Lecture on

“Game Theory”

By

Dr. D. B. Naik (Ph.D. - Mech. Engg.),Professor & Head, Training & Placement Section,Sardar Vallabhbhai National Institute of Technology, Surat, Gujarat, India

Topics to be covered :

(1) What is a game ?

(2) “Two Person Zero Sum” game

(3) Solution of game means ?

(4) Saddle Point

(5) Pure Strategy and Mixed Strategy

(6) Dominance Rule

(7) Value of game for given Mixed Strategies of both

players

(8) Algebraic Method for 2x2 Mixed Strategy game

problems

(9) Methods to solve 2xN or Nx2 games :

[a] Method of sub-games

[b] Graphical Method

(10) Application of LP to solve game problem

(11) Iterative / Approximate Method to solve game

problem

[ 1 ] What is a game ?

B

1 2 ……………. n

1

2

A .

.

m

Pij

Aim of A is to get maximum

A’s payoffs / gains

(Payoff Matrix)

Aim of B is to see that A gets minimum

Game Problems are problems of “Decision Making Under Conflicts” or “Decision Making Under Competitive Situations”.

[ 2 ] “Two Person Zero Sum” game( i ) Two competitors.

( ii ) Finite number of strategies by both players.

( iii ) Play of game results when each competitor

chooses a single strategy or combination of

strategies.

( iv ) After both have chosen strategy / strategies,

their respective gains are finite.

( v ) Gain of one will be loss of other.

( vi ) Gain of a competitor depends on his action

as well as action of opponent.

[ 3 ] Solution of game means ?

( 1 ) Optimal (Best) strategies by both players.

( 2 ) Value of game when both players choose

optimal strategies.

V * = Optimal value of game

= Expected gain of A if both the players use

their best strategies.

[ 4 ] Saddle Point : B

1 2 3 4

1

A 2

3

−5 3 1 20

5 5 4 6

−4 −2 0 −6

Minima

−5

4

−6

5 Maxima 5 4 20

Maximin

Minimax

(Maximum out of Minimums.)

(Minimum out of Maximums.)

Saddle Point is P23

V* = 4

Saddle Point definition :

It is an element of a matrix that is both the lowest

element in its row and the highest element in its

column.

It can be defined as an element of matrix that is at

once the largest of row minima and the smallest of

column maxima.

OR

[ 5 ] Pure Strategy and Mixed Strategy

1 2

2 3

1

2

2 3

V = 2

Pure Strategy Case

1 8

6 4

1

4

6 8

4 ≤ V ≤ 6

Mixed Strategy Case

[ 6 ] Dominance Rule Dominance Rule is applied to reduce the size of game problem.

j

i

k

If Pij ≥ Pkj for each j then

kth strategy of A is deleted.

≤

j k

i

If Pij ≥ Pik for each i then

jth strategy of B is deleted.≥

[ 7 ] Value of game for given Mixed

Strategies of both players

B

(y1) (y2) (y3) (y4)

1 2 3 4

(x1) 1

A (x2) 2

(x3) 3

Pij

Mixed Strategy 3x4 game problem :

x1 + x2 + x3 = 1

y1 + y2 + y3 + y4 = 1

B

(y1) (y2) (y3) (y4)

1 2 3 4

(x1) 1

A (x2) 2

(x3) 3

Pij

V = P11 x1 y1 + P12 x1 y2 + P13 x1 y3 + P14 x1 y4

+ P21 x2 y1 + P22 x2 y2 + P23 x2 y3 + P24 x2 y4

+ P31 x3 y1 + P32 x3 y3 + P33 x3 y3 + P34 x3 y4

[ 8 ] Algebraic Method for 2x2 Mixed

Strategy game problems

P11 P12

P21 P22

y1 y2

x1

x2

P11 x1 + P21 x2 ≥ V

P12 x1 + P22 x2 ≥ V

x1 + x2 = 1

P11 y1 + P12 y2 ≤ V

P21 y1 + P22 y2 ≤ V

y1 + y2 = 1

12212211

12212211

PPPP

PPPPV

1211

2122

2

1

PP

PP

x

x

2111

1222

2

1

PP

PP

y

y

2 −1

−2 1 2 1

Solve following Game Problem :

−1

−2

−1 ≤ V ≤ 1

Mixed Strategy

2112

2112

V

12

21

2

1

x

x

22

11

2

1

y

y

05

0

1

1

3

3 x1 = ½, x2 = ½

2

1

4

2 y1 = 1/3, y2 = 2/3

[ 9 ] Methods to solve 2xN or Nx2 games :

[a] Method of sub-games

[b] Graphical Method

[a] Method of sub-games :

1

2

3

1 2

2 4

3 2

-2 6

2

2

-2

3 6

2 ≤ V ≤ 3 Mixed Strategy

Dominance Rule not applicable.

1

2

3

1 2

2 4

3 2

-2 6

1 2 1 2 1 21

2

1

3

2

3

Sub games :

2 4

3 2

2

2

3 4

2 ≤ V ≤ 3

2 4

-2 6

2

-2

2 6

V = 2

3 2

-2 6

2

-2

3 6

2 ≤ V ≤ 3

1 2

1 2

1 2

1

2

1

3

2

3

2 4

3 2

2

2

3 4

2 4

-2 6

2

-2

2 6

3 2

-2 6

2

-2

3 6

66.2

3

8

4322

4322

V

V = 2

44.2

9

22

2263

2263

V

V* = 2.66

Hence,

1 21

2

2 4 2

2

3 4 3 2

V* = 2.66

0,3

2,3

1

2

1

42

32321

2

1

xxxx

x

3

1,3

2

1

2

32

4221

2

1

yyy

y

[b] Graphical Method

1

2

3

1 2

2 4

3 2

-2 6

2y1 + 4y2 ≤ V

3y1 + 2y2 ≤ V

−2y1 + 6y2 ≤ V

y1 + y2 = 1

2y1 + 4 (1−y1) ≤ V

3y1 + 2 (1−y1) ≤ V

−2y1 + 6 (1−y1) ≤ V

V + 2y1 ≥ 4

V − y1 ≥ 2 V + 8y1 ≥ 6

0−2

2

4

6

8

−2

2 3

V

y1

V* = 2.66

y1 = 2/3

I

II

III

V* = 2.66

y1 = 2/3, y2 = 1/3

x3 = 0

[ 10 ] Application of LP to solve game problem

−2 3 4

−1 4 −3

3 −4 5

3 4 5

−2

−3

−4

−2 ≤ V ≤ 3

Add +3 to each element

1 6 7

2 7 0

6 −1 8

1 ≤ V ≤ 6

y1 y2 y3

1y1 + 6y2 + 7y3 ≤ V

2y1 + 7y2 + 0y3 ≤ V

6y1 − 1y2 + 8y3 ≤ V

y1 + y2 + y3 = 1

176 321 V

y

V

y

V

y

1072 321 V

y

V

y

V

y

1816 321 V

y

V

y

V

y

VV

y

V

y

V

y 1321

Max (1/V) = Y1 + Y2 + Y3

thenV

yYIf ,

s/t Y1 + 6Y2 + 7Y3 ≤ 1

2Y1 + 7Y2 + 0Y3 ≤ 1

6Y1 − 1Y2 + 8Y3 ≤ 1 Y1, Y2, Y3 ≥ 0

Solving this LPP we shall get :

Y1 = ___ Y2 = ___ Y3 = ___ 1/V = ___ V = ___ y1 = ___ y2 = ___ y3 = ___ V = ___

Vactual = V − 3

Now X1, X2, X3 will be dual variables of the primal problem.

The values of X1, X2, X3 will be obtained from final table of simplex of primal problem.

X1 = ___ X2 = ___ X3 = ___

x1 = ___ x2 = ___ x3 = ___

[ 11 ] Approximate method / Method of Iteration

1 6 7

2 7 0

6 −1 8

1 6 7

2 7 0

6 −1 8

1

2

66 1 8

13

16

4

7

9

5

20

16

12

21

18

18

22

20

24

23

22

30

29

29

29

30

31

35

36

38

34

3

3

4

8 6 8 10 13 8 11 19 15 12 25 22 18 24 30 24 23 38 25 29 45 31 28 53 33 35 53 5 4 1

Hence, 33/10 V* 38/10

x1 : x2 : x3 = 3 : 3 : 4

y1 : y2 : y3 = 5 : 4 : 1

Thank youThank youFor any Query or suggestion :

Contact :Dr. D. B. Naik Professor & Head, Training & Placement (T&P)S. V. National Institute of Technology (SVNIT), Ichchhanath, Surat – 395 007 (Gujarat) INDIA.

Email ID : dbnaik@gmail.comdbnaik@svnit.ac.indbnaik_svr@yahoo.com

Phone No. : 0261-2201540 (D), 2255225 (O)