# Imbalanced Assignment Problem Linear Programming Graphical Method

Received 9 December 2015; accepted 24 January 2016; published 29 January 2016

ABSTRACT

In this paper, we discuss a new approach for solving an unbalanced assignment problem. A Lexi- search algorithm is used to assign all the jobs to machines optimally. The results of new approach are compared with existing approaches, and this approach outperforms other methods. Finally, numerical example (Table 1) has been given to show the efficiency of the proposed methodology.

**Keywords:**

Assignment Problem, Lexi-Search Algorithm, Jobs Clubbing Method

1. Introduction

Consider a problem which consists of a set of “n” machines. A set of “m” jobs which are to be considered assign for execution on “n” available machines. The execution cost of each job on all the machines are known and mentioned in the matrix, namely assigned cost matrix (ACM) of order, where. The unbalanced assignment problem is a special type of linear programming problem in which our objective is to assign number of salesmen to number of areas at a minimum cost (time). The mathematical formulation of the problem suggests that this is a 0 - 1 programming problem. It is highly degenerate all the algorithms developed to find optimal solution of transportation problem, applicable to unbalanced assignment problem. However, due to its highly degeneracy nature a specially designed algorithm, widely known as Hungarian method proposed by Kuhn [1] , is used for its solution, and Kadhirvel and Balamurugan [2] solved the unbalanced assignment problems using triangular fuzzy Numbers. Different methods have been presented for Assignment Problem and various articles have been published on the subject [3] -[7] .

The objectives are to determine the optimal assignment cost, in such a way that all the jobs are to be allotted on the available machines in an optimum way. The mathematical formulation of the assignment problem [8] [9] is as follows.

Table 1. Assigned cost matrix (ACM).

2. Model Construction of Simple Assignment Problem

Minimize (Maximize):

Subject to

; for

; for

where

Problem definition:

Also, if the numbers of jobs are not equal to number of machines, then it is known as an unbalanced assignment problem. Now consider the assumptions of choosing an unbalanced assignment problem as:

• The completion of a program from computational point of view means that the all jobs are assigned to various machines and final optimal assignment cost has been obtained.

• The number of jobs are more than number of machines.

The variants of assignment problem are considered by various researchers like Kagade & Bajaj [10] and Avanish Kumar [11] . From the work of these authors, they found that the approach of clubbing the costs of the jobs was implemented for multi objective problems and single objective problems, where as this paper considers the clubbing of jobs for an assignment problem by the exact solution problem with Lexi-search approach [12] [13] .

3. Methodology

To determine the assignment cost as well as combination of job (s) Vs machine (s) of an unbalanced assignment problem for a set of “n” machines. A set of “m” jobs which are to be considered as assigned for execution on “n” available machines with an execution cost, where and are mentioned in the ACM of order, where m > n. First of all, we obtain the sum of each row and each column of the ACM store and the results should be arranged in the array, namely, and. Then we select the first m rows (jobs) on the basis of that is, starting with the most minimum to next minimum to the array and deleting rows (jobs) correspo- nding to the remaining (m-n) jobs. Store results in the new array that will be the array for the first sub problem (Table 2). Repeating this process until the remaining jobs become less than “n” machines, when remaining jobs are less than n then deleting (n-m) columns (machines) on the basis of. That is, corresponding to value (s) most maximum to next maximum to form the last sub problem (Table 3). Store the results in the new array that shall be the array for the last sub problem. which are now balanced assignment problems, in this way for the

defined assignment problem.Now we apply the Lexi-search approach to obtain the exact optimum solution of each sub problem (Tables 4-7). Finally, add the total assignment cost of each sub problem to obtain the optimal assignment cost along with assignment sets. And also we check the assignment cost for jobs clubbing problem (Table 8) through Lexi-search approach (Tables 9-11), getting the same value. To solve this problem we follow the following algorithm.

Algorithm

Step-1: Consider “m” jobs on “n” machines costs given as a matrix (ACM), which is an unbalanced assign- ment problem where.

Step-2:

Step-2.1: Obtain the sum of each row and column of the ACM and the store the results in the arrays namely and.

Step-2.2: Select the first m rows (jobs) on the basis of. That is, starting with the most minimum to next minimum to the array and deleting rows (jobs) corresponding to the remaining (m-n) jobs. Store the results in the new array that shall be the array for the first sub problem.

Step-2.2.1: If there is no remaining jobs, i.e., (m-n = 0), then go to step-3.

Step-2.2.2: If the remaining (m-n) jobs are still more than n, then repeat step-2.2 for the remaining jobs to form next sub-problem (s), else, step-2.3.

Step-2.3: If remaining jobs are less than n then deleting (n-m) columns (machines) on the basis of . That is corresponding to value (s) most maximum to next maximum to form the last sub problem. Store the results in the new array that shall be the array for the last sub problem.

Step-3: If the total effectiveness of ACM is to be maximized, change the sign of each cost element in the effectiveness matrix and go to step-4, otherwise go directly to step-5 if ACM has the total value as minimum.

Step-4: Arrange all the jobs according to their cost (i.e. available jobs). This arrangement consists of n columns and m rows. Each column represents a machine, and the elements in that column are the costs arranged in increasing order according to their jobs.

Step-5: Include the job from the first machine in the partial solution value (psv) “w”. If the cost itself is greater than or equal to trial value (TRV) then stop. Otherwise go to next step.

Step-6: Calculate the bound.

Step-7: If the sum of bound and psv is greater than or equal to TRV then drop the job added in step 5, and go to step 5. Otherwise go to next step, i.e. go to Sub block (GS).

Step-8: Include the next available job (from the last job included in the partial solution “w”) into the partial solution.

Table 8. Jobs clubbing modified problem is.

Step-9: If partial solution value is greater than or equal to the TRV then drop the job added in step-8, and go to step-7. Otherwise go to step-10.

Step-10: If the sum of bound and psv is greater than or equal to TRV then drop the newly added job in step-8, and go to step -7.otherwise go to step 11.

Step-11: If the partial solution contains n − 1 jobs add the dummy job to the partial solution if it is greater than or equal to TRV then drop the dummy job and last two jobs from the partial solution. That is Jump out to the next higher order blocks (JO). If “w” contains only one job, go to step-5, otherwise go to step-8. Otherwise go to the next step.

Step-12: Now calculate the bound.

Step-13: If the sum of bound and psv is greater than or equal to TRV then drop the dummy job and also last job from “w”, and go to step-8. Otherwise go to step-14.

Step-14: Include the latest possible job from the dummy job in“w”

Step-15: If psv is greater than or equal to TRV then drop the last dummy job and also the job from which the dummy job was assigned, and go to step-8. Otherwise go to next step.

Step-16: Now calculate the bound.

Step-17: If sum of bound and psv is greater than or equal to TRV then drop the recently added job in “w” and go to step-14. Otherwise go to next step.

Step-18: Include the latest available job from the last job in “w”

Step-19: Now calculate the bound.

Step-20: If the sum of bound and psv is greater than or equal to TRV then drop the latest job, and go to step- 18. Otherwise go to next step.

Step-21: If the number of elements in “w” is less than “n” go to step-18. Otherwise go to next step.

Step-22: Replace TRV by partial solution value and trial solution by w. Now go to step-18.

4. Illustration

A company is faced with the problem of assigning five different machines to eight different jobs (Table 1). The costs are estimated as follows (in hundreds of rupees):

Solve the problem assuming that the objective is to minimize the total cost. Now obtain the sum of each row and column of, i.e., the sum of each row and each column is as follows:

We partition the matrix to define the first sub-problem by selecting rows corresp- onding and second sub problem by selecting rows corresponding to the jobs and by deleting columns corresponding to, then the modified matrices are as follows:

Sub-Problem-I:

Sub-Problem-II:

4.1. Now apply the Lexi-search method for Sub-Problem-I:

RP: Repitition, GNSB or JB: Go to next super block (JB), A.B or TRV = Absolute bound or Trail bound.

The final optimal assignments of as follows:

, , , ,

4.2. Now Apply the Lexi-Search Method for Sub-Problem-II:

The final optimal assignments is:, ,

The final optimal assignments assigned cost matrix (ACM) is :, , , , , , ,. The Hungarian method gives us total assignment cost as 890 along with the other one job assigned to dummy machine, in other words the job that is assigned to dummy machine under the Hungarian method was ignored for further processing. While, the original problem was divided into two sub problems, which are balanced assignment problem in nature. Now for the two sub problems with the use of Lexi-search approach, the total cost 870 is recorded for the sub problem-I along with none of the jobs assigned to dummy machine, and the total cost 680 was recorded for the second sub problem-II along with none of the jobs assigned to dummy machine.Now the total cost of the assigned cost matrix (ACM) is 870 + 670 = 1550.

5. Job Clubbing Method

Jobs Clubbing Modified Problem Is: Lexi-Search Approach

The final optimal assignments as follows:, , , ,

Total assignment cost = 1550.

6. Conclusion

The above illustration was taken by the defined algorithm and implemented on several sizes of the problems to test the effectiveness of the algorithm. This approach was implemented on different sizes of unbalanced assignment problems. From the above, we notice that the standard Hungarian method uses the dummy assignment which may not be possible in some applications, whereas this new approach never assigns the dummy machine in getting the optimum value. The time complexity with the Lexi-search method is verified and found that they are the same in getting optimum. Here, the optimum value of the original unbalanced assignment problems varies from that of balanced assignment problems either in Hungarian method or Lexi-search approach. The only advantage is that the Lexi-search method gives an exact optimum value with the same time complexity. Therefore the present paper suggests a new approach of clubbing the jobs for solving the unbalanced assignment problem with Lexi-search methodology.

Cite this paper

VentepakaYadaiah,V. V.Haragopal, (2016) A New Approach of Solving Single Objective Unbalanced Assignment Problem. *American Journal of Operations Research*,**06**,81-89. doi: 10.4236/ajor.2016.61011

References

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http://dx.doi.org/10.9790/3021-02810115 - 8. Gillett Billy, E. (2000) Introduction to Operations Research—A Computer Oriented Algorithm Approach. Tata McGraw Hill, New Delhi.
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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: