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Operations Research




                    Notes          The given problem can be formulated as an LPP from A’s and B’s point of view as follows:
                                   Let x , x  and x  be the probabilities with which A chooses respectively the strategies a , a  and a
                                       1  2   3                                                        1  2   3
                                   and y , y  and y  be the probabilities in respect of B choosing b , b  and b  respectively. From A’s
                                       1  2    3                                    1  2    3
                                   point of view we
                                                          
                                                Minimise         = x  + x  + x
                                                              1   2  3
                                   Subject to    8x  + 2x  + 4x    1
                                                   1    2   3
                                                  9x  + 5x  + x    1
                                                    1   2   3
                                                 3x  + 6x  + 7x    1
                                                   1    2   3
                                   and              x  + x  + x   = 1
                                                     1  2   3
                                                      x , x , x    0
                                                       1  2  3
                                                               Where, X  =
                                                                       i
                                   From B’s point of view, we have

                                   Maximise                   = y  + y  + y
                                                                1   2   3
                                   Subject to    8y  + 9y  + 3y    1
                                                   1   2    3
                                                 2y  + 5y  + 6y    1
                                                   1   2    3
                                                  4y  + y  + 7y    1
                                                    1  2    3
                                                      y , y , y    0;  and Y  =
                                                       1  2  3        1
                                   To calculate the required values we can solve either of these LPPs and read solution to the other
                                   from it as each one is the dual of the other. We shall solve the game from B’s point of view.
                                   Introducing slack variables the objective function can be written as,
                                   Maximise  1 = y  + y  + y  + OS  + OS  + OS
                                               1   2  3    1    2    3
                                   Subject to  8y  + 9y  = 3y  + S   = 1
                                               1    2   3   1
                                             2y  + 5y  + 6y  + S   = 1
                                               1    2   3   2
                                              4y  + y  + 7y  + S   = 1
                                                1   2   3   3
                                                               y , y , y   0; s , s , s   0
                                                                1  2  3    1  2  3
                                   Simplex Table 1
                                       BV       CB     XB      Y1     Y2    Y3      Min. Ratio   S1   S2    S3
                                       S1       0       1      8      9      3     1/8 = 0.125   1    0     0
                                       S2       0       1      2      5      6      ½ = 0.50    0     1     0
                                       S3       0       1      4      1      7      ¼ = 0.25    0     0     1
                                      ZB = 0            Zj     0      0      0                  0     0     0
                                                        Cj     1      1      1                  0     0     0
                                                      Zj – Cj   –1    –1    –1                  0     0     0












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