208 Operations Research (MTH601) 209 Table x x 3 0x x 0x 1 5 0 0x 6 5 0x All the above markings can be done in a single matrix itself. We need not write the matrix Remark: repeatedly. This is done only to clarify the presentation. Hungarian algorithm in the two examples above, the first one gave a solution leaving no zeros. It was a case of no ambiguity and in the second, make we had more zeros and the tie was broken arbitrarily. Sometimes if we proceed in the steps explained above, we get a maximal assignment, which does not contain an assignment in every row or column. We are faced with a question of how to solve the problem. In such a case, the effectiveness matrix has to be modified, so that after a finite number of iterations an optimal assignment will be in sight. The following is the algorithm to solve a problem of this kind and this is known as Hungarian algorithm.
The result is as shown in table. Table x 5 4 7 0x x x 0x Now we writing have two zeros in rows 3 and 4 in columns 3 and they occupy the corners of a square. An arbitrary assignment has to be made. If we make an assignment in (3, 3) and delete the remaining zero in row 3 and in column 3, this leaves one zero in the position (4, 5) and an assignment is made there. Thus we have a solution to the problem as in table. Table x x 3 4 0x 0 0 1 0x x 6 5 0 One more assignment (as a solution) is possible in this problem. (i.e.) we could have made an assignment at (3, 5) deleting other zero in the row 3 and zero in column 5 and making the last assignment at (4, 3). This is shown below in table.
This is shown in table. Table x x x 0x Row 2 has a single zero in the first column. So make an assignment and delete the remaining zeros in column 1 as shown in table. Table x x Operations Research (MTH601) 208 4 0x Row 3, 4 and 5 have more than a single zero. So we skip these rows and examine the columns. Columns 3 has three zeros and so omit. Column 4 has a single zero in row. So we make an assignment, deleting the remaining zeros in row.
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If there is a tie among zeros defer the decision. Next consider columns, for single unmarked zero, mark them ( ) and. Mark (X) any other unmarked zero in their rows. Repeat (c) and (d) successively until one of the two occurs. 205 Operations handwriting Research (MTH601) 206 1) There are no zeros left unmarked. 2) The remaining unmarked zeros lie atleast two in each row and column. E., they occupy corners of a square.
If the outcome is (1 we have a maximal assignment. In the outcome (2) we use arbitrary assignments. This process may yield multiple solutions. 206 Operations Research (MTH601) 207 multiple solutions example 1 given the following matrix, find the optimal assignment. Solution: Note that all the rows and columns have at least one zero. Row 1 has a single zero in column. So make an assignment, delete (mark X) the second zero in 0 denotes assignment column.
Hence cross (3, D). All the assignments made in this way are as shown in table. Table 5 Worker Job d 204 Operations Research (MTH601) step 4 : Now having assigned certain jobs to certain workers we proceed to the column. Since there is an assignment in this column, we proceed to the second column. There is only one zero in the cell (3, b we assign the jobs 3 to worker. Thus all the four jobs have been assigned to four workers.
Thus we obtain the solution to the problem as shown in the table. Table 6 Worker Job d the assignments are job to worker 1 A 2 C 3 B 4 d we summarise the above procedure as a set of following rules:. Subtract the minimum element in each row from all the elements in its row to make sure that at least we get one zero in that row. Subtract the minimum element in each column from all the elements in its column in the above reduced matrix, to make sure that we get at least one zero in each column. Having obtained at least one zero in each row and atleast one zero in each column, examine rows successively until a row with exactly one unmarked zero is found and mark ( ) this zero, indicating that assignment in made there. Mark (X) all other zeros in the same column, to show that they cannot be used to make other assignments. Proceed in this way until all rows have been examined.
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Hence we assign the job 2 to worker c, indicating by a square. Any other zero in this column is crossed (X). Proceed to the third row. Here we have two zeros corresponding to (3, B) and (3, D). Since there is a tie for the job 3, go to the next row deferring the decision for the present. Proceeding to the fourth row, we have only one zero in (4, D). Hence we assign job 4 to worker. Now the column D has a zero in the third row.writing
and each column, we assign starting from first row. In the first row, we have a zero in (1, A). Hence we assign job 1 to the worker. This assignment is indicated by a square. All other zeros in the column are crossed (X) to show that the other jobs cannot be assigned to worker a as he has already been assigned. In the above problem we do not have other zeros in the first column. Proceed to the second row. We have a zero in (2, C).
Depending on the efficiency and the capacity of the individual the times taken by each differ as shown in the table. How should the tasks be assigned one jot to a worker so as to minimize the total man-hours? Table 2 Worker Job d world operations Research (MTH601) The following steps are followed to find an optimal solution. Solution: step 1 :Consider each row. Select the minimum element in each row. Subtract this smallest element form all the elements in that row. This results in the table. Table 3 Worker Job d step 2 : we subtract the minimum element in each column from all the elements in its column.
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Operations, research (MTH601) 203, subject to restrictions, row restrictions x 11 x12 x13 x14 1 for job 1 x 21 x22 x23 x24 1 for job 2 x 31 x32 x33 x34 1 for job 3 x 41 x42 x43 x44 1 for job. xin 1, for i 1, 2,., n x 1j x2j. xnj 1, for j 1, 2,., n and When compared with a transportation problem, we see that a i 1 and bj 1 for all rows and columns, xij 0. We shall not attempt simplex algorithem or the transportation algorithm to get a solution to an assignment problem. Certain systematic procedure has been devised so as to obtain the solution to the problem with ease. Solution of an assignment problem the solution to an assignment problem is based on the following theorem. Theorem : If in an assignment problem we add a constant to every element of a row or column in the effectiveness matrix then an assignment that minimizes the total effectiveness in one matrix also minimizes the total effectiveness in the other improve matrix. Example : A works manager has to allocate four different jobs to four workmen.