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TRANSPORTATION SIMPLEX METHOD: A SPECIAL-PURPOSE SOLUTION PROCEDURE  297



                                        Table 7.14 MODI Evaluation of Each Cell in Solution
                                                                              v j

                                                        u i      3        2       –1       0
                                                                   3        2        7        6
                                                        0       3 500    1 500     8        6


                                                                   7        5        2        3

                                                        3       1        2 500    2 000    1 500

                                                                   2        5        4        5

                                                       –1       2 500     4        6        6






                                         Table 7.16 shows the initial feasible solution obtained using the minimum cost
                                      method for a transportation problem involving m ¼ 3 origins and n ¼ 3 destina-
                                      tions. To use the MODI method for this problem, we must have m + n   1 ¼ 3+
                                      3   1 ¼ 5 occupied cells. Since the initial feasible solution has only four occupied
                                      cells, the solution is degenerate.
                                         Suppose that we try to use the MODI method to calculate row and column
                                      indexes to begin phase II for this problem. Setting u 1 ¼ 0 and calculating the
                                      column indexes for each occupied cell in row 1, we obtain v 1 ¼ 3 and v 2 ¼ 6 (see
                                      Table 7.16). Continuing, we then calculate the row indexes for all occupied cells in
                                      columns 1 and 2. Doing so gives u 2 ¼ 5   6 ¼ 1. At this point, we cannot calculate
                                      any more row and column indexes because no cells in columns 1 and 2 of row 3 and
                                      no cells in rows 1 or 2 of column 3 are occupied.
                                         To calculate all the row and column indexes when fewer than m + n   1cells
                                      areoccupied,wemustcreateoneor more‘artificially’ occupied cells with a flow of
                                      zero. In Table 7.16 we must create one artificially occupied cell to have five
                                      occupied cells. Any currently unoccupied cell can be made an artificially occupied
                                      cell if doing so makes it possible to calculate the remaining row and column
                                      indexes. For instance, treating the cell in row 2 and column 3 of Table 7.16 as an



                                        Table 7.15 Optimal Solution to the Foster Electronics Transportation Problem

                                                 Route
                                        From           To         Units Shipped  Cost per Unit, E  Total Cost, E

                                        Czech Republic  Boston        3 500            3            10 500
                                        Czech Republic  Dubai         1 500            2             3 000
                                        Brazil         Dubai          2 500            5            12 500
                                        Brazil         Singapore      2 000            2             4 000
                                        Brazil         London         1 500            3             4 500
                                        China          Boston         2 500            2             5 000
                                                                                                    39 500






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