A novel method for solving constrained matrix games with fuzzy payoffs

Abstract

In last few years, lots of researchers have proposed different methods to solve the constrained matrix games with fuzzy payoffs. In this paper, it has been shown that the mathematical programming problem of constrained matrix games with fuzzy payoffs, considered by researchers, is mathematically invalid and hence the method, proposed by researchers to obtain the complete solution (minimum expected gain of Player I, maximum expected loss of Player II and their corresponding optimal strategies) of constrained matrix games with fuzzy payoffs by solving the mathematical programming problem with fuzzy payoffs, are also invalid. Further, in the present paper, a new method has been proposed to find the complete solution of matrix games with fuzzy payoffs. To illustrate the proposed method, some existing numerical problems of constrained matrix games with fuzzy payoffs have been solved by the proposed method.

Keywords

Matrix games fuzzy payoffs mathematical programming problem

1 Introduction

Game theory [15] is a mathematical tool to describe strategic interactions among multiple decision makers who behave rationale. It has many applications in broad areas such as strategic welfares, economic or social problems, animal behavior, political voting systems etc. Although, the concept of game theory was started with Von Neumann’s study on zero-sum games [17], in which he proved the famous minimax theorem for zero-sum games. But, game theory was significantly advanced at Princeton University through the work of Nash [14].

There are several kinds of games which deals with antagonistic decision problems [15]. From conceptual and applications point of view, one of the class of games is constrained matrix games [5], which is the extension of matrix games.

Charnes [4] pointed out that in two person zero sum games (matrix games), it is assumed that only following constraints should be satisfied for strategies {x_i, i = 1, 2, … , m } and {y_j, j = 1, 2, … , n } of Player I and Player II respectively:

$\begin{matrix} (i) x_{i} ⩾ 0, i = 1, 2, \dots, m & (ii) \sum_{i = 1}^{m} x_{i} = 1 \\ (iii) y_{j} ⩾ 0, j = 1, 2, \dots, n & (iv) \sum_{j = 1}^{n} y_{j} = 1 \end{matrix}$

However, in real life, there are certain matrix games where the strategies of the players are also constrained to satisfy general linear inequalities not only the above mentioned constraints. To deal with such real life problems [5], Charnes [4] extended the matrix games into constrained matrix games and thereafter it has been further extended by Kawaguchi and Maruyama [8].

In real life situations, the players cannot estimate the payoffs in game exactly due to want of adequate information [1–3 , 18]. Therefore, fuzzy set [19] has been proposed to overcome this shortcoming.

Li and Cheng [11] pointed out that there is no method in the literature to solve constrained matrix games with fuzzy payoffs and proposed a method for the same. Later, Li [12, 13] with other co-authors proposed different methods for solving constrained matrix games with fuzzy payoffs.

In this paper, the flaws of the existing methods [12, 13] for solving constrained matrix games with fuzzy payoffs have been pointed out. To resolve these flaws, a new method is also proposed to obtain the optimal strategies as well as minimum expected gain of Player I and maximum expected loss of Player II for constrained matrix games with fuzzy payoffs. The proposed method has been illustrated by some existing numerical problems of constrained matrix games with fuzzy payoffs. Since, in [11, 12] triangular fuzzy numbers have used for the payoffs which is a special case of trapezoidal fuzzy number so trapezoidal fuzzy numbers are used in the present paper.

2 Preliminaries

In this section, some basic definitions and arithmetic operations of fuzzy numbers have been presented [5] which will be used in the sequel.

2.1 Some basic definitions

In this section, some basic definitions are reviewed [5].

Definition 2.1 A fuzzy number $\tilde{A} = (a^{L} (0), a (1), a^{R} (0))$ is called a triangular fuzzy number if its membership function $μ_{\tilde{A}}$ is given by: $μ_{\tilde{A}} (x) = {\begin{matrix} \frac{x - a^{L} (0)}{a (1) - a^{L} (0)}, a^{L} (0) ⩽ x < a (1), \\ \frac{x - a^{R} (0)}{a (1) - a^{R} (0)}, a (1) ⩽ x ⩽ a^{R} (0), \\ 0, x < a^{L} (0), x > a^{R} (0) . \end{matrix}$

Further, the α- cut of the triangular fuzzy number $\tilde{A}$ is the closed interval

$A (α) = [(a (1) - a^{L} (0)) α + a^{L} (0), (a (1) - a^{R} (0)) α + a^{R} (0)]$

Definition 2.2 A fuzzy number $\tilde{A} = (a^{L} (0), a^{L} (1), a^{R} (1), a^{R} (0))$ is called a trapezoidal fuzzy number if its membership function $μ_{\tilde{A}}$ is given by $μ_{\tilde{A}} (x) = {\begin{matrix} \frac{x - a^{L} (0)}{a^{L} (1) - a^{L} (0)}, & a^{L} (0) ⩽ x < a^{L} (1), \\ 1, & a^{L} (1) ⩽ x ⩽ a^{R} (1), \\ \frac{x - a^{R} (0)}{a^{R} (1) - a^{R} (0)}, & a^{R} (1) < x ⩽ a^{R} (0), \\ 0, & x < a^{L} (0), x > a^{R} (0) . \end{matrix}$

Further, the α- cut of the trapezoidal fuzzy number $\tilde{A}$ is the closed interval

$A^{α} = [(a^{L} (1) - a^{L} (0)) α + a^{L} (0), (a^{R} (1) - a^{R} (0)) α + a^{R} (0)]$

2.2 Arithmetic operations of trapezoidal fuzzy numbers

In this section, some arithmetic operations of two trapezoidal fuzzy numbers, defined on universal set of real numbers $R$ , have been presented.

(i) Let ${\tilde{A}}_{1} = (a_{1}^{L} (0), a_{1}^{L} (1), a_{1}^{R} (1), a_{1}^{R} (0))$ and ${\tilde{A}}_{2} = (a_{2}^{L} (0), a_{2}^{L} (1), a_{2}^{R} (1), a_{2}^{R} (0))$ be two trapezoidal fuzzy numbers. Then, $\begin{matrix} {\tilde{A}}_{1} + {\tilde{A}}_{2} = (a_{1}^{L} (0) + a_{2}^{L} (0), a_{1}^{L} (1) + a_{2}^{L} (1), a_{1}^{R} (1) \\ + a_{2}^{R} (1), a_{1}^{R} (0) + a_{2}^{R} (0)) \\ {\tilde{A}}_{1} - {\tilde{A}}_{2} = (a_{1}^{L} (0) - a_{2}^{R} (0), a_{1}^{L} (1) - a_{2}^{R} (1), a_{1}^{R} (1) \\ - a_{2}^{L} (1), a_{1}^{R} (0) - a_{2}^{L} (0)) \end{matrix}$

(ii) Let ${\tilde{A}}_{1} = (a_{1}^{L} (0), a_{1}^{L} (1), a_{1}^{R} (1), a_{1}^{R} (0))$ be a trapezoidal fuzzy number and λ be a real number. Then, $λ {\tilde{A}}_{1} = {\begin{matrix} (λ a_{1}^{L} (0), λ a_{1}^{L} (1), λ a_{1}^{R} (1), λ a_{1}^{R} (0)), λ ⩾ 0, \\ (λ a_{1}^{R} (0), λ a_{1}^{R} (1), λ a_{1}^{L} (1), λ a_{1}^{L} (0)), λ < 0 . \end{matrix}$

3 Existing mathematical formulation of constrained matrix games with fuzzy payoffs

In the literature [15], the mathematical formulation of constrained matrix games with crisp payoffs is obtained by replacing the constraint set of strategies $X = {x_{i}, i = 1, 2, \dots, m | x_{i} ⩾ 0, i = 1, 2, \dots, m; \sum_{i = 1}^{m} x_{i} = 1}$ for Player I with $\begin{matrix} X = & {x_{i}, i = 1, 2, \dots, m | x_{i} ⩾ 0, i = 1, 2, \dots, m; \\ \sum_{i = 1}^{m} b_{il} x_{i} ⩽ c_{l}, l = 1, 2, \dots, p} \end{matrix}$ where $\sum_{i = 1}^{m} b_{il} x_{i} ⩽ c_{l}$ includes $\sum_{i = 1}^{m} x_{i} = 1$ which is latter is equivalent to $\sum_{i = 1}^{m} x_{i} ⩽ 1, - \sum_{i = 1}^{m} x_{i} ⩽ - 1$ and $Y = {y_{j}, j = 1, 2, \dots, n | y_{j} ⩾ 0, j = 1, 2, \dots, n; \sum_{j = 1}^{n} y_{j} = 1}$ or Player II with $\begin{matrix} Y = & {y_{j}, j = 1, 2, \dots, n | y_{j} ⩾ 0, j = 1, 2, \dots, n; \\ \sum_{j = 1}^{n} e_{kj} y_{j} ⩾ d_{k}, k = 1, 2, \dots, q} \end{matrix}$ where $\sum_{j = 1}^{n} e_{kj} y_{j} ⩾ d_{k}$ includes $\sum_{j = 1}^{n} y_{j} = 1$ which is latter equivalent to $\sum_{j = 1}^{n} y_{j} ⩾ 1, - \sum_{j = 1}^{n} y_{j} ⩾ - 1$ . The minimum expected gain of Player I, maximum expected loss of Player II and their corresponding strategies for constrained matrix games with crisp payoffs can be obtained by solving Problem 3.1 and Problem 3.2.

Problem 3.1 $Minimize {\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i} a_{ij} y_{j}} .$

Subject to $\begin{matrix} \sum_{i = 1}^{m} b_{il} x_{i} ⩽ c_{l}, l = 1, 2, \dots, p; \\ \sum_{j = 1}^{n} e_{kj} y_{j} ⩾ d_{k}, k = 1, 2, \dots, q; \\ x_{i} ⩾ 0, i = 1, 2, \dots, m; y_{j} ⩾ 0, j = 1, 2, \dots, n . \end{matrix}$

Problem 3.2 $Minimize {\sum_{i = 1}^{m} \sum_{j = 1}^{n} x_{i} a_{ij} y_{j}}$

Subject to

Constraints of Problem 3.1

Further, in the literature [15], it is pointed out that as in the Problem 3.1 only y_j and in Problem 3.2 only x_i have been considered as decision variables so, Problem 3.1 and Problem 3.2 are linear programming problems. Therefore, the optimal value of Problem 3.1 and Problem 3.2 is same as the optimal values of its corresponding dual problem i.e., Problem 3.3 and Problem 3.4 respectively.

Problem 3.3 $Minimize {\sum_{k = 1}^{q} d_{k} z_{k}}$

Subject to $\begin{matrix} \sum_{k = 1}^{q} e_{kj} z_{k} ⩽ \sum_{i = 1}^{m} a_{ij} x_{i}, j = 1, 2, \dots, n; \end{matrix}$ $\begin{matrix} \sum_{i = 1}^{m} b_{il} x_{i} \\ ⩽ c_{l}, l = 1, 2, \dots, p; x_{i} ⩾ 0, i = 1, 2, \dots, m; \end{matrix}$ $\begin{matrix} z_{k} \\ ⩾ 0, k = 1, 2, \dots, q . \end{matrix}$

Problem 3.4 $Minimize {\sum_{l = 1}^{p} c_{l} t_{l}}$

Subject to $\begin{matrix} \sum_{l = 1}^{p} b_{il} t_{l} ⩾ \sum_{j = 1}^{n} a_{ij} y_{j}, i = 1, 2, \dots, m; \end{matrix}$ $\begin{matrix} \sum_{j = 1}^{n} e_{kj} y_{j} ⩾ \\ d_{k}, k = 1, 2, \dots, q; t_{l} ⩾ 0, l = 1, 2, \dots, p; \end{matrix}$ $\begin{matrix} y_{j} ⩾ 0, \\ j = 1, 2, \dots, n . \end{matrix}$

On the same direction, Li and Cheng [11] used the following method to obtain the mathematical formulation of constrained matrix games with fuzzy payoffs.

Step 1: Replacing the parameters a_ij, b_il, e_kj, d_k and c_l with trapezoidal fuzzy numbers $(a_{ij}^{L} (0), a_{ij}^{L} (1), a_{ij}^{R} (1), a_{ij}^{R} (0)), (b_{il}^{L} (0), b_{il}^{L} (1), b_{il}^{R} (1), b_{il}^{R} (0))$ , $(e_{kj}^{L} (0), e_{kj}^{L} (1), e_{kj}^{R} (1), e_{kj}^{R} (0)), (d_{k}^{L} (0), d_{k}^{L} (1), d_{k}^{R} (1), d_{k}^{R} (0))$ and $(c_{l}^{L} (0), c_{l}^{L} (1), c_{l}^{R} (1), c_{l}^{R} (0))$ respectively, Problem 3.1 and Problem 3.2 are transformed into Problem 3.5 and Problem 3.6 respectively.

Problem 3.5

$Minimize {\sum_{j = 1}^{n} (\sum_{i = 1}^{m} (a_{ij}^{L} (0), a_{ij}^{L} (1), a_{ij}^{R} (1), a_{ij}^{R} (0)) x_{i}) y_{j}}$

Subject to $\begin{matrix} \sum_{i = 1}^{m} (b_{il}^{L} (0), b_{il}^{L} (1), b_{il}^{R} (1), b_{il}^{R} (0)) x_{i} \underline{≺} (c_{l}^{L} (0), c_{l}^{L} (1), c_{l}^{R} (1), \\ c_{l}^{R} (0)), l = 1, 2, \dots, p; \\ \sum_{j = 1}^{n} (e_{kj}^{L} (0), e_{kj}^{L} (1), e_{kj}^{R} (1), e_{kj}^{R} (0)) y_{j} \underline{≻} (d_{k}^{L} (0), d_{k}^{L} (1), \\ d_{k}^{R} (1), d_{k}^{R} (0)), k = 1, 2, \dots, q; \\ x_{i} ⩾ 0, i = 1, 2, \dots, m; y_{j} ⩾ 0, j = 1, 2, \dots, n . \end{matrix}$

Problem 3.6

$Minimize {\sum_{i = 1}^{m} (\sum_{j = 1}^{n} (a_{ij}^{L} (0), a_{ij}^{L} (1), a_{ij}^{R} (1), a_{ij}^{R} (0)) y_{j}) x_{i}}$

Subject to

Constraints of Problem 3.5.

Step 2: Using the property, $\sum_{i = 1}^{n} λ (a_{i}^{L} (0), a_{i}^{L} (1), a_{i}^{R} (1)$ , $a_{i}^{R} (0)) = (\sum_{i = 1}^{n} λ a_{ij}^{L} (0), \sum_{i = 1}^{n} λ a_{ij}^{L} (1), \sum_{i = 1}^{n} λ a_{ij}^{R} (1), \sum_{i = 1}^{n} λ a_{ij}^{L} (0))$ , λ ⩾ 0, Problem 3.5 and Problem 3.6 can be transformed into Problem 3.7 and Problem 3.8 respectively.

Problem 3.7

$Minimize {\sum_{j = 1}^{n} (\begin{matrix} \sum_{i = 1}^{m} x_{i} a_{ij}^{L} (0), \sum_{i = 1}^{m} x_{i} a_{ij}^{L} (1), \\ \sum_{i = 1}^{m} x_{i} a_{ij}^{R} (1), \sum_{i = 1}^{m} x_{i} a_{ij}^{R} (0) \end{matrix}) y_{j}}$

Subject to $\begin{matrix} (\sum_{i = 1}^{m} b_{il}^{L} (0) x_{i}, \sum_{i = 1}^{m} b_{il}^{L} (1) x_{i}, \sum_{i = 1}^{m} b_{il}^{R} (1) x_{i}, \sum_{i = 1}^{m} b_{il}^{R} (0) x_{i}) \underline{≺} \\ (c_{l}^{L} (0), c_{l}^{L} (1), c_{l}^{R} (1), c_{l}^{R} (0)), l = 1, 2, \dots, p; \\ (\sum_{j = 1}^{n} e_{kj}^{L} (0) y_{j}, \sum_{j = 1}^{n} e_{kj}^{L} (1) y_{j}, \sum_{j = 1}^{n} e_{kj}^{R} (1) y_{j}, \sum_{j = 1}^{n} e_{kj}^{R} (0) y_{j}) \underline{≻} \\ (d_{k}^{L} (0), d_{k}^{L} (1), d_{k}^{R} (1), d_{k}^{R} (0)), k = 1, 2, \dots, q; \\ x_{i} ⩾ 0, i = 1, 2, \dots, m; y_{j} ⩾ 0, j = 1, 2, \dots, n . \end{matrix}$

Problem 3.8 $Minimize {\sum_{i = 1}^{m} (\begin{matrix} \sum_{j = 1}^{n} y_{j} a_{ij}^{L} (0), \sum_{j = 1}^{n} y_{j} a_{ij}^{L} (1), \\ \sum_{j = 1}^{n} y_{j} a_{ij}^{R} (1), \sum_{j = 1}^{n} y_{j} a_{ij}^{R} (0) \end{matrix}) x_{i}}$

Subject to

Constraints of Problem 3.7.

Step 3: Using the relation

$(a^{L} (0), a^{L} (1), a^{R} (1), a^{R} (0)) \underline{≺} (b^{L} (0), b^{L} (1), b^{R} (1), b^{R} (0))$ $\Rightarrow a^{L} (0) ⩽ b^{L} (0), a^{L} (1) ⩽ b^{L} (1), a^{R} (1) ⩽ b^{R} (1), a^{R} (0) ⩽$ $b^{R} (0) \Rightarrow λ_{1} a^{L} (0) + λ_{2} a^{L} (0) + λ_{3} a^{R} (1) + λ_{4} a^{R} (0) ⩽ λ_{1} b^{L} (0) +$ $λ_{2} b^{L} (0) + λ_{3} b^{R} (1) + λ_{4} b^{R} (0), where λ_{1}, λ_{2}, λ_{3}, λ_{4} ⩾ 0$

and λ₁ + λ₂ + λ₃ + λ₄ = 1, Problem 3.7 and Problem 3.8 can be transformed into Problem 3.9 and Problem 3.10 respectively.

Problem 3.9

$Minimize {\sum_{j = 1}^{n} (\begin{matrix} λ_{1} \sum_{i = 1}^{m} a_{ij}^{L} (0) x_{i} + λ_{2} \sum_{i = 1}^{m} a_{ij}^{L} (1) x_{i} + \\ λ_{3} \sum_{i = 1}^{m} a_{ij}^{R} (1) x_{i} + λ_{4} \sum_{i = 1}^{m} a_{ij}^{R} (0) x_{i} \end{matrix}) y_{j}}$

Subject to $\begin{matrix} λ_{1} \sum_{i = 1}^{m} b_{il}^{L} (0) x_{i} + λ_{2} \sum_{i = 1}^{m} b_{il}^{L} (1) x_{i} + λ_{3} \sum_{i = 1}^{m} b_{il}^{R} (1) x_{i} + λ_{4} \\ \sum_{i = 1}^{m} b_{il}^{R} (0) x_{i} ⩽ λ_{1} c_{l}^{L} (0) + λ_{2} c_{l}^{L} (1) + λ_{3} c_{l}^{R} (1) + λ_{4} c_{l}^{R} (0), \\ l = 1, 2, \dots, p; λ_{1} \sum_{j = 1}^{n} e_{kj}^{L} (0) y_{j} + λ_{2} \sum_{j = 1}^{n} e_{kj}^{L} (1) y_{j} + λ_{3} \\ \sum_{j = 1}^{n} e_{kj}^{R} (1) y_{j} + λ_{4} \sum_{j = 1}^{n} e_{kj}^{R} (0) y_{j} ⩾ λ_{1} d_{k}^{L} (0) + λ_{2} d_{k}^{L} (1) \\ + λ_{3} d_{k}^{R} (1) + λ_{4} d_{k}^{R} (0), k = 1, 2, \dots, q; x_{i} ⩾ 0, i = 1, 2, \dots, m; \\ y_{j} ⩾ 0, j = 1, 2, \dots, n . \end{matrix}$

Problem 3.10

$Minimize {\sum_{i = 1}^{m} (\begin{matrix} λ_{1} \sum_{j = 1}^{n} a_{ij}^{L} (0) y_{j} + λ_{2} \sum_{j = 1}^{n} a_{ij}^{L} (1) y_{j} + \\ λ_{3} \sum_{j = 1}^{n} a_{ij}^{R} (1) y_{j} + λ_{4} \sum_{j = 1}^{n} a_{ij}^{R} (0) y_{j} \end{matrix}) x_{i}}$

Subject to

Constraints of Problem 3.9.

Step 4: The dual of Problem 3.9 and Problem 3.10 by considering only y_j, j = 1, 2, …, n and x_i, i = 1, 2, …, m as decision variables are Problem 3.11 and Problem 3.12 respectively.

Problem 3.11 $Minimize {\begin{matrix} λ_{1} \sum_{k = 1}^{q} d_{k}^{L} (0) z_{k} + λ_{2} \sum_{k = 1}^{q} d_{k}^{L} (1) z_{k} + \\ λ_{3} \sum_{k = 1}^{q} d_{k}^{R} (1) z_{k} + λ_{4} \sum_{k = 1}^{q} d_{k}^{R} (0) z_{k} \end{matrix}}$

Subject to $\begin{matrix} λ_{1} \sum_{k = 1}^{q} e_{kj}^{L} (0) z_{k} + λ_{2} \sum_{k = 1}^{q} e_{kj}^{L} (1) z_{k} + λ_{3} \sum_{k = 1}^{q} e_{kj}^{R} (1) z_{k} \\ + λ_{4} \sum_{k = 1}^{q} e_{kj}^{R} (0) z_{k} ⩽ λ_{1} \sum_{i = 1}^{m} a_{ij}^{L} (0) x_{i} + λ_{2} \sum_{i = 1}^{m} a_{ij}^{L} (1) x_{i} \end{matrix}$ $\begin{matrix} + λ_{3} \sum_{i = 1}^{m} a_{ij}^{R} (1) x_{i} + λ_{4} \sum_{i = 1}^{m} a_{ij}^{R} (0) x_{i}, j = 1, 2, \dots, n; \\ λ_{1} \sum_{i = 1}^{m} b_{il}^{L} (0) x_{i} + λ_{2} \sum_{i = 1}^{m} b_{il}^{L} (1) x_{i} + λ_{3} \sum_{i = 1}^{m} b_{il}^{R} (1) x_{i} \\ + λ_{4} \sum_{i = 1}^{m} b_{il}^{R} (0) x_{i} ⩽ λ_{1} c_{l}^{L} (0) + λ_{2} c_{l}^{L} (1) + λ_{3} c_{l}^{R} (1) \\ + λ_{4} c_{l}^{R} (0), l = 1, 2, \dots, p; x_{i} ⩾ 0, i = 1, 2, \dots, m; \\ z_{k} ⩾ 0, k = 1, 2, \dots, q . \end{matrix}$

Problem 3.12 $Minimize {\begin{matrix} λ_{1} \sum_{l = 1}^{p} c_{l}^{L} (0) t_{l} + λ_{2} \sum_{l = 1}^{p} c_{l}^{L} (1) t_{l} + \\ λ_{3} \sum_{l = 1}^{p} c_{l}^{R} (1) t_{l} + λ_{4} \sum_{l = 1}^{p} c_{l}^{R} (0) t_{l} \end{matrix}}$

Subject to

$\begin{matrix} λ_{1} \sum_{l = 1}^{p} b_{il}^{L} (0) t_{l} + λ_{2} \sum_{l = 1}^{p} b_{il}^{L} (1) t_{l} \\ + λ_{3} \sum_{l = 1}^{p} b_{il}^{R} (1) t_{l} + λ_{4} \sum_{l = 1}^{p} b_{il}^{R} (0) t_{l} ⩾ \\ {\begin{matrix} λ_{1} \sum_{j = 1}^{n} a_{ij}^{L} (0) y_{j} + λ_{2} \sum_{j = 1}^{n} a_{ij}^{L} (1) y_{j} + \\ λ_{3} \sum_{j = 1}^{n} a_{ij}^{R} (1) y_{j} + λ_{4} \sum_{j = 1}^{n} a_{ij}^{R} (0) y_{j} \end{matrix}}, i = 1, 2, \dots, m; \\ λ_{1} \sum_{j = 1}^{n} e_{kj}^{L} (0) y_{j} + λ_{2} \sum_{j = 1}^{n} e_{kj}^{L} (1) y_{j} + λ_{3} \sum_{j = 1}^{n} e_{kj}^{R} (1) y_{j} \\ + λ_{4} \sum_{j = 1}^{n} e_{kj}^{R} (0) y_{j} ⩾ λ_{1} d_{k}^{L} (0) + λ_{2} d_{k}^{L} (1) + λ_{3} d_{k}^{R} (1) \\ + λ_{4} d_{k}^{R} (0), k = 1, 2, \dots, q; t_{l} ⩾ 0, l = 1, 2, \dots, p; \\ y_{j} ⩾ 0, j = 1, 2, \dots, n . \end{matrix}$

Step 5: Using the relation, λ₁a^L (0) + λ₂a^L (1) + λ₃a^R (1) + λ₄a^R (0) ⩾ λ₁b^L (0) + λ₂b^L (1) + λ₃b^R (1) + λ₄b^R (0) $\Rightarrow (a^{L} (0), a^{L} (1), a^{R} (1), a^{R} (0)) \underline{≻}$ (b^L (0) , b^L (1) , b^R (1) , b^R (0)), Problem 3.11 and Problem 3.12 can be transformed into Problem 3.13 and Problem 3.14 respectively.

Problem 3.13

$Minimize {(\begin{matrix} \sum_{k = 1}^{q} d_{k}^{L} (0) z_{k}, \sum_{k = 1}^{q} d_{k}^{L} (1) z_{k}, \\ \sum_{k = 1}^{q} d_{k}^{R} (1) z_{k}, \sum_{k = 1}^{q} d_{k}^{R} (0) z_{k} \end{matrix})}$

Subject to $\begin{matrix} (\sum_{k = 1}^{q} e_{kj}^{L} (0) z_{k}, \sum_{k = 1}^{q} e_{kj}^{L} (1) z_{k}, \sum_{k = 1}^{q} e_{kj}^{R} (1) z_{k}, \sum_{k = 1}^{q} e_{kj}^{R} (0) z_{k}) \underline{≺} \\ (\sum_{i = 1}^{m} a_{ij}^{L} (0) x_{i}, \sum_{i = 1}^{m} a_{ij}^{L} (1) x_{i}, \sum_{i = 1}^{m} a_{ij}^{R} (1) x_{i}, \sum_{i = 1}^{m} a_{ij}^{R} (0) x_{i}), \\ j = 1, 2, \dots, n; \\ (\sum_{i = 1}^{m} b_{il}^{L} (0) x_{i}, \sum_{i = 1}^{m} b_{il}^{L} (1) x_{i}, \sum_{i = 1}^{m} b_{il}^{R} (1) x_{i}, \sum_{i = 1}^{m} b_{il}^{R} (0) x_{i}) \underline{≺} \\ (c_{l}^{L} (0), c_{l}^{L} (1), c_{l}^{R} (1), c_{l}^{R} (0)), l = 1, 2, \dots, p; \\ x_{i} ⩾ 0, i = 1, 2, \dots, m; z_{k} ⩾ 0, k = 1, 2, \dots, q . \end{matrix}$

Problem 3.14

$Minimize {(\begin{matrix} \sum_{l = 1}^{p} c_{l}^{L} (0) t_{l}, \sum_{l = 1}^{p} c_{l}^{L} (1) t_{l}, \\ \sum_{l = 1}^{p} c_{l}^{R} (1) t_{l}, \sum_{l = 1}^{p} c_{l}^{R} (0) t_{l} \end{matrix})}$

Subject to $\begin{matrix} (\sum_{l = 1}^{p} b_{il}^{L} (0) t_{l}, \sum_{l = 1}^{p} b_{il}^{L} (1) t_{l}, \sum_{l = 1}^{p} b_{il}^{R} (1) t_{l}, \sum_{l = 1}^{p} b_{il}^{R} (0) t_{l}) \underline{≻} \\ (\sum_{j = 1}^{n} a_{ij}^{L} (0) y_{j}, \sum_{j = 1}^{n} a_{ij}^{L} (1) y_{j}, \sum_{j = 1}^{n} a_{ij}^{R} (1) y_{j}, \sum_{j = 1}^{n} a_{ij}^{R} (0) y_{j}), \\ i = 1, 2, \dots, m \\ (\sum_{j = 1}^{n} e_{kj}^{L} (0) y_{j}, \sum_{j = 1}^{n} e_{kj}^{L} (1) y_{j}, \sum_{j = 1}^{n} e_{kj}^{R} (1) y_{j}, \sum_{j = 1}^{n} e_{kj}^{R} (0) y_{j}) \underline{≻} \\ (d_{k}^{L} (0), d_{k}^{L} (1), d_{k}^{R} (1), d_{k}^{R} (0)), k = 1, 2, \dots, q; \\ t_{l} ⩾ 0, l = 1, 2, \dots, p; y_{j} ⩾ 0, j = 1, 2, \dots, n . \end{matrix}$

4 Literature review of constrained matrix games with fuzzy payoffs

In this section, a brief review of the methods, proposed in the literature for solving constrained matrix games with fuzzy payoffs, has been presented.

Li and Cheng [11] solved the Problem 3.11 and Problem 3.12 to find minimum expected gain of Player I and maximum expected loss of Player II respectively.

Li and Hong [13] split the Problem 3.13 and Problem 3.14 into (41), (36), (37), (42) and (43), (38), (39), (44) respectively in their work [13] to obtain the value of game.

5 Flaws of the existing methods

If a^L (0) , a^L (1) , a^R (1) , a^R (0) , b^L (0) , b^L (1) , b^R (1) and b^R (0) are real numbers then λ₁a^L (0) + λ₂a^L (1) + λ₃a^R (1) + λ₄a^R (0) ⩾ λ₁b^L (0) + λ₂b^L (1) + λ₃b^R (1) + λ₄b^R (0) , λ₁ + λ₂ + λ₃ + λ₄ = 1, λ₁, λ₂, λ₃, λ₄ ⩾ 0⇏a^L (0) ⩾ b^L (0) , a^L (1) ⩾ b^L (1), a^R (1) ⩾ b^R (1) , a^R (0) ⩾ b^R (0) e.g., if a^L (0) =1, a^L (1) =2, a^R (1) =3, a^R (0) =4, b^L (0) =0, $b^{L} (1) = 3, b^{R} (1) = 5, b^{R} (0) = 10, λ_{1} = λ_{2} = λ_{3} = λ_{4} = \frac{1}{4}$ then $λ_{1} a^{L} (0) + λ_{2} a^{L} (1) + λ_{3} a^{R} (1) + λ_{4} a^{R} (0) = \frac{10}{4}$ and $λ_{1} b^{L} (0) + λ_{2} b^{L} (1) + λ_{3} b^{R} (1) + λ_{4} b^{R} (0) = \frac{18}{4}$ ⇒λ₁b^L (0)+ λ₂b^L (1) + λ₃b^R (1) + λ₄b^R (0) >λ₁a^L (0) + λ₂a^L (1) + λ₃a^R (1) + λ₄a^R (0) but b^L (0) < a^L (0) , b^L (1) < a^L (1) , b^R (1) < a^R (1) , b^R (0) < a^R (0).

Similarly, if a^L (0) =1, a^L (1) =2, a^R (1) =3, a^R (0) =5, b^L (0) =0, b^L (1) =1, b^R (1) =4, b^R (0) =6, $λ_{1} = λ_{2} = λ_{3} = λ_{4} = \frac{1}{4}$ then $λ_{1} a^{L} (0) + λ_{2} a^{L} (1) + λ_{3} a^{R} (1) + λ_{4} a^{R} (0) = λ_{1} b^{L} (0) + λ_{2} b^{L} (1) + λ_{3} b^{R} (1) + λ_{4} b^{R} (0) = \frac{11}{4}$ . While, b^L (0) < a^L (0) , b^L (1) < a^L (1) , b^R (0) > a^R (0). However, Li and Hong [12,13, 12,13] have used the mathematical incorrect assumption λ₁a^L (0) + λ₂a^L (1) + λ₃a^R (1) + λ₄a^R (0) ⩾ λ₁b^L (0) + λ₂b^L (1) + λ₃b^R (1) + λ₄b^R (0) ; λ₁ + λ₂ + λ₃ + λ₄ = 1; λ₁, λ₂, λ₃, λ₄ ⩾ 0 ⇒ a^L (0) ⩾ b^L (0) , a^L (1) ⩾ b^L (1), a^R (1) ⩾ b^R (1) , a^R (0) ⩾ b^R (0) for transforming Problem 2.13 and Problem 3.14 into (41), (36), (37), (42) and (43), (38), (39), (44) respectively in their work [13].

Li and Hong [13] solved the linear programming problems (41), (36), (37), (42) and (43), (38), (39), (44) to obtain the minimum expected gain of player 1 and maximum expected loss of player 2. Then it is claimed that minimum expected gain of player 1 will be equal to maximum expected loss of player 2 where (V^L (0) , V^L (1) , V^R (1) , V^R (0)) = (d^Tz^L (0) , d^Tz^L (1) , d^Tz^R (1) , d^Tz^R (0)). Li and Hong [13] have solved these linear programming problems (41), (36), (37), (42) independently so V^L (0) , V^L (1) , V^R (1) and V^R (0) were obtained from four different vectors namely z^L (0) , z^L (1) , z^R (1) and z^R (0). It is pertinent to mention here that the maximum value of V^L (0), V^L (1) , V^R (1) and V^R (0) is attained at different values (namely z^L (0), z^L (1) , z^R (1) and z^R (0)) so there is no perception of common optimal value. But, Li and Hong [13] inadvertently transformed the Problem 3.13 into four independent linear programming problems (41), (36), (37), (42) which is invalid. Similar, interpretations hold for the Problem 3.14 and for the work presented in [12].

Remark: In [13], Li and Hong have used the variable y and z for the strategies of Player I and Player II respectively and the variable x is used to obtain value of game. In the present paper, the variables x, y and z are used in place y, z and x respectively.

6 Proposed method

In the existing methods [12,13, 12,13], it is assumed that if $\tilde{a} = (a^{L} (0), a^{L} (1), a^{R} (1), a^{R} (0))$ and $\tilde{b} = (b^{L} (0),$ b^L (1) b^R (1) , b^R (0)) are two trapezoidal fuzzy numbers then $\tilde{a} \underline{≻} \tilde{b}$ if a^L (0)⩾ b^L (0) , a^L (1) ⩾ b^L (1) , a^R (1) ⩾ b^R (1) , a^R (0) ⩾ b^R (0). Based on this comparing method, a new method is proposed in this section to find the minimum expected gain of Player I and corresponding optimal strategies of Player I. Likewise, the maximum expected loss of Player II and their corresponding optimal strategies of Player II can be obtained.

Using the comparing method, $\tilde{a} \underline{≺} \tilde{b}$ if a^L (0) ⩽ b^L (0) , a^L (1) ⩽ b^L (1) , a^R (1) ⩽ b^R (1) , a^R (0) ⩽ b^R (0), the minimum expected gain of Player I and the corresponding optimal strategies can be obtained as follows:

Step 1: Find $(y_{1}^{*}, y_{2}^{*}, \dots, y_{n}^{*}) \in Y$ such that the value of $\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i} (a_{ij}^{L} (0), a_{ij}^{L} (0), a_{ij}^{R} (1), a_{ij}^{R} (0)) y_{j}$ is minimum for all (x₁, x₂, …, x_m) ∈ X i.e., find the optimal solution of Problem 3.5.

Step 2: Using the property, $\sum_{i = 1}^{n} λ (a_{i}^{L} (0), a_{i}^{L} (1), a_{i}^{R} (1), a_{i}^{R} (0))$ = $(\sum_{i = 1}^{n} λ a_{ij}^{L} (0), \sum_{i = 1}^{n} λ a_{ij}^{L} (1), \sum_{i = 1}^{n} λ a_{ij}^{R} (1), \sum_{i = 1}^{n} λ a_{ij}^{L} (0)), λ ⩾ 0$ , the Problem 3.5 can be transformed into Problem 3.7.

Step 3: Using the property (a^L (0) , a^L (1) , a^R (1), $a^{R} (0)) \underline{≺} (b^{L} (0), b^{L} (1), b^{R} (1), b^{R} (0)) \Rightarrow a^{L} (0) ⩽ b^{L} (0)$ , a^L (1) ⩽ b^L (1) , a^R (1) ⩽ b^R (1) , a^R (0) ⩽ b^R (0), the Problem 3.7 can be transformed into Problem 6.1.

Problem 6.1

$Minimize {(\begin{matrix} \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i} a_{ij}^{L} (0) y_{j}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i} a_{ij}^{L} (1) y_{j}, \\ \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i} a_{ij}^{R} (1) y_{j}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i} a_{ij}^{R} (0) y_{j} \end{matrix})}$

Subject to $\begin{matrix} \sum_{i = 1}^{m} b_{il}^{L} (0) x_{i} ⩽ c_{l}^{L} (0), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{L} (1) x_{i} ⩽ c_{l}^{L} (1), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{R} (1) x_{i} ⩽ c_{l}^{R} (1), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{R} (0) x_{i} ⩽ c_{l}^{R} (0), l = 1, 2, \dots, p; \\ \sum_{j = 1}^{n} e_{kj}^{L} (0) y_{j} ⩾ d_{k}^{L} (0), k = 1, 2, \dots, q; \\ \sum_{j = 1}^{n} e_{kj}^{L} (1) y_{j} ⩾ d_{k}^{L} (1), k = 1, 2, \dots, q; \\ \sum_{j = 1}^{n} e_{kj}^{R} (1) y_{j} ⩾ d_{k}^{R} (1), k = 1, 2, \dots, q; \end{matrix}$ $\begin{matrix} \sum_{j = 1}^{n} e_{kj}^{R} (0) y_{j} ⩾ d_{k}^{R} (0), k = 1, 2, \dots, q; \\ x_{i} ⩾ 0, i = 1, 2, \dots, m; y_{j} ⩾ 0, j = 1, 2, \dots, n . \end{matrix}$

Step 4: According to comparing method, to find optimal solution ${y_{j}^{*}, j = 1, 2, \dots, n}$ of Problem 6.1 such that the value of $(\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i} a_{ij}^{L} (0) y_{j}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i} a_{ij}^{L} (1) y_{j},$ $\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i} a_{ij}^{R} (1) y_{j}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i} a_{ij}^{R} (0) y_{j})$ is minimum for all (x₁, x₂, …, x_m) ∈ X is equivalent to find ${y_{j}^{*}, j = 1, 2, \dots, n}$ such that value of $\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i} a_{ij}^{L} (0) y_{j}$ , $\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i} a_{ij}^{L} (1) y_{j}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i} a_{ij}^{R} (1) y_{j}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i} a_{ij}^{R} (0) y_{j}$ is minimum for all (x₁, x₂, … , x_m) ∈ X or if it is not possible to find so, then it is equivalent to find ${y_{j 1}^{*}, j = 1, 2, \dots, n}$ such that value of $\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 1} a_{ij}^{L} (0) y_{j 1}$ is minimum for all (x₁₁, x₂₁, …, x_m1) ∈ X but value of $\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 1} a_{ij}^{L} (1) y_{j 1}$ , and/or $\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 1} a_{ij}^{R} (1) y_{j 1}$ and/or $\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 1} a_{ij}^{R} (0) y_{j 1}$ is not minimum and to find ${y_{j 2}^{*}, j = 1, 2, \dots, n}$ such that value of $\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 2} a_{ij}^{L} (1) y_{j 2}$ is minimum for all (x₁₂, x₂₂, …, x_m2) ∈ X but value of $\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 2} a_{ij}^{L} (0) y_{j 2} and / or \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 2} a_{ij}^{R} (1) y_{j 2}$ and/or $\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 2} a_{ij}^{R} (0) y_{j 2}$ is not minimum and to find ${y_{j 3}^{*}, j = 1, 2, \dots, n}$ such that value of $\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 3} a_{ij}^{R} (1) y_{j 3}$ is minimum for all (x₁₃, x₂₃, …, x_m3) ∈ X but value of $\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 3} a_{ij}^{L} (0) y_{j 3}$ and/or $\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 3} a_{ij}^{L} (1) y_{j 3}$ and/or $\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 3} a_{ij}^{R} (0) y_{j 3}$ is not minimum and to find ${y_{j 4}^{*}, j = 1, 2, \dots, n}$ such that value of $\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 4} a_{ij}^{R} (0) y_{j 4}$ is minimum for all (x₁₄, x₂₄, …, x_m4) ∈ X but value of $\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 4} a_{ij}^{L} (0) y_{j 4} and / or \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 4} a_{ij}^{L} (1) y_{j 4}$ and/or $\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 4} a_{ij}^{R} (1) y_{j 4}$ is not minimum i.e., to find the optimal solution ${y_{j 1}^{*}, j = 1, 2, \dots, n}$ , ${y_{j 2}^{*}, j = 1, 2, \dots, n}$ , ${y_{j 3}^{*}, j = 1, 2, \dots, n}$ , ${y_{j 4}^{*}, j = 1, 2, \dots, n}$ of Problem 6.2, Problem 6.3, Problem 6.4, Problem 6.5 respectively.

Problem 6.2 $Minimize {\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 1} a_{ij}^{L} (0) y_{j 1}}$

Subject to $\begin{matrix} \sum_{i = 1}^{m} b_{il}^{L} (0) x_{i 1} ⩽ c_{l}^{L} (0), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{L} (1) x_{i 1} ⩽ c_{l}^{L} (1), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{R} (1) x_{i 1} ⩽ c_{l}^{R} (1), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{R} (0) x_{i 1} ⩽ c_{l}^{R} (0), l = 1, 2, \dots, p; \\ \sum_{j = 1}^{n} e_{kj}^{L} (0) y_{j 1} ⩾ d_{k}^{L} (0), k = 1, 2, \dots, q; \\ \sum_{j = 1}^{n} e_{kj}^{L} (1) y_{j 1} ⩾ d_{k}^{L} (1), k = 1, 2, \dots, q; \\ \sum_{j = 1}^{n} e_{kj}^{R} (1) y_{j 1} ⩾ d_{k}^{R} (1), k = 1, 2, \dots, q; \\ \sum_{j = 1}^{n} e_{kj}^{R} (0) y_{j 1} ⩾ d_{k}^{R} (0), k = 1, 2, \dots, q; \\ x_{i 1} ⩾ 0, i = 1, 2, \dots, m; y_{j 1} ⩾ 0, j = 1, 2, \dots, n . \end{matrix}$

Problem 6.3 $Minimize {\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 2} a_{ij}^{L} (1) y_{j 2}}$

Subject to $\begin{matrix} \sum_{i = 1}^{m} b_{il}^{L} (0) x_{i 2} ⩽ c_{l}^{L} (0), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{L} (1) x_{i 2} ⩽ c_{l}^{L} (1), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{R} (1) x_{i 2} ⩽ c_{l}^{R} (1), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{R} (0) x_{i 2} ⩽ c_{l}^{R} (0), l = 1, 2, \dots, p; \\ \sum_{j = 1}^{n} e_{kj}^{L} (0) y_{j 2} ⩾ d_{k}^{L} (0), k = 1, 2, \dots, q; \\ \sum_{j = 1}^{n} e_{kj}^{L} (1) y_{j 2} ⩾ d_{k}^{L} (1), k = 1, 2, \dots, q; \\ \sum_{j = 1}^{n} e_{kj}^{R} (1) y_{j 2} ⩾ d_{k}^{R} (1), k = 1, 2, \dots, q; \\ \sum_{j = 1}^{n} e_{kj}^{R} (0) y_{j 2} ⩾ d_{k}^{R} (0), k = 1, 2, \dots, q; \\ x_{i 2} ⩾ 0, i = 1, 2, \dots, m; y_{j 2} ⩾ 0, j = 1, 2, \dots, n . \end{matrix}$

Problem 6.4 $Minimize {\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 3} a_{ij}^{R} (1) y_{j 3}}$

Subject to $\begin{matrix} \sum_{i = 1}^{m} b_{il}^{L} (0) x_{i 3} ⩽ c_{l}^{L} (0), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{L} (1) x_{i 3} ⩽ c_{l}^{L} (1), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{R} (1) x_{i 3} ⩽ c_{l}^{R} (1), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{R} (0) x_{i 3} ⩽ c_{l}^{R} (0), l = 1, 2, \dots, p; \end{matrix}$ $\begin{matrix} \sum_{j = 1}^{n} e_{kj}^{L} (0) y_{j 3} ⩾ d_{k}^{L} (0), k = 1, 2, \dots, q; \\ \sum_{j = 1}^{n} e_{kj}^{L} (1) y_{j 3} ⩾ d_{k}^{L} (1), k = 1, 2, \dots, q; \\ \sum_{j = 1}^{n} e_{kj}^{R} (1) y_{j 3} ⩾ d_{k}^{R} (1), k = 1, 2, \dots, q; \\ \sum_{j = 1}^{n} e_{kj}^{R} (0) y_{j 3} ⩾ d_{k}^{R} (0), k = 1, 2, \dots, q; \\ x_{i 3} ⩾ 0, i = 1, 2, \dots, m; y_{j 3} ⩾ 0, j = 1, 2, \dots, n . \end{matrix}$

Problem 6.5 $Minimize {\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 4} a_{ij}^{R} (0) y_{j 4}}$

Subject to $\begin{matrix} \sum_{i = 1}^{m} b_{il}^{L} (0) x_{i 4} ⩽ c_{l}^{L} (0), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{L} (1) x_{i 4} ⩽ c_{l}^{L} (1), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{R} (1) x_{i 4} ⩽ c_{l}^{R} (1), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{R} (0) x_{i 4} ⩽ c_{l}^{R} (0), l = 1, 2, \dots, p; \\ \sum_{j = 1}^{n} e_{kj}^{L} (0) y_{j 4} ⩾ d_{k}^{L} (0), k = 1, 2, \dots, q; \\ \sum_{j = 1}^{n} e_{kj}^{L} (1) y_{j 4} ⩾ d_{k}^{L} (1), k = 1, 2, \dots, q; \\ \sum_{j = 1}^{n} e_{kj}^{R} (1) y_{j 4} ⩾ d_{k}^{R} (1), k = 1, 2, \dots, q; \\ \sum_{j = 1}^{n} e_{kj}^{R} (0) y_{j 4} ⩾ d_{k}^{R} (0), k = 1, 2, \dots, q; \\ x_{i 4} ⩾ 0, i = 1, 2, \dots, m; y_{j 4} ⩾ 0, j = 1, 2, \dots, n . \end{matrix}$

Step 5: Since, in Problem 6.2, Problem 6.3, Problem 6.4 and Problem 6.5 only y_j1, y_j2, y_j3 and y_j4 respectively have been considered as decision variables. So, Problem 6.2, Problem 6.3, Problem 6.4 and Problem 6.5 are linear programming problems and hence, the optimal value of Problem 6.2, Problem 6.3, Problem 6.4 and Problem 6.5 will be equal to its optimal value of corresponding dual problem i.e., Problem 6.6, Problem 6.7, Problem 6.8 and Problem 6.9 respectively.

Problem 6.6 $Minimize {\sum_{k = 1}^{q} (\begin{matrix} d_{k}^{L} (0) z_{k 1} + d_{k}^{L} (1) u_{k 1} + \\ d_{k}^{R} (1) v_{k 1} + d_{k}^{R} (0) w_{k 1} \end{matrix})}$

Subject to $\begin{matrix} {\sum_{k = 1}^{q} (\begin{matrix} e_{kj}^{L} (0) z_{k 1} + e_{kj}^{L} (1) u_{k 1} + \\ e_{kj}^{R} (1) v_{k 1} + e_{kj}^{R} (0) w_{k 1} \end{matrix})} \\ ⩽ \sum_{i = 1}^{m} a_{ij}^{L} (0) x_{i 1}, j = 1, 2, \dots, n; \\ \sum_{i = 1}^{m} b_{il}^{L} (0) x_{i 1} ⩽ c_{l}^{L} (0), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{L} (1) x_{i 1} ⩽ c_{l}^{L} (1), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{R} (1) x_{i 1} ⩽ c_{l}^{R} (1), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{R} (0) x_{i 1} ⩽ c_{l}^{R} (0), l = 1, 2, \dots, p; \\ x_{i 1} ⩾ 0, i = 1, 2, \dots, m; z_{k 1}, u_{k 1}, v_{k 1}, w_{k 1} ⩾ 0, \\ k = 1, 2, \dots, q . \end{matrix}$

Problem 6.7 $Minimize {\sum_{k = 1}^{q} (\begin{matrix} d_{k}^{L} (0) z_{k 2} + d_{k}^{L} (1) u_{k 2} + \\ d_{k}^{R} (1) v_{k 2} + d_{k}^{R} (0) w_{k 2} \end{matrix})}$

Subject to ${\sum_{k = 1}^{q} (\begin{matrix} e_{kj}^{L} (0) z_{k 2} + e_{kj}^{L} (1) u_{k 2} + \\ e_{kj}^{R} (1) v_{k 2} + e_{kj}^{R} (0) w_{k 2} \end{matrix})}$ $\begin{matrix} ⩽ \sum_{i = 1}^{m} a_{ij}^{L} (1) x_{i 2}, j = 1, 2, \dots, n; \\ \sum_{i = 1}^{m} b_{il}^{L} (0) x_{i 2} ⩽ c_{l}^{L} (0), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{L} (1) x_{i 2} ⩽ c_{l}^{L} (1), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{R} (1) x_{i 2} ⩽ c_{l}^{R} (1), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{R} (0) x_{i 2} ⩽ c_{l}^{R} (0), l = 1, 2, \dots, p; \\ x_{i 2} ⩾ 0, i = 1, 2, \dots, m; z_{k 2}, u_{k 2}, v_{k 2}, w_{k 2} ⩾ 0, \\ k = 1, 2, \dots, q . \end{matrix}$

Problem 6.8 $Minimize {\sum_{k = 1}^{q} (\begin{matrix} d_{k}^{L} (0) z_{k 3} + d_{k}^{L} (1) u_{k 3} + \\ d_{k}^{R} (1) v_{k 3} + d_{k}^{R} (0) w_{k 3} \end{matrix})}$

Subject to $\begin{matrix} {\sum_{k = 1}^{q} (\begin{matrix} e_{kj}^{L} (0) z_{k 3} + e_{kj}^{L} (1) u_{k 3} + \\ e_{kj}^{R} (1) v_{k 3} + e_{kj}^{R} (0) w_{k 3} \end{matrix})} \\ ⩽ \sum_{i = 1}^{m} a_{ij}^{R} (1) x_{i 3}, j = 1, 2, \dots, n; \\ \sum_{i = 1}^{m} b_{il}^{L} (0) x_{i 3} ⩽ c_{l}^{L} (0), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{L} (1) x_{i 3} ⩽ c_{l}^{L} (1), l = 1, 2, \dots, p; \end{matrix}$ $\begin{matrix} \sum_{i = 1}^{m} b_{il}^{R} (1) x_{i 3} ⩽ c_{l}^{R} (1), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{R} (0) x_{i 3} ⩽ c_{l}^{R} (0), l = 1, 2, \dots, p; \\ x_{i 3} ⩾ 0, i = 1, 2, \dots, m; z_{k 3}, u_{k 3}, v_{k 3}, w_{k 3} ⩾ 0, \\ k = 1, 2, \dots, q . \end{matrix}$

Problem 6.9 $Minimize {\sum_{k = 1}^{q} (\begin{matrix} d_{k}^{L} (0) z_{k 4} + d_{k}^{L} (1) u_{k 4} + \\ d_{k}^{R} (1) v_{k 4} + d_{k}^{R} (0) w_{k 4} \end{matrix})}$

Subject to

$\begin{matrix} {\sum_{k = 1}^{q} (\begin{matrix} e_{kj}^{L} (0) z_{k 4} + e_{kj}^{L} (1) u_{k 4} + \\ e_{kj}^{R} (1) v_{k 4} + e_{kj}^{R} (0) w_{k 4} \end{matrix})} \\ ⩽ \sum_{i = 1}^{m} a_{ij}^{R} (0) x_{i 4}, j = 1, 2, \dots, n; \\ \sum_{i = 1}^{m} b_{il}^{L} (0) x_{i 4} ⩽ c_{l}^{L} (0), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{L} (1) x_{i 4} ⩽ c_{l}^{L} (1), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{R} (1) x_{i 4} ⩽ c_{l}^{R} (1), l = 1, 2, \dots, p; \\ \sum_{i = 1}^{m} b_{il}^{R} (0) x_{i 4} ⩽ c_{l}^{R} (0), l = 1, 2, \dots, p; \\ x_{i 4} ⩾ 0, i = 1, 2, \dots, m; z_{k 4}, u_{k 4}, v_{k 4}, w_{k 4} ⩾ 0, \\ k = 1, 2, \dots, q . \end{matrix}$

Step 6: Substitute the optimal solution ${x_{i 1}^{*}, i = 1, 2, \dots, m}$ , ${x_{i 2}^{*}, i = 1, 2, \dots, m}, {x_{i 3}^{*}, i = 1, 2, \dots, m}$ and ${x_{i 4}^{*}, i = 1, 2, \dots, m}$ of Problem 6.6, Problem 6.7, Problem 6.8 and Problem 6.9 in Problem 6.2, Problem 6.3, Problem 6.4 and Problem 6.5 respectively and find all the alternative basic optimal solutions ${y_{j 1}^{g *}, g = 1, 2, \dots, u}, {y_{j 2}^{r *}, r = 1, 2, \dots, f}, {y_{j 3}^{s *}, s = 1, 2, \dots, h}$ and ${y_{j 4}^{t *}, t = 1, 2, \dots, w}$ of Problem 6.2, Problem 6.3, Problem 6.4 and Problem 6.5 respectively.

Step 7: Find $minimum {(\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 1}^{*} a_{ij}^{L} (0) y_{j 1}^{g *},$

$\begin{matrix} \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 1}^{*} a_{ij}^{L} (1) y_{j 1}^{g *}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 1}^{*} a_{ij}^{R} (1) y_{j 1}^{g *}, \end{matrix}$ $\begin{matrix} \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 1}^{*} a_{ij}^{R} (0) y_{j 1}^{g *}); \\ (\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 2}^{*} a_{ij}^{L} (0) y_{j 2}^{r *}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 2}^{*} a_{ij}^{L} (1) y_{j 2}^{r *}, \\ \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 2}^{*} a_{ij}^{R} (1) y_{j 2}^{r *}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 2}^{*} a_{ij}^{R} (0) y_{j 2}^{r *}); \\ (\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 3}^{*} a_{ij}^{L} (0) y_{j 3}^{s *}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 3}^{*} a_{ij}^{L} (1) y_{j 3}^{s *}, \\ \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 3}^{*} a_{ij}^{R} (1) y_{j 3}^{s *}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 3}^{*} a_{ij}^{R} (0) y_{j 3}^{s *}); \\ (\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 4}^{*} a_{ij}^{L} (0) y_{j 4}^{t *}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 4}^{*} a_{ij}^{L} (1) y_{j 4}^{t *}, \\ \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 4}^{*} a_{ij}^{R} (1) y_{j 4}^{t *}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 4}^{*} a_{ij}^{R} (0) y_{j 4}^{t *}), \\ g = 1, 2, \dots, u; r = 1, 2, \dots, f; s = 1, 2, \dots, h; \\ t = 1, 2, \dots, w \end{matrix}$

Step 8: All the trapezoidal fuzzy numbers which will be minimum, represents minimum expected gain of Player I and the optimal strategies for all such minimum which will be obtained corresponding to

$(\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 1}^{*} a_{ij}^{L} (0) y_{j 1}^{g *}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 1}^{*} a_{ij}^{L} (1) y_{j 1}^{g *},$ $\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 1}^{*} a_{ij}^{R} (1) y_{j 1}^{g *}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 1}^{*} a_{ij}^{R} (0) y_{j 1}^{g *}),$ $g = 1, 2, \dots, u; (\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 2}^{*} a_{ij}^{L} (0) y_{j 2}^{r *}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 2}^{*} a_{ij}^{L} (1) y_{j 2}^{r *},$ $\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 2}^{*} a_{ij}^{R} (1) y_{j 2}^{r *}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 2}^{*} a_{ij}^{R} (0) y_{j 2}^{r *}), r = 1, 2, \dots, f;$ $\begin{matrix} (\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 3}^{*} a_{ij}^{L} (0) y_{j 3}^{s *}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 3}^{*} a_{ij}^{L} (1) y_{j 3}^{s *}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} \\ x_{i 3}^{*} a_{ij}^{R} (1) y_{j 3}^{s *}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 3}^{*} a_{ij}^{R} (0) y_{j 3}^{s *}), s = 1, 2, \dots, h; \\ (\sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 4}^{*} a_{ij}^{L} (0) y_{j 4}^{t *}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 4}^{*} a_{ij}^{L} (1) y_{j 4}^{t *}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} \\ x_{i 4}^{*} a_{ij}^{R} (1) y_{j 4}^{t *}, \sum_{j = 1}^{n} \sum_{i = 1}^{m} x_{i 4}^{*} a_{ij}^{R} (0) y_{j 4}^{t *}), t = 1, 2, \dots, w, \end{matrix}$ $will be {x_{i 1}^{*}, i = 1, 2, \dots, m}, {x_{i 2}^{*}, i = 1, 2, \dots, m},$

${x_{i 3}^{*}, i = 1, 2, \dots, m}$ and ${x_{i 4}^{*}, i = 1, 2, \dots, m}$ respectively, which is optimal solution of Problem 6.6, Problem 6.7, Problem 6.8 and Problem 6.9 respectively.

7 Numerical examples

In this section, two examples have been presented to demonstrate the results of Section 6.

7.1 Existing numerical example considered by Li and Hong

Let us consider a simple numerical example which is given by Li and Hong [13]. Suppose the payoff matrix of Player I be

$\tilde{A} = [\begin{matrix} (27, 29, 32, 35) & (- 25, - 19, - 18, - 17) \\ (- 11, - 10, - 8, - 5) & (35, 40, \frac{81}{2}, 41) \end{matrix}],$

where the entries are trapezoidal fuzzy numbers. Let X = { (x₁, x₂) |x₁, x₂ ⩾ 0 ; x₁ + x₂ ⩽ 1 ; - x₁ - x₂ ⩽ -1 ; 80₁+ 50x₂ ⩽ 60 } and Y = { y₁, y₂ | y₁, y₂ ⩾ 0 ; y₁ + y₂ ⩾ 1 ; - y₁ - y₂⩾ -1 ; -40 y₁ - 70y₂ ⩾ -50 } be the set of constraints for Player I and Player II respectively.

According to Step 4 of the proposed method, to find the minimum expected gain of Player I, the mathematical programming problems are Problem 7.1, Problem 7.2, Problem 7.3 and Problem 7.4.

Problem 7.1

$Minimize {27 x_{11} y_{11} - 25 x_{11} y_{21} - 11 x_{21} y_{11} + 35 x_{21} y_{21}}$

Subject to

$\begin{matrix} 80 x_{11} + 50 x_{21} ⩽ 60; - 40 y_{11} - 70 y_{21} ⩾ - 50; \\ x_{11} + x_{21} = 1; y_{11} + y_{21} = 1; x_{11}, x_{21} ⩾ 0; y_{11}, y_{21} ⩾ 0 . \end{matrix}$

Problem 7.2

$Minimize {29 x_{12} y_{12} - 19 x_{12} y_{22} - 10 x_{22} y_{12} + 40 x_{22} y_{22}}$

Subject to

$\begin{matrix} 80 x_{12} + 50 x_{22} ⩽ 60; - 40 y_{12} - 70 y_{22} ⩾ - 50; \\ x_{12} + x_{22} = 1; y_{12} + y_{22} = 1; x_{12}, x_{22} ⩾ 0; y_{12}, y_{22} ⩾ 0 . \end{matrix}$

Problem 7.3

$Minimize {32 x_{13} y_{13} - 18 x_{13} y_{23} - 8 x_{23} y_{13} + \frac{81}{2} x_{23} y_{23}}$

Subject to

$\begin{matrix} 80 x_{13} + 50 x_{23} ⩽ 60; - 40 y_{13} - 70 y_{23} ⩾ - 50; \\ x_{13} + x_{23} = 1; y_{13} + y_{23} = 1; x_{13}, x_{23} ⩾ 0; y_{13}, y_{23} ⩾ 0 . \end{matrix}$

Problem 7.4

$Minimize {35 x_{14} y_{14} - 17 x_{14} y_{24} - 5 x_{24} y_{14} + 41 x_{24} y_{24}}$

Subject to

$\begin{matrix} 80 x_{14} + 50 x_{24} ⩽ 60; - 40 y_{14} - 70 y_{24} ⩾ - 50; \\ x_{14} + x_{24} = 1; y_{14} + y_{24} = 1; x_{14}, x_{24} ⩾ 0; y_{14}, y_{24} ⩾ 0 . \end{matrix}$

Following the Step 5 of the proposed method, the dual of Problem 7.1, Problem 7.2, Problem 7.3 and Problem 7.4 is Problem 7.5, Problem 7.6, Problem 7.7 and Problem 7.8 respectively.

Problem 7.5

Minimize{ - 50z₁₁ + z₂₁ } Subject to $\begin{matrix} - 40 z_{11} + z_{21} ⩽ 27 x_{11} - 11 x_{21}; - 70 z_{11} + z_{21} ⩽ \\ - 25 x_{11} + 35 x_{21}; 80 x_{11} + 50 x_{21} ⩽ 60; x_{11} \\ + x_{21} = 1; x_{11}, x_{21} ⩾ 0, z_{11} ⩾ 0 . \end{matrix}$

Problem 7.6

Minimize{ - 50z₁₂ + z₂₂ } Subject to $\begin{matrix} - 40 z_{12} + z_{22} ⩽ 29 x_{12} - 10 x_{22}; - 70 z_{11} + z_{21} ⩽ \\ - 19 x_{12} + 40 x_{22}; 80 x_{12} + 50 x_{22} ⩽ 60; x_{12} \\ + x_{22} = 1; x_{12}, x_{22} ⩾ 0, z_{12} ⩾ 0 . \end{matrix}$

Problem 7.7

Minimize{ - 50z₁₃ + z₂₃ } Subject to $\begin{matrix} - 40 z_{13} + z_{23} ⩽ 32 x_{13} - 8 x_{23}; - 70 z_{13} + z_{23} ⩽ \\ - 18 x_{13} + \frac{81}{2} x_{23}; 80 x_{13} + 50 x_{23} ⩽ 60; \\ x_{13} + x_{23} = 1; x_{13}, x_{23} ⩾ 0, z_{13} ⩾ 0 . \end{matrix}$

Problem 7.8

Minimize{ - 50z₁₄ + z₂₄ } Subject to $\begin{matrix} - 40 z_{14} + z_{24} ⩽ 35 x_{14} - 5 x_{24}; - 70 z_{14} + z_{24} ⩽ \\ - 17 x_{14} + 41 x_{24}; 80 x_{14} + 50 x_{24} ⩽ 60; x_{14} \\ + x_{24} = 1; x_{14}, x_{24} ⩾ 0, z_{14} ⩾ 0 . \end{matrix}$

Substituting the optimal solution ${x_{11}^{*} = \frac{1}{3}, x_{21}^{*} = \frac{2}{3}}, {x_{12}^{*} = \frac{1}{3}, x_{22}^{*} = \frac{2}{3}}, {x_{13}^{*} = \frac{1}{3}, x_{23}^{*} = \frac{2}{3}}$ and ${x_{14}^{*} = \frac{1}{3}, x_{24}^{*} = \frac{2}{3}}$ of Problem 7.5, Problem 7.6, Problem 7.7 and Problem 7.8 in Problem 7.1, Problem 7.2, Problem 7.3 and Problem 7.4 respectively. The obtained optimal solution of Problem 7.1, Problem 7.2, Problem 7.3 and Problem 7.4 is ${y_{11}^{*} = 1, y_{21}^{*} = 0}, {y_{12}^{*} = 1, y_{22}^{*} = 0}$ , ${y_{13}^{*} = 1, y_{23}^{*} = 0}$ and ${y_{14}^{*} = 1, y_{24}^{*} = 0}$ respectively.

As per the Step 7 of the proposed method, the obtained minimum value is $(\frac{5}{3}, \frac{9}{3}, \frac{16}{3}, \frac{25}{3})$ . This minimum value is obtained corresponding to optimal strategies $\begin{matrix} {x_{11}^{*} = \frac{1}{3}, x_{21}^{*} = \frac{2}{3}}, {x_{12}^{*} = \frac{1}{3}, x_{22}^{*} = \frac{2}{3}}, \\ {x_{13}^{*} = \frac{1}{3}, x_{23}^{*} = \frac{2}{3}} \end{matrix}$ and ${x_{14}^{*} = \frac{1}{3}, x_{24}^{*} = \frac{2}{3}} .$

Therefore, the minimum expected gain of Player I is $(\frac{5}{3}, \frac{9}{3}, \frac{16}{3}, \frac{25}{3})$ . Likewise, following the proposed method, the obtained maximum expected loss of Player II is $(\frac{5}{3}, \frac{9}{3}, \frac{16}{3}, \frac{25}{3})$ and corresponding optimal strategies are ${y_{11}^{*} = 1, y_{21}^{*} = 0}, {y_{12}^{*} = 1, y_{22}^{*} = 0}$ , ${y_{13}^{*} = 1, y_{23}^{*} = 0}$ and ${y_{14}^{*} = 1, y_{24}^{*} = 0}$ .

The minimum expected gain of Player I $(\tilde{ν})$ and maximum expected loss of Player II $(\tilde{ω})$ have trapezoidal fuzzy number values which are depicted in Fig. 1.

Fig. 1

The minimum expected gain of Player I and maximum expected loss of Player II.

7.2 Existing numerical example considered by Li and Cheng

Let us consider a simple numerical example which is given by Li and Cheng [11]. Suppose the payoff matrix of Player I be $\tilde{A} = [\begin{matrix} (18, 20, 23) & (- 21, - 18, - 16) \\ (- 33, - 32, - 27) & (38, 40, 43) \end{matrix}]$ where the entries are triangular fuzzy numbers. Let $X = {(x_{1}, x_{2}) | x_{1}, x_{2} ⩾ 0; x_{1} + x_{2} ⩽ 1; - x_{1} - x_{2} ⩽ - 1; (70, 80, 88) x_{1} + (44, 50, 54) x_{2} \underline{≺} (61, 67, 74)}$ $Y = {y_{1}, y_{2} | y_{1}, y_{2} ⩾ 0; y_{1} + y_{2} ⩾ 1; - y_{1} - y_{2} ⩾ - 1; (- 48, - 40, - 35) y_{1} + (- 79, - 70, - 65) y_{2} \underline{≻} (- 60, - 52, - 50)}$ be the set of constraints for Player I and Player II respectively.

According to the proposed method, the minimum expected gain of Player I and maximum expected loss of Player II is $(- \frac{41}{10}, - \frac{38}{15}, \frac{4}{3})$ . The corresponding optimal strategies for Player I and Player II are ${x_{11}^{*} = \frac{17}{30}, x_{21}^{*} = \frac{13}{30}}, {x_{12}^{*} = \frac{17}{30}, x_{22}^{*} = \frac{13}{30}}$ , ${x_{13}^{*} = \frac{17}{30}, x_{23}^{*} = \frac{13}{30}}$ and ${y_{11}^{*} = 1, y_{21}^{*} = 0}$ , ${y_{12}^{*} = 1, y_{22}^{*} = 0}, {y_{13}^{*} = 1, y_{23}^{*} = 0}$ respectively.

The minimum expected gain of Player I $(\tilde{ν})$ and maximum expected loss of Player II $(\tilde{ω})$ have trapezoidal fuzzy number values which are depicted in Fig. 2.

Fig. 2

The minimum expected gain of Player I and maximum expected loss of Player II.

8 Comparison analysis

On solving the exiting numerical example [13], the obtained optimal solution is ${z_{1}^{L} (0), z_{2}^{L} (0), z_{3}^{L} (0)} =$ ${0, \frac{5}{3}, 0}$ , ${z_{1}^{L} (1), z_{2}^{L} (1), z_{3}^{L} (1)} = {0, 3, 0}, {z_{1}^{R} (1), z_{2}^{R} (1), z_{3}^{R} (1)}$ $= {0, \frac{16}{3}, 0}, {z_{1}^{R} (1), z_{2}^{R} (1), z_{3}^{R} (1)} = {0, \frac{16}{3}, 0}$ and ${z_{1}^{R} (0), z_{2}^{R} (0), z_{3}^{R} (0)} = {0, \frac{25}{3}, 0}$ . Since, to obtain the value of game, Li and Hong [13] have used different set of vectors to obtain the common value of game which is invalid.

On solving the existing numerical example [11] with the existing method [12,13, 12,13], the obtained optimal solution is ${x_{1}^{L} (0) = \frac{17}{26}, x_{2}^{L} (0) = \frac{9}{26}}$ , ${x_{1} (1) = \frac{17}{30},$ $x_{2} (1) = \frac{9}{26}}, {x_{1}^{R} (0) = \frac{10}{17}, x_{2}^{R} (0) = \frac{7}{17}}$ , ${z_{1}^{L} (0) = \frac{12}{403}, z_{2}^{L} (0) = \frac{1431}{806}}, {z_{1} (1) = 0, z_{2} (1) = - \frac{38}{15}}$ and ${z_{1}^{R} (0) = 0, z_{2}^{R} (0) = \frac{41}{17}}$ . The value of game (V^L (0) , V (1) , V^R (0)) = (d^Tz^L (0) , d^Tz (1) , $d^{T} z^{R} (0)) = (- \frac{9}{806}, - \frac{38}{15}, \frac{41}{17})$ . Since, three different values z^L (0) , z (1) , z^R (0) are used to obtain the value of game which is not valid. Moreover, the different set of strategies are obtained through three linear programming problems. So, it cannot lead to common value of game.

9 Conclusion

The concept of fuzzy, introduced by Zadeh [19], is used for the payoffs of constrained matrix games. The strategies of Player I and Player II are obtained for the minimum expected gain and maximum expected loss respectively. Based on the comparing method, the multi objective mathematical programming problem has been transformed into its equivalent four mathematically programming problem and then the solution has been obtained. The proposed method ensures that the minimum expected gain and maximum expected loss will be a fuzzy number and the corresponding strategies will not vary with respect to the α- cuts. There are some flaws in the existing methods [12, 13] and the authors inadvertently used the wrong mathematical programming problem for constrained fuzzy matrix games. The method proposed in this paper overcome the flaws of the existing methods [12,13, 12,13] and is based on the correct mathematical programming problem.

Footnotes

Acknowledgments

The author conveys her sincere thanks to the anonymous reviewers for their insightful comments. The author is also thankful to the Indian Institute of Technology Ropar, Rupnagar, Punjab, India for the financial support given to her for pursuing Post-Doctorate.

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