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226   CHAPTER 5 LINEAR PROGRAMMING: THE SIMPLEX METHOD


                                     Figure 5.2 Feasible Region and Extreme Points for the HighTech Industries Problem


                                                       x 2
                                                     60



                                                     50   Warehouse Capacity


                                                     40
                                                    Portable  30



                                                        5         4            Display Units
                                                     20

                                                                             3
                                                     10        Feasible
                                                                Region        Assembly Time

                                                        1                    2
                                                             10     20     30    40     50   x 1
                                                                       Deskpro



                                     row – that for s 3 equals  6.25. One of the managerial benefits of the Simplex
                                     method is that virtually every part of the tableau can be used to help management
                                     understand the consequences of decisions. Here we see that s 3 takes a negative value
                                     of E6.25. The interpretation of this is straightforward; s 3 has just been taken out of
                                     the solution and is non-basic with a zero value. But this coefficient tells us that if we
                                     were to force s 3 back into the situation at this stage, it would have a negative impact
                                     on the objective function. In other words, if we insisted on having unused warehouse
                                     space (which is what s 3 measures), then each square metre of unused space will
                                     effectively cost HighTech E6.25 in lost profit contribution since the only way of
                                     freeing up warehouse space at this stage is to reduce production of x 1 .
                                       To determine which variable will be removed from the basis when x 2 enters, we
                                     must calculate for each row i the ratio b i /a i2 (remember, though, that we should
                                     calculate this ratio only if a i2 is greater than zero); then we select the variable to
                                     leave the basis that corresponds to the minimum ratio. As before, we will show these
                                     ratios in an extra column of the simplex tableau:


                                                     x 1    x 2     s 1  s 2     s 3
                                                                                                     b
                                      Basis    c B   50   40        0     0        0                a i2
                                                                                                  37:5
                                                0     0    3.125    1     0     0.375      37.5        ¼ 12
                                                                                                  3:125
                                      s 1
                                                                                                    20
                                      s 2       0     0    1        0     1        0       20          ¼ 20
                                                                                                     1
                                                                                                  37:5
                                      x 1      50     1    0.625    0     0     0.125      37.5        ¼ 60
                                                                                                  0:625
                                                     50   31.25     0     0      6.25   1 875
                                           z j
                                                      0    8.75     0     0      6.25
                                         c j – z j

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