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292   CHAPTER 7 TRANSPORTATION, ASSIGNMENT AND TRANSSHIPMENT PROBLEMS


                                     identify an incoming arc. The incoming arc is the currently unused route (unoccu-
                                     pied cell) where making a flow allocation will cause the largest per-unit reduction in
                                     total cost. Flow is then assigned to the incoming arc, and the amounts being shipped
                                     over all other arcs to which flow had previously been assigned (occupied cells) are
                                     adjusted as necessary to maintain a feasible solution. In the process of adjusting the
                                     flow assigned to the occupied cells, we identify and drop an outgoing arc from the
                                     solution. So, at each iteration in phase II, we bring a currently unused arc (unoccu-
                                     pied cell) into the solution, and remove an arc to which flow had previously been
                                     assigned (occupied cell) from the solution.
                                       To show how phase II of the transportation Simplex method works, we must
                                     explain how to identify the incoming arc (cell), how to make the adjustments to the
                                     other occupied cells when flow is allocated to the incoming arc and how to identify
                                     the outgoing arc (cell). We first consider identifying the incoming arc.
                                       As mentioned, the incoming arc is the one that will cause the largest reduction
                                     per unit in the total cost of the current solution. To identify this arc, we must
                                     compute for each unused arc the amount by which total cost will be reduced by
                                     shipping one unit over that arc. The modified distribution or MODI method is a way
                                     to make this computation.
                                       The MODI method requires that we define an index u i for each row of the
                                     tableau and an index v j for each column of the tableau. Calculating these row and
                                     column indexes requires that the cost coefficient for each occupied cell equal u i + v j .
                                     So, since c ij is the cost per unit from origin i to destination j, u i + v j ¼ c ij for each
                                     occupied cell. Let us return to the initial feasible solution which we found using the
                                     minimum cost method (see Table 7.9), and use the MODI method to identify the
                                     incoming arc.
                                       Requiring that u i + v j ¼ c ij for all the occupied cells in the initial feasible solution
                                     leads to a system of six equations and seven indexes, or variables:


                                                  Occupied Cell                      u i + v j ¼ c ij

                                                  Czech Republic–Boston              u 1 + v 1 ¼ 3
                                                  Czech Republic–Dubai               u 1 + v 2 ¼ 2
                                                  Brazil–Boston                      u 2 + v 1 ¼ 7
                                                  Brazil–Singapore                   u 2 + v 3 ¼ 2
                                                  Brazil–London                      u 2 + v 4 ¼ 3
                                                  China–Boston                       u 3 + v 1 ¼ 2



                                       With one more index (variable) than equation in this system, we can freely pick
                                     a value for one of the indexes and then solve for the others. We will always
                                     choose u 1 ¼ 0 and then solve for the values of the other indexes. Setting u 1 ¼ 0,
                                     we obtain:
                                                                     0 þ v 1 ¼ 3
                                                                     0 þ v 2 ¼ 2
                                                                    u 2 þ v 1 ¼ 7
                                                                    u 2 þ v 3 ¼ 2
                                                                    u 2 þ v 4 ¼ 3
                                                                    u 3 þ v 1 ¼ 2








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