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ASSIGNMENT PROBLEM: A SPECIAL-PURPOSE SOLUTION PROCEDURE  311


                                      optimal assignment – in this case, 15 for Terry to client 2, 5 for Karl to client 3
                                      and 6 for Mustafa to client 1. So, we obtain the solution time of
                                      15 + 5 + 6 ¼ 26 days.


                                      Finding the Minimum Number of Lines
                                      Sometimes it is not obvious how the lines should be drawn through rows and
                                      columns of the matrix in order to cover all the zeros with the smallest number of
                                      lines. In these cases, the following heuristic works well. Choose any row or column
                                      with a single zero. If it is a row, draw a line through the column the zero is in; if it is a
                                      column, draw a line through the row the zero is in. Continue in this fashion until you
                                      cover all the zeros.
                      Can you solve an   If you make the mistake of drawing too many lines to cover the zeros in the
                      assignment problem  reduced matrix and thus conclude incorrectly that you have reached an optimal
                      using the Hungarian
                      method? Try Problem 19.  solution, you will be unable to identify a zero-value assignment. So, if you think you
                                      have reached the optimal solution, but cannot find a set of zero-value assignments,
                                      go back to the preceding step and check to see whether you can cover all the zeros
                                      with fewer lines.


                                      Problem Variations
                                      We now discuss how to handle the following problem variations with the Hungarian
                                      method:
                                         1 number of people not equal to number of tasks;
                                         2 maximization objective function;
                                         3 unacceptable assignments.

                                      Number of People Not Equal to Number of Tasks The Hungarian method
                                      requires that the number of rows (people) equal the number of columns (tasks).
                                      Suppose that in the Fowle problem four project leaders had been available for
                                      assignment to the three new clients (tasks). Fowle still faces the same basic
                                      problem, namely, which project leaders should be assigned to which clients to
                                      minimize the total days required. Table 7.25 shows the project completion time
                                      estimates with a fourth project leader.
                                         We know how to apply the Hungarian method when the number of rows and the
                                      number of columns are equal. We can apply the same procedure if we can add a new
                                      client. If we do not have another client, we simply add a dummy column, or a dummy
                                      client. This dummy client is nonexistent, so the project leader assigned to the
                                      dummy client in the optimal assignment solution, in effect, will be the unassigned
                                      project leader.


                                        Table 7.25 Estimated Project Completion Time (Days) for the Fowle Assignment
                                        Problem with Four Project Leaders
                                                                                     Client

                                        Project Leader                  1              2              3
                                        Terry                          10              15             9
                                        Karl                            9              18             5
                                        Mustafa                         6              14             3
                                        Helen                           8              16             6





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