Solving Game Problems Using a Quick Simplex ...

165 Int. J Latest Trend Math

Vol-4 No 1 March, 2014

Solving Game Problems Using a Quick Simplex Algorithm a New Method #1 #1

Mrs. N. V. Vaidya, #2Dr. Mrs. N.N .kasturiwale

Lecturer, Dr. Babasaheb Ambedkar College of Engg and Research Wanadongari , Nagpur 441110 Associate Professor, Department of Statistics, Institute of Science, Nagpur 440 001 , India

[email protected],[email protected]

Abstract—In this paper, a new approach is suggested while solving game theory problems (linear programming problems using simplex method).The method sometimes involves less iteration than in the simplex method or at the most an equal number because the method attempts to replace more than one basic variable simultaneously.

Keywords: basic feasible solution, optimum solution, simplex method, key determinant.

1. Introduction: Many practical problems require decision- making a competitive situation where there are two or more opposing parties with conflicting interests and where the action of one depends upon the one taken by the opponent, for example, candidates for an election, advertising and marketing campaigns by competing business firms, countries involved in military battles, etc, have their conflicting interests. On a competitive situation the courses of action (alternatives) for each competitor may be either finite or infinite. A competitive situation will be called a “Game” if it has the following properties: i) There are a finite number of competitors (participants) called players. ii) Each player has a finite number of strategies (alternatives) available to him. iii) A play of the game takes place when each player employs his strategy. iv) Every game results in an outcome e.g., loss or gain or a draw, usually called pay off, to some players. 1.1. General Solution of m x n Rectangular Games In a rectangular game with an m  n pay off matrix, if there does not exist any saddle point and also it is not possible to reduce the size of the game to m  2 or 2  n pay off matrix by using the concept of dominance, the following methods are generally used to solve the game. a) Linear Programming Method, and b) Interactive method for approximate solution. 1.2. Linear programming Method To illustrate the connection between a game problem and a liner programming problem let us consider an m  n pay off matrix (aij) for player A. Let Sm =

 pi i 1

B1 ,    Bn  q ,      q  n   1

n

m

Where

A1 ,- - - - - - -Am  P ,- - - - - - -P  and Sn = m   1

=

q j

=1

j1

be the mixed strategies for the two players respectively. Then, the expected gain gj (j = j,----n) of player A against B’s is pure strategies will be g1 = a11 P1 + a21 P2 + -----am1 Pm g2 = a12 P2+ a22 P2+-----+ am2 Pm gn= a1n P1 + a2n P2+-----amn Pm

________________________________________________ International Journal of Latest Trends in Mathematics IJLTCM, E-ISSN: 2049 - 2561 Copyright © ExcelingTech, Pub, UK (http://excelingtech.co.uk/)

166 Int. J Latest Trend Math

Vol-4 No 1 March, 2014 1

and the expected losses li (i = 1,----m) of player B against A S pure strategies will be l1 = a11 q1 + q12 q2 + ----- q1n qn l2 = a21 q1 + q22 q2 + -----+ q2nqn lm =am1q1 + am2q2 + -----+ amnqn The objective of player A is to select Pi (i=1,2,---m) such that he can maximize his minimum expected gains; and the player B desires to select qj (j = 1, 2,---n) that will minimize his expected losses. Thus, if we let u = And  = =

Max i

Min j

M

a p i 1

ij i

(j = 1,2,-----n)

M

a i 1

ij

q j (i = 1,2,-----m);

The problem of two players could be written as; Player A

1 = u

Maximize u = Minimize

M

 i 1

Pi u

Subject to the constraints; m

a i 1

ij

p i ≥ u and

 Pi = 1, Pi ≥ 0 (i = 1 , 2,----m)

Player B

1 = ν

Minimize  = Maximize

n

qj

J 1

ν



subject

To the constraints m

a j1

ij

q j ≤  and

q

j

 1 , qj ≥ 0 (j = 1, 2 ----- n).

Assuming that u > o and  = 0, we introduce new variables defined by, P’i =

qj pi and q’j = (i = 1, 2,-----m; j=1,2,-----n). u ν

Then, the pair of linear programming problems can be re-written as : Player A Minimize p0= P’1 + P’2 + -----+ P’m subject to the constraint. aijp’1 + a2jp’2 + ----- amj p’m ≥ 1 p’i ≥ 0 (i = 1,2-----m; j = 1,2,-----n) Player B Maximize q0 = q’1 + q’2 + ----- + q’n subject to constraints; ai1 q’1 + ai2q’2 + ----- + ainq’n ≤ 1 q’j ≥ 0 (i=1,2--m; j=1,2---n) Remarks: 1 It is easy to note that the L P P’, s of the two players represent a primal – dual pair. Therefore, by fundamental theorem of duality one can read off the optimal solution of one player,, just from the optimum simplex table of the opponent. That is, we need to solve just one players L.PP by simplex method. Remarks: 2 Liner programming technique requires all variables to be non-negative and, therefore, to obtain a nonnegative value  of the game; the data to the problem, i.e aij in the payoff table should all be non – negative. If there are some negative elements in the pay off table, a constant to every element in the pay off table must be added so as to make the smallest element zero; the solution to this new game will give an optimal mixed strategy for the original game. The value of the original game then equals the value of the new game minus the constant.

2. Quick Simplex Method:

167 Int. J Latest Trend Math

Vol-4 No 1 March, 2014

1. Replacement of n variables simultaneously and obtaining simplex table after such replacement. 2. Simultaneous replacement of n variables is possible only when pivotal elements in the entering vectors are in different rows. 3. We define key determinant R, which is of order n. 4. R is the determinant of sub matrix of matrix A. 5. This sub matrix is obtained by using rows and columns containing pivotal elements. 𝑷𝟏

𝑷𝟐

𝑷𝟑

𝑷𝟒

𝑷𝟓

𝑷𝟔

𝑷𝟕

𝑷𝟖

Pivot 𝒂𝟏

𝒃𝟏

𝒄𝟏

𝒅𝟏

1

0

0

0

𝒂𝟐

Pivot𝒃𝟐

𝒄𝟐

𝒅𝟐

0

1

0

0

𝒂𝟑

𝒃𝟑

𝐏𝐢𝐯𝐨𝐭 𝒄𝟑

𝒅𝟑

0

0

1

0

𝒂𝟒

𝒃𝟒

𝒄𝟒

𝒅𝟒

0

0

0

1

Here 𝑎1 , 𝑏2 𝑎𝑛𝑑 𝑐3 are the pivotal elements when 𝑃1 , 𝑃2 , 𝑃3 are the entering vectors in initial simplex table, then R= Pivot 𝒂𝟏 𝒃𝟏 𝒄𝟏 𝒂𝟐 Pivot𝒃𝟐 𝒄𝟐 𝒂𝟑 𝒃𝟑 𝐏𝐢𝐯𝐨𝐭 𝒄𝟑 is of order 3 as 3 variables are entered simultaneously. 6. We are giving formula to obtain simplex table after replacement of such n variables. 7. We shall call elements in the new simplex table as * elements. 8. Star elements are obtained by ratio of two determinants. 9. Denominator is nothing but the determinant R. 10. In the rows containing a pivotal element numerator is of order n. 11. Numerators are obtained as follows: Here the column of pivotal element is replaced by the column for which * elements are to be obtained. Numerator of star elements in the rows in which no pivotal element is there is a determinant of order (n+1) 12. It is obtained by adding the corresponding column in the current simplex table and the row of the element to R. 13. In previous case, simplex table after replacing 𝑃5 , 𝑃6 , 𝑃7 by 𝑃1 , 𝑃2 , 𝑃3 will be as follows. 𝑃1

𝑃2

𝑃3

1

0

0

0

1

0

0

0

1

0

0

0

𝑃4 𝑏1 𝑑1∗∗∗ = |𝑏2 𝑏3

𝑐1 𝑐2 𝑐3

𝑑1 𝑎1 𝑑2 | / |𝑎2 𝑑3 𝑎3

𝑃5 𝑏1 𝑏2 𝑏3

𝑎1 𝑑1 𝑐1 |𝑎2 𝑑2 𝑐2 | 𝑎 𝑑3 𝑐3 ∗∗∗ 𝑑2 = 3 𝑎1 𝑏1 𝑐1 |𝑎2 𝑏2 𝑐2 | 𝑎3 𝑏3 𝑐3 𝑎1 𝑏1 𝑑1 |𝑎2 𝑏2 𝑑2 | 𝑎 𝑏3 𝑑3 𝑑3∗∗∗ == 3 𝑎1 𝑏1 𝑐1 |𝑎2 𝑏2 𝑐2 | 𝑎3 𝑏3 𝑐3 𝑎1 𝑏1 𝑐1 𝑑1 𝑎 𝑏2 𝑐2 𝑑2 | 2 | 𝑎3 𝑏3 𝑐3 𝑑3 𝑎 𝑏4 𝑐4 𝑑4 𝑑4∗∗∗ = 4 𝑎1 𝑏1 𝑐1 |𝑎2 𝑏2 𝑐2 | 𝑎3 𝑏3 𝑐3

𝑐1 𝑐2 | 𝑐3

𝑃6

𝑃7

𝑃8 0

0

0

1

168 Int. J Latest Trend Math

Vol-4 No 1 March, 2014

14. One must evaluate 𝑋𝐵 Column first and nth iteration table should be evaluated.

3.

Quick simplex method formulae for entering two variables simultaneously:

Step 1: Use new criteria for entering vector as above. Step 2: Simultaneous exchange of two variables instead of exchanging one basic variable at a time (as done in classical simplex method) Step 3: Both the Pivotal element should not be in same row. Otherwise we cannot use our method. Step 4: Development of the formula to go to third simplex table from first simplex table. Consider Initial simplex table as following: Step 1: Initial simplex table 𝑦1

𝑦2

𝑦3

Pivot 𝑎1

𝑏1

𝑐1

𝑎2

Pivot𝑏2

𝑐2

𝑎3

𝑏3

𝑐3

Here y1 and y2 are entering vectors. Step 2: Second Simplex table 𝑦1

𝑦2

𝑦3

𝑏1 𝑎1 𝑎2 𝑏1 𝑏2∗ = 𝑏2 − 𝑎1 𝑎 3 𝑏1 𝑏3∗ = 𝑏3 − 𝑎1

𝑐1 𝑎1 𝑎2 𝑐1 𝑐2∗ = 𝑐2 − 𝑎1 𝑎 3 𝑐1 𝑐3∗ = 𝑐3 − 𝑎1

𝒚𝟏

𝒚𝟐

𝒚𝟑

1

0

0

1

0

0

𝑐1∗ =

𝑏1∗ =

1 0 0

Step 3:Third Simplex table

𝑐1∗∗ = 𝑐1∗ − 𝑐2∗∗ =

𝑏1∗ 𝑐2∗ 𝑏2∗

𝑐2∗ 𝑏2∗

𝑐3∗∗ = 𝑐3∗ −

𝑏3∗ 𝑐2∗ 𝑏2∗

Here we can find 𝑐1∗∗ , 𝑐2∗∗ and 𝑐3∗∗ using following formula. 𝑏1 𝑏2 𝑅

(−1)1 | 𝑐1∗∗

=

𝑐4∗∗ =

𝑐1 | 𝑐2

𝑎1 |𝑎2 𝑎4

𝑐2∗∗

𝑏1 𝑏2 𝑏4 𝑅

=

𝑎1 (−1)2 |𝑎

2

𝑐1 𝑐2 |

𝑅

𝑐1 𝑐2 | 𝑐4

4. Statement of the Problem: I Solve the following game by liner programming technique:

𝑐3∗∗ =

𝑅=|

𝑎1 𝑎2

𝑏1 | 𝑏2

𝑎1 |𝑎2 𝑎3

𝑏1 𝑏2 𝑏3 𝑅

𝑐1 𝑐2 | 𝑐3

169 Int. J Latest Trend Math

Vol-4 No 1 March, 2014

1  Player A  3 6 

3  5 -3  2 - 2 

-1

4. Solution of the problem: Since some of the entries in the pay off matrix are negative, we add a suitable constant to such of the entries to ensure them all positive. Thus, adding a constant c=4 to each element, we get the following revised pay off matrix :

5 3 7    Player A  7 9 1  10 6 2    Let the strategic of the two players b

A1 A 2 P1 P2

SA = 

A3  B1 , SB =   P3  q1

B2 q2

B3  , q 3 

Where P1 + P2+ P3 = 1 and q1 + q2+ q3 = 1 The linear programming formulation for the two players problems are: For Player A Maximize   Minimize

1 = x1 + x2+ x3 ν

Subject to the constraints: 5x1 + 7x2 + 10x3 ≥1, 3x1 + 9x2 + 6x3 ≥ 1, 7x1 + x2 + 2x3 ≥; xj ≥ 0 (j = 1,2,3) For Player: B Minimize  = Maximize

1 = y1 +y2 +y3 ν

Subject to the constraints; 5y1 + 3y2 + 7y3 ≤ 7y1 + 9y2 + y3 ≤ 1, 10y1 + 6y2 + 2y3 ≤ 1; yj ≥ 0

(j = 1,2,3)

Where xj = pj/u (j = 1,2,3) and yj = qj / v, (j=1,2,3) ; u = minimum expected gain to A and  = minimum expected loss to B Let us now solve the problem for player B, By introducing slack variables x4 ≥ 0, x3 ≥ 0 and x6 ≥ 0;. We can write problem as follows Max Z = x1 + x2 + x3 Subject to the constraints 5x1 + 3x2 + 7x3 + x4 =1 7x1 + 9x2 + x3 + x5 =1 10x1 + 6x2 + 2x3 + x6 =1

170 Int. J Latest Trend Math

Vol-4 No 1 March, 2014

5. Conventional Simplex method Initial iteration Introduce y1, drop y6

1

1

1

0

0

0

Basis

CB

y1

y2

y3

y4

y5

y6

XB

y4

0

5

3

7

1

0

0

1

y5

0

7

9

1

0

1

0

1

y6

0

10

6

2

0

0

1

1

-1

-1

-1

0

0

0



Ratio (1) 1/5

Ratio (2) 1/3

1/7

1/9

1/10

1/6

Ratio (3)

1 Min 7 1 1 1 2



Second Iteration, Introduce y3 and drop y4, 1

1

1

0

0

0

Basis

CB

y1

y2

y3

y4

y5

y6

XB

y4

0

0

0

6

1

0

-1/2

1/2

y5

0

0

24/5

-2/5

0

1

-7/10

3/10

y1

1

1

3/5

1/5

0

0

1/10

1/10

0

-2/5

-4/5

0

0

1/10





Ratio 1/12 -ve 1/2

Third Iteration, Introduce y2 and drop y5, 1

1

1

0

0

0

Basis

CB

y1

y2

y3

y4

y5

y6

XB

Y3

0

0

0

1

1/6

0

-1/12

1/12

y5

0

0

24/5

0

1/15

1

-11/15

1/3

y1

1

1

3/5

0

-1/30

0

7/60

1/12

0

-2/5

0

4/30

0

1/30





Third Iteration, Introduce y6 and drop y1 1

1

1

0

0

0

Ratio

5/72 5/36

171 Int. J Latest Trend Math

Vol-4 No 1 March, 2014

Ratio

Basis

CB

y1

y2

y3

y4

y5

y6

XB

y3

0

0

0

1

1/6

0

-1/12

1/12

y2

0

0

1

0

1/72

5/24

-11/72

5/72

y1

1

1

0

0

-1/24

-1/8

5/24

1/24

0

0

0

5/36

1/12

-2/33



5/72 5/36



Fourth Iteration 1

1

1

0

0

0

Basis

CB

y1

y2

y3

y4

y5

y6

XB

y3

1

2/5

0

1

-3/20

-1/20

0

1/10

y2

1

11/15

1

0

-1/60

7/60

0

1/10

y6

0

24/5

0

0

-1/5

-3/5

1

1/5

17/15

0

0

2/15

1/15

0

all, Zj – Cj ≥ 0 solution exists

6. Quick Simplex method: Since this is a cyclic problem, so we tried to introduce y2, y3 simultaneously and outgoing vectors are y5, y4 we get direct third simplex table using above formulae. 1

1

1

0

0

0

Basis

CB

y1

y2

y3

y4

y5

y6

XB

y4

0

5

𝒃𝟏 =3

𝒂𝟏 =7

1

0

0

1

y5

0

7

𝒃𝟐 =9

𝒂𝟐 =1

0

1

0

1

y6

0

10

𝒃𝟑 =6

𝒂𝟑 =2

0

0

1

1

-1

-1

-1

0

0

0









Ratio (2) 1/3 1/9 1/6

To find new values in 𝑿𝑩 column. Here we can find 𝑐1∗∗ , 𝑐2∗∗ and 𝑐3∗∗ using following formula.

𝑐1∗∗

3 (−1)1 | 9 = 60

1 | 1 = 1 10

𝑐2∗∗

7 1 38

(−1)2 | =

1 | 1 = 1 10

Ratio (3)

1 Min 7 1 1 1 2

172 Int. J Latest Trend Math

𝑐3∗∗

7 |1 = 2

Vol-4 No 1 March, 2014

3 1 9 1| 6 1 =1 60 5

𝑅=|

7 1

3 | = 60 9

1

New 𝑿𝑩 =

10 1 10 1

[5] To find other entries in third simplex table. 5 Column 𝑃1 = [ 7 ] : − 10

𝑐1∗∗ =

3 9 60

(−1)1 |

5 | 7 =2 5

𝑐2∗∗ =

7 1 60

(−1)2 |

5 | 7 = 11 15

𝑐3∗∗

7 |1 = 2

3 5 9 7| 6 10 = 24 60 5

𝑐3∗∗

7 |1 = 2

3 1 9 0| 6 0 = −1 60 5

𝑐3∗∗

7 |1 = 2

3 0 9 1| 6 0 = −3 60 5

2

New Column 𝑃1 =

5 11 15 24

[5] 1 Column 𝑃4 = [0] : − 0

𝑐1∗∗

3 9 60

(−1)1 | =

1 | 0 = 3 20

𝑐2∗∗

7 1 60

(−1)2 | =

1 | 0 = −1 60 3

New Column 𝑃4 =

20 −1 60 −1

[5] 0 Column 𝑃5 = [1] : − 0

𝑐1∗∗

3 9 60

(−1)1 | =

0 | 1 = −1 20

𝑐2∗∗

7 1 60

(−1)2 | =

0 | 1 = 7 60 −1

New Column 𝑃5 =

20 7 60 −3

[5] 0 Column 𝑃6 = [0] : − 1

173 Int. J Latest Trend Math

𝑐1∗∗

3 9 60

(−1)1 | =

Vol-4 No 1 March, 2014

0 | 0 =0

𝑐2∗∗

7 1 60

(−1)2 | =

0 | 0 =0 𝑐3∗∗

7 |1 = 2

3 0 9 0| 6 1 =1 60

0 New Column 𝑃6 = [0] 1

1

1

1

0

0

0

Basis

CB

y1

y2

y3

y4

y5

y6

XB

y3

1

2/5

0

1

-3/20

-1/20

0

1/10

y2

1

11/15

1

0

-1/60

7/60

0

1/10

y6

0

24/5

0

0

-1/5

-3/5

1

1/5

17/15

0

0

2/15

1/15

0

all, Zj – Cj ≥ 0 solution exists Here we arrive direct fourth iteration ,since it is a cyclic problem. The expected value or game, therefore is obtained as 0 =  - 4 = 5 – 4 = 1 Optimum strategies for player B are, q10, = 0, q20 =

1 1 1 1 5= and q30 = x5= 2 10 10 2

Making use of duality, the optimum strategies for player A are obtained as, p0i =

2 1 1 2 x 5 = , p02 = x5= and p03 = 0 15 15 3 3

Hence, optimum solution to the given problem is

SA =

 A1 A 2 A 3  2/3 1/3 0  , SB  

B1 0 

B3  and  = 1 1/2 1/2

B2

7. Statement of the problem - II Solve the following 3 x 3 game :

3 - 2  Player A - 1 4  2 2

4 2  6

8. Solution of the Problem : Since some of the entries in the play off matrix are negative, we add a suitable constant to such of the entries to ensure them all +ve. Thus, adding a constant C= 3 to each element, we get the following revised pay off matrix : Player B

174 Int. J Latest Trend Math

Vol-4 No 1 March, 2014

6  Player A 2  5

7 7 5  5 9

1

Let the strategies of the two players be,

SA =

 A1 P  2

A3  , SB = P3 

A2 P2

B1 q  1

B2 q2

B3  q 3 

Where, P1 + P2 + P3 = 1 and q1 + q2+ q3 = 1 The linear programming formulation for the two players problems are : For players A Maximize  = Minimize

1 = x1 + x2 + x3 ν

Subject to the constraints 6x1 + 2x2 + 5x3 ≥ 1 x1 + 7x2 + 5x3 ≥ 1 7x1 + 5x2 + 9x3 ≥ 1 and xj ≥ 0 (j = 1 ,2,3) For player B Minimize  = Maximize

1 = y1 +y2 + y3 ν

Subject to the constraints : 6y1 + y2 + 7 y3 ≤ 2y1 + 7y2 + 5y3 ≤ 1 5y1 + 5y2 + 9y2 ≤ 1 and yj ≥ 0 (j = 1,2,3) Where xj = pj/u (j=1,2,3) and yj  qj/ (j = 1,2,3) u = minimum expected gain to A and  = minimum expected loss to B Let us now solve the problem for palyer B. By introducing slack variables y4, y5, y6, ≥ = 0 We can write problem as follows Max z = x1 + x2 + x3 Subject to the constraints 6x1 + x2 + 7x3 + x4 = 1 2x1 + 7x2 + 5x3 + x5 = 1 5x1 + 5x2 + 9x3 + x6 = 1 xj ≥ 0

9. Conventional Simplex method Initial iteration Introduce y2, drop y5 1

1

1

0

0

0

Basis

CB

y1

y2

y3

y4

y5

y6

XB

y4 y5 y6

0 0 0

6 2 5

1 7 5

7 5 9

1 0 0

0 1 0

0 0 1

1 1 1

Ratio (1) 1/6 1/2 1/5

Ratio (2) 1/1 1/7 1/5

Ratio (3) 1/7 1/5 1/9

175 Int. J Latest Trend Math Zj-Cj

Vol-4 No 1 March, 2014

-1

-1

-1

0

0

0





2nd Iteration Introduce y3 and drop y6

Basis y4 y5 y3

CB 0 0 1

Zj-Cj

1 y1 19/9 -7/9 5/9

1 y2 -26/9 38/9 5/9

1 y3 0 0 1

0 y4 1 0 0

0 y5 0 1 0

0 y6 -7/9 -5/9 1/9

-4/9

-4/9

0

0

0

1/9



XB 2/9 4/9 1/9

Ratio 2/19 -4/7 1/5

Ratio -2/26 2/19 1/5

1 7



3rd Iteration:

Basis y1 y5 y3

CB 1 0 1 Zj-Cj

1 y1 1 0 0 0

1 y2 -26/9 60/19 25/19 -20/19

1 y3 0 0 1 0



↓

0 y4 9/19 7/19 -5/19 4/19

0 y5 0 1 0 0

0 y6 -7/19 16/19 6/19 -1/19

XB 2/19 10/19 1/19

4th Iteration :

Basis y1 y5 y2

CB 1 0 1 Zj-Cj

1 y1 1 0 0 0

1 y2 0 0 1 0

1 y3 26/25 -12/5 19/25 4/5

0 y4 1/5 1 -1/5 0

0 y5 0 1 0 0

0 y6 -1/25 8/95 6/25 1/5

XB 4/25 2/5 1/25 1/5

 all zj – cj ≥ 0 solution exists.

10. Applying Quick Simplex method. Here we introduce y1 ,y2, y3 table using above formulae.

simultaneously and outgoing vectors are y4, y5, y6 we get direct fourth simplex

1

1

1

0

0

0

Basis

CB

y1

y2

y3

y4

y5

y6

XB

y4

0

𝒃𝟑 =1

0

0

1

0 0

𝒂𝟑 =7 𝒂𝟐 =5

1

y5 y6 Zj-Cj

𝒄𝟑 =6 𝒄𝟐= 2

𝒂𝟏 =9 -1

0 0 0

1 0 0

0 1 0

1 1









𝒄𝟏 =5 -1

𝒃𝟐 =7 𝒃𝟏 =5 -1





Ratio (1) 1/6 1/2 1/5

Ratio (2) 1/1 1/7 1/5

Ratio (3) 1/7 1/5 1/9

176 Int. J Latest Trend Math

Vol-4 No 1 March, 2014

To find new values in 𝑿𝑩 column 9 Here ∴ 𝑅 = |5 7 𝑥1 =

5 |7 1

5 1 2 1| 6 1 𝑅

=

−1 6

, 𝑥2 =

9 |5 7

5 7 1 1 1 1 𝑅

5 2| = 60 6

5 2| 6

1

=

6

, 𝑥3 =

9 |5 7

5 1 7 1| 1 1 𝑅

=

1 3

Since in 𝑋𝐵 column, we get negative value therefore we cannot entered all three vectors simultaneously so we enter two vectors according by considering minimum ratios. Here we introduce y2, y3 above formulae.

simultaneously and outgoing vectors are y5, y6 we get direct third simplex table using

1

1

1

0

0

0 Ratio (1)

Basis

CB

y1

y2

y3

y4

y5

y6

XB

y4

0

6

𝒃𝟑 =1

𝒂𝟑 =7

1

0

0

1

y5

0

2

𝒂𝟐 =5

0

1

0

1

1/6 1/2

y6 Zj-Cj

0

5 -1

𝒃𝟐 =7 𝒃𝟏 =5 -1

𝒂𝟏 =9 -1

0 0

0 0

1 0

1

1/5









9 5 38

1 | 1 = 2 19

Ratio (2) 1/1

Ratio (3) 1/7

1/7 1/5

1/5 1/9

To find new values in 𝑿𝑩 column. Here we can find 𝑐1∗∗ , 𝑐2∗∗ and 𝑐3∗∗ using following formula.

𝑐1∗∗

5 (−1)1 | 7 = 38

𝑐3∗∗

9 |5 = 7

1 | 1 = 1 19

𝑐2∗∗

5 1 7 1| 1 1 = 10 38 19

(−1)2 | =

𝑅=|

9 5

5 | = 38 7

10

New 𝑿𝑩 =

19 2 19 1

[19] To find other entries in third simplex table. 6 Column 𝑃1 = [2] : − 5 𝑐1∗∗

5 7 38

(−1)1 | =

5 | 2 = 25 38

𝑐2∗∗

9 5 38

(−1)2 | =

5 | 2 = −7 38

𝑐3∗∗

9 |5 = 7

5 5 7 2| 1 6 = 30 38 19

177 Int. J Latest Trend Math

Vol-4 No 1 March, 2014 30

New Column 𝑃1 =

19 −7 38 25

[ 38 ] 1 Column 𝑃4 = [0] : − 0 𝑐1∗∗

5 7 38

(−1)1 | =

0 | 0 =0

𝑐2∗∗

9 5 38

(−1)2 | =

0 | 0 =0 𝑐3∗∗

9 |5 = 7

5 0 7 0| 1 1 =1 38

𝑐3∗∗

9 |5 = 7

5 0 7 0| 1 1 = 13 38 19

1 New Column 𝑃4 = [0] 0 0 Column 𝑃5 = [1] : − 0 𝑐1∗∗

5 7 38

(−1)1 | =

0 | 1 = −5 38

𝑐2∗∗

9 5 38

0 | 0 = 9 38

(−1)2 | =

13

New Column 𝑃5 =

19 9 38 −5

[ 38 ] 0 Column 𝑃6 = [0] : − 1 𝑐1∗∗

5 7 38

(−1)1 | =

1 | 0 = 7 38

𝑐2∗∗

9 5 38

1 | 0 = −5 38

(−1)2 | =

𝑐3∗∗

9 |5 = 7

5 1 7 0| 1 0 = −22 38 19

−22

New Column 𝑃5 =

19 −5 38 7

[ 38 ] Here we get 1

1

1

0

0

0

Basis

CB

y1

y2

y3

y4

y5

y6

XB

y4 y2 y3 Zj-Cj

0 1 1

30/19 -7/38

0 1 0 0

0 0

1 0 0 0

13/19 9/38 -5/38 4/38

-22/19 -5/38 7/38 2/38

10/19 2/19 1/19

25/38 -20/38

1 0



Ratio (1) 1/3 -ve 2/25



Introduce y1 and drop y3 we get final table 1 1 1

0

0

0

Basis

CB

y1

y2

y3

y4

y5

y6

XB

Ratio (1)

y4

0

0

0

-12/5

1

1

-8/5

2/5

1/3

178 Int. J Latest Trend Math y2 y1 Zj-Cj

Vol-4 No 1 March, 2014

1 1

0 1 0

1 0 0

7/25 38/25 4/5

0 0 0

1/5 -1/5 0

-2/25 7/25 1/5

3/25 2/25 1/5

-ve 2/25

all, Zj – Cj ≥ 0 solution exists The expected value or game, therefore is obtained as 0 = 5 – 3 = 2 Optimum strategies for player B are, q01 =

2 x 5 = 2/5 25

q02 = 3/35 x 5 = 3/5 q03 = 0 Making use of duality, the optimum strategies for player A are obtained as,

1 x 5 1 5

p01 = 0 x 5 = 0, p02 = 0 x 5 = 0, P30 =

Hence optimum solution to the given problem is

B1 B2 2  , 3 0 5 5 11. Statement of the Problem III :  A1 SA =  0

A2

A3  , SB = 1 

B3   and  = 2 0 

In a town there are only two discount stores ABC and XYZ. Both stores run annual pre–diwali sales during the first week of October. Sales are advertised through local newspapers with the aid of an advertising firm. ABC stores constructed following pay off in units of Rs. 1,00,000. Find the optimum strategies for both stores and the value of the game. Strategies of XYZ

 1 - 2 1 - 1 3 - 2  Strategies of ABC   - 1 - 2 3 12. Solution of the Problem: Since some of the entries in the pay off matrix are negative, we add a suitable constant to such of the entries to ensure them all positive. Thus, adding a constant c = 3 to each element, we get the following revised pay off matrix : Store XYZ

Store ABC

4 2  2

1 4 6 1 1 6 

Let the strategies of the two stores be, SA =

A1 A 2 P P 2  1

A3  , P3 

B1 q1

SB = 

B2 q2

B3  q 3 

Where, P1 + P2 + P3 = 1 and q1 + q2+ q3 = 1 The linear programming formulation for the ‘two stores’– problems are: For stores ABC Maximize  = Minimize

1 = x1 + x2 + x3 ν

179 Int. J Latest Trend Math

Vol-4 No 1 March, 2014

Subject to the constraints 4x1 + 2x2 + 2x2 + 2x3 ≤ 1 x2 + 6x2 + x3 ≤ 1 4x1 + x2 + 6x3 ≤ 1 xj ≥ 0 (j =1, 2,3) For store XYZ Minimize  = Maximize

1 = y1 + y2+ y3 ν

Subject to the constraints, 4y1 + y2 + 4y3 ≤ 2y1 + 6y2 + y3 ≤1 2y1 + y2 + 6 y3 ≤ 1 yj ≥ 0 (j = 1,2,3) Where xj = pj/u, (j = 1,2,3) are yj = qj/ (j = 1,2,3) u = minimum expected gain to A and  = minimum expected loss to B Let us new solve the problem for store XYZ By Introducing slack variable x4 ≥ 0, x5 ≥ 0, x6 ≥ 0 We can write problem as follows Subject to the constraint 4x1 + x2 + 4x3 + x4 = 1 2x1 + 6x2 + x3 + x5 = 1 2x1 + x2 + 6x= + x6 = 1

13. Conventional Simplex method: Initial Iteration Introduce y2 and drop y2. 1

1

1

0

0

0

Basis

CB

y1

y2

y3

y4

y5

y6

XB

y4 y5 y6

0 0 0

4 2 2 -1

1

4 1 6 -1

1 0 0

0 1 0

0 0 1

1 1 1

6 1 -1 

Ratio (1) 1/4 1/2 1/2

Ratio (2) 1 1/6 1/1

Ratio (3) 1/4 1/1 1/6



Ist Iteration Introduce y1 drop y4 Basis y4 y2 y6

CB 0 1 0

Y1 11/3 1/3 5/3 -2/3

Y2 0 1 0 0

Y3 23/6 1/6 35/6 -5/6 

Y4 1 0 0 0

Y5 -1/6 1/6 -1/6 1/6

Y6 0 0 1 0 

XB 5/6 1/6 5/6 1/6

Ratio 0.22 0.5 0.5

180 Int. J Latest Trend Math

Vol-4 No 1 March, 2014

nd

II Iteration Introduce y3 and drop y 6.

Basis

CB

1 y1

1 y2

1 y3

0 y4

0 y5

y4

0

18/7

0

0

1

2/35

y2 y3

1 1

2/7 2/7 -3/7

1 0 0

0

0 0 0

6/35 -1/35 5/35

1 0



0 y6 23/35 -1/35 6/35 5/35

XB

Ratio 1/9

2/7 1/7 1/7

1/2 1/2



IIIrd Iteration :

Basis y1 y2 y3

1 y1 1 0 0 0

CB 1 1 1

1 y2 0 1 0 0

1 y3 0 0 1 0

0 y4 7/18 -1/9 -1/9 3/18

0 y5 1/45 52/315 -11/315 16/105

0 y6 -23/90 2/45 11/45 1/30

XB 1/9 1/9 1/9

Ratio

 all zj – cj ≥ 0, solution exists.

14. .Applying Quick Simplex method. Initial Iteration Introduce y2 and drop y2. 1 1 1

0

0

0

Basis

CB

y1

y2

y3

y4

y5

y6

XB

y4

0

𝒃𝟑 =1

𝒂𝟑 =4

1

0

0

1

y5

0

𝒄𝟑 =4 𝒄𝟐 =2

𝒂𝟐 =1

0

1

0

1

y6

0

𝒄𝟏 =2 -1

𝒃𝟐 =6 𝒃𝟏 =1 -1

Ratio (1) 1/4 1/2

𝒂𝟏 =6 -1

0

0

1

1

1/2







↓



↓

Ratio (2) 1 1/6 1/1

Ratio (3) 1/4 1/1 1/6

Here we introduce y1 ,y2, y3 simultaneously and outgoing vectors are y4, y5, y6 we get direct fourth simplex table using above formulae. 6 1 2 ∴ 𝑅 = |1 6 2| = 90 4 1 3 To find new values in 𝑿𝑩 column. 𝑥1 =

1 |6 1

2 1 2 1| 4 1 𝑅

1

= , 𝑥2 =

6 |1 4

9

1 2 1 2| 1 4 𝑅

=

1 9

, 𝑥3 =

6 |1 4

1 1 6 1| 1 1 𝑅

=

1 9

To find other entries in fourth simplex table. 1 Column 𝑃4 = [0] : − 0 𝑥1 =

1 |6 1

2 0 2 0| 4 1 𝑅

=

−1 9

, 𝑥2 =

6 |1 4

0 0 1 𝑅

2 2| 4

=

−1 9

,𝑥3 =

6 |1 4

1 0 6 0| 1 1 𝑅

=

7 18

181 Int. J Latest Trend Math

Vol-4 No 1 March, 2014

0 Column 𝑃5 = [1] : − 0

𝑥1 =

1 |6 1

2 0 2 1| 4 0 𝑅

−1

=

45

, 𝑥2 =

6 |1 4

0 1 0 𝑅

2 2| 4

=

8 45

, 𝑥3 =

6 |1 4

1 0 6 1| 1 0 𝑅

−1

=

45

0 Column 𝑃6 = [0] : − 1

𝑥1 =

1 |6 1

2 1 2 0| 4 0 𝑅

=

11

, 𝑥2 =

45

6 |1 4

1 2 0 2| 0 4 𝑅

=

2 45

, 𝑥3 =

6 |1 4

1 1 6 0| 1 0 𝑅

=

−23 90

IVTH Iteration :

Basis y1 y2 y3

CB 1 1 1

1 y1 1 0 0

1 y2 0 1 0

1 y3 0 0 1

0 y4 7/18 -1/9 -1/9

0 y5 -1/45 8/45 -1/45

0 y6 -23/90 2/45 11/45

0

0

0

3/18

6/45

1/30

XB 1/9 1/9 1/9

Ratio

1 3

 all zj – cj ≥ 0, solution exists The expected value or game, therefore is obtained as 0 = 3 – 3 = 0 Optimum strategies for store XYZ are q10 = 1/9 x 3 = 1/3 q20 = 1/9 x 3 = 1/3 q30 = 1/9 x 3 = 1/3 Making use of duality the optimum strategies for store ABC are obtained as, P 1o =

1 6 2 1 1 3 x 3  , p30 = x 3  x 3 = , p20 = 45 5 2 30 10 18

Hence the optimum solution to the given problem is

SA =

 A1 A 2 1/2 2/5 

 , SB = 1/10

A3

B1 B 2 B 2  1/3 1/3 1/3 and  = 0  

15. Conclusion : Game problems were successfully solved using Quick simplex a new method . It has been observed that the number of iterations were either less or remained the same when compared to the solution using simplex method. The number of iteration did not increase in any of the cases tried.

References: [1] Gass S. I. (1964) : Linear programming, McGraw-Hill Book Co. Inc., New York.

182 Int. J Latest Trend Math

Vol-4 No 1 March, 2014

[2] Taha Hamdy. A. : Operation Research An Introduction.Published by Dorling Kindersley(India) Pvt. Ltd.,licensees of Pearson Education in South Asia. [3] Prem kumar Gupta and Hira D. S. : Operation Research.,S.Chand and Company Ltd. [4] Sharma Anand : Operation Research., Himalaya Publishing House. [5] Mrs. N. V. Vaidya and N. W. Khobragade (2012):Optimum Solution to the Simplex Method- An Alternative Approach Int. J Latest Trend Math Vol-2 No. 3 September 2012 [6] Mrs. N. V. Vaidya and N. W. Khobragade :Solution of Game Problems Using New Approach, International Journal of Engineering and Innovative Technology (IJEIT), Volume 3, Issue 5, November 2013 [7] Mrs. N. V. Vaidya , N. W. Khobragade and Navneet K. Lamba:Approximation algorithm for optimal solution to the linear programming problem, Int. J. Mathematics in Operational Research, Vol. 6, No. 2, 2014