JNTUK III Mech, OR Unit 4

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GAME THEORY
 Game theory may be defined as “a body of knowledge that deals with making decisions
when two or more intelligent and rational opponents are involved under conditions of
conflict and competition”.
 For example, if two firms are locked up in a war to maintain their market share, then a
price cut by the first firm will invite reaction from the second firm in the nature of a price
cut. This will, in turn, affect the sales and profits of the first firm, which will again have
to develop a counter-strategy to meet the challenge from the second firm. The game will
thus go on.
 Game theory, thus, helps to determine the best course of action for a firm in view of the
expected counter moves from the competitors. Game theory deals with competitive
situations of decision-making under uncertainty
 The term ‘game’ represents a conflict between two or more parties. There can be several
types of games, e.g. two-person and n-person games, zero-sum and non-zero-sum games,
constant-sum games, co-operative and non-co-operative games, pure strategy games and
mixed strategy games, etc.
(i) When there are two competitors playing a game, it is called a two-person game.
(ii) If the number of competitors are N (N > 2), it is known as an N person game.
(iii) When the sum of amounts won by all winners is equal to the sum of the amounts lost by
all losers, it is called a zero-sum game.
(iv) In a zero-game or a constant-sum game, the sum of gains and losses of the game is zero.
As opposed to this, if the sum of gains or losses in a game is not equal to zero, it is called a
non-zero-sum game.
(v) When the best strategy for each player is to play one particular strategy throughout the
game, it is known as a pure strategy game. In case the optimum plan for each player is to
employ different strategies at different times, it is called a mixed strategy game. When there
is communication between the participants they may reach an agreement and increase their
pay-off through some forms of co-operative game, otherwise it is a non-co-operative game.
Rules of the Game
Game theory is applicable to situations that satisfy the following conditions:
(a) The number of competitors is finite.
(b) The players act rationally and intelligently.
(c). Each player has available to him a finite set of possible courses of action.
(d) There is a conflict of interests between the participants.
(e) The players make individual decisions without directly communicating.
(f) The rules governing the choice are specified and known to the players.
(g) The players simultaneously select their respective courses of action.
(h) The payoff (outcome) is fixed and determined in advance.

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Basic Terminology of Game Theory
(1) Strategy
A strategy for a player has been defined as a set of rules or alternative courses
of action available to him in advance, by which player decides the courses of
action that he should adopt. There are two types:
(i) Pure Strategy
If the player selects the same strategy each time, then it is a pure strategy. In this case each
player knows exactly what the other is going to do, i.e. there is a deterministic situation and the
objective of the players is to maximize gains or to minimize losses.
(ii) MixedStrategy
When the players use a combination of strategies and each player is always kept guessing as to
which course of action is to be selected by the other, then it is known as a mixed strategy. Thus,
there is a probabilistic situation and the objective of the player is to maximize expected gains or
to minimize losses. Thus, mixed strategy is a selection among pure strategies with fixed
possibilities.
(2) Optimal Strategy
A course of action which puts the player in the most preferred position irrespective of the
strategy of his competitors. Any deviation from this strategy results in a decreased pay-off for the
player.
(3) Value of the Game
The expected pay-off of the game when all the players of the game follow their optimum
strategies. The game is called fair if the value of the game is zero and unfair if it is non-zero.
(4) Two-person zero-sum game
There are two types of Two-person zero-sum games. In one, the most preferred position is
achieved by adopting a single strategy and therefore the game is known as the pure strategy
game. The second type requires the adoption by both players of a combination of different
strategies in order to achieve the most preferred position, and is thus referred to as the mixed-
strategy game.
(5) Pay-off Matrix
A two-person zero-sum game is conveniently represented by a matrix. The matrix which shows
the outcome of the game as the players select their particular strategies is known as the pay-off
matrix. It is important to assume that each player knows not only his own list of possible courses
of action but also that of his opponent.
Let player A have m courses of action (A1, A2, A3, …, Am) and player B have n courses of action
(B1, B2, B3, …, Bn). (The numbers m and n need not to be equal). The possible number of
outcomes is therefore (mxn).

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The outcomes are shown in the pay-off matrix:
B1 B2 B3
A1 a11 a12 a13
A2 a21 a22 a23
A3 a31 a32 a33
The term aij represents the gain for the player A (and loss suffered by B) when A adopts the ith
strategy and B adopts the jth strategy.
Pure strategies (games with a saddle point)
The minimum value in each row represents the least gain (pay-off) guaranteed to player A, if he
plays his particular strategy. These are indicated in the matrix by row-minima. Player A will then
select the strategy that maximizes the minimum gains. Player A’s selection is called the Maximin
strategy and his corresponding game is called the Maximin value of the game.
Player B on the other hand, would like to minimize his losses. The maximum in each column
represents the maximum losses to player B if he plays his particular strategy. These are indicated
in the matrix by column maxima. Player B will then select the strategy that minimizes his
maximum losses. Player B’s selection is called the minimax strategy and his corresponding loss
is called the minimax value of the matrix.
Saddle Point
If the maximin value equals the minimax value, then the game is said to have saddle point and
the corresponding strategies are called optimum strategies. The amount of payoff at the
equilibrium point is known as the value of the game. It may be noted that if player A adopts
minimax criterion, then player B has adopted maximin criterion, as it is a two-person zero-sum
game.
Rules to determine the Saddle Point
(1) Select the minimum element in each row of the pay-off matrix and write them under ‘row
minima’ heading. Then select a largest element among these elements and enclose it in a
rectangle. Call it Maximin.
2. Select the maximum element in each column of the pay-off matrix and write them under
‘column maxima’ heading. Then select a lowest element among these elements and enclose it in
a circle. Call it Minimax.
3. Find the element(s) which is the same in the circle as well as the rectangle and mark the
position of such element(s) in the matrix. This element represent the value of the game and is
called the saddle (or equilibrium) point.

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Example:
Player A Player B
B1 B2 B3 Row Minima
A1 -1 2 -2 -2
A2 6 4 - 6 -6
Column
Maxima
6 4 -2 Maximin = -2
Minimax = -2
Here, Maximin = - 2 = Minimax.
So, Saddle point = - 2. Value of the Game = - 2
The Optimal Strategies are: A will adopt A1 strategy. B will adopt B3 strategy.
Example 2:
A company management and the labour union are negotiating a new 3-year settlement.
Each of these has 4 strategies.
I: Hard and aggressive bargaining; II: Reasoning and logical approach; III: Legalistic strategy
and IV: Conciliatory approach.
The costs to the company are given for every pair of strategy choice
Union
Strategies
Company Strategies
I II III IV
I 20 15 12 35
II 25 14 8 10
III 40 2 10 5
IV -5 4 11 0
What strategy will the two sides adopt? Also determine the value of the game.
Solution
Union
Strategies
Company Strategies
I II III IV Row Minima
I 20 15 12 35 12
II 25 14 8 10 8
III 40 2 10 5 2
IV -5 4 11 0 -5
Column
Maxima
40 15 12 35 Maximin = 12
Minimax = 12

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Maximin = 12 = Minimax. Saddle point = 12 = Value of the Game.
Optimal Strategies
Union will adopt Strategy I: Hard & Aggressive Bargaining
Company management will adopt Strategy III: Legalistic Strategy.
Example 3:
Find the range of values of p and q that will render (A2, B2) a saddle point.
Player A Player B
B1 B2 B3
A1 2 4 5
A2 10 7 q
A3 4 p 6
 For (A2, B2) = 7 to be a saddle point, it has to be both Maximin and Minimax.
So, it has to be the minimum in Row A2. So, q > 7.
In column B2, it has to be the maximum. So, p < 7.
Therefore, (A2, B2) will be a saddle point if p < 7 and q > 7.
SOLUTION OF GAMES WITHOUT SADDLE POINTS
 In certain cases when there is no pure strategy solution for games, i.e. no saddle point
exists, that is Maximin is not equal to Minimax.
 In all such cases, both the players must determine an optimum mixture of strategies to
find the value of the game and an optimal strategy.
 The optimum strategy mixture for each player may be determined by assigning to each
strategy its probability of being chosen.
 Since strategies so determined are probabilistic combinations of available choices of
strategy, they are called mixed strategies.
 For solving a 2 X 2 game without a saddle point, the following formula is used:
Player A Player B
B1 B2
A1 a11 a12
A2 a21 a22

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 Then p1 = Probability that A will use strategy A1
p1= (𝑎22 − 𝑎21)/[( 𝑎11+ 𝑎22) − ( 𝑎12 + 𝑎21)]
and p2 = Probability that A will use strategy A2 = (1-p1).
q1 = Probability that B will use strategy B1
q1 = (𝑎22 − 𝑎12)/[( 𝑎11+ 𝑎22) − ( 𝑎12 + 𝑎21)]
and q2 = Probability that B will use strategy B2 = (1-q1).
Value of the Game = V= = (𝑎11. 𝑎22 − 𝑎21. 𝑎12)/[( 𝑎11+ 𝑎22) − ( 𝑎12 + 𝑎21)]
Example
In a game of “matching coins” between two players, suppose
(i) A wins one unit of value when there are two heads,
(ii) wins nothing when there are two tails and
(iii) loses ½ unit of value when there is one head and one tail.
Determine the pay-off matrix, the best strategies for each and the value of the game to A.
Solution
The pay-off matrix is as shown below:
Player A Player B Row Minima
Head Tail
Head 1 -1/2 -1/2
Tail -1/2 0 -1/2
Column Maxima 1 0 Maximin = -1/2
Minimax = 0
Maximin ≠ Minimax. Therefore, saddle point does NOT exist.
Here a11 = 1; a12 = - 1/2;
a21 = - 1/2; a22 =0.
Probability that A will use Strategy I (pick up head) = p1
p1 == (𝑎22 − 𝑎21)/[( 𝑎11+ 𝑎22) − ( 𝑎12 + 𝑎21)]
= (0 – (- ½))/(1 + 0) – (- ½ -1/2)
= ¼.
p2 = Probability that A will use Strategy II (pick up tail) =1-p1= 1-1/4 = ¾.
Probability that B will use Strategy I (pick up head) = q1
q1 == (𝑎22 − 𝑎12)/[( 𝑎11+ 𝑎22) − ( 𝑎12 + 𝑎21)]
= (0 – (- ½))/(1 + 0) – (- ½ -1/2)
= ¼.
q2 = Probability that B will use Strategy II (pick up tail) =1-q1= 1-1/4 = ¾.
Value of the Game = V= = (𝑎11. 𝑎22 − 𝑎21. 𝑎12)/[( 𝑎11+ 𝑎22) − ( 𝑎12 + 𝑎21)
So, V = [(1X0 ) – (-1/2 X -1/2)]/ (1+0) –(-1/2 -1/2) = - 1/8

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Rules of (Principles) Dominance
The concept of dominance can be applied to any two-person zero-sum game with any number of
strategies for each player. For a pay-off matrix of large size, the rule of dominance can be used to
reduce its size by carefully eliminating certain rows and columns prior to the final analysis to
determine the optimum strategy selection for each player. In general the following rules of
dominance are used to reduce the size of the matrix.
Rule 1
If all the elements in a row (say the ith row) of a pay-off matrix are less than or equal to the
corresponding elements of the other row (say the jth row), then the player A will never choose
the ith strategy or in other words, the ith strategy is dominated by the jth strategy.
Rule 2
If all the elements in a column (say the rth column) of a pay-off matrix are greater than or equal
to the corresponding elements of the other column (say the sth column), then the player B will
never choose the rth strategy or in other words, the rth strategy is dominated by the sth strategy.
Rule 3
A pure strategy may be dominated if it is inferior to the average of two or more other pure
strategies.
Solution methods of games without saddle points
In certain cases when there is no pure strategy solution for games, i.e. no saddle point exists,
both the players must determine an optimum mixture of strategies to find a saddle point. The
optimum strategy mixture for each player may be determined by assigning to each strategy its
probability of being chosen. Since strategies so determined are probabilistic combinations of
available choices of strategy, they are mixed strategies. A mixed strategy can be solved by the
following methods:
(1) Arithmetic Method
If the game can be reduced to a 2 X 2 matrix with the help of the rules of dominance, then the
cost matrix looks as follows:
B1 B2
A1 a11 a12
A2 a21 a22
The process is as follows:
(1) Find the difference between the two values in the first row and put it against the second row
of the matrix (neglecting the negative sign, if any).
(2) Find the difference between the two values in the second row and put it against the first row
of the matrix (neglecting the negative sign, if any).
(3) Repeat Steps 1 and 2 for the columns also.
The values obtained by ‘swapping the differences’ represent the optimal relative frequencies of
play for both players’ strategies.
These may be converted to probabilities by dividing each of them by their sum.

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Example for Arithmetic Method
The payoff matrix of the game between Firms A and B is as follows:
Firm A Firm B
No advtg
B1
Medium advtg
B2
Large advtg
B3
Row Minima
No advtg
A1
50 40 28 28
Medium advtg
A2
70 50 45 45
Large advtg
A3
75 47.5 50 47.5
Column
Maxima
75 50 50 Maximin = 47.5
Minimax - 50
Since Maximin ≠ minimax, saddle point does NOT exist.
Using rules of dominance
(1) Row A1 is dominated by A2 and A3 as all elements of A2 are less than or equal to the
corresponding elements in A2 and A3 as can be seen below:
B1 B2 B3
A1 50 40 28
A2 70 50 45
A3 75 47.5 50
So, A will never choose Strategy A1 and thus Row A1 can be deleted from the payoff matrix.
(2) Column B1 is dominated by B2 and B3 as all elements of B2 are more than or equal to the
corresponding elements in B2 and B3 as can be seen below:
B1 B2 B3
A1 50 40 28
A2 70 50 45
A3 75 47.5 50
So, B will never choose Strategy B1 and thus Row B1 can be deleted from the payoff matrix.
The reduced payoff matrix becomes
B2 B3
A2 50 45
A3 47.5 50
This reduced payoff matrix also does NOT have a saddle point as Maximin ≠ Minimax.

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B2 B3
A2 50 45 47.5-50 = 2.5 p1 = (2.5)/(2.5 + 5)
= 1/3
A3 47.5 50 50-45 = 5 p2 = (5)/(2.5 + 5)
= 2/3
45- 50 = 5 50-47.5 = 2.5
q1 = (5)/(2.5 + 5)
= 2/3
q2 = (2.5)/(2.5 +
5)
= 1/3
So, A adopts strategy A2 (Medium advtg) with a probability 1/3.
and adopts strategy A3 (Large advtg) with a probability 2/3.
B adopts strategy B2 (Medium advtg) with a probability 2/3.
and adopts strategy B3 (Large advtg) with a probability 1/3.
Value of the Game = Expected Gain to A (or) Expected loss to B
Expected Gain to A
If B adopts B2, V = 50 X (1/3) + 47.5 X (2/3) = 145/3
If B adopts B3, V = 45X (1/3) + 50 X (2/3) = 145/3
Expected Loss to B
If A adopts A2, V = 50 X (2/3) + 45 X (1/3) = 145/3
If A adopts A3, V = 47.5 X (2/3) + 50 X (1/3) = 145/3.
Therefore, Value of the Game = 145/3

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(2) Algebraic Method
If the game can be reduced to a 2 X 2 matrix with the help of the rules of dominance, then the
cost matrix looks as follows:
B1 B2
A1 a11 a12
A2 a21 a22
The process of algebraic method is as follows:
Expected Gain to A
a11 p1 + a21 p2 ≥ V
a12 p1 + a22 p2 ≥ V
Expected Loss to B
a11 q1 + a12 q2 ≤ V
a21 q1 + a22 q2 ≤ V
Considering these inequities as equations, we get the values of p1, p2 and q1, q2.
From this, the Value of the Game, V can be calculated.
Example for Algebraic Method
The payoff matrix of the game between Firms A and B is as follows:
Firm A Firm B
No advtg
B1
Medium advtg
B2
Large advtg
B3
Row Minima
No advtg
A1
50 40 28 28
Medium advtg
A2
70 50 45 45
Large advtg
A3
75 47.5 50 47.5
Column
Maxima
75 50 50 Maximin = 47.5
Minimax - 50
Since Maximin ≠ minimax, saddle point does NOT exist.

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Using rules of dominance
(1) Row A1 is dominated by A2 and A3 as all elements of A2 are less than or equal to the
corresponding elements in A2 and A3 as can be seen below:
B1 B2 B3
A1 50 40 28
A2 70 50 45
A3 75 47.5 50
So, A will never choose Strategy A1 and thus Row A1 can be deleted from the payoff matrix.
(2) Column B1 is dominated by B2 and B3 as all elements of B2 are more than or equal to the
corresponding elements in B2 and B3 as can be seen below:
B1 B2 B3
A1 50 40 28
A2 70 50 45
A3 75 47.5 50
So, B will never choose Strategy B1 and thus Row B1 can be deleted from the payoff matrix.
The reduced payoff matrix becomes
B2 B3
A2 50 45
A3 47.5 50
This reduced payoff matrix also does NOT have a saddle point as Maximin ≠ Minimax.
Using the algebraic method,
50 p1 + 47.5 p2 ≥ V
45 p1 + 50 p2 ≥ V
Considering the above inequalities as equations,
50 p1 + 47.5 p2 = 45 p1 + 50 p2
So, 5 p1 = 2.5 p2
5 p1 - 2.5 p2 = 0
We know that p1 + p2 = 1

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Therefore, p1 = 1/3 and p2 = 2/3.
Similarly,
50 q1 + 45 q2 ≤ V
47.5 q1 + 50 q2 ≤ V
Considering the above inequalities as equations,
50 q1 + 45 q2 = 47.5 q1 + 50 q2
So, 2.5 q1 = 5 q2
2.5 q1 - 5 q2 = 0
We know that q1 + q2 = 1
Therefore, q1 = 2/3 and q2 = 1/3.
Value of the Game = 50 p1 + 47.5 p2 = 50 (1/3) + 47.5 (2/3) = 145/3
(3) Graphical Method (2 X n or m X 2 games)
 Consider the solution of games where one of the players has only two strategies
available:
When the player A, for example, has only 2 strategies to choose from and the player B
has n, the game shall be of the order 2 x n, whereas in case B has only two strategies
available to him and A has m strategies, the game shall be a m x 2 game.
 The problem may originally be a 2 x n or a m x 2 game or a problem might have been
reduced to such size after applying the dominance rule. In either case, we can use
graphical method to solve the problem.
Procedure for Graphical Method
(1) Draw two vertical, parallel lines representing p1 = 1 and p1 = 0.
(2) Draw the lines representing A’s strategies in response to each of B’s strategies.
(3) A is concerned with his least pay-off when he plays a particular strategy, which is
represented by the lowest value at that point, and wishes to choose p1 so as to maximize this
minimum pay-off.
(4) Identify the point in the figure where the lower envelope (represented by the shaded region),
the lowest of the lines, is the highest.
(5) The value of p1 is the value corresponding to the point where the lower envelope is highest.

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(6) Value of the Game, V, is the value given by the pay-off for A at that point.
We shall illustrate the solution of the 2 x n and m x 2 games in turn with the help of the
following examples.
Payoff matrix
B1 B2 B3 B4
A1 4 3 1 3
A2 0 5 2 1
Here, A has two strategies, A1 and A2. B has four strategies, B1, B2, B3 and B4.
Since this matrix does not have a saddle point, one must resort to mixed strategies for a solution.
If Player A plays strategy A1 with probability p1 and strategy A2 with probability (1 – p1), his
expected pay-off, R, against each of Player II’s strategies are:
B’s Strategy A’s Pay-offs
B1 R1 : 4 p1 + 0 (1-p1) = 4 p1
B2 R2: 3 p1 +5 (1-p1) = 5 – 2 p1
B3 R3 : 1p1 + 2 (1-p1) = 2- p1
B4 R4 : 3p1 + 1 (1-p1) = 2 p1+ 1

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The lines are marked R1, R2, R3 and R4 and they represent the pay-offs for A for the different
strategies adopted by B. For each value of p1, the height of the lines at that point denotes the pay-
offs of each of B's strategies against (p1, 1 - p1) for A. A is concerned with his least pay-off when
he plays a particular strategy, which is represented by the lowest of the four lines at that point,
and wishes to choose p1 so as to maximize this minimum pay-off. This is at the point in the
figure where the lower envelope (represented by the shaded region), the lowest of the lines, is the
highest. This point lies at the intersection of the lines R1 and R3 representing strategies B1 and
B3.
At this intersecting point p1 = 0.40 and Value of the game = 1.60.
(3) Linear Programming Method (Algebraic Method)
This method is used when:
(1) There is no saddle point in the pay-off matrix; and
(2) The pay-off matrix cannot be reduced to a 2 x 2 matrix using dominance property; and
(3) The game is not of the form 2 x n or m x 2.
Procedure for solving Game by Linear Programming Method
Let the pay-off matrix be as follows:
B1 B2 B3 --- Bn
A1 C11 C12 C13 ---- C1N
A2 C21 C22 C23 ---- C2N
A3 C31 C32 C33 --- C3N
---- ---- --- --- --- ----
AM CM1 CM2 CM3 --- CMN
Then, player A’s Linear Program is as follows:
Decision Variables
Let p1 = Probability that A uses strategy A1;
p 2 = Probability that A uses strategy A2;
p3 = Probability that A uses strategy A3;
--------------------------------------------
pm = Probability that A uses strategy AM;
Constraints
C11 p1 +C21 p 2 + C31 p3 + ………………….. + CM1 pm ≥ V

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C12 p1 +C22 p 2 + C32 p3 + ………………….. + CM2 pm ≥ V
C13 p1 +C23 p 2 + C33 p3 + ………………….. + CM3 pm ≥ V
………………………………………………………….
C11 P1 +C21 P 2 + C31 P3 + ………………….. + CM1 PM ≥ V
Where V is Value of the game.
Objective Function
Maximize Z = V.
Thus, the LP will solve for the values of p1, p2, p3, ………….pm that will satisfy the constraints
and maximize the value of the Game.
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UNIVERSITY QUESTIONS FOR UNIT 4
GAME THEORY
Q 1 (a) Explain the maxi-min and mini-max principles used in game theory.
(b) Discuss the rectangular games without saddle points.
Q (2) (a) Explain:
(i) Two-person zero-sum game; (ii) Saddle point ; (iii) Pure and mixed strategies with
reference to game theory
(b) Solve the following two-person-zero-sum game using graphical method
PLAYER A
PLAYER B
B1 B2
A1 2 - 4
A2 -1 6
A3 3 5
A4 4 1
A5 3 4
A6 -7 6
Q (3) In a town there are only two discount stores ABC and XYZ. Both stores run annual pre-
Diwali sales. Sales are advertized through local newspapers with the help of an advertizing firm.
ABC stores constructed following pay off table in units of Rs 1,00,000. Find the optimal
strategies for both stores and the value of the game.

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Store ABC Store XYZ
B1 B2 B3
A1 1 -2 1
A2 -1 3 2
A3 - 1 -2 3