Algorithm for Hungarian Method of Assignment
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The Hungarian assignment algorithm is used to solve assignment problems to maximize total profit or minimize total cost. It involves 9 steps: 1) add dummy rows or columns if needed, 2) subtract smallest row elements from each row, 3) subtract smallest column elements from each column, 4) examine rows/columns for single zeros and cross out other zeros, 5) check if number of assigned cells equals rows/columns, 6) draw lines through zeros, 7) check if minimum lines equals rows/columns, 8) revise matrix by subtracting smallest covered element and adding to line intersections, 9) repeat steps 4-8 until optimum solution is found.
Algorithm for Hungarian Method of Assignment
- 1. By A. Raja Subject: Operations Research Topic : Hungarian Assignment Algorithm
- 2. INTRODUCTION The assignment problem is a particular case of transportation problem in which the number of jobs are equal to the number of facilities. The objective is to maximise total profit of allocation or to minimise the total cost.
- 3. Hungarian Assignment Algorithm Various steps of the computational procedure for obtaining an optimal solution may be summarised as follows. Step1: If the number of rows are not equal to the number of columns and vice versa , then a dummy row or dummy column must be added with zero cost elements.
- 4. Step2: Find the smallest cost in each row of the cost matrix . subtract this smallest cost element from each element in that row . therefore , there will be at least one zero in each row of this new matrix which is called the first reduced cost matrix.
- 5. Step3: In the reduced cost matrix , find the smallest element in each column . subtract the smallest cost element from each element in that column . as a result , there would be at least one zero in each row and column of the second reduced cost matrix.
- 6. Step4: Determine an optimum solution a follows: 1.Examine the rows successively until a row with exactly one zero is found . box around the zero element as an assigned cell and cross out all other zero in its column. Proceed in this manner until all the rows have been examined. If there are more than one zero in any row, then do not consider that row and pass on the next row. 2.Repeat the procedure for the columns of the reduced matrix. If there is no single zero in any row or column of the reduced matrix, then arbitrarily choose a row or column having the minimum number of zeros. Arbitrarily, choose zero in the row or column and cross the remaining zeros in that row or column. Repeat (1) and (2) steps until all zeros are either assigned or crossed out.
- 7. Step5: An optimal assignment is found , if the number assigned cells equals the number of rows and columns. Step6: Draw the number of horizontal or vertical lines through all the zeros.
- 8. Step7: If the minimum number of lines passing through all the zeros is equal to the number of rows or columns the optimum solution is attained by an arbitrary allocation in the position of the zeros not crossed in step3 otherwise go to the next step.
- 9. Step8: Revise the cost matrix as follows: 1. Find the elements that covered by a line. Choose the smallest of this elements and subtract this elements from all the uncrossed elements and add the same at the point of the intersection of the two lines. 2. Other elements crossed by the lines remain unchanged. Step9: Go to step4 and repeat the procedure till an optimum solution is attained.
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