Maximizing a monomial geometric objective function subject to bipolar max-product fuzzy relation constraints

Abstract

In this paper, the problem of maximizing a monomial geometric objective function subject to bipolar max-product fuzzy relation constraints is studied. First of all, it is shown that the bipolar max-product Fuzzy Relation Inequality (FRI) system can equivalently be converted to a bipolar max-product Fuzzy Relation Equation (FRE) system. Hence, the structure of feasible domain of the problem is determined in the case of the bipolar max-product FRE system. It is shown that its solution set is non-convex, in a general case. Some sufficient conditions are proposed for solution existence of its feasible domain. An algorithm is designed to solve the optimization problem with regard to the structure of its feasible domain and the properties of the objective function. Its importance is also illustrated by an application example in the area of economics and covering problem. Some numerical examples are given to illustrate the above points.

Keywords

Bipolar fuzzy relation equation max-product composition non-convex optimization bipolar variables

1 Introduction

The main idea of FRE is based on the composition of fuzzy binary relations. FRE and FRI have been studied by many researchers both from a theoretical standpoint and in view of application since they were first investigated by Sanchez [46]. FRE and FRI have been applied in many areas such as fuzzy optimization, fuzzy decision-making, fuzzy symptom diagnosis, and fuzzy medical diagnosis [13 , 37]. The most important type of FRE is those with max-T composition for finite scenarios, where T is a triangular norm. The solvability criteria of max-min equations were first proposed by Sanchez [46]. They were then extended in [34, 38]. Its structure of solution set was first determined by Sanchez [47]. It was then extended to max-T FRE by Di Nola et al. [14]. The complete solution set of a consistent finite system of max-T FRE can be determined by a maximum solution and a finite number of minimal solutions. The computation of the maximum solution is not difficult. It can be applied to check the consistency of system. However, the detection of all the minimal solutions is equivalent to the covering problem. Hence, it is an NP-hard problem. Various approaches have been designed to compute the minimal solutions [23, 42]. Some of the approaches are based on a type of quasi-characteristic matrix [39, 40]. Some rule-based methods were proposed in [6, 7]. Lichun and Boxing [26] designed an algebraic method to compute all minimal solutions. Luoh et al. [27] applied the matrix pattern to calculate graphically the minimal solutions. The algebraic method was improved by Peeva [43]. Allame and Vatankhahan [8] presented a new iteration algorithm. An algorithm was proposed based on the concept of a fuzzy determinant [9]. Recently, Lichun and Boxing’s method [26] was improved by Yeh [58]. A comprehensive review about analytical approaches of FRE resolution can be found in [15]. Various estimates of the number of the minimal solutions can be considered in [37 , 53]. Good overviews about theory and applications of FRE can also be found in [21 , 35].

The problem of minimizing (or maximizing) a linear objective function subject to a consistent system of max-min FRE was first appeared in the textile industry and investigated by Fang and Li [18]. This problem can be decomposed into two subproblems by separating the nonnegative and negative coefficients of the objective function with the same constraints. The second subproblem obtains its optimum at the maximum solution. The first subproblem is solved by the branch-and-bound method. This problem was improved by Wu et al. [54, 55] by providing the proper upper bounds for the branch-and-bound procedure. Loetamonphong and Fang [30] studied the problem with the max-product composition. They applied the similar idea to [18] for resolution of the problem. Some other generalizations on this issue can be found in [2 , 51]. The linear function optimization problem with max-min FRI was investigated by Zhang et al. [65]. An algorithm was also proposed to accelerate the resolution of this problem by Guo and Xia [22]. However, the fuzzy relation programming with a nonlinear objective function has been developing very slowly. It is difficult to solve this problem using traditional nonlinear optimization methods. The initial research on the nonlinear objective function optimization problem with FRE constraints can be found in [25]. Lu and Fang [25] designed a genetic algorithm to solve nonlinear optimization problems subject to a system of max-min equations. The optimization problems have been studied with the different objective functions in literature [3 , 56]. The geometric programming theory was firstly proposed by Zeneret et al. [17, 44] in 1961. This problem has many applications in different areas such as environmental engineering, business administration, economics, and resource allocation [68]. The fuzziness occurring in the geometric programming problem is categorized as fuzzy geometric programming problem. This problem was investigated by Cao [11]. This problem was applied in problems of power systems [12] and economic management [32]. Biswal [10] and Verma [52] also considered the fuzzy geometric programming problem with multi-objective functions. Further considering the wide applicability of geometric programming, Cao [11] proposed fuzzy geometric programming. Yang and Cao [59 –62] studied another variant of the model, called a fuzzy relational geometric programming subject to max-min and max-product FRE. They decomposed the problem into two subproblems with the negative and nonnegative exponents in the objective function. The first sub-problem is easily solved by the maximum solution and the second subproblem is solved by one of the minimal solutions. The monomial geometric programming problem with FRI constraints was considered by Shivanian and Khorram [45]. The objective function of the problem is a monomial in x₁, …, x_n, i.e., it is a product of form $c \times x_{1}^{α_{1}} \times x_{2}^{α_{2}} \times \dots \times x_{n}^{α_{n}}$ , where α_j ∈ R, j = 1, …, n, and c > 0. With regard to the structure of the problem, the quasi-minimal solutions and the maximum solution of its feasible domain guarantee the resolution of the problem. Their proposed algorithm is based on the structure and the pairwise comparison of quasi-minimal solutions. Wu [57] also studied a geometric objective function subject to max-min fuzzy relational equations. He gave a reduction procedure for solving the problem. Zhou and Ahat [66] considered a geometric programming problem with the max-product fuzzy relational equation constraints. They designed an efficient algorithm to find its optimal solution. Singh et al. [49] studied a geometric objective function subject to a system of max-min fuzzy relation equations. Singh et al. [49] suggested a two-stage procedure to find the optimal solution for the geometric optimization problem. The above mentioned fuzzy relation compositions create FRE or FRI constraints which are increasing with respect to each variable. In some application problems such as economic and covering problems, it is necessary to use variables with a bipolar character [19 , 33]. Freson et al. [19] first investigated the system of bipolar max-min fuzzy relation equations. They also studied the associated linear minimization problem with a potential application of product public awareness in revenue management. They showed that the solution set of the system can be characterized by a finite set of maximal and minimal solution pairs. Li and Jin [31] also showed that the determination of its consistency is NP-complete. Consequently, the resolution of the optimization problem is NP-hard. The max-min operator is usually applied when a system needs conservative solutions such that the goodness of one value cannot compensate the badness of another value. In fact, there exist cases 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 [16 , 67]. Some outlines for choosing an appropriate operator has been presented by Yager [64]. Hence, the study of the system of bipolar max-product fuzzy relation equations or inequalities is motivated. On the other hand, the nonlinear objective function optimization with bipolar FRE and FRI have not been investigated up to now. In this paper, maximizing a monomial geometric objective function with bipolar max-product fuzzy relation constraints is studied. The motivation behind it is an economics and covering problem which is considered in Section 2. The associated geometric programming problem is designed to minimize the investment risk. First of all, it is illustrated how converting the bipolar inequality system to the bipolar equality system. Then the structure of the complete solution set of the bipolar system with the max-product composition is investigated. Also, some sufficient conditions for solution existence of the problem are proposed. An algorithm is designed for resolution of the geometric programming problem with the constraints of the bipolar max-product fuzzy relation equations or inequalities based on the obtained results from the properties of its feasible domain. An application example is expressed in the area of economics and covering problem in the real world. Its main results and contributions are briefly as follows: 1 - The problem with inequality constraints is equivalently converted to a problem with equality constraints. Hence, it is enough to study the problem in the case of equality constraints. 2 - Some sufficient conditions are given for solution existence for the problem. 3 - The complete solution set of its feasible domain is determined. 4 - An algorithm is designed to solve the problem with regard to the structure of the feasible domain and the properties of the objective function. 5 - An application example in the area of economics and covering problem is presented to illustrate the problem. The advantages of the model proposed in this paper with respect to the previous models in fuzzy relation programming [2–5 , 66] are as follows: The given models cannot applied for the real world problems containing variables with a bipolar character. Also, the optimization models with bipolar variables are an extension from the above models.

2 An application example

Consider a city with six educational zones as it has been presented in [1]. A schoolmaster decides to cover six zones 1-6 by enhancing the educational quality of his school (A). He considers five criteria below to convince the parents to select school A. (1) The quality of cultural activities. (2) The quality of laboratories of school. (3) The quality of the athletic-recreational facilities (such as playground, pool, etc.). (4) The educational quality of school A. (5) The quality of cleanliness of school A. He evaluates the quality of the athletic-recreational facilities of school A versus the schools in zones 1-6, separately. Also, the schoolmaster has some plans for each potentially poor criterion. By considering criteria 1-5, he can categorize the parents’ requirements in the five classes: (1) the problem of the cultural activities of students, (2) the problem of equipping laboratories, (3) the athletic-recreational activities of the students, (4) the educational problems of the students, and (5) the cleanliness problem of school A. Now, suppose that $a_{ij}^{+}$ denotes the required quality level of criterion (i) from viewpoint of students’ parents in zone j. This matrix for the six zones and the five criteria is as follows: $A^{+} = (\begin{matrix} 0.12 & 0.25 & 0.3 & 0.36 & 0.24 & 0.88 \\ 0.15 & 0.22 & 0.27 & 0.45 & 0.9 & 0.34 \\ 0.45 & 0.38 & 0.9 & 0.34 & 0.4 & 0.3 \\ 0.25 & 0.39 & 0.75 & 0.26 & 0.36 & 0.25 \\ 0.45 & 0.37 & 0.28 & 0.822857 & 0.45 & 0.4 \end{matrix}) .$ The schoolmaster estimates if he expends cost x_j (x_j has been normalized in [0,1]) to overcome the requirement from kind of (i), then he will obtain quality level $a_{ij}^{+} . x_{j}$ from viewpoint of the parents in zone (j) from criterion (i). On the other hand, the schoolmaster receives the costs from students’ parents. Increasing the costs, the satisfactory of parents will decrease to $a_{ij}^{-} . (1 - x_{j})$ , where $a_{ij}^{-}$ denotes the satisfactory level of parents from criterion (i) from viewpoint of students’ parents in zone (j). The satisfactory levels $a_{ij}^{-}$ are denoted in terms of the matrix A^- as follows: $A^{-} = (\begin{matrix} 0.1 & 0.4 & 0.3 & 0.4 & 0.8 & 0.3 \\ 0.24 & 0.16 & 0.38 & 0.46 & 0.9 & 0.56 \\ 0.4 & 0.35 & 0.42 & 0.33 & 0.24 & 0.27 \\ 0.36 & 0.25 & 0.39 & 0.2 & 0.1 & 0.4 \\ 0.96 & 0.4 & 0.35 & 0.26 & 0.27 & 0.5 \end{matrix}) .$ Also, the schoolmaster estimates at least levels b_i, for i = 1, …, 5, such that if he fulfills at least quality and satisfactory levels b_i, i = 1, …, 5, for criterion (i) at least for students’ parents of one of zones, then he will overcome the difficulties in the requirements from kind of (i). Vector b is as follows: $b = {(\begin{matrix} 0.528 & 0.594 & 0.54 & 0.45 & 0.576 \end{matrix})}^{T} .$ The range of costs that the schoolmaster should pay to fulfill the quality levels b_i, for i = 1, …, 5, is obtained by solving bipolar fuzzy relation equations $A^{+} \circ x \lor A^{-} \circ \bar{x} = b$ , where “∘” and “⋁” denote the operators of max-product and maximum, respectively. Moreover, if the schoolmaster expends cost x_j for zones j = 1, …, 6, to overcome the requirement from kind of (i), i = 1, …, 5, in zone (j), then the investment risk is obtained by the function $z (x) = x_{1}^{3} \times x_{2} \times x_{3}^{2} \times x_{4} \times x_{5}^{2} \times x_{6}^{4}$ . Now, the schoolmaster wants to minimize the investment risk subject to system A⁺ ∘ x ∨ A^- ∘ x = b and x ∈ [0, 1] ⁶. Equivalently, he can maximize $z {(x)}^{- 1} = x_{1}^{- 3} \times x_{2}^{- 1} \times x_{3}^{- 2} \times x_{4}^{- 1} \times x_{5}^{- 2} \times x_{6}^{- 4}$ subject to system A⁺ ∘ x ∨ A^- ∘ x = b and x ∈ [0, 1] ⁶. This problem motivates us to propose an approach to solve such problems. This problem will be solved in Section 6.

Bipolar and unipolar FRE, FRI, and the optimization problems related to them have been applied in many areas such as textile industry, data transmission mechanism in BitTorrent-like Peer-to-Peer file sharing systems [63], covering and economics problems, the product public awareness in revenue management [19], and so on. To find a comprehensive review from the practical problems in this area, the readers can refer to [51]. These problems cannot formulate by other known mathematical models. These models cannot be solved by the traditional method. Hence, it is necessary to propose new methods to solve them. Therefore, the proposed method in this paper can help us to find the bestscenario.

3 Monomial geometric optimization with bipolar max-product fuzzy relation constraints

This section is divided to four subsections. In the first subsection, the monomial geometric optimization problem with bipolar max-product fuzzy relation constraints is formulated. The bipolar max-product FRI is converted to the bipolar max-product FRE in the second subsection. In the third subsection, its objective function is converted to an increasing function. In the fourth subsection, the monomial geometric optimization problem with bipolar max-product FRE constraints isdiscussed.

3.1 Problem formulation

The monomial geometric optimization problem with bipolar max-product fuzzy relation constraints is formulated as follows: $Max z (x) = c \prod_{j = 1}^{n} x_{j}^{α_{j}},$ (1) $\begin{matrix} s . t . max_{j = 1, \dots, n} max (prod (a_{ij}^{+}, x_{j}), prod (a_{ij}^{-}, {\bar{x}}_{j})) *_{i} b_{i}, \\ i = 1, \dots, m, \\ 0 \leq x_{j} \leq 1, j = 1, \dots, n, \end{matrix}$ where $a_{ij}^{+}$ , $a_{ij}^{-}$ , b_i ∈ [0, 1], c > 0, α_j ∈ R, $*_{i} \in {\geq, =, \leq}, {\bar{x}}_{j} = 1 - x_{j},$ and prod (a, b) = a × b. Moreover, the notation $\prod_{j = 1}^{n} x_{j}^{α_{j}}$ denotes $x_{1}^{α_{1}} \times \dots \times x_{n}^{α_{n}}$ . The problem can equivalently rewrite as follows:

$\begin{matrix} Max z (x) = c \times x_{1}^{α_{1}} \times . . . \times x_{n}^{α_{n}}, \\ s . t . max {max (prod (a_{11}^{+}, x_{1}), prod (a_{11}^{-}, {\bar{x}}_{1})), \dots, \\ max (prod (a_{1 n}^{+}, x_{n}), prod (a_{1 n}^{-}, {\bar{x}}_{n}))} \\ {\geq or = or \leq} b_{1}, \\ . \\ . \\ . \\ max {max (prod (a_{m 1}^{+}, x_{1}), prod (a_{m 1}^{-}, {\bar{x}}_{1})), \dots, \\ max (prod (a_{mn}^{+}, x_{n}), prod (a_{mn}^{-}, {\bar{x}}_{n}))} \\ {\geq or = or \leq} b_{m}, \\ 0 \leq x_{j} \leq 1, j = 1, \dots, n . \end{matrix}$ (2)

A vector x satisfying the constraints is called a feasible vector. A feasible vector is called a maximizer, if z (x) is maximum. Its corresponding value z (x) is the maximum value of objective function. If all $a_{ij}^{-}$ ’s equal to zero, then the problem reduces to the max-product fuzzy relation programming problem [45, 60, 66].

The logic of this paper for resolution of problem (2) is as follows: First of all, it is shown that the inequality constraint as ≤ or ≥ can equivalently be converted to the equality constraint. Therefore, the problem (2) can be considered as an optimization problem with the equality constraints, in a general case. The optimization problem is firstly studied in a special case, i.e., with unipolar max-product FRE constraints, in Section 4. Then the optimization problem with bipolar max-product FRE constraints is considered in Section 5. This problem is firstly investigated with a single bipolar max-product FRE constraint in Subsections 5.1 and 5.2. This constraint is decomposed into two systems of unipolar max-product fuzzy relation constraint in Subsection 5.2. The solution sets of the systems is determined by the obtained results in Section 4. Then the system of bipolar max-product FRE constraints is investigated in Subsection 5.3. It can be taken intersection of the solution sets of m single bipolar max-product FRE constraints. Hence, the solution set of the system can be obtained. With regard to the structure of the solution set, an algorithm is designed for finding the optimal solution of the optimization problem with bipolar FRE constraints in Subsection 5.4.

3.2 Converting inequality constraints to equality constraints

The bipolar FRI constraints can be converted to bipolar FRE constraints below.

(i) Lower bound constraints: A lower bound constraint is expressed as follows: $max_{j = 1, \dots, n} max (prod (a_{j}^{+}, x_{j}), prod (a_{j}^{-}, {\bar{x}}_{j})) \geq b .$ (3) The relation (3) can be rewritten as follows:

$\begin{matrix} min (max_{j = 1, \dots, n} max (prod (a_{j}^{+}, x_{j}), prod (a_{j}^{-}, {\bar{x}}_{j})), b) \\ = b, \end{matrix}$ (4) or equivalently, it is concluded that

$\begin{matrix} max_{j = 1, \dots, n} max (min (prod (a_{j}^{+}, x_{j}), b), \\ min (prod (a_{j}^{-}, {\bar{x}}_{j}), b)) = b . \end{matrix}$ (5)

(ii) Upper bound constraints: An upper bound constraint is expressed as follows: $max_{j = 1, \dots, n} max (prod (a_{j}^{+}, x_{j}), prod (a_{j}^{-}, {\bar{x}}_{j})) \leq b .$ (6)

The relation (6) can be rewritten as follows:

$\begin{matrix} max (max_{j = 1, \dots, n} max (prod (a_{j}^{+}, x_{j}), prod (a_{j}^{-}, {\bar{x}}_{j})), b) \\ = b, \end{matrix}$ (7) or equivalently, it is concluded that

$\begin{matrix} max_{j = 1, \dots, n} max (max (prod (a_{j}^{+}, x_{j}), b), \\ max (prod (a_{j}^{-}, {\bar{x}}_{j}), b)) = b . \end{matrix}$ (8) The analysis of the solution set of Equations (5) and (8) is similar to the analysis of bipolar FRE. Therefore, it can be focused on the optimization problem (2) with the equality constraints and investigated to the feasible domain of problem (2) in the equality case.

3.3 Converting the objective function to an increasing function

The monomial geometric objective function z (x) is clearly increasing or decreasing in each of its arguments. The optimization problem would be solved more simple when the type of monotonicity would be the same with respect to each argument. Let A_< ={ j|α_j < 0 } and A_≥ ={ j|α_j ≥ 0 }. Then the objective function can be rewritten as follows:

$\begin{matrix} z (x) & = & c \prod_{j = 1}^{n} {x_{j}}^{α_{j}} = c \prod_{j \in A_{\geq}} {x_{j}}^{α_{j}} \times \prod_{j \in A_{<}} {x_{j}}^{α_{j}} \\ = & c \prod_{j \in A_{\geq}} {x_{j}}^{α_{j}} \times \prod_{j \in A_{<}} {(1 - {\bar{x}}_{j})}^{α_{j}} . \end{matrix}$ (9)

Consider the change of variables as follows: $y_{j} = {\begin{matrix} x_{j} j \in A_{\geq}, \\ {\bar{x}}_{j} j \in A_{<} . \end{matrix}$ (10) Then the objective function is converted as follows: $z (y) = c \prod_{j \in A_{\geq}} {y_{j}}^{α_{j}} \times \prod_{j \in A_{<}} {(1 - y_{j})}^{α_{j}} .$ Therefore, the objective function has been converted to an increasing function with respect to y_j for j = 1, …, n. The effect of the change of variables on the constraints is rather straightforward. A constraint $max_{j = 1, \dots, n} max (prod (a_{ij}^{+}, x_{j}), prod (a_{ij}^{-}, {\bar{x}}_{j})) = b_{i},$ is trivially converted to the equivalent constraint as follows: $\begin{matrix} max ({max}_{j \in A_{\geq}} max (prod (a_{ij}^{+}, y_{j}), prod (a_{ij}^{-}, {\bar{y}}_{j})), \\ max_{j \in A_{<}} max (prod (a_{ij}^{+}, {\bar{y}}_{j}), prod (a_{ij}^{-}, y_{j}))) = b_{i} . \end{matrix}$

3.4 A related problem

In this subsection, the monomial geometric optimization problem (1) with equality constraints is formulated in the set of real numbers R asfollows:

$\begin{matrix} max z (x) = c \prod_{j = 1}^{n} {x_{j}}^{α_{j}}, \\ s . t . max_{j = 1, \dots, n} max (prod (a_{ij}^{+}, x_{j}), prod (a_{ij}^{-}, - x_{j})) = b_{i}, \\ i = 1, \dots, m, \end{matrix}$ (11) where x ∈ Rⁿ, $a_{ij}^{+}, a_{ij}^{-}, b_{i} \in R$ , and c > 0. Now, the aim is converting the problem (11) into an optimization problem of the type considered in this paper. To do this, it can be used the function Ψ : R → [0, 1] as follows: $Ψ (t) = \frac{1}{2} (1 + \frac{t}{1 + | t |}) .$ It can easily be verified that $Ψ (0) = \frac{1}{2},$ Ψ (+ ∞) = 1, Ψ (- ∞) = 0 and Ψ (t) + Ψ (- t) = 1. Moreover, Ψ is strictly increasing, i.e., for t < t′ then Ψ (t) < Ψ (t′). It is now considered the new variable as Y = Ψ (x) defined by y_j = Ψ (x_j). Moreover, let us set $a_{ij}^{+ *} = Ψ (a_{ij}^{+})$ , $a_{ij}^{- *} = Ψ (a_{ij}^{-})$ , and $b_{i}^{*} = Ψ (b_{i})$ . Now, the following problem is concluded:

$\begin{matrix} max z (y) = c \prod_{j = 1}^{n} {y_{j}}^{α_{j}}, \\ s . t . max_{j = 1, \dots, n} max (prod (a_{ij}^{+ *}, y_{j}), prod (a_{ij}^{- *}, {\bar{y}}_{j})) \\ = b_{i}^{*}, i = 1, \dots, m . \end{matrix}$ (12) It is obvious that x ∈ Rⁿ is a maximizer of z (x) subject to the original constraints if and only if Ψ (x) ∈ [0, 1] ⁿ is a maximizer of z (Ψ (x)) subject to the converted constraints.

4 Optimization with unipolar max-product FRE constraints

If all $a_{ij}^{-}$ ’s equal to zero, then the system of bipolar max-product FRE constraints is converted to a system of fuzzy relation equation constraints. Since FRE has been studied extensively, the corresponding optimization problem can be solved as follows. Let us consider the following simplified optimization problem:

$\begin{matrix} max z (x) = c \prod_{i = 1}^{n} {x_{i}}^{α_{i}}, \\ s . t . max_{j = 1, \dots, n} (prod (a_{ij}^{+}, x_{j})) = b_{i}, i = 1, \dots, m, \\ x \in {[0, 1]}^{n}, \end{matrix}$ (13) where $a_{ij}^{+}, b_{i} \in [0, 1]$ and c > 0. The non-empty feasible domain of the optimization problem (13), i.e., the solution set D, is determined by a maximum element g ∈ [0, 1] ⁿ and a finite set S ⊂ [0, 1] ⁿ of minimal elements as follows: $D = ⋃_{s \in S} [s, g] = ⋃_{s \in S} {x \in {[0, 1]}^{n} | s \leq x \leq g} .$ The construction of the maximum element g and the minimal elements in S is performed step-wisely. To do this, each constraint is considered separately. The ith constraint has a greatest solution gⁱ and a set Sⁱ of minimal solutions given by the following relation:

$\begin{matrix} g^{i} & = & {({(g^{i})}_{1}, \dots, {(g^{i})}_{n})}^{T}, where \\ {(g^{i})}_{j} & = & {\begin{matrix} 1 & if a_{ij} \leq b_{i}, \\ \frac{b_{i}}{a_{ij}} & if a_{ij} > b_{i}, \end{matrix} j = 1, \dots, n, \end{matrix}$ (14) and $S^{i} = {s_{k}^{i} | a_{ik} \geq b_{i} > 0},$ where $s_{k}^{i} = {({(s_{k}^{i})}_{1}, \dots, {(s_{k}^{i})}_{n})}^{T},$ (15) with ${(s_{k}^{i})}_{j} = \frac{b_{i}}{a_{ik}} δ_{kj},$ j = 1, …, n for a_ik ≥ b_i > 0, and $s_{k}^{i} = 0$ for a_ik = b_i = 0. Also, δ_kj is kronecker’s delta. The feasible solution set Dⁱ of the ith constraint is then given by the following relation: $D^{i} = ⋃_{s \in S^{i}} [s, g^{i}] .$ (16) Using the structure Dⁱ for each i = 1, …, m, the feasible solution set D of problem (13) is the intersection of all feasible solution sets Dⁱ, i.e., $D = ⋂_{i = 1}^{m} D^{i}$ . Hence, it is concluded that the maximum solution g of the feasible domain of problem (13) is given by the following relation: g = (g₁, …, g_n) ^T where $g_{j} = inf_{i = 1, \dots, m} {(g^{i})}_{j} for j = 1, \dots, n .$ (17) The set of minimal solutions S is given by the minimal elements of the following set:

$\begin{matrix} {t = (t_{1}, \dots, t_{n})^{T} | t_{j} = sup_{i = 1, \dots, m} (t^{i})_{j}, for \\ j = 1, \dots, n and \forall i = 1, \dots, m, t^{i} \in S^{i}} . \end{matrix}$ (18) However, there is a sufficient condition to check the feasibility of problem (13). As it is well-known checking whether g satisfies the constraints of problem (13) or not, it can be determined that D≠ ∅ or D =∅, respectively. On the other hand, a necessary condition for the feasibility of problem (13) is easily obtained. A unipolar max-product function f : [0, 1] ⁿ → [0, 1] defined by $f (x) = max_{j = 1, \dots, n} (prod (a_{j}, x_{j}))$ is obviously increasing in x_j, j = 1, …, n, and $range (f) = [0, max_{j = 1, \dots, n} a_{j}]$ . With regard to this point and the constraints of problem (13), it is concluded that $\forall i = 1, \dots, m, b_{i} \leq max_{j = 1, \dots, n} a_{ij} .$ (19) The statement (19) is a necessary condition to be non-empty D. It is possible that every constraint has a non-empty solution set Dⁱ. But it doesn’t necessitate that $⋂_{i = 1}^{m} D^{i}$ be non-empty, in a general case. Finding a maximizer for optimization problem (13) is simple. If the objective function is increasing in each of its variables, the greatest solution g is a maximizer. Also, a maximizer is to be obtained in S when the objective function is decreasing. In a general case, the problem (13) can be decomposed into two sub-problems with non-negative and negative exponents. The first sub-problem is solved by the maximum solution g and the second sub-problem is solved by one of the minimal solutions of the feasible domain of problem (13). Combining two optimal solutions of two sub-problems, one can find the optimal solution of problem (13) in a general case. Since the computation of the greatest solution g is simple with regard to relation (17), the main problem is reduced to finding the proper s ∈ S to compute the maximizer of the second sub-problem.

5 Optimization with bipolar max-product FRE constraints

Motivated by the unipolar max-product FRE constraints, we will now solve the general optimization problem (2). With regard to the mentioned points in Section 3, the problem (2) can be considered in the following form:

$\begin{matrix} Max z (x) = c \prod_{j = 1}^{n} x_{j}^{α_{j}}, \\ s . t . max_{j = 1, \dots, n} max (prod (a_{ij}^{+}, x_{j}), prod (a_{ij}^{-}, {\bar{x}}_{j})) = b_{i}, \\ i = 1, \dots, m, \\ 0 \leq x_{j} \leq 1, j = 1, \dots, n, \end{matrix}$ (20) where $a_{ij}^{+}, a_{ij}^{-}, b_{i} \in [0, 1], c \in R^{+}$ , and α_j ∈ R⁺. First of all, the feasible domain of problem (20) should be determined.

5.1 Bipolar max-product functions

First of all, the bipolar max-product function is defined as follows:

Definition 1. Suppose that two coefficient vectors $(a_{1}^{+}, \dots, a_{n}^{+})$ and $(a_{1}^{-}, \dots, a_{n}^{-})$ in [0, 1] ⁿ have been given. The function f : [0, 1] ⁿ → [0, 1] defined by $\begin{matrix} f (x) = max_{j = 1, \dots, n} max (prod (a_{j}^{+}, x_{j}), prod (a_{j}^{-}, {\bar{x}}_{j})) . \end{matrix}$ (21) Then the function f is a bipolar max-product function. If for each j = 1, …, n, $a_{j}^{-} = 0$ , the function is called a unipolar max-product function.

Now, the function f (x) of type (21) is studied. The function has a unique decomposition in terms of one-variable functions as follows: $f (x) = max_{j = 1, \dots, n} f_{j} (x_{j}),$ (22) where $f_{j} (x_{j}) = max (prod (a_{j}^{+}, x_{j}), prod (a_{j}^{-}, {\bar{x}}_{j}))$ (23) For each j = 1, …, n, the behavior of the function f_j depends on two coefficients $a_{j}^{+}$ and $a_{j}^{-}$ . The range of values f_j changes its least value until its most value. The function f_j is piecewise linear. The Fig. 1 illustrates the range of changes of f_j. Its range is as follows: $range (f_{j}) = [\frac{a_{j}^{+} \times a_{j}^{-}}{a_{j}^{+} + a_{j}^{-}}, max (a_{j}^{+}, a_{j}^{-})] .$ (24) The value $\frac{a_{j}^{-}}{a_{j}^{-} + a_{j}^{+}}$ is the value x_j of the meeting place of the two lines. The function value for this value is $\frac{a_{j}^{-} \times a_{j}^{+}}{a_{j}^{-} + a_{j}^{+}}$ . This value is the least value for the function f_j (x_j). The value of $\max (a_{j}^{+}, a_{j}^{-})$ is the most value of f_j (x_j).

Therefore, the range of function f becomes as follows:

$\begin{matrix} range (f) \\ = [max_{j = 1, \dots, n} (\frac{a_{j}^{+} \times a_{j}^{-}}{a_{j}^{+} + a_{j}^{-}}), max_{j = 1, \dots, n} (max (a_{j}^{+}, a_{j}^{-}))] . \end{matrix}$ (25) In the next subsection, the feasible domain of problem (20) will be obtained. First of all, the case of a single constraint is studied. It is then generalized to the feasible domain of problem (20).

5.2 The solution set of a single bipolar max-product fuzzy relation equality constraint

The ith equality constraint of problem (20) is considered as follows: $max_{j = 1, \dots, n} max (prod (a_{ij}^{+}, x_{j}), prod (a_{ij}^{-}, {\bar{x}}_{j})) = b_{i} .$ (26) It is concluded that a necessary and sufficient condition for existence of solution of equality (26) is as follows: $b_{i} \in [max_{j = 1, \dots, n} (\frac{a_{ij}^{+} \times a_{ij}^{-}}{a_{ij}^{+} + a_{ij}^{-}}), max_{j = 1, \dots, n} (max (a_{ij}^{+}, a_{ij}^{-}))] .$ (27) The Equation (26) can be decomposed into two systems of unipolar max-product fuzzy relation constraint as follows: ${\begin{matrix} max_{j = 1, \dots, n} (prod (a_{ij}^{+}, x_{j})) = b_{i}, \\ max_{j = 1, \dots, n} (prod (a_{ij}^{-}, {\bar{x}}_{j})) \leq b_{i}, \end{matrix}$ (28) or ${\begin{matrix} max_{j = 1, \dots, n} (prod (a_{ij}^{+}, x_{j})) \leq b_{i}, \\ max_{j = 1, \dots, n} (prod (a_{ij}^{-}, {\bar{x}}_{j})) = b_{i} . \end{matrix}$ The necessary and sufficient conditions for existence of solution are as follows: $b_{i} \leq max_{j = 1, \dots, n} a_{ij}^{+} or b_{i} \leq max_{j = 1, \dots, n} a_{ij}^{-} .$ (29) The above expression is equivalent to the upper bound condition of (27). The lower bound condition of (25) is found by the intersection of solution sets and it is not obtained by the constraints (28), directly. The constraints (28) are studied in four cases as FRE and FRI to find the feasible domain (26) as follows. To do this, the following four sets are considered: $\begin{matrix} S_{i}^{1} & = & {x \in {[0, 1]}^{n} | max_{j = 1, \dots, n} (prod (a_{ij}^{+}, x_{j})) = b_{i}}, \\ S_{i}^{2} & = & {x \in {[0, 1]}^{n} | max_{j = 1, \dots, n} (prod (a_{ij}^{+}, x_{j})) \leq b_{i}}, \\ S_{i}^{3} & = & {x \in {[0, 1]}^{n} | max_{j = 1, \dots, n} (prod (a_{ij}^{-}, {\bar{x}}_{j})) \leq b_{i}}, \\ S_{i}^{4} & = & {x \in {[0, 1]}^{n} | max_{j = 1, \dots, n} (prod (a_{ij}^{-}, {\bar{x}}_{j})) = b_{i}} . \end{matrix}$ If the sets of $S_{i}^{k}$ , k = 1, …, 4 can be determined, then the solution set of D_i of equation (26) is as follows: $D_{i} = (S_{i}^{1} \cap S_{i}^{3}) \cup (S_{i}^{2} \cap S_{i}^{4}) .$ (30) 1- The solution set of $S_{i}^{1}$ is determined by relations (14), (15), and (16) as follows: $S_{i}^{1} = ⋃_{s \in S^{i +}} [s, g^{i +}],$ (31) where gⁱ⁺ and Sⁱ⁺ are computed with regard to relations (14) and (15), respectively.

2- The set of $S_{i}^{2}$ has the same maximum solution to the set of $S_{i}^{1}$ . It has only a single minimal solution as o = (0, 0, …, 0) ^T. Therefore, it is concluded $S_{i}^{2} = [o, g^{i +}] .$ (32) 3- To obtain the solution set of $S_{i}^{4}$ , set $y_{j} = {\bar{x}}_{j}$ , where j = 1, …, n, and determine the solution set of $S_{i}^{4}$ by relations (14), (15), and (16) in terms of variable y_j as follows:⋃_s∈S^i* [s, g^i*], where g^i* and s ∈ S^i* are the maximum solution and the minimal solutions of the set ${y \in {[0, 1]}^{n} | max_{j = 1, \dots, n} (prod (a_{ij}^{-}, y_{j})) = b_{i}}$ . Considering the variable change $y_{j} = {\bar{x}}_{j}$ for j = 1, …, n, a minimal solution will be as $s^{i -} = \bar{g^{i *}}$ and a set $G^{i -} = {g | g = \bar{s}, where s \in S^{i *}}$ from maximal solutions. Therefore, the solution set of $S_{i}^{4}$ is determined as follows: $S_{i}^{4} = ⋃_{g \in G^{i -}} [s^{i -}, g] .$ (33) 4- For set $S_{i}^{3}$ , let $y_{j} = {\bar{x}}_{j}$ , for j = 1, …, n. Then it is concluded ${y \in {[0, 1]}^{n} | max_{j = 1, \dots, n} (prod (a_{ij}^{-}, y_{j})) \leq b_{i}} .$ Its solution set in terms of variable y is as [o, g^i*]. Considering the variable change, it is concluded $S_{i}^{3} = [s^{i -}, \tilde{1}]$ with $\tilde{1} = {(1, 1, \dots, 1)}^{T}$ . With regard to relation (30) and the four cases 1 - 4, it is concluded that

$\begin{matrix} D_{i} & = & ((⋃_{s \in S^{i +}} [s, g^{i +}]) \cap [s^{i -}, \tilde{1}]) \\ \cup ([o, g^{i +}] \cap (⋃_{g \in G^{i -}} [s^{i -}, g])) . \end{matrix}$ (34) The relation (34) can be simplified as follows:

$\begin{matrix} D_{i} & = & (⋃_{s \in S^{i +}} [s \lor s^{i -}, g^{i +}]) \\ \cup (⋃_{g \in G^{i -}} [s^{i -}, g \land g^{i +}]) . \end{matrix}$ (35) With regard to relation (35), the following corollary is concluded.

Corollary 5.1. (i) D_i ⊆ [s^i-, gⁱ⁺]. (ii) If s^i- ≤ gⁱ⁺ is not hold, then the solution set of D_i is empty.

Corollary 5.2. If neither of the decomposed constraints of (28) has a solution then D_i =∅.

We are now ready to study the feasible domain of the m constraints, from type of constraint (26).

5.3 The solution set of a system of bipolar max-product fuzzy relation equality constraints

It can now be considered the m constraints from type of constraint (26). Hence, the solution sets of different equations can be combined and obtained the feasible solution set of the constraints of problem (20). It is concluded that the feasible domain of problem (20) is non-convex. To find the feasible solution set of problem (20), it can be taken intersection of sets (35) for i = 1, …, m. Hence, it is concluded that a necessary condition is as: all these sets are not empty. With regard to (27), it is concluded that for each i = 1, …, m,

$\begin{matrix} b_{i} \\ \in [max_{j = 1, \dots, n} (\frac{a_{j}^{+} \times a_{j}^{-}}{a_{j}^{+} + a_{j}^{-}}), max_{j = 1, \dots, n} (max (a_{j}^{+}, a_{j}^{-}))] . \end{matrix}$ (36) The relation (27) is a necessary and sufficient condition. The relation of (36) is only a necessary condition, because it is possible that each constraint has a non-empty solution set, but they don’t have any common solutions. The feasible solution set of problem (20) can be obtained intersecting sets of D_i of (35) for i = 1, …, m. Its simplified form is as follows:

$\begin{matrix} D & = & ⋂_{i = 1}^{m} D_{i} = ⋂_{i = 1}^{m} [s^{i -}, g^{i +}] \\ \cap ((⋃_{s \in S^{i +}} [s \lor s^{i -}, g^{i +}]) \\ \cup (⋃_{g \in G^{i -}} [s^{i -}, g \land g^{i +}])) . \end{matrix}$ (37) As it is considered, the feasible solution set is bounded to a lower bound s^- and upper bound g⁺. With regard to the structure of D, it can be concluded that $s^{-} = sup_{i = 1, \dots, m} s^{i -}$ and $g^{+} = inf_{i = 1, \dots, m} g^{i +}$ . The following corollaries are the direct results of the relation (36).

Corollary 5.3. If the inequality s^- ≤ g⁺ is not hold or s^- and g⁺ be incomparable, then D =∅.

Corollary 5.4. (i) If g⁺ ∈ D, then it is the maximum solution of set D. (ii) if s^- ∈ D, then it is the minimum solution of set D.

Corollary 5.5. $D = ⋂_{i = 1}^{m} D_{i} = ⋂_{i = 1}^{m}$ ((⋃ _s∈Sⁱ⁺ [s ∨ s^i-, gⁱ⁺]) ∪ (⋃ _g∈G^i- [s^i-, g ∧ gⁱ⁺])) .

We are now ready to find an optimal solution for problem (20).

5.4 An algorithm for finding the optimal solution of problem (20)

First of all, the objective function of problem (20) is converted to an increasing function. Then a maximizer for problem (20) is found. If its feasible domain, i.e., D, equals to empty, then there is no optimal solution for the problem. If g⁺ belongs to D, it is a maximizer and z (g⁺) is a maximum value. If g⁺ is not feasible, a maximizer will be found among the maximal elements of the feasible domain. The maximal solutions of D are either g⁺ or belong to the following set: $G = {sup_{i = 1, \dots, m} g_{i} | (\forall i) (g_{i} \in G^{i -})} .$ In this method, it is firstly required to produce all G^i- and determine set G. However, it is possible to apply a more simple method based on the following observation. Any vector in (37) is built using a finite set of values, i.e., $\begin{matrix} {g^{i +}, s^{i -}} \cup G^{i -} \cup S^{i +} \\ \subseteq ({0, 1} \cup {\frac{b_{i}}{a_{ij}^{+}} | b_{i} < a_{ij}^{+}} \\ {\cup {\bar{(\frac{b_{i}}{a_{ij}^{-}})} | b_{i} < a_{ij}^{-}})}^{n} \end{matrix}$ for i = 1, …, m and j = 1, …, n . Furthermore, if B is defined as follows:

$\begin{matrix} B & = & ({0, 1} \cup ⋃_{j = 1}^{n} ⋃_{i = 1}^{m} {(\frac{b_{i}}{a_{ij}^{+}} | b_{i} < a_{ij}^{+}), \\ {(\frac{{\bar{b}}_{i}}{a_{ij}^{-}} | {\bar{b}}_{i} < a_{ij}^{-})})}^{n} \cap [s^{-}, g^{+}], \end{matrix}$ (38) then B is a sub-lattice of [0, 1] ⁿ and the feasible domain D can be characterized as a union of interval [s, g] with s, g ∈ B. Finally, it is easily seen that any maximizer of the objective function z either belongs to B or is located in between two other maximizers belonging to B. It is therefore sufficient to determine the maximizers belonging to B in order to fully describe all maximizers. To summarize, the optimization problem (20) can be solved by the following algorithm.

Algorithm 1.

Suppose that problem (20) has been given.

Check necessary conditions (36). If it is not fulfilled, then stop and D =∅.

Construct g⁺ and s^-.

Check the sufficient conditions expressed in Corollary 5.3. If it is fulfilled, then stop and D =∅.

Generate all elements of B and keep those that satisfy the constraints of problem (20). If none of them belongs to D, then stop and D =∅.

Select the elements in B ∩ D with the maximum value for the objective function. Collect them in a set M.

With regard to Subsection 3.3, the objective function can be converted to an increasing function. If none of exponents α_j equals 0, then the function becomes strictly increasing function. Therefore, if the function z is such that none of exponents α_j equal 0, the above description can be further simplified. Indeed, in this case z is strictly increasing in each of its variables and the maximizers are nothing else but the maximal elements of the feasible domain set D, or equivalently, of the set B ∩ D. The full set of maximizers is then immediately given by M.

6 Numerical examples

First of all, the application example given in Section 2 is solved. If Algorithm 1 is applied, then its optimal solution is as follows: $x^{*} = {(\begin{matrix} 0.6 & 0 & 0.6 & 0.7 & 0.66 & 0.6 \end{matrix})}^{T},$ where $s^{-} = {(\begin{matrix} 0.6 & 0 & 0.6 & 0.7 & 0.66 & 0.6 \end{matrix})}^{T}$ and g⁺ = (0.6 1 0.60.7 0.66 0.6) ^T. In the application example, the schoolmaster should expend costs x₁ = 0.6, x₂ = 0, x₃ = 0.6, x₄ = 0.7, x₅ = 0.66, x₆ = 0.6 for zones j = 1, 2, 3, 4, 5, 6, respectively, so that the constraints are satisfied and the investment risk is minimized.

Some numerical examples are now presented to illustrate the problem in different cases.

Example 1. Consider the following problem.

Max z (x) = x₁² × x₂³, $s . t . max (max (prod (0.2, x_{1}), prod (0.8, {\bar{x}}_{1})),$ $max (prod (0.7, x_{2}), prod (0.5, {\bar{x}}_{2}))) = 0.4,$ $max (max (prod (0.4, x_{1}), prod (0.2, {\bar{x}}_{1})),$ $max (prod (0.9, x_{2}), prod (0.5, {\bar{x}}_{2}))) = 0.3,$ x ∈ [0, 1] ² .

In this example, it is shown that D =∅. To do this, the following conditions are checked: $i = 1 \Rightarrow {\begin{matrix} b_{1} = 0.4 \leq max_{j = 1, 2} max (a_{1 j}^{+}, a_{1 j}^{-}) = \\ max (0.8, 0.7) = 0.8, \\ b_{1} = 0.4 \geq max_{j = 1, 2} (\frac{a_{1 j}^{+} \times a_{1 j}^{-}}{a_{1 j}^{+} + a_{1 j}^{-}}) = \\ max (0.16, \frac{0.35}{1.2}) = \frac{35}{120}, \end{matrix}$ $i = 2 \Rightarrow {\begin{matrix} b_{2} = 0.3 \leq max_{j = 1, 2} max (a_{2 j}^{+}, a_{2 j}^{-}) = \\ max (0.4, 0.6) = 0.6, \\ b_{2} = 0.3 ≱ max_{j = 1, 2} (\frac{a_{2 j}^{+} \times a_{2 j}^{-}}{a_{2 j}^{+} + a_{2 j}^{-}}) = \\ max (\frac{8}{60}, \frac{45}{140}) = \frac{45}{140} = 0.32 . \end{matrix}$ The necessary condition (36) is not hold for i = 2. Therefore, D =∅.

Example 2. Consider the following problem. $\begin{matrix} Max z (x) = {x_{1}}^{2} \times {x_{2}}^{3}, \\ s . t . max (max (prod (0.23, x_{1}), prod (0.9, {\bar{x}}_{1})), \\ max (prod (0.7, x_{2}), prod (0.4, {\bar{x}}_{2}))) = 0.6, \\ max (max (prod (0.3, x_{1}), prod (0.55, {\bar{x}}_{1})), \\ max (prod (0.4, x_{2}), prod (0.2, {\bar{x}}_{2}))) = 0.4, \\ x \in {[0, 1]}^{2} . \end{matrix}$ In this example, another case is considered where D =∅. To do this, the following conditions are checked: $i = 1 \Rightarrow {\begin{matrix} b_{1} = 0.6 \leq max_{j = 1, 2} max (a_{1 j}^{+}, a_{1 j}^{-}) = \\ max (0.9, 0.7) = 0.9, \\ b_{1} = 0.6 \geq max_{j = 1, 2} (\frac{a_{1 j}^{+} \times a_{1 j}^{-}}{a_{1 j}^{+} + a_{1 j}^{-}}) = \\ max (\frac{0.207}{1.13}, \frac{0.28}{1.1}) = \frac{0.28}{1.1} = 0.2545, \end{matrix}$ $i = 2 \Rightarrow {\begin{matrix} b_{2} = 0.4 \leq max_{j = 1, 2} max (a_{2 j}^{+}, a_{2 j}^{-}) = \\ max (0.55, 0.4) = 0.55, \\ b_{2} = 0.4 \geq max_{j = 1, 2} (\frac{a_{2 j}^{+} \times a_{2 j}^{-}}{a_{2 j}^{+} + a_{2 j}^{-}}) = \\ max (\frac{0.165}{0.85}, \frac{0.08}{0.6}) = \frac{0.165}{0.85} = 0.194 . \end{matrix}$ Hence, the following table is concluded.

With regard to Table 1, it is concluded that $g^{+} = {(1, \frac{6}{7})}^{T}$ and $s^{-} = {(\frac{1}{3}, 0)}^{T}$ . Note that s^- ≤ g⁺. The vector g⁺ does not satisfy the second constraint. It is not a feasible vector. The set B is now generated as follows: $\begin{matrix} B = {{(1, 0)}^{T}, {(1, \frac{6}{7})}^{T}, {(1, \frac{1}{3})}^{T}, {(1, \frac{8}{11})}^{T}, {(\frac{6}{7}, 0)}^{T}, \\ {(\frac{6}{7}, \frac{6}{7})}^{T}, (\frac{6}{7}, \frac{1}{3})^{T}, (\frac{6}{7}, \frac{8}{11})^{T}, {(\frac{1}{3}, 0)}^{T}, {(\frac{1}{3}, \frac{6}{7})}^{T}, \\ {(\frac{1}{3}, \frac{1}{3})}^{T}, {(\frac{1}{3}, \frac{8}{11})}^{T}, {(\frac{8}{11}, 0)}^{T}, {(\frac{8}{11}, \frac{6}{7})}^{T}, \\ {(\frac{8}{11}, \frac{1}{3})}^{T}, {(\frac{8}{11}, \frac{8}{11})}^{T}} . \end{matrix}$

None of the elements of this set satisfies both constraints and hence D =∅.

Example 3. Consider the following problem. $\begin{matrix} Max z (x) = {x_{1}}^{2} \times {x_{2}}^{3}, \\ s . t . max (max (prod (0.8, x_{1}), prod (0.75, {\bar{x}}_{1}))], \\ max (prod (0.1, x_{2}), prod (0.4, {\bar{x}}_{2}))) = 0.5, \\ max (max (prod (0.35, x_{1}), prod (0.6, {\bar{x}}_{1})), \\ max (prod (0.7, x_{2}), prod (0.19, {\bar{x}}_{2}))) = 0.4, \\ x \in {[0, 1]}^{2} . \end{matrix}$ The conditions for existence of solution are firstly checked as follows: $i = 1 \Rightarrow {\begin{matrix} b_{1} = 0.5 \leq max_{j = 1, 2} max (a_{1 j}^{+}, a_{1 j}^{-}) = \\ max (0.8, 0.4) = 0.8, \\ b_{1} = 0.5 \geq max_{j = 1, 2} (\frac{a_{1 j}^{+} \times a_{1 j}^{-}}{a_{1 j}^{+} + a_{1 j}^{-}}) = \\ max (\frac{0.6}{1.55}, \frac{0.04}{0.5}) = \frac{0.6}{1.55} = 0.3870, \end{matrix}$ $i = 2 \Rightarrow {\begin{matrix} b_{2} = 0.4 \leq max_{j = 1, 2} max (a_{2 j}^{+}, a_{2 j}^{-}) = \\ max (0.6, 0.7) = 0.7, \\ b_{2} = 0.4 \geq max_{j = 1, 2} (\frac{a_{2 j}^{+} \times a_{2 j}^{-}}{a_{2 j}^{+} + a_{2 j}^{-}}) = \\ max (\frac{0.21}{0.95}, \frac{0.133}{0.89}) = \frac{0.21}{0.95} = 0.2211 . \end{matrix}$ The necessary condition (36) is hold. Hence, the following table is concluded.

With regard to Table 2, it is concluded that $g^{+} = {(\frac{5}{8}, \frac{4}{7})}^{T}$ and $s^{-} = {(\frac{1}{3}, 0)}^{T}$ . Note that s^- ≤ g⁺. Since g⁺ satisfies both constraints, this vector is the unique maximal element of non-empty feasible domain. Since z (x) is strictly increasing, it is also the unique maximizer.

Example 4. It is now focused on an example where vector g⁺ is not a feasible vector, but the feasible domain is not empty. $\begin{matrix} Max z (x) = {x_{1}}^{2} \times {x_{2}}^{3}, \\ s . t . max (max (prod (0.5, x_{1}), prod (0.7, {\bar{x}}_{1})), \\ max (prod (0.45, x_{2}), prod (0.25, {\bar{x}}_{2}))) = 0.7, \\ max (max (prod (0.2, x_{1}), prod (0.8, {\bar{x}}_{1})), \\ max (prod (0.5, x_{2}), prod (0.4, {\bar{x}}_{2}))) = 0.8, \\ x \in {[0, 1]}^{2} . \end{matrix}$ The conditions for existence of solution are firstly checked as follows: $i = 1 \Rightarrow {\begin{matrix} b_{1} = 0.7 \leq max_{j = 1, 2} max (a_{1 j}^{+}, a_{1 j}^{-}) = \\ max (0.7, 0.45) = 0.7, \\ b_{1} = 0.7 \geq max_{j = 1, 2} (\frac{a_{1 j}^{+} \times a_{1 j}^{-}}{a_{1 j}^{+} + a_{1 j}^{-}}) = \\ max (\frac{0.35}{1.2}, \frac{0.1125}{0.7}) = \frac{0.35}{1.2} = 0.2917, \end{matrix}$ $i = 2 \Rightarrow {\begin{matrix} b_{2} = 0.8 \leq max_{j = 1, 2} max (a_{2 j}^{+}, a_{2 j}^{-}) = \\ max (0.8, 0.5) = 0.8, \\ b_{2} = 0.8 \geq max_{j = 1, 2} (\frac{a_{2 j}^{+} \times a_{2 j}^{-}}{a_{2 j}^{+} + a_{2 j}^{-}}) = \\ max (\frac{0.7}{1.7}, \frac{0.32}{1.2}) = \frac{0.7}{1.7} = 0.412 . \end{matrix}$ The necessary condition (36) is hold.

With regard to Table 3, it is concluded that g⁺ = (1, 1) ^T and s^- = (0, 0) ^T. Since g⁺ is not feasible vector, the set B is created as follows: $\begin{matrix} B = {{(0, 0)}^{T}, {(0, 1)}^{T}, {(1, 1)}^{T}, {(1, 0)}^{T}} \\ \Rightarrow B \cap D = {{(0, 0)}^{T}, {(0, 1)}^{T}} . \end{matrix}$ This problem has optimal solutions as {x|(0, 0) ^T ≤ x ≤ (0, 1) ^T} and D = [(0, 0) ^T, (0, 1) ^T]. If z (x) = x₂³, then there is a unique optimal solution x^* = (0, 1) ^T.

Example 5. Consider the following problem. $\begin{matrix} Max z (x) = 2^{x_{1}} \times 3^{x_{2}}, \\ s . t . max (max (prod (0.4, x_{1}), prod (0.9, {\bar{x}}_{1})), \\ max (prod (0.8, x_{2}), prod (0.9, {\bar{x}}_{2}))) = 0.9, \\ max (max (prod (1, x_{1}), prod (0.7, {\bar{x}}_{1})), \\ max (prod (0.8, x_{2}), prod (0.4, {\bar{x}}_{2}))) = 0.8, \\ x \in {[0, 1]}^{2} . \end{matrix}$ The conditions for existence of solution are firstly checked as follows: $i = 1 \Rightarrow {\begin{matrix} b_{1} = 0.9 \leq max_{j = 1, 2} max (a_{1 j}^{+}, a_{1 j}^{-}) = \\ max (0.9, 0.9) = 0.9, \\ b_{1} = 0.9 \geq max_{j = 1, 2} (\frac{a_{1 j}^{+} \times a_{1 j}^{-}}{a_{1 j}^{+} + a_{1 j}^{-}}) = \\ max (\frac{0.36}{1.3}, \frac{0.72}{1.7}) = \frac{0.72}{1.7} = 0.4235, \end{matrix}$ $i = 2 \Rightarrow {\begin{matrix} b_{2} = 0.8 \leq max_{j = 1, 2} max (a_{2 j}^{+}, a_{2 j}^{-}) = \\ max (1, 0.8) = 0.8, \\ b_{2} = 0.8 \geq max_{j = 1, 2} (\frac{a_{2 j}^{+} \times a_{2 j}^{-}}{a_{2 j}^{+} + a_{2 j}^{-}}) = \\ max (\frac{0.7}{1.7}, \frac{0.32}{1.2}) = \frac{0.7}{1.7} = 0.412 . \end{matrix}$ The necessary condition (36) is hold. The set B for this example is as follows: $\begin{matrix} B = {{(0, 0)}^{T}, {(0, 1)}^{T}, {(0, 0.8)}^{T}, {(1, 0)}^{T}, {(1, 1)}^{T}, \\ {(1, 0.8)}^{T}, (0.8, 0)^{T}, {(0.8, 1)}^{T}, {(0.8, 0.8)}^{T}} . \\ B \cap D = {{(0, 1)}^{T}, {(0.8, 0)}^{T}} \Rightarrow g_{1} = {(0, 1)}^{T}, \\ g_{2} = {(0.8, 0)}^{T} . \end{matrix}$

Therefore, it is concluded: $\begin{matrix} {\begin{matrix} z (g_{1}) = z ((0, 1)^{T}) = 2^{0} \times 3^{1} = 3, \\ z (g_{2}) = z ((0.8, 0)^{T}) = 2^{0.8} \times 3^{0} = 2^{0.8}, \end{matrix} \Rightarrow \\ z (g_{1}) > z (g_{2}) \Rightarrow x^{*} = g_{1} . \end{matrix}$

7 Conclusions

In this paper, the optimization problem of a geometric objective function subject to the bipolar max-product fuzzy relation constraints was studied. An application example in the area of economics and covering problem was given to express the important of the problem. The problem with inequality constraints was equivalently converted to a problem with equality constraints. Some sufficient conditions were presented for solution existence of the problem. The solution set of its feasible domain was completely determined. Then, an algorithm was designed to solve the problem with regard to the structure of the feasible domain and the properties of the objective function. The advantages of the model proposed in this paper with respect to the previous models in fuzzy relation programming were expressed. Also, the optimization models with bipolar variables are an extension from the previous models. In the future work, a data transmission mechanism in BitTorrent-like peer-to-peer file sharing system will be studied with separable functions.

Footnotes

Acknowledgments

The authors are very grateful to the anonymous reviewers for their comments and suggestions which have been very helpful in improving the paper.

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