Assignment Problem: Maximization

There are problems where certain facilities have to be assigned to a number of jobs, so as to maximize the overall performance of the assignment.

The Hungarian Method can also solve such assignment problems, as it is easy to obtain an equivalent minimization problem by converting every number in the matrix to an opportunity loss.

The conversion is accomplished by subtracting all the elements of the given matrix from the highest element. It turns out that minimizing opportunity loss produces the same assignment solution as the original maximization problem.

example Example: Maximization In An Assignment Problem

At the head office of www.universalteacherpublications.com there are five registration counters. Five persons are available for service.

Person
Counter A B C D E
1 30 37 40 28 40
2 40 24 27 21 36
3 40 32 33 30 35
4 25 38 40 36 36
5 29 62 41 34 39

How should the counters be assigned to persons so as to maximize the profit?

Solution.

Here, the highest value is 62. So we subtract each value from 62. The conversion is shown in the following table.

Table

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Person
Counter A B C D E
1 32 25 22 34 22
2 22 38 35 41 26
3 22 30 29 32 27
4 37 24 22 26 26
5 33 0 21 28 23

Now the above problem can be easily solved by Hungarian method. After applying steps 1 to 3 of the Hungarian method, we get the following matrix.

Table

Person
Counter A B C D E
1 10 3 8
2 16 13 15 4
3 8 7 6 5
4 15 2 4
5 33 21 24 23

Draw the minimum number of vertical and horizontal lines necessary to cover all the zeros in the reduced matrix.

Table

Assignment Problem, Maximization Example

Select the smallest element from all the uncovered elements, i.e., 4. Subtract this element from all the uncovered elements and add it to the elements, which lie at the intersection of two lines. Thus, we obtain another reduced matrix for fresh assignment. Repeating step 3, we obtain a solution which is shown in the following table.

Final Table: Maximization Problem

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Person
Counter A B C D E
1 14 3 8
2 12 9 11
3 4 3 2 1
4 19 2 4
5 37 21 24 23

The total cost of assignment = 1C + 2E + 3A + 4D + 5B

Substituting values from original table:
40 + 36 + 40 + 36 + 62 = 214.

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