Having performed the step 1 and step 2, we will be getting at least one zero in each column in the reduced cost table. Now, the assignments are made for the reduced table in following manner.
(i) Rows are examined successively, until the row with exactly single (one) zero is found.
Now, this smallest element is subtracted form each element of that row.
So, we will be getting at least one zero in each row of this new table. Having constructed the table (as by step-1) take the columns of the table.
(iii) Step 3, (i) and 3 (ii) are repeated till all the zeros are either marked or crossed out.
Now, if the number of marked zeros or the assignments made are equal to number of rows or columns, optimum solution has been achieved. At this stage, draw the minimum number of lines (horizontal and vertical) necessary to cover all zeros in the matrix obtained in step 3, Following procedure is adopted: (i) Tick mark () all rows that do not have any assignment.
He will have to take decision regarding which job should be given to which worker. Each facility or say worker can perform each job, one at a time.
But there should be certain procedure by which assignment should be made so that the profit is maximized or the cost or time is minimized. It maybe noted here that this is a special case of transportation problem when the number of rows is equal to number of columns.
Consider a facility location problem with four facilities (and four locations).
One possible assignment is shown in the figure below: facility 2 is assigned to location 1, facility 1 is assigned to location 2, facility 4 is assigned to location 3, and facility 4 is assigned to location 3.