Tamsui Oxford Journal of Mathematical Sciences 22(1) (2006) 95-115
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-721 102, 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
∗
E-mail: [email protected]
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 = [aL , aR ] = {x ∈ < : aL ≤ x ≤ aR ; <, be the set of all real numbers }.
The numbers aL , aR on the real line are called respectively the lower and upper
limits of the interval A. When aL = aR then A = [aL , aL ] is an ordinary real
number. Interval number A alternatively represented as A = hm(A), w(A)i,
where m(A) and w(A) are the mid-point and width of A, i.e.,
1
1
mean(A) = (aL + aR ) and width(A) = (aR − aL ).
2
2
The set of all interval numbers is denoted by I(<).
2.1. Interval arithmetic
Let A = [a1 , a2 ] and B = [b1 , b2 ] be two interval numbers. If a1 ≥ 0,A is
called a non-negative interval. The arithmetic and multiplication by a real
number c are defined as follows :
Solution of Rectangular Interval Games Using Graphical Method
97
(i) The sum of these two interval numbers is interval number, i.e.,
A + B = [a1 + b1 , a2 + b2 ].
(ii) The subtraction of these two numbers is also an interval
number, i.e., A − B = [a1 − b2 , a2 − b1 ].
(iii) If c 6= 0 be a scalar then, cA = [ca1 , ca2 ] if c ≥ 0, and
= [ca2 , ca1 ] if c < 0.
(iv) The product of these two interval numbers is given by
AB = [min{a1 b1 , a1 b2 , a2 b1 , a2 b2 }, max{a1 b1 , a1 b2 , a2 b1 , a2 b2 }].
(v) The division of these two interval numbers with 0 6∈ B is
A/B = [min{ ab11 , ab21 , ab12 , ab22 }, max{ ab11 , ab21 , ab12 , ab22 }].
(vi) A − A 6= 0, (vii) A/A 6= 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 = [ 41 , 23 ].
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.
2.2 . 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 = [aL , aR ], B = [bL , bR ] be two
interval numbers within I(<). Moore [?] defined transitive order relations over
intervals as : A is strictly less than B if and only if aR < bL and this is denoted
by A < B. This relation is an extension of ‘ <’ on the real line. This relation
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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 = [aL , aR ], B = [bL , bR ] be two
intervals in I(<) such that aL ≤ bL < bR ≤ aR . 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,
B,
if the player is optimistic
if the player is pessimistic.
The notation A ∨ B represents the maximum among the interval numbers A
and B. Similarly
(
A∧B =
B,
A,
if the player is optimistic
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 ≤M W are introduced in
[?] as
A ≤LR B iff aL ≤ bL and aR ≤ bR
and A ≤M W B iff m1 ≤ m2 and w1 ≥ w2
along with the strict order relations <LR and <M W as
A <LR B iff A ≤LR B and A 6= B
Solution of Rectangular Interval Games Using Graphical Method
99
and A <M W B iff A ≤M W B and A 6= 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 = hm1 , w1 i and B = hm2 , w2 i be two interval
numbers in mean-width (center-radius) notation.
Definition 2. For m1 ≤ m2 , the acceptability index (AI) of the premise
A ≺ B is defined by
m2 − m1
,
AI(A ≺ B) =
w1 + w2
where w1 + w2 6= 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 i.e., if A, B ∈ I(<), 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 m1 < m2 and aR ≤ bL , which refer to Case 1
(ii) 0 < AI(A ≺ B) < 1 when m1 < m2 and aL > bL
(iii) AI(A ≺ B) = 0 when m1 = m2 . 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 m1 < m2 and aR ≤ bL . In this case,
A is preferred over B (i.e., 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 m1 < m2 and aR > bL , 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 m1 = m2 . Now, if w1 = w2 then
there is no question of comparison as A is identical with B. But, if w1 6= w2 ,
then the intervals A and B are non-inferior to each other, i.e., 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-
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Prasun Kumar Nayak and Madhumangal Pal
tance. Again, if the DM is pessimistic then he/she will pay more attention on
more uncertainty i.e., 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] = h25, 5i and B = [34, 38] = h36, 2i be two
= 1.57 > 1. Hence the DM accept the
intervals. Then AI(A ≺ B) = 36−25
5+2
decision that ‘A is less than B’ with full satisfaction.
Example 5. Let A = [20, 26] = h23, 3i and B = [24, 28] = h26, 2i then
AI(A ≺ B) = 26−23
= 0.6 < 1. Here ‘A is less than B’ with grade of satisfac3+2
tion 0.6.
Example 6. Let A = [8, 16] = h12, 4i and B = [6, 18] = h12, 6i be two intervals. Here m1 = m2 = 12 and w1 < w2 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 i.e., 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,
A ∨ B =  A,

B,
if AI(A ≺ B) > 0
if AI(A ≺ B) = 0 and w1 < w2 and DM is pessimistic
if AI(A ≺ B) = 0 and w1 < w2 and DM is optimistic.
The notation A ∨ B represents the maximum among the interval numbers A
and B. Similarly, if m1 ≥ m2 and w1 ≥ w2 , then there also exist a strict
preference relation between A and B. Thus similar observations can be put
Solution of Rectangular Interval Games Using Graphical Method
101
into a compact form as


 B,
A ∧ B =  A,

B,
if AI(B ≺ A) > 0
if AI(B ≺ A) = 0 and w1 > w2 and DM is optimistic
if AI(B ≺ A) = 0 and w1 > w2 and DM is pessimistic.
The notation A ∧ B represents the minimum among the interval numbers A
and B.
Example 7.
W
{[−2, 0], [−1, 5]} = [−1, 5] and {[−3, −1], [−2, 4]} = [−3, −1].
V
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, i.e., 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(<)
AI(A ≺ B) =
m3 − m2
m2 − m1
≥ 0, AI(B ≺ C) =
≥ 0.
w1 + w2
w2 + w3
Now, m3 − m1 = (m3 − m2 ) + (m2 − m1 ) ≥ 0
⇒
m3 − m1
≥ 0 ⇒ AI(A ≺ C) ≥ 0.
w1 + w3
(iv) For any intervals A and B (A 6= B) in I(<), 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
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) i.e., he/she will try to make maximum profit and B is minimization
(pessimistic) player (column player) i.e., 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
B1
B2 · · ·
Bn


A1
[a11 , b11 ] [a12 , b12 ] · · · [a1n , b1n ]
A2 
[a22 , b22 ] · · · [a2n , b2n ] 
 [a21 , b21 ]


.

··· 
···
···
···
Am
[am1 , bm1 ] [am2 , bm2 ] · · · [amn , bmn ]
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 Ai and the player B selects strategy Bj it results in a payoff of the closed interval [aij , bij ] to the player A with bij − aij > 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 = ([aij , bij ]); i = 1, 2, ..., m; j = 1, 2, ..., n be an m × n
pay-off matrix for a two person IG, where [aij , bij ] are interval numbers. Then
W V
V W
the following inequality is satisfied { [aij , bij ]} ≤ { [aij , bij ]}.
i
j
j
i
Definition 3. Saddle Point. The concept of saddle point is introduced
Solution of Rectangular Interval Games Using Graphical Method
103
by Neumann [?]. The (k, r)th position of the pay-off matrix will be called a
saddle point, if and only if,
[akr , bkr ] = { [aij , bij ]} = { [aij , bij ]}.
W V
i
j
V W
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
A1
A2
where
B1
B2
!
[a11 , b11 ] [a12 , b12 ]
,
[a21 , b21 ] [a22 , b22 ]
{ [aij , bij ]} 6=
W V
i
j
{ [aij , bij ]}. In such IG the principle of saddle point
V W
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
B1 B2
!
[-1,1] [1,3]
A1
A2
[0,2] [-2,0]
because
{ [aij , bij ]} = [−1, 1] 6= [0, 2] = { [aij , bij ]}.
W V
i
j
V W
j
i
6. Solution of IG Without Saddle Point
6.1 . Solution of 2 × 2 IG
Consider the following IG with the given pay-off matrix for which there is
no saddle point
104
Prasun Kumar Nayak and Madhumangal Pal
B1
B2
!
[α11 , β11 ] [α12 , β12 ]
,
[α21 , β21 ] [α22 , β22 ]
A1
A2
where
{ [αij , βij ]} 6=
W V
i
j
{ [αij , βij ]}. Let λij = βij − αij > 0 for i, j = 1, 2.
V W
j
i
The normalized pay-off matrix is then
11 β11
12 α12
, λ11 ] [ αλ12
, λ12 ]
[ αλ11
α21 α21
α22 α22
[ λ21 , λ21 ] [ λ22 , λ22 ]
where
{ [aij , bij ]} 6=
W V
i
j
!
=
[a11 , b11 ] [a12 , b12 ]
[a21 , b21 ] [a22 , b22 ]
!
,
{ [aij , bij ]}. We are to determine the probabilities
V W
j
i
with which each strategy of A and B will be played to get an optimal solution.
Let xi and yj be the probabilities with which A chooses his ith strategy
and B chooses his j th strategy respectively. Then for this problem, the mixed
strategies of A and B are respectively, X = (x1 , x2 ) and Y = (y1 , y2 ) such that
x1 + x2 = 1; x1 , x2 ≥ 0 and y1 + y2 = 1; y1 , y2 ≥ 0. The expected gains
for A when B chooses B1 and B2 are respectively [a11 , b11 ]x1 + [a21 , b21 ]x2 and
[a12 , b12 ]x1 + [a22 , b22 ]x2 . 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, i.e.,
[a11 , b11 ]x1 + [a21 , b21 ]x2 = [a12 , b12 ]x1 + [a22 , b22 ]x2
or, a11 x1 + a21 x2 = a12 x1 + a22 x2 and b11 x1 + b21 x2 = b12 x1 + b22 x2
or, a11 x1 + a21 (1 − x1 ) = a12 x1 + a22 (1 − x1 )
and b11 x1 + b21 (1 − x1 ) = b12 x1 + b22 (1 − x1 ).
The relations satisfied by x1 is
x1 =
b22 − b21
a22 − a21
and x1 =
.
a11 + a22 − a12 − a21
b11 + b22 − b12 − a21
As a whole, it follows that a solution exists only when the aij ’s and bij ’s satisfy
b22 − b21
a22 − a21
=
a11 + a22 − a12 − a21
b11 + b22 − b12 − a21
and according to the relation bij = aij + 1 for i, j = 1, 2 this equality holds,
when all the intervals are of same length i.e., βij = αij + µ. Thus x1 =
a22 −a21
11 −a12
22 −a12
and x2 = a11 +aa22
. Similarly y1 = a11 +aa22
, y2 =
a11 +a22 −a12 −a21
−a12 −a21
−a12 −a21
a11 −a21
, which are crisp numbers and the value of the game can be easa11 +a22 −a12 −a21
a22 −a12 a21
ily computed as V = [ a11a11
, b11 b22 −b12 b22 ].
+a22 −a12 −a21 b11 +b22 −b12 −b21
105
Solution of Rectangular Interval Games Using Graphical Method
Example 8. Consider, for example, the IG whose pay-off matrix is given
below and which has no saddle point.
B1
B2
!
A1
[-3,-1] [4,6]
.
A2
[6,8] [-7,-5]
13
, x2 =
Hence the required probabilities are x1 = 20
3 43
value of the game is [ 20
, 20 ].
7
, y1
20
=
11
, y2
20
=
9
20
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 i.e., 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.
7.1 . General rule to draw a graph
Consider the following two 2 × n and m × 2 IG without saddle point:
B1
B2 · · ·
Bn
(a)
!
[a11 , b11 ] [a12 , b12 ] · · · [a1n , b1n ]
A1
A2
[a21 , b21 ] [a22 , b22 ] · · · [a2n , b2n ]
(b)
A1
A2
···
Am





B1
B2
[a11 , b11 ] [a12 , b12 ]
[a21 , b21 ] [a22 , b22 ]
···
···
[am1 , bm1 ] [am2 , bm2 ]



.

(i) Draw two parallel vertical lines one unit distance apart and mark a scale
on each.
106
Prasun Kumar Nayak and Madhumangal Pal
(ii) For the case (a), the two strategies A1 , A2 of A are represented by these
two straight lines.
(iii) For the case (b) the two strategies B1 , B2 of B are represented by these
two straight lines.
(iv) For the case (a), since the player A has two strategies,
let the mixed strat!
A 1 A2
egy for the player A is given by SA =
where x1 +x2 = 1; x1 , x2 ≥
x1 x2
0. 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
B1
B2
..
.
Bn
A’s expected pay-off E(x)
[a11 , b11 ]x1 + [a21 , b21 ](1 − x1 ) = [a11 − b21 , b11 − a21 ]x1 + [a21 , b21 ]
[a12 , b12 ]x1 + [a22 , b22 ](1 − x1 ) = [a12 − b22 , b12 − a22 ]x1 + [a22 , b22 ]
..
..
..
.
.
.
[a1n , b1n ]x1 + [a2n , b2n ](1 − x1 ) = [a1n − b2n , b1n − a2n ]x1 + [a2n , b2n ].
This shows that the player A’s expected pay-off varies linearly with x1 . According to the min-max criterion for the mixed strategy games, the player A
should select the value of x1 so as to minimize his minimum expected payoffs. Thus the player B would like to choose that pure moves Bj against
SA for which Ej (x) is a minimum for j = 1, 2, . . . , n. Let us denote this
minimum expected pay-off for A by v = min{Ej (x); j = 1, 2, . . . , n}. The
objective of player A is to select x1 and hence x2 in such a way that v is as
large as possible. This may be done by plotting the regions R1 , R2 , . . . , Rn
plotted by the pair of parallel straight lines as {y = (a11 − b21 )x1 + a21 , y =
(b11 − a21 )x1 + b21 }, {y = (a12 − b22 )x1 + a22 , y = (b12 − a22 )x1 + b22 }, . . . ,
{y = (a1n − b2n )x1 + a2n , y = (b1n − a2n )x1 + b2n } drawn to represent the gains
of A corresponding to B1 , B2 , . . . , Bn respectively of B on the line representing
A1 with on the line representing A2 .
(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.
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 B2
mixed strategy for the player A is given by SB =
it follows that
y1 y2
y2 = 1 − y1 ; y1 , y2 ≥ 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
A1
A2
..
.
An
B’s expected pay-off E(y)
[a11 , b11 ]y1 + [a12 , b12 ](1 − y1 ) = [a11 − b12 , b11 − a12 ]y1 + [a12 , b12 ]
[a21 , b21 ]y1 + [a22 , b22 ](1 − y1 ) = [a21 − b22 , b21 − a22 ]y1 + [a22 , b22 ]
..
..
..
.
.
.
[am1 , bm1 ]y1 + [am2 , bm2 ](1 − y1 ) = [am1 − bm2 , bm1 − am2 ]y1 +[am2 , bm2 ].
This shows that the player B’s expected pay-off varies linearly with y1 . According to the max-min criterion for the mixed strategy games, the player B
should select the value of y1 so as to minimize his maximum expected pay-offs.
This may be done by plotting the regions L1 , L2 , . . . , Lm plotted by the pair
of parallel straight lines as {y = (a11 − b12 )y1 + a12 , y = (b11 − a12 )y1 + b12 },
{y = (a21 −b22 )y1 +a22 , y = (b21 −a22 )y1 +b22 }, . . . , {y = (am1 −bm2 )y1 +am2 , y =
(bm1 − am2 )y1 + bm2 } drawn to represent the gains of B corresponding to
A1 , A2 , . . . , Am respectively of A on the line representing B1 with on the line
representing B2 . 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 x1 , x2 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
W V
A would try to obtain { [aij , bij ]} while for the player B the objective is
j
i
{ [aij , bij ]}. If the real matrix ([aij , bij ])m×n has no saddle points mixed
V W
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 xi which minimize the smallest expected gains in columns and the minimizing player chooses his probabilities yj which minimize the least expected
gains of rows. Obviously the players obtain their optimal strategies from
W V
V W
{ [aij , bij ]xi yj } and { [aij , bij ]xi yj }. This problem becomes classical game
i
j
j
i
108
Prasun Kumar Nayak and Madhumangal Pal
and we are to choose xi , yj such that
{ [aij , bij ]xi yj } =
W V
i
j
{ [aij , bij ]xi yj }.
V W
j
i
Let V the value of the game given by the pay-off matrix [aij , bij ]2×n as
B1
B2 · · ·
Bn
!
A1
[a11 , b11 ] [a12 , b12 ] · · · [a1n , b1n ]
A2
[a21 , b21 ] [a22 , b22 ] · · · [a2n , b2n ]
without any saddle point. Let the mixed strategies used by A be X = (x1 , x2 )
and B be Y = (y1 , y2 , . . . , yn ), in which
P
P
xi = 1 and xi ≥ 0; yj = 1 and yj ≥ 0.
i
j
Then the expected gain of the player A given that B plays his pure strategy
Bj is given by
Ej (X) = [a1j , b1j ]x1 + [a2j , b2j ]x2
= [a1j , b1j ]x1 + [a2j , b2j ](1 − x1 ), j = 1, 2, . . . , n.
Now both x1 , x2 must lie in the open interval (0, 1), because if either x1 or
x2 = 1 the game is of pure strategy. Hence Ej (X) is a linear function of either
x1 or x2 . Considering Ej (X) as a linear function of x1 , we have from the
limiting values (0, 1) of x1 ;
(
Ej (X) =
[a2j , b2j ], x1 = 0
[a1j , b1j ], x1 = 1
Hence Ej (X) represents a region joining the points (0, a2j ), (0, b2j ) and (1, a1j ), (1, b1j ).
Now A expects a least possible gain V so that Ej (X) ≥ V , for all j. The main
objective of A is to select x1 , x2 in such a way that x1 + x2 = 1 and both lie
in the open interval (0, 1). Consider the following 2 × 2 pay-off matrix, where
Bk and Bl are two critical moves of B,
Bk
Bl
!
[a1k , b1k ] [a1l , b1l ]
A1
.
A2
[a2k , b2k ] [a2l , b2l ]
Then the expectation of A is
= [a1k , b1k ]x1 y1 + [a1l , b1l ]x1 y2 + [a2k , b2k ]x2 y1 + [a2l , b2l ]x2 y2
2l −a1l
2k
}{y1 − a1k +aa2l
}
= [a1k +a2l −a1l −a2k , b1k +b2l −b1l −b2k ]{x1 − a1k +aa2l2l−a
−a1l −a2k
−a1l −a2k
b1k b2l −b1l b2k
a1k a2l −a1l a2k
+ [ a1k +a2l −a1l −a2k , b1k +b2l −b1l −b2k ].
Player A would always try to mix his moves with such probabilities so as to
2k
maximize his expected gain. This shows that if A chooses x1 = a1k +aa2l2l−a
−a1l −a2k
Solution of Rectangular Interval Games Using Graphical Method
109
−a1l a2k
2l −b1l b2k
then he can ensure that his expectation will be at least [ a1ka1k+aa2l
, b1kb1k+bb2l
].
−b1l −b2k
2l −a1l −a2k
This choice of x1 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 x1 . 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 xi
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 x1 and x2 .
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 y1 , y2 .
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.
A1
A2
B1 B2 B3
B4
!
[0,2] [2,4] [-1,1] [1,3]
.
[2,4] [-1,1] [0,2] [-2,0]
!
A1 A2
Let the player A play the mixed strategy SA =
against player B,
x1 x2
where x2 = 1 − x1 , x1 , x2 ≥ 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)
B1
[−4x1 + 2, 4]
B2
[x1 − 1, 5x1 + 1]
B3
[−3x1 , x1 + 2]
B4
[x1 − 2, 5x1 ].
110
Prasun Kumar Nayak and Madhumangal Pal
We draw two parallel axes, the strategies A1 ,A2 of A are represented by these
two straight lines. These expected pay-off equations are plotted as functions
of x1 as shown in the Figure 1. Now since the player A wishes to maximize
6
6
5
5
4
4
3 @@ 3
@
2@
2
R
@
@
@
@1
S1 @
@
@
@
@
0 @ @
0Q
@
−1 P
@ @ −1
@
@ −2
−2
A1
A2
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 B3 and B4 as it is the region of intersection of B3 and
B4 . The solution to the original 2 × 4 game, therefore, reduces to that of the
simpler game with the 2 × 2 pay-off matrix.
B3 B4
!
A1
[-1,1] [1,3]
.
A2
[0,2] [-2,0]
The abscissa of the points P and R are 21 and the ordinates of P and R are
− 21 , 32 . Therefore the optimal strategies are for A( 21 , 12 ) and the value of the
game (V ) is [− 12 , 32 ]. The probabilities for B3 and B4 are obtained from graph
as (0, 0, 43 , 14 ) and in this case also the value of the game (V ) is [− 12 , 23 ]. Here
B1 and B2 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 B1 . In this
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.
B1 B2


A1
[-4,-2] [1,3]

A2 
 [-2,0] [-1,1]  .
A3
[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 B1 and B2 on the figure 2.
6
6
5 LL
5
4 L
4
3 LL LL
3
2 L LR 2
1 SLL L
1
L
0 L LQ0
L −1
−1 PL
L
−2
L −2
L −3
−3 −4
−4
B1
B2
Figure 2: Graph of example 10
!
B1 B2
Let the player B play the mixed strategy SB =
with y2 =
y1 y2
1 − y1 ; y1 , y2 ≥ 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)
A1
[−7y1 + 1, −3y1 + 3]
A2
[−3y1 − 1, y1 + 1]
A3
[5y1 − 3, 9y1 − 1]
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 A1 and A3 as it is the region of intersection L1 and L3 . 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
B1 B2
!
A1
[-4,-2] [1,3]
.
A3
[4,6] [-3,-1]
The abscissa of the points P and Q are − 23 , 43 and the ordinates are 31 . Therefore the optimal strategies are for B is ( 13 , 23 ) and the value (V ) of the game is
= [− 23 , 43 ].
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
B1
B2 B3 B4
!
A1
[5,7] [4,6] [1,3] [2,4]
.
A2
[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 (B1 , B4 ) and (B3 , B4 ) 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 B4 .
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.
Solution of Rectangular Interval Games Using Graphical Method
113
9
9
8
8
7
7
6TT 6
5@@
5
T
T
R
4TS
4
@
@T T@
3
T T@ 3
@
T 2
2 T@
T@
P
Q
1
T@ 1
T 0
0
−1
−1
A1
A2
Figure 3: Graph of example 11
B1 B2 B3 !
[-2,0] [1,3] [2,4]
A1
.
A2
[5,7] [2,4] [1,3]
The graphical representation of game is shown in the figure 4.
8
8
7
7
6
6
5
5
4PP
4
R Q3
P
P
3P
2PP 2
S PP
1
P P 1
0
0
−1
−1
−2
−2
A1
A2
Figure 4: Graph of example 12
It is seen that the three gain regions of A corresponding to B1 , B2 , B3 have
a common maximum of minimum region P QRS. Solving the 2 × 2 sub game
involving B1 and B2 of B we get the optimal solution, for A optimal strategy
114
Prasun Kumar Nayak and Madhumangal Pal
is A( 21 , 12 ) and the value of the game is [ 32 , 72 ]. The value of the game is [ 32 , 72 ]
for the two 2 × 2 sub game with (B1 , B3 ) and (B2 , B3 ) also. Thus the game
can be reduced to 2 × 2 in the following manner:
A1
A2
B1 B2 !
[-2,0] [1,3]
,
[5,7] [2,4]
A1
A2
B3
B4 !
[1,3] [2,4]
.
[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
B1 B2
!
A1
[ 21 , 32 ] [ 12 , 32 ]
[ 21 , 32 ] [ 12 , 32 ]
A2
optimal solution which will mixes three strategies B1 , B2 , B3 . 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 [ 23 , 27 ]. So the value of the IG to the player A is [ 32 , 72 ].
9. 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.
Solution of Rectangular Interval Games Using Graphical Method
115
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