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242   CHAPTER 5 LINEAR PROGRAMMING: THE SIMPLEX METHOD


                                     Alternative Optimal Solutions
                                     A linear programme with two or more optimal solutions is said to have alternative
                                     optimal solutions. When using the Simplex method, we cannot recognize that a
                                     linear programme has alternative optimal solutions until the final tableau is reached.
                                     Then if the linear programme has alternative optimal solutions, c j – z j will equal zero
                                     for one or more non-basic variables.
                                       To illustrate the case of alternative optimal solutions when using the Simplex
                                     method, consider changing the objective function for the HighTech problem from
                                     50x 1 +40x 2 to 30x 1 +50x 2 ; in doing so, we obtain the revised linear programme:
                                                               Max  30x 1 þ 50x 2
                                                               s:t:
                                                                    3x 1 þ 5x 2   150
                                                                          1x 2   20
                                                                    8x 1 þ 5x 2   300
                                                                      x 1 ; x 2   0


                                       The final tableau for this problem is shown here:


                                                      x 1    x 2    s 1       s 2       s 3
                                      Basis    c B    30     50      0        0         0

                                      x 2      50      0      1      0        1         0           20
                                      s 3       0      0      0     2.6667    8.3333    1           66.6667
                                      x 1      30      1      0      0.3333   1.6667    0           16.6667
                                                      30     50     10        0         0         1 500
                                           z j
                                                       0      0    10         0         0
                                          c j – z j


                                       All values in the net evaluation row are less than or equal to zero, indicating that
                                     an optimal solution has been found. This solution is given by x 1 ¼ 16.6667, x 2 ¼ 20,
                                     s 1 ¼ 0, s 2 ¼ 0 and s 3 ¼ 66.6667. The value of the objective function is 1500.
                                       In looking at the net evaluation row in the optimal simplex tableau, we see that
                                     the c j – z j value for non-basic variable s 2 is equal to zero. This indicates that the
                                     linear programme may have alternative optimal solutions. In other words, because
                                     the net evaluation row entry for s 2 is zero, we can introduce s 2 into the basis without
                                     changing the value of the solution. The tableau obtained after introducing s 2 follows:



                                                         x 1     x 2      s 1       s 2     s 3
                                      Basis      c B     30      50        0        0        0
                                                 50       0      1         0.32     0        0.12       12
                                      x 2
                                                  0       0      0        0.32      1        0.12        8
                                      s 3
                                                 30       1      0        0.20      0        0.20       30
                                      x 1
                                            z j          30      50       10        0        0        1 500
                                          c j – z j       0      0       10         0        0






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