Solution of Rectangular Interval Games Using Graphical Method

Transcription

1 Tamsui Oxford Journal of Mathematical Sciences 22(1 ( Aletheia University Solution of Rectangular Interval Games Using Graphical Method Prasun Kumar Nayak and Madhumangal Pal Department of Applied Mathematics with Oceanology, and Computer Programming, Vidyasagar University, Midnapore , India Received March 17, 2005, Accepted March 8, 2006 Abstract A new approach for solving m 2 or 2 n rectangular games based on imprecise number instead of a real number such as interval number takes into account is introduced To reduce m 2 or 2 n rectangular interval game into simpler 2 2 interval game, we introduce here the graphical method which works well in interval numbers under consideration The interval graphical method is a state of the art technique to deal with the games to an inexact environment The approach is illustrated by numerical examples showing that a m 2 or 2 n interval games can be reduced to 2 2 interval games Keywords and Phrases: Crisp game, Interval numbers, Interval games 1 Interval The theory of fuzzy sets, proposed by Zadeh [?], has gained successful applications in various fields Interval game (IG theory which is a special case madhumangal@lycoscom

2 96 Prasun Kumar Nayak and Madhumangal Pal of fuzzy game theory, is an important content in interval fuzzy mathematics IG theory has been widely applied to manufacturing company, economics, management etc, where two decision makers are present or two possible conflicting objectives should be taken into account in order to reach optimality In [?], it is shown that how crisp game can be fuzzified and how 2 2 sub games involving interval number can be solved by different approaches, mainly interval number ranking process This approach is suitable when each player has chosen a procedure such that it is always possible to give the maximum or minimum of two interval numbers But reduction a rectangular m 2 or 2 n IG without saddle point to a 2 2 interval sub-game, is the basic problem in Interval Game theory In dominance method [?], if the convex combination of any two rows (columns of a pay-off matrix is dominated by the third row (column which indicates that the third move of the row (column player will be an optimal move but we are not certain which one of the first two moves will be an optimal one To determine it, we shall have to use the graphical method The purpose of this paper is to introduce a graphical method for reducing an m 2 or 2 n IG to 2 2 interval sub games 2 Interval Number Definition 1 An interval number A is defined by the ordered pairs in brackets [?] as A = [a L, a R ] = {x R : a L x a R ; R, be the set of all real numbers } The numbers a L, a R on the real line are called respectively the lower and upper limits of the interval A When a L = a R then A = [a L, a L ] is an ordinary real number Interval number A alternatively represented as A = m(a, w(a, where m(a and w(a are the mid-point and width of A, ie, mean(a = 1 2 (a L + a R and width(a = 1 2 (a R a L The set of all interval numbers is denoted by I(R 21 Interval arithmetic Let A = [a 1, a 2 ] and B = [b 1, b 2 ] be two interval numbers If a 1 0,A is called a non-negative interval The arithmetic and multiplication by a real number c are defined as follows :

3 Solution of Rectangular Interval Games Using Graphical Method 97 (i The sum of these two interval numbers is interval number, ie, A + B = [a 1 + b 1, a 2 + b 2 ] (ii The subtraction of these two numbers is also an interval number, ie, A B = [a 1 b 2, a 2 b 1 ] (iii If c 0 be a scalar then, ca = [ca 1, ca 2 ] if c 0, and = [ca 2, ca 1 ] if c < 0 (iv The product of these two interval numbers is given by AB = [min{a 1 b 1, a 1 b 2, a 2 b 1, a 2 b 2 }, max{a 1 b 1, a 1 b 2, a 2 b 1, a 2 b 2 }] (v The division of these two interval numbers with 0 B is A/B = [min{ a 1 b 1, a 1 b 2, a 2 b 1, a 2 b 2 }, max{ a 1 b 1, a 1 b 2, a 2 b 1, a 2 b 2 }] (vi A A 0, (vii A/A 1 Example 1 If A = [1, 2] and B = [3, 4] are two interval numbers then A + B = [4, 6], A B = [ 3, 1], AB = [3, 8], A/B = [ 1, 2] 4 3 Acceptability index defines how much higher is one interval from another in terms of interpreter s grade of satisfaction To get the maximum or the minimum interval number from a given set of interval numbers it is necessary to compare two interval numbers An overview on comparison between interval numbers is presented in the next section 22 Comparison between interval numbers Here we consider the order relation between intervals Comparison of two intervals is very important problem in interval analysis Case 1 (Disjoint subintervals : Let A = [a L, a R ], B = [b L, b R ] be two interval numbers within I(R Moore [?] defined transitive order relations over intervals as : A is strictly less than B if and only if a R < b L and this is denoted by A < B This relation is an extension of < on the real line This relation

4 98 Prasun Kumar Nayak and Madhumangal Pal seems to be strict order relation that A is smaller than B Example 2 Let A = [ 2, 0] and B = [3, 5] be two non overlapping subintervals So here A < B Case 2 (Nested subintervals : Let A = [a L, a R ], B = [b L, b R ] be two intervals in I(R such that a L b L < b R a R Then B is contained in A and this is denoted by B A which is the extension of the concept of the set inclusion [?] The extension of the set inclusion here only describes the condition that the interval A is nested in B but it can not order A and B in terms of value In this case, we define the ranking order of A and B as A B = { A, if the player is optimistic B, if the player is pessimistic The notation A B represents the maximum among the interval numbers A and B Similarly A B = { B, if the player is optimistic A, if the player is pessimistic The notation A B represents the minimum among the interval numbers A and B Example 3 Let A = [0, 8] and B = [4, 6] be two nested subintervals For optimistic player A B = A and A B = B For pessimistic case A B = B and A B = A Case 3 (Overlapping subintervals : The above mentioned order relations introduced by Moore [?] can not explain ranking between two overlapping closed intervals Two more order relations LR and MW are introduced in [?] as A LR B iff a L b L and a R b R and A MW B iff m 1 m 2 and w 1 w 2 along with the strict order relations < LR and < MW as A < LR B iff A LR B and A B

5 Solution of Rectangular Interval Games Using Graphical Method 99 and A < MW B iff A MW B and A B Different order relation to compare two intervals are widely investigated by several researchers [?,?,?,?] Some probabilistic view to order relation is incorporated in [?] Out of them, order relations which seems to more significant is introduced in [?] Let A = m 1, w 1 and B = m 2, w 2 be two interval numbers in mean-width (center-radius notation Definition 2 For m 1 m 2, the acceptability index (AI of the premise A B is defined by AI(A B = m 2 m 1 w 1 + w 2, where w 1 + w 2 0 and this is the value judgement by which A is inferior to B or precisely A is less than B For any sort of value judgement the AI index consistently satisfies the decision maker ie, if A, B I(R, then either AI(A B > 0 or AI(B A > 0 or AI(A B = AI(B A = 0 holds Thus, on the basis of comparative position of mean and width of intervals A, B the values of AI(A B of the premise A B are given by (i AI(A B 1 when m 1 < m 2 and a R b L, which refer to Case 1 (ii 0 < AI(A B < 1 when m 1 < m 2 and a L > b L (iii AI(A B = 0 when m 1 = m 2 Using the AI index we have presented the ordering for closed intervals reflecting decision maker s preference as (i When AI(A B 1 we have m 1 < m 2 and a R b L In this case, A is preferred over B (ie, A is less than B with acceptability index greater than or equal to 1 and so the decision maker (DM accepts it with absolute satisfaction (ii When 0 < AI(A B < 1 we have m 1 < m 2 and a R > b L, then A is preferred over B with different grades of satisfaction lying between 0 and 1, excluding 0 and 1 (iii If AI(A B = 0 then obviously m 1 = m 2 Now, if w 1 = w 2 then there is no question of comparison as A is identical with B But, if w 1 w 2, then the intervals A and B are non-inferior to each other, ie, acceptability index becomes insignificant In this case, DM has to negotiate with the widths of A and B Let A and B be two cost intervals and minimum cost interval is to be chosen If the DM is optimistic then he/she will prefer the interval with maximum width along with the risk of more uncertainty giving less impor-

6 100 Prasun Kumar Nayak and Madhumangal Pal tance Again, if the DM is pessimistic then he/she will pay more attention on more uncertainty ie, on the right end points of the intervals and will choose the interval with minimum width The case will be reverse when A and B represent profit intervals Example 4 Let A = [20, 30] = 25, 5 and B = [34, 38] = 36, 2 be two intervals Then AI(A B = = 157 > 1 Hence the DM accept the 5+2 decision that A is less than B with full satisfaction Example 5 Let A = [20, 26] = 23, 3 and B = [24, 28] = 26, 2 then AI(A B = = 06 < 1 Here A is less than B with grade of satisfaction Example 6 Let A = [8, 16] = 12, 4 and B = [6, 18] = 12, 6 be two intervals Here m 1 = m 2 = 12 and w 1 < w 2 and so AI(A B = 0 Hence both the intervals are non-inferior to each other In this case, from the optimistic point of view the DM will prefer the interval B instead of A Because, if A and B are both the profit intervals then DM will pay more attention on the highest possible profit of 18 unit ignoring the risk of minimum profit of 6 unit In the same manner if A and B are cost intervals then the DM will pay his attention on the minimum cost of 6 units ie, the left end points of both the cost intervals A and B and select B instead of A as 6 < 8 Again, when both the intervals are profit intervals then the DM with a pessimistic outlook will prefer the profit interval A because his attention will be drawn to the fact that the minimum profit of 8 unit will never be decreased, whereas his choice of B might cost him the loss of 2 unit profit and this apprehension will determine from selecting the interval B Similar is the explanation when A and B are cost intervals The above observations can be put into a compact form as follows B, if AI(A B > 0 A B = A, if AI(A B = 0 and w 1 < w 2 and DM is pessimistic B, if AI(A B = 0 and w 1 < w 2 and DM is optimistic The notation A B represents the maximum among the interval numbers A and B Similarly, if m 1 m 2 and w 1 w 2, then there also exist a strict preference relation between A and B Thus similar observations can be put

7 Solution of Rectangular Interval Games Using Graphical Method 101 into a compact form as B, if AI(B A > 0 A B = A, if AI(B A = 0 and w 1 > w 2 and DM is optimistic B, if AI(B A = 0 and w 1 > w 2 and DM is pessimistic The notation A B represents the minimum among the interval numbers A and B Example 7 {[ 2, 0], [ 1, 5]} = [ 1, 5] and {[ 3, 1], [ 2, 4]} = [ 3, 1] Theorem 1 The interval ordering defines a partial order relation between the intervals Proof (i If AI(A B = 0 and w(a = w(b then A is identical with B, ie, A B and we interpret that A and B are non-inferior to each other Hence the proposed AI index is reflexive (ii For any two intervals A and B, AI(A B > 0, AI(B A > 0 A = B Hence AI index is antisymmetric (iii Our proposed method is transitive, for any three intervals A, B, C I(R AI(A B = m 2 m 1 w 1 + w 2 0, AI(B C = m 3 m 2 w 2 + w 3 0 Now, m 3 m 1 = (m 3 m 2 + (m 2 m 1 0 m 3 m 1 w 1 + w 3 0 AI(A C 0 (iv For any intervals A and B (A B in I(R, AI(A B or AI(B A holds (never conflict From the conditions (i, (ii, (iii the relations are connected with the axiom of order relation and not totally, but partially ordered 3 Two Person Interval Game Interval game theory deals with making decisions under conflict caused by opposing interests An IG involving 2 players is called a 2 person IG Here A is maximizing (row or optimistic player and B is the minimizing (column or

8 102 Prasun Kumar Nayak and Madhumangal Pal pessimistic player Here we consider a two persons non-zero sum IG with single pay-off matrix We assume that A is maximization (optimistic player (row player ie, he/she will try to make maximum profit and B is minimization (pessimistic player (column player ie, he/she will try to minimize the loss 4 Pay-off Matrix The IG can be considered as a natural extension of classical game The pay-off of the game are affected by various sources of fuzziness The course of the IG is determined by the desire of A to maximize his/her gain and that of restrict his/her loss to a minimum Interval number ranking process is suitable when each player has chosen a procedure such that it is always possible to give the maximum or minimum of two interval numbers The table showing how payments should be made at the end of the game is called a payoff matrix If the player A has m strategies available to him and the player B has n strategies available to him, then the pay-off for various strategies combinations is represented by an m n pay-off matrix Here we consider the pay-off as interval numbers The pay-off matrix can be written in the matrix form as B 1 B 2 B n A 1 A 2 A m [a 11, b 11 ] [a 12, b 12 ] [a 1n, b 1n ] [a 21, b 21 ] [a 22, b 22 ] [a 2n, b 2n ] [a m1, b m1 ] [a m2, b m2 ] [a mn, b mn ] Intervals of real numbers enable one to describe the uncertainty about the actual values of the numerical variable Here it is assumed that when player A chooses the strategy A i and the player B selects strategy B j it results in a payoff of the closed interval [a ij, b ij ] to the player A with b ij a ij > 0 Intervals do not admit of comparison among themselves Here we provide an order relation for the entries in the pay-off matrix, so that comparison of intervals is possible Theorem 2 Let A = ([a ij, b ij ]; i = 1, 2,, m; j = 1, 2,, n be an m n pay-off matrix for a two person IG, where [a ij, b ij ] are interval numbers Then the following inequality is satisfied { [a ij, b ij ]} { [a ij, b ij ]} i j j i Definition 3 Saddle Point The concept of saddle point is introduced

9 Solution of Rectangular Interval Games Using Graphical Method 103 by Neumann [?] The (k, rth position of the pay-off matrix will be called a saddle point, if and only if, [a kr, b kr ] = { [a ij, b ij ]} = { [a ij, b ij ]} i j j i 5 IG Without Saddle Point There are interval games having no saddle point Consider a simple 2 2 IG with no saddle point with the pay-off matrix A 1 A 2 B 1 B ( 2 [a11, b 11 ] [a 12, b 12 ], [a 21, b 21 ] [a 22, b 22 ] where { [a ij, b ij ]} = { [a ij, b ij ]} In such IG the principle of saddle point i j j i solution breaks down and the players do not have a single best plan as their best strategy The IG in this case is said to be unstable To solve such IG, Neumann [?] introduced the concept of mixed strategy in classical form In all such cases to solve games, both the players must determine an optimal mixture of strategies to find a saddle point An example of 2 2 IG without saddle point is B 1 B ( 2 A 1 [-1,1] [1,3] A 2 [0,2] [-2,0] because { [a ij, b ij ]} = [ 1, 1] [0, 2] = { [a ij, b ij ]} i j j i 6 Solution of IG Without Saddle Point 61 Solution of 2 2 IG Consider the following IG with the given pay-off matrix for which there is no saddle point

10 104 Prasun Kumar Nayak and Madhumangal Pal A 1 A 2 B 1 B ( 2 [α11, β 11 ] [α 12, β 12 ], [α 21, β 21 ] [α 22, β 22 ] where { [α ij, β ij ]} = { [α ij, β ij ]} Let λ ij = β ij α ij > 0 for i, j = 1, 2 i j j i The normalized pay-off matrix is then ( [ α 11 λ 11, β 11 λ 11 ] [ α 12 λ 12, α 12 λ 12 ] [ α 21 λ 21, α 21 λ 21 ] [ α 22 λ 22, α = 22 λ 22 ] ( [a11, b 11 ] [a 12, b 12 ] [a 21, b 21 ] [a 22, b 22 ] where { [a ij, b ij ]} = { [a ij, b ij ]} We are to determine the probabilities i j j i with which each strategy of A and B will be played to get an optimal solution Let x i and y j be the probabilities with which A chooses his i th strategy and B chooses his j th strategy respectively Then for this problem, the mixed strategies of A and B are respectively, X = (x 1, x 2 and Y = (y 1, y 2 such that x 1 + x 2 = 1; x 1, x 2 0 and y 1 + y 2 = 1; y 1, y 2 0 The expected gains for A when B chooses B 1 and B 2 are respectively [a 11, b 11 ]x 1 + [a 21, b 21 ]x 2 and [a 12, b 12 ]x 1 + [a 22, b 22 ]x 2 The values of X has to be selected in such a way that the gain for A remains the same whatever be the strategy chosen by B, ie, [a 11, b 11 ]x 1 + [a 21, b 21 ]x 2 = [a 12, b 12 ]x 1 + [a 22, b 22 ]x 2 or, a 11 x 1 + a 21 x 2 = a 12 x 1 + a 22 x 2 and b 11 x 1 + b 21 x 2 = b 12 x 1 + b 22 x 2 or, a 11 x 1 + a 21 (1 x 1 = a 12 x 1 + a 22 (1 x 1 and b 11 x 1 + b 21 (1 x 1 = b 12 x 1 + b 22 (1 x 1 The relations satisfied by x 1 is x 1 = a 22 a 21 a 11 + a 22 a 12 a 21 and x 1 = b 22 b 21 b 11 + b 22 b 12 a 21 As a whole, it follows that a solution exists only when the a ij s and b ij s satisfy a 22 a 21 a 11 + a 22 a 12 a 21 = b 22 b 21 b 11 + b 22 b 12 a 21 and according to the relation b ij = a ij + 1 for i, j = 1, 2 this equality holds, when all the intervals are of same length ie, β ij = α ij + µ Thus x 1 = a 22 a 21 a a 11 +a 22 a 12 a 21 and x 2 = 11 a 12 a a 11 +a 22 a 12 a 21 Similarly y 1 = 22 a 12 a 11 +a 22 a 12 a 21, y 2 = a 11 a 21 a 11 +a 22 a 12 a 21, which are crisp numbers and the value of the game can be easily computed as V = [ 11 a 22 a 12 a 21 b a a 11 +a 22 a 12 a 21, 11 b 22 b 12 b 22 b 11 +b 22 b 12 b 21 ],

11 Solution of Rectangular Interval Games Using Graphical Method 105 Example 8 Consider, for example, the IG whose pay-off matrix is given below and which has no saddle point B 1 B ( 2 A 1 [-3,-1] [4,6] A 2 [6,8] [-7,-5] Hence the required probabilities are x 1 = 13, x 20 2 = 7, y 20 1 = 11, y 20 2 = 9 value of the game is [ 3, 43] and the 7 Graphical Method If the graphical method is applied for a particular problem, then the same reasoning can be used to solve any IG with mixed strategies that has only two non-terminated pure strategies for one of the players This method is useful for the IG where the pay-off matrix is of the size 2 n or m 2 ie, the IG with mixed strategies that has only two pure strategies for one of the players in the IG Optimal strategies for both the players assign non-zero probabilities to the same number of pure strategies It is clear that if one player has only two strategies, the other will also use the same number of strategies Hence, graphical method is useful to find out which of the two strategies can be used 71 General rule to draw a graph Consider the following two 2 n and m 2 IG without saddle point: (a B 1 B 2 B ( n A 1 [a11, b 11 ] [a 12, b 12 ] [a 1n, b 1n ] A 2 [a 21, b 21 ] [a 22, b 22 ] [a 2n, b 2n ] (b B 1 B 2 A 1 [a 11, b 11 ] [a 12, b 12 ] A 2 [a 21, b 21 ] [a 22, b 22 ] A m [a m1, b m1 ] [a m2, b m2 ] (i Draw two parallel vertical lines one unit distance apart and mark a scale on each

12 106 Prasun Kumar Nayak and Madhumangal Pal (ii For the case (a, the two strategies A 1, A 2 of A are represented by these two straight lines (iii For the case (b the two strategies B 1, B 2 of B are represented by these two straight lines (iv For the case (a, since the player ( A has two strategies, let the mixed strategy for the player A is given by S A = 2 A1 A where x x 1 x 1 +x 2 = 1; x 1, x Thus for each of the pure strategies available to the player B, the expected pay-off for the player A, would be as follows: B s pure move A s expected pay-off E(x B 1 [a 11, b 11 ]x 1 + [a 21, b 21 ](1 x 1 = [a 11 b 21, b 11 a 21 ]x 1 + [a 21, b 21 ] B 2 [a 12, b 12 ]x 1 + [a 22, b 22 ](1 x 1 = [a 12 b 22, b 12 a 22 ]x 1 + [a 22, b 22 ] B n [a 1n, b 1n ]x 1 + [a 2n, b 2n ](1 x 1 = [a 1n b 2n, b 1n a 2n ]x 1 + [a 2n, b 2n ] This shows that the player A s expected pay-off varies linearly with x 1 According to the min-max criterion for the mixed strategy games, the player A should select the value of x 1 so as to minimize his minimum expected payoffs Thus the player B would like to choose that pure moves B j against S A for which E j (x is a minimum for j = 1, 2,, n Let us denote this minimum expected pay-off for A by v = min{e j (x; j = 1, 2,, n} The objective of player A is to select x 1 and hence x 2 in such a way that v is as large as possible This may be done by plotting the regions R 1, R 2,, R n plotted by the pair of parallel straight lines as {y = (a 11 b 21 x 1 + a 21, y = (b 11 a 21 x 1 + b 21 }, {y = (a 12 b 22 x 1 + a 22, y = (b 12 a 22 x 1 + b 22 },, {y = (a 1n b 2n x 1 + a 2n, y = (b 1n a 2n x 1 + b 2n } drawn to represent the gains of A corresponding to B 1, B 2,, B n respectively of B on the line representing A 1 with on the line representing A 2 (v Now the two strategies of player B corresponding to those regions which pass through the maximum region can be determined It helps in reducing the size of the IG to 2 2

13 Solution of Rectangular Interval Games Using Graphical Method 107 (vi Similarly, for the case (b, since the player B( has two strategies, let the B1 B mixed strategy for the player A is given by S B = 2 it follows that y 1 y 2 y 2 = 1 y 1 ; y 1, y 2 0 Thus for each of the pure strategies available to the player A, the expected pay-off for the player B, would be as follows A s pure moves B s expected pay-off E(y A 1 [a 11, b 11 ]y 1 + [a 12, b 12 ](1 y 1 = [a 11 b 12, b 11 a 12 ]y 1 + [a 12, b 12 ] A 2 [a 21, b 21 ]y 1 + [a 22, b 22 ](1 y 1 = [a 21 b 22, b 21 a 22 ]y 1 + [a 22, b 22 ] A n [a m1, b m1 ]y 1 + [a m2, b m2 ](1 y 1 = [a m1 b m2, b m1 a m2 ]y 1 +[a m2, b m2 ] This shows that the player B s expected pay-off varies linearly with y 1 According to the max-min criterion for the mixed strategy games, the player B should select the value of y 1 so as to minimize his maximum expected pay-offs This may be done by plotting the regions L 1, L 2,, L m plotted by the pair of parallel straight lines as {y = (a 11 b 12 y 1 + a 12, y = (b 11 a 12 y 1 + b 12 }, {y = (a 21 b 22 y 1 +a 22, y = (b 21 a 22 y 1 +b 22 },, {y = (a m1 b m2 y 1 +a m2, y = (b m1 a m2 y 1 + b m2 } drawn to represent the gains of B corresponding to A 1, A 2,, A m respectively of A on the line representing B 1 with on the line representing B 2 We illustrate this method by different types of numerical examples Theorem 3 A region which will gives the bounds of maximize the minimum expected gain of B will give an optimal value of the probability x 1, x 2 and the region will refer two courses of action taken by A for the purpose Proof According to the general theory of games, in a first step the player A would try to obtain { [a ij, b ij ]} while for the player B the objective is i j { [a ij, b ij ]} If the real matrix ([a ij, b ij ] m n has no saddle points mixed j i strategies must be used The mixed strategy problem is solved by applying the min-max criterion in which the maximizing player chooses the probabilities x i which minimize the smallest expected gains in columns and the minimizing player chooses his probabilities y j which minimize the least expected gains of rows Obviously the players obtain their optimal strategies from { [a ij, b ij ]x i y j } and { [a ij, b ij ]x i y j } This problem becomes classical game i j j i

14 108 Prasun Kumar Nayak and Madhumangal Pal and we are to choose x i, y j such that { [a ij, b ij ]x i y j } = { [a ij, b ij ]x i y j } i j j i Let V the value of the game given by the pay-off matrix [a ij, b ij ] 2 n as B 1 B 2 B ( n A 1 [a11, b 11 ] [a 12, b 12 ] [a 1n, b 1n ] A 2 [a 21, b 21 ] [a 22, b 22 ] [a 2n, b 2n ] without any saddle point Let the mixed strategies used by A be X = (x 1, x 2 and B be Y = (y 1, y 2,, y n, in which x i = 1 and x i 0; y j = 1 and y j 0 i j Then the expected gain of the player A given that B plays his pure strategy B j is given by E j (X = [a 1j, b 1j ]x 1 + [a 2j, b 2j ]x 2 = [a 1j, b 1j ]x 1 + [a 2j, b 2j ](1 x 1, j = 1, 2,, n Now both x 1, x 2 must lie in the open interval (0, 1, because if either x 1 or x 2 = 1 the game is of pure strategy Hence E j (X is a linear function of either x 1 or x 2 Considering E j (X as a linear function of x 1, we have from the limiting values (0, 1 of x 1 ; E j (X = { [a2j, b 2j ], x 1 = 0 [a 1j, b 1j ], x 1 = 1 Hence E j (X represents a region joining the points (0, a 2j, (0, b 2j and (1, a 1j, (1, b 1j Now A expects a least possible gain V so that E j (X V, for all j The main objective of A is to select x 1, x 2 in such a way that x 1 + x 2 = 1 and both lie in the open interval (0, 1 Consider the following 2 2 pay-off matrix, where B k and B l are two critical moves of B, A 1 A 2 B k B l ( [a1k, b 1k ] [a 1l, b 1l ] [a 2k, b 2k ] [a 2l, b 2l ] Then the expectation of A is = [a 1k, b 1k ]x 1 y 1 + [a 1l, b 1l ]x 1 y 2 + [a 2k, b 2k ]x 2 y 1 + [a 2l, b 2l ]x 2 y 2 a = [a 1k +a 2l a 1l a 2k, b 1k +b 2l b 1l b 2k ]{x 1 2l a 2k a 1k +a 2l a 1l a 2k }{y 1 a + [ 1k a 2l a 1l a 2k a 1k +a 2l a 1l a 2k, b 1k b 2l b 1l b 2k b 1k +b 2l b 1l b 2k ] a 2l a 1l a 1k +a 2l a 1l a 2k } Player A would always try to mix his moves with such probabilities so as to maximize his expected gain This shows that if A chooses x 1 = a 2l a 2k a 1k +a 2l a 1l a 2k

15 Solution of Rectangular Interval Games Using Graphical Method 109 a 1k a 2l a 1l a 2k a 1k +a 2l a 1l a 2k, b 1k b 2l b 1l b 2k b 1k +b 2l b 1l b 2k ] then he can ensure that his expectation will be at least [ This choice of x 1 will thus be optimal to the player A Thus the lower bound of the regions will give the minimum expected gain of A as a function of x 1 Hence with the help of graphical method, we shall be in a position to find two particular moves which will gives the bounds of maximum pay-off among the choosing minimum expected pay-off on the lower bounds of B Thus the mixed strategy problem is solved by applying the min-max criterion in which the maximizing player chooses the probabilities x i which minimize the smallest expected gains in columns Thus we are to find a region which will gives the bounds of maximum pay-off among the minimum expected pay-off on the lower region of B For this any two pair of lines having opposite signs for their slopes will define an alternative optimal solution region and the optimal value of probability x 1 and x 2 Theorem 4 A region which will give the bounds of minimize the maximum expected loss of A, and the lowest bound of all regions will give the minimum expected pay-off (min-max value and the optimal value of the probability y 1, y 2 Proof The proof of this theorem is similar to Theorem 3 8 Particular Cases and Numerical Examples Example 9 Consider the 2 4 IG problem whose pay-off matrix is given below The game has no saddle point B 1 B 2 B 3 B ( 4 A 1 [0,2] [2,4] [-1,1] [1,3] A 2 [2,4] [-1,1] [0,2] [-2,0] ( A1 A Let the player A play the mixed strategy S A = 2 against player B, x 1 x 2 where x 2 = 1 x 1, x 1, x 2 0 The A s expected pay-off s against B s pure moves are given by B s pure move A s expected pay-off E(x B 1 [ 4x 1 + 2, 4] B 2 [x 1 1, 5x 1 + 1] B 3 [ 3x 1, x 1 + 2] B 4 [x 1 2, 5x 1 ]

16 110 Prasun Kumar Nayak and Madhumangal Pal We draw two parallel axes, the strategies A 1,A 2 of A are represented by these two straight lines These expected pay-off equations are plotted as functions of x 1 as shown in the Figure 1 Now since the player A wishes to maximize R 2 S Q 1 P A 1 A 2 Figure 1: Graph of example 9 his minimum expected pay-off, we consider the highest region of intersection on the lower region of A s expected pay-off equations We bound the figure so obtain from above by drawing a region, P R of this bounds refers to the maximum of minimum gains At this region B has used his two courses of action and they are B 3 and B 4 as it is the region of intersection of B 3 and B 4 The solution to the original 2 4 game, therefore, reduces to that of the simpler game with the 2 2 pay-off matrix B 3 B ( 4 A 1 [-1,1] [1,3] A 2 [0,2] [-2,0] The abscissa of the points P and R are 1 and the ordinates of P and R are 2 1, 3 Therefore the optimal strategies are for A( 1, 1 and the value of the game (V is [ 1, 3] The probabilities for B and B 4 are obtained from graph as (0, 0, 3, 1 and in this case also the value of the game (V is [ 1, 3 ] Here B 1 and B 2 are dominated strategies as is obvious from the graph as also from the pay-off matrix The method is equally applicable to n 2 IG But in this case the graph stands for B s loss against the probability y with which B plays B 1 In this

17 Solution of Rectangular Interval Games Using Graphical Method 111 case to minimize the maximum loss of B, we are to bound the figure from above The lowest region of this bound will refer to the two courses of the action taken by A for the purpose Example 10 We consider the IG whose pay-off matrix is given below along with graph B 1 B 2 A 1 A 2 A 3 [-4,-2] [1,3] [-2,0] [-1,1] [4,6] [-3,-1] The given IG problem does not possess any saddle point In the graph we have plotted the A s pure strategies on the corresponding strategies of B shown by two parallel lines B 1 and B 2 on the figure R 2 1 S 1 0 Q0 1 P B 1 B 2 Figure 2: Graph of example 10 Let the player B play the mixed strategy S B = ( B1 B 2 y 1 y 2 with y 2 = 1 y 1 ; y 1, y 2 0 against player A Then B s expected pay-off against A s pure moves are given by A s pure move B s expected pay-off E(x A 1 [ 7y 1 + 1, 3y 1 + 3] A 2 [ 3y 1 1, y 1 + 1] A 3 [5y 1 3, 9y 1 1]

18 112 Prasun Kumar Nayak and Madhumangal Pal Since the player B wishes to minimize his maximum expected pay-off, we consider the lowest region of intersection on the upper sections of B s expected pay-off equations At this region A has used his two courses of action and they are A 1 and A 3 as it is the region of intersection L 1 and L 3 The region represents the minimum expected value of the game for player B We bound the figure so obtained from above by drawing a region, P R of this bounds refers to the minimum of the maximum losses So the 2 2 simpler sub matrix which provides the optimal solution of the given game is B 1 B ( 2 A 1 [-4,-2] [1,3] A 3 [4,6] [-3,-1] The abscissa of the points P and Q are 2, 4 and the ordinates are 1 Therefore the optimal strategies are for B is ( 1, 2 and the value (V of the game is = [ 2, 4] 3 3 It may sometimes so happen that we get two pairs of strategies having the same value for the game giving two pairs of optimal strategies Example 11 Let us consider the IG whose pay-off matrix is given below B 1 B 2 B 3 B ( 4 A 1 [5,7] [4,6] [1,3] [2,4] A 2 [0,2] [1,3] [5,7] [2,4] Graphical representation of the game as represented by the given matrix is given in the figure 3, the manner of computation being the same as in the previous examples From the graph it is seen that the player B has two pairs of supporting strategies, for example (B 1, B 4 and (B 3, B 4 giving the same value [2, 4], for the sub games The player A has infinite number of mixed strategies in the region P QRS This can be found either by solving the two 2 2 sub games or from the abscissae of points It can be shown by solving the two 2 2 sub games formed by two different pairs of B s strategies that the optimal strategy for B is his pure strategy B 4 It may so happen in some pay-off matrix that three gain regions have the min-max or max-min region This will be explained with the graph given below Example 12 Consider, for example, the IG whose pay-off is given by the adjacent matrix

19 Solution of Rectangular Interval Games Using Graphical Method S R P Q A 1 A 2 Figure 3: Graph of example 11 B 1 B 2 B ( 3 A 1 [-2,0] [1,3] [2,4] A 2 [5,7] [2,4] [1,3] The graphical representation of game is shown in the figure R 4 3 Q S P A 1 A 2 Figure 4: Graph of example 12 It is seen that the three gain regions of A corresponding to B 1, B 2, B 3 have a common maximum of minimum region P QRS Solving the 2 2 sub game involving B 1 and B 2 of B we get the optimal solution, for A optimal strategy

20 114 Prasun Kumar Nayak and Madhumangal Pal is A( 1 2, 1 2 and the value of the game is [ 3 2, 7 2 ] The value of the game is [ 3 2, 7 2 ] for the two 2 2 sub game with (B 1, B 3 and (B 2, B 3 also Thus the game can be reduced to 2 2 in the following manner: A 1 A 2 B 1 B 2 B 3 B ( ( 4 [-2,0] [1,3] A, 1 [1,3] [2,4] [5,7] [2,4] A 2 [2,4] [1,3] It is also note that the average of above two pay-off matrices will also be the additional possibility of reducing the game to 2 2 Thus the additional possibility of 2 2 game will also yield new matrix B 1 B ( 2 A 1 [ 1, 3] [ 1, 3] A 2 [ 1, 3] [ 1, 3] optimal solution which will mixes three strategies B 1, B 2, B 3 Thus 2 2 game is solved by solving the governing simultaneous equations Solution of game with reduced matrix in additional possibility can be obtained easily, because it has a saddle point [ 3, 7] So the value of the IG to the player A is [ 3, 7] Conclusion In this paper, we present an application of graphical method for finding the solution of two person non-zero sum IG for fixed strategies Numerical examples are presented to illustrate the methodology Our proposed method is simple and more accurate to deal with the general two person of IG problems Hence the proposed method is suitable for solving the practical IG problems in real applications Application of IG takes place in civil engineering, lunching advertisement camping for competing products, planning war strategies for opposing armies and other areas Acknowledgments Thanks are due to both anonymous referee for their fruitful comments, careful corrections and valuable suggestions to improve the presentation of the paper

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