Fuzzy Solution Concepts for Non-cooperative Games. Interval, Fuzzy and Intuitionistic Fuzzy Payoffs 978-3-030-16162-0

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Tina Verma Amit Kumar •

Fuzzy Solution Concepts for Non-cooperative Games Interval, Fuzzy and Intuitionistic Fuzzy Payoffs

123

Tina Verma Department of Mathematics Indian Institute of Technology Ropar Rupnagar, Punjab, India

Amit Kumar School of Mathematics Thapar Institute of Engineering and Technology Patiala, Punjab, India

ISSN 1434-9922 ISSN 1860-0808 (electronic) Studies in Fuzziness and Soft Computing ISBN 978-3-030-16161-3 ISBN 978-3-030-16162-0 (eBook) https://doi.org/10.1007/978-3-030-16162-0 Library of Congress Control Number: 2019935157 © Springer Nature Switzerland AG 2020

Preface

In the last decades, several methods have been proposed in the literature to find the solution of non-cooperative games with interval/fuzzy/intuitionistic fuzzy payoffs. However, after a deep study, it is observed that some mathematically incorrect assumptions have been considered in all these methods. Therefore, it is scientifically incorrect to use the existing methods to find the solution of real-life non-cooperative games with interval/fuzzy/intuitionistic fuzzy payoffs. The aim of this book is to provide the valid methods for solving different types of non-cooperative games with interval/fuzzy/intuitionistic fuzzy payoffs and to make the researchers aware about those mathematically incorrect assumptions which are considered in the existing methods. The contents of the book are divided into six chapters. In Chap. 1, a new method (named as Gaurika method) is proposed to obtain the optimal strategies as well as minimum expected gain of Player I and maximum expected loss of Player II for matrix games (or two-person zero-sum games) with interval payoffs (matrix games in which payoffs are represented by intervals). Furthermore, to illustrate the proposed Gaurika method, some existing numerical problems of matrix games with interval payoffs are solved by the proposed Gaurika method. In Chap. 2, the method (named as Mehar method) to obtain the optimal strategies as well as minimum expected gain of Player I and maximum expected loss of Player II for matrix games with fuzzy payoffs (matrix games in which payoffs are represented as fuzzy numbers) is proposed. Furthermore, to illustrate the proposed Mehar method, the existing numerical problems of matrix games with fuzzy payoffs are solved by the proposed Mehar method. In Chap. 3, a new method (named as Vaishnavi method) is 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 (constrained matrix games in which payoffs are represented by fuzzy numbers). In Chap. 4, new methods (named as Ambika method-I, Ambika method-II, Ambika method-III and Ambika method-IV) are proposed to obtain the optimal strategies as well as minimum expected gain of Player I and maximum expected loss of Player II for matrix games with intuitionistic fuzzy payoffs (matrix games in

which payoffs are represented by intuitionistic fuzzy numbers). Furthermore, to illustrate proposed Ambika methods, some existing numerical problems of matrix games with intuitionistic fuzzy payoffs are solved by proposed Ambika methods. In Chap. 5, a new method (named as Mehar method) is proposed for solving such bimatrix games or two-person non-zero sum games (matrix games in which gain of one player is not equal to the loss of other player) in which payoffs are represented by intuitionistic fuzzy numbers. In Chap. 6, based on the present study future work has been suggested. Rupnagar, India Patiala, India

Tina Verma Amit Kumar

Contents

1 Matrix Games with Interval Payoffs . . . . . . . . . . . . . . . . . . . . . 1.1 Matrix Games with Interval Payoffs . . . . . . . . . . . . . . . . . . 1.2 Existing Mathematical Formulation of Matrix Games with Interval Payoffs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Literature Review of Matrix Game with Interval Payoffs . . . 1.4 Arithmetic Operations over Intervals . . . . . . . . . . . . . . . . . 1.5 Flaws of the Existing Methods . . . . . . . . . . . . . . . . . . . . . 1.6 Invalidity of Existing Mathematical Formulation of Matrix Games with Interval Payoffs . . . . . . . . . . . . . . . . . . . . . . . 1.6.1 Existing Method to Obtain Mathematical Formulation of Matrix Games with Interval Payoffs . 1.6.2 Mathematically Incorrect Assumptions Considered in the Existing Method . . . . . . . . . . . . . . . . . . . . . . 1.7 Minimum and Maximum of Intervals . . . . . . . . . . . . . . . . . 1.7.1 Minimum of Intervals . . . . . . . . . . . . . . . . . . . . . . 1.7.2 Maximum of Intervals . . . . . . . . . . . . . . . . . . . . . . 1.8 Proposed Gaurika Method . . . . . . . . . . . . . . . . . . . . . . . . . 1.8.1 Minimum Expected Gain of Player I . . . . . . . . . . . . 1.8.2 Maximum Expected Loss of Player II . . . . . . . . . . . 1.9 Numerical Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.9.1 Existing Numerical Example Considered by Nayak and Pal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.9.2 Existing Numerical Example Considered by Li et al. 1.10 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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26 30 35 36

2 Matrix Games with Fuzzy Payoffs . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Matrix Games with Fuzzy Payoffs . . . . . . . . . . . . . . . . . . . . . . .

37 37

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2.2

Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Some Basic Definitions . . . . . . . . . . . . . . . . . . 2.2.2 Arithmetic Operations of Trapezoidal Fuzzy Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.3 Comparison of Fuzzy Numbers . . . . . . . . . . . . . 2.3 Existing Mathematical Formulation of Matrix Games with Fuzzy Payoffs . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Literature Review of Matrix Games with Fuzzy Payoffs 2.5 Flaws of the Existing Methods . . . . . . . . . . . . . . . . . . 2.6 Invalidity of Existing Mathematical Formulation of Matrix Games with Fuzzy Payoffs . . . . . . . . . . . . . . 2.7 Proposed Mehar Method . . . . . . . . . . . . . . . . . . . . . . . 2.7.1 Minimum Expected Gain of Player I . . . . . . . . . 2.7.2 Maximum Expected Loss of Player II . . . . . . . . 2.8 Numerical Example . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8.1 Minimum Expected Gain of Player I . . . . . . . . . 2.8.2 Maximum Expected Loss of Player II . . . . . . . . 2.9 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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49 49 50 53 57 57 60 62 62

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65 65

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66

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71 74 74 75 80 85

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86

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92 98 98

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3 Constrained Matrix Games with Fuzzy Payoffs . . . . . . . . . . . . . 3.1 Constrained Matrix Games with Fuzzy Payoffs . . . . . . . . . . . 3.2 Existing Mathematical Formulation of Constrained Matrix Games with Fuzzy Payoffs . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Literature Review of Constrained Matrix Games with Fuzzy Payoffs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Flaws of the Existing Methods . . . . . . . . . . . . . . . . . . . . . . 3.5 Proposed Vaishnavi Method . . . . . . . . . . . . . . . . . . . . . . . . 3.5.1 Minimum Expected Gain of Player I . . . . . . . . . . . . . 3.5.2 Maximum Expected Loss of Player II . . . . . . . . . . . . 3.6 Numerical Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.1 Existing Numerical Example Considered by Li and Hong . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.2 Existing Numerical Example Considered by Li and Cheng . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Matrix Games with Intuitionistic Fuzzy Payoffs . . . . . . . . . . . . 4.1 Matrix Games with Intuitionistic Fuzzy Payoffs . . . . . . . . . 4.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.2 Arithmetic Operations over Trapezoidal Vague Sets . 4.3 Existing Mathematical Formulation of Matrix Games with Intuitionistic Fuzzy Payoffs . . . . . . . . . . . . . . . . . . . .

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101 101 102 102 104

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4.4

Literature Review of Matrix Games with Intuitionistic Fuzzy Payoffs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Flaws of the Existing Methods . . . . . . . . . . . . . . . . . 4.6 Proposed Ambika Methods . . . . . . . . . . . . . . . . . . . . 4.6.1 Ambika Method-I . . . . . . . . . . . . . . . . . . . . . 4.6.2 Ambika Method-II . . . . . . . . . . . . . . . . . . . . . 4.6.3 Ambika Method-III . . . . . . . . . . . . . . . . . . . . 4.6.4 Ambika Method-IV . . . . . . . . . . . . . . . . . . . . 4.7 Numerical Examples . . . . . . . . . . . . . . . . . . . . . . . . . 4.7.1 Existing Numerical Example Considered by Nan et al. . . . . . . . . . . . . . . . . . . . . . . . . . 4.7.2 Existing Numerical Example Considered by Li et al. . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7.3 Existing Numerical Example Considered by Nan et al. . . . . . . . . . . . . . . . . . . . . . . . . . 4.7.4 Existing Numerical Example Considered by Nan et al. . . . . . . . . . . . . . . . . . . . . . . . . . 4.8 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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105 112 117 117 121 126 130 136

. . . . . . . . 136 . . . . . . . . 140 . . . . . . . . 144 . . . . . . . . 146 . . . . . . . . 149 . . . . . . . . 149

5 Bimatrix Games with Intuitionistic Fuzzy Payoffs . . . . . . . . . . . . 5.1 The Difference-Index Based Ranking Method . . . . . . . . . . . . 5.2 Maximum of Trapezoidal Intuitionistic Fuzzy Numbers . . . . 5.3 Flaws in the Existing Mathematical Formulation of Bimatrix Games with Intuitionistic Fuzzy Payoffs . . . . . . . . . . . . . . . . 5.3.1 Mathematical Formulation of Bimatrix Games with Intuitionistic Fuzzy Payoffs . . . . . . . . . . . . . . . . . . . 5.3.2 Mathematically Incorrect Assumption Considered by Li and Yang . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Exact Solution of Bimatrix Games with Intuitionistic Fuzzy Payoffs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.1 Exact Mathematical Formulation of Bimatrix Games with Intuitionistic Fuzzy Payoffs . . . . . . . . . . . . . . . . 5.4.2 Proposed Mehar Method . . . . . . . . . . . . . . . . . . . . . 5.4.3 Convergence of the Proposed Mehar Method . . . . . . 5.5 Numerical Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . 151 . . . 151 . . . 152 . . . 152 . . . 152 . . . 155 . . . 156 . . . . . .

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157 158 158 159 163 163

6 Future Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

Chapter 1

Matrix Games with Interval Payoffs

In this chapter, flaws of the existing methods [3–7, 11] for solving matrix games (or two person zero sum games) with interval payoffs (matrix games in which payoffs are represented by intervals) are pointed out. To resolve these flaws, a new method (named as Gaurika 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 of matrix games with interval payoffs. To illustrate the proposed Gaurika method, some existing numerical problems of matrix games with interval payoffs are solved by the proposed Gaurika method.

1.1 Matrix Games with Interval Payoffs Game theory [12] 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 welfare, 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 [14], in which he proved the famous minimax theorem for zero-sum games. It was also basis for the book “Theory of Games and Economic Behavior” by Von Neumann and Morgenstern [15]. But, Game theory was significantly advanced at Princeton University through the work of Nash [9]. Games are roughly be classified into two major categories: Cooperative games and Non-cooperative games [12]. Cooperation of players may be assumed in games, since it often exists in reality, though in many, if not most cases, the existence of non-cooperation is more attractive because it is often more realistic, especially in the presence of competition between players. In non-cooperative games, an important, from a conceptual and applications point of view, class of games are matrix games.

2

1 Matrix Games with Interval Payoffs

In matrix games, there are only two players Player I and Player II, and it can be denoted by an m × n matrix A of real numbers. Thus, such a game is called matrix game. A mixed strategy of PlayerI is a probability distribution x over the m  xi = 1, xi ≥ 0, rows of A, i.e., an element of the set X = x = (x1 , x2 , ..., xm ) | i=1  i = 1, 2, ..., m . Similarly, a strategy of Player II is a probability distribution y over  n  the columns of A i.e., an element of the set Y = y = (y1 , y2 , ..., yn ) | y j = 1, j=1  y j ≥ 0, j = 1, 2, ..., n . A strategy x of Player I is called pure if xi = 1 for some i = 1, 2, ..., m. Similarly, a strategy y of Player II is called pure if y j = 1 for some j = 1, 2, ..., n. If Player I plays over row i(i.e., pure strategy x = (0,  0, ..., xi = 1, 0, ..., 0)) and Player II plays over column j(i.e., pure strategy y = 0, 0, ..., y j = 1, 0, ..., 0 ), then Player I receives ai j and Player II pays ai j , where ai j is the entry in row i and column j of matrix A. If Player I plays strategy x and Player II plays strategy y, then Player I n m   ai j xi y j . receives the expected payoff E(x, y) = j=1

i=1

A position (i, j) is called saddle point if ai j ≥ ak j ∀ k = 1, 2, ..., m and ai j ≤ ail ∀ l = 1, 2, ..., n i.e., if ai j is maximal in its column j and minimal in its row i. Evidently, if (i, j) is a saddle point, then ai j must be the value of the game. In the classical (or crisp) game theory, usually the payoffs of players are represented by crisp values i.e. real numbers. But in real life, there is need to represent the players’ payoffs by their subjective judgments (or opinions) about competitive situations (or outcomes) instead of real numbers. For example, the decision problem in which two companies try to improve some products sales in some target market may be regarded as a game problem. In this scenario, the payoffs of players (i.e., companies) are represented by the company managers subjective judgments (or opinions) of the product shares in the target market at various situations. Such subjective judgments may be expressed with terms of linguistic variables such as “very large”,“larger”, “medium” and“small” as well as “smaller”. Obviously, these judgments usually involve some fuzziness or uncertainty due to the bounded rationality of players and behavior complexity. One way to describe impreciseness in the payoffs is to represent the payoffs by intervals. In the literature [1–7, 10, 11, 13], such matrix games in which payoffs are represented by intervals are named as matrix games with interval payoffs. Theorem 1.1 ([12]) If ai , i = 1, 2, ..., n are n real numbers then minimum  n n   ai xi | xi = 1, xi ≥ 0, i = 1, 2..., n = minimum{ai , i = 1, 2, ..., n}. i=1

i=1

Proof Let minimum{ai , i = 1, 2, ..., n} = ak . Then, ak ≤ ai ∀i = 1, 2, ..., n ⇒ ak xi ≤ ai xi , xi ≥ 0 ∀i = 1, 2, ..., n

1.1 Matrix Games with Interval Payoffs

⇒

n 

ak x i ≤

i=1

⇒ ak

n 

xi ≤

i=1

⇒ ak ≤

n 

n 

3

ai xi

i=1 n 

ai xi

i=1

ai xi

n    ⇒ ak ≤ minimum ai xi | xi = 1, xi ≥ 0, i = 1, 2..., n i=1

i=1

⇒ minimum {ai , i = 1, 2, ..., n} ≤ minimum

 n

ai xi |

i=1

n

 xi = 1, xi ≥ 0, i = 1, 2..., n

i=1

(1.1.1) Further, ak = a1 ×0 + a2 × 0 + ... + ak × 1 + ... + an × 0 n   ⇒ ak ≥ minimum ai xi | xi = 1, xi ≥ 0, i = 1, 2..., n i=1

⇒ minimum {ai , i = 1, 2, ..., n} ≥ minimum

 n

ai xi |

i=1

 From (1.1.1) and (1.1.2) minimum

n 

ai xi |

n

 xi = 1, xi ≥ 0, i = 1, 2..., n

i=1



(1.1.2)  xi = 1, xi ≥ 0, i = 1, 2..., n

i=1

= minimum{ai , i = 1, 2, ..., n}. L R Theorem 1.2 ([6]) If ai , ai , i = 1, 2, ..., n are n intervals then minimum n n 

 L R  xi = 1, xi ≥ 0, i = 1, 2..., n = minimum aiL , aiR , i = 1, 2, ai , ai xi | i=1

i=1

..., n}.

 L R 

Proof Let minimum ai , ai , i = 1, 2, ..., n = akL , akR . Then,

ak ≤ aiL , aiR ∀i = 1, 2, ..., n

⇒ akL , akR xi ≤ aiL , aiR xi , xi ≥ 0 ∀i = 1, 2, ..., n n n  L R  L R ⇒ ak , ak x i ≤ ai , ai xi i=1

i=1

n n



 aiL , aiR xi ⇒ akL , akR xi ≤ i=1

i=1

n



aiL , aiR xi ⇒ akL , akR ≤ i=1  n 

 L R  ai , ai xi | xi = 1, xi ≥ 0, i = 1, 2..., n ⇒ akL , akR ≤ minimum i=1

⇒ minimum

 n  n   



 aiL , aiR xi | xi = 1, xi ≥ 0, i = 1, 2..., n aiL , aiR , i = 1, 2, ..., n ≤ minimum i=1

Further,

+ anL , anR × 0

i=1





(1.1.3) L R L R ak , ak = a1 , a1 × 0 + a2L , a2R × 0 + ... + akL , akR × 1 + ...

4

1 Matrix Games with Interval Payoffs

  n n

  ⇒ akL , akR ≥ minimum ai xi | xi = 1, xi ≥ 0, i = 1, 2..., n i=1

⇒ minimum

i=1

 n  n   



 aiL , aiR xi | xi = 1, xi ≥ 0, i = 1, 2..., n aiL , aiR , i = 1, 2, ..., n ≥ minimum i=1

i=1

(1.1.4) From (1.1.3) and (1.1.4)  n n  L R  ai , ai xi | xi = 1, xi ≥ 0, i = 1, 2..., n minimum i=1 i=1  L R  = minimum ai , ai , i = 1, 2, ..., n . n  L R  Similarly, it can be proved that maximum ai , ai xi | xi = 1, xi ≥ 0, i=1   L R i = 1, 2..., n} = maximum ai , ai , i = 1, 2, ..., n .

1.2 Existing Mathematical Formulation of Matrix Games with Interval Payoffs Let Player I and Player II be two players and A = (ai j )m×n be the payoff matrix for Player I. Let S1 = {δ1 , δ2 , ..., δm } and S2 = {η1 , η2 , ..., ηn } be set of course of actions available to Player I and Player II respectively and x1 , x2 , ..., xm and y1 , y2 , ..., yn be the probabilities forselecting the course of action δ1 , δ2 , ..., δm and η1 , η2 , ...,  ηn m  respectively. Let X = x = (x1 , x2 , ..., xm )| xi = 1, xi ≥ 0, i = 1, 2, ..., m and i=1   n  Y = y = (y1 , y2 , ..., yn )| y j = 1, y j ≥ 0, j = 1, 2, ..., n be the set of stratej=1

gies for Player I and Player II respectively. Then, the optimal value of Problem 1.2.1 and Problem 1.2.2 represents the minimum expected gain of Player I and maximum expected loss of Player II respectively [12]. Further, the optimal solution {x1 , x2 , ..., xm } and {y1 , y2 , ..., yn } of Problem 1.2.1 and Problem 1.2.2 represents the optimal strategies of Player I and Player II respectively. Problem 1.2.1 Maximize{υ} Subject to m  ai j xi ≥ υ, j = 1, 2, ..., n; i=1 m 

xi = 1; xi ≥ 0, i = 1, 2, ..., m.

i=1

Problem 1.2.2 Minimize{ω} Subject to

1.2 Existing Mathematical Formulation of Matrix Games with Interval Payoffs n  j=1 n 

5

ai j y j ≤ ω, i = 1, 2, ..., m; y j = 1; y j ≥ 0, j = 1, 2, ..., n.

j=1

On the same direction, in the literature [3–7, 11], it is assumed that if payoffs are represented by intervals [aiLj , aiRj ] instead of real numbers ai j then to find the minimum expected gain of Player I and maximum expected loss of Player II is equivalent to find the optimal value of Problem 1.2.3 and Problem 1.2.4 respectively. Further, the optimal solution {x1 , x2 , ..., xm } and {y1 , y2 , ..., yn } of Problem 1.2.3 and Problem 1.2.4 represents the optimal strategies of Player I and Player II respectively. Problem 1.2.3 Maximize{[υ L , υ R ]} Subject to m  [aiLj , aiRj ]xi ≥ [υ L , υ R ], j = 1, 2, ..., n;

i=1 m 

xi = 1; xi ≥ 0, i = 1, 2, ..., m.

i=1

Problem 1.2.4 Minimize{[ω L , ω R ]} Subject to n  [aiLj , aiRj ]y j ≤ [ω L , ω R ], i = 1, 2, ..., m; j=1 n 

y j = 1; y j ≥ 0, j = 1, 2, ..., n.

j=1

Remark 1.1 It can be easily verified that the pair of Problems 1.2.1 and 1.2.2 is a primal-dual pair of crisp linear programming problems. So, the minimum expected gain of Player I will be equal to the maximum expected loss of Player II.

1.3 Literature Review of Matrix Game with Interval Payoffs In this section, a brief review of the methods, proposed in the literature in last ten years for solving matrix games with interval payoffs, is presented. Nayak and Pal [10] proposed a graphical method for solving 2 × n or m × 2 matrix games with interval payoffs. This method cannot be used for solving m × n matrix games with interval payoffs. Liu and Kao [7] transformed the Problem 1.2.3 into Problem 1.3.1 and Problem 1.3.2 for evaluating the right end point (υ R ) and left end point (υ L ) respectively of the minimum expected gain [υ L , υ R ] of Player I as well as Problem 1.2.4 into

6

1 Matrix Games with Interval Payoffs

Problem 1.3.3 and Problem 1.3.4 respectively for evaluating the left end point ω L and right end point ω R of the maximum expected loss [ω L , ω R ] of Player II. Liu and Kao [7] used the optimal values υ L , υ R and ω L , ω R to obtain the interval [υ L , υ R ] and [ω L , ω R ], representing the minimum expected gain of Player I and maximum expected loss of Player II respectively. Problem 1.3.1 υ R = Maximize{ν} Subject to m  pi j ≥ ν, j = 1, 2, ..., n; i=1 aiLj xi m 

≤ pi j ≤ aiRj xi , i = 1, 2, ..., m, j = 1, 2, ..., n;

xi = 1; xi ≥ 0, i = 1, 2, ..., m.

i=1

Problem 1.3.2 υ L = Minimize{u} Subject to n  qi j ≤ u, i = 1, 2, ..., m; j=1

aiLj y j ≤ qi j ≤ aiRj y j , i = 1, 2, ..., m j = 1, 2, ..., n; n  y j = 1; y j ≥ 0, j = 1, 2, ..., n. j=1

Problem 1.3.3 ω L = Minimize{u} Subject to n  si j ≤ u, i = 1, 2, ..., m; j=1

aiLj y j ≤ si j ≤ aiRj y j , i = 1, 2, ..., m, j = 1, 2, ..., n; n  y j = 1; y j ≥ 0, j = 1, 2, ..., n. j=1

Problem 1.3.4 ω R = Maximize{v} Subject to m  ti j ≥ v, j = 1, 2, ..., m; i=1 aiLj xi m  i=1

≤ ti j ≤ aiRj xi , i = 1, 2, ..., m j = 1, 2, ..., n;

xi = 1; xi ≥ 0, i = 1, 2, ..., m.

1.3 Literature Review of Matrix Game with Interval Payoffs

7

Liu and Kao [7] themselves pointed out that their method cannot be used to obtain the optimal strategy of matrix games with interval payoffs and have left the issue for future research work. Collins and Hu [3] proposed a method for finding the optimal solution of matrix games with interval payoffs by assuming the minimum expected gain of Player I and the maximum expected loss of Player II as real numbers. Collins and Hu [3] themselves pointed out that as the payoffs are intervals so the minimum expected gain of Player I and maximum expected loss of Player II will be a linear combination of intervals. Therefore, it is unrealistic to assume the minimum expected gain of Player I and maximum expected loss of Player II as real numbers. Collins and Hu [3] also pointed out that to propose a method for finding the optimal solution of matrix games with interval payoffs without considering this unrealistic assumption is left for future research work. Using Moore’s concept of set inclusion [8], Nayak and Pal [11] transformed the Problem 1.2.3 and Problem 1.2.4 into Problem 1.3.5 and Problem 1.3.6 for a fixed parameter α = 1 for evaluating left end point υ L of minimum expected gain [υ L , υ R ] of Player I and left end point ω L of maximum expected loss [ω L , ω R ] of Player II respectively as well as the corresponding optimal strategies. Problem 1.3.5 Maximize{υ L } Subject to m  aiLj xi ≥ υ L , j = 1, 2, ..., n;

i=1 m 

xi = 1; xi ≥ 0, i = 1, 2, ..., m.

i=1

Problem 1.3.6 Minimize{ω L } Subject to n  aiRj y j ≤ ω L , i = 1, 2, ..., m; j=1 n 

y j = 1; y j ≥ 0, j = 1, 2, ..., n.

j=1

Li [5] pointed out that Nayak and Pal [11], have wrongly employed Moore’s concept of set inclusions [8] to transform the Problem 1.2.4 into Problem 1.3.6. Li [5] claimed that using Moore’s concept of set inclusions [8], Problem 1.2.4 will be transformed into Problem 1.3.7 instead of Problem 1.3.6. Problem 1.3.7 Minimize{ω R } Subject to n n   aiRj y j ≤ ω R , i = 1, 2, ..., m; y j = 1; y j ≥ 0, j = 1, 2, ..., n. j=1

j=1

8

1 Matrix Games with Interval Payoffs

Li [5] further pointed out that minimum expected gain of Player I and the corresponding optimal strategy, obtained by solving Problem 1.3.5, depend only on the left limits/bounds of the interval [aiLj , aiRj ] whereas the maximum expected loss of Player II and the corresponding optimal strategy, obtained by solving Problem 1.3.7, depend only on the right limits/bounds of the interval [aiLj , aiRj ]. Li [5] also pointed out that the right end point υ R of minimum expected gain of Player I cannot be obtained by solving Problem 1.3.5 and left end point ω L of maximum expected loss of Player II cannot be obtained by solving Problem 1.3.7. Therefore, the minimum expected gain of Player I and maximum expected loss of Player II will not be bounded intervals, which shows that the method, proposed by Nayak and Pal [11], cannot be employed to solve a generic m × n matrix games with interval payoffs. To resolve these flaws of the method, proposed by Nayak and Pal [11], Li [5] transformed the Problem 1.2.3 and Problem 1.2.4 into Problem 1.3.8 and Problem 1.3.9 respectively.  υL + υR Maximize υ , 2 Subject to m  aiLj xi ≥ υ L , j = 1, 2, ..., n; Problem 1.3.8  L

i=1

(1 − α)

m  i=1

aiRj xi + α

m  i=1

aiLj xi ≥ (1 − α)υ R + αυ L , j = 1, 2, ..., n;

υ R − υ L ≥ 0; m  xi = 1; xi ≥ 0, i = 1, 2..., m. i=1

Problem  1.3.9 Minimize ω R ,

ωL + ωR 2



Subject to n  aiRj y j ≤ ω R , i = 1, 2, ..., m; j=1 n 

α

j=1 ωR − n 

aiRj y j + (1 − α)

n  j=1

aiLj y j ≤ αω R + (1 − α)ω L , i = 1, 2, ..., m;

ω L ≥ 0;

y j = 1; y j ≥ 0, j = 1, 2..., n.

j=1

Further, Li [5] transformed the Problem 1.3.8 into Problem 1.3.10 and Problem 1.3.11 as well as transformed the Problem 1.3.9 into Problem 1.3.12 and Problem 1.3.13. Finally, Li [5] used the optimal solution of Problem 1.3.11 to obtain the minimum expected gain and optimal strategy for Player I and used optimal solution of Problem 1.3.13 to obtain the maximum expected loss and optimal strategy of Player II.

1.3 Literature Review of Matrix Game with Interval Payoffs

Problem 1.3.10   Maximize υ L Subject to m  aiLj xi ≥ υ L , j = 1, 2, ..., n;

i=1

(1 − α)

m 

i=1 L

aiRj xi + α

m  i=1

aiLj xi ≥ (1 − α)υ R + αυ L , j = 1, 2, ..., n;

υ R − υ ≥ 0; m  xi = 1; xi ≥ 0, i = 1, 2..., m. i=1

Problem 1.3.11   L υ + υR Maximize 2 Subject to m  aiLj xi ≥ υ L , j = 1, 2, ..., n;

i=1

(1 − α)

m 

i=1 L

aiRj xi + α

m  i=1

aiLj xi ≥ (1 − α)υ R + αυ L , j = 1, 2, ..., n;

υ R − υ ≥ 0; υ L ≥ Optimal value of Problem 1.3.10; m  xi = 1; xi ≥ 0, i = 1, 2..., m. i=1

Problem 1.3.12  Minimize ω R Subject to n  aiRj y j ≤ ω R , i = 1, 2, ..., m; j=1 n 

α

j=1 R

aiRj y j + (1 − α)

n  j=1

aiLj y j ≤ αω R + (1 − α)ω L , i = 1, 2, ..., m;

ω − ω L ≥ 0; n 

y j = 1; y j ≥ 0, j = 1, 2..., n.

j=1

Problem  1.3.13  ωL + ωR Minimize 2 Subject to n  aiRj y j ≤ ω R , i = 1, 2, ..., m; j=1 n 

α

j=1

aiRj y j + (1 − α)

n  j=1

aiLj y j ≤ αω R + (1 − α)ω L , i = 1, 2, ..., m;

9

10

1 Matrix Games with Interval Payoffs

ω R − ω L ≥ 0; ω R ≥ Optimal value of Problem 1.3.12; n  y j = 1; y j ≥ 0, j = 1, 2..., n. j=1

Li et al. [6] transformed the Problem 1.3.8 into Problem 1.3.14, instead of transforming Problems 1.3.10 and 1.3.11 as well as transformed Problem 1.3.9 into Problem 1.3.15, instead of transforming Problems 1.3.12 and 1.3.13. Finally, Li et al. [6] used optimal solution of Problem 1.3.14 to obtain the minimum expected gain and optimal strategy for Player I and used optimal solution of Problem 1.3.15 to obtain the maximum expected loss and optimal strategy for Player II. Problem 1.3.14   L 3υ + υ R Maximize 4 Subject to m  aiLj xi ≥ υ L , j = 1, 2, ..., n;

i=1

(1 − α)

m 

i=1 L

aiRj xi + α

m  i=1

aiLj xi ≥ (1 − α)υ R + αυ L , j = 1, 2, ..., m;

υ R − υ ≥ 0; m  xi = 1; xi ≥ 0, i = 1, 2..., m. i=1

Problem  1.3.15  ω L + 3ω R Minimize 4 Subject to n  aiRj y j ≤ ω R , i = 1, 2, ..., m; j=1 n 

α

j=1 ωR − n 

aiRj y j + (1 − α)

n  j=1

aiLj y j ≤ αω R + (1 − α)ω L , i = 1, 2, ..., m;

ω L ≥ 0;

y j = 1; y j ≥ 0, j = 1, 2..., n.

j=1

Li [4] pointed out that on solving matrix games with interval payoffs, the obtained minimum expected gain of Player I, represented by interval, should be equal to the maximum expected loss of Player II. While, for the minimum expected gain for Player I and maximum expected loss of Player II, obtained on solving matrix games with interval payoffs by using existing methods [5, 6], this condition is not satisfying. To resolve this fundamental error, Li [4] split Problem 1.2.3 into Problem 1.3.16 and Problem 1.3.17 as well as split Problem 1.2.4 into Problem 1.3.18 and Problem 1.3.19. Finally, Li [4] used the optimal value of Problem 1.3.16 and Problem 1.3.17 to obtain the minimum expected gain of Player I and used the optimal value of Problems 1.3.18 and 1.3.19 to obtain maximum expected loss of Player II.

1.3 Literature Review of Matrix Game with Interval Payoffs

11

Problem 1.3.16   Maximize υ L Subject to m  aiLj xi ≥ υ L , j = 1, 2, ..., n;

i=1 m 

xi = 1; xi ≥ 0, i = 1, 2..., m.

i=1

Problem 1.3.17   Maximize υ R Subject to m  aiRj xi ≥ υ R , j = 1, 2, ..., n;

i=1 m 

xi = 1; xi ≥ 0, i = 1, 2..., m.

i=1

Problem 1.3.18  Minimize ω L Subject to n  aiLj y j ≤ ω L , i = 1, 2, ..., m; j=1 n 

y j = 1; y j ≥ 0, j = 1, 2..., n.

j=1

Problem 1.3.19  Minimize ω R Subject to n  aiRj y j ≤ ω R , i = 1, 2, ..., m; j=1 n 

y j = 1; y j ≥ 0, j = 1, 2..., n.

j=1

Akyar and Akyar [1] proposed a graphical method for solving 2 × n or m × 2 matrix games with interval payoffs. This method cannot be used for m × n matrix games with interval payoffs.

12

1 Matrix Games with Interval Payoffs

Akyar et al. [2] proposed an iterative method for solving crisp matrix games to matrix game with interval payoffs. Akyar et al. [2] themselves pointed out that by using this method only approximate solution is obtained while by using linear programming exact solution is obtained.

1.4 Arithmetic Operations over Intervals In this section, some arithmetic operations over intervals are presented [8]. Let A = [a L , a R ] and B = [b L , b R ] be two intervals. Then, (i) The addition of two intervals is defined as A + B = [a L + b L , a R + b R ]. (ii) The subtraction of two intervals is defined as A − B = [a L − b R , a R − b L ].  L [λa , λa R ] λ ≥ 0 (iii) If λ is a scalar then λA = R λa , λa L λ < 0 (iv) The multiplication of two intervals is defined as AB = [minimum{a L b L , a L b R , a R b L , a R b R }, maximum{a L b L , a L b R , a R b L , a R b R }].

1.5 Flaws of the Existing Methods In this section, flaws of the existing methods [3–7, 11] are pointed out. 1. Liu and Kao [7] ⎡ considered a matrix game with following payoff matrix ⎤ for 10 −30 20 10 [−30, −25] ⎢ [24, 30] 30 [−40, −30] 30 [−38, −34] ⎥ ⎥ Player I. A = ⎢ ⎣ [−3, 3] [35, 40] ⎦ −10 −30 20 −40 0 −20 −10 [23, 26] Liu and Kao [7] claimed that the minimum expected gain of Player I and maximum expected loss of Player II for the considered problem can be obtained by solving Problem 1.5.1 [7, Sect. 4, p. 1699] and Problem 1.5.2 [7, Sect. 4, p. 1699] respectively. Problem 1.5.1 [7, Sect. 4, p. 1699] Maximize{[υ L , υ R ]} Subject to 10x1 + [24, 30]x2 + [−3, 3]x3 − 40x4 ≥ [υ L , υ R ]; −30x1 + 30x2 + [35, 40]x3 ≥ [υ L , υ R ]; 20x1 + [−40, −30]x2 − 10x3 − 20x4 ≥ [υ L , υ R ]; [−30, −25]x1 + 30x2 − 30x3 + [23, 26]x4 ≥ [υ L , υ R ]; 10x1 + [−38, −34]x2 + 20x3 − 10x4 ≥ [υ L , υ R ]; x1 + x2 + x3 + x4 = 1; x1 , x2 , x3 , x4 ≥ 0.

1.5 Flaws of the Existing Methods

13

Problem 1.5.2 [7, Sect. 4, p. 1699] Minimize{[ω L , ω R ]} Subject to 10y1 − 30y2 + 20y3 + [−30, −25]y4 + 10y5 ≤ [ω L , ω R ]; [24, 30]y1 + 30y2 + [−40, −30]y3 + 30y4 + [−38, −34]y5 ≤ [ω L , ω R ]; [−3, 3]y1 + [35, 40]y2 − 10y3 − 30y4 + 20y5 ≤ [ω L , ω R ]; −40y1 − 20y3 + [23, 26]y4 − 10y5 ≤ [ω L , ω R ]; y1 + y2 + y3 + y4 + y5 = 1; y1 , y2 , y3 , y4 , y5 ≥ 0. In the optimal solution of the Problem 1.5.1 [7], the variables x1 , x2 , x3 and x4 should always be real numbers i.e., lower bound and upper bound of variables x1 , x2 , x3 and x4 should be equal. However, it is obvious from the optimal values of variables x1 , x2 , x3 and x4 , obtained by Liu and Kao [7] which are shown in Table 1.1, that the lower bound and upper bound of the variables x1 , x2 , x3 and x4 are not equal i.e., the optimal values of variables x1 , x2 , x3 and x4 , obtained by Liu and Kao [7], are not real numbers. Similarly, in the optimal solution of the Problem 1.5.2 [7] the variables y1 , y2 , y3 , y4 and y5 should always be real numbers i.e., lower bound and upper bound of the variables y1 , y2 , y3 , y4 and y5 should be equal. However, it is obvious from the optimal values of variables y1 , y2 , y3 , y4 and y5 , obtained by Liu and Kao [7] which are shown in Table 1.2, that the lower bound and upper bound of the variables y1 , y2 , y3 , y4 and y5 are not equal i.e., the optimal values of variables y1 , y2 , y3 , y4 and y5 obtained Liu and Kao [7], are not real numbers. Hence, neither the optimal solution, obtained by Liu and Kao [7] which are shown in Table 1.1, is the optimal solution of Problem 1.5.1 [7] nor the optimal solution, obtained by Liu and Kao [7] which are shown in Table 1.2, is the optimal solution of the Problem 1.5.2 [7]. 2. Li [5] and Li et al. [6] considered that if [a L , a R ] and [b L , b R ] are two intervals then for a particular value of α ∈ [0, 1), (i) [a L , a R ] ≤ [b L , b R ] i.e., minimum{[a, L , a R ], [b L , b R ]} = [a L , a R ] iff a R ≤ b R and (1 − α)a L + αa R ≤ (1 − α)b L + αb R . (ii) [a L , a R ] ≥ [b L , b R ] i.e., maximum{[a, L , a R ], [b L , b R ]} = [a L , a R ] iff a L ≥ b L and αa L + (1 − α)a R ≥ αb L + (1 − α)b R . Using this relation, Li [5] and Li et al. [6] claimed that to find the minimum expected gain of Player I i.e.,

Table 1.1 Lower bound and upper bound of variables [7]

Lower bound (xiL ) x1L x2L x3L x4L

= 0.4328 = 0.1517 = 0.1086 = 0.3069

Upper bound (xiR ) x1R x2R x3R x4R

= 0.4583 = 0.2186 = 0.1114 = 0.2117

14

1 Matrix Games with Interval Payoffs

Table 1.2 Lower bound and upper bound of variables [7]

Lower bound (y Lj )

Upper bound (y Rj )

y1L = 0.0715

y1R = 0

y2L = 0.2312

y2R = 0.1915

= 0.4933

y3R = 0.4362

= 0.2580

y4R = 0.2697

=0

y5R = 0.10126

y3L y4L y5L

m m  L R  [υ L , υ R ] = minimum [ai j , ai j ]xi , j = 1, 2, ..., n| xi = 1, xi ≥ 0, i = 1, i=1

i=1

2, ..., m} and to find the  maximum expected loss of Player II i.e., n n   [ω L , ω R ] = minimum [aiLj , aiRj ]y j , i = 1, 2, ..., m| y j = 1, y j ≥ 0, j = j=1

j=1

1, 2, ..., n} is equivalent to find the optimal solution of Problem 1.2.3 and Problem 1.2.4 respectively. Further, using the relation, [a L , a R ] ≤ [b L , b R ] ⇒ a R ≤ b R and (1 − α)a L + αa R ≤ (1 − α)b L + αb R and [a L , a R ] ≥ [b L , b R ] ⇒ a L ≥ b L and αa L + (1 − α)a R ≥ αb L + (1 − α)b R , Li [5] and Li et al. [6] transformed the Problem 1.2.3 into Problem 1.5.3. Problem 1.5.3 Maximize{[υ L , υ R ]} Subject to m  aiLj ≥ υ L , j = 1, 2, ..., n;

i=1 m 

α

i=1 m 

aiLj xi + (1 − α)

m  i=1

aiRj xi ≥ αυ L + (1 − α)υ R , j = 1, 2, ..., n;

xi = 1, xi ≥ 0, i = 1, 2, ..., m.

i=1

Now, if S = {xi , i = 1, 2, ..., m; υ L , υ R } is the set of all feasible solutions of ∗ ∗ Problem 1.5.3 then a feasible solution {xi∗ , i = 1, 2, ..., m; υ L , υ R } will be an ∗ ∗ ∗ optimal solution of Problem 1.5.3 if υ L ≤ υ L ∀υ L ∈ S and αυ L + (1 − α)υ R ≤ αυ L + (1 − α)υ R ∀υ L , υ R ∈ S. ∗ ∗ ∗ ∗ However, if υ1L , υ1R and υ2L , υ2R are two feasible solutions of Problem 1.5.3 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ such that υ1L ≥ υ2L but αυ1L + (1 − α)υ1R ≤ αυ2L + (1 − α)υ2R or υ1L ≤ υ2L but ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ αυ1L + (1 − α)υ1R ≥ αυ2L + (1 − α)υ2R . Then, both [υ1L , υ1R ] and [υ1L , υ1R ] will be the optimal solution of Problem 1.5.3. For example, if [1, 5] and [2, 3] are feasible solutions of Problem 1.5.3 at α = 21 then neither [1, 5] > [2, 3] nor [2, 3] < [1, 5]. So, both [1, 5] and [2, 3] will represent maximum {[2, 3], [1, 5]} and hence both [1, 5] and [2, 3] will be optimal value of Problem 1.5.3. To obtain the unique optimal value of Problem 1.5.3, Li [5] and Li et al. [6] used the following methods:

1.5 Flaws of the Existing Methods

15

(a) Li [5] used the following method to find the optimal solution {xi∗ , i = 1, 2, ..., m, ∗ ∗ υ L , υ R } from the set of all the feasible solutions S = {xi , i = 1, 2, ..., m, L R υ , υ }. Find minimum{υ L |υ L ∈ S}. Case 1: If minimum{υ L } occurs corresponding to only one feasible solution then that feasible solution is optimal solution of Problem 1.5.3. to two or more than two feasible Case 2: If minimum{υ L } occurs  Lcorresponding υ +υ R L R ; υ , υ are those feasible solution corresolutions then find minimum 2 sponding to which minimum{υ L } exists } . exists, The feasible solutions corresponding to which minimum value of υ +υ 2 will be the optimal solution of Problem 1.5.3. It is obvious that Li [5] has transformed the constraints of Problem 1.2.3 into constraints of Problem 1.5.3 by using the relation [a L , a R ] ≥ [b L , b R ] ⇒ a L ≥ b L and αa L + (1 − α)a R ≥ αb L + (1 − α)b R . While, Li [5] has used the relation L R L R > b +b for choosing the [a L , a R ] ≥ [b L , b R ] ⇒ a L > b L or if a L = b L then a +a 2 2 optimal solution of Problem 1.5.3 from all the possible feasible solutions of Problem 1.5.3 i.e., Li [5] has used two different methods simultaneously for solving Problem 1.5.3, which is mathematically incorrect. Similarly, it can be easily verified that Li [5] has used two different methods [a L , a R ] ≤ [b L , b R ] ⇒ a R ≤ b R and (1 − α)a L + αa R ≤ (1 − α)b L + αb R as well L R L R < b +b simultaneas [a L , a R ] ≤ [b L , b R ] ⇒ a R < b R or if a R = b R then a +a 2 2 ously for solving Problem 1.2.4, which is mathematically incorrect. L

R

(b) Li et al. [6] used the following method to find the optimal solution {xi∗ , i = ∗ ∗ 1, 2, ..., m, υ L , υ R } from the set of all the feasible solutions S = {xi , i = 1, 2, ..., m, υ L , υ R }. L R Step 1: Find maximum{ 3υ 4+υ |υ L , υ R ∈ S}. Step 2: All the feasible solution {xi , i = 1, 2, ..., m, υ L , υ R } corresponding to L R L R which value of 3υ 4+υ will be equal to maximum{ 3υ 4+υ |υ L , υ R ∈ S} will be optimal solution of Problem 1.5.3. It is obvious that Li et al. [6] have transformed the constraints of Problem 1.2.3 into constraints of Problem 1.5.3 by using the relation [a L , a R ] ≥ [b L , b R ] ⇒ a L ≥ b L and αa L + (1 − α)a R ≥ αb L + (1 − α)b R . While, Li et al. [6] have used the L R L R relation [a L , a R ] ≥ [b L , b R ] ⇒ 3a 4+a ≥ 3b 4+b for choosing the optimal solution of Problem 1.5.3 from all the possible feasible solutions of Problem 1.5.3 i.e., Li et al. [6] have used two different methods simultaneously for solving Problem 1.5.3, which is mathematically incorrect. Similarly, it can be easily verified that Li et al. [6] have used two different methods [a L , a R ] ≤ [b L , b R ] ⇒ a R ≤ b R and (1 − α)a L + αa R ≤ (1 − α)b L + αb R as L R L R ≤ b +3b simultaneously for solving Problem well as [a L , a R ] ≤ [b L , b R ] ⇒ a +3a 2 2 1.2.4, which is mathematically incorrect.

16

1 Matrix Games with Interval Payoffs

3. It is obvious that the Problem 1.3.16 and the Problem 1.3.17 are independent problems. Therefore, on solving the Problem 1.3.16 and the Problem 1.3.17 different optimal strategies will be obtained. Hence, the optimal strategy of the Problem 1.2.3 cannot be obtained with the help of optimal strategies of the Problem 1.3.16 and the Problem 1.3.17. Similarly, the optimal strategy of the Problem 1.2.4 cannot be obtained with the help of optimal strategies of the Problems 1.3.18 and 1.3.19. However, in the existing method [4], the minimum expected gain of Player I and maximum expected loss of Player II are obtained by transforming Problem 1.2.3 into Problems 1.3.16 and 1.3.17 and Problem 1.2.4 into Problems 1.3.18 and 1.3.19, which is mathematically incorrect.

1.6 Invalidity of Existing Mathematical Formulation of Matrix Games with Interval Payoffs In all the existing methods [3–7, 11], Problem 1.2.3 and Problem 1.2.4 are solved by different ways to find the minimum expected gain of Player I and maximum expected loss of Player II respectively. In this section, firstly the method, followed in the existing methods [3–7, 11], to obtain Problems 1.2.3 and 1.2.4 is presented. Then, it is shown that the authors have considered several mathematically incorrect assumptions to obtain Problems 1.2.3 and 1.2.4 and hence, Problems 1.2.3 and 1.2.4 as well as existing methods [3–7, 11], based on Problems 1.2.3 and 1.2.4, are not valid.

1.6.1 Existing Method to Obtain Mathematical Formulation of Matrix Games with Interval Payoffs Let Player I and Player II be two players and A = (ai j )m×n be the payoff matrix for Player I. Let S1 = {δ1 , δ2 , ..., δm } and S2 = {η1 , η2 , ..., ηn } be set of course of actions available to Player I and Player II respectively and x1 , x2 , ..., xm and y1 , y2 , ..., yn be the probabilities forselecting the course of action δ1 , δ2 , ..., δm and η1 , η2 , ...,  ηn m  respectively. Let X = x = (x1 , x2 , ..., xm )| xi = 1, xi ≥ 0, i = 1, 2, ..., m and i=1   n  Y = y = (y1 , y2 , ..., yn )| y j = 1, y j ≥ 0, j = 1, 2, ..., n be the set of stratej=1

gies for Player I and Player II respectively. Then, the optimal value of Problem 1.2.1 and Problem 1.2.2 represents the minimum expected gain of Player I and maximum expected loss of Player II respectively [12]. Then, to find minimum expected gain of n  m  Player I there is need to find {y1∗ , y2∗ , ..., yn∗ } so that the value of ( [aiLj , aiRj ]xi )y ∗j j=1 i=1

is minimum ∀{x1 , x2 , ..., xm } ∈ X and to find maximum expected loss of Player II

1.6 Invalidity of Existing Mathematical Formulation of Matrix Games …

there is need to find {x1∗ , x2∗ , ..., xm∗ } so that the value of

17

m  n  ( [aiLj , aiRj ]y j )xi∗ is

i=1 j=1

maximum ∀(y1 , y2 , ..., yn ) ∈ Y i.e., to find minimum expected gain of Player I there is need to find the optimal solution {y ∗j , j = 1, 2, ..., n} of Problem 1.6.1 and to find maximum expected loss of Player II, there is need to find the optimal solution {xi∗ , i = 1, 2, ..., m} of Problem 1.6.2. Problem  1.6.1  n  m  Minimize ( [aiLj , aiRj ]xi )y j j=1 i=1

Subject to m  xi = 1; xi ≥ 0, i = 1, 2, ..., m; i=1 n 

y j = 1; y j ≥ 0, j = 1, 2, ..., n.

j=1

Problem 1.6.2   m  n  L R Maximize ( [ai j , ai j ]y j )xi i=1 j=1

Subject to m  xi = 1; xi ≥ 0, i = 1, 2, ..., m; i=1 n 

y j = 1; y j ≥ 0, j = 1, 2, ..., n.

j=1

In the literature [3–7, 11], Problem 1.6.1 and Problem 1.6.2 have been transformed into Problem 1.6.3 and Problem 1.6.4 respectively and hence, can be transformed into Problem 1.6.5 and Problem 1.6.6 respectively. Problem  1.6.3  m  Minimize [aiLj , aiRj ]xi , j = 1, 2, ..., n i=1

Subject to m  xi = 1; xi ≥ 0, i = 1, 2, ..., m. i=1

Problem 1.6.4   n  Maximize [aiLj , aiRj ]y j , i = 1, 2, ..., m j=1

Subject to n  y j = 1; y j ≥ 0, j = 1, 2, ..., n. j=1

Problem 1.6.5 Maximize{[υ L , υ R ]} Subject to

18 m  i=1 m 

1 Matrix Games with Interval Payoffs

[aiLj , aiRj ]xi ≥ [υ L , υ R ], j = 1, 2, ..., n; xi = 1; xi ≥ 0, i = 1, 2, ..., m.

i=1

Problem 1.6.6 Minimize{[ω L , ω R ]} Subject to n  [aiLj , aiRj ]y j ≤ [ω L , ω R ], i = 1, 2, ..., m; j=1 n 

y j = 1; y j ≥ 0, j = 1, 2, ..., n.

j=1

In the literature [3–7, 11], the following method has been used to transform Problem 1.6.1 and Problem 1.6.2 into Problems 1.6.3 and Problem 1.6.4 and hence into Problem 1.6.5 and Problem 1.6.6 respectively. In the literature [3–7, that if [aiL , aiR ], i = 1, 2, ...,  n 11], it is proved  n are n intern  L R  vals then minimum [ai , ai ]xi | xi = 1, xi ≥ 0, i = 1, 2, ..., n = minimum i=1  i=1  L R [ai , ai ], i = 1, 2, ...,n . Using this result, the objective function minimum n  m n   ( [aiLj , aiRj ]xi )y j where, y j = 1; y j ≥ 0, j = 1, 2, ..., n of Problem 1.6.1 j=1 i=1 j=1 m  L R [ai j , ai j ]xi , can be transformed into the objective function minimum i=1  j = 1, 2, ..., n of Problem 1.6.3 and the objective function maximum   m  n m   L R ( [ai j , ai j ]y j )xi where, xi = 1; xi ≥ 0, i = 1, 2, ..., m of Problem 1.6.2 i=1 j=1 i=1  n  [aiLj , aiRj ]y j , i = 1, 2, can be transformed into the objective function maximum j=1

..., m} of Problem 1.6.4.

1.6.2 Mathematically Incorrect Assumptions Considered in the Existing Method If [1, 5] and [3, 4] are two intervals then on using the comparing method used in the existing methods [3–7, 11], minimum{[1, 5], [3, 4]} = [1, 5] and [3, 4]. While, | x1 + x2 = 1, x1 , x2 ≥ 0} = λ[1, 5] + (1 − λ)[3,4], minimum {[1, 5]x1 + [3, 4]x2 n n   λ ∈ [0, 1]. Hence, minimum [aiL , aiR ]xi | xi = 1, xi ≥ 0, i = 1, 2, ..., n = i=1 i=1  L R minimum [ai , ai ], i = 1, 2, ..., n . This clearly indicates that in the existing proof

1.6 Invalidity of Existing Mathematical Formulation of Matrix Games …



19

 n  [aiL , aiR ]xi | xi = 1, xi ≥ 0, i = 1, 2, ..., n = i=1 i=1  minimum [aiL , aiR ], i = 1, 2, ..., n ., presented in Theorem 1.2, and hence, in existing method to transform the Problem 1.6.1 and Problem 1.6.2 into Problem 1.6.3 and Problem 1.6.4 respectively, some mathematically incorrect assumptions have been considered. In this section, these mathematically incorrect assumptions are pointed out.

of

the

result

minimum

n 

(i) In the existing proof [6], presented in Theorem 1.2, it is assumed that minimum {[aiL , aiR ], i = 1, 2, ..., n} will be unique interval [a pL , a pR ] such that [a pL , a pR ] ≤ [aiL , aiR ]∀i = 1, 2, ..., n. However, this assumption is not valid. e.g., for the intervals [a L , a R ] = [3, 4] and [b L , b R ] = [1, 5], the condition a R < b R and (1 − α)a L + αa R > (1 − α)b L + αb R is satisfying at α = 21 . So, according to existing methods [3–7, 11] both the intervals [3, 4] and [1, 5] will represent minimum{[1, 5], [3, 4]}. Hence, to assume that minimum [aiL , aiR ], i = 1, 2, ..., n} will be a unique interval [a pL , a pR ] is mathematically invalid. (ii) In the existing proof, presented in Theorem 1.2, it is assumed that if [a pL , a pR ]   is an element of the set [aiL , aiR ], i = 1, 2, ..., n then [a pL , a pR ] ≤ minimum n  n  L R  [ai , ai ]xi | xi = 1, xi ≥ 0, i = 1, 2, ..., n . However, this assumption is i=1

i=1

not valid. e.g., [3, 4] is an element of the set {[1, 5], [3, 4]} but [3, 4] is neither smaller nor equal to [1, 5] which is obtained by substituting x1 = 0, x2 = 1 in [1, 5]x1 + [3, 4]x2 .

1.7 Minimum and Maximum of Intervals The methods, used for comparing intervals in the existing methods are not discussed in the published papers [3–7, 11] in detailed manner. So, in this section, the methods to find the minimum and maximum of intervals, used in the existing methods [3–7, 11], are presented in detailed manner.

1.7.1 Minimum of Intervals If [aiL , aiR ], i = 1, 2, ..., n are n intervals then minimum{[aiL , aiR ], i = 1, 2, ..., n}, obtained by using the approaches used in all the existing methods [3–7, 11], will be either two intervals [x L , x R ] and [y L , y R ] or will be a unique interval [x L , x R ] = [y L , y R ]. The intervals [x L , x R ] and [y L , y R ] can be obtained as follows: Step 1: Choose a real number α ∈ [0, 1]. Step 2: Find minimum{(1 − α)aiL + αaiR , i = 1, 2, ..., n} for chosen value of α. Case 1: If minimum{(1 − α)aiL + αaiR , i = 1, 2, ..., n} exist corresponding to unique value of i(say p) then x L = a pL and x R = a pR .

20

1 Matrix Games with Interval Payoffs

Case 2: If minimum{(1 − α)aiL + αaiR , i = 1, 2, ..., n} exist corresponding to more than one values of i(sayi = 1, 2, ..., k) then find minimum{aiR , i = 1, 2, ..., k}. Let, minimum{aiR , i = 1, 2, ..., k} exist corresponding to i = q then x L = aqL and x R = aqR . Step 3: Find minimum{aiR , i = 1, 2, ..., n}. Case 1: If minimum{aiR , i = 1, 2, ..., n} exist corresponding to unique value of i(say p) then y L = a pL and y R = a pR . Case 2: If minimum{aiR , i = 1, 2, ..., n} exist corresponding to more than one value of i(sayi = 1, 2, ..., k) then find minimum{(1 − α)aiL + αaiR , i = 1, 2, ..., k}. Let, minimum{(1 − α)aiL + αaiR , i = 1, 2, ..., k} exist corresponding to i = q then y L = aqL and y R = aqR .

1.7.2 Maximum of Intervals If [aiL , aiR ], i = 1, 2, ..., n are n intervals then maximum{[aiL , aiR ], i = 1, 2, ..., n}, obtained by using the approaches used in all the existing methods [3–7, 11], will be either two intervals [x L , x R ] and [y L , y R ] or will be a unique interval [x L , x R ] = [y L , y R ]. The intervals [x L , x R ] and [y L , y R ] can be obtained as follows: Step 1: Choose a real number α ∈ [0, 1]. Step 2: Find maximum{αaiL + (1 − α)aiR , i = 1, 2, ..., n} for chosen value of α. Case 1: If maximum{αaiL + (1 − α)aiR , i = 1, 2, ..., n} exist corresponding to unique value of i(say p) then x L = a pL and x R = a pR . Case 2: If maximum{αaiL + (1 − α)aiR , i = 1, 2, ..., n} exist corresponding to more than one values of i(sayi = 1, 2, ..., k) then find maximum{aiL , i = 1, 2, ..., k}. Let, maximum{aiL , i = 1, 2, ..., k} exist corresponding to i = q then x L = aqL and x R = aqR . Step 3: Find maximum{aiL , i = 1, 2, ..., n}. Case 1: If maximum{aiL , i = 1, 2, ..., n} exist corresponding to unique value of i(say p) then y L = a pL and y R = a pR . Case 2: If maximum{aiL , i = 1, 2, ..., n} exist corresponding to more than one value of i(sayi = 1, 2, ..., k) then find maximum{αaiL + (1 − α)aiR , i = 1, 2, ..., k}. Let, maximum{αaiL + (1 − α)aiR , i = 1, 2, ..., k} exist corresponding to i = q then y L = aqL and y R = aqR .

1.8 Proposed Gaurika Method It is obvious from Sect. 1.6 that it is not genuine to use Problems 1.2.3 and 1.2.4 to obtain the minimum expected gain of Player I and maximum expected loss of Player II as Problems 1.2.3 and 1.2.4 are obtained by considering some mathematically

1.8 Proposed Gaurika Method

21

incorrect assumptions. Hence, it is not genuine to use any of the existing methods [3–7, 11] in which the minimum expected gain of Player I and maximum expected loss of Player II is obtained with the help of optimal solution of Problem 1.2.3 and Problem 1.2.4 respectively. Furthermore, in the existing methods [3–7, 11], it is assumed that if [a L , a R ] and L [b , b R ] are two intervals then for a chosen value of α ∈ [0, 1), [a L , a R ] ≥ [b L , b R ] if a L ≥ b L and αa L + (1 − α)a R ≥ αb L + (1 − α)b R and [a L , a R ] ≤ [b L , b R ] if a R ≤ b R and (1 − α)a L + αa R ≤ (1 − α)b L + αb R . In this section, a new method (named as Gaurika method), on the basis of this comparing method, is proposed to find minimum expected gain of Player I, maximum expected loss of Player II and their corresponding optimal strategies.

1.8.1 Minimum Expected Gain of Player I Using the comparing method, [a L , a R ] ≤ [b L , b R ] if a R ≤ b R and (1 − α)a L + αa R ≤ (1 − α)b L + αb R , the minimum expected gain of Player I and corresponding optimal strategies can be obtained as follows: n  m  ( [aiLj , aiRj ]xi )y j is miniStep 1: Find (y1∗ , y2∗ , ..., yn∗ ) ∈ Y such that value of j=1 i=1

mum for all {x1 , x2 , .., xm } ∈ X i.e., find the optimal solution {y ∗j , j = 1, 2, ..., n} of Problem 1.8.1. Problem  1.8.1  n  m  L R Minimize ( [ai j , ai j ]xi )y j j=1 i=1

Subject to m n   xi = 1; y j = 1; xi ≥ 0, i = 1, 2, ..., m; y j ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 2: Using the property,

n  i=1

λ[aiL , aiR ] =

n  i=1

[λaiL , λaiR ], λ ≥ 0 the Problem 1.8.1

can be transformed into Problem 1.8.2. Problem  1.8.2  n  m  L R [ai j xi y j , ai j xi y j ] Minimize j=1 i=1

Subject to m n   xi = 1; y j = 1; xi ≥ 0, i = 1, 2, ..., m; y j ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 3: According to comparing method, to find the optimal solution {y1∗ , y2∗ , ..., yn∗ } n  m  of Problem 1.8.2 such that value of [aiLj xi y j , aiRj xi y j ] is minimum for all j=1 i=1

22

1 Matrix Games with Interval Payoffs

(x1 , x2 , ..., xm ) is equivalent to find {y1∗ , y2∗ , ..., yn∗ } such that value of (1 − α) aiLj xi y j + α

n  m  j=1 i=1

aiRj xi y j and

n  m  j=1 i=1

n  m  j=1 i=1

aiRj xi y j is minimum for all (x1 , x2 , ..., xm ) ∈

X or if it is not possible to find {y1∗ , y2∗ , ..., yn∗ } for which the value of (1 − n  n  n  m m m    α) aiLj xi y j + α aiRj xi y j and aiRj xi y j is minimum then find such j=1 i=1

j=1 i=1

j=1 i=1

{y ∗j1 , j = 1, 2, ..., n} for which the value of (1 − α) xi1 y j1 is minimum but value of

n  m  j=1 i=1

n  m  j=1 i=1

aiLj xi1 y j1 + α

n  m  j=1 i=1

aiRj xi1 y j1 is not minimum for all (x11 , x21 , ...,

xm1 ) ∈ X and to find such {y ∗j2 , j = 1, 2, ..., n} for which the value of (1 − α) aiLj xi2 y j2

+α

n  m  j=1 i=1

aiRj xi2 y j2

aiRj

is not minimum but value of

n  m 

aiRj xi2 y j2

n  m  j=1 i=1

is mini-

j=1 i=1 i.e., to find optimal solution {y ∗j1 ,

j = 1, 2, ..., n} mum for all (x12 , x22 , ..., xm2 ) ∈ X and {y ∗j2 , j = 1, 2, ..., n} of Problem 1.8.3 and Problem 1.8.4 respectively. Problem  1.8.3 Minimize (1 − α)

n  m  j=1 i=1

aiLj xi1 y j1

+α

n  m  j=1 i=1

 aiRj xi1 y j1

Subject to m n   xi1 = 1; y j1 = 1; xi1 ≥ 0, i = 1, 2, ..., m; y j1 ≥ 0, j = 1, 2, ..., n. i=1

j=1

Problem  1.8.4  n  m  R Minimize ai j xi2 y j2 j=1 i=1

Subject to m n   xi2 = 1; y j2 = 1; xi2 ≥ 0, i = 1, 2, ..., m; y j2 ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 4: Since, in Problems 1.8.3 and 1.8.4 only y j1 and y j2 respectively have been considered as decision variables. So, Problem 1.8.3 and Problem 1.8.4 are linear programming problem and hence, the optimal value of Problem 1.8.3 and Problem 1.8.4 will be equal to optimal value of its corresponding dual problem i.e., Problem 1.8.5 and Problem 1.8.6 respectively. Problem 1.8.5 Maximize{υ1 } Subject to m m   (1 − α) aiLj xi1 + α aiRj xi1 ≥ υ1 , j = 1, 2, ..., n; m  i=1

i=1

i=1

xi1 = 1; xi1 ≥ 0, i = 1, 2, ..., m.

1.8 Proposed Gaurika Method

23

Problem 1.8.6 Maximize{υ2 } Subject to m  aiRj xi2 ≥ υ2 , j = 1, 2, ..., n;

i=1 m 

xi2 = 1; xi2 ≥ 0, i = 1, 2, ..., m.

i=1 ∗ ∗ Step 5: Substitute the value of {xi1 , i = 1, 2, ..., m} and {xi2 , i = 1, 2, ..., m} of Problem 1.8.5 and Problem 1.8.6 in Problem 1.8.3 and Problem 1.8.4 respectively and ∗ find all the alternative basic optimal solution {y kj1 , j = 1, 2, ..., n; k = 1, 2, ..., l} q∗

and {y j2 , j = 1, 2, ..., n; q = 1, 2, ..., h} of Problem 1.8.3 and Problem 1.8.4 respectively.  n  n m m  ∗   q∗ ∗ ∗ Step 6: Find minimum xi1 [aiLj , aiRj ]y kj1 , xi1 [aiLj , aiRj ]y j2 , k = 1, 2, ..., j=1 i=1 j=1 i=1  l; q = 1, 2, ..., h .  Step 7: All the intervals which will be minimum 

n  m  j=1 i=1

∗

∗ xi1 [aiLj , aiRj ]y kj1 ,

n  m  j=1 i=1

q∗

∗ xi1 [aiLj , aiRj ]y j2 , k = 1, 2, ..., l; q = 1, 2, ..., h , that represent minimum expected

gain of Player I and the optimal strategies for all such minimum which will be n  m  ∗ ∗ ∗ obtained corresponding to xi1 [aiLj , aiRj ]y kj1 , k = 1, 2, ..., l will be {xi1 ,i = j=1 i=1

1, 2, ..., m} and for all such minimum which will be obtained corresponding to n  m  q∗ ∗ ∗ xi1 [aiLj , aiRj ]y j2 , q = 1, 2, ..., h will be {xi2 , i = 1, 2, ..., m}. j=1 i=1

1.8.2 Maximum Expected Loss of Player II Using the comparing method, [a L , a R ] ≥ [b L , b R ] if a L ≥ b L and αa L + (1 − α)a R ≥ αb L + (1 − α)b R , the maximum expected loss of Player II and corresponding optimal strategies can be obtained as follows: m  n  Step 1: Find (x1∗ , x2∗ , ..., xm∗ ) ∈ X such that value of ( [aiLj , aiRj ]y j )xi is maxii=1 j=1

mum for all (y1 , y2 , ..., yn ) ∈ Y i.e., find the optimal solution {xi∗ , i = 1, 2, ..., m} of Problem 1.8.7. Problem 1.8.7   m  n  L R Maximize ( [ai j , ai j ]y j )xi i=1 j=1

24

1 Matrix Games with Interval Payoffs

Subject to m n   xi = 1; y j = 1; xi ≥ 0, i = 1, 2, ..., m; y j ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 2: Using the property,

n  i=1

λ[aiL , aiR ] =

n  i=1

[λaiL , λaiR ], λ ≥ 0 the Problem 1.8.7

can be transformed into Problem 1.8.8. Problem 1.8.8   m  n  L R Maximize [ai j xi y j , ai j xi y j ] i=1 j=1

Subject to m n   xi = 1; y j = 1; xi ≥ 0, i = 1, 2, ..., m; y j ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 3: According to comparing method, to find the optimal solution of Problem 1.8.8 m  n  [aiLj xi y j , aiRj xi y j ] is maximum for all (y1 , y2 , ..., yn ) ∈ Y such that value of i=1 j=1

is equivalent to find {x1∗ , x2∗ , ..., xm∗ } such that value of α α)

n  m  j=1 i=1

aiRj xi y j

and

n  m  j=1 i=1

aiLj xi y j

n  m  j=1 i=1

aiRj xi y j and

for which the value but value of

n  m  j=1 i=1

n  m 

n  m  j=1 i=1

n  m  j=1 i=1

aiLj xi y j + (1 −

∗ aiLj xi y j is maximum then find such {xi1 , i = 1, 2, ..., m} j=1 i=1 n  n  m m   aiLj xi1 y j1 + (1 − α) aiRj xi1 y j1 is maximum of α j=1 i=1 j=1 i=1

aiLj xi1 y j1 is not maximum for all (y11 , y21 , ..., yn1 ) ∈ Y and

∗ , i = 1, 2, ..., m} for which the value of α to find such {xi2

α)

j=1 i=1

aiLj xi y j + (1 −

is maximum for all (y1 , y2 , ..., yn ) ∈ Y or if it

is not possible to find {x1∗ , x2∗ , ..., xm∗ } for which the value of α α)

n  m 

aiRj xi2 y j2 is not maximum but value of

n  m  j=1 i=1

n  m  j=1 i=1

aiLj xi2 y j2 + (1 −

aiLj xi2 y j2 is maximum for

∗ , i = 1, 2, ..., m} and all (y12 , y22 , ..., yn2 ) ∈ Y i.e., to find optimal solution {xi1 ∗ {xi2 , i = 1, 2, ..., m} of Problem 1.8.9 and Problem 1.8.10 respectively.

Problem 1.8.9   n  n  m m   Maximize α aiLj xi1 y j1 + (1 − α) aiRj xi1 y j1 j=1 i=1

j=1 i=1

Subject to m n   xi1 = 1; y j1 = 1; xi1 ≥ 0, i = 1, 2, ..., m; y j1 ≥ 0, j = 1, 2, ..., n. i=1

j=1

1.8 Proposed Gaurika Method

25

Problem 1.8.10   n  m  L Maximize ai j xi2 y j2 j=1 i=1

Subject to m n   xi2 = 1; y j2 = 1; xi2 ≥ 0, i = 1, 2, ..., m; y j2 ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 4: Since, in Problem 1.8.9 and Problem 1.8.10 only xi1 and xi2 , where i = 1, 2 respectively have been considered as decision variables. So, Problem 1.8.9 and Problem 1.8.10 are linear programming problem and hence, the optimal value of Problem 1.8.9 and Problem 1.8.10 will be equal to optimal value of its corresponding dual problem i.e., Problem 1.8.11 and Problem 1.8.12 respectively. Problem 1.8.11 Minimize{ω1 } Subject to n n   aiLj y j1 + (1 − α) aiRj y j1 ≤ ω1 , i = 1, 2, ..., m; α j=1

n 

j=1

y j1 = 1; y j1 ≥ 0, j = 1, 2, ..., n.

j=1

Problem 1.8.12 Minimize{ω2 } Subject to n  aiLj y j1 ≤ ω2 , i = 1, 2, ..., m; j=1 n 

y j2 = 1; y j2 ≥ 0, j = 1, 2, ..., n.

j=1

Step 5: Substitute the value of {y ∗j1 , j = 1, 2, ..., n} and {y ∗j2 , j = 1, 2, ..., n} of Problem 1.8.11 and Problem 1.8.12 in Problem 1.8.9 and Problem 1.8.10 respectively and k∗ , i = 1, 2, ..., m, k = 1, 2, ..., l} and find all the alternative basic optimal solution {xi1 q∗ {xi2 , i = 1, 2, ..., m, q = 1, 2, ..., h} of Problem 1.8.9 and Problem 1.8.10 respectively.  n  n  m m   q∗ k∗ L Step 6: Find maximum xi1 [ai j , aiRj ]y ∗j1 , xi2 [aiLj , aiRj ]y ∗j2 , k = 1, 2, ..., j=1 i=1

l; q = 1, 2, ..., h}.

j=1 i=1



n  n  m m   k∗ L Step 7: All the intervals which will be maximum xi1 [ai j , aiRj ]y ∗j1 , j=1 i=1 j=1 i=1  q∗ L R ∗ xi2 [ai j , ai j ]y j2 , k = 1, 2, ..., l; q = 1, 2, ..., h , will represent maximum expected loss of Player II and the optimal strategies for all such maximum which will be n  m  k∗ L xi1 [ai j , aiRj ]y ∗j1 , k = 1, 2, ..., l will be {y ∗j1 , j = obtained corresponding to j=1 i=1

26

1 Matrix Games with Interval Payoffs

1, 2, ..., n} and for all such maximum which will be obtained corresponding to n  m  q∗ xi2 [aiLj , aiRj ]y ∗j2 , q = 1, 2, ..., h will be {y ∗j2 , j = 1, 2, ..., n}. j=1 i=1

1.9 Numerical Examples In this section, some existing numerical examples have been solved by proposed Gaurika method.

1.9.1 Existing Numerical Example Considered by Nayak and Pal 

 [−3, −1] [4, 6] , [6, 8] [−7, −5] chosen by Nayak and Pal [11], is solved by the proposed Gaurika method.

In this section, matrix games with interval payoffs A =

1.9.1.1

Minimum Expected Gain of Player I

Using the proposed Gaurika method, minimum expected gain of Player I and corresponding optimal strategies, can be obtained as follows: Step 1: Find (y1∗ , y2∗ ) ∈ Y such that value of ([−3, −1]x1 y1 + [4, 6]x1 y2 + [6, 8] x2 y1 + [−7, −5]x2 y2 ) is minimum for all (x1 , x2 ) ∈ X i.e., find the optimal solution {y1∗ , y2∗ } of Problem 1.9.1. Problem 1.9.1 Minimize{[−3, −1]x1 y1 + [4, 6]x1 y2 + [6, 8]x2 y1 + [−7, −5]x2 y2 } Subject to x1 + x2 = 1; y1 + y2 = 1; x1 , x2 ≥ 0; y1 , y2 ≥ 0. Step 2: Using the property, [

n 

i=1

aiL ,

n  i=1

n  i=1

λ[aiL , aiR ] =

n  i=1

[λaiL , λaiR ], λ ≥ 0 and

n  i=1

[aiL , aiR ] =

aiR ], the Problem 1.9.1 can be transformed into Problem 1.9.2.

Problem 1.9.2 Minimize{[−3x1 y1 + 4x1 y2 + 6x2 y1 − 7x2 y2 , −1x1 y1 + 6x1 y2 + 8x2 y1 − 5x2 y2 ]} Subject to x1 + x2 = 1; y1 + y2 = 1; x1 , x2 ≥ 0; y1 , y2 ≥ 0. Step 3: According to comparing method, to find the optimal solution {y1∗ , y2∗ } of Problem 1.9.2 such that value of [−3x1 y1 + 4x1 y2 + 6x2 y1 − 7x2 y2 , −1x1 y1 + 6x1 y2 +

1.9 Numerical Examples

27

8x2 y1 − 5x2 y2 ] is minimum for all (x1 , x2 ) ∈ X is equivalent to find {y1∗ , y2∗ } such that value of (1 − α)(−3x1 y1 + 4x1 y2 + 6x2 y1 − 7x2 y2 ) + α(−1x1 y1 + 6x1 y2 + 8x2 y1 − 5x2 y2 ) and (−1x1 y1 + 6x1 y2 + 8x2 y1 − 5x2 y2 ) is minimum for all (x1 , x2 ) ∈ X or if it is not possible to find {y1∗ , y2∗ } for which the value of (1 − α)(−3x1 y1 + 4x1 y2 + 6x2 y1 − 7x2 y2 ) + α(−1x1 y1 + 6x1 y2 + 8x2 y1 − 5x2 y2 ) and ∗ ∗ , y21 } for which the −1x1 y1 + 6x1 y2 + 8x2 y1 − 5x2 y2 is minimum then find such {y11 value of (1 − α)(−3x11 y11 + 4x11 y21 + 6x21 y11 − 7x21 y21 ) + α(−1x11 y11 + 6x11 y21 + 8x21 y11 − 5x21 y21 ) is minimum but value of (−1x11 y11 + 6x11 y21 + ∗ ∗ , y22 } 8x21 y11 − 5x21 y21 ) is not minimum for all (x11 , x21 ) ∈ X and to find such {y12 for which the value of (1 − α)(−3x12 y12 + 4x12 y22 + 6x22 y12 − 7x22 y22 ) + α(−1x12 y12 + 6x12 y22 + 8x22 y12 − 5x22 y22 ) is not minimum but value of (−1x12 y12 + 6x12 y22 + 8x22 y12 − 5x22 y22 ) is minimum for all (x12 , x22 ) ∈ X i.e., ∗ ∗ ∗ ∗ , y21 } and {y12 , y22 } of Problem 1.9.3 and Problem 1.9.4 find optimal solution {y11 respectively. Problem 1.9.3   (1 − α)(−3x11 y11 + 4x11 y21 + 6x21 y11 − 7x21 y21 )+ Minimize α(−1x11 y11 + 6x11 y21 + 8x21 y11 − 5x21 y21 ) Subject to x11 + x21 = 1; y11 + y21 = 1; x11 , x21 ≥ 0; y11 , y21 ≥ 0. Problem 1.9.4 Minimize{(−1x12 y12 + 6x12 y22 + 8x22 y12 − 5x22 y22 )} Subject to x12 + x22 = 1; y12 + y22 = 1; x12 , x22 ≥ 0; y12 , y22 ≥ 0. Step 4: Since, in Problem 1.9.3 and Problem 1.9.4 only y j1 and y j2 , where j = 1, 2 respectively have been considered as decision variables. So, Problem 1.9.3 and Problem 1.9.4 are linear programming problem and hence, the optimal value of Problem 1.9.3 and Problem 1.9.4 will be equal to optimal value of its corresponding dual problem i.e., Problem 1.9.5 and Problem 1.9.6 respectively. Problem 1.9.5 Maximize{υ1 } Subject to (1 − α)(−3x11 + 6x21 ) + α(−1x11 + 8x21 ) ≥ υ1 ; (1 − α)(4x11 − 7x21 ) + α(6x11 − 5x21 ) ≥ υ1 ; x11 + x21 = 1; x11 , x21 ≥ 0. Problem 1.9.6 Maximize{υ2 } Subject to −1x12 + 8x22 ≥ υ2 ; 6x12 − 5x22 ≥ υ2 ; x12 + x22 = 1; x12 , x22 ≥ 0.

28

1 Matrix Games with Interval Payoffs

 13 Step 5: The optimal solution of Problem 1.9.5 for all α ∈ [0, 1] is x11 = 20 ,    7 13 7 x21 = 20 and the optimal solution of Problem 1.9.6 is x12 = 20 , x22 = 20 . On  13 solving Problem 1.9.3 and Problem 1.9.4 after substituting these values x11 = 20 ,    7 13 7 x21 = 20 and x12 = 20 , x22 = 20 respectively in Problem 1.9.3 and Problem 1.9.4, the obtained alternative basic optimal solutions of Problem 1.9.3 and Prob1 1 2 2 1 1 2 = 1, y21 = 0}, {y11 = 0, y21 = 1} and {y12 = 1, y22 = 0}, {y12 = lem 1.9.4 are {y11 2 0, y22 = 1} respectively. 1 1 1 + [4, 6]x11 y21 + [6, 8]x21 y11 + [−7, −5] Step 6: Now, minimum{[−3, −1]x11 y11 1 2 2 2 2 , [−3, −1] x21 y21 , [−3, −1]x11 y11 + [4, 6]x11 y21 + [6, 8]x21 y11 + [−7, −5]x21 y21 1 1 1 1 2 + [4, 6]x12 y22 + [6, 8]x22 y12 + [−7, −5]x22 y22 −1]x x12 y12 12 + [4, 6]

312 y43  , [−3, 3 43 3 43 2 2 2 , 20 , x12 y22 + [6, 8]x22 y12 + [−7, −5]x22 y22 } = minimum 20 , 20 , 20 , 20 , 20 3 43  . , 20 20 3 43 Step 7: Since, minimum value, obtained in Step 6, is 20 , 20 . So, the minimum 3 43 expected gain of Player I is 20 , 20 and the optimal strategies for Player I is     13 7 13 7 x11 = 20 = x12 = 20 . , x21 = 20 , x22 = 20 1.9.1.2

Maximum Expected Loss of Player II

Using the proposed Gaurika method maximum expected loss of Player II and corresponding optimal strategies, can be obtained as follows: Step 1: Find (x1∗ , x2∗ ) ∈ X such that value of [−3, −1]x1 y1 + [4, 6]x1 y2 + [6, 8] x2 y1 + [−7, −5]x2 y2 is maximum for all (y1 , y2 ) ∈ Y i.e., find the optimal solution of Problem 1.9.7. Problem 1.9.7 Maximize{[−3, −1]x1 y1 + [4, 6]x1 y2 + [6, 8]x2 y1 + [−7, −5]x2 y2 } Subject to x1 + x2 = 1; y1 + y2 = 1; x1 , x2 ≥ 0; y1 , y2 ≥ 0. Step 2: Using the property, [

n 

i=1

aiL ,

n  i=1

n  i=1

λ[aiL , aiR ] =

n  i=1

[λaiL , λaiR ], λ ≥ 0 and

n  i=1

[aiL , aiR ] =

aiR ], the Problem 1.9.7 can be transformed into Problem 1.9.8.

Problem 1.9.8 Maximize{[−3x1 y1 + 4x1 y2 + 6x2 y1 − 7x2 y2 , −1x1 y1 + 6x1 y2 + 8x2 y1 − 5x2 y2 ]} Subject to x1 + x2 = 1; y1 + y2 = 1; x1 , x2 ≥ 0; y1 , y2 ≥ 0. Step 3: According to comparing method, to find the optimal solution {x1∗ , x2∗ } of Problem 1.9.7 such that value of [−3x1 y1 + 4x1 y2 + 6x2 y1 − 7x2 y2 , −1x1 y1 + 6x1 y2 + 8x2 y1 − 5x2 y2 ] is maximum for all (y1 , y2 ) ∈ Y is equivalent to find {x1∗ , x2∗ } such that value of α(−3x1 y1 + 4x1 y2 + 6x2 y1 − 7x2 y2 ) + (1 − α)(−1x1 y1 + 6x1 y2 + 8x2 y1 − 5x2 y2 ) and (−3x1 y1 + 4x1 y2 + 6x2 y1 − 7x2 y2 ) is maximum for all

1.9 Numerical Examples

29

(y1 , y2 ) ∈ Y or if it is not possible to find {x1∗ , x2∗ } for which the value of α(−3x1 y1 + 4x1 y2 + 6x2 y1 − 7x2 y2 ) + (1 − α)(−1x1 y1 + 6x1 y2 + 8x2 y1 − 5x2 y2 ) and (−3x1 ∗ ∗ y1 + 4x1 y2 + 6x2 y1 − 7x2 y2 ) is maximum then find such {x11 , x21 } for which the value of α(−3x11 y11 + 4x11 y21 + 6x21 y11 − 7x21 y21 ) + (1 − α)(−1x11 y11 + 6x11 y21 + 8x21 y11 − 5x21 y21 ) is maximum but value of (−3x11 y11 + 4x11 y21 + 6x21 y11 − ∗ ∗ , x22 } for which 7x21 y21 ) is not maximum for all (y11 , y21 ) ∈ Y and to find such {x12 the value of α(−3x12 y12 + 4x12 y22 + 6x22 y12 − 7x22 y22 ) + (1 − α)(−1x12 y12 + 6x12 y22 + 8x22 y12 − 5x22 y22 ) is not maximum but value of (−3x12 y12 + 4x12 y22 + 6x22 y12 − 7x22 y22 ) is maximum for all (y12 , y22 ) ∈ Y i.e., find optimal solution ∗ ∗ ∗ ∗ , y21 } and {y12 , y22 } of Problem 1.9.9 and Problem 1.9.10 respectively. {y11 Problem 1.9.9   α(−3x11 y11 + 4x11 y21 + 6x21 y11 − 7x21 y21 )+ Maximize (1 − α)(−1x11 y11 + 6x11 y21 + 8x21 y11 − 5x21 y21 ) Subject to x11 + x21 = 1; y11 + y21 = 1; x11 , x21 ≥ 0; y11 , y21 ≥ 0. Problem 1.9.10 Maximize{(−3x12 y12 + 4x12 y22 + 6x22 y12 − 7x22 y22 )} Subject to x12 + x22 = 1; y12 + y22 = 1; x12 , x22 ≥ 0; y12 , y22 ≥ 0. Step 4: Since, in Problem 1.9.9 and Problem 1.9.10 only xi1 and xi2 respectively have been considered as decision variables. So, Problem 1.9.9 and Problem 1.9.10 are linear programming problem and hence, the optimal value of Problem 1.9.9 and Problem 1.9.10 will be equal to optimal value of its corresponding dual problem i.e., Problem 1.9.11 and Problem 1.9.12 respectively. Problem 1.9.11 Minimize{ω1 } Subject to α(−3y11 + 4y21 ) + (1 − α)(−1y11 + 6y21 ) ≤ ω1 ; α(6y11 − 7y21 ) + (1 − α)(8y11 − 5y21 ) ≤ ω1 ; y11 + y21 = 1; y11 , y21 ≥ 0. Problem 1.9.12 Minimize{ω2 } Subject to −3y12 + 4y22 ≤ ω2 ; 6y12 − 7y22 ≤ ω2 ; y12 + y22 = 1; y12 , y22 ≥ 0.

 11 , Step 5: The optimal solution of Problem 1.9.11 for all α ∈ [0, 1] is y11 = 20    9 11 9 y21 = 20 and the optimal solution of Problem 1.9.12 is y12 = 20 , y22 = 20 . On  11 solving Problem 1.9.9 and Problem 1.9.10 after substituting these values y11 = 20 ,    9 11 9 y21 = 20 and y12 = 20 , y22 = 20 respectively in Problem 1.9.9 and Problem 1.9.10, the obtained alternative basic optimal solutions of Problem 1.9.9 and Problem

30

1 Matrix Games with Interval Payoffs

1 1 2 2 1 1 2 1.9.10 are {x11 = 1, x21 = 0}, {x11 = 0, x21 = 1} and {x12 = 1, x22 = 0}, {x12 = 0, 2 x22 = 1} respectively. 1 1 1 y11 + [4, 6]x11 y21 + [6, 8]x21 y11 + [−7, −5] Step 6: Now, maximum{[−3, −1]x11 1 2 2 2 2 x21 y21 , [−3, −1]x11 y11 + [4, 6]x11 y21 + [6, 8]x21 y11 + [−7, −5]x21 y21 , [−3, −1] 1 1 1 1 2 2 y12 + [4, 6]x12 y22 + [6, 8]x22 y12 + [−7, −5]x22 y22 , [−3, −1]x 6]x 12 x12  3 43 312 y4312 + [4, 3 43 2 2 y22 + [6, 8]x22 y12 + [−7, −5]x22 y22 } = maximum 20 , 20 , 20 , 20 , 20 , 20 , 3 43  . , 20 20 3 43 Step 7: Since, maximum value, obtained in Step 6, is 20 , 20 . So, the maximum 3 43 expected loss of Player II is 20 , 20 and the optimal strategies for Player I is     11 9 11 9 y11 = 20 = y12 = 20 . , y21 = 20 , y22 = 20

1.9.2 Existing Numerical Example Considered by Li et al. 

 [175, 190] [120, 158] In this section, matrix games with interval payoffs A = , [80, 100] [180, 190] chosen by Li et al. [6], is solved by the proposed Gaurika method.

1.9.2.1

Minimum Expected Gain of Player I

Using the proposed Gaurika method minimum expected gain of Player I and corresponding optimal strategies, can be obtained as follows: Step 1: Find (y1∗ , y2∗ ) ∈ Y such that value of [175, 190]x1 y1 + [120, 158]x1 y2 + [80, 100]x2 y1 + [180, 190]x2 y2 is minimum for all (x1 , x2 ) ∈ X i.e., find the optimal solution (y1∗ , y2∗ ) of Problem 1.9.13. Problem 1.9.13 Minimize{[175, 190]x1 y1 + [120, 158]x1 y2 + [80, 100]x2 y1 + [180, 190]x2 y2 } Subject to x1 + x2 = 1; y1 + y2 = 1; x1 , x2 ≥ 0; y1 , y2 ≥ 0. Step 2: Using the property, [

n 

i=1

aiL ,

n  i=1

n  i=1

λ[aiL , aiR ] =

n  i=1

[λaiL , λaiR ], λ ≥ 0 and

n  i=1

[aiL , aiR ] =

aiR ], the Problem 1.9.13 can be transformed into Problem 1.9.14.

Problem 1.9.14 Minimize{[175x1 y1 + 120x1 y2 + 80x2 y1 + 180x2 y2 , 190x1 y1 + 158x1 y2 + 100x2 y1 + 190x2 y2 ]} Subject to x1 + x2 = 1; y1 + y2 = 1; x1 , x2 ≥ 0; y1 , y2 ≥ 0. Step 3: According to comparing method, to find the optimal solution {y1∗ , y2∗ } of Problem 1.9.14 such that value of [175x1 y1 + 120x1 y2 + 80x2 y1 + 180x2 y2 , 190x1 y1 + 158x1 y2 + 100x2 y1 + 190x2 y2 ] is minimum for all (x1 , x2 ) ∈ X , is equivalent to

1.9 Numerical Examples

31

find {y1∗ , y2∗ } such that value of (1 − α)(175x1 y1 + 120x1 y2 + 80x2 y1 + 180x2 y2 ) + α(190x1 y1 + 158x1 y2 + 100x2 y1 + 190x2 y2 ) and (190x1 y1 + 158x1 y2 + 100x2 y1 + 190x2 y2 ) is minimum for all (x1 , x2 ) ∈ X or if it is not possible to find {y1∗ , y2∗ } for which the value of (1 − α)(175x1 y1 + 120x1 y2 + 80x2 y1 + 180x2 y2 ) + α(190x1 y1 + 158x1 y2 + 100x2 y1 + 190x2 y2 ) and 190x1 y1 + 158x1 y2 + 100x2 y1 + 190x2 ∗ ∗ y2 is minimum then find such {y11 , y21 } for which the value of (1 − α)(175x11 y11 + 120x11 y21 + 80x21 y11 + 180x21 y21 ) + α(190x11 y11 + 158x11 y21 + 100x21 y11 + 190x21 y21 ) is minimum but value of (190x11 y11 + 158x11 y21 + 100x21 y11 + 190x21 ∗ ∗ y21 ) is not minimum for all (x11 , x21 ) ∈ X and to find such {y12 , y22 } for which the value of (1 − α)(175x12 y12 + 120x12 y22 + 80x22 y12 + 180x22 y22 ) + α(190x12 y12 + 158x12 y22 + 100x22 y12 + 190x22 y22 ) is not minimum but value of (190x12 y12 + 158x12 y22 + 100x22 y12 + 190x22 y22 ) is minimum for all (x12 , x22 ) ∈ X i.e., find ∗ ∗ ∗ ∗ , y21 } and {y12 , y22 } of Problem 1.9.15 and Problem 1.9.16 optimal solution {y11 respectively. Problem 1.9.15   (1 − α)(175x11 y11 + 120x11 y21 + 80x21 y11 + 180x21 y21 )+ Minimize α(190x11 y11 + 158x11 y21 + 100x21 y11 + 190x21 y21 ) Subject to x11 + x21 = 1; y11 + y21 = 1; x11 , x21 ≥ 0; y11 , y21 ≥ 0. Problem 1.9.16 Minimize{(190x12 y12 + 158x12 y22 + 100x22 y12 + 190x22 y22 )} Subject to x12 + x22 = 1; y12 + y22 = 1; x12 , x22 ≥ 0; y12 , y22 ≥ 0. Step 4: Since, in Problem 1.9.15 and Problem 1.9.16 only y j1 and y j2 , where j = 1, 2 respectively have been considered as decision variables. So, Problem 1.9.15 and Problem 1.9.16 are linear programming problem and hence, the optimal value of Problem 1.9.15 and Problem 1.9.16 will be equal to optimal value of its corresponding dual problem i.e., Problem 1.9.17 and Problem 1.9.18 respectively. Problem 1.9.17 Maximize{υ1 } Subject to (1 − α)(175x11 + 80x21 + α(190x11 + 100x21 ) ≥ υ1 ; (1 − α)(120x11 + 180x21 ) + α(158x11 + 190x21 ) ≥ υ1 ; x11 + x21 = 1; x11 , x21 ≥ 0. Problem 1.9.18 Maximize{υ2 } Subject to 190x12 + 100x22 ≥ υ2 ; 158x12 + 190x22 ≥ υ2 ; x12 + x22 = 1; x12 , x22 ≥ 0.

32

1 Matrix Games with Interval Payoffs

Step 5: The optimal solution of Problem 1.9.17 and Problem 1.9.18 for α = 0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.55, 0.56, 0.58, 0.6, 0.7, 0.8 and 0.9 are shown in second and fourth column of Table 1.3. On solving the Problem 1.9.15 after substituting the optimal solution, shown in second column of Table 1.3, the obtained alternative basic optimal solutions of Problem 1.9.15 for all chosen values of α are 1 1 2 2 = 1, y21 = 0}, {y11 = 0, y21 = 1} and on solving the Problem 1.9.16 after sub{y11 stituting the optimal solution, shown in fourth column of Table 1.3, the obtained alternative basic optimal solutions of Problem 1.9.16 for all chosen values of α are 1 1 2 2 = 1, y21 = 0}, {y11 = 0, y21 = 1}. {y11 α 1 α 1 α 1 y11 + [120, 158]x11 y21 + [80, 100]x21 y11 + Step 6: Now, minimum{[175, 190]x11 α 1 α 2 α 2 α 2 [180, 190]x21 y21 , [175, 190]x11 y11 + [120, 158]x11 y21 + [80, 100]x21 y11 + [180, α 2 α 1 α 1 α 1 α y21 , [175, 190]x12 y12 + [120, 158]x12 y22 + [80, 100]x22 y12 + [180, 190]x22 190]x21 1 α 2 α 2 α 2 α 2 y22 , [175, 190]x12 y12 + [120, 158]x12 y22 + [80, 100]x22 y12 + [180, 190]x22 y22 } at α = 0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.55, 0.56, 0.58, 0.6, 0.7, 0.8 and 0.9 are shown in Table 1.3. Step 7: The intervals, shown in third column of Table 1.3, represent the minimum expected gain of Player I corresponding to the optimal strategies for Player I are shown in second column of Table 1.3 and the intervals, shown in fifth column of Table 1.3, represents the minimum expected gain of Player I corresponding to the optimal strategies for Player I are shown in fourth column of Table 1.3.

1.9.2.2

Maximum Expected Loss of Player II

Using the proposed Gaurika method maximum expected loss of Player II and corresponding optimal strategies, can be obtained as follows: Step 1: Find (x1∗ , x2∗ ) ∈ X such that value of [175, 190]x1 y1 + [120, 158]x1 y2 + [80, 100]x2 y1 + [180, 190]x2 y2 is maximum for all (y1 , y2 ) ∈ Y i.e., find the optimal solution of Problem 1.9.19. Problem 1.9.19 Maximize{[175, 190]x1 y1 + [120, 158]x1 y2 + [80, 100]x2 y1 + [180, 190]x2 y2 } Subject to x1 + x2 = 1; y1 + y2 = 1; x1 , x2 ≥ 0; y1 , y2 ≥ 0. Step 2: Using the property, [

n 

i=1

aiL ,

n  i=1

n  i=1

λ[aiL , aiR ] =

n  i=1

[λaiL , λaiR ], λ ≥ 0 and

n  i=1

[aiL , aiR ] =

aiR ], the Problem 1.9.19 can be transformed into Problem 1.9.20.

Problem 1.9.20 Maximize{[175x1 y1 + 120x1 y2 + 80x2 y1 + 180x2 y2 , 190x1 y1 + 158x1 y2 + 100x2 y1 + 190x2 y2 ]} Subject to x1 + x2 = 1; y1 + y2 = 1; x1 , x2 ≥ 0; y1 , y2 ≥ 0.

1.9 Numerical Examples

33

Table 1.3 Minimum expected gain of Player I and corresponding optimal strategies α

{x1∗ , x2∗ }

[υ1L , υ1R ]

0

 20

4380

 0.1  0.2 0.3



11 31 , 31

990 527 1517 , 1517 35 18 53 , 53

 480

0.5

 190

0.55

 270

229 709 , 709 87 277 , 277

 0.56



0.7

121 391 , 391





31

,

7565 8450 53 , 53

208230 102320 40210 



930 389 1319 , 1319

277





990890 





,

232400 1451

,

114100 709

44800 277

,





0.9

 130



26670

95140 105700 643 , 643 179

,



29600 179



193870 215600 1319 , 1319



1103200 6793

49365 27475 338 , 169

 460



497240 553700 3413 , 3413 6793





56930 63400 391 , 391

0.8

49 179 , 179

,





215410 240800 1517 , 1517

1451



2083 6793 , 6793

183 643 , 643

4900 31

709

2360 1053 3413 , 3413

235 103 338 , 338

 



 4710

0.6





970 481 1451 , 1451

0.4

0.58







{x1∗ , x2∗ }   45 16 61 , 61   45 16 61 , 61   45 16 61 , 61   45 16 61 , 61   45 16 61 , 61   45 16 61 , 61   45 16 61 , 61   45 16 61 , 61   45 16 61 , 61   45 16 61 , 61   45 16 61 , 61   45 16 61 , 61   45 16 61 , 61

[υ2L , υ2R ]   8280 10150 61 , 61   8280 10150 61 , 61   8280 10150 61 , 61   8280 10150 61 , 61   8280 10150 61 , 61   8280 10150 61 , 61   8280 10150 61 , 61   8280 10150 61 , 61   8280 10150 61 , 61   8280 10150 61 , 61   8280 10150 61 , 61   8280 10150 61 , 61   8280 10150 61 , 61

Step 3: According to comparing method, to find the optimal solution {x1∗ , x2∗ } of Problem 1.9.20 such that value of [175x1 y1 + 120x1 y2 + 80x2 y1 + 180x2 y2 , 190x1 y1 + 158x1 y2 + 100x2 y1 + 190x2 y2 ] is maximum for all (y1 , y2 ) ∈ Y is equivalent to find {x1∗ , x2∗ } such that value of α(175x1 y1 + 120x1 y2 + 80x2 y1 + 180x2 y2 ) + (1 − α)(190x1 y1 + 158x1 y2 + 100x2 y1 + 190x2 y2 ) and (175x1 y1 + 120x1 y2 + 80x2 y1 + 180x2 y2 ) is maximum for all (y1 , y2 ) ∈ Y or if it is not possible to find {x1∗ , x2∗ } for which the value of α(175x1 y1 + 120x1 y2 + 80x2 y1 + 180x2 y2 ) + (1 − α)(190x1 y1 + 158x1 y2 + 100x2 y1 + 190x2 y2 ) and (175x1 y1 + 120x1 y2 + 80x2 y1 + 180x2 ∗ ∗ y2 ) is maximum then find such {x11 , x21 } for which the value of α(175x11 y11 + 120x11 y21 + 80x21 y11 + 180x21 y21 ) + (1 − α)(190x11 y11 + 158x11 y21 + 100x21 y11 + 190x21 y21 ) is maximum but value of (175x11 y11 + 120x11 y21 + 80x21 y11 + 180x21 ∗ ∗ y21 ) is not maximum for all (y11 , y21 ) ∈ Y and to find such {x12 , x22 } for which the value of α(175x12 y12 + 120x12 y22 + 80x22 y12 + 180x22 y22 ) + (1 − α)(190x12 y12 + 158x12 y22 + 100x22 y12 + 190x22 y22 ) is not maximum but value of (175x12 y12 + 120x12 y22 + 80x22 y12 + 180x22 y22 ) is maximum for all (y12 , y22 ) ∈ Y i.e., find opti-

34

1 Matrix Games with Interval Payoffs

∗ ∗ ∗ ∗ mal solution {y11 , y21 } and {y12 , y22 } of Problem 1.9.21 and Problem 1.9.22 respectively.

Problem 1.9.21   α(175x11 y11 + 120x11 y21 + 80x21 y11 + 180x21 y21 )+ Maximize (1 − α)(190x11 y11 + 158x11 y21 + 100x21 y11 + 190x21 y21 ) Subject to x11 + x21 = 1; y11 + y21 = 1; x11 , x21 ≥ 0; y11 , y21 ≥ 0. Problem 1.9.22 Maximize{175x12 y12 + 120x12 y22 + 80x22 y12 + 180x22 y22 } Subject to x12 + x22 = 1; y12 + y22 = 1; x12 , x22 ≥ 0; y12 , y22 ≥ 0. Step 4: Since, in Problem 1.9.21 and Problem 1.9.22 only xi1 and xi2 , where i = 1, 2 respectively have been considered as decision variables. So, Problem 1.9.21 and Problem 1.9.22 are linear programming problem and hence, the optimal value of Problem 1.9.21 and Problem 1.9.22 will be equal to optimal value of its corresponding dual problem i.e., Problem 1.9.23 and Problem 1.9.24 respectively. Problem 1.9.23 Minimize{ω1 } Subject to α(175y11 + 120y21 + (1 − α)(190y11 + 158y21 ) ≤ ω1 ; α(80y11 + 180y21 ) + (1 − α)(100y11 + 190y21 ) ≤ ω1 ; y11 + y21 = 1; y11 , y21 ≥ 0. Problem 1.9.24 Minimize{ω2 } Subject to 175y12 + 120y22 ≤ ω2 ; 80y12 + 180y22 ≤ ω2 ; y12 + y22 = 1; y12 , y22 ≥ 0. Step 5: The optimal solution of Problems 1.9.23 and 1.9.24 for α = 0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.55, 0.56, 0.58, 0.6, 0.7, 0.8 and 0.9 are shown in second and fourth column of Table 1.4. On solving the Problem 1.9.21 after substituting the optimal solution, shown in second column of Table 1.4, the alternative basic optimal solutions of 1 1 2 2 = 1, x21 = 0}, {x11 = 0, x21 = 1} Problem 1.9.21 for all chosen values of α are {x11 and on solving the Problem 1.9.22 after substituting the optimal solutions, shown in fourth column of Table 1.4, the alternative basic optimal solutions of Problem 1.9.22 1 1 2 2 = 1, x21 = 0}, {x11 = 0, x21 = 1}. for all chosen values of α are {x11 1 y α + [120, 158]x 1 y α + [80, 100]x 1 y α + [180, 190] Step 6: Now, maximum {[175, 190]x11 11 11 21 21 11 1 y α , [175, 190]x 2 y α + [120, 158]x 2 y α + [80, 100]x 2 y α + [180, 190]x 2 y α , [175, 190]x 1 x21 21 11 11 11 21 21 11 21 21 12 α + [120, 158]x 1 y α + [80, 100]x 1 y α + [180, 190]x 1 y α , [175, 190]x 2 y α + [120, 158]x 2 y12 12 22 22 12 22 22 12 12 12

1.9 Numerical Examples

35

Table 1.4 Maximum expected loss of Player II and corresponding optimal strategies α 0 0.1 0.2

{y1∗ , y2∗ }   16 45 61 , 61   348 905 1253 , 1253   188 455 643 , 643

[ω1L , ω1R ]   9380 10150 61 , 61   190740 206750 1253 , 1253   96940 105250 643 , 643

{y1∗ , y2∗ }





0.3  0.4  0.5  0.55 0.56

54 115 169 , 169 92 185 277 , 277

  0.6  0.7  0.8 

290 439 , 439









 





572 945 1517 , 1517



 





 12



 12



 12



 12







 12







 12





 12





 12





 12







 12







 12





 12





 12





197020 214250 1319 , 1319 25020 27250 169 , 169 40660 44350 277 , 277

409740 447250 2803 , 2803

64120



516 935 1451 , 1451 136 235 371 , 371





2412 4645 7057 , 7057 244 465 709 , 709





948 1855 2803 , 2803

 149

0.58

0.9

404 915 1319 , 1319

439

,

70000 439

19 31 , 31 19 31 , 31 19 31 , 31



19 31 , 31 19 31 , 31 19 31 , 31



19 31 , 31

1029060 1123750 7057 , 7057 103220 112750 709 , 709 209580 229250 1451 , 1451 53180 58250 371 , 371

[ω2L , ω2R ]   4380 5282 31 , 31   4380 5282 31 , 31   4380 5282 31 , 31



215860 236750 1517 , 1517





19 31 , 31 19 31 , 31 19 31 , 31 19 31 , 31 19 31 , 31 19 31 , 31

4380 5282 31 , 31 4280 5282 31 , 31 4280 5282 31 , 31 4280 5282 31 , 31 4280 5282 31 , 31 4280 5282 31 , 31 4280 5282 31 , 31 4280 5282 31 , 31 4280 5282 31 , 31 4280 5282 31 , 31

         

α + [80, 100]x 2 y α + [180, 190]x 2 y α } at α = 0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.55, 0.56, 0.58, 0.6, y22 22 12 22 22 0.7, 0.8 and 0.9 are shown in Table 1.4.

Step 7: The intervals, shown in third column of Table 1.4, represent the maximum expected loss of Player II corresponding to the optimal strategies for Player II are shown in second column of Table 1.4 and the intervals, shown in fifth column of Table 1.4, represent the maximum expected loss of Player II corresponding to the optimal strategies for Player II are shown in fourth column of Table 1.4.

1.10 Conclusion On the basis of present study, it can be concluded that some mathematically incorrect assumptions have been considered in the existing methods [3–7, 11] for solving matrix games with interval payoffs. Therefore, it is not genuine to use these methods. Furthermore, to resolve flaws of the existing methods [3–7, 11], a new method (named as Gaurika method) is proposed for solving matrix games with interval payoffs.

36

1 Matrix Games with Interval Payoffs

References 1. Akyar, H., Akyar, E.: A graphical method for solving interval matrix games. Abstr. Appl. Anal. 2011, 1–17 (2011) 2. Akyar, E., Akyar, H., Duzce, S.A.: Brown-Robinson method for interval matrix games. Soft Comput. 15, 2057–2064 (2011) 3. Collins, W.D., Hu, C.Y.: Studying interval valued matrix games with fuzzy logic. Soft Comput. 12, 147–155 (2008) 4. Li, D.F.: Linear programming approach to solve interval-valued matrix games. Omega 39, 655–666 (2011) 5. Li, D.F.: Notes on “Linear programming technique to solve two-person matrix games with interval pay-offs”. Asia-Pacific J. Oper. Res. 28, 705–737 (2011) 6. Li, D.F., Nan, J.X., Zhang, M.J.: Interval programming models for matrix games with interval payoffs. Optim. Methods Softw. 27, 1–16 (2012) 7. Liu, S.T., Kao, C.: Matrix games with interval data. Comput. Ind. Eng. 56, 1697–1700 (2009) 8. Moore, R.E.: Methods and Applications of Interval Analysis. Siam, Philadelphia (1979) 9. Nash, J.: Non-cooperative games. Ann. Math. 54, 286–295 (1951) 10. Nayak, P.K., Pal, M.: Solution of rectangular interval games using graphical method. Tamsui Oxf. J. Math. Sci. 22, 95–115 (2006) 11. Nayak, P.K., Pal, M.: Linear programming technique to solve two person matrix games with interval payoffs. Asia-Pacific J. Oper. Res. 26, 285–305 (2009) 12. Owen, G.: Game Theory, 2nd edn. Academic Press, New York (1982) 13. Shashikhin, V.N.: Antagonistic game with interval payoff function. Cybern. Syst. Anal. 40, 556–564 (2004) 14. Von Neumann, J.: Zur theorie der Gesellschaftssoiele. Mathematische Annalen 100, 295–320 (1928) 15. Von Neumann, J., Morgenstern, O.: Theory of Games and Economic Behavior. Princeton University Press (1944)

Chapter 2

Matrix Games with Fuzzy Payoffs

In this chapter, flaws of the existing methods [5, 10–12, 14] for solving matrix game with fuzzy payoffs (matrix games in which payoffs are represented as fuzzy numbers) are pointed out. To resolve these flaws, a new method (named as Mehar 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 matrix games with fuzzy payoffs. To illustrate the proposed Mehar method, the existing numerical problems of matrix games with fuzzy payoffs are solved by the proposed Mehar method.

2.1 Matrix Games with Fuzzy Payoffs In an interval the membership value of every element belongs to the set is equal so it is not realistic to represent the payoffs by intervals e.g., Let, ‘A’ grade has been assigned to those students who secured marks between 90 and 100 in an examination, so the range of grade ‘A’ if represented in the interval will be [90,100], which indicates that the student who secured 90.5 marks and who secured 99.5 marks are equally efficient. But in general, it is not so. The student who secured 99.5 marks is more efficient than the student who secured 90.5 marks. Zadeh [17] proposed an interesting generalization of interval called fuzzy sets to capture this aspect of human behavior. In the literature [1–5, 7, 10–14] fuzzy numbers have been used to represent the payoffs and such matrix games in which payoffs are represented in matrix games by fuzzy numbers are named as matrix games with fuzzy payoffs.

38

2 Matrix Games with Fuzzy Payoffs

2.2 Preliminaries In this section, some basic definitions, arithmetic operations of fuzzy numbers and method for comparing fuzzy numbers is presented [6].

2.2.1 Some Basic Definitions In this section, some basic definitions are reviewed [6]. Definition 2.1 Let, X be a classical set of objects. Then, the set of ordered pairs A˜ = {(x, μ A˜ (x)) : x ∈ X }, where μ A˜ : X → [0, 1], is called a fuzzy set in X . The evaluation function μ A˜ is called the membership function. Definition 2.2 Let, A˜ be a fuzzy set in X and α ∈ (0, 1] be a real number. Then, α−cut of the fuzzy set A˜ is the crisp set A(α) = {x ∈ X : μ A˜ (x) ≥ α}. Definition 2.3 Let, A˜ be a fuzzy set in X . Then, the support of A˜ is the crisp set given by {x ∈ X : μ A˜ (x) > 0}. Definition 2.4 Let, A˜ be a fuzzy set in X . Then, the fuzzy set A˜ is said to be normal if supx ∈ X μ A˜ (x) = 1. Definition 2.5 A fuzzy set A˜ in set of real numbers  is called a fuzzy number if it satisfies the following conditions: (i) A˜ is normal, (ii) A˜ is a closed interval for every α ∈ (0, 1], (iii) The support of A˜ is bounded. Definition 2.6 A fuzzy number A˜ = (a L (0), a(1), a R (0)) is called a triangular fuzzy ⎧ number if its membership function μ A˜ is given by x−a L (0) L ⎪ ⎨ a(1)−a L (0) a (0) ≤ x < a(1) R x−a (0) μ A˜ = a(1) ≤ x ≤ a R (0) R ⎪ ⎩ a(1)−a (0) 0 x < a L (0), x > a R (0) Further, the α−cut of the triangular fuzzy number A˜ is the closed interval [a L (0) + α(a(1) − a L (0)), a R (0) + α(a(1) − a R (0))]. Definition 2.7 A fuzzy number A˜ = (a L (0), a L (1), a R (1), a R (0)) is called a trapezoidal ⎧ fuzzy number if its membership function μ A˜ is given by x−a L (0) ⎪ a L (0) ≤ x < a L (1) ⎪ a L (1)−a L (0) ⎪ ⎨ 1 a L (1) ≤ x ≤ a R (1) μ A˜ = R x−a (0) ⎪ a R (1) < x ≤ a R (0) R R ⎪ ⎪ ⎩ a (1)−a (0) 0 x < a L (0), x > a R (0) Further, the α−cut of the trapezoidal fuzzy number A˜ is the closed interval [a L (0) + α(a L (1) − a L (0)), a R (0) + α(a R (1) − a R (0))].

2.2 Preliminaries

39

2.2.2 Arithmetic Operations of Trapezoidal Fuzzy Numbers In this section, some arithmetic operations of two trapezoidal fuzzy numbers, defined on universal set of real numbers , are presented [8, 9]. Let A˜ 1 = (a1L (0), a1L (1), a1R (1), a1R (0)) and A˜2 = (a2L (0), a2L (1), a2R (1), a2R (0)) be two trapezoidal fuzzy numbers then A˜ 1 + A˜ 1 −

A˜ 2 = (a1L (0) + a2L (0), a1L (1) + a2L (1), a1R (1) + a2R (1), a1R (0) + a2R (0)). R R A˜ 2 = (a1L (0) − a2R (0), a1L (1) − a2L (1), a1R (0) − a2L (0)).  − a2L (1), a1 (1) L (λa1 (0), λa1 (1), λa1R (1), λa1R (0)) λ ≥ 0 3. If λ is a real number then λ A˜ 1 = (λa1R (0), λa1R (1), λa1L (1), λa1L (0)) λ < 0 1. 2.

2.2.3 Comparison of Fuzzy Numbers If a and b are two distinct real numbers then it can be easily verified that a > b or a < b. However, if A˜ 1 and A˜ 2 are two fuzzy numbers then there is no unique way to verify A˜ 1  A˜ 2 or A˜ 1 ≺ A˜ 2 . In this section, the method, used by the authors [5, 10–12, 14] to compare two trapezoidal fuzzy numbers, is presented. Let A˜ 1 = (a1L (0), a1L (1), a1R (1), a1R (0)) and A˜ 2 = (a2L (0), a2L (1), a2R (1), a2R (0)) be two trapezoidal fuzzy numbers. Then, 1. 2. 3.

A˜ 1  A˜ 2 ⇔ a1L (0) > a2L (0), a1L (1) > a2L (1), a1R (1) > a2R (1), a1R (0) > a2R (0). A˜ 1 ≺ A˜ 2 ⇔ a1L (0) < a2L (0), a1L (1) < a2L (1), a1R (1) < a2R (1), a1R (0) < a2R (0). A˜ 1 ≈ A˜ 2 ⇔ a1L (0) = a2L (0), a1L (1) = a2L (1), a1R (1) = a2R (1), a1R (0) = a2R (0).

2.3 Existing Mathematical Formulation of Matrix Games with Fuzzy Payoffs In the literature, Problems 2.3.1 and 2.3.2 have been used to obtain the minimum expected gain of Player I and maximum expected loss of Player II as well as their corresponding optimal strategies. These problems have been obtained by using the same procedure, discussed in Sect. 1.6.1 of Chapt. 1, just by replacing the intervals [aiLj , aiRj ], [υ L , υ R ] and [ω L , ω R ] with fuzzy numbers a˜ i j , υ˜ and ω˜ respectively. Problem 2.3.1 Maximize{υ} ˜ Subject to m  a˜ i j xi  υ, ˜ j = 1, 2, ..., n; i=1 m  i=1

xi = 1; xi ≥ 0, i = 1, 2, ..., m.

40

2 Matrix Games with Fuzzy Payoffs

Problem 2.3.2 Minimize{ω} ˜ Subject to n  a˜i j y j ω, ˜ i = 1, 2, ..., m; j=1 n 

y j = 1; y j ≥ 0, j = 1, 2, ..., n.

j=1

2.4 Literature Review of Matrix Games with Fuzzy Payoffs In this section, a brief review of the methods, proposed in the literature in last ten years for solving matrix games with triangular/trapezoidal fuzzy payoffs, is presented. Liu and Kao [14] firstly transformed the Problem 2.3.1 and Problem 2.3.2 into Problem 2.4.1 and Problem 2.4.2 respectively by replacing the trapezoidal fuzzy numbers a˜ i j , υ˜ and ω˜ with its α−cuts, [aiLj (α), aiRj (α)], [υ L (α), υ R (α)] and [ω L (α), ω R (α)] respectively. Problem 2.4.1 Maximize{[υ L (α), υ R (α)]} Subject to m  [aiLj (α), aiRj (α)]xi ≥ [υ L (α), υ R (α)], j = 1, 2, ..., n;

i=1 m 

xi = 1; xi ≥ 0, i = 1, 2, ..., m.

i=1

Problem 2.4.2 Minimize{[ω L (α), ω R (α)]} Subject to n  [aiLj (α), aiRj (α)]y j ≤ [ω L (α), ω R (α)], i = 1, 2, ..., m; j=1 n 

y j = 1; y j ≥ 0, j = 1, 2, ..., n.

j=1

Then, Liu and Kao [14] used the existing method [15], discussed in Sect. 1.2 of Chapt. 1, to obtain the optimal solution {xi , i = 1, 2, ..., m; υ L (α), υ R (α)} and {y j , j = 1, 2, ..., n; ω L (α), ω R (α)} of Problem 2.4.1 and Problem 2.4.2 respectively corresponding to different value of α. Finally, Liu and Kao [14] used the optimal ˜ repvalues υ L (α), υ R (α) and ω L (α), ω R (α) to obtain the fuzzy numbers υ˜ and ω, resenting the minimum expected gain of Player I and maximum expected loss of Player II respectively.

2.4 Literature Review of Matrix Games with Fuzzy Payoffs

41

Since, on solving matrix games with fuzzy payoffs by using the existing method [15], optimal strategies of Player I and Player II are not obtained. So, it is not appropriate to use Liu and Kao’ [14] method to find the solution of matrix games with fuzzy payoffs. Li [10] transformed the Problem 2.3.1 and Problem 2.3.2 into Problem 2.4.3 and Problem 2.4.4 respectively. Problem 2.4.3   L υ (0) + υ R (0) Maximize 2 Subject to m  aiLj (0)xi ≥ υ L (0), j = 1, 2, ..., n;

i=1 m  i=1 m  i=1 L

ai j (1)xi ≥ υ(1), j = 1, 2, ..., n; aiRj (0)xi ≥ υ R (0), j = 1, 2, ..., n; 0

υ ≥ υ L (0); 0 υ R ≥ υ R (0); m  xi = 1; xi ≥ 0, i = 1, 2, ..., m.

i=1

Problem  2.4.4  ω L (0) + ω R (0) Minimize 2 Subject to n  aiLj (0)y j ≤ ω L (0), i = 1, 2, ..., m; j=1 n  j=1 n  j=1 L

ai j (1)y j ≤ ω(1), i = 1, 2, ..., m; aiRj (0)y j ≤ ω R (0), i = 1, 2, ..., m; 0

ω ≤ ω L (0); 0 ω R ≤ ω R (0); n  y j = 1; y j ≥ 0, j = 1, 2, ..., n. j=1 0

0

where, υ L (0), υ 0 (1) and υ R (0) are the optimal solution of Problem 2.4.5 and 0 0 ω L (0), ω 0 (1) and ω R (0) are the optimal solution of Problem 2.4.6. Problem 2.4.5 Maximize{υ(1)} Subject to m  aiLj (0)xi ≥ υ L (0), j = 1, 2, ..., n;

i=1 m  i=1

ai j (1)xi ≥ υ(1), j = 1, 2, ..., n;

42 m  i=1 m 

2 Matrix Games with Fuzzy Payoffs

aiRj (0)xi ≥ υ R (0), j = 1, 2, ..., n; xi = 1; xi ≥ 0, i = 1, 2, ..., m.

i=1

Problem 2.4.6 Minimize{ω(1)} Subject to n  aiLj (0)y j ≤ ω L (0), i = 1, 2, ..., m; j=1 n  j=1 n  j=1 n 

ai j (1)y j ≤ ω(1), i = 1, 2, ..., m; aiRj (0)y j ≤ ω R (0), i = 1, 2, ..., m; y j = 1; y j ≥ 0, j = 1, 2, ..., n.

j=1

Li [10] claimed that the optimal solution {xi , i = 1, 2, ..., m} of Problem 2.4.3 will be the optimal strategies of Player I and using the optimal solution {υ L (0), υ(1), υ R (0)} of Problem 2.4.3, the triangular fuzzy number (υ L (0), υ(1), υ R (0)), representing the minimum expected gain of Player I, can be obtained. Also, the optimal solution {y j , j = 1, 2, ..., n} of Problem 2.4.4 will be the 0 0 optimal strategies of Player II and using the optimal solution {ω L (0), ω 0 (1), ω R (0)} L0 0 R0 of Problem 2.4.4, the triangular fuzzy number (ω (0), ω (1), ω (0)), representing the maximum expected loss of Player II, can be obtained. Clemente et al. [5] transformed the Problem 2.3.1 and Problem 2.3.2 into Problem 2.4.7 and Problem 2.4.8 respectively. Problem 2.4.7

 Maximize υ L (0), υ(1), υ R (0) Subject to m  aiLj (0)xi ≥ υ L (0), j = 1, 2, ..., n;

i=1 m  i=1 m  i=1 m 

ai j (1)xi ≥ υ(1), j = 1, 2, ..., n; aiRj (0)xi ≥ υ R (0), j = 1, 2, ..., n; xi = 1; xi ≥ 0, i = 1, 2, ..., m.

i=1

Problem

2.4.8  Minimize ω L (0), ω(1), ω R (0) Subject to n  aiLj (0)y j ≤ ω L (0), i = 1, 2, ..., m; j=1

2.4 Literature Review of Matrix Games with Fuzzy Payoffs n  j=1 n  j=1 n 

43

ai j (1)y j ≤ ω(1), i = 1, 2, ..., m; aiRj (0)y j ≤ ω R (0), i = 1, 2, ..., m; y j = 1; y j ≥ 0, j = 1, 2, ..., n.

j=1

Clemente et al. [5] claimed that the Pareto optimal solution {xi , i = 1, 2, ..., m} of Problem 2.4.7 represents the Pareto optimal security strategy and the corresponding triangular fuzzy number υ L (0), υ(1), υ R (0) represents the security level (minimum expected gain of Player I). Also, the Pareto optimal solution {y j , j = 1, 2, ..., n} of Problem 2.4.8 represents the Pareto optimal security strategy and the corresponding  triangular fuzzy number ω L (0), ω(1), ω R (0) represents the security level (maximum expected loss of Player II). Li [11] pointed out that on solving matrix games with fuzzy payoffs, the obtained minimum expected gain of Player I, represented by a fuzzy number, should be equal to the obtained maximum expected loss of Player II. While, for the minimum expected gain of Player I and maximum expected loss of Player II, obtained on solving matrix games with fuzzy payoffs by using the existing methods [1, 10, 14], this condition is not satisfying. To resolve this fundamental error of the existing methods [1, 10, 14], Li [11] split Problem 2.3.1 into three independent problems, Problem 2.4.9, 2.4.10 and 2.4.11 as well as split Problem 2.3.2 into three independent problems, Problem 2.4.12, 2.4.13 and 2.4.14. Problem 2.4.9 Maximize υ L (0) Subject to m  aiLj (0)xi ≥ υ L (0), j = 1, 2, ..., n;

i=1 m 

xi = 1; xi ≥ 0, i = 1, 2, ..., m.

i=1

Problem 2.4.10 Maximize{υ(1)} Subject to m  ai j (1)xi ≥ υ(1), j = 1, 2, ..., n; i=1 m 

xi = 1; xi ≥ 0, i = 1, 2, ..., m.

i=1

Problem 2.4.11 Maximize υ R (0) Subject to m  aiRj (0)xi ≥ υ R (0), j = 1, 2, ..., n;

i=1

44 m 

2 Matrix Games with Fuzzy Payoffs

xi = 1; xi ≥ 0, i = 1, 2, ..., m.

i=1

Problem 2.4.12 Minimize ω L (0) Subject to n  aiLj (0)y j ≤ ω L (0), i = 1, 2, ..., m; j=1 n 

y j = 1; y j ≥ 0, j = 1, 2, ..., n.

j=1

Problem 2.4.13 Minimize{ω(1)} Subject to n  ai j (1)y j ≤ ω(1), i = 1, 2, ..., m; j=1 n 

y j = 1; y j ≥ 0, j = 1, 2, ..., n.

j=1

Problem 2.4.14 Minimize ω R (0) Subject to n  aiRj (0)y j ≤ ω R (0), i = 1, 2, ..., m; j=1 n 

y j = 1; y j ≥ 0, j = 1, 2, ..., n.

j=1 ∗

∗

∗

∗

Li [11] claimed that if υ L (0), υ ∗ (1), υ R (0), ω L (0), ω ∗ (1) and ω R (0) are the optimal values of Problem 2.4.9, Problem 2.4.10, Problem 2.4.11, Problem 2.4.12, ∗ ∗ Problem 2.4.13 and Problem 2.4.14 respectively then (υ L (0), υ ∗ (1), υ R (0)) and L∗ ∗ R∗ (ω (0), ω (1), ω (0)) represents the minimum expected gain of Player I and the maximum expected loss of Player II respectively. Dutta and Gupta [7] extended the existing models [16] by representing payoffs as trapezoidal fuzzy numbers instead of triangular fuzzy numbers as well as proposed the existence of equilibrium strategies for these models. To find the solution of matrix games with trapezoidal fuzzy numbers, Dutta and Gupta [7] firstly

L transformed a matrix games with fuzzy payoffs ai j (0), aiLj (1), aiRj (1), aiRj (0) into a parametric bi-matrix games with (1 − λ)aiLj (0) + λaiLj (1) as gains of Player I and (1 − μ)aiRj (1) + μaiRj (0) as losses of Player II, where λ, μ ∈ [0, 1]. Then, Dutta and Gupta [7] obtained the Pareto Nash equilibrium strategies {xi , i = 1, 2, ..., m} and {y j , j = 1, 2, ..., n} of Player I and Player II by solving the n m  n   L L (0) + λa pj (1)]y j ≤ xi [(1 − λ)aiLj (0) + λaiLj (1)]y j , inequalities [(1−λ)a pj j=1

p = 1, 2, ..., m,

m  n  i=1 j=1

i=1 j=1

xi [(1 −

μ)aiRj (1)

+ μaiRj (0)]y j ≤

m  i=1

xi [(1 − μ)aqRj (1) +

2.4 Literature Review of Matrix Games with Fuzzy Payoffs

μaqRj (0)], q = 1, 2, ..., n and the expected payoffs by using the expressions xi [(1 − λ)aiLj (0) + λaiLj (1)]y j and

m  n  i=1 j=1

45 m  n  i=1 j=1

xi [(1 − μ)aiRj (1) + μaiRj (0)]y j .

Li [12] used the existing method [11] for solving such matrix games in which payoffs are represented by trapezoidal fuzzy numbers.

2.5 Flaws of the Existing Methods In this section, flaws of the existing methods [5, 10–12, 14] are pointed out. 1. Liu and Kao [14] solved the Problems 2.5.1 [14] and 2.5.2 [14] to illustrate their proposed method. Problem 2.5.1 [14] Maximize{υ} ˜ Subject to (26, 27, 28, 30)x1 − 28x2 + (−27, −25, −25, −22)x3 + 30x4 ≥ υ; ˜ ˜ −10x1 + 20x2 + (10, 11, 12, 14)x3 + (−38, −36, −35, −34)x4 ≥ υ; ˜ −20x1 − 10x2 + 20x3 + (−34, −32, −32, −30)x4 ≥ υ; ˜ (−3, 0, 0, 3)x1 + (32, 34, 35, 37)x2 − 30x3 + (32, 33, 34, 36)x4 ≥ υ; ˜ x1 + x2 + x3 + x4 ≥ υ; x1 , x2 , x3 , x4 ≥ 0, υ˜ unrestricted in sign. Problem 2.5.2 [14] Maximize{u} ˜ Subject to (26, 27, 28, 30)y1 − 10y2 − 20y3 + (−3, 0, 0, 3)y4 ≤ u; ˜ ˜ −28y1 + 20y2 − 10y3 + (32, 34, 35, 37)y4 ≤ u; ˜ (−27, −25, −25, −22)y1 + (10, 11, 12, 14)y2 + 20y3 − 30y4 ≤ u; 30y1 + (−38, −36, −35, −34)y2 + (−34, −32, −32, −30)y3 + (32, 33, 34, 36) ˜ y4 ≤ u; y1 + y2 + y3 + y4 = 1; y1 , y2 , y3 , y4 ≥ 0, u˜ unrestricted in sign. In the optimal solution of the Problem 2.5.1 [14], the variables x1 , x2 , x3 and x4 should always be real numbers i.e., lower bound and upper bound of the α−cut of variables x1 , x2 , x3 and x4 corresponding to different α should be equal. However, it is obvious from the optimal values of variables x1 , x2 , x3 and x4 , obtained by Liu and Kao [14] which are shown in Table 2.1, that the lower bound and upper bound of α−cut of the variables x1 , x2 , x3 and x4 corresponding to α = 0 and α = 1 are not equal i.e., the optimal values of variables x1 , x2 , x3 and x4 , obtained by Liu and Kao [14], are not real numbers. Similarly, in the optimal solution of the Problem 2.5.2 [14] the variables y1 , y2 , y3 and y4 should always be real numbers i.e., lower bound and upper bound of α− cut of

46

2 Matrix Games with Fuzzy Payoffs

Table 2.1 Lower bound and upper bound of variables [14] Lower bound of α−cut (xiL (α))

Upper bound of α−cut (xiR (α))

x1L (0) = 0.2045

x1R (0) = 0.1331

x2L (0) = 0.1144

x2R (0) = 0.0925

x3L (0) = 0.4589

x3R (0) = 0.5045

x4L (0) x1L (1) x2L (1) x3L (1) x4L (1)

= 0.2222

x4R (0) = 0.2699

= 0.1890

x1R (1) = 0.1690

= 0.1092

x2R (1) = 0.0977

= 0.4713

x3R (1) = 0.4843

= 0.2305

x4R (1) = 0.2490

Table 2.2 Lower bound and upper bound of variables [14] Lower bound of α−cut (y Lj (α))

Upper bound of α− cut (y Rj (α)

y1L (0) = 0.2621

y1R (0) = 0.2514

y2L (0) = 0.0750

y2R (0) = 0.1047

y3L (0) y4L (0) y1L (1) y2L (1) y3L (1) y4L (1)

= 0.4521

y3R (0) = 0.4357

= 0.2108

y4R (0) = 0.2082

= 0.2580

y1R (1) = 0.2568

= 0.0887

y2R (1) = 0.0880

= 0.4451

y3R (1) = 0.4476

= 0.2082

y4R (1) = 0.2076

the variables y1 , y2 , y3 and y4 corresponding to different α should be equal. However, it is obvious from the optimal values of variables y1 , y2 , y3 and y4 , obtained by Liu and Kao [14] which are shown in Table 2.2, that the lower bound and upper bound of α−cut of the variables y1 , y2 , y3 and y4 corresponding to α = 0 and α = 1 are not equal i.e., the optimal values of variables y1 , y2 , y3 and y4 obtained Liu and Kao [14], are not real numbers. Hence, neither the optimal solution, obtained by Liu and Kao [14] which are shown in Table 2.1, is the optimal solution of Problem 2.5.1 [14] nor the optimal solution, obtained by Liu and Kao [14] which are shown in Table 2.2, is the optimal solution of the Problem 2.5.2 [14]. 

2. Clemente et al. [5] have used the relation, a L (0), a(1), a R (0) b L (0), b(1),  b R (0) ⇒ a L (0) ≤ b L (0), a(1) ≤ b(1), a R (0) ≤ b R (0) for transforming the fuzzy constraints of Problem 2.3.1 into crisp constraints of Problem 2.4.7. Therefore, the same relation should be used to find the optimal solution of

2.5 Flaws of the Existing Methods

47

Problem 2.4.7. However, Clemente et al. [5] have claimed that the Problem 2.4.7 is a multi-objective problem and can be solved by any of the existing methods. As there are several methods for solving multi-objective linear programming problem and it is not necessary that for the optimal solution

L∗ ∗ υ (0), υ ∗ (1), υ R (0) of Problem  2.4.7,

obtained by all the  existing methods, the ∗ L∗ condition υ (0), υ ∗ (1), υ R (0) υ L (0), υ(1), υ R (0) , where υ L (0), υ(1), υ R (0) is a feasible solution of Problem 2.4.7, will be satisfied. Also, as the constraints υ L (0) ≤ υ(1) ≤ υ R (0) are not considered in Problem 2.4.7 so for the optimal solution of Problem 2.4.7, the condition υ L (0) ≤ υ(1) ≤ υ R (0) may or may not be satisfied i.e., it is not always possible to obtain a triangular fuzzy number, representing the optimal value of Problem 2.4.7, by using the optimal solution 3725 , υ(1) = 0, υ R = 0, x1 = of Problem 2.4.7. e.g., it can be verified that υ L = 24 5 19 , x2 = is an efficient solution of existing Problem 2.5.3. However, in the 24 24 efficient solution the inequalities υ(1) − υ L (0) ≥ 0 and υ R (0) −υ(1) ≥ 0 are 

L  3725 R , 0, 0 not satisfying i.e., the value of game υ˜ = υ (0), υ(1), υ (0) = 24 is not a triangular fuzzy number. Problem 2.5.3

 Maximize υ L (0), υ(1), υ R (0) Subject to 150x1 + 175x2 ≥ υ L (0); 175x1 + 80x2 ≥ υ L (0); R 158x1 + 190x2 ≥ υ R (0); 190x1 + 100x2 ≥ υ (0); 156x1 + 158x2 ≥ υ L (0); x1 + x2 = 1; x1 , x2 ≥ 0 180x1 + 90x2 ≥ υ(1); 3. Li [11] considered that if (a L (0), a(1), a R (0)) and (b L (0), b(1), b R (0)) are two triangular fuzzy numbers then (a L (0), a(1), a R (0)) (b L (0), b(1), b R (0)) i.e., minimum{(a L (0), a(1), a R (0)), (b L (0), b(1), b R (0))} = (a L (0), a(1), a R (0)) iff a L (0) ≤ b L (0), a(1) ≤ b(1) and a R (0) ≤ b R (0). Using this relation, Li [11] claimed that to find the minimum expected gain m  of Player I i.e., (υ L (0), υ(1), υ R (0)) = minimum{ (aiLj (0), ai j (1), aiRj (0))xi , j = 1, 2, ..., n|

m 

i=1

xi = 1, xi ≥ 0, i = 1, 2, ..., m} and to find the maximum expected loss

i=1

of Player II i.e., (ω L (0), ω(1), ω R (0)) = maximum { 1, 2, ..., m|

n 

n  j=1

(aiLj (0), ai j (1), aiRj (0))y j , i =

y j = 1, y j ≥ 0, j = 1, 2, ..., n} is equivalent to find the optimal solu-

j=1

tion of Problem 2.3.1 and Problem 2.3.2 respectively. Furthermore, using the relation, (a L (0), a(1), a R (0))  (b L (0), b(1), b R (0)) ⇒ L a (0) ≥ b L (0), a(1) ≥ b(1), a R (0) ≥ b R (0), Li [11] transformed the Problem 2.3.1 into Problem 2.5.4.

48

2 Matrix Games with Fuzzy Payoffs

Problem 2.5.4 Maximize (υ L (0), υ(1), υ R (0)) Subject to m  aiLj (0)xi ≥ υ L (0), j = 1, 2, ..., n;

i=1 m  i=1

aiRj (0)xi ≥ υ R (0), j = 1, 2, ..., n;

m  i=1 m  i=1

ai j (1)xi ≥ υ(1), j = 1, 2, ..., n; xi = 1; xi ≥ 0, i = 1, 2, ..., m.

Now, if S = xi , i = 1, 2..., m; υ L (0), υ(1), υ R (0) is the set of all feasible solu∗ tions of Problem 2.5.4 then a feasible solution xi∗ , i = 1, 2..., m; υ L (0), υ ∗ (1), ∗ ∗ υ R (0) will be an optimal solution of Problem 2.5.4 if υ L (0) ≤ υ L (0), υ ∗ (1) ≤ R∗ R L R υ(1), υ (0) ≤ υ (0) ∀υ (0), υ(1), υ (0) ∈ S. However, if (υ1L (0), υ1 (1), υ1R (0)) and (υ2L (0), υ2 (1), υ2R (0)) are two feasible solutions of Problem 2.5.4 such that υ1L (0) ≤ υ2L (0), υ1 (1) ≤ υ2 (1) and υ1R (0) ≤ υ2R (0) are not satisfied. Then both (υ1L (0), υ1 (1), υ1R (0)) and (υ2L (0), υ2 (1), υ2R (0)) will be optimal solution of Problem 2.5.4. For example, if (1, 4, 5) and (2, 3, 4) are feasible solutions of Problem 2.5.4 then neither (1, 4, 5)  (2, 3, 4) nor (2, 3, 4)  (1, 4, 5). So, both (1, 4, 5) and (2, 3, 4) will represent maximum{(1,4,5),(2,3,4)} and hence, (1, 4, 5) and (2, 3, 4) will be optimal solution of Problem 2.5.4. To tackle used the following method to find the optimal this problem, Li [10] ∗ ∗ the set of all the feasible solution xi∗ , i = 1, 2..., m; υ L (0), υ ∗ (1), υ R (0) from solutions S = xi , i = 1, 2..., m; υ L (0), υ(1), υ R (0) . Find maximum{υ(1)|υ(1) ∈ S}. Case 1: If maximum{υ(1)} occurs corresponding to only one feasible solution then that feasible solution is optimal solution of Problem 2.5.4. Case 2: If maximum{υ(1)} occurs to two or more than two feasi corresponding υ L (0) + υ R (0) L ble solutions then find maximum ; υ (0), υ R (0) are those feasible 2 solution corresponding to which maximum{υ(1)} exists } . The feasible solutions υ L (0) + υ R (0) will exist, will be the corresponding to which maximum value of 2 optimal solution of Problem 2.5.4. It is obvious that Li [10] has transformed the constraints of Problem 2.3.1 into constraints of Problem 2.5.4 by using the relation (a L (0), a(1), a R (0))  (b L (0), b(1), b R (0)) ⇒ a L (0) ≥ b L (0), a(1) ≥ b(1), a R (0) ≥ b R (0). While, Li [10] has used the relation (a L (0), a(1), a R (0))  (b L (0), b(1), b R (0)) ⇒ a L (0) > b L (0) or if b L (0) + b R (0) a L (0) + a R (0) > for choosing the optimal a L (0) = b L (0) then 2 2 solution of Problem 2.5.4 from all the possible feasible solutions of Problem 2.5.4 i.e., Li [10] has used two different methods simultaneously for solving Problem 2.3.1, which is mathematically incorrect. Similarly, it can be easily verified that Li [10] has used two different methods (a L (0), a(1), a R (0)) (b L (0), b(1), b R (0)) ⇒ a L (0) ≤ b L (0), a(1) ≤ b(1), a R (0) ≤ b R (0) as well as (a L (0), a(1), a R (0)) (b L (0), b(1), b R (0)) ⇒ a L (0) < b L (0) + b R (0) a L (0) + a R (0) < simultaneously b L (0) or if a L (0) = b L (0) then 2 2 for solving Problem 2.3.2, which is mathematically incorrect.

2.6 Invalidity of Existing Mathematical Formulation of Matrix Games with Fuzzy Payoffs

49

2.6 Invalidity of Existing Mathematical Formulation of Matrix Games with Fuzzy Payoffs In Sect. 1.6 Chap. 1, it has been proved that  of  n n   minimum [aiL , aiR ]xi | xi = 1, xi ≥ 0, i = 1, 2, ..., n =minimum [aiL , aiR ], i=1

i=1

i = 1, 2, ..., n}. Similarly, it can be easily verified and proved that the comparing method  for n n   used in the existing methods [5, 10–12, 14] , minimum a˜ i xi | xi = 1, xi ≥ 0, i=1

i=1

i = 1, 2, ..., n} = minimum{a˜ i , i = 1, 2, ..., n}. While, to obtain the mathematical formulation 2.3.1 and 2.3.2, the mathematically incorrect assumptions  Problems m  n m n     minimum a˜ i j xi y j | xi =1, xi ≥0, i = 1, 2, ..., m; y j = 1, y j ≥ 0, j=1 i=1 i=1 j=1 m m   a˜ i j xi , j = 1, 2, ..., n| xi = 1, xi ≥ 0, i = 1, j = 1, 2, ..., n} = minimum i=1 i=1    m n n    2, ..., m} and maximum a˜ i j y j xi | xi = 1; xi ≥ 0, i = 1, 2, ..., m; i=1 j=1 j=1   n n n    y j = 1; y j ≥ 0, j = 1, 2, ..., n = maximum a˜ i j y j , i=1, 2, ..., m| j=1 j=1 j=1  y j = 1, y j ≥0, j = 1, 2, ..., n respectively have been considered. Therefore, the Problems 2.3.1 and 2.3.2 are not valid and hence, the existing methods [5, 10–12, 14] in which the minimum expected gain of Player I and maximum loss of Player II are obtained by solving these problems are also not valid.

2.7 Proposed Mehar Method

In the existing methods [5, 10–12, 14] , it is assumed that if a L (0), a L (1), a R (1),

 R R b L (0), b L (1), bR (1),

bL (0) Lare twoR trapezoidal  fuzzy LnumbersLthen

a L(0) and L R R R b (0) if a (0) ≥ b (0), a (0), a (1), a (1), a (0)  b (0), b (1), b (1),

 a L (1) ≥ b L (1), a R (1) ≥ b R (1), a R (0) ≥ b R (0) and a L (0), a L (1), a R (1), a R (0) b L (0), b L (1), b R (1), b R (0) 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).

50

2 Matrix Games with Fuzzy Payoffs

In this section, a new method (named as Mehar method), on the basis of this comparing method, is proposed to find minimum expected gain of Player I and maximum expected loss of Player II and their corresponding optimal strategies.

2.7.1 Minimum Expected Gain of Player I



Using the comparing method, a L (0), a L (1), a R (1), a R (0) b L (0), b L (1), b R (1), b R (0) 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 corresponding optimal strategies can be obtained as follows: m n   L ∗ ∗ ∗ (ai j (0), aiLj (1), aiRj (1), Step 1: Find (y1 , y2 , ..., yn ) ∈ Y such that value of j=1 i=1  aiRj (0))xi y j is minimum for all (x1 , x2 , ..., xm ) ∈ X i.e., find the solution {y ∗j , j = 1, 2, ..., n} of Problem 2.7.1. Problem  2.7.1    n m   L L R R Minimize (ai j (0), ai j (1), ai j (1), ai j (0))xi y j j=1

i=1

Subject to m n   xi = 1; y j = 1; xi ≥ 0, i = 1, 2, ..., m; y j ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 2: Using the property,

n  i=1

λ(aiL (0), aiL (1), aiR (1), aiR (0)) =

n  i=1

(λaiL (0), λaiL (1),

λaiR (1), λaiR (0)), the Problem 2.7.1 can be transformed into Problem 2.7.2. Problem  2.7.2  n  m  L L R R Minimize (ai j (0)xi y j , ai j (1)xi y j , ai j (1)xi y j , ai j (0)xi y j ) j=1 i=1

Subject to m n   xi = 1; y j = 1; xi ≥ 0, i = 1, 2, ..., m; y j ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 3: According to comparing method, to find the optimal solution {y1∗ , y2∗ , ..., yn∗ } n  m  of Problem 2.7.2 such that value of (aiLj (0)xi y j , aiLj (1)xi y j , aiRj (1)xi y j , aiRj (0) j=1 i=1

xi y j ) is minimum for all (x1 , x2 , ..., xm ) ∈ X is equivalent to find {y1∗ , y2∗ , ..., yn∗ } such n  n  n  n  m m m m     aiLj (0)xi y j , aiLj (1)xi y j , aiRj (1)xi y j and aiRj (0) that value of j=1 i=1

j=1 i=1

j=1 i=1

j=1 i=1

xi y j is minimum for all (x1 , x2 , ..., xm )∈X or if it is not possible to find {y1∗ , y2∗ , ..., yn∗ } n  n  n  m m m    aiLj (0)xi y j , aiLj (1)xi y j , aiRj (1)xi y j and for which the value of j=1 i=1

j=1 i=1

j=1 i=1

2.7 Proposed Mehar Method n  m 

51

aiRj (0)xi y j is minimum then find such {y ∗j1 , j = 1, 2, ..., n} for which the value j=1 i=1 n  n  n  m m m    aiLj (0)xi1 y j1 is minimum but value of aiLj (1)xi1 y j1 , aiRj (1)xi1 of j=1 i=1 j=1 i=1 j=1 i=1 n  m  y j1 and aiRj (0)xi1 y j1 is not minimum for all (x11 , x21 , ..., xm1 ) ∈ X and find j=1 i=1 n  m  ∗ aiLj (1)xi2 y j2 is minimum but such {y j2 , j = 1, 2, ..., n} for which the value of j=1 i=1 n  n  n  m m m    aiLj (0)xi2 y j2 , aiRj (1)xi2 y j2 and aiRj (0)xi2 y j2 is not minvalue of j=1 i=1 j=1 i=1 j=1 i=1 imum for all (x12 , x22 , ..., xm2 ) ∈ X and to find such {y ∗j3 , j = 1, 2, ..., n} for which n  n  n  m m m    aiRj (1)xi3 y j3 is minimum but value of aiLj (0)xi3 y j3 , the value of j=1 i=1 j=1 i=1 j=1 i=1 n  m  aiLj (1)xi3 y j3 and aiRj (0)xi3 y j3 is not minimum for all (x13 , x23 , ..., xm3 ) ∈ X j=1 i=1 n  m  aiRj (0)xi4 y j4 is miniand find such {y ∗j4 , j = 1, 2, ..., n} for which the value of j=1 i=1 n  n  n  m m m    aiLj (0)xi4 y j4 , aiLj (1)xi4 y j4 and aiRj (1)xi4 y j4 are mum but value of j=1 i=1 j=1 i=1 j=1 i=1 not minimum for all (x14 , x24 , ..., xm4 ) ∈ X i.e., to find optimal solution {y ∗j1 , j = 1, 2, ..., n}, {y ∗j2 , j = 1, 2, ..., n}, {y ∗j3 , j = 1, 2, ..., n} and {y ∗j4 , j = 1, 2, ..., n} of Problem 2.7.3, Problem 2.7.4, Problem 2.7.5 and Problem 2.7.6 respectively. Problem  2.7.3  n  m  L Minimize ai j (0)xi1 y j1 j=1 i=1

Subject to m n   xi1 = 1; y j1 = 1; i=1

j=1

xi1 ≥ 0, i = 1, 2, ..., m; y j1 ≥ 0, j = 1, 2, ..., n.

Problem  2.7.4  n  m  L Minimize ai j (1)xi2 y j2 j=1 i=1

Subject to m n   xi2 = 1; y j2 = 1; i=1

j=1

xi2 ≥ 0, i = 1, 2, ..., m; y j2 ≥ 0, j = 1, 2, ..., n.

Problem  2.7.5  n  m  R Minimize ai j (1)xi3 y j3 j=1 i=1

52

2 Matrix Games with Fuzzy Payoffs

Subject to m n   xi3 = 1; y j3 = 1; i=1

j=1

xi3 ≥ 0, i = 1, 2, ..., m; y j3 ≥ 0, j = 1, 2, ..., n.

Problem  2.7.6  n  m  R Minimize ai j (0)xi4 y j4 j=1 i=1

Subject to m n   xi4 = 1; y j4 = 1; i=1

j=1

xi4 ≥ 0, i = 1, 2, ..., m; y j4 ≥ 0, j = 1, 2, ..., n.

Step 4: Since, in Problem 2.7.3, Problem 2.7.4, Problem 2.7.5 and Problem 2.7.6 only y j1 , y j2 , y j3 and y j4 respectively have been considered as decision variables. So, Problem 2.7.3, Problem 2.7.4, Problem 2.7.5 and Problem 2.7.6 are linear programming problems and hence, the optimal value of Problem 2.7.3, Problem 2.7.4, Problem 2.7.5 and Problem 2.7.6 will be equal to optimal value of its corresponding dual problem i.e., Problem 2.7.7, Problem 2.7.8, Problem 2.7.9 and Problem 2.7.10 respectively. Problem 2.7.7 Maximize{υ1 } Subject to m  aiLj (0)xi1 ≥ υ1 , j = 1, 2, ..., n;

i=1 m 

xi1 = 1; xi1 ≥ 0, i = 1, 2, ..., m

i=1

Problem 2.7.8 Maximize{υ2 } Subject to m  aiLj (1)xi2 ≥ υ2 , j = 1, 2, ..., n;

i=1 m 

xi2 = 1; xi2 ≥ 0, i = 1, 2, ..., m.

i=1

Problem 2.7.9 Maximize{υ3 } Subject to m  aiRj (1)xi3 ≥ υ3 , j = 1, 2, ..., n;

i=1 m  i=1

xi3 = 1; xi3 ≥ 0, i = 1, 2, ..., m

2.7 Proposed Mehar Method

53

Problem 2.7.10 Maximize{υ4 } Subject to m  aiRj (0)xi4 ≥ υ4 , j = 1, 2, ..., n;

i=1 m 

xi4 = 1; xi4 ≥ 0, i = 1, 2, ..., m.

i=1 ∗ ∗ ∗ Step 5: Substitute the value of {xi1 , i = 1, 2, ..., m}, {xi2 , i = 1, 2, ..., m}, {xi3 ,i = ∗ 1, 2, ..., m} and {xi4 , i = 1, 2, ..., m} of Problem 2.7.7, Problem 2.7.8, Problem 2.7.9 and Problem 2.7.10 in Problem 2.7.3, Problem 2.7.4, Problem 2.7.5 and Problem 2.7.6 respectively and find all the alternative basic optimal solutions ∗ ∗ ∗ {y kj1 , j = 1, 2, ..., n, k = 1, 2, ..., l}, {y rj2 , j = 1, 2, ..., n, r = 1, 2, ..., u}, {y sj3 , j = q∗

1, 2, ..., n, s = 1, 2, ..., e} and {y j4 , j = 1, 2, ..., n, q = 1, 2, ..., h} of Problem 2.7.3, Problem 2.7.4, Problem 2.7.5 and Problem 2.7.6 respectively. Step 6: Find ⎧ n m ⎫  L ⎪ ⎪ L R R ∗ k∗ ⎪ (ai j (0), ai j (1), ai j (1), ai j (0))xi1 y j1 ; k = 1, 2, ..., l, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ j=1 i=1 ⎪ ⎪ ⎪ ⎪ n  m ⎪ ⎪  ⎪ ⎪ ∗ L L R R ∗ r ⎪ ⎪ ⎪ ⎪ (a (0), a (1), a (1), a (0))x y ; r = 1, 2, ..., u, ⎨ ⎬ ij ij ij ij i2 j2 j=1 i=1 minimum  n  m ⎪ ∗ s∗ ⎪ ⎪ (aiLj (0), aiLj (1), aiRj (1), aiRj (0))xi3 y j3 ; s = 1, 2, ..., e, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ j=1 i=1 ⎪ ⎪ ⎪ ⎪ n m ⎪ ⎪ ∗   ⎪ ⎪ q L L R R ∗ ⎪ ⎪ ⎪ (ai j (0), ai j (1), ai j (1), ai j (0))xi4 y j4 ; q = 1, 2, ..., h ⎪ ⎩ ⎭ j=1 i=1

Step 7: All the minimum trapezoidal fuzzy numbers will represent minimum ∗ expected gain of Player I. The optimal strategies will be {xi1 , i = 1, 2, ..., m} if n  m  ∗ k∗ (aiLj (0), aiLj (1), aiRj (1), aiRj (0))xi1 y j1 ; k = 1, 2, ..., l; minimum corresponds to j=1 i=1

∗ , i = 1, 2, ..., m} if minimum corresponds to {xi2

n  m  j=1 i=1

(aiLj (0), aiLj (1), aiRj (1), aiRj (0))

∗

∗ r ∗ y j2 ; r = 1, 2, ..., u ; {xi1 , i = 1, 2, ..., m} if minimum corresponds to xi2 ∗

n  m  j=1 i=1

∗ s ∗ (aiLj (0), aiLj (1), aiRj (1), aiRj (0))xi3 y j3 ; s=1, 2, ..., e and {xi4 , i = 1, 2, ..., m} if minn  m ∗  ∗ q (aiLj (0), aiLj (1), aiRj (1), aiRj (0))xi4 y j4 ; q = 1, 2, ..., h. imum corresponds to j=1 i=1

2.7.2 Maximum Expected Loss of Player II



Using the comparing method, a L (0), a L (1), a R (1), a R (0)  b L (0), b L (1), b R (1), b R (0) 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 maximum expected loss of Player II and corresponding optimal strategies can be obtained as follows:

54

2 Matrix Games with Fuzzy Payoffs

m n   L Step 1: Find (x1∗ , x2∗ , ..., xm∗ ) ∈ X such that value of (ai j (0), aiLj (1), aiRj (1), j=1 i=1  aiRj (0))xi y j is maximum for all (y1 , y2 , ..., yn ) ∈ Y i.e., find the solution {xi∗ , i = 1, 2, ..., m} of Problem 2.7.11. Problem 2.7.11     n m   Maximize (aiLj (0), aiLj (1), aiRj (1), aiRj (0))xi y j j=1

i=1

Subject to m n   xi = 1; y j = 1; xi ≥ 0, i = 1, 2, ..., m; y j ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 2: Using the property,

n  i=1

λ(aiL (0), aiL (1), aiR (1), aiR (0)) =

n  i=1

(λaiL (0),

λaiL (1), λaiR (1), λaiR (0)), the Problem 2.7.11 can be transformed into Problem 2.7.12. Problem 2.7.12   n  m  Maximize (aiLj (0)xi y j , aiLj (1)xi y j , aiRj (1)xi y j , aiRj (0)xi y j ) j=1 i=1

Subject to m n   xi = 1; y j = 1; xi ≥ 0, i = 1, 2, ..., m; y j ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 3: According to comparing method, to find the optimal solution {x1∗ , x2∗ , ..., xm∗ } n  m  of Problem 2.7.12 such that value of (aiLj (0)xi y j , aiLj (1)xi y j , aiRj (1)xi y j , aiRj (0) j=1 i=1

xi y j ) is maximum for all (y1 , y2 , ..., yn ) ∈ Y is equivalent to find {x1∗ , x2∗ , ..., xm∗ } such n  n  n  n  m m m m     aiLj (0)xi y j , aiLj (1)xi y j , aiRj (1)xi y j and aiRj (0) that value of j=1 i=1

j=1 i=1

j=1 i=1

j=1 i=1

xi y j is maximum for all (y1 , y2 , ..., yn )∈Y or if it is not possible to find {x1∗ , x2∗ , ..., xm∗ } n  n  n  m m m    aiLj (0)xi y j , aiLj (1)xi y j , aiRj (1)xi y j and for which the value of n  m  j=1 i=1

j=1 i=1

aiRj (0)xi y j

value of

n  m  j=1 i=1

j=1 i=1

is maximum then find such

∗ {xi1 ,

j=1 i=1

j = 1, 2, ..., m} for which the

aiLj (0)xi1 y j1 is maximum but value of

aiRj (1)xi1 y j1 and

n  m  j=1 i=1

j=1 i=1

aiLj (1)xi1 y j1 ,

n  m  j=1 i=1

aiRj (0)xi1 y j1 is not maximum for all (y11 , y21 , ..., yn1 ) ∈ Y

∗ , i = 1, 2, ..., m} for which the value of and find such {xi2

imum but value of

n  m 

n  m  j=1 i=1

aiLj (0)xi2 y j2 ,

n  m  j=1 i=1

n  m  j=1 i=1

aiRj (1)xi2 y j2 and

aiLj (1)xi2 y j2 is max-

n  m  j=1 i=1

aiRj (0)xi2 y j2 is

2.7 Proposed Mehar Method

55

∗ not maximum for all (y12 , y22 , ..., yn2 ) ∈ X and to find such {xi3 , i = 1, 2, ..., m} for n n  m m  R  ai j (1)xi3 y j3 is maximum but value of aiLj (0)xi3 y j3 , which the value of n  m  j=1 i=1

aiLj (1)xi3 y j3

j=1 i=1 n  m 

and

j=1 i=1

j=1 i=1

aiRj (0)xi3 y j3

is not maximum for all (y13 , y23 , ..., yn3 )

∗ , i = 1, 2, ..., m} for which the value of ∈ X and find such {xi4

maximum but value of

n  m  j=1 i=1

aiLj (0)xi4 y j4 ,

n  m  j=1 i=1

n  m  j=1 i=1

aiLj (1)xi4 y j4 and

aiRj (0)xi4 y j4 is

n  m  j=1 i=1

aiRj (1)xi4

y j4 is not maximum for all (x14 , x24 , ..., xm4 ) ∈ X i.e., to find optimal solution ∗ ∗ ∗ ∗ , i = 1, 2, ..., m}, {xi2 , i = 1, 2, ..., m}, {xi3 , i = 1, 2, ..., m} and {xi4 ,i = {xi1 1, 2, ..., m} of Problem 2.7.13, Problem 2.7.14, Problem 2.7.15 and Problem 2.7.16 respectively. Problem 2.7.13   n  m  L Maximize ai j (0)xi1 y j1 j=1 i=1

Subject to m n   xi1 = 1; y j1 = 1; i=1

j=1

xi1 ≥ 0, i = 1, 2, ..., m; y j1 ≥ 0, j = 1, 2, ..., n.

Problem 2.7.14   n  m  L Maximize ai j (1)xi2 y j2 j=1 i=1

Subject to m n   xi2 = 1; y j2 = 1; i=1

j=1

xi2 ≥ 0, i = 1, 2, ..., m; y j2 ≥ 0, j = 1, 2, ..., n.

Problem 2.7.15   n  m  R Maximize ai j (1)xi3 y j3 j=1 i=1

Subject to m n   xi3 = 1; y j3 = 1; i=1

j=1

xi3 ≥ 0, i = 1, 2, ..., m; y j3 ≥ 0, j = 1, 2, ..., n.

Problem 2.7.16   n  m  R Maximize ai j (0)xi4 y j4 j=1 i=1

56

2 Matrix Games with Fuzzy Payoffs

Subject to m n   xi4 = 1; y j4 = 1; i=1

j=1

xi4 ≥ 0, i = 1, 2, ..., m; y j4 ≥ 0, j = 1, 2, ..., n.

Step 4: Since, in Problem 2.7.13, Problem 2.7.14, Problem 2.7.15 and Problem 2.7.16 only xi1 , xi2 , xi3 and xi4 respectively have been considered as decision variables. So, Problem 2.7.13, Problem 2.7.14, Problem 2.7.15 and Problem 2.7.16 are linear programming problems and hence, the optimal value of Problem 2.7.13, Problem 2.7.14, Problem 2.7.15 and Problem 2.7.16 will be equal to optimal value of its corresponding dual problem i.e., Problem 2.7.17, Problem 2.7.18, Problem 2.7.19 and Problem 2.7.20 respectively. Problem 2.7.17 Minimize{ω1 } Subject to n  aiLj (0)y j1 ≤ ω1 , i = 1, 2, ..., m; j=1 n 

y j1 = 1; y j1 ≥ 0, j = 1, 2, ..., n

j=1

Problem 2.7.18 Minimize{ω2 } Subject to n  aiLj (1)y j2 ≤ ω2 , i = 1, 2, ..., m; j=1 n 

y j2 = 1; y j2 ≥ 0, j = 1, 2, ..., n.

j=1

Problem 2.7.19 Minimize{ω3 } Subject to n  aiRj (1)y j3 ≤ ω3 , i = 1, 2, ..., m; j=1 n 

y j3 = 1; y j3 ≥ 0, j = 1, 2, ..., n

j=1

Problem 2.7.20 Minimize{ω4 } Subject to n  aiRj (0)y j4 ≤ ω4 , i = 1, 2, ..., m; j=1 n  j=1

y j4 = 1; y j4 ≥ 0, j = 1, 2, ..., n.

2.7 Proposed Mehar Method

57

Step 5: Substitute the value of {y ∗j1 , j = 1, 2, ..., n}, {y ∗j2 , j = 1, 2, ..., n}, {y ∗j3 , j = 1, 2, ..., n} and {y ∗j4 , j = 1, 2, ...n} of Problem 2.7.17, Problem 2.7.18, Problem 2.7.19 and Problem 2.7.20 in Problem 2.7.13, Problem 2.7.14, Problem 2.7.15 and Problem 2.7.16 respectively and find all the alternative basic optimal solutions k∗ s∗ r∗ , i = 1, 2, ..., m, k = 1, 2, ..., l}, {xi2 , i = 1, 2, ..., m, r = 1, 2, ..., u}, {xi3 ,i = {xi1 q∗ 1, 2, ..., m, s = 1, 2, ..., e} and {xi4 , i = 1, 2, ..., m, q = 1, 2, ..., h} of Problem 2.7.13, Problem 2.7.14, Problem 2.7.15 and Problem 2.7.16 respectively. Step 6: Find ⎧ n m ⎫ ⎪  L ⎪ k∗ ∗ L R R ⎪ (ai j (0), ai j (1), ai j (1), ai j (0))xi1 y j1 ; k = 1, 2, ..., l, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ j=1 i=1 ⎪ ⎪ ⎪ ⎪ n m ⎪ ⎪   ⎪ ⎪ ∗ L L R R r ∗ ⎪ ⎪ ⎪ ⎪ (a (0), a (1), a (1), a (0))x y ; r = 1, 2, ..., u, ⎨ ⎬ i2 j2 ij ij ij ij j=1 i=1 maximum  n  m ⎪ ⎪ s∗ ∗ L L R R ⎪ ⎪ ⎪ ⎪ j=1 i=1(ai j (0), ai j (1), ai j (1), ai j (0))xi3 y j3 ; s = 1, 2, ..., e, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ n  m ⎪ ⎪ ∗ ⎪ ⎪ q ∗ L L R R ⎪ ⎪ ⎪ ⎪ (a (0), a (1), a (1), a (0))x y ; q = 1, 2, ..., h ⎩ ⎭ i4 j4 ij ij ij ij j=1 i=1

Step 7: All the maximum trapezoidal fuzzy numbers will represent maximum expected loss of Player II. The optimal strategies will be {y ∗j1 , j = 1, 2, ..., n} if n  m  k∗ ∗ (aiLj (0), aiLj (1), aiRj (1), aiRj (0))xi1 y j1 ; k = 1, 2, ..., l; maximum corresponds to j=1 i=1

{y ∗j2 ,

j = 1, 2, ..., n} if maximum corresponds to

n  m  j=1 i=1

(aiLj (0), aiLj (1), aiRj (1), aiRj (0))

∗

r ∗ y j2 ; r = 1, 2, ..., u; {y ∗j3 , j = 1, 2, ..., n} if maximum corresponds xi2 ∗

n  m  j=1 i=1

(aiLj (0),

s ∗ aiLj (1), aiRj (1), aiRj (0))xi3 y j3 ; s = 1, 2, ..., e and {y ∗j4 , j = 1, 2, ..., n} if maximum n m  L q∗ (ai j (0), aiLj (1), aiRj (1), aiRj (0))xi4 y ∗j4 ; t = 1, 2, ..., h. corresponds to j=1 i=1

2.8 Numerical Example In this section, matrix games with fuzzy payoffs A˜ =



(175, 180, 190) (150, 156, 158) (80, 90, 100) (175, 180, 190)



chosen by Li [11], is solved by the proposed Mehar method.

2.8.1 Minimum Expected Gain of Player I Using the proposed Mehar method, the minimum expected gain of Player I and corresponding optimal strategies can be obtained as follows:

58

2 Matrix Games with Fuzzy Payoffs

Step 1: Find (y1∗ , y2∗ ) ∈ Y such that value of (175, 180, 190)x1 y1 + (150, 156, 158) x1 y2 + (80, 90, 100)x2 y1 + (175, 180, 190)x2 y2 is minimum for all (x1 , x2 ) ∈ X i.e., find the solution {y ∗j , j = 1, 2} of Problem 2.8.1. Problem 2.8.1 Minimize{(175, 180, 190)x1 y1 + (150, 156, 158)x1 y2 + (80, 90, 100)x2 y1 + (175, 180, 190)x2 y2 } Subject to x1 + x2 = 1; y1 + y2 = 1; x1 , x2 ≥ 0; y1 , y2 ≥ 0. Step 2: Using the property, λaiR (1), λaiR (0)) n  i=1

aiR (1),

n  i=1

and

n  i=1 n  i=1

λ(aiL (0), aiL (1), aiR (1), aiR (0)) =

(aiL (0), aiL (1), aiR (1), aiR (0)) = (

n 

i=1 n 

i=1

(λaiL (0), aiL (1),

aiL (0),

n  i=1

aiL (1),

aiR (0)) the Problem 2.8.1 can be transformed into Problem 2.8.2.

Problem 2.8.2  (175x y + 150x y + 80x y + 175x y , 180x1 y1 + 156x1 y2 + 90x2 y1 + 180x2 y2 , Minimize 190x 1y 1+ 158x 1y 2+ 100x2 y1 + 190x2 y2 ) 1 1

1 2

2 1

2 2

Subject to x1 + x2 = 1; y1 + y2 = 1; x1 , x2 ≥ 0; y1 , y2 ≥ 0. Step 3: Find optimal solution {y ∗j1 , j = 1, 2}, {y ∗j2 , j = 1, 2} and {y ∗j3 , j = 1, 2} of Problem 2.8.3, Problem 2.8.4 and Problem 2.8.5 respectively. Problem 2.8.3 Minimize{175x11 y11 + 150x11 y21 + 80x21 y11 + 175x21 y21 } Subject to x11 + x21 = 1; y11 + y21 = 1; x11 , x21 ≥ 0; y11 , y21 ≥ 0. Problem 2.8.4 Minimize{180x12 y12 + 156x12 y22 + 90x22 y12 + 180x22 y22 } Subject to x12 + x22 = 1; y12 + y22 = 1; x12 , x22 ≥ 0; y12 , y22 ≥ 0. Problem 2.8.5 Minimize{190x13 y13 + 158x13 y23 + 100x23 y13 + 190x23 y23 } Subject to x13 + x23 = 1; y13 + y23 = 1; x13 , x23 ≥ 0; y13 , y23 ≥ 0. Step 4: Since, in Problem 2.8.3, Problem 2.8.4 and Problem 2.8.5 only y j1 , y j2 and y j3 respectively have been considered as decision variables. So, Problem 2.8.3, Problem 2.8.4 and Problem 2.8.5 are linear programming problems and hence, the optimal value of Problem 2.8.3, Problem 2.8.4 and Problem 2.8.5 will be equal to optimal value of its corresponding dual problem i.e., Problem 2.8.6, Problem 2.8.7 and Problem 2.8.8 respectively.

2.8 Numerical Example

59

Problem 2.8.6 Maximize{υ1 } Subject to 175x11 + 80x21 ≥ υ1 ; 150x11 + 175x21 ≥ υ1 ; x11 + x21 = 1; x11 , x21 ≥ 0. Problem 2.8.7 Maximize{υ2 } Subject to 180x12 + 90x22 ≥ υ2 ; 156x12 + 180x22 ≥ υ2 ; x12 + x22 = 1; x12 , x22 ≥ 0. Problem 2.8.8 Maximize{υ3 } Subject to 190x13 + 100x23 ≥ υ3 ; 158x13 + 190x23 ≥ υ3 ; x13 + x23 = 1; x13 , x23 ≥ 0.

 ∗ x11

   19 5 15 4 ∗ , x21 = , x12 = , x22 = = 24 24 19 19

Step 5: Substituting the value of   45 16 ∗ , x23 = of Problem 2.8.6, Problem 2.8.7 and Problem 2.8.8 in and x13 = 61 61 Problem 2.8.3, Problem 2.8.4 and Problem 2.8.5 respectively and the obtained alternative of Problem Problem 2.8.4 Problem 2.8.5 solutions 2.8.3, and 1 basic1 optimal 2 2 1 1 2 2 = 0, y = 1 ; y = 1, y = 0 , y = 0, y22 =1 y11 = 1, y21 = 0 , y 11 21 12 22 12 1 1 2 2 = 1, y23 = 0 , y13 = 0, y23 = 1 respectively. and y13 Step 6: Now, ⎧       ⎫ ⎪ ⎪ ⎨

3725 ,  24 9155 ⎪ ⎪ , ⎩ 61

minimum 

645 685 494 3725 3060 3250 , , 161, , , , 155, 4 4 3 19  24   19 9540 10150 9550 9900 10150 , , , , 61 61 61 61    61

,

2950 3060 3130 , , 19 19 19



⎪ ⎪ ⎬ ⎪ ⎪ ⎭

494 2950 3060 3130 9155 9540 10150 3725 , 161, , , , , , , . 24 3 19 19 19 61 61 61  3725 494 2950 3060 3130 Step 7: The triangular fuzzy numbers , 161, , , , 24 3 19 19 19   9155 9540 10150 , , represent minimum expected gain of Player I. The optiand 61 61 61     3725 494 19 5 ∗ mal strategies corresponding to , 161, are x11 , x21 = , cor= 24 3 24 24     2950 3060 3130 15 4 ∗ responding to , , are x12 , x22 = and corresponding = 19 19 19 19 19     9155 9540 10150 45 16 ∗ to , , are x13 . = , x23 = 61 61 61 61 61

=

60

2 Matrix Games with Fuzzy Payoffs

2.8.2 Maximum Expected Loss of Player II Using the proposed Mehar method, the maximum expected loss of Player II and corresponding optimal strategies can be obtained as follows: Step 1: Find (x1∗ , x2∗ ) ∈ X such that value of (175, 180, 190)x1 y1 + (150, 156, 158) x1 y2 + (80, 90, 100)x2 y1 + (175, 180, 190)x2 y2 is maximum for all (y1 , y2 ) ∈ Y i.e., find the solution {xi∗ , i = 1, 2} of Problem 2.8.9. Problem 2.8.9 Maximize{(175, 180, 190)x1 y1 + (150, 156, 158)x1 y2 + (80, 90, 100)x2 y1 + (175, 180, 190)x2 y2 } Subject to x1 + x2 = 1; y1 + y2 = 1; x1 , x2 ≥ 0; y1 , y2 ≥ 0. Step 2: Using the property, λaiR (1), λaiR (0)) n  i=1

aiR (1),

n  i=1

and

n  i=1 n  i=1

λ(aiL (0), aiL (1), aiR (1), aiR (0)) =

(aiL (0), aiL (1), aiR (1), aiR (0)) = (

n 

i=1 n 

i=1

(λaiL (0), aiL (1),

aiL (0),

n  i=1

aiL (1),

aiR (0)) the Problem 2.8.9 can be transformed into Problem 2.8.10.

Problem 2.8.10   1 y1 + 150x 1 y2 + 80x 2 y1 + 175x 2 y2 , 180x 1 y1 + 156x 1 y2 + 90x 2 y1 + 180x 2 y2 , Maximize (175x 190x y + 158x y + 100x y + 190x y ) 1 1

1 2

2 1

2 2

Subject to x1 + x2 = 1; y1 + y2 = 1; x1 , x2 ≥ 0; y1 , y2 ≥ 0. ∗ ∗ ∗ Step 3: Find optimal solution {xi1 , i = 1, 2}, {xi2 , i = 1, 2} and {xi3 , j = 1, 2} of Problem 2.8.11, Problem 2.8.12 and Problem 2.8.13 respectively.

Problem 2.8.11 Maximize{175x11 y11 + 150x11 y21 + 80x21 y11 + 175x21 y21 } Subject to x11 + x21 = 1; y11 + y21 = 1; x11 , x21 ≥ 0; y11 , y21 ≥ 0. Problem 2.8.12 Maximize{180x12 y12 + 156x12 y22 + 90x22 y12 + 180x22 y22 } Subject to x12 + x22 = 1; y12 + y22 = 1; x12 , x22 ≥ 0; y12 , y22 ≥ 0. Problem 2.8.13 Maximize{190x13 y13 + 158x13 y23 + 100x23 y13 + 190x23 y23 } Subject to x13 + x23 = 1; y13 + y23 = 1; x13 , x23 ≥ 0; y13 , y23 ≥ 0.

2.8 Numerical Example

61

Step 4: Since, in Problem 2.8.3, Problem 2.8.4 and Problem 2.8.5 only xi1 , xi2 and xi3 respectively have been considered as decision variables. So, Problem 2.8.11, Problem 2.8.12 and Problem 2.8.13 are linear programming problems and hence, the optimal value of Problem 2.8.11, Problem 2.8.12 and Problem 2.8.13 will be equal to optimal value of its corresponding dual problem i.e., Problem 2.8.14, Problem 2.8.15 and Problem 2.8.16 respectively. Problem 2.8.14 Minimize{ω1 } Subject to 175y11 + 150y21 ≤ ω1 ; 80y11 + 175y21 ≤ ω1 ; y11 + y21 = 1; y11 , y21 ≥ 0. Problem 2.8.15 Minimize{ω2 } Subject to 180y12 + 156y22 ≤ ω2 ; 90y12 + 180y22 ≤ ω2 ; y12 + y22 = 1; y12 , y22 ≥ 0. Problem 2.8.16 Minimize{ω3 } Subject to 190y13 + 158y23 ≤ ω3 ; 100y13 + 190y23 ≤ ω3 ; y13 + y23 = 1; y13 , y23 ≥ 0.

 ∗ y11

   5 19 4 15 ∗ , y21 = , y12 = , y22 = = 24 24 19 19

Step 5: Substituting the value of   16 45 ∗ , y23 = of Problem 2.8.14, Problem 2.8.15 and Problem 2.8.16 and y13 = 61 61 in Problem 2.8.11, Problem 2.8.12 and Problem 1 2.8.131respectively 2 and the 2obtained alternative basic optimal solutions x = 1, x = 0 , x 11 = 0, x21 = 1 ; 11 21 1 1 2 2 1 1 2 2 = 0 , x12 = 0, x22 = 1 and x13 = 1, x23 = 0 , x13 = 0, x23 =1 x12 = 1, x22 of Problem 2.8.11, Problem 2.8.12 and Problem 2.8.13 respectively. Step 6: Now, ⎧       ⎫ ⎪ ⎪ ⎨

3725 , 24 maximum  9155 ⎪ ⎪ , ⎩ 61 

645 685 3725 494 3060 3250 , , , 161, , 155, , 4 4 24 3 19    19 9540 10150 9550 9900 10150 , , , , 61 61 61 61  61 

,

2950 3060 3130 , , 19 19 19

⎪ ⎪ ⎬ ⎪ ⎪ ⎭

 3060 3250 9550 9900 10150 3725 645 685 , , , 155, , , , , . 24 4 4 19  19 61  61 61  3725 645 685 3060 3250 Step 7: The triangular fuzzy numbers , , , 155, , and 24 4 4 19 19   9550 9900 10150 represent maximum expected loss of Player II. The optimal , , 61 61 61

=

62

2 Matrix Games with Fuzzy Payoffs

    3725 645 685 5 19 ∗ , , are y11 , y21 = , correstrategies corresponding to = 24 4 24   4 24 3060 3250 4 15 ∗ sponding to 155, , are y12 , y22 = , and corresponding to = 19 19 19 19     9550 9900 10150 16 45 ∗ , , are y13 , y23 = . = 61 61 61 61 61

2.9 Conclusion On the basis of present study, it can be concluded that some mathematically incorrect assumptions have been considered in the existing methods [5, 10–12, 14] for solving matrix games with fuzzy payoffs. Therefore, it is not genuine to use these methods. Furthermore, to resolve flaws of the existing methods [5, 10–12, 14] , a new method (named as Mehar method) is proposed for solving matrix games with fuzzy payoffs.

References 1. Bector, C.R., Chandra, S.: Fuzzy Mathematical Programming and Fuzzy Matrix Games. Springer, Berlin (2005) 2. Campos, L.: Fuzzy linear programming models to solve fuzzy matrix games. Fuzzy Sets Syst. 32, 275–289 (1989) 3. Campos, L., Gonzalez, A.: Fuzzy matrix games considering the criteria of the players. Kybernetes 20, 17–23 (1991) 4. Campos, L., Gonzalez, A., Vila, M.A.: On the use of the ranking function approach to solve fuzzy matrix games in a direct way. Fuzzy Sets Syst. 49, 193–203 (1992) 5. Clemente, M., Fernandez, F.R., Puerto, J.: Pareto-optimal security in matrix games with fuzzy payoffs. Fuzzy Sets Syst. 176, 36–45 (2011) 6. Dubois, D., Prade, H.: Fuzzy Sets and Systems Theory and Applications. Academic Press, New York (1980) 7. Dutta, B., Gupta, S.K.: On Nash equilibrium strat-egy of two person zero sum games with trapezoidal fuzzy payoffs. Fuzzy Inf. Eng. 6, 299–314 (2014) 8. Kaufmann, A., Gupta, M.M: Introduction to Fuzzy Arithmetic Theory and Applications. Van Nostrand Publishing Co (1991) 9. Kaufmann, A., Gupta, M.M.: Fuzzy Mathematical Models in Engineering and Management Science. New York, USA (1988) 10. Li, D.F.: Lexicographic method for matrix games with payoffs of triangular fuzzy numbers. Int. J. Uncertain. Fuzziness Knowledge-Based Syst. 16, 371–389 (2008) 11. Li, D.F.: A fast approach to compute fuzzy values of matrix games with payoffs of triangular fuzzy numbers. Eur. J. Oper. Res. 223, 421–429 (2012) 12. Li, D.F.: An effective methodology for solving matrix games with fuzzy payoffs. IEEE Trans. Cybern. 43, 610–621 (2013) 13. Li, D.F.: Linear Programming Models and Methods of Matrix Games with Payoffs of Triangular Fuzzy Numbers. Springer, Berlin (2015)

References

63

14. Liu, S.T., Kao, C.: Solution of fuzzy matrix games: an application of the extension principle. Int. J. Intell. Syst. 22, 891–903 (2007) 15. Liu, S.T., Kao, C.: Matrix games with interval data. Comput. Ind. Eng. 56, 1697–1700 (2009) 16. Maeda, T.: On characterization of equilibrium strategy of two-person zero-sum games with fuzzy payoffs. Fuzzy Sets Syst. 139, 283–296 (2003) 17. Zadeh, L.A.: Fuzzy sets. Inf. Control. 8, 338–353 (1965)

Chapter 3

Constrained Matrix Games with Fuzzy Payoffs

In this chapter, flaws of the existing methods [4–7] for solving constrained matrix games with fuzzy payoffs (constrained matrix games in which payoffs are represented by fuzzy numbers) are pointed out. To resolve these flaws, a new method (named as Vaishnavi 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. To illustrate the proposed Vaishnavi method, some existing numerical problems of constrained matrix games with fuzzy payoffs are solved.

3.1 Constrained Matrix Games with Fuzzy Payoffs Charnes [1] pointed out that in two person zero sum games (matrix games), it is assumed that for strategies {xi , i = 1, 2, . . . , m} and {y j , j = 1, 2, . . . , n} of Player I and Player II respectively only the following constraints should be satisfied. (i)xi ≥ 0, i = 1, 2, . . . , m (ii)

m  i=1

xi = 1 (iii)y j ≥ 0, j = 1, 2, . . . , n (iv)

n 

yj = 1

j=1

However, there are certain matrix games in real life where the strategies of the players are constrained to satisfy general linear inequalities rather than the only above mentioned constraints. To handle such real life problem [2], Charnes [1] extended the matrix games into constrained matrix games and then later in some what more generally by Kawaguchi and Maruyama [3]. Li and Cheng [5] pointed out that there is no method in the literature to solve such constrained matrix games in which payoffs are represented by fuzzy numbers and proposed a method for the same. Later, Li [6, 7] with other co-authors proposed different methods for solving constrained matrix games with fuzzy payoffs.

66

3 Constrained Matrix Games with Fuzzy Payoffs

3.2 Existing Mathematical Formulation of Constrained Matrix Games with Fuzzy Payoffs In the literature [8], the mathematical formulation of constrained matrix games with crisp  payoffs is obtained by replacing the constraint set of strategies  m  xi = 1 of Player I with X = xi , i = 1, 2, . . . , m|xi ≥ 0, i = 1, 2, . . . , m;  X = xi , i = 1, 2, . . . , m|xi ≥ 0, i = 1, 2, . . . , m;



m 

xi ≤ 1, −

i=1

m 

i=1

xi ≤ −1,

i=1

and Y = y j , j = 1, 2, . . . , n|y j ≥ 0, j = 1, 2, . . . , m;

i=1

n 

 bil xi ≤ cl , l = 1, 2, . . . , p



y j = 1 of Player II with

j=1

 Y =

m 

y j , j = 1, 2, . . . , n|y j ≥ 0, j = 1, 2, . . . , m;

n  j=1

y j ≥ 1, −

n  j=1

y j ≥ −1,

n  j=1

 ek j y j ≥ dk , k = 1, 2, . . . , q

i.e., the minimum expected gain of Player I, the maximum expected loss of Player II and their corresponding strategies for constrained matrix games with crisp payoffs can be obtained by solving Problems 3.2.1 and 3.2.2. Problem  3.2.1  n  m  Minimize ai j xi y j j=1 i=1

Subject to m  bil xi ≤ cl , l = 1, 2, . . . , p; i=1 n  j=1 m  i=1 n 

ek j y j ≥ dk , k = 1, 2, . . . , q; xi ≤ 1; −

m 

xi ≤ −1;

i=1 n 

y j ≥ 1; −

j=1

y j ≥ −1

j=1

xi ≥ 0, i = 1, 2, . . . , m; y j ≥ 0, j = 1, 2, . . . , n. Problem 3.2.2   m  n  Maximize ai j y j xi i=1 j=1

Subject to m  bil xi ≤ cl , l = 1, 2, . . . , p; i=1 n  j=1 m  i=1

ek j y j ≥ dk , k = 1, 2, . . . , q; xi ≤ 1; −

m  i=1

xi ≤ −1;

3.2 Existing Mathematical Formulation of Constrained Matrix Games with Fuzzy Payoffs n 

n 

y j ≥ 1, −

j=1

67

y j ≥ −1

j=1

xi ≥ 0, i = 1, 2, . . . , m; y j ≥ 0, j = 1, 2, . . . , n. Furthermore, in the literature [8], it is pointed out that only y j and xi have been considered as decision variables in the Problem 3.2.1 and in Problem 3.2.2 respectively, so Problems 3.2.1 and 3.2.2 are linear programming problems. Therefore, the optimal value of Problems 3.2.1 and 3.2.2 is same as the optimal values of its corresponding dual problem i.e., Problem 3.2.3 and Problem 3.2.4 respectively. Problem 3.2.3   q  Maximize dk z k + z q+1 − z q+2 k=1

Subject to q m   ek j z k + z q+1 − z q+2 ≤ ai j xi , j = 1, 2, . . . , n; k=1 m  i=1 m  i=1

i=1

bil xi ≤ cl , l = 1, 2, . . . , p; m 

xi ≤ 1; −

xi ≤ −1;

i=1

xi ≥ 0, i = 1, 2, . . . , m; z k ≥ 0, k = 1, 2, . . . , q + 2.

Problem  3.2.4  p  Minimize cl tl + t p+1 − t p+2 l=1

Subject to p n   bil tl + t p+1 − t p+2 ≥ ai j y j , i = 1, 2, . . . , m; l=1 n  j=1 n  j=1

j=1

ek j y j ≥ dk , k = 1, 2, . . . , q; y j ≥ 1, −

n 

y j ≥ −1

j=1

y j ≥ 0, j = 1, 2, . . . , n; tl ≥ 0, l = 1, 2, . . . , p + 2. On the same direction, Li and Cheng [5] used the following method to obtain the mathematical formulation of constrained matrix games with fuzzy payoffs.Step 1:

68

3 Constrained Matrix Games with Fuzzy Payoffs

Replacing the parameters ai j ,bil , ek j , dk and cl with trapezoidal fuzzy numbers    L ai j (0), aiLj (1), aiRj (1), aiRj (0) , bilL (0), bilL (1), bilR (1), bilR (0) , ekLj (0), ekLj (1),      ekRj (1), ekRj (0) , dkL (0), dkL (1), dkR (1), dkR (0), and clL (0), clL (1), clR (1), clR (0) respectively, Problem 3.2.1 and Problem 3.2.2 are transformed into Problem 3.2.5 and Problem 3.2.6 respectively. Problem  3.2.5 n 

Minimize

j=1

 m

 aiLj (0), aiLj (1), aiRj (1), aiRj (0) xi



 yj

i=1

Subject to



 m

 bilL (0), bilL (1), bilR (1), bilR (0) xi  clL (0), clL (1), clR (1), clR (0) , l = 1, 2, . . . , p; i=1

 n    ekLj (0), ekLj (1), ekRj (1), ekRj (0) y j  dkL (0), dkL (1), dkR (1), dkR (0) , k = 1, 2, . . . , q; j=1 m 

m 

xi ≤ 1; −

i=1

xi ≤ −1;

i=1

n 

y j ≥ 1; −

j=1

n 

y j ≥ −1;

j=1

xi ≥ 0, i = 1, 2, . . . , m; y j ≥ 0, j = 1, 2, . . . , n.

Problem 3.2.6    m n

  aiLj (0), aiLj (1), aiRj (1), aiRj (0) y j xi Maximize i=1

j=1

Subject to Constraints of Problem 3.2.5.

n n  L    L L R R λ ai (0), ai (1), ai (1), ai (0) = λai (0), Step 2:Using the property, i=1  i=1 n n n    λaiL (1), λaiR (1), λaiR (0) , λ ≥ 0, Problem 3.2.5 and Problem 3.2.6 can i=1

i=1

i=1

be transformed into Problem 3.2.7 and Problem 3.2.8 respectively. Problem  3.2.7   n m m m m      Minimize aiLj (0)xi , aiLj (1)xi , aiRj (1)xi , aiRj (0)xi y j j=1

i=1

i=1

i=1

i=1

Subject to 

m m m m       bilL (0), bilL (1), bilR (1), bilR (0) xi  clL (0), clL (1), clR (1), clR (0) , l = 1, 2, . . . , p; i=1 i=1 i=1 i=1 n n n n       ekLj (0), ekLj (1), ekRj (1), ekRj (0) y j  dkL (0), dkL (1), dkR (1), dkR (0) , k = 1, 2, . . . , q; j=1

m  i=1

j=1 m 

xi ≤ 1; −

i=1

j=1

xi ≤ −1;

n  j=1

j=1

y j ≥ 1; −

n  j=1

y j ≥ −1;

xi ≥ 0, i = 1, 2, . . . , m; y j ≥ 0, j = 1, 2, . . . , n.

3.2 Existing Mathematical Formulation of Constrained Matrix Games with Fuzzy Payoffs

69

Problem 3.2.8   m n n n n      L L R R Maximize ai j (0)y j , ai j (1)y j , ai j (1)y j , ai j (0)y j xi i=1

j=1

j=1

j=1

j=1

Subject to Constraints of Problem 3.2.7.    Step 3: Using the relation a L (0), a L (1), a R (1), a R (0)  b L (0), b L (1), b R (1),  b R (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) ⇒ λ1 a L (0) + λ2 a L (1) + λ3 a R (1) + λ4 a R (0) ≤ λ1 b L (0) + λ2 b L (1) + λ3 b R (1) + λ4 b R (0), where λ1 + λ2 + λ3 + λ4 = 1 and λ1 , λ2 , λ3 , λ4 ≥ 0, Problem 3.2.7 and Problem 3.2.8 can be transformed into Problem 3.2.9 and Problem 3.2.10 respectively. Problem  3.2.9 Minimize

n 

λ1

j=1

i=1

Subject to λ1

m 

m 

m 

+ λ2

m  i=1

i=1

bilL (1)xi + λ3

m 

aiLj (1)xi

+ λ3

m  i=1

bilR (1)xi + λ4

aiRj (1)xi

+ λ4

m  i=1



 aiRj (0)xi

yj

m 

bilR (0)xi ≤ i=1 i=1 λ1 clL (0) + λ2 clL (1) + λ3 clR (1) + λ4 clR (0), l = 1, 2, . . . , p; n n n n     ekLj (0)y j + λ2 ekLj (1)y j + λ3 ekRj (1)y j + λ4 ek j y Rj (0)y j ≥ λ1 j=1 j=1 j=1 j=1 λ1 dkL (0) + λ2 dkL (1) + λ3 dkR (1) + λ4 dkR (0), k = 1, 2, . . . , q; m m n n     xi ≤ 1; − xi ≤ −1; y j ≥ 1; − y j ≥ −1; i=1 i=1 j=1 j=1 i=1

bilL (0)xi + λ2

aiLj (0)xi

xi ≥ 0, i = 1, 2, . . . , m; y j ≥ 0, j = 1, 2, . . . , n.

Problem 3.2.10  Maximize

m 

i=1

λ1

n  j=1

aiLj (0)y j

+ λ2

n  j=1

aiLj (1)y j

+ λ3

n  j=1

aiRj (1)y j

+ λ4

n  j=1

aiRj (0)y j

 xi

Subject to Constraints of Problem 3.2.9. Step 4: The dual of Problem 3.2.9 by considering only y j , j = 1, 2, . . . , n as decision variables and dual of Problem 3.2.10 by considering only xi , i = 1, 2, . . . , m as decision variables are Problem 3.2.11 and Problem 3.2.12 respectively. Problem 3.2.11  q  Maximize λ1 dkL (0)z k + λ2 k=1

q  k=1

dkL (1)z k + λ3

q  k=1

dkR (1)z k + λ4

q  k=1

 dkR (0)z k + z q+1 − z q+2

Subject to λ1

q q q q     ekLj (0)z k + λ2 ekLj (1)z k + λ3 ekRj (1)z k + λ4 ekRj (0)z k + z q+1 − z q+2 ≤

k=1 m 

k=1 k=1 k=1 m m m    L L R λ1 ai j (0)xi + λ2 ai j (1)xi + λ3 ai j (1)xi + λ4 aiRj (0)xi , j = 1, 2, . . . , n; i=1 i=1 i=1 i=1 m m m m     λ1 bilL (0)xi + λ2 bilL (1)xi + λ3 bilR (1)xi + λ4 bilR (0)xi ≤ i=1 i=1 i=1 i=1 λ1 clL (0) + λ2 clL (1) + λ3 clR (1) + λ4 clR (0), l = 1, 2, . . . , p;

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3 Constrained Matrix Games with Fuzzy Payoffs

m 

m 

xi ≤ 1; −

xi ≤ −1; i=1 i=1 xi ≥ 0, i = 1, 2, . . . , m; z k ≥ 0, k = 1, 2, . . . , q + 2.

Problem 3.2.12  p p p p     Minimize λ1 clL (0)tl + λ2 clL (1)tl + λ3 clR (1)tl + λ4 clR (0)tl + t p+1 − t p+2 l=1

Subject to p 

λ1

p 

l=1 n 

bilL (1)tl + λ3

l=1

p 

l=1 n 

bilR (1)tl + λ4

l=1

p 

bilR (0)tl + t p+1 − t p+2 ≥ l=1 n n   aiLj (0)y j + λ2 aiLj (1)y j + λ3 aiRj (1)y j + λ4 aiRj (0)y j , i = 1, 2, . . . , m; λ1 j=1 j=1 j=1 j=1 n n n n     λ1 ekLj (0)y j + λ2 ekLj (1)y j + λ3 ekRj (1)y j + λ4 ekRj (0)y j ≥ j=1 j=1 j=1 j=1 λ1 dkL (0) + λ2 dkL (1) + λ3 dkR (1) + λ4 dkR (0), k = 1, 2, . . . , q; n n   y j ≥ 1; − y j ≥ −1; j=1 j=1 l=1

bilL (0)tl + λ2

l=1

y j ≥ 0, j = 1, 2, . . . , n; tl ≥ 0, l = 1, 2, . . . , p + 2.

L Step 5: Using the relation, λ1 a L (0)+ λ2 a L (1) + λ3 a R (1) + λ4a R (0)  L≤ λ1 b (0) L R R L L R R + λ2 b (1) + λ3 b (1)  + λ4 b (0) ⇒ a (0), a (1), a (1), a (0)  b (0), b L (1), b R (1), b R (0) , Problem 3.2.11 and Problem 3.2.12 can be transformed into Problem 3.2.13 and Problem 3.2.14 respectively.

Problem 3.2.13  q   q q q  L    Maximize dk (0)z k , dkL (1)z k , dkR (1)z k , dkR (0)z k + z q+1 − z q+2 k=1

Subject to

k=1

k=1

k=1

 q  ekRj (1)z k , ekRj (0)z k + z q+1 − z q+2  k=1 k=1 k=1 k=1  m m m m     aiLj (0)xi , aiLj (1)xi , aiRj (1)xi , aiRj (0)xi , j = 1, 2, . . . , n; i=1 i=1  i=1 i=1

m m m m    L    bil (0)xi , bilL (1)xi , bilR (1)xi , bilR (0)xi  clL (0), clL (1), clR (1), clR (0), clL (0) , l = 1, 2, . . . , p; q 

ekLj (0)z k ,

i=1

m 

ekLj (1)z k ,

i=1

m 

xi ≤ 1; −

i=1 xi ≥

q 

q 

i=1

i=1

xi ≤ −1;

i=1

0, i = 1, 2, . . . , m; z k ≥ 0, k = 1, 2, . . . , q + 2.

Problem  3.2.14   p p p p     clL (0)tl , clL (1)tl , clR (1)tl , clR (0)tl + t p+1 − t p+2 Minimize l=1

l=1

l=1

l=1

Subject to

 p p p p     bilL (0)tl , bilL (1)tl , bilR (1)tl , bilR (0)tl + t p+1 − t p+2  l=1 l=1 l=1 l=1 n n n n     L L R R ai j (0)y j , ai j (1)y j , ai j (1)y j , ai j (0)y j , i = 1, 2, . . . , m; j=1 j=1 j=1 j=1 n n n n       ekLj (0)y j , ekLj (1)y j , ekRj (1)y j , ekRj (0)y j  dkL (0), dkL (1), dkR (1), dkR (0) , k = 1, 2, . . . , q; j=1

n  j=1

y j ≥ 1; −

j=1

n  j=1

j=1

j=1

y j ≥ −1;

y j ≥ 0, j = 1, 2, . . . , n; tl ≥ 0, l = 1, 2, . . . , p + 2.

3.3 Literature Review of Constrained Matrix Games with Fuzzy Payoffs

71

3.3 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, is presented. Li and Cheng [5] solved the Problem 3.2.11 and Problem 3.2.12 to find minimum expected gain of Player I and maximum expected loss of Player II respectively. Li and Hong [6, 7] split the Problem 3.2.11 into Problems 3.3.1, 3.3.2, 3.3.3 and 3.3.4 and obtained the minimum expected gain of Player I by using the optimal values of these problems as well as split the Problem 3.2.12 into Problems 3.3.5, 3.3.6, 3.3.7 and 3.3.8 and obtained the maximum expected loss of Player I by using the optimal values of these problems. Problem 3.3.1   q  L dk (0)z k + z q+1 − z q+2 Maximize k=1

Subject to q m   ekLj (0)z k + z q+1 − z q+2 ≤ aiLj (0)xi , j = 1, 2, . . . , n;

k=1 m  i=1 m 

i=1

bilL (0)xi

≤

m 

xi ≤ 1; −

i=1

clL (0), l

= 1, 2, . . . , p;

xi ≤ −1;

i=1

xi ≥ 0, i = 1, 2, . . . , m; z k ≥ 0, k = 1, 2, . . . , q + 2.

Problem 3.3.2   q  L Maximize dk (1)z k + z q+1 − z q+2 k=1

Subject to q m   ekLj (1)z k + z q+1 − z q+2 ≤ aiLj (1)xi , j = 1, 2, . . . , n;

k=1 m  i=1 m  i=1

i=1

bilL (1)xi

≤

xi ≤ 1; −

clL (1), l m 

i=1

= 1, 2, . . . , p;

xi ≤ −1;

xi ≥ 0, i = 1, 2, . . . , m; z k ≥ 0, k = 1, 2, . . . , q + 2.

72

3 Constrained Matrix Games with Fuzzy Payoffs

Problem 3.3.3   q  R dk (1)z k + z q+1 − z q+2 Maximize k=1

Subject to q m   ekRj (1)z k + z q+1 − z q+2 ≤ aiRj (1)xi , j = 1, 2, . . . , n;

k=1 m  i=1 m 

i=1

bilR (1)xi

≤

xi ≤ 1; −

i=1

clR (1), l

m 

= 1, 2, . . . , p;

xi ≤ −1;

i=1

xi ≥ 0, i = 1, 2, . . . , m; z k ≥ 0, k = 1, 2, . . . , q + 2.

Problem 3.3.4  q   R Maximize dk (0)z k + z q+1 − z q+2 k=1

Subject to q m   ekRj (0)z k + z q+1 − z q+2 ≤ aiRj (0)xi , j = 1, 2, . . . , n;

k=1 m  i=1 m 

i=1

bilR (0)xi

≤

xi ≤ 1; −

i=1

clR (0), l

m 

= 1, 2, . . . , p;

xi ≤ −1;

i=1

xi ≥ 0, i = 1, 2, . . . , m; z k ≥ 0, k = 1, 2, . . . , q + 2.

Problem  3.3.5  p  L Minimize cl (0)tl + t p+1 − t p+2 l=1

Subject to p n   bilL (0)tl + t p+1 − t p+2 ≥ aiLj (0)y j , i = 1, 2, . . . , m;

l=1 n 

j=1 n  j=1

j=1

ekLj (0)y j

≥

y j ≥ 1; −

dkL (0), k

n  j=1

= 1, 2, . . . , q;

y j ≥ −1;

y j ≥ 0, j = 1, 2, . . . , n; tl ≥ 0, l = 1, 2, . . . , p + 2.

3.3 Literature Review of Constrained Matrix Games with Fuzzy Payoffs

Problem  3.3.6  p  clL (1)tl + t p+1 − t p+2 Minimize l=1

Subject to p n   bilL (1)tl + t p+1 − t p+2 ≥ aiLj (1)y j , i = 1, 2, . . . , m;

l=1 n 

j=1 n 

j=1

ekLj (1)y j ≥ dkL (1), k = 1, 2, . . . , q; y j ≥ 1; −

j=1

n 

y j ≥ −1;

j=1

y j ≥ 0, j = 1, 2, . . . , n; tl ≥ 0, l = 1, 2, . . . , p + 2. Problem  3.3.7  p  R Minimize cl (1)tl + t p+1 − t p+2 l=1

Subject to p n   bilR (1)tl + t p+1 − tl+2 ≥ aiRj (1)y j , i = 1, 2, . . . , m;

l=1 n 

j=1 n 

j=1

ekRj (1)y j

≥

y j ≥ 1; −

j=1

dkR (1), k

n 

= 1, 2, . . . , q;

y j ≥ −1;

j=1

y j ≥ 0, j = 1, 2, . . . , n; tl ≥ 0, l = 1, 2, . . . , p + 2. Problem  3.3.8  p  R Minimize cl (0)tl + t p+1 − t p+2 l=1

Subject to p n   bilR (0)tl + t p+1 − t p+2 ≥ aiRj (0)y j , i = 1, 2, . . . , m;

l=1 n 

j=1 n  j=1

j=1

ekRj (0)y j

≥

y j ≥ 1; −

dkR (0), k

n  j=1

= 1, 2, . . . , q;

y j ≥ −1;

y j ≥ 0, j = 1, 2, . . . , n; tl ≥ 0, l = 1, 2, . . . , p + 2.

73

74

3 Constrained Matrix Games with Fuzzy Payoffs

3.4 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 λ1 a L (0) + λ2 a L (1) + λ3 a R (1) + λ4 a R (0) ≥ λ1 b L (0) + λ2 b L (1) + λ3 b R (1) + λ4 b 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 1 (0) = 4, b L (0) = 0, b L (1) = 3, b R (1) = 5, b R (0) = 10, λ1 = λ2 = λ3 = λ4 = 4 10 and λ1 b L (0) + λ2 b L (1) + λ3 then λ1 a L (0) + λ2 a L (1) + λ3 a R (1) + λ4 a R (0) = 4 18 ⇒ λ1 b L (0) + λ2 b L (1) + λ3 b R (1) + λ4 b R (0) > λ1 a L (0)+ b R (1) + λ4 b R (0) = 4 λ2 a L (1) + λ3 a R (1) + λ4 a 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, 1 b R (1) = 4, b R (0) = 6, λ1 = λ2 = λ3 = λ4 = then λ1 a L (0) + λ2 a L (1) + λ3 a R 4 11 while b L (0) (1) + λ4 a R (0) = λ1 b L (0) + λ2 b L (1) + λ3 b R (1) + λ4 b R (0) = 4 < a L (0), b L (1) < a L (1), b R (1) > a R (1), b R (0) > a R (0). However, Li and Hong [6, 7] have used the mathematically incorrect assumption λ1 a L (0) + λ2 a L (1) + λ3 a R (1) + λ4 a R (0) ≥ λ1 b L (0) + λ2 b L (1) + λ3 b R (1) + λ4 b 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 3.2.11 and Problem 3.2.12 into Problem 3.3.1, Problem 3.3.2, Problem 3.3.3, Problem 3.3.4 and Problem 3.3.5, Problem 3.3.6, Problem 3.3.7, Problem 3.3.8 respectively. Therefore, the existing method [5– 7], in which the minimum expected gain of Player I obtained by splitting 3.2.11 into Problem 3.3.1, Problem 3.3.2, Problem 3.3.3, Problem 3.3.4 as well as maximum expected loss of Player II obtained by splitting Problem 3.3.12 into Problem 3.3.5, Problem 3.3.6, Problem 3.3.7, Problem 3.3.8 respectively, is not valid.

3.5 Proposed Vaishnavi Method   In the existing methods [6, 7], it is assumed that if a L (0), a L (1), a R (1), a R (0) L b L (1), b R (1), bR (0) are two trapezoidal fuzzy numbers then andL b (0), L R L L a (0), a (1), a R (1), a R (0)  b L (0), b L (1), b R (1),  Lb (0) L if aR (0) ≥R b (0),  L L R R R R a (1) ≥ b (1), a (1) ≥ b (1), a (0) ≥ b (0) and a (0), a (1), a (1), a (0)  b L (0), b L (1), b R (1), b R (0) 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). In this section, a new method (named as Vaishnavi method), on the basis of this comparing method, is proposed to find the minimum expected gain of Player I, maximum expected loss of Player II and their corresponding optimal strategies.

3.5 Proposed Vaishnavi Method

75

3.5.1 Minimum Expected Gain of Player I    Using the comparing method, a L (0), a L (1), a R (1), a R (0)  b L (0), b L (1), b R (1), b R (0) 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 corresponding optimal strategies can be obtained as follows:

m n   L ∗ ∗ ∗ (ai j (0), aiLj (1), Step 1: Find (y1 , y2 , . . . , yn ) ∈ Y such that value of j=1 i=1  aiRj (1), aiRj (0))xi y j is minimum for all (x1 , x2 , . . . , xm ) ∈ X i.e., find the solution {y ∗j , j = 1, 2, . . . , n} of Problem 3.2.5.

n n   L λ(aiL (0), aiL (1), aiR (1), aiR (0)) = λai (0), Step 2: Using the property, i=1

n  i=1

λaiL (1),

n  i=1

λaiR (1),

n  i=1

i=1



λaiR (0) , λ ≥ 0, the Problem 3.2.5 can be transformed

into Problem 3.2.7.    Step 3: Usingthe comparing method, a L (0), a L (1), a R (1), a R (0)  b L (0), b L (1), b R (1), b R (0) 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 Problem 3.2.7 can be transformed into Problem 3.5.1. Problem  3.5.1 Minimize

n  n  n  n  m m m m     aiLj (0)xi y j , aiLj (1)xi y j , aiRj (1)xi y j , aiRj (0)xi y j

j=1 i=1

j=1 i=1

j=1 i=1

Subject to m  bilL (0)xi ≤ clL (0), l = 1, 2, . . . , p;

i=1 m 

i=1 n  j=1 n  j=1 m 

m  i=1 m 

bilR (1)xi ≤ clR (1), l = 1, 2, . . . , p;

i=1 n 

ekLj (0)y j ≥ dkL (0), k = 1, 2, . . . , q;

j=1 n 

ekRj (1)y j ≥ dkR (1), k = 1, 2, . . . , q; xi ≤ 1;

i=1

−

m 

n 

xi ≤ −1;

i=1

j=1



j=1 i=1

bilL (1)xi ≤ clL (1), l = 1, 2, . . . , p; bilR (0)xi ≤ clR (0), l = 1, 2, . . . , p;

ekLj (1)y j ≥ dkL (1), k = 1, 2, . . . , q; ekRj (0)y j ≥ dkR (0), k = 1, 2, . . . , q;

y j ≥ 1;

−

j=1

xi ≥ 0, i = 1, 2, . . . , m; y j ≥ 0, j = 1, 2, . . . , n.

n 

y j ≥ −1;

j=1

Step 4: According to comparing method, to find the optimal solution {y1∗ , y2∗ , . . . , yn∗ } of Problem 3.5.1 such that value of

n  m 

j=1 i=1

(aiLj (0)xi y j , aiLj (1)xi y j , aiRj (1)xi y j , aiRj (0)xi y j )

is minimum for all (x1 , x2 , . . . , xm ) ∈ X is equivalent to find {y1∗ , y2∗ , . . . , yn∗ } such that value of

n  m 

j=1 i=1

aiLj (0)xi y j ,

n  m 

j=1 i=1

aiLj (1)xi y j ,

n  m 

j=1 i=1

aiRj (1)xi y j and

n  m 

j=1 i=1

aiRj (0)xi y j

76

3 Constrained Matrix Games with Fuzzy Payoffs

is minimum for all (x1 , x2 , . . . , xm ) ∈ X or if it is not possible to find {y1∗ , y2∗ , . . . , yn∗ } n  n  n  m m m    for which the value of aiLj (0)xi y j , aiLj (1)xi y j , aiRj (1)xi y j and n  m  j=1 i=1

j=1 i=1

aiRj (0)xi y j

value of and

n  m 

j=1 i=1

is minimum then find such

{y ∗j1 ,

aiLj (0)xi1 y j1 is minimum but value of

j=1 i=1 n  m  aiRj (0)xi1 y j1 j=1 i=1

j=1 i=1

j = 1, 2, . . . , n} for which the n  m  j=1 i=1

aiLj (1)xi1 y j1 ,

n  m  j=1 i=1

aiRj (1)xi1 y j1

is not minimum for all (x11 , x21 , . . . , xm1 ) ∈ X and find such n  m 

{y ∗j2 , j = 1, 2, . . . , n} for which the value of

aiLj (1)xi2 y j2 is minimum but j=1 i=1 n  n  n  m m m    aiLj (0)xi2 y j2 , aiRj (1)xi2 y j2 and aiRj (0)xi2 y j2 is not minivalue of j=1 i=1 j=1 i=1 j=1 i=1 mum for all (x12 , x22 , . . . , xm2 ) ∈ X and to find such {y ∗j3 , j = 1, 2, . . . , n} for which n  m  m m n  n    aiRj (1)xi3 y j3 is minimum but value of aiLj (0)xi3 y j3 , aiLj (1)xi3 y j3 the value of j=1 i=1 j=1 i=1 j=1 i=1 m n   and aiRj (0)xi3 y j3 is not minimum for all (x 13 , x 23 , . . . , x m3 ) ∈ X and find such j=1 i=1

n  m 

{y ∗j4 , j = 1, 2, . . . , n} for which the value of

aiRj (0)xi4 y j4 is minimum but j=1 i=1 n  n  n  m m m    aiLj (0)xi4 y j4 , aiLj (1)xi4 y j4 and aiRj (1)xi4 y j4 are not minvalue of j=1 i=1 j=1 i=1 j=1 i=1 imum for all (x14 , x24 , . . . , xm4 ) ∈ X i.e., to find optimal solution {y ∗j1 , j = 1, 2, . . . , n}, {y ∗j2 , j = 1, 2, . . . , n}, {y ∗j3 , j = 1, 2, . . . , n} and {y ∗j4 , j = 1, 2, . . . , n} of Problem 3.5.2, Problem 3.5.3, Problem 3.5.4 and Problem 3.5.5 respectively. Problem  3.5.2  n  m  L Minimize ai j (0)xi1 y j1 j=1 i=1

Subject to m  bilL (0)xi1 ≤ clL (0), l = 1, 2, . . . , p;

i=1 m 

i=1 n  j=1 n  j=1 m  i=1

bilR (1)xi1 ≤ clR (1), l = 1, 2, . . . , p; ekLj (0)y j1

≥

dkL (0), k

= 1, 2, . . . , q;

ekRj (1)y j1 ≥ dkR (1), k = 1, 2, . . . , q; xi1 ≤ 1;

−

m  i=1

xi1 ≤ −1;

n  j=1

m 

bilL (1)xi1 ≤ clL (1), l = 1, 2, . . . , p;

i=1 m  i=1

n 

j=1 n  j=1

bilR (0)xi1 ≤ clR (0), l = 1, 2, . . . , p;

ekLj (1)y j1 ≥ dkL (1), k = 1, 2, . . . , q;

ekRj (0)y j1 ≥ dkR (0), k = 1, 2, . . . , q;

y j1 ≥ 1;

xi1 ≥ 0, i = 1, 2, . . . , m; y j1 ≥ 0, j = 1, 2, . . . , n.

−

n  j=1

y j1 ≥ −1;

3.5 Proposed Vaishnavi Method

77

Problem  3.5.3  n  m  L Minimize ai j (1)xi2 y j2 j=1 i=1

Subject to m  bilL (0)xi2 ≤ clL (0), l = 1, 2, . . . , p;

i=1 m 

i=1 n  j=1 n  j=1 m 

m 

bilR (1)xi2 ≤ clR (1), l = 1, 2, . . . , p; ekLj (0)y j2

≥

dkL (0), k

i=1

n 

= 1, 2, . . . , q;

j=1 n 

ekRj (1)y j2 ≥ dkR (1), k = 1, 2, . . . , q; xi2 ≤ 1;

−

i=1

m 

xi2 ≤ −1;

i=1

n 

bilL (1)xi2 ≤ clL (1), l = 1, 2, . . . , p;

i=1 m 

bilR (0)xi2 ≤ clR (0), l = 1, 2, . . . , p;

ekLj (1)y j2 ≥ dkL (1), k = 1, 2, . . . , q; ekRj (0)y j2 ≥ dkR (0), k = 1, 2, . . . , q;

j=1

y j2 ≥ 1;

−

j=1

n 

y j2 ≥ −1;

j=1

xi2 ≥ 0, i = 1, 2, . . . , m; y j2 ≥ 0, j = 1, 2, . . . , n.

Problem  3.5.4  n  m  R Minimize ai j (1)xi3 y j3 j=1 i=1

Subject to m  bilL (0)xi3 ≤ clL (0), l = 1, 2, . . . , p;

i=1 m 

i=1 n  j=1 n  j=1 m 

bilR (1)xi3 ≤ clR (1), l = 1, 2, . . . , p; ekLj (0)y j3 ≥ dkL (0), k = 1, 2, . . . , q; ekRj (1)y j3 ≥ dkR (1), k = 1, 2, . . . , q; xi3 ≤ 1;

−

i=1

m 

xi3 ≤ −1;

i=1

n 

m 

bilL (1)xi3 ≤ clL (1), l = 1, 2, . . . , p;

i=1 m  i=1 n  j=1 n 

bilR (0)xi3 ≤ clR (0), l = 1, 2, . . . , p;

ekLj (1)y j3 ≥ dkL (1), k = 1, 2, . . . , q;

j=1

ekRj (0)y j3 ≥ dkR (0), k = 1, 2, . . . , q;

y j3 ≥ 1;

j=1

xi3 ≥ 0, i = 1, 2, . . . , m; y j3 ≥ 0, j = 1, 2, . . . , n.

−

n 

y j3 ≥ −1;

j=1

Problem  3.5.5  n  m  R Minimize ai j (0)xi4 y j4 j=1 i=1

Subject to m  bilL (0)xi4 ≤ clL (0), l = 1, 2, . . . , p;

i=1 m  i=1

bilR (1)xi4 ≤ clR (1), l = 1, 2, . . . , p;

m  i=1 m  i=1

bilL (1)xi4 ≤ clL (1), l = 1, 2, . . . , p; bilR (0)xi4 ≤ clR (0), l = 1, 2, . . . , p;

78 n  j=1 n  j=1 m 

3 Constrained Matrix Games with Fuzzy Payoffs

ekLj (0)y j4 ≥ dkL (0), k = 1, 2, . . . , q; ekRj (1)y j4 ≥ dkR (1), k = 1, 2, . . . , q; xi4 ≤ 1;

−

i=1

m 

xi4 ≤ −1;

i=1

n 

n  j=1 n  j=1

ekLj (1)y j4 ≥ dkL (1), k = 1, 2, . . . , q;

ekRj (0)y j4 ≥ dkR (0), k = 1, 2, . . . , q;

y j4 ≥ 1;

j=1

xi4 ≥ 0, i = 1, 2, . . . , m; y j4 ≥ 0, j = 1, 2, . . . , n.

−

n 

y j4 ≥ −1;

j=1

Step 5: Since, in Problem 3.5.2, Problem 3.5.3, Problem 3.5.4 and Problem 3.5.5 only y j1 , y j2 , y j3 and y j4 respectively have been considered as decision variables. So, these problems are linear programming problems and hence, the optimal value of Problem 3.5.2, Problem 3.5.3, Problem 3.5.4 and Problem 3.5.5 will be equal to optimal value of its corresponding dual problem i.e., Problem 3.5.6, Problem 3.5.7, Problem 3.5.8 and Problem 3.5.9 respectively. Problem 3.5.6   q  L Maximize dk (0)z k1 + z (q+1)1 − z (q+2)1 k=1

Subject to q m   ekLj (0)z k1 + z (q+1)1 − z (q+2)2 ≤ aiLj (0)xi1 , j = 1, 2, . . . , n;

k=1 m  i=1 m 

i=1

bilL (0)xi1

≤

xi1 ≤ 1; −

i=1

clL (0), l

m 

= 1, 2, . . . , p;

xi1 ≤ −1;

i=1

xi1 ≥ 0, i = 1, 2, . . . , m; z k1 ≥ 0, k = 1, 2, . . . , q + 2.

Problem 3.5.7   q  L Maximize dk (1)z k2 + z (q+1)2 − z (q+2)2 k=1

Subject to q m   ekLj (1)z k2 + z (q+1)2 − z (q+2)2 ≤ aiLj (1)xi2 , j = 1, 2, . . . , n;

k=1 m  i=1 m 

i=1

bilL (1)xi2

≤

xi2 ≤ 1; −

i=1

clL (1), l m 

= 1, 2, . . . , p;

xi2 ≤ −1;

i=1

xi2 ≥ 0, i = 1, 2, . . . , m; z k2 ≥ 0, k = 1, 2, . . . , q + 2.

Problem 3.5.8  q   R Maximize dk (1)z k3 + z (q+1)3 − z (q+2)3 k=1

Subject to

3.5 Proposed Vaishnavi Method q  k=1 m  i=1 m 

ekRj (1)z k3 + z (q+1)3 − z (q+2)3 ≤ bilR (1)xi3

≤

xi3 ≤ 1; −

i=1

clR (1), l

m 

79 m  i=1

aiRj (1)xi3 , j = 1, 2, . . . , n;

= 1, 2, . . . , p;

xi3 ≤ −1;

i=1

xi3 ≥ 0, i = 1, 2, . . . , m; z k3 ≥ 0, k = 1, 2, . . . , q + 2.

Problem 3.5.9   q  R Maximize dk (0)z k4 + z (q+1)4 − z (q+2)4 k=1

Subject to q m   ekRj (0)z k4 + z (q+1)4 − z (q+2)4 ≤ aiRj (0)xi4 , j = 1, 2, . . . , n;

k=1 m  i=1 m 

i=1

bilR (0)xi4

≤

xi4 ≤ 1; −

i=1

clR (0), l

m 

= 1, 2, . . . , p;

xi4 ≤ −1;

i=1

xi4 ≥ 0, i = 1, 2, . . . , m; z k4 ≥ 0, k = 1, 2, . . . , q + 2.

∗ , i = 1, 2, . . . , m}, {x ∗ , i = 1, 2, . . . , m}, {x ∗ , i = Step 5: Substitute the value of {xi1 i2 i3 ∗ 1, 2, . . . , m} and {xi4 , i = 1, 2, . . . , m} of Problem 3.5.6, Problem 3.5.7, Problem 3.5.8 and Problem 3.5.9 in Problem 3.5.2, Problem 3.5.3, Problem 3.5.4 and Problem g∗ 3.5.5 respectively and find all the alternative basic optimal solutions {y j1 , j = ∗ ∗ 1, 2, . . . , n, g = 1, 2, . . . , u}, {y rj2 , j = 1, 2, . . . , n, r = 1, 2, . . . , f }, {y sj3 , j = ∗ 1, 2, . . . , n, s = 1, 2, . . . , h} and {y tj4 , j = 1, 2, . . . , n, t = 1, 2, . . . , w} of Problem 3.5.2, Problem 3.5.3, Problem 3.5.4 and Problem 3.5.5 respectively.

Step 6: Find ⎧ n m ∗ ⎪  L ∗ g ⎪ (ai j (0), aiLj (1), aiRj (1), aiRj (0))xi1 y j1 ; g = 1, 2, . . . , u, ⎪ ⎪ ⎪ j=1 i=1 ⎪ ⎪ n  m ⎪  ⎪ ∗ r∗ ⎪ ⎪ (aiLj (0), aiLj (1), aiRj (1), aiRj (0))xi2 y j2 ; r = 1, 2, . . . , f, ⎨ j=1 i=1 minimum  n  m ⎪ ∗ s∗ ⎪ (aiLj (0), aiLj (1), aiRj (1), aiRj (0))xi3 y j3 ; s = 1, 2, . . . , h, ⎪ ⎪ ⎪ j=1 i=1 ⎪ ⎪ n  m ⎪ ⎪ ∗ t∗ ⎪ ⎪ (aiLj (0), aiLj (1), aiRj (1), aiRj (0))xi4 y j4 ; q = 1, 2, . . . , w ⎩ j=1 i=1

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

Step 7: All the minimum trapezoidal fuzzy numbers will represent minimum ∗ expected gain of Player I.The optimal strategies for Player I will be {xi1 ,i = 1, 2, . . . , m}if minimum corresponds to 

n  m  j=1 i=1

∗ ∗ yg ; g (aiLj (0), aiLj (1), aiRj (1), aiRj (0))xi1 j1

 ∗ , i = 1, 2, . . . , m} = 1, 2, . . . , u , will be {xi2

80

3 Constrained Matrix Games with Fuzzy Payoffs

if minimum corresponds to 



n  m 

∗

j=1 i=1

∗ y r ; r = 1, 2, . . . , f , will be {x ∗ , i = 1, 2, . . . , m} (aiLj (0), aiLj (1), aiRj (1), aiRj (0))xi2 j2 i3

if minimum corresponds to 

m n   j=1 i=1

 ∗

∗ y s ; s = 1, 2, . . . , h and will be {x ∗ , i = 1, 2, . . . , m} (aiLj (0), aiLj (1), aiRj (1), aiRj (0))xi3 i4 j3

if minimum corresponds to 

n  m  j=1 i=1

 ∗

∗ y t ; t = 1, 2, . . . , w , which is optimal solution (aiLj (0), aiLj (1), aiRj (1), aiRj (0))xi4 j4

of Problem 3.5.6, Problem 3.5.7, Problem 3.5.8 and Problem 3.5.9 respectively.

3.5.2 Maximum Expected Loss of Player II 







Using the comparing method, a L (0), a L (1), a R (1), a R (0)  b L (0), b L (1), b R (1), b R (0) 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 maximum expected loss of Player II and corresponding optimal strategies can obtained as follows:

be n m   ∗ ∗ ∗ (aiLj (0), aiLj (1), aiRj (1), Step 1: Find (x1 , x2 , . . . , xm ) ∈ X such that value of j=1 i=1  aiRj (0))xi y j is maximum for all (y1 , y2 , . . . , yn ) ∈ Y i.e., find the solution {xi∗ , i = 1, 2, . . . , m} of Problem 3.2.6.

n n n   L  Step 2: Using the property, λ(aiL (0), aiL (1), aiR (1), aiR (0)) = λai (0), i=1 i=1 i=1  n n   λaiL (1), λaiR (1), λaiR (0) , λ ≥ 0, the Problem 3.2.6 can be transformed into i=1

i=1

Problem 3.2.8.     Step 3: Using the comparing method, a L (0), a L (1), a R (1), a R (0)  b L (0), b L (1), b R (1), b R (0) 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 Problem 3.2.8 can be transformed into Problem 3.5.10. Problem 3.5.10  Maximize

m n  

j=1 i=1

aiLj (0)xi y j ,

m n   j=1 i=1

aiLj (1)xi y j ,

Subject to m  bilL (0)xi ≤ clL (0), l = 1, 2, . . . , p;

i=1 m  i=1

bilR (1)x1 ≤ clR (1), l = 1, 2, . . . , p;

m n   j=1 i=1

m  i=1 m  i=1

aiRj (1)xi y j ,

m n   j=1 i=1

 aiRj (0)xi y j

bilL (1)xi ≤ clL (1), l = 1, 2, . . . , p; bilR (0)xi ≤ clR (0), l = 1, 2, . . . , p;

3.5 Proposed Vaishnavi Method n  j=1 n  j=1 n 

81 n 

ekLj (0)y j ≥ dkL (0), k = 1, 2, . . . , q;

j=1 n 

ekRj (1)y j ≥ dkR (1), k = 1, 2, . . . , q; y j ≤ 1;

−

j=1

n 

n 

y j ≤ −1;

j=1

j=1

ekLj (1)y j ≥ dkL (1), k = 1, 2, . . . , q; ekRj (0)y j ≥ dkR (0), k = 1, 2, . . . , q;

y j ≥ 1;

−

j=1

n 

y j ≥ −1;

j=1

xi ≥ 0, i = 1, 2, . . . , m; y j ≥ 0, j = 1, 2, . . . , n.

Step 4: According to comparing method, to find the optimal solution {x1∗ , x2∗ , . . . , xm∗ } n  m  of Problem 3.5.10 such that value of (aiLj (0)xi y j , aiLj (1)xi y j , aiRj (1)xi y j , aiRj j=1 i=1

(0)xi y j ) is maximum for all (y1 , y2 , . . . , yn ) ∈ Y is equivalent to find {x1∗ , x2∗ , . . . , xm∗ } n  n  n  m m m    aiLj (0)xi y j , aiLj (1)xi y j , aiRj (1)xi y j and such that value of n  m  j=1 i=1

j=1 i=1

aiRj (0)xi y j

j=1 i=1

is maximum for all (y1 , y2 , . . . , yn ) ∈ Y or if it is not possible

to find {x1∗ , x2∗ , . . . , xm∗ } for which the value of n  m  j=1 i=1

aiRj (1)xi y j and

n  m  j=1 i=1

. . . , m} for which the value of (1)xi1 y j1 ,

n  m  j=1 i=1

j=1 i=1

n  m  j=1 i=1

aiLj (0)xi y j ,

n  m  j=1 i=1

aiLj (1)xi y j ,

∗ aiRj (0)xi y j is maximum then find such {xi1 , i = 1, 2, n  m 

aiLj (0)xi1 y j1 is maximum but value of

j=1 i=1 n  m 

aiRj (1)xi1 y j1 and

j=1 i=1 ∗ {xi2 ,i

n  m  j=1 i=1

aiLj

aiRj (0)xi1 y j1 is not maximum for all (y11 ,

= 1, 2, . . . , m} for which the value of y21 , . . . , yn1 ) ∈ Y and find such n  n  n  m m m    L ai j (1)xi2 y j2 is maximum but value of aiLj (0)xi2 y j2 , aiRj (1)xi2 y j2 j=1 i=1 n  m 

and

j=1 i=1

j=1 i=1

j=1 i=1

aiRj (0)xi2 y j2 is not maximum for all (y12 , y22 , . . . , yn2 ) ∈ X and to find

∗ , i = 1, 2, . . . , m} for which the value of such {xi3 n  m 

n  m 

j=1 i=1

j=1 i=1

j=1 i=1

aiLj (1)xi3 y j3 and

aiRj (1)xi3 y j3 is maximum n  m 

aiRj (0)xi3 y j3 is not j=1 i=1 ∗ , i = 1, 2, . . . , m} for maximum for all (y13 , y23 , . . . , yn3 ) ∈ X and find such {xi4 n  n  m m   R ai j (0)xi4 y j4 is maximum but value of aiLj (0)xi4 y j4 , which the value of j=1 i=1 j=1 i=1 n  n  m m   aiLj (1)xi4 y j4 and aiRj (1)xi4 y j4 is not maximum for all (x14 , x24 , j=1 i=1 j=1 i=1 ∗ ∗ , i = 1, 2, . . . , m}, {xi2 , i = 1, 2, . . . , xm4 ) ∈ X i.e., to find optimal solution {xi1 ∗ ∗ , i = 1, 2, . . . , m} and {xi4 , i = 1, 2, . . . , m} of Problem 3.5.11, Prob. . . , m}, {xi3 but value of

aiLj (0)xi3 y j3 ,

n  m 

lem 3.5.12, Problem 3.5.13, Problem 3.5.14 respectively.

82

3 Constrained Matrix Games with Fuzzy Payoffs

Problem 3.5.11   n  m  L Maximize ai j (0)xi1 y j1 j=1 i=1

Subject to m  bilL (0)xi1 ≤ clL (0), l = 1, 2, . . . , p;

i=1 m 

i=1 n  j=1 n  j=1 n 

m 

bilR (1)xi1 ≤ clR (1), l = 1, 2, . . . , p; ekLj (0)y j1

≥

dkL (0), k

i=1

n 

= 1, 2, . . . , q;

j=1 n 

ekRj (1)y j1 ≥ dkR (1), k = 1, 2, . . . , q; y j1 ≤ 1;

−

j=1

n 

y j1 ≤ −1;

j=1

bilL (1)xi1 ≤ clL (1), l = 1, 2, . . . , p;

i=1 m 

n 

bilR (0)xi1 ≤ clR (0), l = 1, 2, . . . , p;

ekLj (1)y j1 ≥ dkL (1), k = 1, 2, . . . , q; ekRj (0)y j1 ≥ dkR (0), k = 1, 2, . . . , q;

j=1

y j1 ≥ 1;

−

j=1

n 

y j1 ≥ −1;

j=1

xi1 ≥ 0, i = 1, 2, . . . , m; y j1 ≥ 0, j = 1, 2, . . . , n. Problem 3.5.12   n  m  L ai j (1)xi2 y j2 Maximize j=1 i=1

Subject to m  bilL (0)xi2 ≤ clL (0), l = 1, 2, . . . , p;

i=1 m 

i=1 n  j=1 n  j=1 n 

m 

bilR (1)xi2 ≤ clR (1), l = 1, 2, . . . , p;

i=1 n 

ekLj (0)y j2 ≥ dkL (0), k = 1, 2, . . . , q;

j=1 n 

ekRj (1)y j2 ≥ dkR (1), k = 1, 2, . . . , q; y j2 ≤ 1;

−

j=1

n 

yi2 ≤ −1;

j=1

bilL (1)xi2 ≤ clL (1), l = 1, 2, . . . , p;

i=1 m 

n 

bilR (0)xi2 ≤ clR (0), l = 1, 2, . . . , p;

ekLj (1)y j2 ≥ dkL (1), k = 1, 2, . . . , q; ekRj (0)y j2 ≥ dkR (0), k = 1, 2, . . . , q;

j=1

y j2 ≥ 1;

j=1

xi2 ≥ 0, i = 1, 2, . . . , m; y j2 ≥ 0, j = 1, 2, . . . , n.

−

n 

y j2 ≥ −1;

j=1

Problem 3.5.13   n  m  aiRj (1)xi3 y j3 Maximize j=1 i=1

Subject to m  bilL (0)xi3 ≤ clL (0), l = 1, 2, . . . , p;

i=1 m 

i=1 n  j=1 n  j=1

bilR (1)xi3 ≤ clR (1), l = 1, 2, . . . , p; ekLj (0)y j3 ≥ dkL (0), k = 1, 2, . . . , q; ekRj (1)y j3 ≥ dkR (1), k = 1, 2, . . . , q;

m 

bilL (1)xi3 ≤ clL (1), l = 1, 2, . . . , p;

i=1 m  i=1 n  j=1 n  j=1

bilR (0)xi3 ≤ clR (0), l = 1, 2, . . . , p;

ekLj (1)y j3 ≥ dkL (1), k = 1, 2, . . . , q; ekRj (0)y j3 ≥ dkR (0), k = 1, 2, . . . , q;

3.5 Proposed Vaishnavi Method n 

y j3 ≤ 1;

−

j=1

n 

83 n 

y j3 ≤ −1;

j=1

y j3 ≥ 1;

−

j=1

n 

y j3 ≥ −1;

j=1

xi3 ≥ 0, i = 1, 2, . . . , m; y j3 ≥ 0, j = 1, 2, . . . , n. Problem 3.5.14   n  m  R Maximize ai j (0)xi4 y j4 j=1 i=1

Subject to m  bilL (0)xi4 ≤ clL (0), l = 1, 2, . . . , p;

i=1 m 

i=1 n  j=1 n  j=1 n 

m 

bilR (1)xi4 ≤ clR (1), l = 1, 2, . . . , p; ekLj (0)y j4

≥

dkL (0), k

i=1

n 

= 1, 2, . . . , q;

j=1 n 

ekRj (1)y j4 ≥ dkR (1), k = 1, 2, . . . , q; y j4 ≤ 1;

−

j=1

n 

n 

y j4 ≤ −1;

j=1

bilL (1)xi4 ≤ clL (1), l = 1, 2, . . . , p;

i=1 m 

j=1

j=1

bilR (0)xi4 ≤ clR (0), l = 1, 2, . . . , p;

ekLj (1)y j4 ≥ dkL (1), k = 1, 2, . . . , q; ekRj (0)y j4 ≥ dkR (0), k = 1, 2, . . . , q;

y j4 ≥ 1;

−

n 

y j4 ≥ −1;

j=1

xi4 ≥ 0, i = 1, 2, . . . , m; y j4 ≥ 0, j = 1, 2, . . . , n. Step 5: Since, in Problem 3.5.11, Problem 3.5.12, Problem 3.5.13 and Problem 3.5.14 only xi1 , xi2 , xi3 and xi4 respectively have been considered as decision variables. So, these are linear programming problems and hence, the optimal value of Problem 3.5.11, Problem 3.5.12, Problem 3.5.13 and Problem 3.5.14 will be equal to optimal value of its corresponding dual problem i.e., Problem 3.5.15, Problem 3.5.16, Problem 3.5.17 and Problem 3.5.18 respectively. Problem  3.5.15  p  L Minimize cl (0)tl1 + t( p+1)1 − t( p+2)1 l=1

Subject to p n   bilL (0)tl1 + t( p+1)1 − t( p+2)1 ≥ aiLj (0)y j1 , i = 1, 2, . . . , m;

l=1 n 

j=1 n  j=1

j=1

ekLj (0)y j1

≥

y j1 ≥ 1; −

dkL (0), k

n  j=1

= 1, 2, . . . , q;

y j1 ≥ −1;

y j1 ≥ 0, j = 1, 2, . . . , n; tl1 ≥ 0, l = 1, 2, . . . , p + 2.

84

3 Constrained Matrix Games with Fuzzy Payoffs

Problem  3.5.16  p  clL (1)tl2 + t( p+1)2 − t( p+2)2 Minimize l=1

Subject to p n   bilL (1)tl2 + t( p+1)2 − t( p+2)2 ≥ aiLj (1)y j2 , i = 1, 2, . . . , m;

l=1 n 

j=1 n 

j=1

ekLj (1)y j2 ≥ dkL (1), k = 1, 2, . . . , q; y j2 ≥ 1; −

j=1

n 

y j2 ≥ −1;

j=1

y j2 ≥ 0, j = 1, 2, . . . , n; tl2 ≥ 0, l = 1, 2, . . . , p + 2. Problem  3.5.17  p  R Minimize cl (1)tl3 + t( p+1)3 − t( p+2)3 l=1

Subject to p n   bilR (1)tl3 + t( p+1)3 − t( p+2)3 ≥ aiRj (1)y j3 , i = 1, 2, . . . , m;

l=1 n 

j=1 n 

j=1

ekRj (1)y j3

≥

y j3 ≥ 1; −

j=1

dkR (1), k

n 

= 1, 2, . . . , q;

y j3 ≥ −1;

j=1

y j3 ≥ 0, j = 1, 2, . . . , n; tl4 ≥ 0, l = 1, 2, . . . , p + 2. Problem  3.5.18  p  R Minimize cl (0)tl4 + t( p+1)4 − t( p+2)4 l=1

Subject to p n   bilR (0)tl4 + t( p+1)4 − t( p+2)4 ≥ aiRj (0)y j4 , i = 1, 2, . . . , m;

l=1 n 

j=1 n  j=1

j=1

ekRj (0)y j4

≥

y j4 ≥ 1; −

dkR (0), k

n  j=1

= 1, 2, . . . , q;

y j4 ≥ −1;

y j4 ≥ 0, j = 1, 2, . . . , n; tl4 ≥ 0, l = 1, 2, . . . , p + 2.

3.5 Proposed Vaishnavi Method

85

Step 5: Substitute the value of {y ∗j1 , j = 1, 2, . . . , n}, {y ∗j2 , j = 1, 2, . . . , n}, {y ∗j3 , j = 1, 2, . . . , n} and {y ∗j4 , j = 1, 2, . . .n} of Problem 3.5.15, Problem 3.5.16, Problem 3.5.17 and Problem 3.5.18 in Problem 3.5.11, Problem 3.5.12, Problem 3.5.13 and Problem 3.5.14 respectively and find all the alternative basic optimal solug∗ r ∗ , i = 1, 2, . . . , m, r = 1, 2, . . . , f }, {x s ∗ , , i = 1, 2, . . . , m, g = 1, 2, . . . , u}, {xi2 tions {xi1 i3 ∗ t , i = 1, 2, . . . , m, t = 1, 2, . . . , w} of Probi = 1, 2, . . . , m, s = 1, 2, . . . , h} and {xi4 lem 3.5.11, Problem 3.5.12, Problem 3.5.13 and Problem 3.5.14 respectively. Step 6: Find ⎧ n m ⎫  L ⎪ ⎪ g∗ ∗ L R R ⎪ ⎪ (a (0), a (1), a (1), a (0))x y ; g = 1, 2, . . . , u, ⎪ ⎪ i1 j1 ij ij ij ij ⎪ ⎪ ⎪ ⎪ j=1 i=1 ⎪ ⎪ ⎪ ⎪ n m ⎪ ⎪   ⎪ ⎪ ∗ L L R R r ∗ ⎪ ⎪ ⎪ ⎪ (a (0), a (1), a (1), a (0))x y ; r = 1, 2, . . . , f, ⎨ ⎬ i2 j2 ij ij ij ij j=1 i=1 maximum  n  m ⎪ ⎪ s∗ ∗ L L R R ⎪ ⎪ ⎪ ⎪ j=1 i=1(ai j (0), ai j (1), ai j (1), ai j (0))xi3 y j3 ; s = 1, 2, . . . , h, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ n  m ⎪ ⎪ ∗ ⎪ ⎪ t L L R R ∗ ⎪ ⎪ ⎪ ⎪ (a (0), a (1), a (1), a (0))x y ; t = 1, 2, . . . , w ⎩ ⎭ i4 j4 ij ij ij ij j=1 i=1

Step 7: All the maximum trapezoidal fuzzy numbers will represent maximum expected loss of Player II. The optimal strategies for Player II will be {y ∗j1 , j = 1, 2, . . . , n} if maximum corresponds to n  m  g∗ (aiLj (0), aiLj (1), aiRj (1), aiRj (0))xi1 y ∗j1 ; g = 1, 2, . . . , u}, will be {y ∗j2 , j = 1, 2, { j=1 i=1

. . . , n} if maximum is corresponding to n  m  r∗ ∗ (aiLj (0), aiLj (1), aiRj (1), aiRj (0))xi2 y j2 ; r = 1, 2, . . . , f }, will be {y ∗j3 , j = 1, 2, { j=1 i=1

. . . , n} if maximum is corresponding to n  m  s∗ ∗ (aiLj (0), aiLj (1), aiRj (1), aiRj (0))xi3 y j3 ; s = 1, 2, . . . , h} and will be {y ∗j4 , j = { j=1 i=1

1, 2, . . . , n} if maximum is obtained corresponding to n  m  t∗ ∗ (aiLj (0), aiLj (1), aiRj (1), aiRj (0))xi4 y j4 ; t = 1, 2, . . . , w}. { j=1 i=1

3.6 Numerical Examples In this section, some existing numerical examples have been solved by proposed Vaishnavi method.

86

3 Constrained Matrix Games with Fuzzy Payoffs

3.6.1 Existing Numerical Example Considered by Li and Hong In this section, constrained matrix games with fuzzy payoffs  A˜ =

   (−25, (x1 , x2 )|x1 , x2 ≥ 0, x1 + x2 ≤ 1,

−19, −18, −17)  , X = (−11, −10, −8, −5) 35, 40, 81 −x1 − x2 ≤ −1, 80x1 + 50x2 ≤ 60 2 , 41 (27, 29, 32, 35)

and Y = {(y1 , y2 )|y1 , y2 ≥ 0, y1 + y2 ≥ 1, −y1 − y2 ≥ −1, −40y1 − 70y2 ≥ −50} chosen by Li and Hong [6], is solved by the proposed Vaishnavi method.

3.6.1.1

Minimum Expected Gain of Player I

Using the proposed Vaishnavi method, minimum expected gain of Player I and corresponding optimal strategies, can be obtained as follows: Step 1: Find (y1∗ , y2∗ ) ∈ Y such that the value of ((27, 29, 32, 35)x  1 y1 +  (−25, −19, 81 −18, −17)x1 y2 + (−11, −10, −8, −5)x2 y1 + 35, 40, , 41 x2 y2 is minimum 2 for all (x1 , x2 ) ∈ X i.e., find the optimal solution of Problem 3.6.1. Problem 3.6.1  Minimize

(27,

29, 32, 35)x1 y1 + (−25, −19, −18, −17)x1 y2 + (−11, −10, −8, −5)x2 y1 + 81 35, 40, , 41 x2 y2 2



Subject to 80x1 + 50x2 ≤ 60; −40y1 − 70y2 ≥ −50; x1 + x2 ≤ 1; −x1 − x2 ≤ −1; y1 + y2 ≥ 1; −y1 − y2 ≥ −1; x1 , x2 ≥ 0; y1 , y2 ≥ 0. n n n    Step 2: Using the property, λ(aiL (0), aiL (1), aiR (1), aiR (0)) = ( λaiL (0), λaiL (1),

n  i=1

λaiR (1),

n  i=1

i=1

i=1

i=1

λaiR (0)), λ ≥ 0, the Problem 3.6.1 can be transformed into Prob-

lem 3.6.2. Problem 3.6.2  Minimize

27x1 y1 − 25x1 y2 − 11x2 y1 + 35x2 y2 , 29x1 y1 − 19x1 y2 − 10x2 y1 + 40x2 y2 , 81 32x1 y1 − 18x1 y2 − 8x2 y1 + x2 y2 , 35x1 y1 − 17x1 y2 − 5x2 y1 + 41x2 y2 2



Subject to 80x1 + 50x2 ≤ 60; −40y1 − 70y2 ≥ −50; x1 + x2 ≤ 1; −x1 − x2 ≤ −1; y1 + y2 ≥ 1; −y1 − y2 ≥ −1; x1 , x2 ≥ 0; y1 , y2 ≥ 0.

Step 3: According to comparing method, to find optimal solution {y ∗j , j = 1, 2} of Problem 3.6.2 such that the value of

3.6 Numerical Examples



87

27x1 y1 − 25x1 y2 − 11x2 y1 + 35x2 y2 , 29x1 y1 − 19x1 y2 − 10x2 y1 + 40x2 y2 , 81 is 32x1 y1 − 18x1 y2 − 8x2 y1 + x2 y2 , 35x1 y1 − 17x1 y2 − 5x2 y1 + 41x2 y2 2 minimum for all (x1 , x2 ) ∈ X is equivalent to find {y ∗j , j = 1, 2} such that value of (27x1 y1 − 25x1 y2 − 11x2 y1 + 35x2 y2 ), (29x1 y1 − 19x1 y2 − 10x2 y1 + 40x2 y2 ), 81 (32x1 y1 − 18x1 y2 − 8x2 y1 + x2 y2 ) and (35x1 y1 − 17x1 y2 − 5x2 y1 + 41x2 y2 ) is 2 minimum for all (x1 , x2 ) ∈ X or to find {y ∗j1 , j = 1, 2} such that value of (27x11 y11 − 25x11 y21 − 11x21 y11 + 35x21 y21 ) is minimum for all (x11 , x21 ) ∈ X but value of and/or (32x11 y11 − 18x11 y21 (29x11 y11 − 19x11 y21 − 10x21 y11 + 40x21 y21 ) 81 − 8x21 y11 + x21 y21 ) and/or (35x11 y11 − 17x11 y21 − 5x21 y11 + 41x21 y21 ) is not 2 minimum and to find {y ∗j2 , j = 1, 2} such that value of (29x12 y12 − 19x12 y22 − 10x22 y12 + 40x22 y22 ) is minimum for all (x12 , x22 ) ∈ X but value of and/or (32x12 y12 − 18x12 y22 (27x12 y12 − 25x12 y22 − 11x22 y12 + 35x22 y22 ) 81 − 8x22 y12 + x22 y22 ) and/or (35x12 y12 − 17x12 y22 − 5x22 y12 + 41x22 y22 ) is not 2 minimum and to find {y ∗j3 , j = 1, 2} such that value of (32x13 y13 − 18x13 y23 − 81 8x23 y13 + x23 y23 ) is minimum for all (x13 , x23 ) but value of (27x13 y13 − 25x13 y23 2 − 11x23 y13 + 35x23 y23 ) and/or (29x13 y13 − 19x13 y23 − 10x23 y13 + 40x23 y23 ) and/or (35x13 y13 − 17x13 y23 − 5x23 y13 + 41x23 y23 ) is not minimum and to find {y ∗j4 , j = 1, 2} such that value of (35x14 y14 − 17x14 y24 − 5x24 y14 + 41x24 y24 ) is minimum for all (x14 , x24 ) ∈ X but value of (27x14 y14 − 25x14 y24 − 11x24 y14 + 35x24 y24 ) and/or 29x14 y14 − 19x14 y24 − 10x24 y14 + 40x24 y24 and/or (32x14 y14 − 18x14 y24 − 81 8x24 y14 + x24 y24 ) is not minimum i.e., to find the optimal solution {y ∗j1 , j = 2 1, 2}, {y ∗j2 , j = 1, 2}, {y ∗j3 , j = 1, 2}, {y ∗j4 , j = 1, 2} of Problem 3.6.3, Problem 3.6.4, Problem 3.6.5, Problem 3.6.6 respectively. Problem 3.6.3 Minimize{27x11 y11 − 25x11 y21 − 11x21 y11 + 35x21 y21 } Subject to 80x11 + 50x21 ≤ 60; −40y11 − 70y21 ≥ −50; x11 + x21 ≤ 1; −x11 − x21 ≤ −1; y11 + y21 ≥ 1; −y11 − y21 ≥ −1; x11 , x21 ≥ 0; y11 , y21 ≥ 0. Problem 3.6.4 Minimize{29x12 y12 − 19x12 y22 − 10x22 y12 + 40x22 y22 } Subject to 80x12 + 50x22 ≤ 60; −40y12 − 70y22 ≥ −50; x12 + x22 ≤ 1; −x12 − x22 ≤ −1; y12 + y22 ≥ 1; −y12 − y22 ≥ −1; x12 , x22 ≥ 0; y12 , y22 ≥ 0.

88

3 Constrained Matrix Games with Fuzzy Payoffs

Problem  3.6.5 Minimize 32x13 y13 − 18x13 y23 − 8x23 y13 +

81 x23 y23 2



Subject to 80x13 + 50x23 ≤ 60; −40y13 − 70y23 ≥ −50; x13 + x23 ≤ 1; −x13 − x23 ≤ −1; y13 + y23 ≥ 1; −y13 − y23 ≥ −1; x13 , x23 ≥ 0; y13 , y23 ≥ 0. Problem 3.6.6 Minimize{35x14 y14 − 17x14 y24 − 5x24 y14 + 41x24 y24 } Subject to 80x14 + 50x24 ≤ 60; −40y14 − 70y24 ≥ −50; x14 + x24 ≤ 1; −x14 − x24 ≤ −1; y14 + y24 ≥ 1; −y14 − y24 ≥ −1; x14 , x24 ≥ 0; y14 , y24 ≥ 0. Step 4: Since, in Problem 3.6.3, Problem 3.6.4, Problem 3.6.5 and Problem 3.6.6 only y j1 , y j2 , y j3 and y j4 respectively have been considered as decision variables. So, these are linear programming problems and hence, the optimal value of Problem 3.6.3, Problem 3.6.4, Problem 3.6.5 and Problem 3.6.6 will be equal to optimal value of its corresponding dual problem i.e., Problem 3.6.7, Problem 3.6.8, Problem 3.6.9 and Problem 3.6.10 respectively. Problem 3.6.7 Maximize{−50z 11 + z 21 } Subject to −40z 11 + z 21 ≤ 27x11 − 11x21 ; −70z 11 + z 21 ≤ −25x11 + 35x21 ; 80x11 + 50x21 ≤ 60; x11 + x21 = 1; x11 , x21 ≥ 0; z 11 ≥ 0. Problem 3.6.8 Maximize{−50z 12 + z 22 } Subject to −40z 12 + z 22 ≤ 29x12 − 10x22 ; −70z 12 + z 22 ≤ −19x12 + 40x22 ; 80x12 + 50x22 ≤ 60; x12 + x22 = 1; x12 , x22 ≥ 0; z 12 ≥ 0. Problem 3.6.9 Maximize{−50z 13 + z 23 } Subject to −40z 13 + z 23 ≤ 32x13 − 8x23 ; 81 −70z 13 + z 23 ≤ −18x13 + x23 ; 2 80x13 + 50x23 ≤ 60; x13 + x23 = 1; x13 , x23 ≥ 0; z 13 ≥ 0.

3.6 Numerical Examples

89

Problem 3.6.10 Maximize{−50z 14 + z 24 } Subject to −40z 14 + z 24 ≤ 35x14 − 5x24 ; −70z 14 + z 24 ≤ −17x14 + 41x24 ; 80x14 + 50x24 ≤ 60; x14 + x24 = 1; x14 , x24 ≥ 0; z 14 ≥ 0.  1 ∗ 2 , Step 5: Substituting the optimal solution = , x21 = 3 3       1 ∗ 2 1 ∗ 2 1 ∗ 2 ∗ ∗ ∗ x12 , x13 and x14 in Problem = , x22 = = , x23 = = , x24 = 3 3 3 3 3 3 3.6.3, Problem 3.6.4, Problem 3.6.5 and Problem 3.6.6 respectively, the obtained optimal solution ofProblem 3.6.3, Problem 3.6.4, Problem 3.6.5 and Problem 3.6.6 is  ∗ ∗ y = 1, y21 = 0 ,  11 ∗ ∗ = 1, y y 22 = 0 ,  ∗   12 ∗ ∗ ∗ = 1, y23 = 0 and y14 = 1, y24 = 0 respectively. y13 Step

 ⎫ ⎧ 6: Now, minimum ∗ y ∗ + (−25, −19, −18, −17)x ∗ y ∗ + (−11, −10, −8, −5)x ∗ y ∗ + 35, 40, 81 , 41 x ∗ y ∗ ⎪ (27, 29, 32, 35)x11 ⎪ 11 11 21 21 11 21 21 ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪

 ⎪ ⎪ ⎪ ∗ y ∗ + (−25, −19, −18, −17)x ∗ y ∗ + (−11, −10, −8, −5)x ∗ y ∗ + 35, 40, 81 , 41 x ∗ y ∗ ⎪ ⎬ ⎨ (27, 29, 32, 35)x12 12 12 22 22 12 22 22 

∗ x11

2



∗ y ∗ + (−25, −19, −18, −17)x ∗ y ∗ + (−11, −10, −8, −5)x ∗ y ∗ + 35, 40, 81 , 41 x ∗ y ∗ ⎪ ⎪ (27, 29, 32, 35)x13 ⎪ 13 13 23 23 13 23 23 ⎪ ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪

 ⎪ ⎪ 81 ⎩ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ⎭

(27, 29, 32, 35)x14 y14 + (−25, −19, −18, −17)x14 y24 + (−11, −10, −8, −5)x24 y14 + 35, 40,



=minimum

5 9 16 25 , , , 3 3 3 3

, 41 x24 y24

2     5 9 16 25 5 9 16 25 5 9 16 25 5 9 16 25 , , , , , , , , , , , , , , , = 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 

 5 9 16 25 , , , and corStep 7: The minimum expected gain of Player I is   3 3 3 3  1 ∗ 2 1 ∗ 2 ∗ ∗ responding optimal strategies are x11 = , x21 = = x12 = , x22 = = 3 3 3 3    1 ∗ 2 1 ∗ 2 ∗ ∗ x13 = x14 . = , x23 = = , x24 = 3 3 3 3 3.6.1.2

Maximum Expected Loss of Player II

Using the proposed Vaishnavi method maximum expected loss of Player II and corresponding optimal strategies, can be obtained as follows: Step 1: Find (x1∗ , x2∗ ) ∈ X such that the value of



 (27, 29, 32, 35)x1 y1 + (−25, −19, −18, −17)x1 y2 + (−11, −10, −8, −5)x2 y1 + 35, 40,

81 , 41 x2 y2 2

is maximum for all (y1 , y2 ) ∈ Y i.e., find the optimal solution of Problem 3.6.11.

90

3 Constrained Matrix Games with Fuzzy Payoffs

Problem 3.6.11  Maximize

(27, 29, 32, 35)x1 y1 + (−25, −19, −18, −17)x1 y2 + (−11, −10, −8, −5)x2 y1 +

81 35, 40, , 41 x2 y2 2



Subject to 80x1 + 50x2 ≤ 60; −40y1 − 70y2 ≥ −50; x1 + x2 ≤ 1; −x1 − x2 ≤ −1; y1 + y2 ≥ 1; −y1 − y2 ≥ −1; x1 , x2 ≥ 0; y1 , y2 ≥ 0. n n   Step 2: Using the property, λ(aiL (0), aiL (1), aiR (1), aiR (0)) = ( λaiL (0), n  i=1

λaiL (1),

n  i=1

λaiR (1),

n  i=1

i=1

λaiR (0)), λ

i=1

≥ 0, the Problem 3.6.11 can be transformed

into Problem 3.6.12. Problem 3.6.12  Maximize

27x1 y1 − 25x1 y2 − 11x2 y1 + 35x2 y2 , 29x1 y1 − 19x1 y2 − 10x2 y1 + 40x2 y2 , 81 32x1 y1 − 18x1 y2 − 8x2 y1 + x2 y2 , 35x1 y1 − 17x1 y2 − 5x2 y1 + 41x2 y2 2



Subject to 80x1 + 50x2 ≤ 60; −40y1 − 70y2 ≥ −50; x1 + x2 ≤ 1; −x1 − x2 ≤ −1; y1 + y2 ≥ 1; −y1 − y2 ≥ −1; x1 , x2 ≥ 0; y1 , y2 ≥ 0. ∗ ∗ ∗ , i = 1, 2}, {xi2 , i = 1, 2}, {xi3 , i = 1, 2}, Step 3: Find the optimal solution {xi1 ∗ {xi4 , i = 1, 2} of Problem 3.6.13, Problem 3.6.14, Problem 3.6.15, Problem 3.6.16 respectively. Problem 3.6.13 Maximize{27x11 y11 − 25x11 y21 − 11x21 y11 + 35x21 y21 } Subject to 80x11 + 50x21 ≤ 60; −40y11 − 70y21 ≥ −50; x11 + x21 ≤ 1; −x11 − x21 ≤ −1; y11 + y21 ≥ 1; −y11 − y21 ≥ −1; x11 , x21 ≥ 0; y11 , y21 ≥ 0. Problem 3.6.14 Maximize{29x12 y12 − 19x12 y22 − 10x22 y12 + 40x22 y22 } Subject to 80x12 + 50x22 ≤ 60; −40y12 − 70y22 ≥ −50; x12 + x22 ≤ 1; −x12 − x22 ≤ −1; y12 + y22 ≥ 1; −y12 − y22 ≥ −1; x12 , x22 ≥ 0; y12 , y22 ≥ 0. Problem 3.6.15 

 81 Maximize 32x13 y13 − 18x13 y23 − 8x23 y13 + x23 y23 2 Subject to 80x13 + 50x23 ≤ 60; −40y13 − 70y23 ≥ −50; x13 + x23 ≤ 1; −x13 − x23 ≤ −1; y13 + y23 ≥ 1; −y13 − y23 ≥ −1; x13 , x23 ≥ 0; y13 , y23 ≥ 0.

3.6 Numerical Examples

91

Problem 3.6.16 Maximize{35x14 y14 − 17x14 y24 − 5x24 y14 + 41x24 y24 } Subject to 80x14 + 50x24 ≤ 60; −40y14 − 70y24 ≥ −50; x14 + x24 ≤ 1; −x14 − x24 ≤ −1; y14 + y24 ≥ 1; −y14 − y24 ≥ −1; x14 , x24 ≥ 0; y14 , y24 ≥ 0. Step 4: Since, in Problem 3.6.13, Problem 3.6.14, Problem 3.6.15 and Problem 3.6.16 only xi1 , xi2 , xi3 and xi4 respectively have been considered as decision variables. So, these are linear programming problems and hence, the optimal value of Problem 3.6.13, Problem 3.6.14, Problem 3.6.15 and Problem 3.6.16 will be equal to optimal value of its corresponding dual problem i.e., Problem 3.6.17, Problem 3.6.18, Problem 3.6.19 and Problem 3.6.20 respectively. Problem 3.6.17 Minimize{60t11 + t21 } Subject to 80t11 + t21 ≥ 27y11 − 25y21 ; 50t11 + t21 ≥ −11y11 + 35y21 ; −40y11 − 70y21 ≥ −50; y11 + y21 = 1; y11 , y21 ≥ 0; t11 ≥ 0. Problem 3.6.18 Minimize{60t12 + t22 } Subject to 80t12 + t22 ≥ 29y12 − 19y22 ; 50t12 + t22 ≥ −10y12 + 40y22 ; −40y12 − 70y22 ≥ −50; y12 + y22 = 1; y12 , y22 ≥ 0; t12 ≥ 0. Problem 3.6.19 Minimize{60t13 + t23 } Subject to 80t13 + t23 ≥ 32y13 − 18y23 ; 81 y23 ; 50t13 + t23 ≥ −8y13 + 2 −40y13 − 70y23 ≥ −50; y13 + y23 = 1; y13 , y23 ≥ 0; t13 ≥ 0.

92

3 Constrained Matrix Games with Fuzzy Payoffs

Problem 3.6.20 Minimize{60t14 + t24 } Subject to 80t14 + t24 ≥ 35y14 − 17y24 ; 50t14 + t24 ≥ −5y14 + 41y24 ; −40y14 − 70y24 ≥ −50; y14 + y24 = 1; y14 , y24 ≥ 0; t14 ≥ 0.

 ∗   ∗  ∗ ∗ Step the optimal solution y11 = 1, y21 = 0 , y12 = 1, y22 =0 ,   ∗ 5: Substituting ∗ ∗ ∗ = 0 and y14 = 1, y24 = 0 in Problem 3.6.13, Problem 3.6.14, Proby13 = 1, y23 lem 3.6.15 and Problem 3.6.16 respectively, the obtained optimal   solution of Problem 1 ∗ 2 ∗ , 3.6.13, Problem 3.6.14, Problem 3.6.15 and Problem 3.6.16 is x11 = , x21 = 3      3 1 ∗ 2 1 ∗ 2 1 ∗ 2 ∗ ∗ ∗ x12 , x13 and x14 respectively. = , x22 = = , x23 = = , x24 = 3 3 3 3 3 3 Step

 ⎫ ⎧ 6: Now, maximum ∗ y ∗ + (−25, −19, −18, −17)x ∗ y ∗ + (−11, −10, −8, −5)x ∗ y ∗ + 35, 40, 81 , 41 x ∗ y ∗ ⎪ (27, 29, 32, 35)x11 ⎪ 11 11 21 21 11 21 21 ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪

 ⎪ ⎪ 81 ⎪ ∗ y ∗ + (−25, −19, −18, −17)x ∗ y ∗ + (−11, −10, −8, −5)x ∗ y ∗ + 35, 40, ∗ y∗ ⎪ ⎬ ⎨ (27, 29, 32, 35)x12 , 41 x 12 12 22 22 12 22 22

2



∗ y ∗ + (−25, −19, −18, −17)x ∗ y ∗ + (−11, −10, −8, −5)x ∗ y ∗ + 35, 40, 81 , 41 x ∗ y ∗ ⎪ ⎪ (27, 29, 32, 35)x13 ⎪ 13 13 23 23 13 23 23 ⎪ ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪

 ⎪ ⎭ ⎩ (27, 29, 32, 35)x ∗ y ∗ + (−25, −19, −18, −17)x ∗ y ∗ + (−11, −10, −8, −5)x ∗ y ∗ + 35, 40, 81 , 41 x ∗ y ∗ ⎪ 14 14



= maximum

5 9 16 25 , , , 3 3 3 3

14 24

24 14

24 24

2     5 9 16 25 5 9 16 25 5 9 16 25 5 9 16 25 , , , , , , , , , , , , , , , = 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 

 5 9 16 25 , , , and correspondStep 7: The maximum expected loss of Player II is 3 3  3 3  ∗  ∗ ∗ ∗ ing strategies are y11 = 1, y21 = 0 = y12 = 1, y22 =0 =     ∗ optimal ∗ ∗ ∗ = 0 = y14 = 1, y24 =0 . y13 = 1, y23

3.6.2 Existing Numerical Example Considered by Li and Cheng In this section, constrained matrix with fuzzy payoffs    games (18, 20, 23) (−21, −18, −16) (x1 , x2 )|x1 , x2 ≥ 0, x1 + x2 ≤ 1, −x1 − x2 ≤ −1, ˜ and A = (−33, −32, −27) (38, 40, 43) , X = (70, 80, 88)x1 + (44, 50, 54)x2  (61, 67, 74)   (y1 , y2 )|y1 , y2 ≥ 0, y1 + y2 ≥ 1, −y1 − y2 ≥ −1, chosen by Y = (−48, −40, −35)y1 + (−79, −70, −65)y2  (−60, −52, −50) Li and Cheng [5], is solved by the proposed Vaishnavi method.

3.6 Numerical Examples

3.6.2.1

93

Minimum Expected Gain of Player I

Using the proposed Vaishnavi method minimum expected gain of Player I and corresponding optimal strategies, can be obtained as follows: Step 1: Find (y1∗ , y2∗ ) ∈ Y such that the value of ((18, 20, 23)x1 y1 + (−21, −18, −16)x1 y2 + (−33, −32, −27)x2 y1 + 38, 40, 43)x2 y2 ) is minimum for all (x1 , x2 ) ∈ X i.e., find the optimal solution of Problem 3.6.21. Problem 3.6.21 Minimize {(18, 20, 23)x1 y1 + (−21, −18, −16)x1 y2 + (−33, −32, −27)x2 y1 + (38, 40, 43)x2 y2 } Subject to (70, 80, 88)x1 + (44, 50, 54)x2  (61, 67, 74); (−48, −40, −35)y1 + (−79, −70, −65)y2  (−60, −52, −50); x1 + x2 ≤ 1; −x1 − x2 ≤ −1; y1 + y2 ≥ 1; −y1 − y2 ≥ −1; x1 , x2 ≥ 0; y1 , y2 ≥ 0. n n n    Step 2: Using the property, λ(aiL (0), aiL (1), aiR (1), aiR (0)) = ( λaiL (0), λaiL (1),

n  i=1

λaiR (1),

n  i=1

i=1

λaiR (0)), λ

i=1

i=1

≥ 0, the Problem 3.6.21 can be transformed into

Problem 3.6.22. Problem 3.6.22  Minimize

18x1 y1 − 21x1 y2 − 33x2 y1 + 38x2 y2 , 20x1 y1 − 18x1 y2 − 32x2 y1 + 40x2 y2 , 23x1 y1 − 16x1 y2 − 27x2 y1 + 43x2 y2



Subject to (70x1 + 44x2 , 80x1 + 50x2 , 88x1 + 54x2 )  (61, 67, 74);; (−48y1 − 79y2 , −40y1 − 70y2 , −35y1 − 65y2 )  (−60, −52, −50); x1 + x2 ≤ 1; −x1 − x2 ≤ −1; y1 + y2 ≥ 1; −y1 − y2 ≥ −1; x1 , x2 ≥ 0; y1 , y2 ≥ 0.    Step 3: Using the comparing method, a L (0), a L (1), a R (1), a R (0)  b L (0), b L (1),  ab R (1), b R (0) 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 Problem 3.6.22 can be transformed into Problem 3.6.23. Problem 3.6.23  Minimize

18x1 y1 − 21x1 y2 − 33x2 y1 + 38x2 y2 , 20x1 y1 − 18x1 y2 − 32x2 y1 + 40x2 y2 , 23x1 y1 − 16x1 y2 − 27x2 y1 + 43x2 y2



Subject to 70x1 + 44x2 ≤ 61; 80x1 + 50x2 ≤ 67; 88x1 + 54x2 ≤ 74; −48y1 − 79y2 ≥ −60; −40y1 − 70y2 ≥ −52; −35y1 − 65y2 ≥ −50; x1 + x2 ≤ 1; −x1 − x2 ≤ −1; y1 + y2 ≥ 1; −y1 − y2 ≥ −1; x1 , x2 ≥ 0; y1 , y2 ≥ 0. Step 4: Find the optimal solution {y ∗j1 , j = 1, 2}, {y ∗j2 , j = 1, 2}, {y ∗j3 , j = 1, 2} of Problem 3.6.24, Problem 3.6.25, Problem 3.6.26 respectively.

94

3 Constrained Matrix Games with Fuzzy Payoffs

Problem 3.6.24 Minimize{18x11 y11 − 21x11 y21 − 33x21 y11 + 38x21 y21 } Subject to 70x11 + 44x21 ≤ 61; 80x11 + 50x21 ≤ 67; 88x11 + 54x21 ≤ 74; −48y11 − 79y21 ≥ −60; −40y11 − 70y21 ≥ −52; −35y11 − 65y21 ≥ −50; x11 + x21 ≤ 1; −x11 − x21 ≤ −1; y11 + y21 ≥ 1; −y11 − y21 ≥ −1; x11 , x21 ≥ 0; y11 , y21 ≥ 0. Problem 3.6.25 Minimize{20x12 y12 − 18x12 y22 − 32x22 y12 + 40x22 y22 } Subject to 70x12 + 44x22 ≤ 61; 80x12 + 50x22 ≤ 67; 88x12 + 54x22 ≤ 74; −48y12 − 79y22 ≥ −60; −40y12 − 70y22 ≥ −52; −35y12 − 65y22 ≥ −50; x12 + x22 ≤ 1; −x12 − x22 ≤ −1; y12 + y22 ≥ 1; −y12 − y22 ≥ −1; x12 , x22 ≥ 0; y12 , y22 ≥ 0. Problem 3.6.26 Minimize{23x13 y13 − 16x13 y23 − 27x23 y13 + 43x23 y23 } Subject to 70x13 + 44x23 ≤ 61; 80x13 + 50x23 ≤ 67; 88x13 + 54x23 ≤ 74; −48y13 − 79y23 ≥ −60; −40y13 − 70y23 ≥ −52; −35y13 − 65y23 ≥ −50; x13 + x23 ≤ 1; −x13 − x23 ≤ −1; y13 + y23 ≥ 1; −y13 − y23 ≥ −1; x13 , x23 ≥ 0; y13 , y23 ≥ 0. Step 5: Since, in Problem 3.6.24, Problem 3.6.25 and Problem 3.6.26 only y j1 , y j2 and y j3 respectively have been considered as decision variables. So, these are linear programming problems and hence, the optimal value of Problem 3.6.24, Problem 3.6.25 and Problem 3.6.26 will be equal to optimal value of its corresponding dual problem i.e., Problem 3.6.27, Problem 3.6.28 and Problem 3.6.29 respectively. Problem 3.6.27 Maximize{−60z 11 − 52z 21 − 50z 31 + z 41 } Subject to −48z 11 − 40z 21 − 35z 31 + z 41 ≤ 18x11 − 33x21 ; −79z 11 − 70z 21 − 65z 31 + z 41 ≤ −21x11 + 38x21 ; 70x11 + 44x21 ≤ 61; 80x11 + 50x21 ≤ 67; 88x11 + 54x21 ≤ 74; x11 + x21 = 1; x11 , x21 ≥ 0; z 11 , z 21 , z 31 ≥ 0. Problem 3.6.28 Maximize{−60z 12 − 52z 22 − 50z 32 + z 42 } Subject to −48z 12 − 40z 22 − 35z 32 + z 42 ≤ 20x12 − 32x22 ; −79z 12 − 70z 22 − 65z 32 + z 42 ≤ −18x12 + 40x22 ; 70x12 + 44x22 ≤ 61;

3.6 Numerical Examples

95

80x12 + 50x22 ≤ 67; 88x12 + 54x22 ≤ 74; x12 + x22 = 1; x12 , x22 ≥ 0; z 12 , z 22 , z 32 ≥ 0. Problem 3.6.29 Minimize{−60z 13 − 52z 23 − 50z 33 + z 43 } Subject to −48z 13 − 40z 23 − 35z 33 + z 43 ≤ 23x13 − 27x23 ; −79z 13 − 70z 23 − 65z 33 + z 43 ≤ −16x13 + 43x23 ; 70x13 + 44x23 ≤ 61; 80x13 + 50x23 ≤ 67; 88x13 + 54x23 ≤ 74; x13 + x23 = 1; x13 , x23 ≥ 0; z 13 , z 23 , z 33≥ 0. 17

13

 

17

13



∗ = ∗ = Step 6: Substituting the optimal solution x11 , x∗ = , x12 , x∗ = 30 21 30 30 22 30   17 ∗ 13 ∗ ,x = in Problem 3.6.24, Problem 3.6.25 and Problem 3.6.26 and x13 = 30 23 30 respectively, the obtained optimal solution 3.6.24, Problem 3.6.25 and  ∗   ∗ of Problem   ∗ ∗ ∗ ∗ Problem 3.6.26 is y11 = 1, y21 = 0 , y12 = 1, y22 = 0 and y13 = 1, y23 =0 respectively. Step ⎧ 7: Now, ∗minimum ⎫ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ⎪ ⎨ (18, 20, 23)x11 y11 + (−21, −18, −16)x11 y21 + (−33, −32, −27)x21 y11 + (38, 40, 43)x21 y21 ⎪ ⎬ ∗ y ∗ + (−21, −18, −16)x ∗ y ∗ + (−33, −32, −27)x ∗ y ∗ + (38, 40, 43)x ∗ y ∗ (18, 20, 23)x12 12 12 22 22 12 22 22 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ (18, 20, 23)x ⎪ 13 y13 + (−21, −18, −16)x 13 y23 + (−33, −32, −27)x 23 y13 + (38, 40, 43)x 23 y23 ⎪ ⎩ ∗ ∗ ∗ ∗ ∗ ∗ ∗ y∗ ⎭ (18, 20, 23)x14 y14 + (−21, −18, −16)x14 y24 + (−33, −32, −27)x24 y14 + (38, 40, 43)x24 24



=minimum

    −41 −38 4 −41 −38 4 −41 −38 4 −41 −38 4 , , , , , , , , = , , 10 15 3 10 15 3  10 15 3

10 15 3

−41 −38 4 , , and correspondStep 8: The minimum expected gain of Player I is 10   15 3  17 ∗ 13 17 ∗ 13 ∗ ∗ ing optimal strategies are x11 = x12 = = = , x21 = , x22 = 30 30 30 30   17 ∗ 13 ∗ x13 ,x = . = 30 23 30 3.6.2.2

Maximum Expected Loss of Player II

Using the proposed Vaishnavi method, maximum expected loss of Player II and corresponding optimal strategies, can be obtained as follows: Step 1: Find (x1∗ , x2∗ ) ∈ X such that the value of ((18, 20, 23)x1 y1 + (−21, −18, −16)x1 y2 + (−33, −32, −27)x2 y1 + (38, 40, 43) x2 y2 ) is maximum for all (y1 , y2 ) ∈ Y i.e., find the optimal solution of Problem 3.6.30.

96

3 Constrained Matrix Games with Fuzzy Payoffs

Problem 3.6.30 Maximize{(18, 20, 23)x1 y1 + (−21, −18, −16)x1 y2 + (−33, −32, −27)x2 y1 + (38, 40, 43)x2 y2 } Subject to (70, 80, 88)x1 + (44, 50, 54)x2  (61, 67, 74); (−48, −40, −35)y1 + (−79, −70, −65)y2  (−60, −52, −50); x1 + x2 ≤ 1; −x1 − x2 ≤ −1; y1 + y2 ≥ 1; −y1 − y2 ≥ −1; x1 , x2 ≥ 0; y1 , y2 ≥ 0. n n n    Step 2: Using the property, λ(aiL (0), aiL (1), aiR (1), aiR (0)) = ( λaiL (0), λaiL (1),

n  i=1

λaiR (1),

n  i=1

i=1

λaiR (0)), λ

i=1

i=1

≥ 0, the Problem 3.6.30 can be transformed into

Problem 3.6.31. Problem 3.6.31  Maximize

18x1 y1 − 21x1 y2 − 33x2 y1 + 38x2 y2 , 20x1 y1 − 18x1 y2 − 32x2 y1 + 40x2 y2 , 23x1 y1 − 16x1 y2 − 27x2 y1 + 43x2 y2



Subject to (70x1 + 44x2 , 80x1 + 50x2 , 88x1 + 54x2 )  (61, 67, 74); (−48y1 − 79y2 , −40y1 − 70y2 , −35y1 − 65y2 )  (−60, −52, −50); x1 + x2 ≤ 1; −x1 − x2 ≤ −1; y1 + y2 ≥ 1; −y1 − y2 ≥ −1; x1 , x2 ≥ 0; y1 , y2 ≥ 0.    Step 3: Usingthe comparing method, a L (0), a L (1), a R (1), a R (0)  b L (0), b L (1), b R (1), b R (0) 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 Problem 3.6.31 can be transformed into Problem 3.6.32. Problem 3.6.32  Maximize

18x1 y1 − 21x1 y2 − 33x2 y1 + 38x2 y2 , 20x1 y1 − 18x1 y2 − 32x2 y1 + 40x2 y2 , 23x1 y1 − 16x1 y2 − 27x2 y1 + 43x2 y2



Subject to 70x1 + 44x2 ≤ 61; 80x1 + 50x2 ≤ 67; 88x1 + 54x2 ≤ 74; −48y1 − 79y2 ≥ −60; −40y1 − 70y2 ≥ −52; −35y1 − 65y2 ≥ −50; x1 + x2 ≤ 1; −x1 − x2 ≤ −1; y1 + y2 ≥ 1; −y1 − y2 ≥ −1; x1 , x2 ≥ 0; y1 , y2 ≥ 0. ∗ ∗ ∗ , i = 1, 2}, {xi2 , j = 1, 2}, {xi3 , j = 1, 2} of Step 4: Find the optimal solution {xi1 Problem 3.6.33, Problem 3.6.34, Problem 3.6.35 respectively. Problem 3.6.33 Maximize{18x11 y11 − 21x11 y21 − 33x21 y11 + 38x21 y21 } Subject to 70x11 + 44x21 ≤ 61; 80x11 + 50x21 ≤ 67; 88x11 + 54x21 ≤ 74; −48y11 − 79y21 ≥ −60; −40y11 − 70y21 ≥ −52; −35y11 − 65y21 ≥ −50; x11 + x21 ≤ 1; −x11 − x21 ≤ −1; y11 + y21 ≥ 1; −y11 − y21 ≥ −1; x11 , x21 ≥ 0; y11 , y21 ≥ 0.

3.6 Numerical Examples

97

Problem 3.6.34 Maximize{20x12 y12 − 18x12 y22 − 32x22 y12 + 40x22 y22 } Subject to 70x12 + 44x22 ≤ 61; 80x12 + 50x22 ≤ 67; 88x12 + 54x22 ≤ 74; −48y12 − 79y22 ≥ −60; −40y12 − 70y22 ≥ −52; −35y12 − 65y22 ≥ −50; x12 + x22 ≤ 1; −x12 − x22 ≤ −1; y12 + y22 ≥ 1; −y12 − y22 ≥ −1; x12 , x22 ≥ 0; y12 , y22 ≥ 0. Problem 3.6.35 Maximize{23x13 y13 − 16x13 y23 − 27x23 y13 + 43x23 y23 } Subject to 70x13 + 44x23 ≤ 61; 80x13 + 50x23 ≤ 67; 88x13 + 54x23 ≤ 74; −48y13 − 79y23 ≥ −60; −40y13 − 70y23 ≥ −52; −35y13 − 65y23 ≥ −50; x13 + x23 ≤ 1; −x13 − x23 ≤ −1; y13 + y23 ≥ 1; −y13 − y23 ≥ −1; x13 , x23 ≥ 0; y13 , y23 ≥ 0. Step 5: Since, in Problem 3.6.33, Problem 3.6.34 and Problem 3.6.35 only xi1 , xi2 and xi3 respectively have been considered as decision variables. So, these are linear programming problems and hence, the optimal value of Problem 3.6.33, Problem 3.6.34 and Problem 3.6.35 will be equal to optimal value of its corresponding dual problem i.e., Problem 3.6.36, Problem 3.6.37 and Problem 3.6.38 respectively. Problem 3.6.36 Minimize{61t11 + 67t21 + 74t31 + t41 } Subject to 70t11 + 80t21 + 88t31 + t41 ≥ 18y11 − 21y21 ; 44t11 + 50t21 + 54t31 + t41 ≥ −33y11 + 38y21 ; −48y11 − 79y21 ≥ −60; −40y11 − 70y21 ≥ −52; −35y11 − 65y21 ≥ −50; y11 + y21 = 1; y11 , y21 ≥ 0; t11 , t21 , t31 ≥ 0. Problem 3.6.37 Minimize{61t12 + 67t22 + 74t32 + t42 } Subject to 70t12 + 80t22 + 88t32 + t42 ≥ 20y12 − 18y22 ; 44t12 + 50t22 + 54t32 + t42 ≥ −32y12 + 40y22 ; −48y12 − 79y22 ≥ −60; −40y12 − 70y22 ≥ −52; −35y12 − 65y22 ≥ −50; y12 + y22 = 1; y12 , y22 ≥ 0; t12 , t22 , t32 ≥ 0.

98

3 Constrained Matrix Games with Fuzzy Payoffs

Problem 3.6.38 Minimize{61t13 + 67t23 + 74t33 + t43 } Subject to 70t13 + 80t23 + 88t33 + t43 ≥ 23y13 − 16y23 ; 44t13 + 50t23 + 54t33 + t43 ≥ −27y13 + 43y23 ; −48y13 − 79y23 ≥ −60; −40y13 − 70y23 ≥ −52; −35y13 − 65y23 ≥ −50; y13 + y23 = 1; y13 , y23 ≥ 0; t13 , t23 t33 ≥ 0.   ∗   ∗ ∗ ∗ = 1, y21 = 0 , y12 = 1, y22 =0 Step5: Substituting the optimal solution y11  ∗ ∗ = 1, y23 = 0 in Problem 3.6.33, Problem 3.6.34 and Problem 3.6.35 and y13 respectively, the obtained optimal solution  of Problem 3.6.33, Problem   3.6.34 and  17 13 17 13 17 ∗ ∗ ∗ ∗ ∗ , x21 , x12 , x22 and x13 , Problem 3.6.35 is x11 = = = = = 30 30 30 30 30  13 ∗ x23 respectively. = 30 Step ⎧ 6: Now,∗ maximum ∗ + (−21, −18, −16)x ∗ y ∗ + (−33, −32, −27)x ∗ y ∗ + (38, 40, 43)x ∗ y ∗ ⎫ y11 11 21 21 11 21 21 ⎬ ⎨ (18, 20, 23)x11 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ (18, 20, 23)x12 y12 + (−21, −18, −16)x12 y22 + (−33, −32, −27)x22 y12 + (38, 40, 43)x22 y22 ∗

∗

∗

∗

∗

∗

∗

∗

y13 + (−21, −18, −16)x13 y23 + (−33, −32, −27)x23 y13 + (38, 40, 43)x23 y23 ⎭ ⎩ (18, 20, 23)x13 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ (18, 20, 23)x14 y14 + (−21, −18, −16)x14 y24 + (−33, −32, −27)x24 y14 + (38, 40, 43)x24 y24

     −41 −38 4 −41 −38 4 −41 −38 4 −41 −38 4 , , , , , , , , = , , =maximum 10 15 3 10 15 3

10 15 3  10 15 3

−41 −38 4 , , and correspondStep 7: The maximum expected loss of Player II is 10 15 3    ∗   ∗ ∗ ∗ ∗ ing optimal strategies are y11 = 1, y21 = 0 = y12 = 1, y22 = 0 = y13 = 1,  ∗ y23 = 0 .

3.7 Conclusion On the basis of present study, it can be concluded that some mathematically incorrect assumptions have been considered in the existing methods [4–7] for solving constrained matrix games with fuzzy payoffs. Therefore, it is not genuine to use these methods. Furthermore, to resolve flaws of the existing methods [6, 7], a new method (named as Vaishnavi method) is proposed for solving constrained matrix games with fuzzy payoffs.

References 1. Charnes, A.: Constrained games and linear programming. Proc. Natl. Acad. Sci. USA 39, 639 (1953) 2. Dresher, M.: Games of Strategy Theory and Applications. Prentice-Hall, New York (1961)

References

99

3. Kawaguchi, T., Maruyama, Y.: A note on minimax (maximin) programming. Manag. Sci. 670– 676 (1976) 4. Li, D.F.: Fuzzy constrained matrix games with fuzzy payoffs. J. Fuzzy Math. 7, 907–912 (1999) 5. Li, D.F., Cheng, C.T.: Fuzzy multiobjective programming methods for fuzzy constrained matrix games with fuzzy numbers. Int. J. Uncertain. Fuzziness Knowl. Based Syst. 10, 385–400 (2002) 6. Li, D.F., Hong, F.X.: Solving constrained matrix games with payoffs of triangular fuzzy numbers. Comput. Math. Appl. 64, 432–446 (2012) 7. Li, D.F., Hong, F.X.: Alfa-cut based linear programming methodology for constrained matrix games with payoffs of trapezoidal fuzzy numbers. Fuzzy Optim. Decis. Mak. 12, 191–213 (2013) 8. Owen, G.: Game Theory, 2nd edn. Academic Press, New York (1982)

Chapter 4

Matrix Games with Intuitionistic FuzzyPayoffs

In this chapter, flaws of the existing methods [6, 8–10] for solving matrix games with intuitionistic fuzzy payoffs (matrix games in which payoffs are represented by intuitionistic fuzzy numbers) are pointed out. To resolve these flaws, new methods (named as Ambika methods) are also proposed to obtain the optimal strategies as well as minimum expected gain of Player I and maximum expected loss of Player II for matrix games with intuitionistic fuzzy payoffs. To illustrate proposed Ambika methods, some existing numerical problems of matrix games with intuitionistic fuzzy payoffs are solved by proposed Ambika methods.

4.1 Matrix Games with Intuitionistic Fuzzy Payoffs Fuzzy sets are designed to handle uncertainties by attributing a degree, called the membership degree, to which an object belongs to a set. The degree to which it does not belong to the same set is taken as one minus the membership degree, and is termed as the non-membership degree. However, in real decision making problems there are instances where not only the degree to which an object belongs to a set is known but in addition a degree to which the same object does not belong to the set is also known. For example, when several experts evaluate a product, they can give arguments for the product being ‘good’ as well as arguments for the product being ‘bad’. The ‘goodness’ and ‘badness’ of the product can be quantified in terms of pre-decided thresholds. The arguments in favor of ‘good’ describe the membership degree that the product belongs to a ‘set of good products’, while the arguments in favor of ‘badness’ describe the non-membership degree of the product not being in the ‘set of good products’. An important point to observe is that an expert may have his/her own reservations in classifying the object in one of these two categories. In other words, it amounts to say that the two degrees do not necessarily add up to one. Atanassov [1] proposed an interesting generalization of fuzzy sets called an intuitionistic fuzzy sets to capture

102

4 Matrix Games with Intuitionistic Fuzzy Payoffs

this aspect of human behavior. Intuitionistic fuzzy sets are characterized by two membership functions, one for the degree of belongingness and the other for the degree of non-belongingness. These membership functions are defined such that for each element of the universe the sum of their degrees is less than or equal to one rather than being one as in classical fuzzy set theory. Bustince and Burillo [2] pointed out that notion of intuitionistic fuzzy set is same as that of vague set [4] i.e., intuitionistic fuzzy set and vague sets are same only their representations is different. The only difference between the representation of intuitionistic fuzzy set and vague se is that in the intuitionistic fuzzy set, the degree of non-membership is mentioned while, in the vague set instead of degree of nonmembership i.e., 1−(degree of non-membership) is mentioned [3]. In the literature [5–10], such games in which payoffs are represented by intuitionistic fuzzy number (vague set) are named as matrix games with intuitionistic fuzzy payoffs.

4.2 Preliminaries In this section, some basic definitions and arithmetic operations of vague sets (intuitionistic fuzzy sets) are presented [3].

4.2.1 Basic Definitions In this section, some basic definitions are presented.     Definition 4.1 A vague set A˜ = < x, μ A˜ (x), 1 − ν A˜ (x) |x ∈ X > , defined on the universal set X , is characterized by a truth membership function μ A˜ , μ A˜ : X → [0, 1] and complement of a false membership function (1 − ν A˜ ), (1 − ν A˜ ) : X → [0, 1]. The values μ A˜ (x) and 1 − ν A˜ (x) represents the degree of membership and complement of non-membership for x ∈ X and always satisfies the condition μ A˜ (x) ≤ 1 − ν A˜ (x) ∀x ∈ X . The value (1 − μ A˜ (x) − ν A˜ (x)) represents the degree of hesitation for x ∈ X .  Definition 4.2 Let A˜ be vague set then A(α, β) = x ∈ X |μ A˜ (x) ≥ α,  ˜ (1 − ν A˜ (x)) ≥ β , α, β ∈ [0, 1] is said to be an (α, β)−cut of A. ˜ defined on the universal set of real numbers , Definition 4.3 A vague set A, denoted as A˜ =< (a L (0), a(1), a R (0)); w A˜ , 1 − u A˜ >, where a L (0) ≤ a(1) ≤ a R (0) and w A˜ ≤ 1 − u A˜ , is said to be a triangular vague set if the membership function, w A˜ and complement of non-membership function, 1 − u A˜ are given as

4.2 Preliminaries

μ A˜ =

103

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

w A˜ (x−a L (0) a(1)−a L (0)

a L (0) ≤ x < a(1)

w A˜

x = a(1)

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

w A˜ (x−a R (0) a(1)−a R (0)

0

a(1) ≤ x
be a triangular ˜ α ∈ [0, w A˜ ], β ∈ [0, 1 − u A˜ ] for the triangular vague vague set then (α, β)-cut of A, ˜ set A can be defined as A˜ =

 a L (0) + (a(1) − a L (0)) wα , a R (0) − (a R (0) − a(1)) wα ; A˜ A˜ 

β β . , a R (0) − (a R (0) − a(1)) 1−u a L (0) + (a(1) − a L (0)) 1−u A˜

A˜

˜ defined on the universal set of real numbers , denoted Definition 4.5 A vague set A, as A˜ =< (a L (0), a L (1), a R (1), a R (0)); w A˜ , 1 − u A˜ >, where a L (0) ≤ a L (1) ≤ a R (1) ≤ a R (0) and w A˜ ≤ 1 − u A˜ , is said to be a trapezoidal vague set if the membership function, μ A˜ and complement of non-membership function, 1 − ν A˜ are given as ⎧ w (x−a L (0)) ⎧ ˜ ⎪ ⎪ ⎪ a LA(1)−a L (0) a L (0) ≤ x < a L (1) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎨ a L (1) < x < a R (1) w A˜ and 1 − ν μ A˜ = = ˜ A R ⎪ w A˜ (x−a (0)) R ⎪ ⎪ ⎪ ⎪ ⎪ a (1) ≤ x < a R (0) ⎪ ⎪ a R (1)−a R (0) ⎪ ⎪ ⎪ ⎪ ⎩ ⎩ 0 other wise

(1−u A˜ )(x−a L (0)) a L (1)−a L (0)

a L (0) ≤ x < a L (1)

(1 − u A˜ )

a L (1) < x < a R (1)

)(x−a R (0))

(1−u A˜ a R (1)−a R (0)

0

a R (1) ≤ x < a R (0) other wise

where, w A˜ = supremum{μ A˜ (x) : x ∈ } and 1 − u A˜ = supremum{1 − ν A˜ (x) : x ∈ }. Definition 4.6 Let, A˜ =< (a L (0), a L (1), a R (1), a R (0)); w A˜ , 1 − u A˜ > be a trape˜ α ∈ [0, w A˜ ], β ∈ [0, 1 − u A˜ ] for the trapezoidal vague set then (α, β)-cut of A, ˜ zoidal vague set A can be defined as A˜ =



 a L (0) + (a L (1) − a L (0)) wα , a R (0) − (a R (0) − a R (1)) wα ; A˜

A˜  β β a L (0) + (a L (1) − a L (0)) 1−u . , a R (0) − (a R (0) − a R (1)) 1−u A˜

A˜

Remark 4.1 If in the triangular vague set A˜ =< (a L (0), a(1), a R (0)); w A˜ , 1 − u A˜ >, 1 − u A˜ is replaced by u A˜ then A˜ =< (a L (0), a(1), a R (0)); w A˜ , u A˜ > is called triangular intuitionistic fuzzy number. Remark 4.2 If in the trapezoidal vague set A˜ =< (a L (0), a L (1), a R (1), a R (0)); w A˜ , 1 − u A˜ >, 1 − u A˜ is replaced by u A˜ then A˜ =< (a L (0), a L (1), a R (1), a R (0)); w A˜ , u A˜ > is called trapezoidal intuitionistic fuzzy number.

104

4 Matrix Games with Intuitionistic Fuzzy Payoffs

4.2.2 Arithmetic Operations over Trapezoidal Vague Sets In this section, arithmetic operations over trapezoidal vague sets, defined on universal set of real numbers , are presented. Let, A˜ =< (a L (0), a L (1), a R (1), a R (0)); w A˜ , 1 − u A˜ > and B˜ =< (b L (0), b L (1), b R (1), b R (0)); w B˜ , 1 − u B˜ > be two trapezoidal vague sets (trapezoidal intuitionistic fuzzy numbers) then  (i) A˜ + B˜ = (a L (0) + b L (0), a L (1) + b L (1), a R (1) + b R (1), a R (0) + b R (0));  min{w A˜ , w B˜ }, min{1 − u A˜ , 1 − u B˜ } .  (ii) A˜ − B˜ = (a L (0) − b R (0), a L (1) − b R (1), a R (1) − b L (1), a R (0) − b L (0));  min{w A˜ , w B˜ }, min{1 − u A˜ , 1 − u B˜ } .  < (λa L (0), λa L (1), λa R (1), λa R (0)); w A˜ , 1 − u A˜ > λ ≥ 0 (iii) λ A˜ = < (λa R (0), λa R (1), λa L (1), λa L (0)); w A˜ , 1 − u A˜ > λ < 0 Remark 4.3 If in the arithmetic operations presented in Sect. 4.2.2, min{1 − u A˜ , 1 − u B˜ } is replaced by max{u A˜ , u B˜ } then these arithmetic operations will be arithmetic operations over trapezoidal intuitionistic fuzzy numbers.

4.3 Existing Mathematical Formulation of Matrix Games with Intuitionistic Fuzzy Payoffs In the literature, Problems 4.3.1 and 4.3.2 are used to obtain the minimum expected gain of Player I and maximum expected loss of Player II as well as their corresponding optimal strategies. These problems can be obtained by using the same procedure, discussed in Sect. 1.6.1 of Chap. 1, by replacing the intervals [aiLj , aiRj ], [υ L , υ R ] and [ω L , ω R ] with triangular/trapezoidal intuitionistic fuzzy numbers a˜ i j , υ˜ and ω˜ respectively. Problem 4.3.1 Maximize{υ} ˜ Subject to m  a˜ i j xi ≥ υ, ˜ j = 1, 2, ..., n; i=1 m  i=1

xi = 1; xi ≥ 0, i = 1, 2, ..., m.

4.3 Existing Mathematical Formulation of Matrix Games with Intuitionistic …

105

Problem 4.3.2 Minimize{ω} ˜ Subject to n  a˜i j y j ≤ ω, ˜ i = 1, 2, ..., m; j=1 n 

y j = 1; y j ≥ 0, j = 1, 2, ..., n.

j=1

4.4 Literature Review of Matrix Games with Intuitionistic Fuzzy Payoffs In this section, a brief review of the methods, proposed in the literature in last ten years for solving matrix games with intuitionistic fuzzy payoffs (matrix games in which payoffs are either represented by triangular intuitionistic numbers or trapezoidal intuitionistic fuzzy numbers) is presented. Nan et al. [8] proposed a method to find the minimum expected gain of Player I, maximum expected loss of Player II and their corresponding strategies for matrix games with triangular intuitionistic fuzzy payoffs. In this method, firstly authors have transformed Problem 4.3.1 and Problem 4.3.2 into Problem 4.4.1 and Problem 4.4.2 respectively then it is claimed that the optimal solution {xi , i = 1, 2, ..., m} and optimal value of Problem 4.4.3 represents the optimal strategies and minimum expected gain of Player I as well as the optimal solution {y j , j = 1, 2, ..., n} and optimal value of Problem 4.4.5 represents the optimal strategies and maximum expected loss of Player II respectively. Problem 4.4.1

⎧ ⎫ ⎪  L  L  ⎪ ⎨ υ (0) + 2υ(1) + υ R (0) υ (0) + 2υ(1) + υ R (0) ⎬ Maximize min {wa˜ i j } , 1 − max {u a˜ i j } ⎪ ⎪ 1≤i≤m 4 4 ⎩1≤i≤m ⎭ 1≤ j≤n

1≤ j≤n

Subject to m 

  min {wa˜ i j } aiLj (0)+2ai j (1)+aiRj (0)

1≤i≤m

4

i=1 m  i=1

xi ≥ min {wa˜ i j }

   1− max {u a˜ i j } aiLj (0)+2ai j (1)+aiRj (0) 1≤i≤m

4

υ L (0) ≤ υ(1); υ(1) ≤ υ R (0); m  xi = 1; xi ≥ 0, i = 1, 2, ..., m. i=1

1≤i≤m 1≤ j≤n



υ L (0)+2υ(1)+υ R (0) 4

xi ≥ 1 − max {u a˜ i j } 1≤i≤m 1≤ j≤n





; j = 1, 2, ..., n;

υ L (0)+2υ(1)+υ R (0) 4



; j = 1, 2, ..., n;

106

4 Matrix Games with Intuitionistic Fuzzy Payoffs

Problem 4.4.2

⎧ ⎫ ⎪  L   L ⎪ ⎨ R R ω (0) + 2ω(1) + ω (0) ω (0) + 2ω(1) + ω (0) ⎬ Minimize min {wa˜ i j } , 1 − max {u a˜ i j } ⎪ ⎪ 1≤i≤m 1≤i≤m 4 4 ⎩ ⎭ 1≤ j≤n

1≤ j≤n

Subject to n 

  min {wa˜ i j } aiLj (0)+2ai j (1)+aiRj (0)

1≤ j≤n

4

j=1  n 

1− max {u a˜ i j } 1≤ j≤n

y j ≤ min {wa˜ i j }

  aiLj (0)+2ai j (1)+aiRj (0) 4

j=1

1≤i≤m 1≤ j≤n



ω L (0)+2ω(1)+ω R (0) 4

y j ≤ 1 − max {u a˜ i j } 1≤i≤m 1≤ j≤n





; i = 1, 2, ..., m;

ω L (0)+2ω(1)+ω R (0) 4



; i = 1, 2, ..., m;

ω L (0) ≤ ω(1); ω(1) ≤ ω R (0); n  y j = 1; y j ≥ 0, j = 1, 2, ..., n. j=1

Problem 4.4.3 Maximize{υ2 } Subject to   m min {wa˜ i j } aiLj (0)+2ai j (1)+aiRj (0)  1≤i≤m

xi ≥ 4 i=1    1− max {u a˜ i j } aiLj (0)+2ai j (1)+aiRj (0) m  1≤i≤m

4

i=1

υ1 ; j = 1, 2, ..., n;

xi ≥ υ2 ; j = 1, 2, ..., n;

υ2 ≥ υ1 ; υ1 ≥ υ10 ; υ2 ≥ υ20 ; m  xi = 1; xi ≥ 0, i = 1, 2, ..., m.

i=1

where, υ10 and υ20 are optimal solution of Problem 4.4.4. Problem 4.4.4 Maximize{υ1 } Subject to   m min {wa˜ i j } aiLj (0)+2ai j (1)+aiRj (0)  1≤i≤m

xi ≥ 4 i=1    1− max {u a˜ i j } aiLj (0)+2ai j (1)+aiRj (0) m  1≤i≤m

i=1 m 

4

υ1 ; j = 1, 2, ..., n;

xi ≥ υ2 ; j = 1, 2, ..., n;

xi = 1; xi ≥ 0, i = 1, 2, ..., m.   L υ (0) + 2υ(1) + υ R (0) , υ2 = 1 − max {u a˜ i j } where, υ1 = min {wa˜ i j } 1≤i≤m 1≤i≤m 4 1≤ j≤n 1≤ j≤n   L υ (0) + 2υ(1) + υ R (0) . 4

i=1

4.4 Literature Review of Matrix Games with Intuitionistic Fuzzy Payoffs

Problem 4.4.5 Minimize{ω2 } Subject to   n min {wa˜ i j } aiLj (0)+2ai j (1)+aiRj (0)  1≤ j≤n j=1

4 

n 

1− max {u a˜ i j } 1≤ j≤n

y j ≤ ω1 ; i = 1, 2, ..., m;

  aiLj (0)+2ai j (1)+aiRj (0) 4

j=1

107

y j ≤ ω2 ; i = 1, 2, ..., m;

ω2 ≥ ω1 ; ω1 ≤ ω10 ; ω2 ≤ ω20 n  y j = 1; y j ≥ 0, j = 1, 2, ..., n. j=1

where, ω10 and ω20 are optimal solution of Problem 4.4.6. Problem 4.4.6 Minimize{ω1 } Subject to   n min {wa˜ i j } aiLj (0)+2ai j (1)+aiRj (0)  1≤ j≤n j=1 n  j=1 n 



yj ≤ 4   1− max {u a˜ i j } aiLj (0)+2ai j (1)+aiRj (0) 1≤ j≤n

4

ω1 ; i = 1, 2, ..., m;

y j ≤ ω2 ; i = 1, 2, ..., m;

y j = 1; y j ≥ 0, j = 1, 2, ..., n.   L ω (0) + 2ω(1) + ω R (0) , ω2 = 1 − max {u a˜ i j } where, ω1 = min {wa˜ i j } 1≤i≤m 1≤i≤m 4 1≤ j≤n 1≤ j≤n   L ω (0) + 2ω(1) + ω R (0) . 4 j=1

Li et al. [6] proposed a method to find the minimum expected gain of Player I, maximum expected loss of Player II and their corresponding optimal strategies for matrix games with triangular intuitionistic fuzzy payoffs. In this method, firstly authors have transformed Problems 4.3.1 and Problem 4.3.2 into Problem 4.4.1 and Problem 4.4.2 respectively. Then, it is claimed that the optimal solution {xi∗ , i = 1, 2, ..., m} and {y ∗j , j = 1, 2, ..., n} of Problem 4.4.7 and Problem 4.4.8 respectively represent the optimal strategies of Player I and Player II. The minimum expected gain of Player I n  m  xi∗ a˜ i j y ∗j . and maximum expected loss of Player II is obtained by j=1 i=1

108

4 Matrix Games with Intuitionistic Fuzzy Payoffs

Problem 4.4.7 Maximize{λυ1 + (1 − λ)υ2 } Subject to Constraints of Problem 4.4.1.   L υ (0) + 2υ(1) + υ R (0) , υ2 = 1 − max {u a˜ i j } where, υ1 = min {wa˜ i j } 1≤i≤m 1≤i≤m 4 1≤ j≤n 1≤ j≤n   L υ (0) + 2υ(1) + υ R (0) and λ ∈ [0, 1]. 4 Problem 4.4.8 Minimize{λω1 + (1 − λ)ω2 } Subject to Constraints of Problem 4.4.2.   L ω (0) + 2ω(1) + ω R (0) , where, ω1 = min {wa˜ i j } 1≤i≤m 4 1≤ j≤n

ω2 = 1 − max {u a˜ i j } 

1≤i≤m 1≤ j≤n

ω L (0) + 2ω(1) + ω R (0) 4

 and λ ∈ [0, 1].

Nan et al. [9] proposed a method to find the minimum expected gain of Player I, maximum expected loss of Player II and corresponding optimal strategies for matrix games with triangular intuitionistic fuzzy payoffs. In this method, firstly the authors have transformed Problem Problem 4.3.1 and Problem 4.3.2 into Problem 4.4.9 and Problem 4.4.10 respectively. Then, it is claimed that the optimal solution {xi∗ , i = 1, 2, ..., m} and {y ∗j , j = 1, 2, ..., n} of Problem 4.4.11 and Problem 4.4.13 respectively represent the optimal strategies of Player I and Player II. The minimum expected gain of Player I and maximum expected loss of Player II is obtained by n  m  xi∗ a˜ i j y ∗j . j=1 i=1

Problem 4.4.9 Maximize{V1 }, Minimize{A1 } Subject to   2   m   xi L R 2 ai j (0) + 4ai j + ai j (0) λ min {wa˜ i j } + (1 − λ) 1 − max {u ai j } 1≤i≤m 1≤i≤m 6 i=1 ≥ V1 , j = 1, 2, ..., n;  2   m   xi R L 2 ≤ A1 , ai j (0) − ai j (0) λ min {wai j } + (1 − λ) 1 − max {u a˜ i j } 1≤i≤m 1≤i≤m 6 i=1 j = 1, 2, ..., n; m  xi = 1; xi ≥ 0, i = 1, 2, ..., m. i=1

4.4 Literature Review of Matrix Games with Intuitionistic Fuzzy Payoffs

 where, V1 = ⎛

υ L (0) + 4υ(1) + υ R (0) 6 ⎛

 ⎞2 ⎞

⎜ ⎟ ⎟ ⎜ ⎝λ min {wa˜ i j }2 + (1 − λ) ⎝1 − max {u a˜ i j }⎠ ⎠ and A1 = ⎛

1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

⎛

109



υ R (0) − υ L (0) 6



⎞2 ⎞

⎜ ⎜ ⎟ ⎟ ⎝λ min {wa˜ i j }2 + (1 − λ) ⎝1 − max {u a˜ i j }⎠ ⎠. 1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

Problem 4.4.10 Minimize{V2 }, Maximize{A2 } Subject to   2   n   yj L R 2 ai j (0) + 4ai j + ai j (0) λ min {wa˜ i j } + (1 − λ) 1 − max {u a˜ i j } 1≤ j≤n 1≤ j≤n 6 j=1 ≤ V2 , i = 1, 2, ..., m;  2   n   yj R L 2 ≥ A2 , ai j (0) − ai j (0) λ min {wa˜ i j } + (1 − λ) 1 − max {u a˜ i j } 1≤ j≤n 1≤ j≤n 6 j=1 i = 1, 2, ..., m; n  y j = 1; y j ≥ 0, j = 1, 2, ..., n. j=1



where, V2 =  and A2 =

 ω L (0) + 4ω(1) + ω R (0) 6



⎞2 ⎞

⎛

⎛

⎜ ⎟ ⎟ ⎜ ⎝λ min {wa˜ i j }2 + (1 − λ) ⎝1 − max {u a˜ i j }⎠ ⎠ 1≤i≤m 1≤ j≤n

⎛

⎛

1≤i≤m 1≤ j≤n

⎞2 ⎞

ω (0) − ω (0) ⎜ ⎜ ⎟ ⎟ ⎝λ min {wa˜ i j }2 + (1 − λ) ⎝1 − max {u a˜ i j }⎠ ⎠. 1≤i≤m 1≤i≤m 6 R

L

1≤ j≤n

Problem 4.4.11 Minimize{A1 } Subject to Constraints of Problem 4.4.9 with the additional constraints V1 ≥ V10 ; A1 ≤ A01 . where, V10 and A01 are optimal solution of Problem 4.4.12. Problem 4.4.12 Maximize{V1 } Subject to Constraints of Problem 4.4.9. Problem 4.4.13 Maximize{A2 } Subject to

1≤ j≤n

110

4 Matrix Games with Intuitionistic Fuzzy Payoffs

Constraints of Problem 4.4.10 with the additional constrains V2 ≤ V20 ; A2 ≥ A02 . where, V20 and A02 are optimal solution of Problem 4.4.14. Problem 4.4.14 Minimize{V2 } Subject to Constraints of Problem 4.4.10. Nan et al. [10] proposed a method to find the minimum expected gain of Player I, maximum expected loss of Player II and their corresponding optimal strategies for matrix games with trapezoidal intuitionistic fuzzy payoffs. In this method, firstly the authors have transformed Problem 4.3.1 and Problem 4.3.2 into Problem 4.4.15 and Problem 4.4.16 respectively. Problem 4.4.15 Maximize{V1 }, Minimize{A1 } Subject to   m   L L R R ai j (0) + 2ai j (1) + 2ai j (1) + ai j (0) (1 − λ) min {wa˜ i j } + λ(1 − max {u a˜ i j }) 1≤i≤m

i=1 xi

1≤i≤m

≥ V1 , j = 1, 2, ..., n; 6    m  aiRj (0)aiLj (0) + 2aiRj (1) − 2aiLj (1) λ min {wa˜ i j } + (1 − λ)(1 − max {u a˜ i j })

i=1 xi

6 m 

1≤i≤m

1≤i≤m

≤ A1 , j = 1, 2, ..., n;

xi = 1; xi ≥ 0, i = 1, 2, ..., m.   L υ (0) + 2υ L (1) + 2υ R (1) + υ R (0) where, V1 = 6 ⎛ ⎞

i=1

⎜ ⎟ ⎝(1 − λ) min {wa˜ i j } + λ(1 − max {u a˜ i j })⎠ and  A1 = ⎛

1≤i≤m 1≤ j≤n R

1≤i≤m 1≤ j≤n

 υ (0) − υ L (0) + 2υ R (1) − 2υ L (1) 6 ⎞

⎜ ⎟ ⎝λ min {wa˜ i j } + (1 − λ)(1 − max {u a˜ i j })⎠. 1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

Problem 4.4.16 Minimize{V2 }, Maximize{A2 } Subject to   m   aiLj (0) + 2aiLj (1) + 2aiRj (1) + aiRj (0) (1 − λ) min {wa˜ i j } + λ(1 − max {u a˜ i j }) j=1 xi

6

1≤ j≤n

≤ V2 , j = 1, 2, ..., n;

1≤ j≤n

4.4 Literature Review of Matrix Games with Intuitionistic Fuzzy Payoffs

111

  m   aiRj (0)aiLj (0) + 2aiRj (1) − 2aiLj (1) λ min {wa˜ i j } + (1 − λ)(1 − max {u a˜ i j }) j=1 xi

6 m 

1≤ j≤n

1≤ j≤n

≥ A2 , j = 1, 2, ..., n;

y j = 1; y j ≤ 0, j = 1, 2, ..., n.   L ω (0) + 2ω L (1) + 2ω R (1) + ω R (0) where, V2 = 6 ⎛ ⎞ j=1

⎜ ⎟ ⎝(1 − λ) min {wa˜ i j } + λ(1 − max {u a˜ i j })⎠ and  A2 = ⎛

1≤i≤m 1≤ j≤n R

1≤i≤m 1≤ j≤n

ω (0) − ω L (0) + 2ω R (1) − 2ω L (1) 6 ⎞



⎜ ⎟ ⎝λ min {wa˜ i j } + (1 − λ)(1 − max {u a˜ i j })⎠. 1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

Then, it is claimed that the optimal solution {xi∗ , i = 1, 2, ..., m} and {y ∗j , j = 1, 2, ..., n} of Problem 4.4.17 and Problem 4.4.18 respectively represent the optimal strategies of Player I and Player II. The minimum expected gain of Player I and n  m  xi∗ a˜ i j y ∗j . maximum expected loss of Player II is obtained by j=1 i=1

Problem 4.4.17 Maximize{V1 − A1 } Subject to Constraints of Problem 4.4.15. Problem 4.4.18 Minimize{V2 − A2 } Subject to Constraints of Problem 4.4.16. Seikh et al. [11] proposed a method to find the minimum expected gain of Player I, maximum expected loss of Player II and their corresponding optimal strategies for matrix games with triangular intuitionistic fuzzy payoffs. In this method, the authors have transformed Problem 4.3.1 and Problem 4.3.2 into Problem 4.4.19 and Problem 4.4.20 respectively. Problem 4.4.19   υ1 + υ2 Maximize 2 Subject to ⎛ ⎞ m a L (0)+4a (1)+a R (0) m a R (0)−a L (0)   ij ij ij ij ij 1 1 min {w } x − min {1 − u } x + i a˜ i j i ⎟ 6wa˜ i j 2 1≤i≤m 3(1−u a˜ i j ) ⎜ 2 1≤i≤m a˜ i j i=1 i=1 ⎜ ⎟ m m L R R L  ai j (0)+4ai j (1)+ai j (0)  ai j (0)−ai j (0) ⎠ ⎝ 1 1 min {1 − u } x − min {w } x a˜ i j i a˜ i j i 2 6(1−u a˜ ) 2 3wa˜ 1≤i≤m

i=1

ij

1≤i≤m

i=1

ij

112

4 Matrix Games with Intuitionistic Fuzzy Payoffs

υ 1 + υ2 , j = 1, 2, ..., n; 2 υ1 ≤ υ2 ; m  xi = 1; xi ≥ 0, i = 1, 2, ..., m.

≥

i=1

υ L (0) + 4υ(1) + υ R (0) υ R (0) − υ L (0) wυ˜ − wυ˜ and 6 3 υ L (0) + 4υ(1) + υ R (0) υ R (0) − υ L (0) υ2 = (1 − u υ˜ ) − (1 − u υ˜ ). 6 3

where, υ1 =

Problem  4.4.20  ω1 + ω2 Minimize 2 Subject to ⎛ ⎞ n a L (0)+4a (1)+a R (0) n a R (0)−a L (0)   ij ij ij ij ij 1 1 min {w } x − min {1 − u } x + a ˜ i a ˜ i i j i j 6wa˜ i j 2 1≤ j≤n 3(1−u a˜ i j ) ⎜ 2 1≤ j≤n ⎟ j=1 j=1 ⎜ ⎟ n a L (0)+4a (1)+a R (0) n a R (0)−a L (0)   ⎝ 1 ⎠ ij ij ij ij ij 1 min {1 − u } x − min {w } x a ˜ i a ˜ i ij ij 2 6(1−u a˜ ) 2 3wa˜ 1≤ j≤n

j=1

ij

ω1 + ω2 ≤ , i = 1, 2, ..., m; 2 ω1 ≤ ω2 ; n  y j = 1; y j ≥ 0, j = 1, 2, ..., n.

1≤ j≤n

j=1

ij

j=1

ω L (0) + 4ω(1) + ω R (0) ω R (0) − ω L (0) wω˜ − wω˜ and 6 3 L R R L ω (0) + 4ω(1)1 + ω (0) υ (0) − υ (0) (1 − u ω˜ ) − (1 − u ω˜ ). ω2 = 6 3 Then, it is claimed that optimal solution {xi , i = 1, 2, ..., m} of Problem 4.4.19 represents the optimal strategies of Player I as well as optimal solution {y j , j = 1, 2, ..., n} of Problem 4.4.22 represents the optimal strategies of Player II and $the #   n  m  value of game is xi aiLj (0), ai j (1) + aiRj (0) y j , min {wa˜ i j }, max {u a˜ i j } . where, ω1 =

j=1 i=1

1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

4.5 Flaws of the Existing Methods In this section, flaws of the existing methods [6, 8–10] are pointed out. 1. In the existing method [8], Problem 4.4.1 is solved to obtain the minimum ex˜ pected gain of Player I which is obtained by assuming that a˜ b˜ if Sμ (a) ˜ ≤ Sμ (b) ˜ and Sν (a) ˜ ≤ Sν (b), which contradicts the claim of the authors [8, Definition 7] that ˜ or if Sμ (a) ˜ then Sν (a) ˜ Similarly, the aua˜ b˜ if Sμ (a) ˜ < Sμ (b) ˜ = Sμ (b) ˜ < Sν (b). ˜ thors [8] have obtained the Problem 4.4.2 by assuming that a˜ b˜ if Sμ (a) ˜ ≥ Sμ (b) ˜ which contradicts the claim of the authors [8, Definition 7] and Sν (a) ˜ ≥ Sν (b),

4.5 Flaws of the Existing Methods

113

˜ or if Sμ (a) ˜ then Sν (a) ˜ where, that a˜ b˜ if Sμ (a) ˜ > Sμ (b) ˜ = Sμ (b) ˜ > Sν (b), a L (0)+a(1)+a R (0) Sμ = wa˜ and 4 L R (0) Sν = (1 − u a˜ ) a (0)+a(1)+a . 4   2. In the existing method [6], authors have assumed if a˜ = a L (0), a(1), a R (0) ,  L   wa˜ , u a˜  and b˜ = b (0), b(1), b R (0) , wb˜ , u b˜ are two triangular vague sets (triangular intuitionistic fuzzy numbers) then ˜ and Sν (a) ˜ (i) a˜ b˜ if Sμ (a) ˜ ≤ Sμ (b) ˜ ≤ Sν (b). ˜ ˜ ˜ (ii) a˜ = b if Sμ (a) ˜ = Sμ (b) and Sν (a) ˜ = Sν (b). It is obvious that to find the minimum expected gain of Player I of those matrix games in which %payoffs are represented by triangular intuitionistic fuzzy numbers & ' n  m m   i.e., minimum xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j | xi = 1, j=1 i=1 i=1 ( n  y j = 1, xi ≥ 0, i = 1, 2, ..., m, y j ≥ 0, j = 1, 2, ..., n , there is need to find j=1

& n  m  such y ∗j ∈ Y corresponding to which value of xi (aiLj (0), ai j (1), aiRj (0)), j=1 i=1  wa˜ i j , u a˜ i j y j is minimum ∀(xi , x2 , ..., xm ) ∈ X i.e., there is need to find such y ∗j ∈ Y corresponding to which value of ) * & ' n  m  L R Sμ xi (ai j (0), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y j as well as ) j=1 i=1 * & ' n  m  xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j will be minimum i.e., there Sν j=1 i=1

is need to find the optimal solution {y ∗j } of Problem 4.5.1. Problem % 4.5.1) Minimize Sμ

n  m  j=1 i=1

%

)

Minimize Sν

n  m  j=1 i=1

& ' xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j & ' xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j

*( *(

Subject to m n   xi = 1; y j = 1; i=1

j=1

xi ≥ 0, i = 1, 2, ..., m; y j ≥ 0, j = 1, 2, ..., n. Since, Problem 4.5.1 is a multi-objective problem. So, optimal solution of Problem 4.5.1 can be obtained by transforming into a single objective problem i.e., Problem 4.5.2 to find the optimal solution of Problem 4.5.1 is equivalent to find the optimal solution of Problem 4.5.2.

114

4 Matrix Games with Intuitionistic Fuzzy Payoffs

Problem% 4.5.2 ) ) Minimize λ Sμ

** & ' + xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j j=1 i=1 ) ) **( & ' n  m  L R (1 − λ) Sν xi (ai j (0), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y j m n  

j=1 i=1

Subject to m n   xi = 1; y j = 1; i=1

j=1

xi ≥ 0, i = 1, 2, ..., m; y j ≥ 0, j = 1, 2, ..., n. Since, in Problem 4.5.2 only y j are considered as decision variables so Problem 4.5.2 is a linear programming problem and hence, the optimal value of Problem 4.5.2 is equal to the optimal value of its corresponding dual problem i.e., Problem 4.5.3. Problem 4.5.3   Maximize λSμ (υ) ˜ + (1 − λ)Sν (υ) ˜ Subject ⎧ ⎛ to ⎛ ⎞⎞ $ # ⎪ ⎨ m ⎜ ⎜ ⎟⎟ λ ⎝ Sμ ⎝ xi (aiLj (0), ai j (1), aiRj (0)), min {wa˜ i j }, max {u a˜ i j } ⎠⎠ + ⎪ 1≤i≤m 1≤i≤m i=1 ⎩ 1≤ j≤n 1≤ j≤n ⎞⎞⎫ ⎛ ⎛ $ # ⎪ ⎬ m ⎟⎟ ⎜ ⎜ L R (1 − λ) ⎝ Sν ⎝ xi (ai j (0), ai j (1), ai j (0)), min {wa˜ i j }, min {u a˜ i j } ⎠⎠ ≥ ⎪ 1≤i≤m 1≤i≤m i=1 ⎭ 1≤ j≤n 1≤ j≤n   ˜ + (1 − λ)Sν (υ) ˜ , j = 1, 2, ..., n λSμ (υ) m  xi = 1; xi ≥ 0, i = 1, 2, ..., m.

i=1

Since, λx1 + (1 − λ)x2 ≥ λy1 + (1 − λ)y2  x1 ≥ y1 and x2 ≥ y2 . So, Problem 4.5.3 cannot be transformed into Problem 4.5.4.

Problem 4.5.4   Maximize λSμ (υ) ˜ + (1 − λ)Sν (υ) ˜ Subject to ⎛

⎞ # $ m  ⎜ ⎟ Sμ ⎝ xi (aiLj (0), ai j (1), aiRj (0)), min {wa˜ i j }, max {u a˜ i j } ⎠ ≥ Sμ (υ), ˜ j = 1, 2, ..., n; ⎛ ⎜ Sν ⎝

i=1 m 

i=1

m 

#

1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

$

⎞

⎟ xi (aiLj (0), ai j (1), aiRj (0)), min {wa˜ i j }, min {u a˜ i j } ⎠ ≥ Sν (υ), ˜ j = 1, 2, ..., n; 1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

xi = 1; xi ≥ 0, i = 1, 2, ..., m.

i=1

While, in the existing method [6], authors have transformed Problem 4.5.3 into Problem 4.5.4, which is mathematically incorrect. Similarly, in the existing method [6], mathematically wrong assumption is considered to find the maximum expected loss of Player II.

4.5 Flaws of the Existing Methods

115

3. In the existing method [9], Problem 4.4.11 is solved to obtain the minimum ex˜ ˜ ≤ Vλ (b) pected gain of Player I which is obtained by assuming that a˜ b˜ if Vλ (a) ˜ which contradicts the claim of the authors [9, Definition 7] ˜ ≥ Aλ (b), and Aλ (a) ˜ or if Vλ (a) ˜ then Aλ (a) ˜ Simithat a˜ b˜ if Vλ (a) ˜ < Vλ (b) ˜ = Vλ (b) ˜ > Aλ (b). larly, the authors have obtained the Problem 4.4.15 by assuming that a˜ b˜ ˜ and Aλ (a) ˜ This contradicts the claim of the au˜ ≥ Vλ (b) ˜ ≤ Aλ (b). if Vλ (a) ˜ or if Vλ (a) ˜ then ˜ > Vλ (b) ˜ = Vλ (b) thors [9, Definition 7] that a˜ b˜ if Vλ (a) ˜ where, ˜ < Aλ (b), Aλ (a) (λwa2˜ + (1 − λ)(1 − u a˜ )2 )(a L (0) + 2a(1) + a R (0)) and Vλ = 6 2 2 R L (λwa˜ + (1 − λ)(1 − u a˜ ) )(a (0) − a (0)) Aλ = . 6  4. In the existing method [10], authors have assumed that if a˜ = a L (0), a L (1),      a R (1), a R (0) , wa˜ , u a˜ and b˜ = b L (0), b L (1), b L (1), b R (0) , wb˜ , u b˜ are two ˜ and Aλ (a) ˜ ≤ Vλ (b) ˜ ≥ trapezoidal intuitionistic fuzzy numbers then a˜ b˜ if Vλ (a) ˜ Aλ (b), where, (λ(1 − u a˜ ) + (1 − λ)wa˜ )(a L (0) + 2a L (1) + 2a R (1) + a R (0)) Vλ = and 6 R L R L (λwa˜ + (1 − λ)(1 − u a˜ ))(a (0) − a (0) + 2a (1) − 2a (1)) Aλ = . 3 It is obvious that to find the minimum expected gain of Player I of those matrix games in which %payoffs are represented by triangular intuitionistic fuzzy numbers & ' n  m m   i.e., minimum xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j | xi = 1, j=1 i=1 ( i=1 n  y j = 1, xi ≥ 0, i = 1, 2, ..., m, y j ≥ 0, j = 1, 2, ..., n , there is need to find j=1

& n  m  such y ∗j ∈ Y corresponding to which value of xi (aiLj (0), ai j (1), aiRj (0)), j=1 i=1  wa˜ i j , u a˜ i j y j is minimum ∀(xi , x2 , ..., xm ) ∈ X i.e., there is need to find such y ∗j ∈ Y corresponding to which value of ) * & ' n  m  L R Vλ xi (ai j (0), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y j is minimum and value of j=1 i=1 ) * & ' n  m  L R xi (ai j (0), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y j is maximum ∀(x1 , x2 , ..., xm ) Aλ j=1 i=1

∈ X i.e., there is need to find the optimal solution {y ∗j } of Problem 4.5.5. Problem % 4.5.5)

n  m 

&

xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j

'

*(

yj Minimize Vλ % )j=1 i=1 *( & ' n  m  Maximize Aλ xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j j=1 i=1

Subject to

116 m 

4 Matrix Games with Intuitionistic Fuzzy Payoffs

xi = 1;

i=1

n 

y j = 1;

j=1

xi ≥ 0, i = 1, 2, ..., m; y j ≥ 0, j = 1, 2, ..., n. Further, to find optimal solution of Problem 4.5.5, it is equivalent to find the optimal solution of Problem 4.5.6. Problem 4.5.6 ⎧ ) * & ' n  m ⎪  ⎪ L R ⎪ xi (ai j (0), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y j − ⎪ ⎨Vλ j=1 i=1 ) * Minimize & ' n  m ⎪  ⎪ ⎪ xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j ⎪ ⎩ Aλ j=1 i=1

⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭

Subject to m n   xi = 1; y j = 1; i=1

j=1

xi ≥ 0, i = 1, 2, ..., m; y j ≥ 0, j = 1, 2, ..., n. Since, in Problem 4.5.6, only y j has been considered as decision variable. So, Problem 4.5.6 is a linear programming problem and hence, the optimal value of Problem 4.5.6 will be equal to optimal value of its dual problem i.e., Problem 4.5.7. Problem 4.5.7 Maximize{Vλ (υ) ˜ − Aλ (υ)} ˜ Subject to ⎧ ⎛ ⎞ $ # ⎪ ⎪ m ⎪ ⎪ ⎜ ⎟ ⎪ Vλ ⎝ xi (aiLj (0), ai j (1), aiRj (0)), min {wa˜ i j }, max {u a˜ i j } ⎠ − ⎪ ⎪ ⎪ 1≤i≤m 1≤i≤m i=1 ⎨ 1≤ j≤n 1≤ j≤n ⎛ ⎞ $ # ⎪ ⎪ m ⎪ ⎪ ⎜ ⎟ ⎪ Aλ ⎝ xi (aiLj (0), ai j (1), aiRj (0)), min {wa˜ i j }, min {u a˜ i j } ⎠ ⎪ ⎪ ⎪ 1≤i≤m 1≤i≤m i=1 ⎩ 1≤ j≤n

m 

1≤ j≤n

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

≥

Vλ (υ) ˜ − Aλ (υ), ˜ j = 1, 2, ..., n xi = 1; xi ≥ 0, i = 1, 2, ..., m.

i=1

Since, x1 − x2 ≥ y1 − y2  x1 ≥ y1 and x2 ≤ y2 . So, Problem 4.5.7 cannot be transformed into Problem 4.5.8. Problem 4.5.8 ˜ − Aλ (υ)} ˜ Maximize{Vλ (υ) Subject to ⎛ ⎞ $ # m ⎜ ⎟ Vλ ⎝ xi (aiLj (0), ai j (1), aiRj (0)), min {wa˜ i j }, max {u a˜ i j } ⎠ i=1

1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

˜ j = 1, 2, ..., n; ≥ V⎛ λ (υ), ⎞ $ # m  ⎜ ⎟ Aλ ⎝ xi (aiLj (0), ai j (1), aiRj (0)), min {wa˜ i j }, min {u a˜ i j } ⎠ i=1

1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

4.5 Flaws of the Existing Methods

117

≤ Aλ (υ), ˜ j = 1, 2, ..., n; m  xi = 1; xi ≥ 0, i = 1, 2, ..., m. i=1

While, in the existing method [10], authors have transformed Problem 4.5.7 into Problem 4.5.8, which is mathematically incorrect. Similarly, in the existing method [10], mathematically wrong assumption is considered to find the maximum expected loss of Player II.

4.6 Proposed Ambika Methods It is obvious from Sect. 4.5 that different methods for comparing triangular/trapezoidal intuitionistic fuzzy numbers are used in published papers [6, 8–10]. In this section, new methods (named as Ambika methods), corresponding to different existing methods, are proposed for finding the minimum expected gain of Player I, maximum expected loss of Player II and their corresponding optimal strategies of matrix games with triangular/trapezoidal intuitionistic fuzzy payoffs.

4.6.1 Ambika Method-I In the existing method [8], it is assumed that if a˜ =< (a L (0), a(1), a R (0)), wa˜ , u a˜ > and b˜ =< (b L (0), b(1), b R (0)), wa˜ , u a˜ > are two triangular intuitionistic fuzzy num˜ and if Sμ (a) ˜ then Sν (a) ˜ where, ˜ ≤ Sμ (b) ˜ = Sμ (b) ˜ ≤ Sν (b) bers then a˜ b˜ if Sμ (a) L R L w ˜ (a (0) + 2a(1) + a (0)) (1 − u A˜ )(a (0) + 2a(1) + a R (0)) and Sν = . Sμ = A 4 4 In this section, a new method, based on this comparing method, is proposed to find the minimum expected gain of Player I, maximum expected loss of Player II and their corresponding optimal strategies.

4.6.1.1

Minimum Expected Gain of Player I

˜ and if Sμ (a) ˜ then ˜ ≤ Sμ (b) ˜ = Sμ (b) Using the comparing method, a˜ b˜ if Sμ (a) ˜ the minimum expected gain of Player I and the corresponding optiSν (a) ˜ ≤ Sν (b), mal strategies can be obtained as follows: Step 1: According to comparing method, to find the optimal solution {y1∗ , y2∗ , ..., yn∗ } & ' n  m  such that value of xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j is minimum for j=1 i=1

all (x1 , x2 , ..., xm ) ∈ X , firstly it is equivalent to find {y1∗ , y2∗ , ..., yn∗ } such that

118

4 Matrix Games with Intuitionistic Fuzzy Payoffs

) value of Sμ

n  m  j=1 i=1

* & ' L R xi (ai j (0), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y j , is minimum for all

(x1 , x2 , ..., xm ) ∈ X i.e., find optimal solution {y ∗j , j = 1, 2, ..., n} of Problem 4.6.1. Problem % 4.6.1) Minimize Sμ

n  m  j=1 i=1

& ' xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j

*(

Subject to m n   xi = 1; xi ≥ 0, i = 1, 2, ..., m; y j = 1; y j ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 2: Using the property, λ < (aiL (0), ai (1), aiR (0)), wa˜ i , u a˜ i >=< (λaiL (0), λai (1), λaiR (0)), wa˜ i , u a˜ i >, n n n    < (aiL (0), ai (1), aiR (0)), wa˜ i , u a˜ i >=< ( aiL (0), ai (1), λ ≥ 0 and n  i=1

i=1

aiR (0)),

i=1

i=1

min wa˜ i , max u a˜ i >, Problem 4.6.1 can be transformed into Problem

1≤i≤n

1≤i≤n

4.6.2. Problem 4.6.2 Minimize ⎧ ⎞⎫ ⎛ $ ⎪ # ⎪ ⎬ ⎨ m m m n n n     ⎟ ⎜   xi aiLj (0)y j , xi ai j (1)y j , xi aiRj (0))y j , min {wa˜ i j }, max {u a˜ i j } ⎠ Sμ ⎝ ( ⎪ ⎪ 1≤i≤m 1≤i≤m j=1 i=1 j=1 i=1 j=1 i=1 ⎭ ⎩ 1≤ j≤n

1≤ j≤n

Subject to m n   xi = 1; xi ≥ 0, i = 1, 2, ..., m; y j = 1; y j ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 3: Since, in Problem 4.6.2, only y j have been considered as decision variables. So, Problem 4.6.2 is a linear programming problem and hence, the optimal value of Problem 4.6.2 will be equal to optimal value of its corresponding dual problem i.e., Problem 4.6.3. Problem 4.6.3 Maximize{υ} Subject ⎛ to ⎞ $ # m m m   ⎜  ⎟ xi ai j (1), xi aiRj (0)), min {wa˜ i j }, max {u a˜ i j } ⎠ ≥ υ, Sμ ⎝ ( xi aiLj (0), i=1

j = 1, 2, ..., n;

i=1

m 

i=1

1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

xi = 1; xi ≥ 0, i = 1, 2, ..., m.

i=1

Step 4: Find the optimal solution of Problem 4.6.3. Step 5: Substitute the optimal solution {xi∗ , i = 1, 2, ..., m} of Problem 4.6.3 in Problem 4.6.1 and then find the optimal solution {y ∗j , j = 1, 2, ..., n} of Problem 4.6.1.

4.6 Proposed Ambika Methods

119

Case 1: If there exist a unique optimal solution {y ∗j , j = 1, 2, ..., n} of Problem 4.6.1 & n  m  xi∗ (aiLj (0), ai j (1), aiRj (0)), then the minimum expected gain of Player I is j=1 i=1  wa˜ i j , u a˜ i j y ∗j and the corresponding optimal strategy for Player I will be {xi∗ , i = 1, 2, ..., m}, which is optimal solution of Problem 4.6.3. Case 2: If there exist more than one basic optimal solution {y j1 , j = 1, 2, ..., n}, 1, 2,)..., n}, ..., {y j p , j = 1, 2, ..., n} of Problem 4.6.1 then find {y j2 , j = % *( & ' n  m  minimum Sν xi∗ (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y ∗js , s = 1, 2, ..., p . j=1 i=1 ) * & ' n  m  ∗ L R ∗ xi (ai j (0), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y jr represents the minimum If Sν j=1 i=1

& n  m  value then minimum expected gain of Player I is xi∗ (aiLj (0), ai j (1), aiRj (0)), j=1 i=1  wa˜ i j , u a˜ i j y ∗jr and the corresponding optimal strategy for Player I will be {xi∗ , i = 1, 2, ..., m}, which is optimal solution of Problem 4.6.3.

4.6.1.2

Maximum Expected Loss of Player II

˜ and if Sμ (a) ˜ then Using the comparing method, a˜ b˜ if Sμ (a) ˜ ≥ Sμ (b) ˜ = Sμ (b) ˜ the maximum expected loss of Player II and the corresponding optiSν (a) ˜ ≥ Sν (b), mal strategies can be obtained as follows: Step 1: According to comparing method, to find the optimal solution {x1∗ , x2∗ , ..., xm∗ } & ' n  m  such that value of xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j is maximum for all j=1 i=1

, y2 , ..., yn ) ∈ Y , firstly it is equivalent to find {x*1∗ , x2∗ , ..., xm∗ } such that value of (y1) & ' n  m  xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j , is maximum for all Sμ j=1 i=1

(y1 , y2 , ..., yn ) ∈ Y i.e., find optimal solution {xi∗ , i = 1, 2, ..., m} of Problem 4.6.4. Problem 4.6.4 % ) Maximize Sμ

n  m  j=1 i=1

& ' xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j

*(

Subject to m n   xi = 1; xi ≥ 0, i = 1, 2, ..., m; y j = 1; y j ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 2: Using the property, λ < (aiL (0), ai (1), aiR (0)), wa˜ i , u a˜ i >=< n  < (aiL (0), ai (1), (λaiL (0), λai (1), λaiR (0)), wa˜ i , u a˜ i >, λ ≥ 0 and i=1

120

4 Matrix Games with Intuitionistic Fuzzy Payoffs

aiR (0)), wa˜ i , u a˜ i >=< (

n 

i=1

aiL (0),

n 

ai (1),

i=1

n  i=1

aiR (0)), min wa˜ i , max u a˜ i >, Prob1≤i≤n

1≤i≤n

lem 4.6.4 can be transformed into Problem 4.6.5. Problem 4.6.5 Maximize ⎧ ⎛

⎞⎫ $ ⎪ ⎬ ⎜ ⎟ Sμ ⎝ ( xi aiLj (0)y j , xi ai j (1)y j , xi aiRj (0))y j , min {wa˜ i j }, max {u a˜ i j } ⎠ ⎪ ⎪ 1≤i≤m 1≤i≤m j=1 i=1 j=1 i=1 j=1 i=1 ⎩ ⎭ ⎪ ⎨

#

n  m 

n  m 

n  m 

1≤ j≤n

1≤ j≤n

Subject to m n   xi = 1; xi ≥ 0, i = 1, 2, ..., m; y j = 1; y j ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 3: Since, in Problem 4.6.5, only xi have been considered as decision variables. So, Problem 4.6.5 is a linear programming problem and hence, the optimal value of Problem 4.6.5 will be equal to optimal value of its corresponding dual problem i.e., Problem 4.6.6. Problem 4.6.6 Minimize{ω} Subject ⎛ to ⎞ $ # n m m   ⎜  ⎟ Sμ ⎝ ( aiLj (0)y j , ai j (1)y j , aiRj (0))y j , min {wa˜ i j }, max {u a˜ i j } ⎠ ≤ ω, j=1

i=1

i=1

1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

i = 1, 2, ..., m; n  y j = 1; y j ≥ 0, j = 1, 2, ..., n. j=1

Step 4: Find the optimal solution of Problem 4.6.6. Step 5: Substitute the optimal solution {y ∗j , j = 1, 2, ..., n} of Problem 4.6.6 in 4.6.5 and then find the optimal solution {xi∗ , i = 1, 2, ..., m} of Problem 4.6.4. Case 1: If there exist a unique optimal solution {xi∗ , i = 1, 2, ..., m} of Problem 4.6.4 & n  m  xi∗ (aiLj (0), ai j (1), aiRj (0)), then the maximum expected loss of Player II is j=1 i=1  wa˜ i j , u a˜ i j y ∗j and the corresponding optimal strategy for Player II will be {y ∗j , j = 1, 2, ..., n}, which is optimal solution of Problem 4.6.6. Case 2: If there exist more than one basic optimal solution {xi1 , i = 1, 2, ..., m}, {xi2 , i = 1, %2, ..., *( ) m}, ..., {xi p , j = 1, 2, ..., m} of Problem 4.6.4 then find & ' n  m  ∗ L R ∗ maximum Sν xis (ai j (0), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y j , s = 1, 2, ..., p . j=1 i=1 ) * & ' n  m  ∗ L R ∗ xir (ai j (0), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y j represents the maximum If Sν j=1 i=1

& n  m  value then maximum expected loss of Player II is xir∗ (aiLj (0), ai j (1), aiRj (0)), j=1 i=1  wa˜ i j , u a˜ i j y ∗j and the corresponding optimal strategy for Player II will be {y ∗j , j = 1, 2, ..., n}, which is optimal solution of Problem 4.6.6.

4.6 Proposed Ambika Methods

121

4.6.2 Ambika Method-II In the existing method [6], authors have assumed if a˜ =< (a L (0), a(1), a R (0)), wa˜ , u a˜ > and b˜ =< (b L (0), b(1), b R (0)), wa˜ , u a˜ > are two triangular intuitionistic ˜ and Sν (a) ˜ where, ˜ ≤ Sμ (b) ˜ ≤ Sν (b) fuzzy numbers then a˜ b˜ if Sμ (a) w A˜ (a L (0) + 2a(1) + a R (0)) (1 − u A˜ )(a L (0) + 2a(1) + a R (0)) Sμ = and Sν = . 4 4 In this section, a new method, based on this comparing method [6], is proposed to find the minimum expected gain of Player I, maximum expected loss of Player II and their corresponding optimal strategies.

4.6.2.1

Minimum Expected Gain of Player I

˜ and Sν (a) ˜ the min˜ ≤ Sμ (b) ˜ ≤ Sν (b), Using the comparing method, a˜ b˜ if Sμ (a) imum expected gain of Player I and the corresponding optimal strategies can be obtained as follows: Step 1: According to comparing method, to find the optimal solution {y1∗ , y2∗ , ..., yn∗ } & ' n  m  such that value of xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j is minimum for all j=1 i=1

∗ , x2 , ..., xm ) ∈ X is equivalent to find {y1∗ , y2∗ , ..., (x1) * yn } such that values of & ' n  m  xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j and Sμ j=1 i=1 ) * & ' n  m  L R xi (ai j (0), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y j is minimum for all (x1 , x2 , ..., Sν j=1 i=1

∗ xm ))∈ X or if it is not possible to find {y1∗ , y2∗ , ..., y* n } for which the value of & ' n  m  xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j and Sμ j=1 i=1 ) * & ' n  m  L R xi (ai j (0), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y j is minimum then find such Sν j=1 i=1 ) & n  m  xi1 (aiLj (0), ai j (1), aiRj (0)), {y ∗j1 , j = 1, 2, ..., n} for which the value of Sμ ) j=1 i=1 & n  m    xi1 (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j1 is minimum but value of Sν j=1 i=1   wa˜ i j , u a˜ i j y j1 is not minimum for all (x11 , x21 , ..., xm1 ) ∈ X and find such {y ∗j2 , j = 1, 2, ) ..., n} for which the value of * & ' n  m  L R Sν xi2 (ai j (0), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y j2 is minimum but value of ) j=1 i=1 * & ' n  m  xi2 (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j2 is not minimum for all Sμ j=1 i=1

122

4 Matrix Games with Intuitionistic Fuzzy Payoffs

(x12 , x22 , ..., xm2 ) ∈ X i.e., find optimal solution {y ∗j1 , j = 1, 2, ..., n} and {y ∗j2 , j = 1, 2, ..., n} of Problem 4.6.7 and 4.6.8 respectively. Problem % 4.6.7) Minimize Sμ

n  m  j=1 i=1

& ' xi1 (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j1

*(

Subject to m n   xi1 = 1; xi1 ≥ 0, i = 1, 2, ..., m; y j1 = 1; y j1 ≥ 0, j = 1, 2, ..., n. i=1

j=1

Problem % 4.6.8) Minimize Sν

n  m  j=1 i=1

& ' xi2 (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j2

*(

Subject to m n   xi2 = 1; xi2 ≥ 0, i = 1, 2, ..., m; y j2 = 1; y j2 ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 2: Using the property, λ < (aiL (0), ai (1), aiR (0)), wa˜ i , u a˜ i >=< (λaiL (0), n  < (aiL (0), ai (1), aiR (0)), wa˜ i , u a˜ i >=< λai (1), λaiR (0)), wa˜ i , u a˜ i >, λ ≥ 0 and (

n 

i=1

aiL (0),

n 

ai (1),

i=1

n  i=1

i=1

aiR (0)),

min wa˜ i , max u a˜ i >, Problem 4.6.7 and Problem

1≤i≤n

1≤i≤n

4.6.8 can be transformed into Problem 4.6.9 and Problem 4.6.10 respectively. Problem 4.6.9 Minimize ⎧ ⎛

⎞⎫ $ ⎪ # ⎬ m m m n  n  n    ⎟ ⎜  L R xi1 ai j (0)y j1 , xi1 ai j (1)y j1 , xi1 ai j (0))y j1 , min {wa˜ i j }, max {u a˜ i j } ⎠ Sμ ⎝ ( ⎪ ⎪ 1≤i≤m 1≤i≤m j=1 i=1 j=1 i=1 j=1 i=1 ⎩ ⎭ ⎪ ⎨

1≤ j≤n

1≤ j≤n

Subject to m n   xi1 = 1; xi1 ≥ 0, i = 1, 2, ..., m; y j1 = 1; y j1 ≥ 0, j = 1, 2, ..., n. i=1

j=1

Problem 4.6.10 Minimize ⎧ ⎞⎫ ⎛ $ ⎪ # ⎪ ⎨ ⎬ n  n  n  m m m    ⎟ ⎜ xi2 aiLj (0)y j2 , xi2 ai j (1)y j2 , xi2 aiRj (0))y j2 , min {wa˜ i j }, max {u a˜ i j } ⎠ Sν ⎝ ( ⎪ ⎪ 1≤i≤m 1≤i≤m j=1 i=1 j=1 i=1 j=1 i=1 ⎩ ⎭ 1≤ j≤n

1≤ j≤n

Subject to m n   xi2 = 1; xi2 ≥ 0, i = 1, 2, ..., m; y j2 = 1; y j2 ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 3: Since, in Problems 4.6.9 and 4.6.10, only y j1 and y j2 have been considered as decision variables. So, Problem 4.6.9 and Problem 4.6.10 are linear programming problems and hence, the optimal value of Problem 4.6.9 and Problem 4.6.10 will be equal to optimal value of its corresponding dual problem i.e., Problem 4.6.11 and Problem 4.6.12 respectively.

4.6 Proposed Ambika Methods

123

Problem 4.6.11 Maximize{υ1 } Subject to ⎛ m ⎜

⎞

⎟ (aiLj (0), ai j (1), aiRj (0)), min {wa˜ i j }, max {u a˜ i j } xi1 ⎠ ≥ υ1 , j = 1, 2, ..., n;

Sμ ⎝

1≤i≤m 1≤ j≤n

i=1

m 

$

#

1≤i≤m 1≤ j≤n

xi1 = 1; xi1 ≥ 0, i = 1, 2, ..., m.

i=1

Problem 4.6.12 Maximize{υ2 } Subject to ⎛ m ⎜

#

Sν ⎝

i=1

m 

$

⎞

⎟ (aiLj (0), ai j (1), aiRj (0)), min {wa˜ i j }, max {u a˜ i j } xi2 ⎠ ≥ υ2 , j = 1, 2, ..., n; 1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

xi2 = 1; xi2 ≥ 0, i = 1, 2, ..., m.

i=1

∗ ∗ } and {xi2 } of Problem 4.6.11 and Problem Step 4: Find the optimal solution {xi1 4.6.12 respectively. ∗ ∗ , i = 1, 2, ..., m} ans {xi2 , i = 1, 2, ..., m} Step 5: Substitute the optimal solution {xi1 of Problem 4.6.11 and Problem 4.6.12 in Problem 4.6.9 and Problem 4.6.10 respectively and then find all the alternative basic optimal solutions {y 1∗ j1 , j = r∗ 1∗ , j = 1, 2, ..., n}, ..., {y , j = 1, 2, ..., n} and {y , j = 1, 2, ..., n}, 1, 2, ..., n},{y 2∗ j1 jn j2 s∗ 2∗ {y j2 , j = 1, 2, ..., n}, ..., {y j1 , j = 1, 2, ..., n} of Problem 4.6.9 and Problem 4.6.10 respectively. Step ⎫ ⎧ n 6:mFind&minimum '  ⎪ ⎪ ⎪ ⎪ xi1 (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y h∗ , h = 1, 2, ..., r, ⎬ ⎨ j1 j=1 i=1 . & ' n  m  ⎪ ⎪ ⎪ ⎪ xi2 (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y k∗ , k = 1, 2, ..., s ⎭ ⎩ j2 j=1 i=1

Step 7: All the minimum triangular intuitionistic fuzzy numbers will represent the ∗ minimum expected gain of Player I. The optimal strategy for Player I will be {xi1 ,i = 1, 2, ..., m}, which is optimal solution of Problem 4.6.11 if minimum is obtained cor& ' n  m  xi1 (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y h∗ responding to j1 , h = 1, 2, ..., r and j=1 i=1

∗ , i = 1, 2, ..., m}, which is optimal solution of Problem 4.6.12 if miniwill be {xi2 & ' n  m  xi2 (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y k∗ mum is obtained corresponding to j2 ,

k = 1, 2, ..., s.

j=1 i=1

124

4 Matrix Games with Intuitionistic Fuzzy Payoffs

4.6.2.2

Maximum Expected Loss of Player II

˜ and Sν (a) ˜ the max˜ ≥ Sμ (b) ˜ ≥ Sν (b), Using the comparing method, a˜ b˜ if Sμ (a) imum expected loss of Player II and the corresponding optimal strategies can be obtained as follows: Step 1: According to comparing method, to find the optimal solution {x1∗ , x2∗ , ..., xm∗ } & ' m n   such that value of xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j is maximum for all j=1 i=1

, y2 , ..., yn ) ∈ Y is equivalent to find {x1∗ , x2∗ , ...,*xm∗ } such that values of (y1) & ' n  m  xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j and Sμ ) j=1 i=1 * & ' n  m  L R xi (ai j (0), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y j is maximum for all Sν j=1 i=1

∗ ∗ ∗ (y1 , y) 2 , ..., yn ) ∈ Y or if it is not possible to find {x 1 , x* 2 , ..., x m } for which the value & ' n m  xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j and of Sμ j=1 i=1 ) * & ' n  m  xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j is maximum then find such Sν j=1 i=1

∗ {xi1 ), i = 1, 2, ..., m} for which the value of * & ' n  m  L R Sμ xi1 (ai j (0), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y j1 is maximum but value of ) j=1 i=1 * & ' n  m  xi1 (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j1 is not maximum for all Sν j=1 i=1

∗ (y11), y21 , ..., yn1 ) ∈ Y and find such {xi2 , i = 1, 2, ..., *m} for which the value of & ' n  m  Sν xi2 (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j2 is maximum but value of j=1 i=1 ) * & ' n  m  L R xi2 (ai j (0), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y j2 is not maximum for all Sμ j=1 i=1

∗ ∗ (y12 , y22 , ..., yn2 ) ∈ Y i.e., find optimal solution {xi1 , i = 1, 2, ..., m} and {xi2 ,i = 1, 2, ..., n} of Problem 4.6.13 and Problem 4.6.14 respectively.

Problem 4.6.13 % ) Maximize Sμ

n  m  j=1 i=1

&

xi1 (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j

*(

' y j1

Subject to m n   xi1 = 1; xi1 ≥ 0, i = 1, 2, ..., m; y j1 = 1; y j1 ≥ 0, j = 1, 2, ..., n. i=1

j=1

4.6 Proposed Ambika Methods

Problem 4.6.14 % ) Maximize Sν

n  m  j=1 i=1

125

&

xi2 (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j

*(

' y j2

Subject to m n   xi2 = 1; xi2 ≥ 0, i = 1, 2, ..., m; y j2 = 1; y j2 ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 2: Using the property, λ < (aiL (0), ai (1), aiR (0)), wa˜ i , u a˜ i >=< (λaiL (0), n  < (aiL (0), ai (1), aiR (0)), wa˜ i , u a˜ i >=< λai (1), λaiR (0)), wa˜ i , u a˜ i >, λ ≥ 0 and (

n 

i=1

aiL (0),

n 

ai (1),

i=1

n  i=1

i=1

aiR (0)),

min wa˜ i , max u a˜ i >, Problem 4.6.13 and Problem

1≤i≤n

1≤i≤n

4.6.14 can be transformed into Problem 4.6.15 and Problem 4.6.16 respectively. Problem 4.6.15 Maximize ⎧ ⎛ ⎞⎫ # $ ⎪ ⎪ ⎬ ⎨ n  n  n  m m m   ⎜  ⎟ Sμ ⎝ ( xi1 aiLj (0)y j1 , xi1 ai j (1)y j1 , xi1 aiRj (0))y j1 , min {wa˜ i j }, max {u a˜ i j } ⎠ ⎪ ⎪ 1≤i≤m 1≤i≤m j=1 i=1 j=1 i=1 j=1 i=1 ⎭ ⎩ 1≤ j≤n

1≤ j≤n

Subject to m n   xi1 = 1; xi1 ≥ 0, i = 1, 2, ..., m; y j1 = 1; y j1 ≥ 0, j = 1, 2, ..., n. i=1

j=1

Problem 4.6.16 Maximize ⎧ ⎛ ⎞⎫ # $ ⎪ ⎪ ⎬ ⎨ n  n  n  m m m    ⎜ ⎟ Sν ⎝ ( xi2 aiLj (0)y j2 , xi2 ai j (1)y j2 , xi2 aiRj (0))y j2 , min {wa˜ i j }, max {u a˜ i j } ⎠ ⎪ ⎪ 1≤i≤m 1≤i≤m j=1 i=1 j=1 i=1 j=1 i=1 ⎭ ⎩ 1≤ j≤n

1≤ j≤n

Subject to m n   xi2 = 1; xi2 ≥ 0, i = 1, 2, ..., m; y j2 = 1; y j2 ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 3: Since, in Problems 4.6.15 and 4.6.16, only xi1 and xi2 have been considered as decision variables. So, Problem 4.6.15 and Problem 4.6.16 are linear programming problems and hence, the optimal value of Problem 4.6.15 and Problem 4.6.16 will be equal to optimal value of its corresponding dual problem i.e., Problem 4.6.17 and Problem 4.6.18 respectively. Problem 4.6.17 Minimize{ω1 } Subject to ⎛ n ⎜

Sμ ⎝ n  j=1

j=1

#

$

⎞

⎟ (aiLj (0), ai j (1), aiRj (0)), min {wa˜ i j }, max {u a˜ i j } y j1 ⎠ ≤ ω1 , i = 1, 2, ..., m; 1≤i≤m 1≤ j≤n

y j1 = 1; y j1 ≥ 0, j = 1, 2, ..., n.

1≤i≤m 1≤ j≤n

126

4 Matrix Games with Intuitionistic Fuzzy Payoffs

Problem 4.6.18 Minimize{ω2 } Subject to ⎛ n ⎜

Sν ⎝ n 

#

j=1

$

⎞

⎟ (aiLj (0), ai j (1), aiRj (0)), min {wa˜ i j }, max {u a˜ i j } y j2 ⎠ ≤ ω2 , i = 1, 2, ..., m; 1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

y j2 = 1; y j2 ≥ 0, j = 1, 2, ..., n.

j=1

Step 4: Find the optimal solution {y ∗j1 } and {y ∗j2 } of Problem 4.6.17 and Problem 4.6.18 respectively. Step 5: Substitute the optimal solution {y ∗j1 , j = 1, 2, ..., n} and {y ∗j2 , j = 1, 2, ..., n} of Problem 4.6.17 and Problem 4.6.18 in Problem 4.6.15 and Problem 4.6.16 1∗ ,i = respectively and then find all the alternative basic optimal solutions {xi1 q∗ 2∗ 1∗ 1, 12, ..., m},{xi1 , i = 1, 12, ..., m}, ..., {xi1 , i = 1, 12, ..., m} and {xi2 , i = 1, l∗ 2∗ , i = 1, 12, ..., m}, ..., {xi1 , i = 1, 12, ..., m} of Problem 4.6.15 and 12, ..., m}, {xi2 Problem 4.6.16 respectively. Step ⎫ ⎧ n 6:mFind &maximum ' ⎪ ⎪   x t∗ (a L (0), a (1), a R (0)), w , u y , t = 1, 2, ..., q, ⎪ ⎪ ij a˜ i j a˜ i j j1 ⎬ ⎨ ij ij i1 j=1 i=1 . & ' n m   w∗ ⎪ ⎪ ⎪ xi2 (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j2 , w = 1, 2, ..., l ⎪ ⎭ ⎩ j=1 i=1

Step 7: All the maximum triangular intuitionistic fuzzy numbers will represent the maximum expected loss of Player II. The optimal strategy for Player II will be {y ∗j1 , j = 1, 2, ..., n}, which is optimal solution of Problem 4.6.17, if maximum is obtained corresponding to & ' n  m  t∗ (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j1 , t = 1, 2, ..., q and will be {y ∗j2 , j = xi1 j=1 i=1

1, 2, ..., n}, which is optimal solution of Problem 4.6.18, if maximum os obtained & ' n  m  w∗ (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j2 , w = 1, 2, ..., l. xi2 corresponding to j=1 i=1

4.6.3 Ambika Method-III In the existing method [9], it is assumed that if if a˜ =< (a L (0), a(1), a R (0)), wa˜ , u a˜ > and b˜ =< (b L (0), b(1), b R (0)), wa˜ , u a˜ > are two triangular intuitionistic fuzzy ˜ and if Vλ (a) ˜ then Aλ (a) ˜ ˜ ≤ Vλ (b) ˜ = Vλ (b) ˜ ≥ Aλ (b), numbers then a˜ b˜ if Vλ (a) L R (0) + a(1) + a (0)) (a and where, Vλ = (λwa2˜ + (1 − λ)(1 − u a˜ )2 ) 6 R L (a (0) − a (0)) . Aλ = (λwa2˜ + (1 − λ)(1 − u a˜ )2 ) 6

4.6 Proposed Ambika Methods

127

In this section, a new method, based on this comparing method, is proposed to find the minimum expected gain of Player I, maximum expected loss of Player II and their corresponding optimal strategies.

4.6.3.1

Minimum Expected Gain of Player I

˜ and if Vλ (a) ˜ then Using the comparing method, a˜ b˜ if Vλ (a) ˜ ≤ Vλ (b) ˜ = Vλ (b) ˜ the minimum expected gain of Player I and the corresponding op˜ ≥ Aλ (b), Aλ (a) timal strategies can be obtained as follows: Step 1: According to comparing method, to find the optimal solution {y1∗ , y2∗ , ..., yn∗ } & ' n  m  such that value of xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j is minimum for all j=1 i=1

(x1 , x2 , ..., xm)) ∈ X , firstly it is equivalent to find {y1∗ , y2∗ , ...,*yn∗ } such that & ' n  m  xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j , is minimum for all value of Vλ j=1 i=1

(x1 , x2 , ..., xm ) ∈ X i.e., find optimal solution {y ∗j , j = 1, 2, ..., n} of Problem 4.6.19. Problem % 4.6.19 ) Minimize Vλ

n  m  j=1 i=1

&

xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j

*(

' yj

Subject to m n   xi = 1; xi ≥ 0, i = 1, 2, ..., m; y j = 1; y j ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 2: Using the property, λ < (aiL (0), ai (1), aiR (0)), wa˜ i , u a˜ i >=< (λaiL (0), n  < (aiL (0), ai (1), aiR (0)), wa˜ i , u a˜ i >=< λai (1), λaiR (0)), wa˜ i , u a˜ i >, λ ≥ 0 and (

n 

i=1

aiL (0),

n 

ai (1),

i=1

n  i=1

i=1

aiR (0)),

min wa˜ i , max u a˜ i >, Problem 4.6.19 can be trans-

1≤i≤n

1≤i≤n

formed into Problem 4.6.20. Problem 4.6.20 Minimize ⎧ ⎛ ⎞⎫ # $ ⎪ ⎪ ⎨ ⎬ n n n m m m       ⎜ ⎟ Vλ ⎝ ( xi aiLj (0)y j , xi ai j (1)y j , xi aiRj (0))y j , min {wa˜ i j }, max {u a˜ i j } ⎠ ⎪ ⎪ 1≤i≤m 1≤i≤m ⎩ ⎭ j=1 i=1 j=1 i=1 j=1 i=1 1≤ j≤n

1≤ j≤n

Subject to m n   xi = 1; xi ≥ 0, i = 1, 2, ..., m; y j = 1; y j ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 3: Since, in Problem 4.6.20, only y j have been considered as decision variables. So, Problem 4.6.20 is a linear programming problem and hence, the optimal value of Problem 4.6.20 will be equal to optimal value of its corresponding dual problem i.e., Problem 4.6.21.

128

4 Matrix Games with Intuitionistic Fuzzy Payoffs

Problem 4.6.21 Maximize{υ} Subject ⎛ to ⎞ $ # m m m   ⎜  ⎟ xi ai j (1), xi aiRj (0)), min {wa˜ i j }, max {u a˜ i j } ⎠ ≥ υ, Vλ ⎝ ( xi aiLj (0), i=1

i=1

i=1

1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

j = 1, 2, ..., n; m  xi = 1; xi ≥ 0, i = 1, 2, ..., m. i=1

Step 4: Find the optimal solution of Problem 4.6.21. Step 5: Substitute the optimal solution {xi∗ , i = 1, 2, ..., m} of Problem 4.6.21 in Problem 4.6.19 and then find the optimal solution {y ∗j , j = 1, 2, ..., n} of Problem 4.6.19. Case 1: If there exist a unique optimal solution {y ∗j , j = 1, 2, ..., n} of Problem 4.6.19 & n  m  xi∗ (aiLj (0), ai j (1), aiRj (0)), then the minimum expected gain of Player I is j=1 i=1  wa˜ i j , u a˜ i j y ∗j and the corresponding optimal strategy for Player I will be {xi∗ , i = 1, 2, ..., m}, which is optimal solution of Problem 4.6.21. Case 2: If there exist more than one basic optimal solution {y j1 , j = 1, 2, ..., n}, {y j2 , j = 1, % 2, ..., *( ) n}, ..., {y j p , j = 1, 2, ..., n} of Problem 4.6.1 then find & ' n  m  ∗ L R ∗ maximum Aλ xi (ai j (0), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y js , s = 1, 2, ..., p . j=1 i=1 ) * & ' n  m  ∗ L R ∗ xi (ai j (0), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y jr represents the maximum If Aλ j=1 i=1

& n  m  value then minimum expected gain of Player I is xi∗ (aiLj (0), ai j (1), aiRj (0)), j=1 i=1  wa˜ i j , u a˜ i j y ∗jr and the corresponding optimal strategy for Player I will be {xi∗ , i = 1, 2, ..., m}, which is optimal solution of Problem 4.6.21.

4.6.3.2

Maximum Expected Loss of Player II

˜ and if Vλ (a) ˜ then Using the comparing method, a˜ b˜ if Vλ (a) ˜ ≥ Vλ (b) ˜ = Vλ (b) ˜ the maximum expected loss of Player II and the corresponding ˜ ≤ Aλ (b), Aλ (a) optimal strategies can be obtained as follows: Step 1: According to comparing method, to find the optimal solution {x1∗ , x2∗ , ..., xm∗ } & ' n  m  such that value of xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j is maximum for all j=1 i=1

∗ ∗ ∗ (y1 , y2 , ..., y) n ) ∈ Y , firstly it is equivalent to find {x 1 , x 2 , ...,*x m } such that & ' n  m  xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j , is maximum for all value of Vλ j=1 i=1

4.6 Proposed Ambika Methods

129

(y1 , y2 , ..., yn ) ∈ Y i.e., find optimal solution {xi∗ , i = 1, 2, ..., m} of Problem 4.6.22. Problem 4.6.22 % ) Maximize Vλ

n  m  j=1 i=1

& ' xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j

*(

Subject to m n   xi = 1; xi ≥ 0, i = 1, 2, ..., m; y j = 1; y j ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 2: Using the property, λ < (aiL (0), ai (1), aiR (0)), wa˜ i , u a˜ i >=< (λaiL (0), n  < (aiL (0), ai (1), aiR (0)), wa˜ i , u a˜ i >=< λai (1), λaiR (0)), wa˜ i , u a˜ i >, λ ≥ 0 and (

n 

i=1

aiL (0),

n 

ai (1),

i=1

n  i=1

i=1

aiR (0)),

min wa˜ i , max u a˜ i >, Problem 4.6.22 can be trans-

1≤i≤n

1≤i≤n

formed into Problem 4.6.23. Problem 4.6.23 Maximize ⎧ ⎛

⎞⎫ # $ ⎪ ⎬ n  n  n  m m m   ⎜  ⎟ Vλ ⎝ ( xi aiLj (0)y j , xi ai j (1)y j , xi aiRj (0))y j , min {wa˜ i j }, max {u a˜ i j } ⎠ ⎪ ⎪ 1≤i≤m 1≤i≤m ⎩ ⎭ j=1 i=1 j=1 i=1 j=1 i=1 ⎪ ⎨

1≤ j≤n

1≤ j≤n

Subject to m n   xi = 1; xi ≥ 0, i = 1, 2, ..., m; y j = 1; y j ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 3: Since, in Problem 4.6.23, only xi have been considered as decision variables. So, Problem 4.6.23 is a linear programming problem and hence, the optimal value of Problem 4.6.23 will be equal to optimal value of its corresponding dual problem i.e., Problem 4.6.24. Problem 4.6.24 Minimize{ω} Subject ⎛ to ⎞ $ # n m m   ⎜  ⎟ xi ai j (1)y j , xi aiRj (0))y j , min {wa˜ i j }, max {u a˜ i j } ⎠ ≤ ω, Vλ ⎝ ( aiLj (0)y j , j=1

i=1

i=1

1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

i = 1, 2, ..., m; n  y j = 1; y j ≥ 0, j = 1, 2, ..., n. j=1

Step 4: Find the optimal solution of Problem 4.6.24. Step 5: Substitute the optimal solution {y j ∗, j = 1, 2, ..., n} of Problem 4.6.24 in Problem 4.6.22 and then find the optimal solution {xi∗ , i = 1, 2, ..., m} of Problem 4.6.22. Case 1: If there exist a unique optimal solution {xi∗ , i = 1, 2, ..., m} of Problem 4.6.22 & n  m  xi∗ (aiLj (0), ai j (1), aiRj (0)), then the maximum expected loss of Player II is j=1 i=1

130

4 Matrix Games with Intuitionistic Fuzzy Payoffs

 wa˜ i j , u a˜ i j y ∗j and the corresponding optimal strategy for Player II will be {y ∗j , j = 1, 2, ..., n}, which is optimal solution of Problem 4.6.24. Case 2: If there exist more than one basic optimal solution {xi1 , i = 1, 2, ..., m}, {xi2 , i = 1, % 2, ..., *( ) m}, ..., {xi p , j = 1, 2, ..., m} of Problem 4.6.22 then find & ' n  m  minimum Aλ xis∗ (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y ∗j , s = 1, 2, ..., p . j=1 i=1 ) * & ' n  m  ∗ L R ∗ xir (ai j (0), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y j represents the minimum If Aλ j=1 i=1

& n  m  value then maximum expected loss of Player II is xir∗ (aiLj (0), ai j (1), aiRj (0)), j=1 i=1  wa˜ i j , u a˜ i j y ∗j and the corresponding optimal strategy for Player II will be {y ∗j , j = 1, 2, ..., n}, which is optimal solution of Problem 4.6.24.

4.6.4 Ambika Method-IV In the existing method [10], authors have assumed if a˜ =< (a L (0), a L (1), a R (1)a R (0)), wa˜ , u a˜ > and b˜ =< (b L (0), b L (1), b R (1), b R (0)), wa˜ , u a˜ > are two trapezoidal intuitionistic fuzzy ˜ and Aλ (a) ˜ where, ˜ ≤ Vλ (b) ˜ ≥ Aλ (b) numbers then a˜ b˜ if Vλ (a) (a L (0) + 2a L (1) + 2a R (1) + a R (0)) Vλ = (λ(1 − u a˜ ) + (1 − λ)wa˜ ) and 6 R L R L (a (0) − a (0) + 2a (1) − 2a (1)) . Aλ = (λwa˜ + (1 − λ)(1 − u a˜ )) 3 In this section, a new method, based on this comparing method [10], is proposed to find the minimum expected gain of Player I, maximum expected loss of Player II and their corresponding optimal strategies.

4.6.4.1

Minimum Expected Gain of Player I

˜ and Aλ (a) ˜ ˜ ≤ Vλ (b) ˜ ≥ Aλ (b), Using the comparing method, a˜ b˜ if a˜ b˜ if Vλ (a) the minimum expected gain of Player I and the corresponding optimal strategies can be obtained as follows: Step 1: According to comparing method, to find the optimal solution {y1∗ , y2∗ , ..., yn∗ } & ' n  m  such that value of xi (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), wa˜ i j , u a˜ i j y j is minimum j=1 i=1

for ) all (x1 , x2 , ..., xm ) ∈ X is equivalent to find {y1∗ , y2∗ , ...,*yn∗ } such that values of & ' n  m  xi (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), wa˜ i j , u a˜ i j y j is minimum and Vλ j=1 i=1

4.6 Proposed Ambika Methods

) Aλ

n  m  j=1 i=1

131

&

xi (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), wa˜ i j , u a˜ i j

*

' yj

is maximum for all

(x1 , x2), ..., xm ) ∈ X or if it is not possible to find {y1∗ , y2∗ , ..., y*n∗ } for which the value & ' n  m  of Vλ xi (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), wa˜ i j , u a˜ i j y j is minimum and ) j=1 i=1 * & ' n  m  L L R R xi (ai j (0), ai j (1), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y j is maximum then find Aλ j=1 i=1 ) & n  m  ∗ xi1 (aiLj (0), aiLj (1), such {y j1 , j = 1, 2, ..., n} for which the value of Vλ j=1 i=1 ) '  & n  m  xi1 (aiLj (0), aiRj (1), aiRj (0)), wa˜ i j , u a˜ i j y j1 is minimum but value of Aλ j=1 i=1 '  L R R ai j (1), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y j1 is not maximum for all (x11 , x21 , ..., xm1 ) ∈ X and find such {y ∗j2 , j = 1, 2, ..., n} for which the value of ) * & ' n  m  L L R R Aλ xi2 (ai j (0), ai j (1), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y j2 is maximum but value j=1 i=1 ) * & ' n  m  L L R R xi2 (ai j (0), ai j (1), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y j2 is not minimum of Vλ j=1 i=1

for all (x12 , x22 , ..., xm2 ) ∈ X i.e., find optimal solution {y ∗j1 , j = 1, 2, ..., n} and {y ∗j2 , j = 1, 2, ..., n} of Problem 4.6.25 and Problem 4.6.26 respectively. Problem % 4.6.25 ) Minimize Vλ

n  m  j=1 i=1

& ' xi1 (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), wa˜ i j , u a˜ i j y j1

*(

Subject to m n   xi1 = 1; xi1 ≥ 0, i = 1, 2, ..., m; y j1 = 1; y j1 ≥ 0, j = 1, 2, ..., n. i=1

j=1

Problem 4.6.26 % ) Maximize Aλ

n  m  j=1 i=1

& ' xi2 (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), wa˜ i j , u a˜ i j y j2

*(

Subject to m n   xi2 = 1; xi2 ≥ 0, i = 1, 2, ..., m; y j2 = 1; y j2 ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 2: Using the property, λ < (aiL (0), ai (1), aiR (0)), wa˜ i , u a˜ i >=< (λaiL (0), n  < (aiL (0), ai (1), aiR (0)), wa˜ i , u a˜ i >=< λai (1), λaiR (0)), wa˜ i , u a˜ i >, λ ≥ 0 and (

n 

i=1

aiL (0),

n  i=1

ai (1),

n  i=1

i=1

aiR (0)),

min wa˜ i , max u a˜ i >, Problem 4.6.25 and Problem

1≤i≤n

1≤i≤n

4.6.26 can be transformed into Problem 4.6.27 and Problem 4.6.28 respectively.

132

4 Matrix Games with Intuitionistic Fuzzy Payoffs

Problem 4.6.27 ⎧ )# n  n  n  m m m ⎪    ⎪ L L ⎪ V x a (0)y , x a (1)y , xi1 aiRj (1)y j1 ( ⎪ λ i1 j1 i1 j1 ij ij ⎪ ⎪ j=1 i=1 j=1 i=1 j=1 i=1 ⎨ ⎞ $ Minimize n m ⎪  ⎪ ⎟ ⎪ xi1 aiRj (0))y j1 , min {wa˜ i j }, max {u a˜ i j } ⎠ ⎪ ⎪ ⎪ 1≤i≤m 1≤i≤m j=1 i=1 ⎩ 1≤ j≤n

1≤ j≤n

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

Subject to m n   xi1 = 1; xi1 ≥ 0, i = 1, 2, ..., m; y j1 = 1; y j1 ≥ 0, j = 1, 2, ..., n. i=1

j=1

Problem 4.6.28 ⎧ )# n  n  n  m m m ⎪    ⎪ ⎪ Aλ ( xi2 aiLj (0)y j2 , xi2 aiLj (1)y j2 , xi2 aiRj (1)y j2 ⎪ ⎪ ⎪ j=1 i=1 j=1 i=1 j=1 i=1 ⎨ ⎞ $ Maximize n m ⎪  ⎪ ⎟ ⎪ xi2 aiRj (0))y j2 , min {wa˜ i j }, max {u a˜ i j } ⎠ ⎪ ⎪ ⎪ 1≤i≤m 1≤i≤m j=1 i=1 ⎩ 1≤ j≤n

1≤ j≤n

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

Subject to m n   xi2 = 1; xi2 ≥ 0, i = 1, 2, ..., m; y j2 = 1; y j2 ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 3: Since, in Problems 4.6.27 and 4.6.28, only y j1 and y j2 have been considered as decision variables. So, Problem 4.6.27 and Problem 4.6.28 are linear programming problems and hence, the optimal value of Problem 4.6.27 and Problem 4.6.28 will be equal to optimal value of its corresponding dual problem i.e., Problem 4.6.29 and Problem 4.6.30 respectively. Problem 4.6.29 Maximize{υ1 } Subject ⎛ to ⎞ $ # m ⎜ ⎟ Vλ ⎝ (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), min {wa˜ i j }, max {u a˜ i j } xi1 ⎠ ≥ υ1 , i=1

1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

j = 1, 2, ..., n; m  xi1 = 1; xi1 ≥ 0, i = 1, 2, ..., m. i=1

Problem 4.6.30 Minimize{υ2 } Subject ⎛ to ⎞ $ # m ⎜ ⎟ Aλ ⎝ (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), min {wa˜ i j }, max {u a˜ i j } xi2 ⎠ ≤ υ2 , i=1

j = 1, 2, ..., n; m  xi2 = 1; xi2 ≥ 0, i = 1, 2, ..., m. i=1

1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

4.6 Proposed Ambika Methods

133

∗ ∗ Step 4: Find the optimal solution {xi1 } and {xi2 } of Problem 4.6.29 and Problem 4.6.30 respectively. ∗ ∗ , i = 1, 2, ..., m} ans {xi2 , i = 1, 2, ..., m} Step 5: Substitute the optimal solution {xi1 of Problem 4.6.29 and Problem 4.6.30 in Problem 4.6.27 and Problem 4.6.28 respectively and then find all the alternative basic optimal solutions {y 1∗ j1 , j = 1, 12, ..., 2∗ r∗ 1∗ n},{y j1 , j = 1, 12, ..., n}, ..., {y jn , j = 1, 12, ..., n} and {y j2 , j = 1, 12, ..., n}, s∗ {y 2∗ j2 , j = 1, 12, ..., n}, ..., {y j1 , j = 1, 12, ..., n} of Problem 4.6.27 and Problem 4.6.28 respectively. Step ⎫ ⎧ n 6:mFind&minimum '  h∗ L L R R ⎪ ⎪ ⎪ ⎪ (a y x (0), a (1), a (1), a (0)), w , u , h = 1, 2, ..., r, i1 a ˜ a ˜ ⎬ ⎨ ij ij j1 ij ij ij ij j=1 i=1 . & ' n m  ⎪ ⎪ ⎪ ⎪ xi2 (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), wa˜ i j , u a˜ i j y k∗ ⎩ j2 , k = 1, 2, ..., s ⎭ j=1 i=1

Step 7: All the minimum trapezoidal intuitionistic fuzzy numbers will represent the minimum expected gain of Player I. The optimal strategy for Player I will be ∗ {xi1 , i = 1, 2, ..., m}, which is optimal solution of Problem 4.6.29, if minimum is obtained corresponding to & ' n  m  xi1 (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), wa˜ i j , u a˜ i j y h∗ j1 , h = 1, 2, ..., r and will be j=1 i=1

∗ , i = 1, 2, ..., m}, which is optimal solution of Problem 4.6.30, if minimum is {xi2 obtained corresponding to & ' n  m  xi2 (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), wa˜ i j , u a˜ i j y k∗ j2 , k = 1, 2, ..., s. j=1 i=1

4.6.4.2

Maximum Expected Loss of Player II

˜ and Aλ (a) ˜ the max˜ ≥ Vλ (b) ˜ ≤ Aλ (b), Using the comparing method, a˜ b˜ if Vλ (a) imum expected loss of Player II and the corresponding optimal strategies can be obtained as follows: Step 1: According to comparing method, to find the optimal solution {x1∗ , x2∗ , ..., xm∗ } & ' n  m  such that value of xi (aiLj (0), ai j (1), aiRj (0)), wa˜ i j , u a˜ i j y j is maximum for all j=1 i=1

(y1 , y2), ..., yn ) ∈ Y is equivalent to find {x1∗ , x2∗ , ..., xm∗ } such*that value & ' n  m  xi (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), wa˜ i j , u a˜ i j y j is maximum and of Vλ ) j=1 i=1 * & ' n  m  xi (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), wa˜ i j , u a˜ i j y j is minimum for all Aλ j=1 i=1

(y1 , y2), ..., yn ) ∈ Y or if it is not possible to find {x1∗ , x2∗ , ..., x*m∗ } for which the value & ' n  m  xi (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), wa˜ i j , u a˜ i j y j is maximum and of Vλ j=1 i=1

134

4 Matrix Games with Intuitionistic Fuzzy Payoffs

) Aλ

n  m 

&

xi (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), wa˜ i j , u a˜ i j

*

' yj

is minimum then find

j=1 i=1 ∗ such {x ) i1 , i = n  m 

1, 2, ..., m} for which the value of * & ' Vλ xi1 (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), wa˜ i j , u a˜ i j y j1 is maximum but value j=1 i=1 ) * & ' n  m  L L R R xi1 (ai j (0), ai j (1), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y j1 is not minimum for of Aλ j=1 i=1

∗ all (y11) , y21 , ..., yn1 ) ∈ Y and find such {xi2 , i = 1, 2, ..., m} for*which the value & ' n m  of Aλ xi2 (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), wa˜ i j , u a˜ i j y j2 is minimum but j=1 i=1 ) * & ' n  m  L L R R xi2 (ai j (0), ai j (1), ai j (1), ai j (0)), wa˜ i j , u a˜ i j y j2 is not maxvalue of Vλ j=1 i=1

∗ imum for all (y12 , y22 , ..., yn2 ) ∈ Y i.e., find optimal solution {xi1 , i = 1, 2, ..., m} ∗ , i = 1, 2, ..., n} of Problem 4.6.31 and Problem 4.6.32 respectively. and {xi2

Problem 4.6.31 % ) Maximize Vλ

n  m  j=1 i=1

&

xi1 (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), wa˜ i j , u a˜ i j

*(

' y j1

Subject to m n   xi1 = 1; xi1 ≥ 0, i = 1, 2, ..., m; y j1 = 1; y j1 ≥ 0, j = 1, 2, ..., n. i=1

j=1

Problem % 4.6.32) Minimize Aλ

n  m  j=1 i=1

&

'

*(

xi2 (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), wa˜ i j , u a˜ i j y j2

Subject to m n   xi2 = 1; xi2 ≥ 0, i = 1, 2, ..., m; y j2 = 1; y j2 ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 2: Using the property, λ < (aiL (0), ai (1), aiR (0)), wa˜ i , u a˜ i >=< (λaiL (0), n  < (aiL (0), ai (1), aiR (0)), wa˜ i , u a˜ i >=< λai (1), λaiR (0)), wa˜ i , u a˜ i >, λ ≥ 0 and (

n 

i=1

aiL (0),

n  i=1

ai (1),

n  i=1

i=1

aiR (0)),

min wa˜ i , max u a˜ i >, Problem 4.6.31 and Problem

1≤i≤n

1≤i≤n

4.6.32 can be transformed into Problem 4.6.33 and Problem 4.6.34 respectively. Problem 4.6.33 ⎧ )# n  n  n  m m m ⎪    ⎪ ⎪ Vλ ( xi1 aiLj (0)y j1 , xi1 aiLj (1)y j1 , xi1 aiRj (1)y j1 ⎪ ⎪ ⎪ j=1 i=1 j=1 i=1 j=1 i=1 ⎨ ⎞ $ Maximize n  m ⎪  ⎪ ⎟ ⎪ xi1 aiRj (0))y j1 , min {wa˜ i j }, max {u a˜ i j } ⎠ ⎪ ⎪ ⎪ 1≤i≤m 1≤i≤m j=1 i=1 ⎩ 1≤ j≤n

1≤ j≤n

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

4.6 Proposed Ambika Methods

135

Subject to m n   xi1 = 1; xi1 ≥ 0, i = 1, 2, ..., m; y j1 = 1; y j1 ≥ 0, j = 1, 2, ..., n. i=1

j=1

Problem 4.6.34 ⎧ )# n  n  n  m m m ⎪    ⎪ ⎪ Aλ ( xi2 aiLj (0)y j2 , xi2 aiLj (1)y j2 , xi2 aiRj (1)y j2 ⎪ ⎪ ⎪ j=1 i=1 j=1 i=1 j=1 i=1 ⎨ ⎞ $ Minimize n  m ⎪  ⎪ ⎟ ⎪ xi2 aiRj (0))y j2 , min {wa˜ i j }, max {u a˜ i j } ⎠ ⎪ ⎪ ⎪ 1≤i≤m 1≤i≤m j=1 i=1 ⎩ 1≤ j≤n

1≤ j≤n

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

Subject to m n   xi2 = 1; xi2 ≥ 0, i = 1, 2, ..., m; y j2 = 1; y j2 ≥ 0, j = 1, 2, ..., n. i=1

j=1

Step 3: Since, in Problems 4.6.33 and 4.6.34, only xi1 and xi2 have been considered as decision variables. So, Problem 4.6.33 and Problem 4.6.34 are linear programming problems and hence, the optimal value of Problem 4.6.33 and Problem 4.6.34 will be equal to optimal value of its corresponding dual problem i.e., Problem 4.6.35 and Problem 4.6.36 respectively. Problem 4.6.35 Minimize{ω1 } Subject ⎛ to ⎞ $ # n ⎜ ⎟ Vλ ⎝ (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), min {wa˜ i j }, max {u a˜ i j } y j1 ⎠ ≤ ω1 , j=1

1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

i = 1, 2, ..., m; n  y j1 = 1; y j1 ≥ 0, j = 1, 2, ..., n. j=1

Problem 4.6.36 Maximize{ω2 } Subject ⎛ to ⎞ $ # n  ⎜ ⎟ Sμ ⎝ (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), min {wa˜ i j }, max {u a˜ i j } y j2 ⎠ ≥ ω2 , j=1

1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

i = 1, 2, ..., m; n  y j2 = 1; y j2 ≥ 0, j = 1, 2, ..., n. j=1

Step 4: Find the optimal solution {y ∗j1 } and {y ∗j2 } of Problem 4.6.35 and Problem 4.6.36 respectively. Step 5: Substitute the optimal solution {y ∗j1 , j = 1, 2, ..., n} and {y ∗j2 , j = 1, 2, ..., n} of Problem 4.6.35 and Problem 4.6.36 in Problem 4.6.33 and Problem 4.6.34 respec1∗ , i = 1, 12, ..., m}, tively and then find all the alternative basic optimal solutions {xi1 q∗ 2∗ 1∗ {xi1 , i = 1, 12, ..., m}, ..., {xi1 , i = 1, 12, ..., m} and {xi2 , i = 1, 12, ..., m},

136

4 Matrix Games with Intuitionistic Fuzzy Payoffs

l∗ 2∗ {xi2 , i = 1, 12, ..., m}, ..., {xi1 , i = 1, 12, ..., m} of Problem 4.6.33 and Problem 4.6.34 respectively. Step ⎫ ⎧ n 6:mFind &maximum ' ⎪   x t∗ (a L (0), a L (1), a R (1), a R (0)), w , u y , t = 1, 2, ..., q, ⎪ ⎪ ⎪ a˜ i j a˜ i j j1 ⎬ ⎨ ij ij ij ij i1 j=1 i=1 . & ' n m   w∗ ⎪ ⎪ ⎪ xi2 (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), wa˜ i j , u a˜ i j y j2 , w = 1, 2, ..., l ⎪ ⎭ ⎩ j=1 i=1

Step 7: All the maximum trapezoidal intuitionistic fuzzy numbers will represent the maximum expected loss of Player II. The optimal strategy for Player II will be {y ∗j1 , j = 1, 2, ..., n}, which is optimal solution of Problem 4.6.35, if maximum is obtained corresponding to & ' n  m  t∗ (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), wa˜ i j , u a˜ i j y j1 , t = 1, 2, ..., q and will be xi1 j=1 i=1

{y ∗j2 , j = 1, 2, ..., n}, which is optimal solution of Problem 4.6.36, if maximum is obtained corresponding to & ' n  m  w∗ (aiLj (0), aiLj (1), aiRj (1), aiRj (0)), wa˜ i j , u a˜ i j y j2 , w = 1, 2, ..., l. xi2 j=1 i=1

4.7 Numerical Examples In this section, some existing numerical examples are solved by proposed Ambika methods.

4.7.1 Existing Numerical Example Considered by Nan et al. In this section, matrix games with intuitionistic fuzzy payoffs  < (175, 180, 190); 0.6, 0.2 > < (150, 156, 158); 0.6, 0.2 > A= , chosen by < (80, 90, 100); 0.9, 0.1 > < (175, 180, 190); 0.6, 0.2 > Nan et al. [8], is solved by the proposed Ambika method-I.

4.7.1.1

Minimum Expected Gain of Player I

Using the proposed Ambika method-I minimum expected gain of Player I and corresponding optimal strategies, can be obtained as follows: Step 1: Find {y ∗j , j = 1, 2} such that value of Sμ (< (175, 180, 190); 0.6, 0.2 > x1 y1 + < (150, 156, 158); 0.6, 0.2 > x1 y2 + < (80, 90, 100); 0.9, 0.1 > x2 y1 + < (175, 180, 190); 0.6, 0.2 > x2 y2 ), is minimum for all (x1 , x2 ) ∈ X i.e., find optimal solution {y1∗ , y2∗ } of Problem 4.7.1.

4.7 Numerical Examples

137

Problem 4.7.1 Minimize    < (175, 180, 190); 0.6, 0.2 > x1 y1 + < (150, 156, 158); 0.6, 0.2 > x1 y2 + Sμ < (80, 90, 100); 0.9, 0.1 > x2 y1 + < (175, 180, 190); 0.6, 0.2 > x2 y2 Subject to x1 + x2 = 1; x1 , x2 ≥ 0; y1 + y2 = 1; y1 , y2 ≥ 0. Step 2: Using the property, λ < (aiL (0), ai (1), aiR (0)), wa˜ i , u a˜ i >=< (λaiL (0), n  < (aiL (0), ai (1), aiR (0)), wa˜ i , u a˜ i >= λai (1), λaiR (0)), wa˜ i , u a˜ i >, λ ≥ 0 and , Problem 4.7.1 can be trans-

1≤i≤n

1≤i≤n

formed into Problem 4.7.2. Problem 4.7.2 Minimize  + Sμ

175x1 y1 + 150x1 y2 + 80x2 y1 + 175x2 y2 , 180x1 y1 + 156x1 y2 + 90x2 y1 + 180x2 y2 , 190x1 y1 + 158x1 y2 + 100x2 y1 + 190x2 y2



, ; 0.6, 0.2

Subject to x1 + x2 = 1; x1 , x2 ≥ 0; y1 + y2 = 1; y1 , y2 ≥ 0. Step 3: Since, in Problem 4.7.2, only y j has been considered as decision variables. So, Problem 4.7.2 is a linear programming problem and hence, the optimal value of Problem 4.7.2 will be equal to optimal value of its corresponding dual problem i.e., Problem 4.7.3. Problem 4.7.3 Maximize{υ} Subject to Sμ ( (175x1 + 80x2 , 180x1 + 90x2 , 190x1 + 100x2 ); 0.6, 0.2) ≥ υ; Sμ ( (150x1 + 175x2 , 156x1 + 158x2 , 158x1 + 190x2 ); 0.6, 0.2) ≥ υ; x1 + x2 = 1; x1 , x2 ≥ 0.   73 ∗ 21 ∗ ,x = . Step 4: The optimal solution of Problem 4.7.3 is x1 = 94 2 94  73 ∗ 21 Step 5: Substituting the optimal solution x1∗ = , x2 = of Problem 4.7.3 in 94 94 Problem 4.7.1, the obtained alternative basic optimal solutions of Problem 4.7.1 are {y11∗ = 1, y22∗ = 0} and {y12∗ = 0, y22∗ = 1}. Since, there exists more than one optimal solution {y11∗ = 1, y22∗ = 0} and {y12∗ = 0, y22∗ = 1} of Problem 4.7.1 so find minimum ⎧ )#) * $* ⎫ ⎪ ⎪ ⎪ ⎪ ⎨ Sν ⎪ ⎪ ⎪ ⎪ ⎩

172x1∗ y11∗ + 150x1∗ y21∗ + 80x2∗ y11∗ + 175x2∗ y21∗ , 180x1∗ y11∗ + 156x1∗ y21∗ +

; 0.6, 0.2

⎪ ⎪ ⎪ ⎪

⎬ ∗ 1∗ ∗ 1∗ ∗ 1∗ ∗ 1∗ ∗ 1∗ ∗ 1∗ )#) 90x2∗y12∗ + 180x2∗y22∗, 190x1∗y12∗+ 158x1∗y22∗+ 100x∗2 y2∗1 + 190x∗2 y2∗2 * $* 172x1 y1 + 150x1 y2 + 80x2 y1 + 175x2 y2 , 180x1 y1 + 156x1 y2 + ⎪ ⎪ Sν ; 0.6, 0.2 ⎪ ⎪ ⎭ ∗ 2∗ ∗ 2∗ ∗ 2∗ ∗ 2∗ ∗ 2∗ ∗ 2∗ 90x2 y1 + 180x2 y2 , 190x1 y1 + 158x1 y2 + 100x2 y1 + 190x2 y2  

= minimum

12097 12097 , 94 94

=

12097 . 94

138

4 Matrix Games with Intuitionistic Fuzzy Payoffs )#)

Since,

Sν

)#)

and Sν

172x1∗ y11∗ + 150x1∗ y21∗ + 80x2∗ y11∗ + 175x2∗ y21∗ , 180x1∗ y11∗ + 156x1∗ y21∗ +

*

90x2∗ y11∗ + 180x2∗ y21∗ , 190x1∗ y11∗ + 158x1∗ y22∗ + 100x2∗ y11∗ + 190x2∗ y21∗

172x1∗ y12∗ + 150x1∗ y22∗ + 80x2∗ y12∗ + 175x2∗ y22∗ , 180x1∗ y12∗ + 156x1∗ y22∗ +

$* ; 0.6, 0.2

*

90x2∗ y12∗ + 180x2∗ y22∗ , 190x1∗ y12∗ + 158x1∗ y22∗ + 100x2∗ y12∗ + 190x2∗ y22∗

$* ; 0.6, 0.2

both represents the minimum value so minimum expected gain of Player I is #)

172x1∗ y11∗ + 150x1∗ y21∗ + 80x2∗ y11∗ + 175x2∗ y21∗ , 180x1∗ y11∗ + 156x1∗ y21∗ +

*

∗ 1∗ ∗ 1∗ ∗ 1∗ ∗ 1∗ ∗ 1∗ ∗ 1∗  1 y1 + 158x , 1 y2 + 100x2 y1 + 190x2 y2 +90x2 y1 + 180x2 y2 , 190x 14455 7517 7985 , , ; 0.6, 0.2 and = 94 47 47 #)

$

; 0.6, 0.2

172x1∗ y12∗ + 150x1∗ y22∗ + 80x2∗ y12∗ + 175x2∗ y22∗ , 180x1∗ y12∗ + 156x1∗ y22∗ + 90x ∗ y 2∗ + 180x ∗ y 2∗ , 190x ∗ y 2∗ + 158x ∗ y 2∗ + 100x ∗ y 2∗ + 190x ∗ y 2∗

*

$ ; 0.6, 0.2

2 2  1 1 1 2 2 1 2 2 , + 2 1 14625 7584 7762 , , ; 0.6, 0.2 respectively and the corresponding optimal = 94 47 47   73 ∗ 21 ∗ strategy for Player I will be x1 = , which is optimal solution of ,x = 94 2 94 Problem 4.7.3.

4.7.1.2

Maximum Expected Loss of Player II

Using the proposed Ambika method-I maximum expected loss of Player II and corresponding optimal strategies, can be obtained as follows: Step 1: Find {xi∗ , i = 1, 2} such that value of Sμ (< (175, 180, 190); 0.6, 0.2 > x1 y1 + < (150, 156, 158); 0.6, 0.2 > x1 y2 + < (80, 90, 100); 0.9, 0.1 > x2 y1 + < (175, 180, 190); 0.6, 0.2 > x2 y2 ), is maximum for all (y1 , y2 ) ∈ Y i.e., find optimal solution {x1∗ , x2∗ } of Problem 4.7.4. Problem 4.7.4 Maximize    < (175, 180, 190); 0.6, 0.2 > x1 y1 + < (150, 156, 158); 0.6, 0.2 > x1 y2 + Sμ < (80, 90, 100); 0.9, 0.1 > x2 y1 + < (175, 180, 190); 0.6, 0.2 > x2 y2 Subject to x1 + x2 = 1; x1 , x2 ≥ 0; y1 + y2 = 1; y1 , y2 ≥ 0. Step 2: Using the property, λ < (aiL (0), ai (1), aiR (0)), wa˜ i , u a˜ i >=< (λaiL (0), n  < (aiL (0), ai (1), aiR (0)), wa˜ i , u a˜ i >= λai (1), λaiR (0)), wa˜ i , u a˜ i >, λ ≥ 0 and , Problem 4.7.4 can be trans-

1≤i≤n

1≤i≤n

formed into Problem 4.7.5. Problem 4.7.5 Maximize + Sμ

175x1 y1 + 150x1 y2 + 80x2 y1 + 175x2 y2 , 180x1 y1 + 156x1 y2 + 90x2 y1 + 180x2 y2 , 190x1 y1 + 158x1 y2 + 100x2 y1 + 190x2 y2

Subject to



, ; 0.6, 0.2

4.7 Numerical Examples

139

x1 + x2 = 1; x1 , x2 ≥ 0; y1 + y2 = 1; y1 , y2 ≥ 0. Step 3: Since, in Problem 4.7.5, only xi has been considered as decision variables. So, Problem 4.7.5 is a linear programming problem and hence, the optimal value of Problem 4.7.5 will be equal to optimal value of its corresponding dual problem i.e., Problem 4.7.6. Problem 4.7.6 Minimize{ω} Subject to Sμ ( (175y1 + 150y2 , 180y1 + 156y2 , 190y1 + 158y2 ); 0.6, 0.2) ≤ ω; Sμ ( (80y1 + 175y2 , 90y1 + 158y2 , 100y1 + 190y2 ); 0.6, 0.2) ≤ ω; y1 + y2 = 1; y1 , y2 ≥ 0.   21 ∗ 21 ∗ Step 4: The optimal solution of Problem 4.7.6 is y1 = ,y = . 94 2  94  21 ∗ 73 Step 5: Substituting the optimal solution y1∗ = , y2 = of Problem 4.7.6 94 94 in Problem 4.7.5, the obtained alternative basic optimal solutions of Problem 4.7.5 are {x11∗ = 1, x22∗ = 0} and {x12∗ = 0, x22∗ = 1}. Since, there exists more than one optimal solution {x11∗ = 1, x22∗ = 0} and {x12∗ = 0, x22∗ = 1} of Problem 4.7.5 so find maximum )#) * $* ⎫ ⎧ ⎪ ⎪ ⎪ S ⎪ ⎨ ν ⎪ ⎪ ⎪ ⎪ ⎩ Sν

172x11∗ y1∗ + 150x11∗ y2∗ + 80x21∗ y1∗ + 175x21∗ y2∗ , 180x11∗ y1∗ + 156x11∗ y2∗ +

)#)

90x21∗ y1∗ + 180x21∗ y2∗ , 190x11∗ y1∗ + 158x11∗ y2∗ + 100x21∗ y1∗ + 190x2∗ y2∗

172x12∗ y1∗ + 150x12∗ y2∗ + 80x22∗ y1∗ + 175x22∗ y2∗ , 180x12∗ y1∗ + 156x12∗ y2∗ +

*

$*

  12097 12097 12097 , = . = maximum 94 94 )#) 94 Sν

)#)

and Sν

⎪ ⎪ ⎪ ⎪ ⎭

; 0.6, 0.2

90x22∗ y1∗ + 180x22∗ y2∗ , 190x12∗ y1∗ + 158x12∗ y2∗ + 100x22∗ y1∗ + 190x22∗ y2∗

Since,

⎪ ⎪ ⎪ ⎪ ⎬

; 0.6, 0.2

172x11∗ y1∗ + 150x11∗ y2∗ + 80x21∗ y1∗ + 175x21∗ y2∗ , 180x11∗ y1∗ + 156x11∗ y2∗ + 90x21∗ y1∗ + 180x21∗ y2∗ , 190x11∗ y1∗ + 158x11∗ y2∗ + 100x21∗ y1∗ + 190x2∗ y2∗

172x12∗ y1∗ + 150x12∗ y2∗ + 80x22∗ y1∗ + 175x22∗ y2∗ , 180x12∗ y1∗ + 156x12∗ y2∗ +

*

$* ; 0.6, 0.2

*

90x22∗ y1∗ + 180x22∗ y2∗ , 190x12∗ y1∗ + 158x12∗ y2∗ + 100x22∗ y1∗ + 190x22∗ y2∗

$* ; 0.6, 0.2

both II is$ #) represents the maximum value so maximum expected loss of Player * 172x11∗ y1∗ + 150x11∗ y2∗ + 80x21∗ y1∗ + 175x21∗ y2∗ , 180x11∗ y1∗ + 156x11∗ y2∗ + 90x21∗ y1∗ + 180x21∗ y2∗ , 190x11∗ y1∗ + 158x11∗ y2∗ + 100x21∗ y1∗ + 190x21∗ y2∗

+ = #)

; 0.6, 0.2

 , 14625 7584 7762 , , ; 0.6, 0.2 and 94 47 47

172x12∗ y1∗ + 150x12∗ y2∗ + 80x22∗ y1∗ + 175x22∗ y2∗ , 180x12∗ y1∗ + 156x12∗ y2∗ + 90x22∗ y1∗ + 180x22∗ y2∗ , 190x12∗ y1∗ + 158x12∗ y2∗ + 100x22∗ y1∗ + 190x22∗ y2∗

*

$ ; 0.6, 0.2

 , 14455 7517 7985 , , ; 0.6, 02 respectively and the corresponding optimal 94 47 47   21 ∗ 73 strategy for Player II will be y1∗ = , y2 = , which is optimal solution of 94 94 Problem 4.7.6. +

=

140

4 Matrix Games with Intuitionistic Fuzzy Payoffs

4.7.2 Existing Numerical Example Considered by Li et al. In this section, matrix games with intuitionistic fuzzy payoffs ⎡

< (175, 180, 190); 0.6, 0.2 > < (150, 156, 158); 0.6, 0.2 > < (80, 90, 100); 0.9, 0.1 >

⎤

⎥ ⎢ A˜ = ⎣ < (80, 90, 100); 0.9, 0.1 > < (175, 180, 190); 0.6, 0.2 > < (180, 185, 190); 0.5, 0.3 > ⎦, < (180, 185, 190); 0.5, 0.3 > < (80, 100, 120); 0.7, 0.2 > < (150, 160, 170); 0.5, 0.1 >

chosen by Li et al. [6], is solved by the proposed Ambika method-II.

4.7.2.1

Minimum Expected Gain of Player I

Using the proposed Ambika method-II, the minimum expected gain of Player I and the corresponding optimal strategies can be obtained as follows: Step 1: Find {y ∗j1 , j = 1, 2, 3} ∈ Y such that value of Sμ (< (175, 180, 190); 0.6, 0.2 > x11 y11 + < (150, 156, 158); 0.6, 0.2 > x11 y21+ < (80, 90, 100); 0.9, 0.1 > x11 y31 + < (80, 90, 100); 0.9, 0.1 > x21 y11 + < (175, 180, 190); 0.6, 0.2 > x21 y21 + < (180, 185, 190); 0.5, 0.3 > x21 y31 + < (180, 185, 190); 0.5, 0.3 > x31 y11 + is < (80, 100, 120); 0.7, 0.2 > x31 y21 + < (150, 156, 170); 0.5, 0.1 > x31 y31 ) minimum for all {xi1 , i = 1, 2, 3} ∈ X as well as find {y ∗j2 , j = 1, 2, 3} ∈ Y such that value of Sν (< (175, 180, 190); 0.6, 0.2 > x12 y12 + < (150, 156, 158); 0.6, 0.2 > x12 y22+ < (80, 90, 100); 0.9, 0.1 > x12 y32 + < (80, 90, 100); 0.9, 0.1 > x22 y12 + < (175, 180, 190); 0.6, 0.2 > x22 y22 + < (180, 185, 190); 0.5, 0.3 > x22 y32 + < (180, 185, 190); 0.5, 0.3 > x32 y12 + < (80, 100, 120); 0.7, 0.2 > x32 y22 + < (150, 156, 170); 0.5, 0.1 > x32 y32 ) is minimum for all {xi2 , i = 1, 2, 3} ∈ X i.e., find optimal solution {y ∗j1 , j = 1, 2, 3} and {y ∗j2 , j = 1, 2, 3} of Problem 4.7.7 and Problem 4.7.8 respectively. Problem ⎧ 4.7.7⎛

⎞⎫ < (175, 180, 190); 0.6, 0.2 > x11 y11 + < (150, 156, 158); 0.6, 0.2 > x11 y21 + ⎪ ⎪ ⎪ ⎜ < (80, 90, 100); 0.9, 0.1 > x y + < (80, 90, 100); 0.9, 0.1 > x y + ⎟⎪ ⎪ ⎜ ⎟⎪ 11 31 21 11 ⎜ ⎟⎬ < (175, 180, 190); 0.6, 0.2 > x21 y21 + < (180, 185, 190); 0.5, 0.3 > x21 y21 ⎟ Minimize Sμ ⎜ ⎜ ⎟ ⎪ ⎜ ⎟⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎝ < (180, 185, 190); 0.5, 0.3 > x31 y11 + < (80, 100, 120); 0.7, 0.2 > x31 y21 + ⎠⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ < (150, 156, 170); 0.5, 0.1 > x31 y31 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

Subject to x11 + x21 + x31 = 1; y11 + y21 + y31 = 1; x11 , x21 , x31 ≥ 0; y11 , y21 , y31 ≥ 0. Problem ⎧ 4.7.8⎛

⎞⎫ < (175, 180, 190); 0.6, 0.2 > x12 y12 + < (150, 156, 158); 0.6, 0.2 > x12 y22 + ⎪ ⎪ ⎜ < (80, 90, 100); 0.9, 0.1 > x y + < (80, 90, 100); 0.9, 0.1 > x y + ⎟⎪ ⎪ ⎜ ⎟⎪ ⎪ 12 32 22 12 ⎜ ⎟⎬ ⎜ ⎟ Minimize Sν ⎜ < (175, 180, 190); 0.6, 0.2 > x22 y22 + < (180, 185, 190); 0.5, 0.3 > x22 y22 ⎟ ⎪ ⎜ ⎟⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎝ < (180, 185, 190); 0.5, 0.3 > x32 y12 + < (80, 100, 120); 0.7, 0.2 > x32 y22 + ⎠⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ < (150, 156, 170); 0.5, 0.1 > x32 y32 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

Subject to x12 + x22 + x32 = 1; y12 + y22 + y32 = 1; x12 , x22 , x32 ≥ 0; y12 , y22 , y32 ≥ 0. Step 2: Since, in Problem 4.7.7 and Problem 4.7.8, only y j1 and y j2 have been considered as decision variables. So, Problem 4.7.7 and Problem 4.7.8 are linear

4.7 Numerical Examples

141

programming problems and hence, the optimal value of Problem 4.7.7 and Problem 4.7.8 will be equal to optimal value of its corresponding dual problem i.e., Problem 4.7.9 and Problem 4.7.10 respectively. Problem 4.7.9 Maximize{υ1 } Subject  to  < (175, 180, 190); 0.5, 0.3 > x11 + < (80, 90, 100); 0.5, 0.3 > x21 + ≥ υ1 ; Sμ < (180, 185, 190); 0.5, 0.3 > x31 

Sμ



< (150, 156, 158); 0.5, 0.3 > x11 + < (175, 180, 190); 0.5, 0.3 > x21 + < (80, 100, 120); 0.5, 0.3 > x31



< (80, 90, 100); 0.5, 0.3 > x11 + < (180, 185, 190); 0.5, 0.3 > x21 + Sμ < (150, 160, 170); 0.5, 0.3 > x31 x11 + x21 + x31 = 1; x11 , x21 , x31 ≥ 0.

≥ υ1 ;



≥ υ1 ;

Problem 4.7.10 Maximize{υ2 } Subject  to  < (175, 180, 190); 0.5, 0.3 > x12 + < (80, 90, 100); 0.5, 0.3 > x22 + ≥ υ2 ; Sν < (180, 185, 190); 0.5, 0.3 > x32   < (150, 156, 158); 0.5, 0.3 > x12 + < (175, 180, 190); 0.5, 0.3 > x22 Sν ≥ υ2 ; < (80, 100, 120); 0.5, 0.3 > x32   < (80, 90, 100); 0.5, 0.3 > x12 + < (180, 185, 190); 0.5, 0.3 > x22 + Sν ≥ υ2 ; < (150, 160, 170); 0.5, 0.3 > x32 x12 + x22 + x32 = 1; x12 , x22 , x32 ≥ 0.  , Step 3: The optimal solution of Problem 4.7.9 and Problem 4.7.10 is x11 = 1236 3995    4544 3733 4544 3733 and x12 = 1236 respectively. x21 = 11985 , x31 = 11985 , x = , x = 22 32 3995 11985 11985   4544 3733 Step 4: Substituting the optimal solution x11 = 1236 and , x21 = 11985 , x31 = 11985 3995   1236 4544 3733 x12 = 3995 , x22 = 11985 , x32 = 11985 of Problem 4.7.9 and Problem 4.7.10 in Problem 4.7.7 and Problem 4.7.8 respectively, the alternative basic optimal solution of 1∗ 1∗ 1∗ 2∗ 2∗ = 1, y21 = 0, y31 = 0}, {y11 = 0, y21 = Problem 4.7.7 and Problem 4.7.8 is {y11 2∗ 3∗ 3∗ 3∗ 1∗ 1∗ 1∗ 2∗ = 1, y31 = 0}, {y11 = 0, y21 = 0, y31 = 1} and {y12 = 1, y22 = 0, y32 = 0}, {y12 2∗ 2∗ 3∗ 3∗ 3∗ = 1, y32 = 0}, {y12 = 0, y22 = 0, y32 = 1} respectively. 0, y22 ∗ h∗ y11 + < (150, 156, 158); Step 5: Now, minimum {< (175, 180, 190); 0.6, 0.2 > x11 ∗ h∗ ∗ h∗ y21 + < (80, 90, 100); 0.9, 0.1 > x11 y31 + < (80, 90, 100); 0.6, 0.2 > x11 ∗ h∗ ∗ h∗ 0.9, 0.1 > x21 y11 + < (175, 180, 190); 0.6, 0.2 > x21 y21 + < (180, 185, 190); ∗ h∗ ∗ h∗ y21 + < (180, 185, 190); 0.5, 0.3 > x31 y11 + < (80, 100, 120); 0.5, 0.3 > x21 ∗ ∗h ∗ h∗ y21 + < (150, 156, 170); 0.5, 0.1 > x31 y31 , h = 1, 2, 3; 0.7, 0.2 > x31 s∗ ∗ ∗ s∗ y12 + < (150, 156, 158); 0.6, 0.2 > x12 y22 + < (175, 180, 190); 0.6, 0.2 > x12 ∗ s∗ ∗ 2∗ < (80, 90, 100); 0.9, 0.1 > x12 y32 + < (80, 90, 100); 0.9, 0.1 > x22 y12 + ∗ s∗ ∗ s∗ y22 + < (180, 185, 190); 0.5, 0.3 > x22 y22 + < (175, 180, 190); 0.6, 0.2 > x22 ∗ s∗ ∗ ∗s 0.7, 0.2 > x y < (180, 185, 190); 0.5, 0.3 > x32 y12 + < (80, 100, 120); 32 2 +   353401 124546 ∗ s∗ y32 , s = 1, 2, 3} = < 19816 , , ; < (150, 156, 170); 0.5, 0.1 > x32 414 2397 799  111634 354328 373754   330008 1769668 1897184  0.5, 0.3 >, < 2397 , 11985 , 11985 ; 0.5, 0.3 >, < 799 , 2397 , 2397 ; 0.5, 0.3 >.

142

4 Matrix Games with Intuitionistic Fuzzy Payoffs

  Step 6: Since, all the triangular intuitionistic fuzzy numbers < 19816 , 353401 , 124546 ; 414 2397 799  330008 1769668 1897184   111634 354328 373754  0.5, 0.3 >, < 2397 , 11985 , 11985 ; 0.5, 0.3 >, < 799 , 2397 , 2397 ; 0.5, 0.3 > represent the minimum value. So, all these triangular intuitionistic fuzzy numbers represent gain of Player I and opti the corresponding  minimum expected 4544 3733 1236 4544 = x , x = , x = = , x = , mal strategy is x11 = 1236 21 31 12 22 3995 11985 11985 3995 11985  3733 x32 = 11985 . 4.7.2.2

Maximum Expected Loss of Player II

Using the proposed Ambika method-II, the maximum expected loss of Player II and the corresponding optimal strategies can be obtained as follows: ∗ , i = 1, 2, 3} ∈ X such that value of Sμ (< (175, 180, 190); 0.6, 0.2 > Step 1: Find {xi1 x11 y11 + < (150, 156, 158); 0.6, 0.2 > x11 y21+ < (80, 90, 100); 0.9, 0.1 > x11 y31 + < (80, 90, 100); 0.9, 0.1 > x21 y11 + < (175, 180, 190); 0.6, 0.2 > x21 y21 + < (180, 185, 190); 0.5, 0.3 > x21 y31 + < (180, 185, 190); 0.5, 0.3 > x31 y11 + < (80, 100, 120); 0.7, 0.2 > x31 y21 + < (150, 156, 170); 0.5, 0.1 > x31 y31 ) is max∗ , i = 1, 2, 3} ∈ X such that imum for all {y j1 , i = 1, 2, 3} ∈ Y as well as find {xi2 value of Sν (< (175, 180, 190); 0.6, 0.2 > x12 y12 + < (150, 156, 158); 0.6, 0.2 > x12 y22+ < (80, 90, 100); 0.9, 0.1 > x12 y32 + < (80, 90, 100); 0.9, 0.1 > x22 y12 + < (175, 180, 190); 0.6, 0.2 > x22 y22 + < (180, 185, 190); 0.5, 0.3 > x22 y32 + < (180, 185, 190); 0.5, 0.3 > x32 y12 + < (80, 100, 120); 0.7, 0.2 > x32 y22 + < (150, 156, 170); 0.5, 0.1 > x32 y32 ) is maximum for all {y j2 , j = 1, 2, 3} ∈ Y i.e., ∗ ∗ , i = 1, 2, 3} and {xi2 , i = 1, 2, 3} of Problem 4.7.11 and find optimal solution {xi1 Problem 4.7.12 respectively. Problem 4.7.11 Maximize ⎧ ⎛ ⎞⎫ < (175, 180, 190); 0.6, 0.2 > x11 y11 + < (150, 156, 158); 0.6, 0.2 > x11 y21 + ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎟⎪ ⎪ ⎨ ⎜ ⎜ < (80, 90, 100); 0.9, 0.1 > x11 y31 + < (80, 90, 100); 0.9, 0.1 > x21 y11 + ⎟⎬ ⎜ ⎟ Sμ ⎜ < (175, 180, 190); 0.6, 0.2 > x21 y21 + < (180, 185, 190); 0.5, 0.3 > x21 y21 ⎟ ⎪ ⎪ ⎪ ⎝ < (180, 185, 190); 0.5, 0.3 > x31 y11 + < (80, 100, 120); 0.7, 0.2 > x31 y21 + ⎠⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ < (150, 156, 170); 0.5, 0.1 > x31 y31 Subject to x11 + x21 + x31 = 1; y11 + y21 + y31 = 1; x11 , x21 , x31 ≥ 0; y11 , y21 , y31 ≥ 0. Problem 4.7.12 Maximize ⎧ ⎛ ⎞⎫ < (175, 180, 190); 0.6, 0.2 > x12 y12 + < (150, 156, 158); 0.6, 0.2 > x12 y22 + ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎟⎪ ⎨ ⎜ ⎜ < (80, 90, 100); 0.9, 0.1 > x12 y32 + < (80, 90, 100); 0.9, 0.1 > x22 y12 + ⎟⎬ ⎜ ⎟ Sν ⎜ < (175, 180, 190); 0.6, 0.2 > x22 y22 + < (180, 185, 190); 0.5, 0.3 > x22 y22 ⎟ ⎪ ⎪ ⎪ ⎝ < (180, 185, 190); 0.5, 0.3 > x32 y12 + < (80, 100, 120); 0.7, 0.2 > x32 y22 + ⎠⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ < (150, 156, 170); 0.5, 0.1 > x32 y32 Subject to x12 + x22 + x32 = 1; y12 + y22 + y32 = 1; x12 , x22 , x32 ≥ 0; y12 , y22 , y32 ≥ 0. Step 2: Since, in Problem 4.7.11 and Problem 4.7.12, only xi1 and xi2 have been considered as decision variables. So, Problem 4.7.11 and Problem 4.7.12 are linear

4.7 Numerical Examples

143

programming problems and hence, the optimal value of Problem 4.7.11 and Problem 4.7.12 will be equal to optimal value of its corresponding dual problem i.e., Problem 4.7.13 and Problem 4.7.14 respectively. Problem 4.7.13 Minimize{ω1 } Subject to  Sμ

Sμ

 

Sμ

< (175, 180, 190); 0.5, 0.3 > y11 + < (150, 156, 158); 0.5, 0.3 > y21 + < (80, 90, 100); 0.5, 0.3 > y31

 ≤ ω1 ;

< (80, 90, 100); 0.5, 0.3 > y11 + < (175, 180, 190); 0.5, 0.3 > y21 + < (180, 185, 190); 0.5, 0.3 > y31 < (180, 185, 190); 0.5, 0.3 > y11 + < (80, 100, 120); 0.5, 0.3 > y21 + < (150, 160, 170); 0.5, 0.3 > y31



≤ ω1 ;



≤ ω1 ;

y11 + y21 + y31 = 1; y11 , y21 , y31 ≥ 0. Problem 4.7.14 Minimize{ω2 } Subject to  Sν

Sν Sν

 

< (175, 180, 190); 0.5, 0.3 > y12 + < (150, 156, 158); 0.5, 0.3 > y22 + < (80, 90, 100); 0.5, 0.3 > y32

 ≤ ω2 ;

< (80, 90, 100); 0.5, 0.3 > y12 + < (175, 180, 190); 0.5, 0.3 > y22 + < (180, 185, 190); 0.5, 0.3 > y32 < (180, 185, 190); 0.5, 0.3 > y12 + < (80, 100, 120); 0.5, 0.3 > y22 + < (150, 160, 170); 0.5, 0.3 > y32





≤ ω2 ;

≤ ω2 ;

y12 + y22 + y32 = 1; y12 , y22 , y32 ≥ 0. Step 3: The optimal solution of Problem 4.7.14 is  4.7.13  and Problem   904 4316 67 904 4316 67 and y12 = 2397 , y21 = 11985 , y31 = 11985 , y22 = 11985 , y32 = 11985 y11 = 2397 respectively.   904 4316 67 and , y21 = 11985 , y31 = 11985 Step 4: Substituting the optimal solution y11 = 2397   904 4316 67 y12 = 2397 , y22 = 11985 , y32 = 11985 of Problem 4.7.13 and Problem 4.7.14 in Problem 4.7.11 and Problem 4.7.12 respectively, the alternative basic optimal so1∗ 1∗ 1∗ = 1, x21 = 0, x31 = 0}, lution of Problem 4.7.13 and Problem 4.7.14 is {x11 2∗ 2∗ 2∗ 3∗ 3∗ 3∗ 1∗ 1∗ = 0, {x11 = 0, x21 = 1, x31 = 0}, {x11 = 0, x21 = 0, x31 = 1} and {x12 = 1, x22 1∗ 2∗ 2∗ 2∗ 3∗ 3∗ 3∗ = 0}, {x12 = 0, x22 = 1, x32 = 0}, {x12 = 0, x22 = 0, x32 = 1} respectively. x32 t∗ ∗ y11 + < (150, 156, 158); Step 5: Now, maximum {< (175, 180, 190); 0.6, 0.2 > x11 t∗ ∗ t∗ ∗ y21 + < (80, 90, 100); 0.9, 0.1 > x11 y31 + < (80, 90, 100); 0.9, 0.1 > 0.6, 0.2 > x11 t∗ ∗ t∗ ∗ x21 y11 + < (175, 180, 190); 0.6, 0.2 > x21 y21 + < (180, 185, 190); 0.5, 0.3 > t∗ ∗ t∗ ∗ y21 + < (180, 185, 190); 0.5, 0.3 > x31 y11 + < (80, 100, 120); 0.7, 0.2 > x21 t∗ ∗ t∗ ∗ y21 + < (150, 156, 170); 0.5, 0.1 > x31 y31 , t = 1, 2, 3; < (175, 180, 190); x31 w∗ ∗ w∗ ∗ y12 + < (150, 156, 158); 0.6, 0.2 > x12 y22 + < (80, 90, 100); 0.6, 0.2 > x12 w∗ ∗ w∗ ∗ y32 + < (80, 90, 100); 0.9, 0.1 > x22 y12 + < (175, 180, 190); 0.9, 0.1 > x12 w∗ ∗ w∗ ∗ y22 + < (180, 185, 190); 0.5, 0.3 > x22 y22 + < (180, 185, 190); 0.6, 0.2 > x22 w∗ ∗ w∗ ∗ 0.5, 0.3 > x32 y12 + < (80, 100, 120); 0.7, 0.2 > x32 y2 + <  (150, 156, 170); 590102 1855628 w∗ ∗ y32 , w = 1, 2, 3} = < 112688 , , ; 0.5, 0.3 >, 0.5, 0.1 > x32 799  326246399535432811985   112248 353249 124690  127470 < 799 , 2397 , 799 ; 0.5, 0.3 >, < 2397 , 2397 , 799 ; 0.5, 0.3 >.

144

4 Matrix Games with Intuitionistic Fuzzy Payoffs

  Step 6: Since, all the triangular intuitionistic fuzzy numbers < 112688 , 590102 , 1855628 ; 799 3995 11985  112248 353249 124690   326246 354328 127470  0.5, 0.3 >, < 799 , 2397 , 799 ; 0.5, 0.3 >, < 2397 , 2397 , 799 ; 0.5, 0.3 > represent the maximum value. So, all these triangular intuitionistic fuzzy numbers represent expected loss of PlayerII and opti the corresponding  maximum 904 4316 67 904 4316 = y12 = 2397 , y21 = 11985 , y31 = 11985 , y22 = 11985 , mal strategy is y11 = 2397  67 y32 = 11985 .

4.7.3 Existing Numerical Example Considered by Nan et al. In this section, matrix games with intuitionistic fuzzy payoffs  < (175, 180, 190); 0.6, 0.2 > < (150, 156, 158); 0.6, 0.1 > A˜ = , < (80, 90, 100); 0.9, 0.1 > < (175, 180, 190); 0.6, 0.2 > chosen by Nan et al. [9], is solved by the proposed Ambika method-III.

4.7.3.1

Minimum Expected Gain of Player I

Using the proposed Ambika method-III, the minimum expected gain of Player I and the corresponding optimal strategies can be obtained as follows: Step 1: Find {y j , j = 1, 2} ∈ Y such that value of Vλ (< (175, 180, 190); 0.6, 0.2 > x1 y1 + < (150, 156, 158); 0.6, 0.1 > x1 y2 + < (80, 90, 100); 0.9, 0.1 > x2 y1 + < (175, 180, 190); 0.6, 0.2 > x2 y2 ) is minimum for all (x1 , x2 ) ∈ X i.e., find optimal solution {y j , j = 1, 2} of Problem 4.7.15. Problem 4.7.15 Minimize    < (175, 180, 190); 0.6, 0.2 > x1 y1 + < (150, 156, 158); 0.6, 0.1 > x1 y2 + Vλ < (80, 90, 100); 0.9, 0.1 > x2 y1 + < (175, 180, 190); 0.6, 0.2 > x2 y2 Subject to x1 + x2 = 1; y1 + y2 = 1; x1 , x2 ≥ 0, y1 , y2 ≥ 0. Step 2: Since, in Problem 4.7.15, only y j has been considered as decision variables. So, Problem 4.7.15 is a linear programming problem and hence, the optimal value of Problem 4.7.15 will be equal to optimal value of its corresponding dual problem i.e., Problem 4.7.16. Problem 4.7.16 Maximize{υ} Subject to Vλ (< (175, 180, 190); 0.6, 0.2 > x1 + < (80, 90, 100); 0.6, 0.2 > x2 ) ≥ υ; Vλ (< (150, 156, 158); 0.6, 0.2 > x1 + < (175, 180, 190); 0.6, 0.2 > x2 ) ≥ υ; x1 + x2 = 1; x1 , x2 ≥ 0.   , x2 = 153 . Step 3: The optimal solution of Problem 4.7.16 is x1 = 545 698 698   545 153 Step 4: Substituting the optimal solution x1 = 698 , x2 = 698 of Problem 4.7.16 in Problem 4.7.15, the basic optimal solutions of Problem 4.7.15 are {y11∗ = 1,

4.7 Numerical Examples

145

y21∗ = 0}, {y12∗ = 0, y22∗ = 1}. Since, there exist more than one optimal solution {y 1∗ = 1, y21∗ = 0}, {y12∗ = 0, y22∗ = 1} of Problem 4.7.15 so find maximum  1  Aλ

< (175, 180, 190); 0.6, 0.2 > x1∗ y1h∗ + < (150, 156, 158); 0.6, 0.1 > x1∗ y2h∗ + < (80, 90, 100); 0.9, 0.1 > x2∗ y1h∗ + < (175, 180, 190); 0.6, 0.2 > x2∗ y2h∗ , h = 1, 2

  =maximum 2996 , 5324 = 2996 . Since, 2996 represents the maximum value so, mini1745 5235 1745 1745 mum expected gain of Player I is < (175, 180, 190); 0.6, 0.2 > x1∗ y11∗ + ∗ 1∗ 90, 100); 0.9, 0.1 < (150, 156, 158); 0.6, 0.1 > x1∗ y21∗ + < (80,  > x2 y1 +  107615 55935 59425 ∗ 1∗ < (175, 180, 190); 0.6, 0.2 > x2 y2 =< 698 , 349 , 349 ; 0.6, 0.2 > and cor  responding optimal strategy for Player I will be x1 = 545 , which is op, x2 = 153 698 698 timal solution of Problem 4.7.16.

4.7.3.2

Maximum Expected Loss of Player II

Using the proposed Ambika method-III, the maximum expected loss of Player II and the corresponding optimal strategies can be obtained as follows: Step 1: Find {xi , i = 1, 2} ∈ X such that value of Vλ (< (175, 180, 190); 0.6, 0.2 > x1 y1 + < (150, 156, 158); 0.6, 0.1 > x1 y2 + < (80, 90, 100); 0.9, 0.1 > x2 y1 + < (175, 180, 190); 0.6, 0.2 > x2 y2 ) is maximum for all (y1 , y2 ) ∈ Y i.e., find optimal solution {xi , i = 1, 2} of Problem 4.7.17. Problem 4.7.17 Maximize    < (175, 180, 190); 0.6, 0.2 > x1 y1 + < (150, 156, 158); 0.6, 0.1 > x1 y2 + Vλ < (80, 90, 100); 0.9, 0.1 > x2 y1 + < (175, 180, 190); 0.6, 0.2 > x2 y2 Subject to x1 + x2 = 1; y1 + y2 = 1; x1 , x2 ≥ 0, y1 , y2 ≥ 0. Step 2: Since, in Problem 4.7.17, only xi has been considered as decision variables. So, Problem 4.7.17 is a linear programming problem and hence, the optimal value of Problem 4.7.17 will be equal to optimal value of its corresponding dual problem i.e., Problem 4.7.18. Problem 4.7.18 Minimize{ω} Subject to Vλ (< (175, 180, 190); 0.6, 0.2 > y1 + < (150, 156, 158); 0.6, 0.2 > y2 ) ≤ ω; Vλ (< (80, 90, 100); 0.6, 0.2 > y1 + < (175, 180, 190); 0.6, 0.2 > y2 ) ≤ ω; y1 + y2 = 1; y1 , y2 ≥ 0.   . , y = 545 Step 3: The optimal solution of Problem 4.7.18 is y1 = 153 698  2 698  545 Step 4: Substituting the optimal solution y1 = 153 of Problem 4.7.18 in , y = 2 698 698 Problem 4.7.17, the basic optimal solutions of Problem 4.7.17 are {x11∗ = 1, x21∗ = 0}, {x12∗ = 0, x22∗ = 1}. Since, there exist more than one optimal solution {y11∗ = 1, y 1∗ = 0}, {y12∗ = 0, y22∗ = 1} of Problem 4.7.15 so find minimum   2 Aλ

< (175, 180, 190); 0.6, 0.2 > x1t∗ y1∗ + < (150, 156, 158); 0.6, 0.1 > x1t∗ y2∗ + < (80, 90, 100); 0.9, 0.1 > x2t∗ y1∗ + < (175, 180, 190); 0.6, 0.2 > x2t∗ y2∗ , t = 1, 2

= minimum

 5324

5235

 = , 2996 1745

5324 . 5235

Since,

5324 5235

represents the minimum value so, maxi-

146

4 Matrix Games with Intuitionistic Fuzzy Payoffs

mum expected loss of Player II is < (175, 180, 190); 0.6, 0.2 > x11∗ y1∗ + 1∗ ∗ 90, 100); 0.9, 0.1 < (150, 156, 158); 0.6, 0.1 > x11∗ y2∗ + < (80,  108525  > x2 y1 + 56280 57590 1∗ ∗ < (175, 180, 190); 0.6, 0.2 > x2 y2 =< 698 , 349 , 349 ; 0.6, 0.2 > and the   corresponding optimal strategy for Player II will be y1 = 153 , y2 = 545 , which 698 698 is optimal solution of Problem 4.7.17.

4.7.4 Existing Numerical Example Considered by Nan et al. In this section, matrix games with intuitionistic fuzzy payoffs  < (175, 180, 180, 190); 0.6, 0.3 > < (150, 156, 156, 158); 0.5, 0.2 > ˜ A= , < (125, 128, 132, 140); 0.9, 0.1 > < (175, 185, 195, 200); 0.5, 0.4 > chosen by Nan et al. [10], is solved by the proposed Ambika method-IV.

4.7.4.1

Minimum Expected Gain of Player I

Using Ambika method-IV, the minimum expected gain of Player I and the corresponding optimal strategies can be obtained as follows: Step 1: Find {y j1 , j = 1, 2} ∈ Y such that value of Vλ (< (175, 180, 180, 190); 0.6, 0.3 > x11 y11 + < (150, 156, 156, 158); 0.5, 0.2 > x11 y21 + < (125, 128, 132, 140); 0.9, 0.1 > x21 y11 + < (175, 185, 195, 200); 0.5, 0.4 > x21 y21 ), is minimum ∀{xi1 , i = 1, 2} ∈ X as well as find {y j2 , j = 1, 2} ∈ Y such that value of Aλ (< (175, 180, 180, 190); 0.6, 0.3 > x12 y12 + < (150, 156, 156, 158); 0.5, 0.2 > x12 y22 + < (125, 128, 132, 140); 0.9, 0.1 > x22 y12 + < (175, 185, 195, 200); 0.5, 0.4 > x22 y22 ), is minimum ∀{xi2 , i = 1, 2} ∈ X , is maximum i.e., find optimal solution {y j1 , j = 1, 2} ∈ Y and {y j2 , j = 1, 2} ∈ Y such that value of of Problem 4.7.19 and Problem 4.7.20 respectively. Problem 4.7.19 Minimize   Vλ

< (175, 180, 180, 190); 0.6, 0.3 > x11 y11 + < (150, 156, 156, 158); 0.5, 0.2 > x11 y21 + < (125, 128, 132, 140); 0.9, 0.1 > x21 y11 + < (175, 185, 195, 200); 0.5, 0.4 > x21 y21



Subject to x11 + x21 = 1; y11 + y21 = 1; x11 , x21 ≥ 0; y11 , y21 , ≥ 0. Problem 4.7.20 Maximize   Aλ

< (175, 180, 180, 190); 0.6, 0.3 > x12 y12 + < (150, 156, 156, 158); 0.5, 0.2 > x12 y22 + < (125, 128, 132, 140); 0.9, 0.1 > x22 y12 + < (175, 185, 195, 200); 0.5, 0.4 > x22 y22



Subject to x12 + x22 = 1; y12 + y22 = 1; x12 , x22 ≥ 0; y12 , y22 , ≥ 0. Step 2: Since, in Problem 4.7.19 and Problem 4.7.20, only {y j1 , j = 1, 2} and {y j1 , j = 1, 2} have been considered as decision variables. So, Problem 4.7.19 and Problem 4.7.20 are linear programming problems and hence, the optimal value of

4.7 Numerical Examples

147

Problem 4.7.19 and Problem 4.7.20 will be equal to optimal value of its corresponding dual problem i.e., Problem 4.7.21 and Problem 4.7.22 respectively. Problem 4.7.21 Maximize{υ1 } Subject to Vλ (< (175, 180, 180, 190); 0.5, 0.4 > x11 + < (125, 128, 132, 140); 0.5, 0.4 > x21 ) ≥ υ1 ; Vλ (< (150, 156, 156, 158); 0.5, 0.4 > x11 + < (175, 185, 195, 200); 0.5, 0.4 > x21 ) ≥ υ1 ; x11 + x21 = 1; x11 , x21 ≥ 0. Problem 4.7.22 Minimize{υ2 } Subject to Aλ (< (175, 180, 180, 190); 0.5, 0.4 > x11 + < (125, 128, 132, 140); 0.5, 0.4 > x21 ) ≤ υ2 ; Aλ (< (150, 156, 156, 158); 0.5, 0.4 > x11 + < (175, 185, 195, 200); 0.5, 0.4 > x21 ) ≤ υ2 ; x12 + x22 = 1; x12 , x22 ≥ 0.  ∗ = 350 , Step 3: The optimal solution of Problem 4.7.21 and Problem 4.7.22 is x11 503    153 ∗ ∗ ∗ x21 = 503 and x12 = 1, x22 = 0 respectively.   ∗  ∗ ∗ = 350 , x21 = 153 and x12 = 1, Step 4: Substituting the optimal solution x11 503 503  ∗ = 0 of Problem 4.7.21 and Problem 4.7.22 in Problem 4.7.19 and Problem x22 1∗ 1∗ 4.7.20 respectively, the alternative basic optimal solutions are {y11 = 1, y21 = 0}, 2∗ 2∗ 1∗ 1∗ {y11 = 0, y21 = 1} and {y12 = 1, y21 = 0 of Problem 4.7.17 and Problem 4.7.18 respectively. ∗ h∗ y11 + < (150, 156, Step 5: Now, minimum{< (175, 180, 180, 190); 0.6, 0.3 > x11 ∗ h∗ ∗ h∗ y11 + 156, 158); 0.5, 0.2 > x11 y21 + < (125, 128, 132, 140); 0.9, 0.1 > x21 ∗ h∗ < (175, 185, 195, 200); 0.5, 0.4 > x21 y21 , h = 1, 2; < (175, 180, 180, 190); ∗ 1∗ ∗ 1∗ y12 + < (150, 156, 156, 158); 0.5, 0.2 > x12 y22 + < (125, 128, 0.6, 0.3 > x12 ∗ 1∗ ∗ 1∗ y + < (175, 185, 195, 200); 0.5, 0.4 > x22 y22 } = 132, 140); 0.9, 0.1 > x 22 12  79275 82905 84435 85900 < 503 , 503 , 503 , 503 ; 0.5, 0.4 >, < (175, 180, 180, 195); 0.5, 0.4 >.   Step 6: The trapezoidal intuitionistic fuzzy numbers < 79275 ; , 82905 , 84435 , 85900 503 503 503 503 0.5, 0.4 > and < (175, 180, 180, 195); 0.5, 0.4 > represent the minimum expected   ∗ ∗ and = 350 , x21 = 153 gain of Player I and corresponding optimal strategies are x11 503 503   ∗ ∗ = 0 which is optimal solution of Problem 4.7.21 and Problem 4.7.22 x12 = 1, x22 respectively.

4.7.4.2

Maximum Expected Loss of Player II

Using Ambika method-IV, the maximum expected loss of Player II and the corresponding optimal strategies can be obtained as follows: Step 1: Find {xi1 , i = 1, 2} ∈ X such that value of Vλ (< (175, 180, 180, 190); 0.6,

148

4 Matrix Games with Intuitionistic Fuzzy Payoffs

0.3 > x11 y11 + < (150, 156, 156, 158); 0.5, 0.2 > x11 y21 + < (125, 128, 132, 140); 0.9, 0.1 > x21 y11 + < (175, 185, 195, 200); 0.5, 0.4 > x21 y21 ), is maximum ∀{y j1 , j = 1, 2} ∈ Y as well as find {xi2 , x = 1, 2} ∈ X such that value of Aλ (< (175, 180, 180, 190); 0.6, 0.3 > x12 y12 + < (150, 156, 156, 158); 0.5, 0.2 > x12 y22 + < (125, 128, 132, 140); 0.9, 0.1 > x22 y12 + < (175, 185, 195, 200); 0.5, 0.4 > x22 y22 ), is maximum ∀{y j2 , j = 1, 2} ∈ Y , is minimum i.e., find optimal solution {xi1 , i = 1, 2} ∈ X and {xi2 , i = 1, 2} ∈ X such that value of of Problem 4.7.23 and Problem 4.7.24 respectively. Problem 4.7.23 Maximize    < (175, 180, 180, 190); 0.6, 0.3 > x11 y11 + < (150, 156, 156, 158); 0.5, 0.2 > x11 y21 + Vλ < (125, 128, 132, 140); 0.9, 0.1 > x21 y11 + < (175, 185, 195, 200); 0.5, 0.4 > x21 y21 Subject to x11 + x21 = 1; y11 + y21 = 1; x11 , x21 ≥ 0; y11 , y21 , ≥ 0. Problem 4.7.24 Minimize   Aλ

< (175, 180, 180, 190); 0.6, 0.3 > x12 y12 + < (150, 156, 156, 158); 0.5, 0.2 > x12 y22 + < (125, 128, 132, 140); 0.9, 0.1 > x22 y12 + < (175, 185, 195, 200); 0.5, 0.4 > x22 y22



Subject to x12 + x22 = 1; y12 + y22 = 1; x12 , x22 ≥ 0; y12 , y22 , ≥ 0. Step 2: Since, in Problem 4.7.23 and Problem 4.7.24, only {xi1 , i = 1, 2} and {xi1 , i = 1, 2} have been considered as decision variables. So, Problem 4.7.23 and Problem 4.7.24 are linear programming problems and hence, the optimal value of Problem 4.7.23 and Problem 4.7.24 will be equal to optimal value of its corresponding dual problem i.e., Problem 4.7.25 and Problem 4.7.26 respectively. Problem 4.7.25 Minimize{ω1 } Subject to Vλ (< (175, 180, 180, 190); 0.5, 0.4 > y11 + < (150, 156, 156, 158); 0.5, 0.4 > y21 ) ≤ ω1 ; Vλ (< (125, 128, 132, 140); 0.5, 0.4 > y11 + < (175, 185, 195, 200); 0.5, 0.4 > y21 ) ≤ ω1 ; y11 + y21 = 1; y11 , y21 ≥ 0. Problem 4.7.26 Maximize{ω2 } Subject to Vλ (< (175, 180, 180, 190); 0.5, 0.4 > y12 + < (150, 156, 156, 158); 0.5, 0.4 > y22 ) ≥ ω2 ; Vλ (< (125, 128, 132, 140); 0.5, 0.4 > y12 + < (175, 185, 195, 200); 0.5, 0.4 > y22 ) ≥ ω2 ; y12 + y22 = 1; y12 , y22 ≥ 0.

4.7 Numerical Examples

149

 ∗ Step 3: The optimal solution of Problem 4.7.25 and Problem 4.7.26 is y11 = 203 , 503    300 ∗ ∗ ∗ y21 = 503 and y12 = 1, y22 = 0 respectively.  ∗   ∗ ∗ Step 4: Substituting the optimal solution y11 and y12 = 203 , y21 = 300 = 1, 503 503  ∗ = 0 of Problem 4.7.25 and Problem 4.7.26 in Problem 4.7.23 and Problem y22 4.7.24 respectively, the alternative basic optimal solutions of Problem 4.7.23 and 1∗ 1∗ 2∗ 2∗ 1∗ 1∗ = 1, x21 = 0}, {x11 = 0, x21 = 1} and {x12 = 1, x21 = 0} Problem 4.7.24 are {x11 respectively. t∗ ∗ y11 + < (150, 156, Step 5: Now, maximum{< (175, 180, 180, 190); 0.6, 0.3 > x11 t∗ ∗ t∗ ∗ y11 + 156, 158); 0.5, 0.2 > x11 y21 + < (125, 128, 132, 140); 0.9, 0.1 > x21 t∗ ∗ < (175, 185, 195, 200); 0.5, 0.4 > x21 y21 , t = 1, 2; < (175, 180, 180, 190); 1∗ ∗ 1∗ ∗ y12 + < (150, 156, 156, 158); 0.5, 0.2 > x12 y22 + < (125, 128, 0.6, 0.3 > x12 1∗ ∗ 1∗ ∗ y + < (175, 185, 195, 200); 0.5, 0.4 > x22 y22 } = 132, 140); 0.9, 0.1 > x 22 12   80525 83340 83340 85970 < 503 , 503 , 503 , 503 ; 0.5, 0.4 >, < (175, 180, 180, 190); 0.5, 0.4 >.   Step 6: The trapezoidal intuitionistic fuzzy numbers < 80525 , 83340 , 83340 , 85970 ; 503 503 503 503 0.5, 0.4 > and < (175, 180, 180, 190); 0.5, 0.4 > represent the maximum expected   ∗ ∗ = 203 , y21 = 300 and loss of Player II and corresponding optimal strategies are y11 503 503   ∗ ∗ = 0 which is optimal solution of Problem 4.7.25 and Problem 4.7.26 y12 = 1, y22 respectively.

4.8 Conclusion On the basis of present study, it can be concluded that some mathematically incorrect assumptions have been considered in the existing methods [6, 8–10] for solving matrix games with intuitionistic fuzzy payoffs. Therefore, it is not desirable to use these methods. Furthermore, to resolve flaws of the existing methods [6, 8–10], new methods (named as Ambika methods) is proposed for solving matrix games with intuitionistic fuzzy payoffs.

References 1. Atanassov, K.T.: Intuitionistic fuzzy sets. Fuzzy Sets Syst. 20, 87–96 (1986) 2. Bustince, H., Burillo, P.: Vague sets are intuitionistic fuzzy sets. Fuzzy Sets Syst. 79, 403–405 (1996) 3. Chang, J.R., Chang, K.H., Liao, S.H., Cheng, C.H.: The reliability vague fault-tree analysis on weapon systems fault diagnosis. Soft Comput. 10, 531–542 (2006) 4. Gau, W.L., Buehrer, D.J.: Vague sets. IEEE Trans. Syst. Man Cybern. 23, 29–41 (1993) 5. Li, D.F.: Decision and Game Theory in Management with Intuitionistic Fuzzy Sets. Springer, Berlin (2014) 6. Li, D.F., Nan, J.X., Tang, Z.P., Chen, K.J., Xiang, X.D., Hong, F.X.: A bi-objective programming approach to solve matrix games with payoffs of Atanassov’s triangular intuitionistic fuzzy numbers. Iran. J. Fuzzy Syst. 9, 93–110 (2012)

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7. Li, D.F., Yang, J.: A difference-index based ranking bilinear programming approach to solving bimatrix games with payoffs of trapezoidal intuitionistic fuzzy numbers. J. Appl. Math. 2013, 1–10 (2013) 8. Nan, J.X., Li, D.F., Zhang, M.J.: A lexicographic method for matrix games with payoffs of triangular intuitionistic fuzzy numbers. Int. J. Comput. Intell. Syst. 3, 280–289 (2010) 9. Nan, J.X., Zhang, M.J., Li, D.F.: A methodology for matrix games with payoffs of triangular intuitionistic fuzzy number. J. Intell. Fuzzy Syst. 26, 2899–2912 (2013) 10. Nan, J.X., Zhang, M.J., Li, D.F.: Intuitionistic fuzzy programming models for matrix games with payoffs of trapezoidal intuitionistic fuzzy numbers. Int. J. Fuzzy Syst. 16, 444–456 (2013) 11. Seikh, M.R., Nayak, P.K., Pal, M.: Matrix games with intuitionistic fuzzy payoffs. J. Inf. Optim. Sci. 36, 159–181 (2015)

Chapter 5

Bimatrix Games with Intuitionistic Fuzzy Payoffs

Li and Yang [2] pointed out that there is no method in the literature for solving such bimatrix games or two person non-zero sum games (matrix games in which gain of one player is not equal to the loss of other player) in which payoffs are represented by intuitionistic fuzzy numbers and proposed a method for the same. In this chapter, it is pointed out that Li and Yang have considered some mathematically incorrect assumptions in their proposed method. For resolving the shortcomings of Li and Yang’s method, a new method (named as Mehar method) is proposed. Also, the exact optimal solution of the numerical problem, solved by Li and Yang by their pro-posed method, is obtained by the proposed Mehar method.

5.1 The Difference-Index Based Ranking Method In this section, existing difference-index based ranking method [1], used by Li and Yang [2] for comparing trapezoidal intuitionistic fuzzy numbers, is presented [2, Sect. 2.3, Definition 8, pp. 4].    If a˜ 1 =< a1L (0), a1L (1), a1R (1), a1R (0) ; wa˜ 1 , u a˜ 1 > and a˜ 2 =< a2L (0), a2L (1), a2R (1), a2R (0) ; wa˜ 2 , u a˜ 2 > are two trapezoidal intuitionistic fuzzy numbers and λ ∈ [0, 1]. Then, (i) a˜ 1  a˜ 1 if and only if Dλ (a˜ 1 ) > Dλ (a˜ 2 ). (ii) a˜ 1 ≺ a˜ 2 if and only if Dλ (a˜ 1 ) < Dλ (a˜ 2 ). (iii) a˜ 1 ≈ a˜ 2 if and only if Dλ (a˜ 1 ) = Dλ (a˜ 2 ). where, Dλ (a˜ i ) = Vλ (a˜ i ) − Aλ (a˜ i )

        =< Vλ < aiL (0), aiL (1), aiR (1), aiR (0) ; wa˜ i , u a˜ i > − Aλ < aiL (0), aiL (1), aiR (1), aiR (0) ; wa˜ i , u a˜ i > 

= (λ(1 − u a˜ i ) + (1 − λ)wa˜ i )

aiL (0)+2aiL (1)+2aiR (1)+aiR (0) 12





− (λwa˜ i + (1 − λ)(1 − u a˜ i ))

aiR (0)−aiL (0)+2aiR (1)−2aiL (1) 12



.

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5 Bimatrix Games with Intuitionistic Fuzzy Payoffs

5.2 Maximum of Trapezoidal Intuitionistic Fuzzy Numbers In this section, the method, used by Li and Yang [2], for finding the maximum of trapezoidal intuitionistic fuzzy numbers,   is presented. It is obvious from Sect. 5.1 that if < aiL (0), aiL (1), aiR (1), aiR (0) ; wa˜ i , u a˜ i>, i = 1, 2, ..., n are n trapezoidal intuitionistic fuzzy numbers and λ ∈ [0, 1]. Then, maximum of these numbers can be obtained as follows: Step 1: Find Dλ (a˜ i ) = Vλ (a˜ i ) − Aλ (a˜ i ).

        =< Vλ < aiL (0), aiL (1), aiR (1), aiR (0) ; wa˜ i , u a˜ i > − Aλ < aiL (0), aiL (1), aiR (1), aiR (0) ; wa˜ i , u a˜ i > 

= (λ(1 − u a˜ ) + (1 − λ)wa˜ ) i i

aiL (0)+2aiL (1)+2aiR (1)+aiR (0) 12





− (λwa˜ + (1 − λ)(1 − u a˜ )) i i

aiR (0)−aiL (0)+2aiR (1)−2aiL (1) 12

Step 2: Find max {Dλ (a˜ i )}.

1≤i≤n   Step 3: If max {Dλ (a˜ i )} = Dλ a˜ p (say). Then, max {a˜ i } = a˜ p . 1≤i≤n

1≤i≤n

5.3 Flaws in the Existing Mathematical Formulation of Bimatrix Games with Intuitionistic Fuzzy Payoffs The aim of this section is to point out that Li and Yang [2] have used some mathematically incorrect assumption to obtain the mathematical formulation of bimatrix games with intuitionistic fuzzy payoffs. To point out the same, it is necessary to explain the procedure followed by Li and Yang [2] to obtain this mathematical formulation. So, firstly the procedure followed by Li and Yang [2], to obtain the mathematical formulation of bimatrix games with intuitionistic fuzzy payoffs, is presented and then the mathematically incorrect assumptions, considered by Li and Yang [2], are pointed out.

5.3.1 Mathematical Formulation of Bimatrix Games with Intuitionistic Fuzzy Payoffs Mangasarian and Stone [3] proved that a Nash equilibrium point (a pair of strategies where the objectives of both the players are fulfilled simultaneously) for bimatrix games (two person non-zero sum games) can be obtained by solving a quadratic programming problem. On the same direction, Li and Yang [2] obtained a quadratic Problem 5.3.1 to obtain a Nash equilibrium point for bimatrix games with intuitionistic fuzzy payoffs. Problem 5.3.1 ([2, Equation 22, Sect. 3.2, pp. 5]) 

 m n   ˜ xi Dλ1 a˜ i j + Dλ2 bi j y j − u − v Maximize i=1 j=1



.

5.3 Flaws in the Existing Mathematical Formulation of Bimatrix Games …

153

Subject to n   Dλ1 a˜ i j y j − u ≤ 0, i = 1, 2, ..., m; j=1

m

i=1 m i=1

 Dλ2 b˜i j xi − v ≤ 0, j = 1, 2, ..., n; xi = 1;

n

y j = 1; xi ≥ 0, i = 1, 2, ..., m; y j ≥ 0, j = 1, 2, ..., n.

j=1

To point out the mathematically incorrect assumption considered by Li and Yang [2], it is necessary to explain the method, followed by Li and Yang [2], to obtain the mathematical formulation i.e. Problem 5.3.1 of bimatrix games with intuitionistic fuzzy payoffs. Therefore, in this section, the same is presented. Let, Player I and Player II have mixed strategies as {xi , i =

1,2, ..., m} and   be intu{y j , j = 1, 2, ..., n} respectively. Let, A˜ i j = a˜ i j m×n and B˜ i j = b˜i j m×n

itionistic fuzzy payoff matrices of Player I and Player   II respectively. Player I maximizes profit over rows of fuzzy matrix A˜ m×n = a˜ i j m×n and Player II maximizes

 profit over columns of fuzzy matrix B˜ m×n = b˜i j . Therefore,  the objective of Player I is to n m Maximize xi a˜ i j y j j=1 i=1

Subject to m xi = 1; xi ≥ 0, i = 1, 2, ..., m. i=1

and the objective of Player II is to  n m Maximize xi b˜i j y j j=1 i=1

Subject to n y j = 1; y j ≥ 0, j = 1, 2, ..., n. j=1

Using Sect. of Player I is to  5.1,  the objective n m Maximize Dλ1 xi a˜ i j y j j=1 i=1

Subject to m xi = 1; xi ≥ 0, i = 1, 2, ..., m. i=1

and the objective  of Player II is to  n m xi b˜i j y j Maximize Dλ2 j=1 i=1

Subject to n y j = 1; y j ≥ 0, j = 1, 2, ..., n. j=1

m×n

154

5 Bimatrix Games with Intuitionistic Fuzzy Payoffs





n m



Li and Yang [2] assumed that to Maximize Dλ1 xi a˜ i j y j is equivalent j=1 i=1      n n m m   ˜ xi Dλ1 a˜ i j y j and to Maximize Dλ2 xi bi j y j is to Maximize j=1 i=1 j=1 i=1  

 m n xi Dλ2 b˜i j y j . equivalent to Maximize j=1 i=1

Therefore,  the objective of Player I is to n m   Maximize xi Dλ1 a˜ i j y j j=1 i=1

Subject to m xi = 1; xi ≥ 0, i = 1, 2, ..., m. i=1

and the objective of Player II is to 

 n m xi Dλ2 b˜i j y j Maximize j=1 i=1

Subject to n y j = 1; y j ≥ 0, j = 1, 2, ..., n.



According to Mangasarian and Stone [3], the point xi0 , y 0j will be a Nash equi-

j=1

librium point (a pair of strategies xi0 , y 0j , where the objectives of both the players are fulfilled simultaneously) if there exist real numbers u 0 , v 0 such that xi0 , y 0j , u 0 , v 0 satisfy the following conditions: m n

   xi0 Dλ1 a˜ i j y 0j − u 0 = 0;

i=1 j=1 m n

i=1 j=1 n

j=1

   Dλ1 a˜ i j y 0j − u 0 ≤ 0, i = 1, 2, ..., m;

m

i=1 m i=1

  xi0 Dλ2 b˜i j y 0j − v 0 = 0;

 xi0 Dλ2 b˜i j − v 0 ≤ 0, j = 1, 2, ..., n;

xi0 = 1;

n j=1

y 0j = 1; xi0 ≥ 0, i = 1, 2, ..., m; y 0j ≥ 0, j = 1, 2, ..., n.

Also, according to Mangasarian and Stone [3], the values of xi0 , y 0j , u 0 , v 0 which will satisfy the above conditions will be optimal solution of Problem 5.3.1.

5.3 Flaws in the Existing Mathematical Formulation of Bimatrix Games …

155

5.3.2 Mathematically Incorrect Assumption Considered by Li and Yang It is obvious from Sect. 5.3.1 that to obtain the mathematical   formulation i.e.,  Problem n m 5.3.1, Li and Yang [2] have assumed that to Maximize Dλ1 xi a˜ i j y j is equivj=1 i=1      n n m m   ˜ xi Dλ1 a˜ i j y j and to Maximize Dλ2 xi bi j y j alent to Maximize j=1 i=1



is equivalent to Maximize

n m j=1 i=1

j=1 i=1



 xi Dλ2 b˜i j y j .

However, ⎛ Dλ ⎝

n 

⎞ a˜ i ⎠

i=1

⎛

= Dλ ⎝

⎞ n 

   aiL (0), aiL (1), aiR (1), aiR (0) ; wa˜ , u a˜ ⎠ i i

i=1

⎛⎛ = Dλ ⎝ ⎝

n 

aiL (0),

n 

aiL (1),

n 

aiR (1),

n 

i=1

i=1

i=1

i=1

n 

n 

n 

n 

⎛⎛ = Dλ ⎝ ⎝

aiL (0),

i=1

aiL (1),

i=1

aiR (1),

i=1

i=1

⎞

⎞

aiR (0)⎠ ; min {wa˜ }, max {u a˜ } ⎠ i i 1≤i≤n

1≤i≤n

⎞

⎞

aiR (0)⎠ ; min {wa˜ }, min {(1 − u a˜ )} ⎠ i 1≤i≤n i 1≤i≤n

  ⎞ n n n n aiL (0), aiL (1), aiR (1), aiR (0) ; min {wa˜ }, min {(1 − u a˜ )} − ⎟ ⎜ Vλ i 1≤i≤n i ⎟ ⎜ 1≤i≤n i=1 i=1 i=1 i=1   ⎟  =⎜ ⎟ ⎜ n n n n ⎠ ⎝ Aλ aiL (0), aiL (1), aiR (1), aiR (0) ; min {wa˜ }, min {(1 − u a˜ )} ⎛



i=1

i=1

i=1

i=1

1≤i≤n

i

1≤i≤n

i

⎞ ⎞ ⎛ n n n n L a (0)+2 aiL (1)+2 aiR (1)+ aiR (0) ⎜ ⎟ ⎟ ⎜ i=1 i i=1 i=1 i=1 ⎜ (λ min {(1 − u )} + (1 − λ) min {w }) ⎜ ⎟− ⎟ a˜ i a˜ i ⎝ ⎜ 1≤i≤n ⎠ ⎟ 12 1≤i≤n ⎟ ⎜ ⎟ ⎜ ⎛ ⎞ ⎟ =⎜ n n n n ⎟ ⎜ R L R L ⎟ ⎜ ai (0)− ai (0)+2 ai (1)−2 ai (1) ⎜ ⎜ ⎟ ⎟ i=1 i=1 i=1 ⎜ (λ min {w } + (1 − λ) min {(1 − u )}) ⎜ i=1 ⎟. ⎟ a˜ i ⎠ ⎝ 1≤i≤n a˜ i ⎝ ⎠ 12 1≤i≤n ⎛

and, n 

Dλ (a˜ i )

i=1

=

n 

Dλ

i=1

 L   ai (0), aiL (1), aiR (1), aiR (0) ; wa˜ i , u a˜ i

⎞    L L R R ⎜ i=1 Vλ ai (0), ai (1), ai (1), ai (0) ; wa˜ i , (1 − u a˜ i ) − ⎟ ⎟ =⎜ n  L   ⎠ ⎝ L R R Aλ ai (0), ai (1), ai (1), ai (0) ; wa˜ i , (1 − u a˜ i ) ⎛

n

i=1

(5.1)

156

5 Bimatrix Games with Intuitionistic Fuzzy Payoffs

=

n  i=1

⎛ ⎝

(λwa˜ i + (1 − λ)(1 −





⎞

aiL (0)+2aiL (1)+2aiR (1)+aiR (0) − 12  ⎠

R ai (0)−aiL (0)+2aiR (1)−2aiL (1) u a˜ i )) 12

(λ(1 − u a˜ i ) + (1 − λ)wa˜ i )





n a˜ i = Dλ (a˜ i ). Li and Yang [2, i=1 i=1  n n a˜ i = Dλ (a˜ i ) if and only Theorem 7, pp. 3] have also pointed out that Dλ

It is obvious from Eqs. 5.1 and 5.2 that Dλ

n

(5.2)

i=1

i=1

if for wa˜ i , i = 1, 2, ..., n and u a˜ i , i = 1, 2, ..., n, the conditions wa˜ 1 = wa˜ 2 = ... = wa˜ n and u a˜ 1 = u a˜ 2 = ... =  u a˜ n  will be  satisfied. This clearly n m indicates that to Maximize Dλ1 xi a˜ i j y j is not equivalent to Maximize j=1 i=1      n n m m   xi Dλ1 a˜ i j y j and to Maximize Dλ2 xi b˜i j y j is not equivalent j=1 i=1



to Maximize

n m



 xi Dλ2 b˜i j y j .

j=1 i=1

j=1 i=1

However, Li and Yang [2] have considered the above mentioned mathematically incorrect assumption for obtaining the mathematical formulation i.e., Problem 5.3.1. Therefore, the mathematical formulation i.e., Problem 5.3.1 and hence the method, proposed by Li and Yang [2] for solving bimatrix games with intuitionistic fuzzy payoffs based on the mathematical formulation i.e., Problem 5.3.1, is not valid.

5.4 Exact Solution of Bimatrix Games with Intuitionistic Fuzzy Payoffs Li and Yang [2] claimed that optimal solution {xi , y j , u, v} of any bimatrix games with intuitionistic fuzzy payoffs can be obtained by solving Problem 5.3.1. However, as discussed in Sect. 5.3.2 that the Problem 5.3.1 is not the exact mathematical formulation of bimatrix games with intuitionistic fuzzy payoffs. Therefore, it is not desirable to use Li and Yang’s method [2] for solving bimatrix games with intuitionistic payoffs. In this section, the exact mathematical formulation of bimatrix games with intuitionistic fuzzy payoffs is proposed. Also, a new method (named as Mehar method) to find the exact solution of bimatrix games with intuitionistic fuzzy payoffs is proposed. The necessary and sufficient condition for the convergence of bimatrix games with intuitionistic fuzzy payoffs is also discussed.

5.4 Exact Solution of Bimatrix Games with Intuitionistic Fuzzy Payoffs

157

5.4.1 Exact Mathematical Formulation of Bimatrix Games with Intuitionistic Fuzzy Payoffs n  n It can be easily verified from Sect. 5.1 that Dλ a˜ i = Mλ (a˜ i ), where, i=1 i=1 

L a (0)+2aiL (1)+2aiR (1)+aiR (0) − (λκ + (1 − λ)ξ) Mλ (a˜ i ) = (λξ + (1 − λ)κ) i 12 

R ai (0)−aiL (0)+2aiR (1)−2aiL (1) , ξ = min {(1 − u a˜ i )}, κ = min {wa˜ i }. 12 1≤i≤n 1≤i≤n n  n a˜ i = Mλ (a˜ i ) in the procedure, for Therefore, using the relation Dλ i=1

i=1

obtaining mathematical formulation of bimatrix games with intuitionistic fuzzy payoffs, described in Sect. 5.3.1, the exact mathematical formulation i.e., Problem 5.4.1 of bimatrix games with intuitionistic fuzzy payoffs is obtained. Problem 5.4.1 

 m n   xi Mλ1 a˜ i j + Mλ2 b˜i j y j − u − v Maximize i=1 j=1

Subject to n   Mλ1 a˜ i j y j − u ≤ 0, i = 1, 2, ..., m; j=1

m

i=1 m

 Mλ2 b˜i j xi − v ≤ 0, j = 1, 2, ..., n; xi = 1;

n

y j = 1; xi ≥ 0, i = 1, 2, ..., m; y j ≥ 0, j = 1, 2, ..., n.

a L (0)+2a L (1)+2a R (1)+a R (0)    ij ij ij where, Mλ1 a˜ i j = (λ1 α + (1 − λ1 )β) i j − (λ1 β + (1 − λ1 )α) 12

a R (0)−a L (0)+2a R (1)−2a L (1) 



b L (0)+2b L (1)+2b R (1)+b R (0)  ij ij ij ij ij ij ij , Mλ2 b˜i j = (λ2 ψ + (1 − λ2 )τ ) i j − 12 12 (λ τ + (1 − λ )ψ) 2

2 

i=1

j=1

biRj (0)−biLj (0)+2biRj (1)−2biLj (1) 12

, α = min {(1 − u a˜ i j )}, κ = min {wa˜ i j }, ψ = min {(1 − u b˜i j )}, τ = min {wb˜i j }. 1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

5.4.2 Proposed Mehar Method In this section, a new method (named as Mehar method) is proposed for solving bimatrix games with intuitionistic fuzzy payoffs. The steps of the proposed Mehar method are as follows: Step 1: Find the values of α = min {(1 − u a˜ i j )}, κ = min {wa˜ i j }, ψ = min {(1 − u b˜i j )}, τ = min {wb˜i j }. 1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

1≤i≤m 1≤ j≤n

158

5 Bimatrix Games with Intuitionistic Fuzzy Payoffs

Step 2: Find the values of

a L (0)+2a L (1)+2a R (1)+a R (0)    ij ij ij − (λ1 β + (1 − λ1 )α) Mλ1 a˜ i j = (λ1 α + (1 − λ1 )β) i j 12

a R (0)−a L (0)+2a R (1)−2a L (1)  ij ij ij ij , 12



b L (0)+2b L (1)+2b R (1)+b R (0)  ij ij ij − (λ2 τ + (1 − λ2 )ψ) Mλ2 b˜i j = (λ2 ψ + (1 − λ2 )τ ) i j 12

b R (0)−b L (0)+2b R (1)−2b L (1)  ij ij ij ij . 12

   Step 3: Put the values of Mλ1 a˜ i j and Mλ1 b˜i j , obtained from Step 2, in Problem 5.4.1. Step 4: Find the optimal solution {xi , y j , u, v} of the problem, obtained in Step 3, by choosing λ1 ∈ [0, 1] and λ2 ∈ [0, 1].

5.4.3 Convergence of the Proposed Mehar Method Mangasarian and Stone [3, Sect. II, pp. 349–350] proved that a bimatrix game with crisp payoffs can be formulated into a crisp quadratic programming problem. Mangasarian and Stone [3, Sect. II, pp. 349–350] also proved that the necessary and sufficient condition for the convergence of a bimatrix game with crisp payoffs is that the global optimal value of the corresponding crisp quadratic programming problem is zero. Following the procedure of Mangasarian and Stone [3, Sect.

II, pp. 349–350]   and replacing crisp payoffs ai j and bi j by Dλ1 a˜ i j and Dλ2 b˜i j respectively, Li and Yang [2] proved that a bimatrix game with intuitionistic fuzzy payoffs can be formulated into a quadratic programming Problem 5.3.1. However, in Sect. 5.3.2, it is pointed out that the Problem 5.3.1 is not the exact mathematical formulation of a bimatrix game with intuitionistic fuzzy payoffs. Also, in Sect. 5.4.1, it is pointed out that the exact mathematical formulation of a bimatrix game with intuitionistic fuzzy payoffs can be obtained by following the procedure of Mangasarian and  Stone  a˜ i j and and b by M [3, Sect. II, pp. 349–350] and replacing crisp payoffs a i j i j λ 1

    Mλ2 b˜i j respectively instead of replacing ai j and bi j by Dλ1 a˜ i j and Dλ2 b˜i j respectively.

   Similarly, replacing crisp payoffs ai j and bi j by Mλ1 a˜ i j and Mλ2 b˜i j respectively in the existing proof [3, Sect. II, pp. 349–350], it can be easily proved that the necessary and sufficient condition for the convergence of a bimatrix games with intuitionistic fuzzy payoffs is that the global optimal value of the corresponding mathematical quadratic programming Problem 5.4.1 is zero.

5.5 Numerical Example

159

5.5 Numerical Example Li and Yang [2, Sect. 4, pp. 6–8] solved a commerce retailer’s strategy choice problem to illustrate their proposed method by considering  (50, 60, 70, 80); 0.8, 0.1 (30, 40, 70, 80); 0.4, 0.3

˜ A= and (20, 30, 40, 50); 0.5, 0.4 (40, 50, 60, 70); 0.6, 0.2

  (40, 50, 60, 70); 0.7, 0.1 (30, 40, 50, 60); 0.7, 0.2

as intuitionistic fuzzy B˜ = (20, 30, 40, 50); 0.5, 0.3 (50, 60, 70, 80); 0.8, 0.1

payoffs matrices of first and second commercial retailer respectively. However, as discussed in Sect. 5.3.2, that the existing method [2] is not valid. So, the intuitionistic fuzzy optimal solution of commerce retailer’s strategy choice problem, obtained by Li and Yang [2], is not exact. Also, it can be easily verified that the value of objective function of Problem 5.3.1 corresponding to the existing optimal solution, obtained by Li and Yang [2] shown in Tables 5.1, 5.2, 5.3, 5.4, 5.5, 5.6, 5.7, 5.8, 5.9, 5.10, 5.11 and 5.12, is not zero. While, according to Mangasarian and Stone [3], an optimal solution of Problem 5.3.1 will be Nash equilibrium if and only if the value of objective function of Problem 5.3.1 corresponding to that optimal solution is zero. This clearly indicates that the optimal solution of commerce retailer’s strategy choice problem, obtained by Li and Yang [2] shown in Tables 5.1, 5.2, 5.3, 5.4, 5.5, 5.6, 5.7, 5.8, 5.9, 5.10, 5.11 and 5.12, are actually not Nash equilibrium points. In this section, the exact optimal solution (Nash equilibrium points) of this existing problem is obtained by the proposed Mehar method. Using the proposed Mehar method, the exact optimal solution of this problem can be obtained as follows: Step 1: Using Step 1 of proposed Mehar method α = min{(1 − 0.1), (1 − 0.3), (1 − 0.4), (1 − 0.2)} = min{0.9, 0.7, .6, 0.8} = 0.6 β = min{0.8, 0.4, 0.5, 0.6} = 0.4 ψ = min{(1 − 0.1), (1 − 0.2), (1 − 0.3), (1 − 0.1)} = min{0.9, 0.8, .7, 0.9} = 0.7 τ = min{0.7, 0.7, 0.5, 0.8} = 0.5. Step 2: Using Step 2 of the proposed Mehar method, Mλ1 (a˜12 ) = Mλ1 ( (30, 40, 70, 80); 0.4, 0.3 ) = Mλ1 (a˜21 ) = Mλ1 ( (20, 30, 40, 50); 0.5, 0.4 ) =





 390 

50 12 − (0.6 − 0.2λ1 ) 12  330   110  (0.2λ1 + 0.4) 12 − (0.6 − 0.2λ1 ) 12

   50 (0.2λ1 + 0.4) 210 12 − (0.6 − 0.2λ1 ) 12

Mλ1 (a˜11 ) = Mλ1 ( (50, 60, 70, 80); 0.8, 0.1 ) = (0.2λ1 + 0.4)

Mλ1 (a˜22 ) = Mλ1 ( (40, 50, 60, 70); 0.6, 0.2 ) = (0.2λ1 + 0.4) Mλ2 (b˜11 ) = Mλ2 ( (40, 50, 60, 70); 0.7, 0.1 ) = (0.2λ2 + 0.5) Mλ2 (b˜12 ) = Mλ2 ( (30, 40, 50, 60); 0.7, 0.2 ) = (0.2λ2 + 0.5) Mλ2 (b˜21 ) = Mλ2 ( (20, 30, 40, 50); 0.5, 0.3 ) = (0.2λ2 + 0.5) Mλ2 (b˜22 ) = Mλ2 ( (50, 60, 70, 80); 0.8, 0.1 ) = (0.2λ2 + 0.5)

 330  12

− (0.6 − 0.2λ1 )

12

− (0.7 − 0.2λ2 )

12

− (0.7 − 0.2λ2 )

12

− (0.7 − 0.2λ2 )

12

− (0.7 − 0.2λ2 )

 330   270   210   390 









50 12 50 12 50 12 50 12 50 12



   

Step 3: Putting the values, obtained in Step 2, in Problem 5.4.1, then Problem 5.4.1 can be transformed into Problem 5.5.1.

160

5 Bimatrix Games with Intuitionistic Fuzzy Payoffs

Problem 5.5.1 ⎧

 Maximize 

   330   110  50 ⎪ (0.2λ1 + 0.4) 390 ⎪ 12 − (0.6 − 0.2λ1 ) 12  y1 z 1 + (0.2λ1 + 0.4) 12 − (0.6 − 0.2λ1 ) 12  y1 z 2 + ⎪

⎪     ⎪ 50 330 50 ⎪ ⎪ (0.2λ1 + 0.4) 210 ⎪ 12 − (0.6 − 0.2λ1 ) 12  y2 z 1 + (0.2λ1 + 0.4) 12 − (0.6 − 0.2λ1 ) 12  y2 z 2 + ⎨

 330   270  50 (0.2λ2 + 0.5) 12 − (0.7 − 0.2λ2 ) 12 y1 z 1 + (0.2λ2 + 0.5) 12 − (0.7 − 0.2λ2 ) 50 y1 z 2 + ⎪



12  ⎪ ⎪ ⎪ (0.2λ + 0.5)  210  − (0.7 − 0.2λ ) 50 y z + (0.2λ + 0.5)  390  − (0.7 − 0.2λ ) 50 y z ⎪ 2 2 2 1 2 2 2 2 ⎪ 12 12 12 12 ⎪ ⎪ ⎩ −u − v

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

Subject to  390   50    z1+ (0.2λ 1 + 0.4)  12 − (0.6 − 0.2λ1 ) 12   ≤ u; − (0.6 − 0.2λ1 ) 110 z (0.2λ1 + 0.4) 330 12   210  5012 2   (0.2λ1 + 0.4) 12330 − (0.6 − 0.2λ1 ) 1250 z 1 + ≤ u; (0.2λ1 + 0.4) 12 − (0.6 − 0.2λ1 ) 12 z 2  330   50    (0.2λ 2 + 0.5) 12 − (0.7 − 0.2λ2 ) 12 y1 +  ≤ v; (0.2λ2 + 0.5) 270 − (0.7 − 0.2λ2 ) 50 y  21012  5012 1   (0.2λ2 + 0.5) 12390 − (0.7 − 0.2λ2 ) 1250 y1 + ≤ v; (0.2λ2 + 0.5) 12 − (0.7 − 0.2λ2 ) 12 y2 y1 + y2 = 1; z 1 + z 2 = 1; y1 , y2 , z 1 , z 2 ≥ 0. Step 4: The exact optimal solution of commerce retailer’s strategy choice problem with intuitionistic fuzzy payoffs obtained by solving Problem 5.5.1 as well as the incorrect optimal solution of the same problem obtained by Li and Yang [2] are shown in Tables 5.1, 5.2, 5.3, 5.4, 5.5, 5.6, 5.7, 5.8, 5.9, 5.10, 5.11 and 5.12.

Table 5.1 Nash equilibrium values and corresponding strategies of first commerce retailer Parameters Li and Yang’s method [2, Table 1, pp. 7] Proposed Mehar method λ1 λ2 x ∗T u ∗ (λ1 ) x ∗T u ∗ (λ1 ) 3 1 15 0 0 (0.484, 0.516) 7.94 4, 4 2 3 1 15 0 0.3 (0.492, 0.508) 7.94 , 4 4 2 3 1 15 0 0.5 (0.498, 0.502) 7.94 , 4 4 2 3 1 15 0 0.8 (0.495, 0.505) 7.94 , 4 4 2 3 1 15 0 1 (0.490, 0.510) 7.94 4, 4 2 Table 5.2 Nash equilibrium values and corresponding strategies of second commerce retailer Parameters Li and Yang’s method [2, Table 1, pp. 7] Proposed Mehar method λ1 λ2 y ∗T v ∗ (λ2 ) y ∗T v ∗ (λ2 ) 1 3 115 0 0 (0.311, 0.689) 10.46 4, 4 12 1 3 34 0 0.3 (0.311, 0.689) 11.98 , 4 4 3 1 3 25 0 0.5 (0.311, 0.689) 12.98 , 4 4 2 1 3 57 0 0.8 (0.311, 0.689) 14.48 , 4 4 4 1 3 185 0 1 (0.311, 0.689) 15.47 , 4 4 12

5.5 Numerical Example

161

Table 5.3 Nash equilibrium values and corresponding strategies of first commerce retailer Parameters Li and Yang’s method [2, Table 2, pp. 7] Proposed Mehar method λ1 λ2 x ∗T u ∗ (λ1 ) x ∗T u ∗ (λ1 ) 3 1 125 0.5 0 (0.484, 0.516) 11.19 4, 4 12 3 1 125 0.5 0.3 (0.492, 0.508) 11.19 , 4 4 12 3 1 125 0.5 0.5 (0.498, 0.502) 11.19 4, 4 12 3 1 125 0.5 0.8 (0.495, 0.505) 11.19 , 4 4 12 3 1 125 0.5 1 (0.490, 0.510) 11.19 , 4 4 12

Table 5.4 Nash equilibrium values and corresponding strategies of second commerce retailer Parameters Li and Yang’s method [2, Table 2, pp. 7] Proposed Mehar method λ1 λ2 y ∗T v ∗ (λ2 ) y ∗T v ∗ (λ2 ) 1 3 115 0.5 0 (0.266, 0.734) 10.46 4, 4 12 1 3 34 0.5 0.3 (0.266, 0.734) 11.98 , 4 4 3 1 3 25 0.5 0.5 (0.266, 0.734) 12.98 4, 4 2 1 3 57 0.5 0.8 (0.266, 0.734) 14.48 4, 4 4 1 3 185 0.5 1 (0.266, 0.734) 15.47 , 4 4 12

Table 5.5 Nash equilibrium values and corresponding strategies of first commerce retailer Parameters Li and Yang’s method [2, Table 3, pp. 8] Proposed Mehar method λ1 λ2 x ∗T u ∗ (λ1 ) x ∗T u ∗ (λ1 ) 3 1 40 1 0 (0.484, 0.516) 14.65 4, 4 3 3 1 40 1 0.3 (0.492, 0.508) 14.65 , 4 4 3 3 1 40 1 0.5 (0.498, 0.502) 14.65 4, 4 3 3 1 40 1 0.8 (0.495, 0.505) 14.65 , 4 4 3 3 1 40 1 1 (0.490, 0.510) 14.65 , 4 4 3

Table 5.6 Nash equilibrium values and corresponding strategies of second commerce retailer Parameters Li and Yang’s method [2, Table 3, pp. 8] Proposed Mehar method λ1 λ2 y ∗T v ∗ (λ2 ) y ∗T v ∗ (λ2 ) 1 3 115 1 0 (0.220, 0.780) 10.46 4, 4 12 1 3 34 1 0.3 (0.220, 0.780) 11.98 , 4 4 3 1 3 25 1 0.5 (0.220, 0.780) 12.98 4, 4 2 1 3 57 1 0.8 (0.220, 0.780) 14.48 4, 4 4 1 3 185 1 1 (0.220, 0.780) 15.47 , 4 4 12

162

5 Bimatrix Games with Intuitionistic Fuzzy Payoffs

Table 5.7 Nash equilibrium values and corresponding strategies of first commerce retailer Parameters Li and Yang’s method [2, Table 4, pp. 8] Proposed Mehar method λ1 λ2 x ∗T u ∗ (λ1 ) x ∗T u ∗ (λ1 ) 3 1 15 0 0 (0.516, 0.484) 7.94 4, 4 2 3 1 37 0.3 0 (0.516, 0.484) 9.86 , 4 4 4 3 1 125 0.5 0 (0.516, 0.484) 11.19 4, 4 12 3 1 73 0.8 0 (0.516, 0.484) 13.24 4, 4 6 3 1 40 1 0 (0.516, 0.484) 14.65 , 4 4 3

Table 5.8 Nash equilibrium values and corresponding strategies of second commerce retailer Parameters Li and Yang’s method [2, Table 4, pp. 8] Proposed Mehar method λ1 λ2 y ∗T v ∗ (λ2 ) y ∗T v ∗ (λ2 ) 1 3 115 0 0 (0.311, 0.689) 10.46 4, 4 12 1 3 115 0.3 0 (0.284, 0.716) 10.46 , 4 4 12 1 3 115 0.5 0 (0.266, 0.734) 10.46 4, 4 12 1 3 115 0.8 0 (0.239, 0.761) 10.46 , 4 4 12 1 3 115 1 0 (0.220, 0.780) 10.46 , 4 4 12

Table 5.9 Nash equilibrium values and corresponding strategies of first commerce retailer Parameters Li and Yang’s method [2, Table 5, pp. 9] Proposed Mehar method λ1 λ2 x ∗T u ∗ (λ1 ) x ∗T u ∗ (λ1 ) 3 1 15 0 0.5 (0.502, 0.498) 7.94 4, 4 2 3 1 37 0.3 0.5 (0.502, 0.498) 9.86 , 4 4 4 3 1 125 0.5 0.5 (0.502, 0.498) 11.19 4, 4 12 3 1 73 0.8 0.5 (0.502, 0.498) 13.24 4, 4 6 3 1 40 1 0.5 (0.502, 0.498) 14.65 , 4 4 3

Table 5.10 Nash equilibrium values and corresponding strategies of second commerce retailer Parameters Li and Yang’s method [2, Table 5, pp. 9] Proposed Mehar method λ1 λ2 y ∗T v ∗ (λ2 ) y ∗T v ∗ (λ2 ) 1 3 25 0 0.5 (0.311, 0.689) 12.99 4, 4 2 1 3 25 0.3 0.5 (0.284, 0.716) 12.99 , 4 4 2 1 3 25 0.5 0.5 (0.266, 0.734) 12.99 4, 4 2 1 3 25 0.8 0.5 (0.239, 0.761) 12.99 4, 4 2 1 3 25 1 0.5 (0.220, 0.780) 12.99 , 4 4 2

5.6 Conclusion

163

Table 5.11 Nash equilibrium values and corresponding strategies of first commerce retailer Parameters Li and Yang’s method [2, Table 6, pp. 9] Proposed Mehar method λ1 λ2 x ∗T u ∗ (λ1 ) x ∗T u ∗ (λ1 ) 3 1 15 0 1 (0.590, 0.410) 7.94 4, 4 2 3 1 37 0.3 1 (0.590, 0.410) 9.86 , 4 4 4 3 1 125 0.5 1 (0.590, 0.410) 11.19 4, 4 12 3 1 73 0.8 1 (0.590, 0.410) 13.24 4, 4 6 3 1 40 1 1 (0.590, 0.410) 14.65 , 4 4 3

Table 5.12 Nash equilibrium values and corresponding strategies of second commerce retailer Parameters Li and Yang’s method [2, Table 6, pp. 9] Proposed Mehar method λ1 λ2 y ∗T v ∗ (λ2 ) y ∗T v ∗ (λ2 ) 1 3 185 0 1 (0.311, 0.689) 15.47 4, 4 12 1 3 185 0.3 1 (0.284, 0.716) 15.47 , 4 4 12 1 3 185 0.5 1 (0.266, 0.734) 15.47 4, 4 12 1 3 185 0.8 1 (0.239, 0.761) 15.47 , 4 4 12 1 3 185 1 1 (0.220, 0.780) 15.47 , 4 4 12

5.6 Conclusion On the basis of the present study, it can be concluded that some mathematically incorrect assumptions are used in the existing method [2] and hence, it is not genuine to use the existing method [2] for solving such bimatrix games in which payoffs are represented by intuitionistic fuzzy numbers. Also, it can be concluded that in the Mehar method, proposed in this paper, no mathematically incorrect assumption is considered and hence, Mehar method should be used for solving these type of problems.

References 1. Li, D.F.: A ratio ranking method for triangular intuitionistic fuzzy numbers and its application to MADM problems. Comput. Math. Appl. 60, 1557–1570 (2010) 2. Li, D.F., Yang, J.: A difference-index based rank-ing bilinear programming approach to solving bimatrix games with payoffs of trapezoidal intuitionistic fuzzy numbers. J. Appl. Math. 2013, 1–10 (2013) 3. Mangasarian, O.L., Stone, H.: Two-person nonzero-sum games and quadratic programming. J. Math. Anal. Appl. 9, 348–355 (1964)

Chapter 6

Future Scope

The following research work may be considered as future research work: 1. In this thesis, only flaws in some existing methods for matrix games/constrained matrix games with inter-val/fuzzy/intuitionistic fuzzy payoffs are pointed out and to resolve flaws of these existing methods, new methods are proposed. The flaws, pointed out in Chaps. 1–5, of this thesis are also occurring in the existing methods [1–16] for solving matrix games with fuzzy payoffs/fuzzy goals/intuitionistic fuzzy goals. In future, it may be tried to resolve flaws of these existing methods [1–16]. 2. In all the methods, proposed in this thesis, the existing methods for comparing intervals/fuzzy numbers/intuitionistic fuzzy numbers are used. It can be easily concluded that the maximum/minimum of intervals/fuzzy numbers/intuitionistic fuzzy numbers, obtained by using the existing methods, is not necessarily a unique interval/fuzzy number/intuitionistic fuzzy number. In future, it may be tried to propose such methods for comparing intervals/fuzzy numbers/intuitionistic fuzzy numbers on applying which the maximum/minimum of intervals/fuzzy numbers/intuitionistic fuzzy numbers is always a unique interval/fuzzy number/intuitionistic fuzzy number. Furthermore, new methods, on the basis of proposed comparing methods, can be proposed for solving cooperative and noncooperative games with intervals/fuzzy/intuitionistic fuzzy payoffs.

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6 Future Scope

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