4 0:7 0:5 0:8 0:3 0:5 0:7 0:4

 0:8 0:8   0:7   0:5   0:8  ; 0:3   0:4   0:7   0:2  0:8

b = (0:48 0:56 0:72 0:56 0:64 0:72 0:42 0:64): Step 1: The maximum solution of the problem is xˆ = (0:8 0:8 0:622 0:6 0:7 0:525 0:7 0:8 0:6 0:8): Step 2: xˆ ◦ A = b. Therefore, we know that the solution set is not empty. Step 3: Consider b1 and the ÿrst column of A, we have [ai1 ]i∈I = (0:6 0:5 0:1 0:1 0:3 0:8 0:4 0:6 0:2 0:1)T : Notice that if aij ¡bj , xˆi cannot satisfy constraint j. In this case, we do not have to consider this xˆi for Ij . Hence, the ÿrst column of A is reduced to [ai1 ]i∈{1; 2; 6; 8} = ( 0:6 0:5 0:8 0:6 )T ; and xˆi · [ai1 ]i∈{1; 2; 6; 8} = (0:8 · 0:6 0:8 · 0:5 0:525 · 0:8 0:8 · 0:6)T = ( 0:48 0:4 0:42 0:48 )T : Since x1 · a11 = x8 · a81 = b1 = 0:48, therefore, we obtain I1 = {1; 8}. Similarly, I2 to I8 can be obtained by applying the same procedure. We have I2 = {3; 5}; I3 = {2}; I4 = {5; 7}; I5 = {2}; I6 = {2}; I7 = {4; 6; 9}; I8 = {1; 2; 10}. Step 4: c0 = (0; 3; 2; 3; 5; 2; 1; 2; 5; 6); c00 = (−4; 0; 0; 0; 0; 0; 0; 0; 0; 0). Step 5: Perform problem reduction. Iˆ = {1}; Jˆ = {1; 8}. We eliminate row i ∈ Iˆ and column j ∈ Jˆ

0:6  0:9   0:6   0:8  0 A =   0:4  0:5   0:3   0:5 0:3

0:9 0:4 0:2 0:8 0:1 0:4 0:4 0:7 0:6

0:5 0:7 0:5 0:8 0:1 0:8 0:3 0:4 0:6

0:8 0:5 0:4 0:8 0:2 0:4 0:1 0:9 0:6

0:9 0:7 0:1 0:5 0:8 0:7 0:2 0:9 0:4

 0:3 0:4   0:7   0:5   0:8  ; 0:3   0:5   0:7  0:4

b0 = ( 0:56 0:72 0:56 0:64 0:72 0:42 ); x1∗ = 0:8 and J 0 = J \Jˆ = {2; 3; 4; 5; 6; 7}. Step 6: Find singleton Ij . I = {2}; J = {3; 5; 6}. We eliminate row i ∈ I and column j ∈ J from A0 and the jth element, j ∈ J, of b0 . Therefore, we obtain 

0:9  0:6   0:8   0:4 A0 =   0:5   0:3   0:5 0:3

0:7 0:5 0:8 0:1 0:8 0:3 0:4 0:6

 0:4 0:7   0:5   0:8  ; 0:3   0:5   0:7  0:4


b0 = 0:56 0:56 0:42 ; x2∗ = 0:8 and J 00 = J 0 \J = {2; 4; 7}. Step 7: Decompose the problem. From J 00 = {2; 4; 7}, we have I2 = {3; 5}; I4 = {5; 7}; I7 = {4; 6; 9}. The problem can be decomposed into 2 subproblems with 1 = {I2 ; I4 }; 2 = {I7 }; 1 = I2 × I4 ; 2 = I7 ; I (1) = {3; 5; 7}; I (2) = {4; 6; 9}; J (1) = {2; 4}; J (2) = {7}. The decomposed problem is shown in Fig. 2. Step 8: Deÿne sub-problems. (a) Sub-problem I. x(1) = [xi ]; A00 = [aij ] and 00 b = [bj ], where i ∈ I (1) and j ∈ J (1) . minimize

Z10 = 2x3 + 5x5 + x7

subject to

x(1) ◦ A00 = b00 ; 06xi 61;

i ∈ I (1) :

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J. Loetamonphong, S.-C. Fang / Fuzzy Sets and Systems 118 (2001) 509–517

Fig. 2. The decomposed problem of the example.

Its corresponding 0–1 integer program is   bj · x3j minimize Z10 = 2 max j∈J (1) a3j   bj · x5j + 5 max j∈J (1) a5j   bj · x7j + max j∈J (1) a7j subject to

Fig. 3. Branch-and-bound for sub-problem I.

x32 + x52 = 1; x54 + x74 = 1; xij = 0 or 1;

Fig. 4. Branch-and-bound for sub-problem II.

∀i ∈ I

(1)

; j∈J

(1)

:

(b) Sub-problem II. x(2) = [xi ]; A00 = [aij ] and b00 = [bj ] where i ∈ I (2) and j ∈ J (2) . minimize

Z20 = 3x4 + 2x6 + 5x9

subject to

x(2) ◦ A00 = b00 ; 06xi 61;

i ∈ I (2) :

Its corresponding 0–1 integer program is   bj · x4j minimize Z20 = 3 max j∈J (2) a4j   bj · x6j + 2 max j∈J (2) a6j   bj · x9j + 5 max j∈J (2) a9j subject to

x47 + x67 + x97 = 1; xij = 0 or 1;

∀i ∈ I (2) ; j ∈ J (2) :

Step 9: Solve the integer programs. (a) Sub-problem I. Suppose we start the B&B from the ÿrst constraint. We choose i from I2 = {3; 5}.

Either x32 or x52 has to be 1. This yields nodes 1 and 2. If x32 = 1, then x3 = b2 =a32 = 0:56=0:9 = 0:622. Therefore, the lower bound of node 1 is 2 × 0:622 = 1:244. Also, if x52 = 1, then x5 = b2 =a52 = 0:56=0:8 = 0:7. And, the lower bound of node 2 is 5 × 0:7 = 3:5. From node 1, we can branch further to either node 3 or node 4 with x54 or x74 = 1, respectively. This is equivalent to adding another constraint. Since this added constraint is the last one, we obtain the exact objective values, instead of the lower bounds. The objective value of node 3 is calculated by 1:244+5 × (0:56=0:8) = 4:744. Similarly, the objective value of node 4 is calculated by 1:244 + (0:56=0:8) = 1:944. Since, Z10 at node 4 is smaller than the lower bound of node 2, we can stop branching to node 2. Moreover, Z10 at node 4 yields the optimal value. Fig. 3 shows the B&B of this sub-problem. (b) Sub-problem II. For sub-problem II, we have only 3 nodes to branch, namely, node 1, 2, and 3. The result is shown in Fig. 4. The optimal value is 1.05 obtained from node 2 with x67 = 1. Step 10: Generate an optimal solution of subproblem. The optimal solution for sub-problem I is x3∗ = 0:622; x7∗ = 0:7; x5∗ = 0. And, the optimal so-

J. Loetamonphong, S.-C. Fang / Fuzzy Sets and Systems 118 (2001) 509–517

lution for sub-problem II is x6∗ = 0:42=0:8 = 0:525; x4∗ = 0; x9∗ = 0. Step 11: Generate an optimal solution. Combine all the results. The optimal solution x∗ is (0:8; 0:8; 0:622; 0; 0; 0:525; 0:7; 0; 0; 0) with an optimal value of 2.194. 6. Conclusions In this paper we have studied the solution set of fuzzy relation equations with max-product composition and an optimization problem with a linear objective function subject to such fuzzy relation equations. By identifying the special properties of the feasible domain, we can determine an optimal solution without explicitly generating the whole set of minimal solutions. The optimization problem can be decomposed into two problems, one with non-negative cost coefÿcients and the other with negative cost coecients. The problem can be further reduced by eliminating the constraints that either have been taken care of or are irrelevant. The reduced problem may be further decomposed into sub-problems which are then solved separately. These sub-problems are transformed into 0 –1 integer programs and solved by the branch-andbound method. Since the levels of the branch-andbound are reduced by the previous decomposition, the procedure is very ecient in the sense that we do not have to enumerate all the possible solutions of the original problem. References [1] G.I. Adamopoulos, C.P. Pappis, Some results on the resolution of fuzzy relation equations, Fuzzy Sets and Systems 60 (1993) 83–88. [2] R.E. Bellman, L.A. Zadeh, Local and fuzzy logics, in: J.M. Dunn, D. Epstein (Eds.), Modern Uses of Multiple Valued Logic, Reidel, Dordrecht, 1977, pp. 103–165.

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