Non-linear Programming Approach to Solve Bi-matrix Games with Payoffs Represented by I-fuzzy Numbers

Abstract

The aim of this paper is to develop a new methodology for solving bi-matrix games with payoffs of Atanassov’s intuitionistic fuzzy (I-fuzzy) numbers. In this methodology, we define the concepts of I-fuzzy numbers, the value-index and ambiguity-index, and develop a difference-index based ranking method. Hereby the parameterized non-linear programming models are derived from a pair of auxiliary I-fuzzy mathematical programming models, which are used to determine solutions of bi-matrix games with payoffs represented by I-fuzzy numbers. Validity and applicability of the models and method proposed in this paper are illustrated with a practical example.

Appendices

Appendix 1: Proof of Theorem 1

According to Eqs. (5) and (7), for any $\alpha \in [0,1]$, we have $(\rho \tilde{A} + \tilde{A}^{{\prime }} )^{\alpha } = \rho \tilde{A}^{\alpha } + \tilde{A}^{{{\prime }\alpha }}$. Hence, we have

$$\begin{aligned} L^{\alpha } (\rho \tilde{A} + \tilde{A}^{\prime}) + R^{\alpha } (\rho \tilde{A} + \tilde{A}^{\prime}) & = L^{\alpha } (\rho \tilde{A}) + L^{\alpha } (\tilde{A}^{\prime}) + R^{\alpha } (\rho \tilde{A}) + R^{\alpha } (\tilde{A}^{\prime}) \\ \, & = \rho (L^{\alpha } (\tilde{A}) + R^{\alpha } (\tilde{A})) + L^{\alpha } (\tilde{A}^{\prime}) + R^{\alpha } (\tilde{A}^{\prime}). \\ \end{aligned}$$

Combining with Eq. (9), we have

$$\begin{aligned} V_{\mu } (\rho \tilde{A} + \tilde{A}^{\prime}) & = \rho \int_{0}^{1} {[(L^{\alpha } (\tilde{A}) + R^{\alpha } (\tilde{A}))/2]f(\alpha ){\text{d}}\alpha } + \int_{0}^{1} {[(L^{\alpha } (\tilde{A}^{\prime}) + R^{\alpha } (\tilde{A}^{\prime}))/2]f(\alpha ){\text{d}}\alpha } \\ \, & = \rho V_{\mu } (w,\tilde{A}) + V_{\mu } (w,\tilde{A}^{\prime}), \\ \end{aligned}$$

i.e. $V_{\mu } (\rho \tilde{A} + \tilde{A}^{\prime}) = \rho V_{\mu } (\tilde{A}) + V_{\mu } (\tilde{A}^{\prime})$.

For any $\beta \in [0,1]$, it easily follows from Eqs. (6) and (8) that $(\rho \tilde{A} + \tilde{A}^{\prime})_{\beta } = \rho \tilde{A}_{\beta } + \tilde{A}^{\prime}_{\beta }$. According to Eq. (10), we can similarly prove that $V_{\upsilon } (\rho \tilde{A} + \tilde{A}^{\prime}) = \rho V_{\upsilon } (\tilde{A}) + V_{\upsilon } (\tilde{A}^{\prime})$.

For any $\alpha \in [0,1]$, if $\rho \ge 0$, it is easily derived from Eqs. (5) and (7) that

$$\begin{aligned} R^{\alpha } (\rho \tilde{A} + \tilde{A}^{\prime}) - L^{\alpha } (\rho \tilde{A} + \tilde{A}^{\prime}) & = R^{\alpha } (\rho \tilde{A}) - L^{\alpha } (\rho \tilde{A}) + R^{\alpha } (\tilde{A}^{\prime}) - L^{\alpha } (\tilde{A}^{\prime}) \\ \, & = \rho (R^{\alpha } (\tilde{A}) - L^{\alpha } (\tilde{A})) + R^{\alpha } (\tilde{A}^{\prime}) - L^{\alpha } (\tilde{A}^{\prime}). \\ \end{aligned}$$

Then, combining with Eq. (11), we have

$$\begin{aligned} W_{\mu } (\rho \tilde{A} + \tilde{A}^{{\prime }} ) & = \rho \int_{0}^{1} {(R^{\alpha } (\tilde{A}) - L^{\alpha } (\tilde{A}))f(\alpha ){\text{d}}\alpha } + \int_{0}^{1} {(R^{\alpha } (\tilde{A}^{{\prime }} ) - L^{\alpha } (\tilde{A}^{{\prime }} ))f(\alpha ){\text{d}}\alpha } \\ \, & = \rho W_{\mu } (\tilde{A}) + W_{\mu } (\tilde{A}^{{\prime }} ). \\ \end{aligned}$$

Likewise, if ρ < 0, then $R^{\alpha } (\rho \tilde{A} + \tilde{A}^{{\prime }} ) - L^{\alpha } (\rho \tilde{A} + \tilde{A}^{{\prime }} ) = \rho (L^{\alpha } (\tilde{A}) - R^{\alpha } (\tilde{A})) + R^{\alpha } (\tilde{A}^{{\prime }} ) - L^{\alpha } (\tilde{A}^{{\prime }} )$. Hereby, we have

$$\begin{aligned} W_{\mu } (\rho \tilde{A} + \tilde{A}^{{\prime }} ) & = \rho \int_{0}^{1} {(L^{\alpha } (\tilde{A}) - R^{\alpha } (\tilde{A}))f(\alpha ){\text{d}}\alpha } + \int_{0}^{1} {(R^{\alpha } (\tilde{A}^{{\prime }} ) - L^{\alpha } (\tilde{A}^{{\prime }} ))f(\alpha ){\text{d}}\alpha } \\ \, & = \rho W_{\mu } (\tilde{A}) + W_{\mu } (\tilde{A}^{{\prime }} ). \\ \end{aligned}$$

Therefore, we have proven that $W_{\mu } (\rho \tilde{A} + \tilde{A}^{\prime}) = \rho W_{\mu } (\tilde{A}) + W_{\mu } (\tilde{A}^{\prime})$ for any $\rho \in {\text{R}}$.

Similarly, according to Eqs. (6), (8) and (12), we can prove that $W_{\upsilon } (u,\rho \tilde{A} + \tilde{A}^{\prime}) = \rho W_{\upsilon } (u,\tilde{A}) + W_{\upsilon } (u,\tilde{A}^{\prime}).$

Appendix 2: Proof of Theorem 2

According to Theorem 1, it is derived from Eq. (13) that

$$\begin{aligned}& V_{\lambda } (\rho \tilde{A} + \tilde{A}^{\prime}) = \lambda V_{\upsilon } (\rho \tilde{A} + \tilde{A}^{\prime}) + (1 - \lambda )V_{\mu } (\rho \tilde{A} + \tilde{A}^{\prime}) \hfill \\ \, =& \rho [\lambda V_{\upsilon } (\tilde{A}) + (1 - \lambda )V_{\mu } (\tilde{A})] + \lambda V_{\upsilon } (\tilde{A}^{\prime}) + (1 - \lambda )V_{\mu } (\tilde{A}^{\prime}) = \rho V_{\lambda } (\tilde{A}) + V_{\lambda } (\tilde{A}^{\prime}) \hfill \\ \end{aligned}$$

i.e. $V_{\lambda } (\rho \tilde{A} + \tilde{A}^{{\prime }} ) = \rho V_{\lambda } (\tilde{A}) + V_{\lambda } (\tilde{A}^{{\prime }} ).$

Likewise, according to Theorem 1 and Eq. (14), we can prove that $W_{\lambda } (\rho \tilde{A} + \tilde{A}^{{\prime }} ) = \rho W_{\lambda } (\tilde{A}) + W_{\lambda } (\tilde{A}^{{\prime }} ).$

2.1 Proof of Theorem 3

According to Theorem 2, it is derived from Eq. (15) that

$$\begin{aligned} D_{\lambda } (\rho \tilde{A} + \tilde{A}^{{\prime }} ) = V_{\lambda } (\rho \tilde{A} + \tilde{A}^{{\prime }} ) - W_{\lambda } (\rho \tilde{A} + \tilde{A}^{{\prime }} ) = (\rho V_{\lambda } (\tilde{A}) + V_{\lambda } (\tilde{A}^{{\prime }} )) - (\rho W_{\lambda } (\tilde{A}) + W_{\lambda } (\tilde{A}^{{\prime }} )) \\ \, = \rho (V_{\lambda } (\tilde{A}) - W_{\lambda } (\tilde{A})) + (V_{\lambda } (\tilde{A}^{{\prime }} ) - W_{\lambda } (\tilde{A}^{{\prime }} )) = \rho D_{\lambda } (\tilde{A}) + D_{\lambda } (\tilde{A}^{{\prime }} ). \\ \end{aligned}$$

Thus, we have completed the proof of Theorem 3.

Appendix 3: Proof of Theorem 4

(P1) For any I-fuzzy number $\tilde{A}$, it directly follows from Eq. (15) that $D_{\lambda } (\tilde{A}) \ge D_{\lambda } (\tilde{A})$ for any $\lambda \in [0,1]$. Hereby, according to Definition 1, we have $\tilde{A} \ge_{\text{IF}} \tilde{A}$.

(P2) For any I-fuzzy numbers $\tilde{A}$ and $\tilde{A}^{{\prime }}$, according to Definition 1, we have $D_{\lambda } (\tilde{A}) \ge D_{\lambda } (\tilde{A}^{\prime})$ and $D_{\lambda } (\tilde{A}^{{\prime }} ) \ge D_{\lambda } (\tilde{A})$ for any $\lambda \in [0,1]$. Thus, $D_{\lambda } (\tilde{A}) = D_{\lambda } (\tilde{A}^{{\prime }} )$. Hereby, we have proven that $\tilde{A} \,=\,_{\text{IF}} \tilde{A}^{{\prime }}$.

(P3) For any I-fuzzy numbers $\tilde{A}$, $\tilde{A}^{{\prime }}$ and $\tilde{A}^{{{\prime \prime }}}$, according to Definition 1, we have $D_{\lambda } (\tilde{A}) \ge D_{\lambda } (\tilde{A}^{{\prime }} )$ and $D_{\lambda } (\tilde{A}^{{\prime }} ) \ge D_{\lambda } (\tilde{A}^{{\prime \prime }} )$ for any $\lambda \in [0,1]$. Hence, $D_{\lambda } (\tilde{A}) \ge D_{\lambda } (\tilde{A}^{\prime\prime})$. Therefore, we have proven that $\tilde{A} \,\ge\,_{\text{IF}}\;\tilde{A}^{{\prime \prime }}$.

(P4) It can be easily seen from Eqs. (9)–(15) that the difference-indices of I-fuzzy numbers $\tilde{A}$ and $\tilde{A}^{{\prime }}$ are completely determined by themselves. Thus, the ranking order of $\tilde{A}$ and $\tilde{A}^{{\prime }}$ completely depends on $D_{\lambda } (\tilde{A})$ and $D_{\lambda } (\tilde{A}^{{\prime }} )$, which have nothing to do with the other I-fuzzy numbers under comparison. Therefore, we have proven that $\tilde{A} >_{\text{IF}}\;\tilde{A}^{\prime}$ on $F_{1}$ if and only if $\tilde{A} >_{\text{IF}}\;\tilde{A}^{\prime}$ on $F_{2}$.

(P5) It is derived from Eqs. (9)–(10), we can obtain

$$V_{\mu } (\tilde{A}) = \int_{0}^{1} {[(L^{\alpha } (\tilde{A}) + R^{\alpha } (\tilde{A}))/2]f(\alpha ){\text{d}}\alpha } \ge \int_{0}^{1} {2\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} } \,\alpha {\text{d}}\alpha = \underline{a}$$

and $V_{\mu } (\tilde{A}') = \int_{0}^{1} {[(L^{\alpha } (\tilde{B}) + R^{\alpha } (\tilde{B}))/2]f(\alpha ){\text{d}}\alpha } \le \int_{0}^{1} {2\bar{a}'} \,\alpha {\text{d}}\alpha = \bar{a}'$

Combining with $\sup p(\tilde{A}) > \sup \sup p(\tilde{A}^{\prime})$, it directly follows that $V_{\mu } (\tilde{A}) > V_{\mu } (\tilde{A}^{\prime})$.

Similarly, it follows that $V_{\upsilon } (\tilde{A}) = \int_{0}^{1} {[(L_{\beta } (\tilde{A}) + R_{\beta } (\tilde{A}))/2]g(\beta ){\text{d}}\beta } \ge \int_{0}^{1} {2\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a} } \,\alpha {\text{d}}\alpha = \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}$ and $V_{\upsilon } (\tilde{A}^{{\prime }} ) = \int_{0}^{1} {[(L_{\beta } (\tilde{B}) + R_{\beta } (\tilde{B}))/2]g(\beta ){\text{d}}\beta } \le \int_{0}^{1} {2\bar{a}^{{\prime }} } \,\alpha {\text{d}}\alpha = \bar{a}^{{\prime }}$. Combining with $\sup p(\tilde{A}) > \sup \sup p(\tilde{A}^{\prime})$, it directly follows that $V_{\upsilon } (\tilde{A}) > V_{\upsilon } (\tilde{A}^{'} )$. Therefore, $\lambda V_{\upsilon } (\tilde{A}) + (1 - \lambda )V_{\mu } (\tilde{A}) > \lambda V_{\upsilon } (\tilde{A}') + (1 - \lambda )V_{\mu } (\tilde{A}')$, i.e. $V_{\lambda } (\tilde{A}) > V_{\lambda } (\tilde{A}^{\prime})$.

In a similar way, we have $\lambda W_{\mu } (\tilde{A}) + (1 - \lambda )W_{\upsilon } (\tilde{A}) > \lambda W_{\mu } (\tilde{A}') + (1 - \lambda )W_{\upsilon } (\tilde{A}')$, i.e. $W_{\lambda } (\tilde{A}) > W_{\lambda } (\tilde{A}^{\prime})$.

According to Definition 1, for any $\lambda \in [0,1]$, we have $D_{\lambda } (\tilde{A}) > D_{\lambda } (\tilde{A}^{\prime})$ if and only if $\tilde{A}$ is larger than $\tilde{A}^{{\prime }}$, i.e. $V(\tilde{A},\lambda ) - W(\tilde{A},\lambda ) > V(\tilde{A}^{{\prime }} ,\lambda ) - W(\tilde{A}^{{\prime }} ,\lambda )$. Hence, $\tilde{A} >_{\text{IF}}\;\tilde{A}^{\prime}$.

For instance, taking $f(\alpha ) = \alpha$ ($\alpha \in [0,1]$) and $g(\beta ) = 1 - \beta$ ($\beta \in [0,1]$), by using Eqs. (9)–(15), the difference-indexes of any I-fuzzy number $\tilde{A}$ can be obtained as follows:

$$\begin{aligned} D_{\lambda } (\tilde{A}) & = V(\tilde{A},\lambda ) - W(\tilde{A},\lambda ) = [\lambda V_{\upsilon } (\tilde{A}) + (1 - \lambda )V_{\mu } (\tilde{A})] - [\lambda W_{\mu } (\tilde{A}) + (1 - \lambda )W_{\upsilon } (\tilde{A})] \\ \, & = {{[(3\lambda - 2)\bar{a}_{2} + (6\lambda - 4)a_{2r} + (4 - 2\lambda )a_{2l} + (2 - \lambda )\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} )]} \mathord{\left/ {\vphantom {{[(3\lambda - 2)\bar{a}_{2} + (6\lambda - 4)a_{2r} + (4 - 2\lambda )a_{2l} + (2 - \lambda )\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} )]} {12}}} \right. \kern-0pt} {12}} \\ \, & \quad + {{[(1 - 3\lambda )\bar{a}_{1} + (2 - 6\lambda )a_{1r} + (2 + 2\lambda )a_{1l} + (1 + \lambda )\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{1} )]} \mathord{\left/ {\vphantom {{[(1 - 3\lambda )\bar{a}_{1} + (2 - 6\lambda )a_{1r} + (2 + 2\lambda )a_{1l} + (1 + \lambda )\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{1} )]} {12}}} \right. \kern-0pt} {12}}, \\ \end{aligned}$$

(23)

where $\underline{a}_{2} \le \underline{a}_{1} \le a_{2l} \le a_{1l} \le a_{1r} \le a_{2r} \le \bar{a}_{1} \le \bar{a}_{2}$. If $\sup p(\tilde{A}) > \sup \sup p(\tilde{A}^{\prime})$, i.e. $\underline{a} '_{1} \le a'_{2l} \le a'_{1l} \le$ $a'_{1r} \le a'_{2r} \le \bar{a}'_{1} \le \bar{a}_{2} ' < \underline{a}_{2} \le \underline{a}_{1} \le a_{2l} \le a_{1l} \le a_{1r} \le a_{2r} \le \bar{a}_{1} \le \bar{a}_{2}$, then it follows from Eq. (23) that

$$\begin{aligned} D_{\lambda } (\tilde{A}) - D_{\lambda } (\tilde{A}') > {{(3\lambda - 2)(\bar{a}_{2} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} ) + (6\lambda - 4)(a_{2r} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} ) + (4 - 2\lambda )(a_{2l} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} )]} \mathord{\left/ {\vphantom {{(3\lambda - 2)(\bar{a}_{2} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} ) + (6\lambda - 4)(a_{2r} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} ) + (4 - 2\lambda )(a_{2l} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} )]} {12}}} \right. \kern-0pt} {12}} \\ & + [{{(1 - 3\lambda )(\bar{a}_{1} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} ) + (2 - 6\lambda )(a_{1r} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} ) + (2 + 2\lambda )(a_{1l} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} ) + (1 + \lambda )(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{1} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} )]} \mathord{\left/ {\vphantom {{(1 - 3\lambda )(\bar{a}_{1} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} ) + (2 - 6\lambda )(a_{1r} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} ) + (2 + 2\lambda )(a_{1l} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} ) + (1 + \lambda )(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{1} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} )]} {12}}} \right. \kern-0pt} {12}} \\ & \ge {{[(7\lambda - 2)(a_{2l} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} ) + (6 - 6\lambda )(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{1} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} )]} \mathord{\left/ {\vphantom {{[(7\lambda - 2)(a_{2l} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} ) + (6 - 6\lambda )(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{1} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} )]} {12}}} \right. \kern-0pt} {12}} \\ & \ge (\lambda + 4)(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{1} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}_{2} ) \ge 0 \\ \end{aligned}.$$

Therefore, we have proven that if $\sup p(\tilde{A}) > \sup \sup p(\tilde{A}^{\prime})$, then $\tilde{A} >_{\text{IF}} \tilde{A}^{\prime}$.

(P6) In the same way to that of (P3), for any $\lambda \in [0,1]$, it follows from Definition 1 that

$$D_{\lambda } (\tilde{A}) \ge D_{\lambda } (\tilde{A}^{\prime}) .$$

(24)

Combining with Theorem 1, we have $D_{\lambda } (\tilde{A} + \tilde{A}^{\prime\prime}) = D_{\lambda } (\tilde{A}) +$ $D_{\lambda } (\tilde{A}^{\prime\prime}) \ge D_{\lambda } (\tilde{A}^{\prime}) + D_{\lambda } (\tilde{A}^{\prime\prime}){ = }D_{\lambda } (\tilde{A}^{\prime}{ + }\tilde{A}^{\prime\prime})$, i.e.

$$D_{\lambda } (\tilde{A} + \tilde{A}^{\prime\prime}) \ge D_{\lambda } (\tilde{A}^{\prime}{ + }\tilde{A}^{\prime\prime}) .$$

(25)

Hence, we have $\tilde{A} + \tilde{A}^{\prime\prime} \,\ge\,_{\text{IF}}\;\tilde{A}^{\prime} + \tilde{A}^{\prime\prime}$.

(P6′) Eq. (24) is a strictly inequality due to $\tilde{A}\, >\,_{\text{IF}}\;\tilde{A}^{\prime}$. Thus, Eq. (25) is also a strictly inequality. According to Definition 1, we have proven that $\tilde{A} + \tilde{A}^{{\prime \prime }}>_{\text{IF}}\;\tilde{A}^{{\prime }} + \tilde{A}^{{\prime \prime }}$.