A Single-Variable Method for Solving the Min–Max Programming Problem with Addition–Overlap Function Composition

Abstract

Min–max programming problems with addition–min constraints have been studied in the literature to model data transfer in BitTorrent-like peer-to-peer file-sharing systems. It is well known that the class of overlap functions contains various operators, including the “min” operator. The aim of this paper is to generalize the above min–max programming problem with addition–overlap function constraints. We demonstrate that this new optimization problem can be transformed into a simplified single-variable optimization problem, which makes it easier to find an optimal solution. The bisection method will be used to find this optimal solution. In addition, when the overlap function is explicitly specified, an iterative method is given to compute the optimal objective value with a polynomial time complexity. A numerical example is provided to illustrate the procedures.

1. Introduction

In fuzzy preference modeling, the relational approach provides a convenient framework for people to acquire preferences between two alternatives. In particular, the t-norm aggregation functions are widely used [1,2,3] and have been extensively studied in the field of fuzzy relational equations which model preference in fuzzy settings [4,5]. Sanchez [6] was the first paper that defined the notion of fuzzy relational equations and used the max–min composition in its model. Since then, many other operators such as “product”, “bolt”, “strictly Archimedean”, etc. [7,8], have been proposed for various applications in the literature [5]. In particular, finding criteria of solvability and characterization for the solution set of the fuzzy relational equation becomes an important task in the study [9,10]. An interesting result is that the solution set of the max–min fuzzy relational equation can be characterized by the unique maximum solution and a finite number of minimal solutions [11]. On the other hand, various average operators [12] outside the class of t-norms are also studied. For max–average fuzzy relational equations, the characterization of their solution sets is similar to that of the max–product fuzzy relational equations [13].

While the t-norms provide a convenient way of preferences between two alternatives, as pointed out in Fodor and Roubens [14], there are situations where the associative property in t-norms is not required. The overlap aggregation functions (to be defined in Section 2) which do not require associativity, have been studied in detail over the last decade. Pereira Dimuro et al. [15] introduced the notion of an additive generator pair for overlap functions, and examined how some properties required of an overlap function can be expressed in terms of its generator pair. Not only do the overlap functions have rich theoretical structures, but they are also used to model the indifference situation in the fuzzy preference modeling. Successful applications can be found in object recognition, image processing, fuzzy decision making, etc. [16,17,18]. We refer to Refs. [4,19] for the idea, deep development, and applications of the overlap functions. Note that the commonly seen “min” and “product” functions are examples of overlap functions. However, there are many other overlap functions that are not t-norms.

The min–max optimization problem has been widely utilized across various fields, in particular, engineers often consider the worst case minimization in some optimal control systems. For instance, Vinter [20] established necessary conditions for optimality in the general minimax optimal control problem, where the unknown vector parameter is defined over an arbitrary compact metric space. Their analysis focuses on an objective function that incorporates a maximum operation related to the system response time, accounting for the most unfavorable values of the unknown parameters. Furthermore, the constraints imposed must be satisfied for all possible values of these parameters. For recent developments and interesting applications in machine learning models and supply chain management, please refer to [21,22,23].

Recently, the following min–max programming problem subject to a system of addition–min fuzzy relational inequalities (FRIs) [24,25] has been proposed to study the data transmission in BitTorrent-like peer-to-peer (BT-P2P) file-sharing systems:

(1)$\begin{array}{ccc}\mathrm{Minimize}\hfill & & Z\left(x\right)=max\{x_1,\dots ,x_n\}\hfill \\ \mathrm{subject} \mathrm{to}\hfill & & g_i\left(x\right)={\displaystyle \sum _{j=1}^n}min\{a_{ij},x_j\}\ge b_i,\forall i\in \mathcal{I},\hfill \\ & & x_j\in [0,1],\forall j\in \mathcal{J},\hfill \end{array}$

where parameters $a_{ij}\in [0,1]$ and $b_i>0$ for any $i\in \mathcal{I}=\{1,2,\dots ,m\},j\in \mathcal{J}=\{1,2,\dots ,n\}$.

Li and Yang [25] provided an algorithm to find a minimal solution of FRI in the model (1). Yang et al. [26] showed that the solution set of the model (1) with addition–min FRIs is completely determined by one maximum solution and a convex set of minimal solutions. Instead of studying the resolution of addition–min FRIs, optimization problems with certain specific objective functions usually provide more meaningful significance. To minimize some costs associated with the transmission quality of BT-P2P file-sharing systems, several studies [27,28,29,30] explored the model (1) with a linear cost objective function $Z\left(x\right)={\sum }_{j\in \mathcal{J}}{c}_j{x}_j$. To consider the objective priority of data transfer, Yang et al. [31] studied the multilevel linear programming problem with an objective function $Z\left(x\right)=x_1\to x_2\to \dots \to x_n$ in the model (1). Yang et al. [32] first proposed an algorithm to generate an optimal solution with equal values for all variables. Chiu et al. [33] showed that when the constraint part of the model (1) has a solution, there is always an optimal solution that has the same value for all variables. According to this interesting property, the model (1) could be reduced to a single-variable optimization model. An analytical method and an iterative method were then proposed to obtain an optimal solution $(y^{*},\dots ,y^{*})$ for the model (1), where $y^{*}$ is the optimal value.

Since the “min” operator is an “instance” of the class of overlap functions, we now extend the min–max programming problem (1) into a “class” of optimization problems where the addition–min composition becomes the addition–overlap function composition.

(2)$\begin{array}{ccc}\mathrm{Minimize}\hfill & & Z\left(x\right)=max\{x_1,\dots ,x_n\}\hfill \\ \mathrm{subject} \mathrm{to}\hfill & & g_i\left(x\right)={\displaystyle \sum _{j=1}^n}O(a_{ij},x_j)\ge b_i,\forall i\in \mathcal{I},\hfill \\ & & x_j\in [0,1],\forall j\in \mathcal{J},\hfill \end{array}$

where parameters $a_{ij}\in [0,1]$ and $b_i>0$ for any $i\in \mathcal{I}=\{1,2,\dots ,m\},j\in \mathcal{J}=\{1,2,\dots ,n\}$. The $O(a_{ij},x_j)$ is an overlap function. An optimization problem involving a generic overlap function in its constraint part (similar to the setting in (2)) is a new research topic. The problem of minimizing a linear objective function, $Z\left(x\right)={\sum }_{j=1}^n{c}_j{x}_j$ subject to a max–overlap function fuzzy relational equation constraint was studied by Fang [34].

We will show that when the constraint part of problem (2) is solvable, there always exists an optimal solution with the same value in all its variables. In other words, the interesting result in Chiu et al. [33] holds for the optimization problem (2) with any specific operator in the class of overlap functions. To achieve this, a single-variable optimization problem is proposed to find the optimal solution with equal values in all its variables. The bisection method could be proposed to find this optimal solution.

The bisection method works for the problems using any overlap function operator in the constraint part. When the overlap function operator used in the constraint part is explicitly specific, one could propose alternative methods to utilize its properties for better computational performance. In this paper, an iterative method that yields a sequence of approximate “solutions” is provided to solve the single-variable optimization problem when a specific overlap function operator is used.

The major contributions of this paper are as follows:

• (1). To the best of our knowledge, this study is the first one to explore the optimization problem (2) that aims to minimize $Z\left(x\right)=max\{x_1,\dots ,x_n\}$ subject to constraints defined by addition–overlap functions. Such an optimization problem has not been addressed in the existing optimization literature. Since the class of overlap functions contains so many operators, our results will be valid for the min–max programming problems using any overlap function operator in their constraints. In other words, this paper has advanced the study from an “addition–min” instance to the class of “addition–overlap functions”.

• (2). The min–max programming problem, constrained by FRI utilizing the addition–min composition, has been applied to model data transfer in BitTorrent-like peer-to-peer file-sharing systems [25]. However, associativity is not always a necessary requirement in many practical scenarios. Overlap functions, which are non-associative binary aggregation functions, have found widespread use in various applications [16,17,18]. Therefore, theoretical investigations into the min–max optimization problem incorporating addition–overlap function composition could yield valuable insights for potential applications in real-world problems.

• (3). Numerous solution methodologies have been developed to address optimization problems involving systems of fuzzy relational equations or FRIs [35,36,37,38,39]. The analysis of various solution methods’ properties under different FRI formulations presents a particularly interesting area of research. To contribute to this field from a theoretical perspective, we investigate the characterization of a novel solution method within the framework of FRI constrained by addition–overlap function composition. This approach extends the traditional addition–min composition, offering potential insights into a broader class of fuzzy optimization problems.

This paper is organized as follows. Section 2 reviews key definitions and properties of the overlap function found in the literature and presents a new overlap function. Section 3 presents a single-variable optimization problem designed to generate an optimal solution to problem (2). The bisection method is proposed to solve this single-variable optimization problem. In Section 4, we propose an iterative method that yields a sequence of approximate “solutions” to the single-variable optimization problem, using the new overlap function given in Section 2 in its constraint part. A numerical example is given to illustrate the procedures. The conclusions are given in Section 5.

2. Overlap Functions

In this section, we review the relevant definitions and properties of the overlap function as found in the literature. This foundational knowledge will aid us in constructing a new overlap function.

Definition 1 

(See [11,19]). A bivariate function $T:{[0,1]}^2\to [0,1]$ is said to be the triangular norm (t-norm for short) for all $x,y,z\in [0,1]$ if it satisfies the following conditions:

(T1) Commutativity/symmetric: $T(x,y)=T(y,x)$;

(T2) Associativity: $T(x,T(y,z\left)\right)=T\left(T\right(x,y),z)$;

(T3) Monotonicity: $T(x,y)\le T(x,z)$ if $y\le z$;

(T4) Boundary conditions: $T(x,0)=0$ and $T(x,1)=x$.

Typical examples of t-norms are “minimum: $T_m(x,y)=min\{x,y\}$”; “product: $T_p(x,y)=xy$”; and “Lukasiewicz: $T_L(x,y)=max\{0,x+y-1\}$”, and so on.

Definition 2 

(See [15,19]). A bivariate function $O:{[0,1]}^2\to [0,1]$ is said to be an overlap function if it satisfies the following conditions:

(O1) $O(x,y)=O(y,x)$ for all $x,y\in [0,1]$;

(O2) $O(x,y)=0$ if and only if $xy=0$;

(O3) $O(x,y)=1$ if and only if $x=y=1$;

(O4) O is non-decreasing;

(O5) O is continuous.

It is well known that any continuous t-norm with non-trivial zero divisors is an overlap function [34]. Many common overlap functions are listed in the following examples, but they are not t-norms because the associativity (T2) is not satisfied. Most of them come from [15,19].

$O_{mM}=min\{x,y\}max\{x^2,y^2\};O_p(x,y)=x^p{y}^p, \mathrm{with} p>0,p\ne 1;$

$O_{Mid}(x,y)=xy{\displaystyle \frac{x+y}{2}}; \mathrm{and} O_S(x,y)={\displaystyle \frac{\sqrt{xy}}{\sqrt{xy}+max\{1-x,1-y\}}}.$

2.1. Some Properties of Overlap Functions

Furthermore, previous studies [4,19] have contributed some important properties that can be used to construct various overlap functions.

Corollary 1 

([19], Corollary 1). Let $O_1,\dots ,O_m$ be overlap functions and $w_1,\dots ,w_m$ be non-negative weights with ${\sum }_{i=1}^m{w}_i=1$. Then, the convex sum $O={\sum }_{i=1}^m{w}_i{O}_i$ is also an overlap function.

Theorem 1 

([19], Theorem 5). The mapping $O:{[0,1]}^2\to [0,1]$ is an overlap function if and only if

$O(x,y)={\displaystyle \frac{f(x,y)}{f(x,y)+h(x,y)}}$

for some $f,h:{[0,1]}^2\to [0,1]$ such that:

• (1) 

  $f(x,y)$ and $h(x,y)$ are symmetric.

• (2) 

  $f(x,y)$ is non-decreasing and $h(x,y)$ is non-increasing.

• (3) 

  $f(x,y)=0$ if and only if $xy=0$.

• (4) 

  $h(x,y)=0$ if and only if $xy=1$.

• (5) 

  $f(x,y)$ and $h(x,y)$ are continuous.

In this paper, taking $f(a,x)=min\{a,x\}$ and $h(a,x)=1-max\{0,x+a-1\}$, we have the construction given in Theorem 1 to obtain a new overlap function as follows:

(3)$\begin{array}{c}O(a,x)={\displaystyle \frac{min\{a,x\}}{min\{a,x\}+(1-max\{0,x+a-1\left\}\right)}}, \mathrm{for} \mathrm{all} a,x\in [0,1].\hfill \end{array}$

It is important to note that the new overlap function (3) does not belong to the class of t-norms, as it fails to satisfy the associativity property (T2). For example, we have $O(0.1,O(0.2,0.3))={\displaystyle \frac{1}{11}}$, which is not equal to $O(O(0.1,0.2),0.3))={\displaystyle \frac{1}{12}}$. This demonstrates that the overlap function (3) does not maintain the required associative property characteristic of t-norms.

2.2. Some Characteristics of the New Overlap Function (3)

To further explain the development of a novel method for solving the min–max programming problem (2) with constraints derived from the overlap function (3), we first examine the characteristics of the overlap function.

The following Corollary 2 is essential for the subsequent content.

Corollary 2. 

Assume that the function $f\left(t\right)=t(2-t)$ is defined for $0\le t\le 1$. If $0\le x_1\le x_2\le 1$, then it follows that $f\left(x_1\right)\le f\left(x_2\right)$.

Proof. 

The first derivative of the function $f\left(t\right)$ is given by $f^{\prime }\left(t\right)=2-2t$. For the interval $0\le t\le 1$, we find that $f^{\prime }\left(t\right)\ge 0$, indicating that $f\left(t\right)$ is a non-decreasing function on this interval. Consequently, if $0\le x_1\le x_2\le 1$, it follows that $f\left(x_1\right)\le f\left(x_2\right)$. □

Let $D=min\{a,x\}+(1-max\{0,x+a-1\left\}\right)$ be the denominator of (3). Analyzing the value of the overlap function $O(a,x)$ for (3), the following results can be obtained.

Lemma 1. 

For (3), there are:

• (i) 

  $O(a,0)=0$ and $O(0,x)=0$.

• (ii) 

  $O(a,1)=a$ and $O(1,x)=x$.

• (iii) 

  $D\ge 1$, and if $ax\ne 0$ and $ax\ne 1$, then $D>1$.

• (iv) 

  $O(a,x)\le min\{a,x\}$.

• (v) 

  If $0<a$ and $0<x_1\le x_2\le 1$, then ${\displaystyle \frac{O(a,x_2)}{O(a,x_1)}}\le {\displaystyle \frac{x_2}{x_1}}$.

Proof. 

(i) and (ii) are obvious.

(iii) Since $a,x\in [0,1]$, if $x+a\le 1$, then there exists $D=min\{a,x\}+1\ge 1$. In addition, if $ax\ne 0$, then $D>1$. On the other hand, if $x+a>1$, then there is $D=min\{a,x\}+2-x-a\ge 1$ and yields $D>1$ if $ax\ne 1$.

(iv) From (iii) above, $O(a,x)={\displaystyle \frac{min\{a,x\}}{D}}\le min\{a,x\}$ holds.

(v) Let $\beta ={\displaystyle \frac{O(a,x_2)}{O(a,x_1)}}$. Since $x_1\le x_2$, we will show that $\beta \le {\displaystyle \frac{x_2}{x_1}}$ for all of the following nine cases.

Case 1-1. $a\le x_1\le x_2$ and $a+x_1\le a+x_2\le 1$. In this scenario, we have $O(a,x_1)=O(a,x_2)={\displaystyle \frac{a}{a+1}}$ to obtain $\beta ={\displaystyle \frac{O(a,x_2)}{O(a,x_1)}}=1\le {\displaystyle \frac{x_2}{x_1}}$.

Case 1-2. $a\le x_1\le x_2$ and $a+x_1\le 1\le a+x_2$.

In this case, we have $O(a,x_1)={\displaystyle \frac{a}{a+1}}$ and $O(a,x_2)={\displaystyle \frac{a}{2-x_2}}$ to obtain $\beta ={\displaystyle \frac{a+1}{2-x_2}}\le {\displaystyle \frac{x_2}{x_1}}$. Because $a+x_1\le 1$, it implies that $a\le 1-x_1$. Using Corollary 2, we deduce that this result gives

${\displaystyle \frac{a+1}{2-x_2}}\le {\displaystyle \frac{x_2}{x_1}}\Rightarrow x_1(a+1)\le x_1(2-x_1)\le x_2(2-x_2).$

Case 1-3. $a\le x_1\le x_2$ and $1\le a+x_1\le a+x_2$.

In this case, we have $O(a,x_1)={\displaystyle \frac{a}{2-x_1}}$ and $O(a,x_2)={\displaystyle \frac{a}{2-x_2}}$ to obtain $\beta ={\displaystyle \frac{2-x_1}{2-x_2}}\le {\displaystyle \frac{x_2}{x_1}}$, according to Corollary 2.

Case 2-1. $x_1\le a\le x_2$ and $a+x_1\le a+x_2\le 1$.

In this case, we have $O(a,x_1)={\displaystyle \frac{x_1}{x_1+1}}$ and $O(a,x_2)={\displaystyle \frac{a}{a+1}}$ to obtain $\beta ={\displaystyle \frac{a(x_1+1)}{x_1(a+1)}}\le {\displaystyle \frac{x_2}{x_1}}$, since $a{x}_1(x_1+1)\le x_2{x}_1(a+1).$

Case 2-2. $x_1\le a\le x_2$ and $a+x_1\le 1\le a+x_2$.

In this case, we have $O(a,x_1)={\displaystyle \frac{x_1}{x_1+1}}$ and $O(a,x_2)={\displaystyle \frac{a}{2-x_2}}$ to obtain $\beta ={\displaystyle \frac{a(x_1+1)}{x_1(2-x_2)}}\le {\displaystyle \frac{x_2}{x_1}}$. Since we have $x_1\le 1-a$, this implies that $x_{1}a(x_1+1)\le x_{1}a(2-a)\le x_1{x}_2(2-x_2)$, according to Corollary 2.

Case 2-3. $x_1\le a\le x_2$ and $1\le a+x_1\le a+x_2$.

In this case, we have $O(a,x_1)={\displaystyle \frac{x_1}{2-a}}$ and $O(a,x_2)={\displaystyle \frac{a}{2-x_2}}$ to obtain $\beta ={\displaystyle \frac{a(2-a)}{x_1(2-x_2)}}\le {\displaystyle \frac{x_2}{x_1}}$. Since we have $x_{1}a(2-a)\le x_1{x}_2(2-x_2)$, according to Corollary 2.

Case 3-1. $x_1\le x_2\le a$ and $a+x_1\le a+x_2\le 1$.

In this case, we have $O(a,x_1)={\displaystyle \frac{x_1}{x_1+1}}$ and $O(a,x_2)={\displaystyle \frac{x_2}{x_2+1}}$ to obtain $\beta ={\displaystyle \frac{x_2(x_1+1)}{x_1(x_2+1)}}\le {\displaystyle \frac{x_2}{x_1}}$, since $x_1{x}_2(x_1+1)\le x_1{x}_2(x_2+1)$.

Case 3-2. $x_1\le x_2\le a$ and $a+x_1\le 1\le a+x_2$.

In this case, we have $O(a,x_1)={\displaystyle \frac{x_1}{x_1+1}}$ and $O(a,x_2)={\displaystyle \frac{x_2}{2-a}}$ to obtain $\beta ={\displaystyle \frac{x_2(x_1+1)}{x_1(2-a)}}\le {\displaystyle \frac{x_2}{x_1}}$. Since we have $x_1\le 1-a$, the following inequality holds:

$x_1{x}_2(x_1+1)\le x_1{x}_2((1-a)+1))=x_1{x}_2(2-a).$

Case 3-3. $x_1\le x_2\le a$ and $1\le a+x_1\le a+x_2$.

In this case, we have $O(a,x_1)={\displaystyle \frac{x_1}{2-a}}$ and $O(a,x_2)={\displaystyle \frac{x_2}{2-a}}$ to obtain $\beta ={\displaystyle \frac{x2(2-a)}{x_1(2-a)}}\le {\displaystyle \frac{x_2}{x_1}}$.

To summarize the findings from the nine cases, we conclude that if $0<a$ and $0<x_1\le x_2\le 1$, then the inequality ${\displaystyle \frac{O(a,x_2)}{O(a,x_1)}}\le {\displaystyle \frac{x_2}{x_1}}$ holds true. □

Lemma 2. 

For (3), the following results hold:

• (i) 

  If $x\le a$ and $x+a\le 1$, then $O(a,x)={\displaystyle \frac{x}{x+1}}$.

• (ii) 

  If $x\le a$ and $x+a\ge 1$, then $O(a,x)={\displaystyle \frac{x}{2-a}}$.

• (iii) 

  If $x\ge a$ and $x+a\le 1$, then $O(a,x)={\displaystyle \frac{a}{a+1}}$.

• (iv) 

  If $x\ge a$ and $x+a\ge 1$, then $O(a,x)={\displaystyle \frac{a}{2-x}}$.

Proof. 

(i) Since $x\le a$ and $x+a\le 1$, there is $min\{a,x\}=x$ such that $O(a,x)={\displaystyle \frac{x}{x+1}}$.

(ii) Since $x\le a$ and $x+a\ge 1$, there is $min\{a,x\}=x$ such that $O(a,x)={\displaystyle \frac{x}{x+[1-(x+a-1\left)\right]}}={\displaystyle \frac{x}{2-a}}$.

(iii) Since $x\ge a$ and $x+a\le 1$, there is $min\{a,x\}=a$ such that $O(a,x)={\displaystyle \frac{a}{a+1}}$.

(iv) Since $x\ge a$ and $x+a\ge 1$, there is $O(a,x)={\displaystyle \frac{a}{a+[1-(x+a-1\left)\right]}}={\displaystyle \frac{a}{2-x}}$. □

According to Lemma 2, when $a\in [0,1]$ is used as a given parameter and $a=0.5$ is the dividing point, then the value of the overlap function $O(a,x)$ in (3) can be obtained as follows. Figure 1 and Figure 2 roughly illustrate these two results.

• (1). If $a\le {\displaystyle \frac{1}{2}}$, then there is

  (4)$\begin{array}{c}O(a,x)=\left\{\begin{array}{ccc}{\displaystyle \frac{x}{x+1}}\hfill & \mathrm{if}\hfill & x\le a,\hfill \\ {\displaystyle \frac{a}{a+1}}\hfill & \mathrm{if}\hfill & a\le x\le 1-a,\hfill \\ {\displaystyle \frac{a}{2-x}}\hfill & \mathrm{if}\hfill & x\ge 1-a.\hfill \end{array}\right.\hfill \end{array}$

• (2). If $a\ge {\displaystyle \frac{1}{2}}$, then there is

  (5)$\begin{array}{c}O(a,x)=\left\{\begin{array}{ccc}{\displaystyle \frac{x}{x+1}}\hfill & \mathrm{if}\hfill & x\le 1-a,\hfill \\ {\displaystyle \frac{x}{2-a}}\hfill & \mathrm{if}\hfill & 1-a\le x\le a,\hfill \\ {\displaystyle \frac{a}{2-x}}\hfill & \mathrm{if}\hfill & a\le x\le 1.\hfill \end{array}\right.\hfill \end{array}$

3. A Single-Variable Optimization Problem for the Min–Max Programming Problem (2)

As mentioned above, we extend the min–max programming problem with addition–min composition in the constraint part of the model (1) to the addition–overlap function composition as problem (2). To solve the optimal value of problem (2), we transform it into a single-variable optimization problem.

Definition 3. 

Let the vector $x={\left(x_j\right)}_{j\in \mathcal{J}}$. The solution set of problem (2) is denoted by $X(A,b):=\{x\in {[0,1]}^n|g_i\left(x\right)={\sum }_{j=1}^{n}O(a_{ij},x_j)\ge b_i,\forall i\in \mathcal{I}\}$.

Definition 4. 

Problem (2) is said to be consistent if $X(A,b)\ne \varnothing$. A solution $x^{*}\in X(A,b)$ is optimal for problem (2) if $Z\left(x^{*}\right)\le Z\left(x\right)$ for all $x\in X(A,b)$.

Based on the above definitions and the properties of the overlap function, we propose a single-variable optimization problem to solve the optimal value of problem (2) with the objective function $Z\left(x\right)=max\{x_1,\dots ,x_n\}$, as follows:

(6)$\begin{array}{ccc}\mathrm{Minimize}\hfill & & Z\left(y\right)=y\hfill \\ \mathrm{subject} \mathrm{to}\hfill & & G_i\left(y\right)={\displaystyle \sum _{j=1}^n}O(a_{ij},y)\ge b_i,\forall i\in \mathcal{I},\hfill \\ & & y\in [0,1],\hfill \end{array}$

where parameters $a_{ij}\in [0,1]$ and $b_i>0$, for all $i\in \mathcal{I}=\{1,2,\dots ,m\},j\in \mathcal{J}=\{1,2,\dots ,n\}$.

Theorem 2. 

Suppose $y^{*}$ is an optimal solution to problem (6) and $x^{*}=(x_1^{*},\dots ,x_n^{*})$ is an optimal solution to problem (2), with optimal value $Z\left(x^{*}\right)=max\{x_1^{*},\dots ,x_n^{*}\}$. Then, $max\{x_1^{*},\dots ,x_n^{*}\}=y^{*}$ holds.

Proof. 

Let ${\alpha }^{*}=max\{x_1^{*},\dots ,x_n^{*}\}$, i.e., ${\alpha }^{*}\ge x_j^{*}$, for all $j\in \mathcal{J}$. We want to show that ${\alpha }^{*}=y^{*}$. Suppose not, and if ${\alpha }^{*}<y^{*}$, then since the overlap function is monotone and ${\alpha }^{*}\ge x_j^{*}$, this implies that ${\sum }_{j=1}^{n}O(a_{ij},{\alpha }^{*})\ge {\sum }_{j=1}^{n}O(a_{ij},x_j^{*})\ge b_i,\forall i\in \mathcal{I}$. In other words, ${\alpha }^{*}$ is a feasible solution to problem (6). This makes ${\alpha }^{*}$ a better optimal value than $y^{*}$, and contradicts that $y^{*}$ is an optimal solution to problem (6).

On the other hand, suppose that ${\alpha }^{*}>y^{*}$. Since $y^{*}$ is an optimal solution to problem (6), we have ${\sum }_{j=1}^{n}O(a_{ij},y^{*})\ge b_i,\forall i\in \mathcal{I}$. In other words, $(y^{*},\dots ,y^{*})$ is a feasible solution to problem (2) with the objective value $y^{*}$. This means that the optimal value ${\alpha }^{*}$ for problem (2) could be further reduced, a contradiction.

Hence, we have proved ${\alpha }^{*}=max\{x_1^{*},\dots ,x_n^{*}\}=y^{*}$. □

Theorem 2 shows that the min–max programming problem (2) constrained by solvable addition–overlap functions can be transformed into a single-variable optimization problem (6) to find the optimal value $y^{*}$. Then, an optimal solution $x^{*}={\left(x_j^{*}\right)}_{j\in \mathcal{J}}$ of problem (2) including all variables with the same value $x_j^{*}=y^{*},j\in \mathcal{J}$ can be obtained as follows:

$x^{*}=(x_1^{*},x_2^{*},\dots ,x_n^{*})=(y^{*},y^{*},\dots ,y^{*}).$

Bisection method: It follows that all overlap functions in the constraint part of problem (6) are single-variable, non-decreasing, and continuous. Since the class of overlap functions is very broad, the well-known bisection method is proposed to solve problem (6) using any overlap function. The applicability of this method is underlined by its ability to systematically narrow the search interval, ensuring both convergence to the optimal solution and a high degree of accuracy. Precisely, we can use the bisection approach to determine the solution for each constraint $G_i\left(y\right)$ in problem (6), denoted as $y_i^{*}$ for $i\in \mathcal{I}$. Then, the optimal solution $y^{*}$ for problem (6) is obtained by taking the maximum value among these constraint solutions, as follows:

$y^{*}=max\{y_1^{*},\dots ,y_m^{*}\}.$

It is well known that the computational complexity of the bisection method is primarily determined by the number of iterations required to achieve the desired level of accuracy. Given an initial interval length L, the interval size after k iterations is ${\displaystyle \frac{L}{2^k}}$. To achieve a final interval length of $\epsilon$, the number of iterations k required is approximately ${log}_2\left({\displaystyle \frac{L}{\epsilon }}\right)$. Therefore, the computational complexity to achieve an accuracy of $\epsilon$ is $O\left({\log }_2\left({\displaystyle \frac{L}{\epsilon }}\right)\right)$.

4. An Iterative Method for Solving Problem (7)

In Section 3, the bisection method was proposed to explore the min–max programming problem (2) using any overlap function in the constraint part. However, if a specific overlap function is used in the constraint part, we could then develop an alternative method for finding an optimal solution to this problem.

(7)$\begin{array}{ccc}\mathrm{Minimize}\hfill & & Z\left(y\right)=y\hfill \\ \mathrm{subject} \mathrm{to}\hfill & & G_i\left(y\right)={\displaystyle \sum _{j=1}^n}O(a_{ij},y)\ge b_i,\forall i\in \mathcal{I},\hfill \\ & & y\in [0,1],\hfill \end{array}$

where

(8)$\begin{array}{c}O(a_{ij},y)={\displaystyle \frac{min\{a_{ij},y\}}{min\{a_{ij},y\}+(1-max\{0,y+a_{ij}-1\})}}.\hfill \end{array}$

To solve the optimal solution of problem (7), the bisection method discussed in the previous section can be employed. This method first addresses each constraint $G_i\left(y\right)=b_i$, for all $i\in \mathcal{I}$. The optimal value $y^{*}$ is then derived from the intersection of the solution sets where $G_i\left(y\right)\ge b_i,\forall i\in \mathcal{I}$. However, as the number of constraints in the problem increases, the computational load of the bisection method may rise, potentially leading to slow convergence. To mitigate this issue, the proposed iterative method takes a different approach. It aims to directly satisfy all constraints $G_i\left(y\right)\ge b_i,\forall i\in \mathcal{I}$ of problem (7) by using an approximation. Specifically, we utilize the approximation $\underline{y}={max}_{i\in \mathcal{I}}\left\{{\displaystyle \frac{b_i}{n}}\right\}$ to verify whether all constraints $G_i\left(\underline{y}\right)\ge b_i,\forall i\in \mathcal{I}$ are satisfied at each iteration. If they are, we can identify an optimal solution (value) for problem (7). Otherwise, the iterative method refines this approximation to bring it closer to the optimal value. During this refinement process, constraints satisfied by the current approximation can be disregarded. For this reason, the iterative method can find the optimal solution to the problem more quickly than the bisection method, without the need to solve each inequality.

Below, we propose several properties relevant to the iterative method for finding the optimal solution to problem (7).

Proposition 1. 

If problem (7) is consistent, $Y(A,b)\ne \varnothing$, then for each $i\in \mathcal{I}$, there is at least $a_{ij}>0$ for $j\in \mathcal{J}$.

Proof. 

Suppose to the contrary that for some $i\in \mathcal{I}$ there is $a_{ij}=0$ for all $j\in \mathcal{J}$. According to Lemma 1-(i), it implies that

$G_i\left(y\right)=\sum _{j=1}^{n}O(a_{ij},y)=\sum _{j=1}^{n}O(0,y)=0<b_i.$

This result contradicts the fact that problem (7) is consistent. □

Proposition 2. 

Suppose $Y(A,b)\ne \varnothing$ and ${\overline{b}}_i={\displaystyle \frac{b_i}{n}}$ in problem (7). If for each $i\in \mathcal{I}$ there is at least $a_{ij}<1,j\in \mathcal{J}$, then $G_i\left({\overline{b}}_i\right)<b_i$ holds for all $i\in \mathcal{I}$.

Proof. 

The problem (7) has the overlap function as (8). Let $D_{ij}=min\{a_{ij},{\overline{b}}_i\}+1-max\{0,{\overline{b}}_i+a_{ij}-1\}$ for all $i\in \mathcal{I},j\in \mathcal{J}$.

For each $i\in \mathcal{I}$, there are two cases to discuss the value of $D_{ij},j\in \mathcal{J}$.

Case 1. ${\overline{b}}_i+a_{ij}\le 1$.

Since ${\overline{b}}_i>0$ and at least $a_{ij}>0$, for some $j\in \mathcal{J}$, according to Proposition 2, it implies that there exist

$D_{ij}=min\{a_{ij},{\overline{b}}_i\}+1>1.$

Case 2. ${\overline{b}}_i+a_{ij}>1$.

Since $Y(A,b)\ne \varnothing$, there is ${\sum }_{j=1}^n{a}_{ij}\ge b_i$, according to Proposition 1, and there is at least $a_{ij}<1,j\in \mathcal{J}$; these results imply that

${\overline{b}}_i={\displaystyle \frac{b_i}{n}}\le {\displaystyle \frac{{\sum }_{j=1}^n{a}_{ij}}{n}}<1.$

Furthermore, if $min\{a_{ij},{\overline{b}}_i\}=a_{ij}$, then we have

$D_{ij}=min\{a_{ij},{\overline{b}}_i\}+1-({\overline{b}}_i+a_{ij}-1)=a_{ij}+2-{\overline{b}}_i-a_{ij}=2-{\overline{b}}_i>1.$

If $min\{a_{ij},{\overline{b}}_i\}={\overline{b}}_i$, then there exists

$D_{ij}=min\{a_{ij},{\overline{b}}_i\}+1-({\overline{b}}_i+a_{ij}-1)={\overline{b}}_i+2-{\overline{b}}_i-a_{ij}=2-a_{ij}>1.$

The above two cases show that for each $i\in \mathcal{I}$, there is $D_{ij}>1$ for some $j\in \mathcal{J}$. This result implies that there are ${\sum }_{j=1}^n{\displaystyle \frac{1}{D_{ij}}}<n$ and

$G_i\left({\overline{b}}_i\right)=\sum _{j=1}^{n}O(a_{ij},{\overline{b}}_i)=\sum _{j=1}^n{\displaystyle \frac{min\{a_{ij},{\overline{b}}_i\}}{D_{ij}}}\le \sum _{j=1}^n{\displaystyle \frac{{\overline{b}}_i}{D_{ij}}}={\overline{b}}_i\sum _{j=1}^n{\displaystyle \frac{1}{D_{ij}}}<n{\overline{b}}_i=b_i, \mathrm{for}\mathrm{all} i\in \mathcal{I}.$

□

Proposition 3 indicates that the value of $y={\overline{b}}_i={\displaystyle \frac{b_i}{n}}$ is often not a solution to the ith constraint $G_i\left(y\right)\ge b_i$ of problem (7), since the case of $a_{ij}<1$ for some $j\in \mathcal{J}$ usually occurs in $G_i\left(y\right)$. However, the value of ${\overline{b}}_i={\displaystyle \frac{b_i}{n}}$ can be considered as the lower bound of the variable y to satisfy the ith constraint of problem (7). To consider all the constraints of problem (7), we define

$\underline{y}=\underset{i\in \mathcal{I}}{max}\left\{{\overline{b}}_i\right\}=\underset{i\in \mathcal{I}}{max}\left\{{\displaystyle \frac{b_i}{n}}\right\}.$

In the process of determining the optimal value for problem (7), we introduce $\underline{y}$ as an approximation. The following Theorem 3 demonstrates that if this approximation is feasible, it constitutes the optimal solution (and thus, the optimal value) for problem (7). In cases where feasibility is not achieved, an iterative refinement process can be employed to progressively approach the optimal value.

Theorem 3. 

Suppose the solution set $Y(A,b)\ne \varnothing$ in problem (7).

• (i) 

  If $y\in Y(A,b)$, then $y\ge \underline{y}={max}_{i\in \mathcal{I}}\left\{{\displaystyle \frac{b_i}{n}}\right\}$.

• (ii) 

  If $G_i\left(\underline{y}\right)={\sum }_{j=1}^{n}O(a_{ij},\underline{y})\ge b_i,\forall i\in \mathcal{I}$, then $\underline{y}$ is the optimal solution of problem (7).

Proof. 

(i) Since $y\in Y(A,b)\ne \varnothing$ and $O(a_{ij},y)\le min\{a_{ij},y\}\le y$ for all $i\in \mathcal{I},j\in \mathcal{J}$ according to Lemma 1-(iv), we have

$ny\ge \sum _{j=1}^{n}O(a_{ij},y)\ge b_i,\forall i\in \mathcal{I}.$

Hence, we have $y\ge \underline{y}={max}_{i\in \mathcal{I}}\left\{{\displaystyle \frac{b_i}{n}}\right\}$.

(ii) Since $G_i\left(\underline{y}\right)={\sum }_{j=1}^{n}O(a_{ij},\underline{y})\ge b_i,\forall i\in \mathcal{I}$, $\underline{y}$ is a feasible solution, i.e., $\underline{y}\in Y(A,b)$. For any solution $y\in Y(A,b)$, we have $y\ge \underline{y}={max}_{i\in \mathcal{I}}\left\{{\displaystyle \frac{b_j}{n}}\right\}$ according to (i). This result leads to $Z\left(\underline{y}\right)\le Z\left(y\right)$ for all $y\in Y(A,b)$. Hence, $\underline{y}$ is an optimal solution of problem (7). □

4.1. The Solution Procedure of the Iterative Method

By integrating the aforementioned properties, we present a systematic solution procedure for the proposed iterative method to determine the optimal solution to problem (2). The steps of this procedure are outlined as follows.

Step 1: Transform problem (2) into the single-variable problem (7) and verify that ${\sum }_{j=1}^n{a}_{ij}\ge b_i$ for all $i\in \mathcal{I}$ holds according to Proposition 1 to check its consistency. If it does not, then stop the procedure.

Step 2: Compute $\underline{y}={max}_{i\in \mathcal{I}}\left\{{\displaystyle \frac{b_i}{n}}\right\}$ and test whether $G_i\left(\underline{y}\right)\ge b_i,i\in \mathcal{I}$ holds. If so, generate the optimal value $y^{*}=\underline{y}$ of problem (7) according to Theorem 3, and then, obtain the optimal solution $x^{*}={\left(x_j^{*}\right)}_{j\in \mathcal{J}}$ of problem (2) with $x_j^{*}=y^{*}$ for all $j\in \mathcal{J}$ according to Theorem 2. Stop the procedure.

Step 3: Let the index sets

$I^{\left(0\right)}=\{i\in \mathcal{I}|G_i\left(\underline{y}\right)<b_i\},I^{*\left(0\right)}=arg\{max\{b_i-G_i\left(\underline{y}\right)\}:i\in I^{\left(0\right)}\}$

and compute

$Distance\left(0\right)=b_i-G_i\left(\underline{y}\right),i\in I^{*\left(0\right)}.$

Step 4: Let $y^{\left(0\right)}=\underline{y}$ and $k=1$.

Step 5: Compute

(9)$\begin{array}{c}{y}^{\left(k\right)}=\underset{i\in I^{*(k-1)}}{max}\left\{{\displaystyle \frac{b_i}{G_i\left(y^{(k-1)}\right)}}{y}^{(k-1)}\right\} \mathrm{and} G_i\left(y^{\left(k\right)}\right),i\in I^{(k-1)}.\hfill \end{array}$

We have $G_i\left(y^{\left(k\right)}\right)\le b_i$ for some $i\in \mathcal{I}$ (see the following Proposition 4) and the index sets

(10)$\begin{array}{c}{I}^{\left(k\right)}=\{i\in \mathcal{I}|G_i\left(y^{\left(k\right)}\right)<b_i\},I^{*\left(k\right)}=arg\{max\{b_i-G_i\left(y^{\left(k\right)}\right)\}:i\in I^{\left(k\right)}\}.\hfill \end{array}$

If there does not exist $G_i\left(y^{\left(k\right)}\right)<b_i$ for some $i\in \mathcal{I}$, i.e., $I^{\left(k\right)}=\varnothing$, then we have $G_i\left(y^{\left(k\right)}\right)\ge b_i$ for all $i\in \mathcal{I}$. Let $y^{*}=y^{\left(k\right)}$ and go to step 9.

If there is $G_i\left(y^{\left(k\right)}\right)<b_i$ for some $i\in \mathcal{I}$, i.e., $I^{\left(k\right)}\ne \varnothing$, then we compute

(11)$\begin{array}{c}Distance\left(k\right)=b_i-G_i\left(y^{\left(k\right)}\right),i\in I^{*\left(k\right)}.\hfill \end{array}$

Step 6: Let $k:=k+1$ and go to step 5 until the value of $Distance\left(k\right)<\epsilon ={10}^{-3}$.

Step 7: Find $y^{\left(k\right)}\in [p_1,p_2]$ such that $G_i\left(p_1\right)\le G_i\left(y^{\left(k\right)}\right)\le G_i\left(p_2\right),i\in I^{*\left(k\right)}$, where $p_1$ and $p_2$ are two piecewise points of $G_i\left(y\right)$ that are closest to $y^{\left(k\right)}$.

Step 8: For $G_i\left(y\right)=b_i,i\in I^{*\left(k\right)}$ in $y\in [p_1,p_2]$, generate the equation $G_i\left(y\right)=b_i$ according to (4) and (5), and use the inverse function method to obtain the solution $y_i^{*}$.

Step 9: Generate the optimal value $y^{*}={max}_{i\in I^{*\left(k\right)}}\left\{y_i^{*}\right\}$ of problem (7) and obtain the optimal solution of problem (2) according to Theorem 2 as follows:

$x^{*}=(x_1^{*},x_2^{*},\dots ,x_n^{*})=(y^{*},y^{*},\dots ,y^{*}).$

It is worth noting that the notation $Distance\left(k\right)$ in (11) indicates that the value of the variable $y^{\left(k\right)}$ obtained at the kth iteration can satisfy all the inequalities of problem (7) when the right-hand side term $b_i$, for $i\in \mathcal{I}$, decreases the value of $Distance\left(k\right)$. In addition, the threshold $\epsilon ={10}^{-3}$ for the iterative method to stop the procedure in step 6 can be adjusted by the accuracy of distance from the optimal value. When the value of $\epsilon$ is smaller, the obtained value $y^{\left(k\right)}$ is closer to the optimal value $y^{*}$ of problem (7).

Proposition 3. 

Assume that $y^{\left(k\right)}={max}_{i\in I^{*(k-1)}}\left\{{\displaystyle \frac{b_i}{G_i\left(y^{(k-1)}\right)}}{y}^{(k-1)}\right\} \mathrm{and} G_i\left(y^{\left(k\right)}\right),i\in I^{(k-1)}$ are shown as in (9), then we have $G_i\left(y^{\left(k\right)}\right)\le b_i$ for some $i\in \mathcal{I}$.

Proof. 

Since $y^{\left(k\right)}={max}_{i\in I^{*(k-1)}}\left\{{\displaystyle \frac{b_i}{G_i\left(y^{(k-1)}\right)}}{y}^{(k-1)}\right\},I^{*(k-1)}\ne \varnothing$, it indicates that there is

$y^{\left(k\right)}={\displaystyle \frac{b_i}{G_i\left(y^{(k-1)}\right)}}{y}^{(k-1)}, \mathrm{for} \mathrm{some} i\in \mathcal{I}.$

In addition, we have $G_i\left(y^{(k-1)}\right)\le b_i$ such that $y^{(k-1)}\le y^{\left(k\right)}$. Moreover, according to Lemma 1-(v), it follows that ${\displaystyle \frac{G_i\left(y^{\left(k\right)}\right)}{G_i\left(y^{(k-1)}\right)}}\le {\displaystyle \frac{y^{\left(k\right)}}{y^{(k-1)}}}$. This implies that the following inequality holds.

${\displaystyle \frac{G_i\left(y^{\left(k\right)}\right)}{G_i\left(y^{(k-1)}\right)}}\le {\displaystyle \frac{y^{\left(k\right)}}{y^{(k-1)}}}={\displaystyle \frac{1}{y^{(k-1)}}}\times {\displaystyle \frac{b_i}{G_i\left(y^{(k-1)}\right)}}{y}^{(k-1)}={\displaystyle \frac{b_i}{G_i\left(y^{(k-1)}\right)}}.$

Consequently, we have demonstrated that $G_i\left(y^{\left(k\right)}\right)\le b_i$ for some $i\in \mathcal{I}$. □

Proposition 4. 

Based on the solution procedure of the above iterative method, if $I^{\left(0\right)}=\{i\in \mathcal{I}|G_i\left(\underline{y}\right)<b_i\}\ne \varnothing$, then the following inequalities hold.

• (i) 

  $y^{\left(k\right)}>y^{(k-1)}$ for $k\ge 1$.

• (ii) 

  $G_i\left(y^{\left(k\right)}\right)<b_i$ for $i\in I^{\left(k\right)}$, and $Distance\left(k\right)=b_i-G_i\left(y^{\left(k\right)}\right)>0$ for $i\in I^{*\left(k\right)}$.

Proof. 

(i) For $k=1$, since $b_i>0,\forall i\in \mathcal{I}$, we have $y^{\left(0\right)}=\underline{y}={max}_{i\in \mathcal{I}}\left\{{\displaystyle \frac{b_i}{n}}\right\}>0$. There is also

$I^{\left(0\right)}=\{i\in \mathcal{I}|G_i\left(y^{\left(0\right)}\right)=G_i\left(\underline{y}\right)<b_i\}\ne \varnothing .$

These results mean that we have

$I^{*\left(0\right)}=arg\{max\{b_i-G_i\left(y^{\left(0\right)}\right)\}:i\in I^{\left(0\right)}\}\ne \varnothing \mathrm{and} {\displaystyle \frac{b_i}{G_i\left(y^{\left(0\right)}\right)}}>1,i\in I^{*\left(0\right)},$

such that

$y^{\left(1\right)}=\underset{i\in I^{*\left(0\right)}}{max}\left\{{\displaystyle \frac{b_i}{G_i\left(y^{\left(0\right)}\right)}}{y}^{\left(0\right)}\right\}>y^{\left(0\right)}.$

Similarly, we can show that $y^{\left(k\right)}>y^{(k-1)}$ holds for $k>1$.

(ii) Based on (9), (10), and (11), there are

$\begin{array}{c}{I}^{\left(k\right)}=\{i\in \mathcal{I}|G_i\left(y^{\left(k\right)}\right)<b_i\},I^{*\left(k\right)}=arg\{max\{b_i-G_i\left(y^{\left(k\right)}\right)\}:i\in I^{\left(k\right)}\},\hfill \end{array}$

and

$Distance\left(k\right)=b_i-G_i\left(y^{\left(k\right)}\right),i\in I^{*\left(k\right)}.$

Obviously, we have $G_i\left(y^{\left(k\right)}\right)<b_i$ for $i\in I^{\left(k\right)}$, and $Distance\left(k\right)=b_i-G_i\left(y^{\left(k\right)}\right)>0$ for $I^{*\left(k\right)}$. □

Computational complexity and convergence: In problem (2), there are m constraints, each containing n variables. The computational complexity and convergence of the above iterative method are mainly determined by steps 1 to 6.

Transforming problem (2) into the single-variable problem (7) and checking the consistency of problem (7) in step 1 requires $2mn+1$ operations. Computing $\underline{y}={max}_{i\in \mathcal{I}}\left\{{\displaystyle \frac{b_i}{n}}\right\}={\displaystyle \frac{1}{n}}{max}_{i\in \mathcal{I}}\left\{b_i\right\}$ and checking whether $G_i\left(\underline{y}\right)\ge b_i,i\in \mathcal{I}$ holds in step 2 need m and $9mn$ operations, respectively. If $\underline{y}$ is a feasible solution to problem (7), then $n+1$ operations are required to obtain the optimal solution to problem (2). Similar to the test $G_i\left(\underline{y}\right)\ge b_i,i\in \mathcal{I}$, step 3 requires $9mn$ operations to find the index set $I^{\left(0\right)}$. It costs $2m$ operations to obtain the index sets $I^{*\left(0\right)}$ and the value of $Distance\left(0\right)$. In step 4, setting $y^{\left(0\right)}=\underline{y}$ and $k=1$ costs 2 operations. Step 5 requires $9mn+2m+3$ operations to compute the values of $y^{\left(k\right)}$ and $G_i\left(y^{\left(k\right)}\right)$, the index sets $I^{\left(k\right)}$ and $I^{*\left(k\right)}$, and the value of $Distance\left(k\right)$. In step 6, setting $k:=k+1$ and checking $Distance\left(k\right)<\epsilon ={10}^{-3}$ takes 2 operations.

As a result, running step 1 to step 6 of the iterative method to yield the optimal value for problem (7) costs

$\begin{array}{ccc}& & (2mn+1)+\left[\right(m+9mn)+(n+1\left)\right]+(9mn+2m)+2+(9mn+2m+3)+2\hfill \\ & & =29mn+5m+n+9\hfill \end{array}$

operations. Hence, the computational complexity of the proposed solution procedure (the iterative method) is $O\left(mn\right)$.

Furthermore, since $G_i\left(y\right)$ is a continuous and non-decreasing function, and each iteration must have $y^{\left(k\right)}>y^{(k-1)}$ according to Proposition 5, these results lead to an incremental increase in $G_i\left(y^{\left(k\right)}\right)$. Precisely, the value of $b_i$ is constant while the value of $G_i\left(y^{\left(k\right)}\right)$ increases at each iteration, so that $Distance\left(k\right)=b_i-G_i\left(y^{\left(k\right)}\right)<\epsilon ={10}^{-3}$ in step 6 is met for a finite number of iterations and the iterative procedure converges.

4.2. A Numerical Example

Example 1. 

Find the optimal solution to the following min–max programming problem that is subjected to a system of addition–overlap functions.

(12) $\begin{array}{ccc}Minimize\hfill & & Z\left(x\right)=max\{x_1,x_2,x_3,x_4\}\hfill \\ subject to\hfill & & g_1\left(x\right)=O(0.4,x_1)+O(0.55,x_2)+O(0.3,x_3)+O(0.5,x_4)\ge 1.5,\hfill \\ & & g_2\left(x\right)=O(0.24,x_1)+O(0.22,x_2)+O(0.25,x_3)+O(0.2,x_4)\ge 0.6,\hfill \\ & & g_3\left(x\right)=O(0.7,x_1)+O(0.6,x_2)+O(0.75,x_3)+O(0.8,x_4)\ge 2.0,\hfill \\ & & g_4\left(x\right)=O(0.45,x_1)+O(0.5,x_2)+O(0.4,x_3)+O(0.95,x_4)\ge 1.8,\hfill \\ & & x_j\in [0,1],j\in \mathcal{J}=\{1,2,3,4\},\hfill \end{array}$

where the operator $O(a_{ij},x_j)$ is an overlap function with

$\begin{array}{c}O(a_{ij},x_j)={\displaystyle \frac{min\{a_{ij},x_j\}}{min\{a_{ij},x_j\}+(1-max\{0,x_j+a_{ij}-1\})}}.\hfill \end{array}$

Apply the iterative method described above to solve the problem using the numerical data provided in problem (12) of Example 1.

Step 1: Transform the problem into a single-variable problem and verify its consistency according to Proposition 1.

According to Theorem 2, we transform problem (12) into the following single-variable problem.

(13) $\begin{array}{ccc}Minimize\hfill & & Z\left(y\right)=y\hfill \\ subject to\hfill & & G_1\left(y\right)=O(0.4,y)+O(0.55,y)+O(0.3,y)+O(0.5,y)\ge 1.5,\hfill \\ & & G_2\left(y\right)=O(0.24,y)+O(0.22,y)+O(0.25,y)+O(0.2,y)\ge 0.6,\hfill \\ & & G_3\left(y\right)=O(0.7,y)+O(0.6,y)+O(0.75,y)+O(0.8,y)\ge 2.0,\hfill \\ & & G_4\left(y\right)=O(0.45,y)+O(0.5,y)+O(0.4,y)+O(0.95,y)\ge 1.8,\hfill \\ & & y\in [0,1],\hfill \end{array}$

Checking the consistency of problem (13) by using ${\sum }_{j=1}^4{a}_{ij}\ge b_i,i\in \mathcal{I}=\{1,2,3,4\}$, there are

$\begin{array}{c}{\displaystyle \sum _{j=1}^4}{a}_{1j}=1.75>b_1=1.5,{\displaystyle \sum _{j=1}^4}{a}_{2j}=0.91>b_2=0.6,\hfill \\ {\displaystyle \sum _{j=1}^4}{a}_{3j}=2.85>b_3=2.0, and {\displaystyle \sum _{j=1}^4}{a}_{4j}=2.3>b_4=1.8.\hfill \end{array}$

According to Proposition 1, problem (13) is consistent.

Step 2: Compute $\underline{y}={max}_{i\in \mathcal{I}}\left\{{\displaystyle \frac{b_i}{n}}\right\}$ and test whether $G_i\left(\underline{y}\right)\ge b_i,i\in \mathcal{I}$ holds.

We have

$\underline{y}=max\{{\displaystyle \frac{1.5}{4}},{\displaystyle \frac{0.6}{4}},{\displaystyle \frac{2.0}{4}},{\displaystyle \frac{1.8}{4}}\}=0.5$

to generate

$G_1\left(\underline{y}\right)=1.19464<b_1=1.5,G_2\left(\underline{y}\right)=0.78054>b_2=0.6,$

$G_3\left(\underline{y}\right)=1.55842<b_3=2.0, and G_4\left(\underline{y}\right)=1.40558<b_4=1.8.$

Note that the second constraint $G_2\left(y\right)$ in problem (13) is already satisfied by $\underline{y}=0.5$. However, the other constraints $G_1\left(y\right),G_3\left(y\right)$, and $G_4\left(y\right)$ are not satisfied.

Step 3: From the result of step 2, we have $I^{\left(0\right)}=\{i\in \mathcal{I}|G_i\left(\underline{y}\right)<b_i\}=\{1,3,4\}$. Computing $b_i-G_i\left(\underline{y}\right)$ for $i\in I^{\left(0\right)}=\{1,3,4\}$ yields

$b_1-G_1\left(\underline{y}\right)=1.5-1.19464=0.30536,b_3-G_3\left(\underline{y}\right)=0.44158,b_4-G_4\left(\underline{y}\right)=0.39442,$

$and I^{*\left(0\right)}=arg\{max\{b_i-G_i\left(\underline{y}\right)\}:i\in I^{\left(0\right)}\}=\left\{3\right\},$

we obtain

$Distance\left(0\right)=b_3-G_3\left(\underline{y}\right)=0.44158.$

Step 4: Let $y^{\left(0\right)}=\underline{y}$ and $k=1$. There is $y^{\left(0\right)}=\underline{y}=0.5$.

Step 5: Based on $y^{\left(0\right)}=\underline{y}=0.5,I^{*\left(0\right)}=\left\{3\right\}$, and following (9), (10), and (11), we compute

$\begin{array}{c}{y}^{\left(1\right)}={\displaystyle \underset{i\in I^{*\left(0\right)}}{max}}\left\{{\displaystyle \frac{b_i}{G_i\left(y^{\left(0\right)}\right)}}{y}^{\left(0\right)}\right\}={\displaystyle \frac{b_3}{G_3\left(y^{\left(0\right)}\right)}}{y}^{\left(0\right)}={\displaystyle \frac{2.0}{1.55842}}\times 0.5=0.64167,\hfill \end{array}$

to yield

$G_1\left(y^{\left(1\right)}\right)=1.29826<b_1=1.5,G_3\left(y^{\left(1\right)}\right)=1.98338<b_3=2.0,G_4\left(y^{\left(1\right)}\right)=1.60499<b_4=1.8.$

Obviously, we can obtain $I^{\left(1\right)}=\{i\in \mathcal{I}|G_i\left(y^{\left(1\right)}\right)<b_i\}=\{1,3,4\}$ and

$b_1-G_1\left(y^{\left(1\right)}\right)=1.5-1.29826=0.20174,b_3-G_3\left(y^{\left(1\right)}\right)=0.01662,b_4-G_4\left(y^{\left(1\right)}\right)=0.19501,$

$I^{*\left(1\right)}=arg\{max\{b_i-G_i\left(y^{\left(1\right)}\right)\}:i\in I^{\left(1\right)}\}=\left\{1\right\},$

and obtain

$Distance\left(1\right)=b_1-G_1\left(y^{\left(1\right)}\right)=0.20174.$

Step 6: Let $k:=k+1$ and go to step 5 until the value of $Distance\left(k\right)<\epsilon ={10}^{-3}$.

Iterate step 5 for $k=2,\dots ,7$ to yield

$y^{\left(2\right)}=0.71439,I^{\left(2\right)}=\{1,4\},I^{*\left(2\right)}=\left\{1\right\},G_1\left(y^{\left(2\right)}\right)=1.39042,Distance\left(2\right)=0.10958;$

$y^{\left(3\right)}=0.79982,I^{\left(3\right)}=\left\{1\right\},I^{*\left(3\right)}=\left\{1\right\},G_1\left(y^{\left(3\right)}\right)=1.45811,Distance\left(3\right)=0.04189;$

$y^{\left(4\right)}=0.82279,I^{\left(4\right)}=\left\{1\right\},I^{*\left(4\right)}=\left\{1\right\},G_1\left(y^{\left(4\right)}\right)=1.48657,Distance\left(4\right)=0.01343;$

$y^{\left(5\right)}=0.83023,I^{\left(5\right)}=\left\{1\right\},I^{*\left(5\right)}=\left\{1\right\},G_1\left(y^{\left(5\right)}\right)=1.49602,Distance\left(5\right)=0.00398;$

$y^{\left(6\right)}=0.83244,I^{\left(6\right)}=\left\{1\right\},I^{*\left(6\right)}=\left\{1\right\},G_1\left(y^{\left(6\right)}\right)=1.49885,Distance\left(6\right)=0.00115;$

and

$y^{\left(7\right)}=0.83308,I^{\left(7\right)}=\left\{1\right\},I^{*\left(7\right)}=\left\{1\right\},G_1\left(y^{\left(7\right)}\right)=1.49967,Distance\left(7\right)=0.00033<\epsilon ={10}^{-3}.$

The threshold $\epsilon ={10}^{-3}$ is met by $Distance\left(7\right)$ at $k=7$ iterations.

Step 7: Find $y^{\left(k\right)}\in [p_1,p_2]$ such that $G_i\left(p_1\right)\le G_i\left(y^{\left(k\right)}\right)\le G_i\left(p_2\right),i\in I^{*\left(k\right)}$, where $p_1$ and $p_2$ are two piecewise points of $G_i\left(y\right)$ that are closest to $y^{\left(k\right)}$.

Since $k=7$ and $I^{*\left(7\right)}=\left\{1\right\}$, the different piecewise points of $G_1\left(y\right)$ are 0, 0.3, 0.4, 0.45, 0.5, 0.55, 0.6, 0.7, and 1, respectively. Obviously, we have $y^{\left(7\right)}=0.83308\in [0.7,1.0]=[p_1,p_2]$ and

$G_1\left(p_1\right)=G_1\left(0.7\right)=1.34615\le G_1\left(y^{\left(7\right)}\right)=G_1\left(0.83308\right)=1.49967\le G_1\left(p_2\right)=G_1\left(1.0\right)=1.75.$

Step 8: For $G_i\left(y\right)=b_i,i\in I^{*\left(k\right)}$ in $y\in [p_1,p_2]$, generate the equation $G_i\left(y\right)=b_i$ according to (4) and (5), and use the inverse function method to obtain the solution $y_i^{*}$.

Based on the result of step 7, for $G_1\left(y\right)=1.5$ in $y\in [0.7,1]$ according to (4) and (5), we have

$G_1\left(y\right)=O(0.4,y)+O(0.55,y)+O(0.3,y)+O(0.5,y)={\displaystyle \frac{0.4}{2-y}}+{\displaystyle \frac{0.55}{2-y}}+{\displaystyle \frac{0.3}{2-y}}+{\displaystyle \frac{0.5}{2-y}}=1.5.$

The value $y={\displaystyle \frac{5}{6}}$ can be obtained to $G_1\left(y\right)=1.5$. It is denoted as $y_1^{*}={\displaystyle \frac{5}{6}}$.

Step 9: Generate the optimal value $y^{*}={max}_{i\in I^{*\left(k\right)}}\left\{y_i^{*}\right\}$ of problem (13) and obtain the optimal solution $x^{*}={\left(x_j^{*}\right)}_{j\in \mathcal{J}}$ with $x_j^{*}=y^{*},j\in \mathcal{J}$ of problem (12) according to Theorem 2.

Since $I^{*\left(7\right)}=\left\{1\right\}$ with $y_1^{*}={\displaystyle \frac{5}{6}}$, the optimal value generated by the iterative method for problem (13) is

$y^{*}=\underset{i\in I^{*\left(7\right)}}{max}\left\{y_i^{*}\right\}=max\left\{y_1^{*}\right\}={\displaystyle \frac{5}{6}},$

and the optimal solution of problem (12) is

$\begin{array}{c}{x}^{*}=(x_1^{*},x_2^{*},x_3^{*},x_4^{*})=(y^{*},y^{*},y^{*},y^{*})=({\displaystyle \frac{5}{6}},{\displaystyle \frac{5}{6}},{\displaystyle \frac{5}{6}},{\displaystyle \frac{5}{6}}).\hfill \end{array}$

Comparison with the bisection method: Let us consider the number of iterations and results using the bisection method to solve Example 1.

For the first inequality $G_1\left(y\right)=b_1=1.5$, it takes nine iterations to find the solution $y_1^{*}=0.833984375$, with an error of $G_1\left(y_1^{*}\right)-b_1=0.0008375209<\epsilon ={10}^{-3}$. For the second inequality $G_2\left(y\right)=b_2=0.6$, it requires 10 iterations to obtain the solution $y_2^{*}=0.1767578125$, with an error of $G_2\left(y_2^{*}\right)-b_2=0.0008298755$. The third inequality $G_3\left(y\right)=b_3=2.0$ also takes 10 iterations, yielding the solution $y_3^{*}=0.6474609375$, with an error of $G_3\left(y_3^{*}\right)-b_3=-0.0008234854$. Finally, for the fourth inequality $G_4\left(y\right)=b_4=1.8$, 10 iterations are needed to find $y_4^{*}=0.7529296875$, with an error of $G_4\left(y_4^{*}\right)-b_4=-0.0003869106$.

Then, the optimal solution $y^{*}$ for problem (13) using the bisection approach is obtained by taking the maximum value among these constraint solutions, as follows:

$y^{*}=max\{y_1^{*},y_2^{*},y_3^{*},y_4^{*}\}=0.833984375\approx {\displaystyle \frac{5}{6}},$

and the optimal solution of problem (12) is

$\begin{array}{c}{x}^{*}=(x_1^{*},x_2^{*},x_3^{*},x_4^{*})=(y^{*},y^{*},y^{*},y^{*})=({\displaystyle \frac{5}{6}},{\displaystyle \frac{5}{6}},{\displaystyle \frac{5}{6}},{\displaystyle \frac{5}{6}}).\hfill \end{array}$

Note that the computational process for example 1 requires a total of 39 iterations to obtain the optimal solution using the bisection method. In contrast, an alternative iterative method achieves the same result in only seven iterations.

5. Conclusions

In this paper, we studied the min–max programming problem subject to a system of addition–overlap functions. We showed that, when it is solvable, there exists an optimal solution with the same value in all its variables. This property allowed us to transform the min–max programming problem (2) into a single-variable optimization problem (6). The bisection method can be employed to obtain the optimal value for the problem using any overlap function in its constraints.

When an overlap function is explicitly specified, one could propose alternative methods utilizing its properties for better computational efficiency. In this paper, we proposed an iterative method that yields approximate solutions converging to an optimal solution. The proposed iterative method, while more efficient than the bisection method for Example 1, has limitations in its applicability to overlap function operators. Unlike the bisection method, which can be universally applied, the iterative approach relies on the overlap function operator being explicitly specified. Therefore, future research should focus on developing a more robust and comprehensive approach to improve upon the iterative method.

Author Contributions

Conceptualization, S.-M.G. and Y.-C.C.; Methodology, Y.-K.W. All authors have read and agreed to the published version of the manuscript.

Data Availability Statement

The original contributions presented in the study are included in the article, further inquiries can be directed to the corresponding author.

Conflicts of Interest

The authors declare no conflict of interest.

Footnotes

Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content.

Figures

Figure 1. The value of [Forumla omitted. See PDF.] with [Forumla omitted. See PDF.].

Figure 2. The value of [Forumla omitted. See PDF.] with [Forumla omitted. See PDF.].

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