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Imbalanced Assignment Problem Linear Programming Examples

Unbalanced Assignment Problem

In the previous section, the number of persons and the number of jobs were assumed to be the same. In this section, we remove this assumption and consider a situation where the number of persons is not equal to the number of jobs. In all such cases, fictitious rows and/or columns are added in the matrix to make it a square matrix.

Then, we apply the usual Hungarian algorithm to this resulting balanced assignment problem. We provide the following example to illustrate the solution of an unbalanced assignment problem.

Example: Unbalanced Assignment Problem

Job
Person 1 2 3 4
A 20 25 22 28
B 15 18 23 17
C 19 17 21 24

Solution

Since the number of persons is less than the number of jobs, we introduce a dummy person (D) with zero values. The revised assignment problem is given below:

Table

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Job
Person 1 2 3 4
A 20 25 22 28
B 15 18 23 17
C 19 17 21 24
D (dummy) 0 0 0 0

Now use the Hungarian method to obtain the optimal solution yourself.
Ans. = 20 + 17 + 17 + 0 = 54.

Whenever the cost matrix of an assignment problem is not a square matrix, that is, whenever the number of sources is not equal to the number of destinations, the assignment problem is called an unbalanced assignment problem. In such problems, dummy rows (or columns) are added in the matrix so as to complete it to form a square matrix. The dummy rows or columns will contain all costs elements as zeroes. The Hungarian method may be used to solve the problem.

Example : A company has five machines that are used for four jobs. Each job can be assigned to one and only one machine. The cost of each job on each machine is given in the following Table.

Assignment Problem

Solution: Convert the 4 × 5 matrix into a square matrix by adding a dummy row D5.

Dummy Row D5 Added

Row-wise Reduction of the Matrix

Column-wise reduction is not necessary since all columns contain a single zero. Now, draw minimum number of lines to cover all the zeros, as shown in Table.

All Zeros in the Matrix Covered

Number of lines drawn ≠ Order of matrix. Hence not optimal.

Select the least uncovered element, i.e., 1, subtract it from other uncovered elements, add to the elements at intersection of lines and leave the elements that are covered with single line unchanged as shown in Table.

Subtracted or Added to Elements

Number of lines drawn ≠ Order of matrix. Hence not optimal.

Again Added or Subtracted 1 from Elements

Number of lines drawn = Order of matrix. Hence optimality is reached. Now assign the jobs to machines, as shown in Table.

Assigning Jobs to Machines

Example : In a plant layout, four different machines M1, M2, M3 and M4 are to be erected in a machine shop. There are five vacant areas A, B, C, D and E. Because of limited space, Machine M2 cannot be erected at area C and Machine M4 cannot be erected at area A. The cost of erection of machines is given in the Table.

Assignment Problem

Find the optimal assignment plan.

Solution: As the given matrix is not balanced, add a dummy row D5 with zero cost values. Assign a high cost H for (M2, C) and (M4, A). While selecting the lowest cost element neglect the high cost assigned H, as shown in Table below.

Dummy Row D5 Added

- Row-wise reduction of the matrix is shown in Table.

Matrix Reduced Row-wise

Note: Column-wise reduction is not necessary, as each column has at least one single zero. Now, draw minimum number of lines to cover all the zeros, see Table.

Lines Drawn to Cover all Zeros

Number of lines drawn ≠ Order of matrix. Hence not Optimal. Select the smallest uncovered element, in this case 1. Subtract 1 from all other uncovered element and add 1 with the elements at the intersection. The element covered by single line remains unchanged. These changes are shown in Table. Now try to draw minimum number of lines to cover all the zeros.

Added or Subtracted 1 from Elements

Now number of lines drawn = Order of matrix, hence optimality is reached. Optimal assignment of machines to areas are shown in Table.

Optimal Assignment

Hence, the optimal solution is: