describe unbalanced assignment problem

Assignment Problem: Meaning, Methods and Variations | Operations Research

After reading this article you will learn about:- 1. Meaning of Assignment Problem 2. Definition of Assignment Problem 3. Mathematical Formulation 4. Hungarian Method 5. Variations.

Meaning of Assignment Problem:

An assignment problem is a particular case of transportation problem where the objective is to assign a number of resources to an equal number of activities so as to minimise total cost or maximize total profit of allocation.

The problem of assignment arises because available resources such as men, machines etc. have varying degrees of efficiency for performing different activities, therefore, cost, profit or loss of performing the different activities is different.

Thus, the problem is “How should the assignments be made so as to optimize the given objective”. Some of the problem where the assignment technique may be useful are assignment of workers to machines, salesman to different sales areas.

Definition of Assignment Problem:

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Suppose there are n jobs to be performed and n persons are available for doing these jobs. Assume that each person can do each job at a term, though with varying degree of efficiency, let c ij be the cost if the i-th person is assigned to the j-th job. The problem is to find an assignment (which job should be assigned to which person one on-one basis) So that the total cost of performing all jobs is minimum, problem of this kind are known as assignment problem.

The assignment problem can be stated in the form of n x n cost matrix C real members as given in the following table:

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How to Solve the Assignment Problem: A Complete Guide

Table of Contents

Assignment problem is a special type of linear programming problem that deals with assigning a number of resources to an equal number of tasks in the most efficient way. The goal is to minimize the total cost of assignments while ensuring that each task is assigned to only one resource and each resource is assigned to only one task. In this blog, we will discuss the solution of the assignment problem using the Hungarian method, which is a popular algorithm for solving the problem.

Understanding the Assignment Problem

Before we dive into the solution, it is important to understand the problem itself. In the assignment problem, we have a matrix of costs, where each row represents a resource and each column represents a task. The objective is to assign each resource to a task in such a way that the total cost of assignments is minimized. However, there are certain constraints that need to be satisfied – each resource can be assigned to only one task and each task can be assigned to only one resource.

Solving the Assignment Problem

There are various methods for solving the assignment problem, including the Hungarian method, the brute force method, and the auction algorithm. Here, we will focus on the steps involved in solving the assignment problem using the Hungarian method, which is the most commonly used and efficient method.

Step 1: Set up the cost matrix

The first step in solving the assignment problem is to set up the cost matrix, which represents the cost of assigning a task to an agent. The matrix should be square and have the same number of rows and columns as the number of tasks and agents, respectively.

Step 2: Subtract the smallest element from each row and column

To simplify the calculations, we need to reduce the size of the cost matrix by subtracting the smallest element from each row and column. This step is called matrix reduction.

Step 3: Cover all zeros with the minimum number of lines

The next step is to cover all zeros in the matrix with the minimum number of horizontal and vertical lines. This step is called matrix covering.

Step 4: Test for optimality and adjust the matrix

To test for optimality, we need to calculate the minimum number of lines required to cover all zeros in the matrix. If the number of lines equals the number of rows or columns, the solution is optimal. If not, we need to adjust the matrix and repeat steps 3 and 4 until we get an optimal solution.

Step 5: Assign the tasks to the agents

The final step is to assign the tasks to the agents based on the optimal solution obtained in step 4. This will give us the most cost-effective or profit-maximizing assignment.

Solution of the Assignment Problem using the Hungarian Method

The Hungarian method is an algorithm that uses a step-by-step approach to find the optimal assignment. The algorithm consists of the following steps:

• Subtract the smallest entry in each row from all the entries of the row.
• Subtract the smallest entry in each column from all the entries of the column.
• Draw the minimum number of lines to cover all zeros in the matrix. If the number of lines drawn is equal to the number of rows, we have an optimal solution. If not, go to step 4.
• Determine the smallest entry not covered by any line. Subtract it from all uncovered entries and add it to all entries covered by two lines. Go to step 3.

The above steps are repeated until an optimal solution is obtained. The optimal solution will have all zeros covered by the minimum number of lines. The assignments can be made by selecting the rows and columns with a single zero in the final matrix.

Applications of the Assignment Problem

The assignment problem has various applications in different fields, including computer science, economics, logistics, and management. In this section, we will provide some examples of how the assignment problem is used in real-life situations.

Applications in Computer Science

The assignment problem can be used in computer science to allocate resources to different tasks, such as allocating memory to processes or assigning threads to processors.

Applications in Economics

The assignment problem can be used in economics to allocate resources to different agents, such as allocating workers to jobs or assigning projects to contractors.

Applications in Logistics

The assignment problem can be used in logistics to allocate resources to different activities, such as allocating vehicles to routes or assigning warehouses to customers.

Applications in Management

The assignment problem can be used in management to allocate resources to different projects, such as allocating employees to tasks or assigning budgets to departments.

Let’s consider the following scenario: a manager needs to assign three employees to three different tasks. Each employee has different skills, and each task requires specific skills. The manager wants to minimize the total time it takes to complete all the tasks. The skills and the time required for each task are given in the table below:

The assignment problem is to determine which employee should be assigned to which task to minimize the total time required. To solve this problem, we can use the Hungarian method, which we discussed in the previous blog.

Using the Hungarian method, we first subtract the smallest entry in each row from all the entries of the row:

Next, we subtract the smallest entry in each column from all the entries of the column:

We draw the minimum number of lines to cover all the zeros in the matrix, which in this case is three:

Since the number of lines is equal to the number of rows, we have an optimal solution. The assignments can be made by selecting the rows and columns with a single zero in the final matrix. In this case, the optimal assignments are:

• Emp 1 to Task 3
• Emp 2 to Task 2
• Emp 3 to Task 1

This assignment results in a total time of 9 units.

I hope this example helps you better understand the assignment problem and how to solve it using the Hungarian method.

Solving the assignment problem may seem daunting, but with the right approach, it can be a straightforward process. By following the steps outlined in this guide, you can confidently tackle any assignment problem that comes your way.

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Operations Research

1 Operations Research-An Overview

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3 Linear Programming-Simplex Method

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4 Transportation Problem

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5 Assignment Problem

• Solution of the Assignment Problem
• Unbalanced Assignment Problem
• Problem with some Infeasible Assignments
• Maximisation in an Assignment Problem
• Crew Assignment Problem

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Hungarian Algorithm for Assignment Problem | Set 1 (Introduction)

• Hungarian Algorithm for Assignment Problem | Set 2 (Implementation)
• Introduction to Exact Cover Problem and Algorithm X
• Greedy Approximate Algorithm for Set Cover Problem
• Job Assignment Problem using Branch And Bound
• Implementation of Exhaustive Search Algorithm for Set Packing
• Channel Assignment Problem
• Chocolate Distribution Problem | Set 2
• Transportation Problem | Set 1 (Introduction)
• OLA Interview Experience | Set 11 ( For Internship)
• Top 20 Greedy Algorithms Interview Questions
• Job Sequencing Problem - Loss Minimization
• Prim's Algorithm (Simple Implementation for Adjacency Matrix Representation)
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• For each row of the matrix, find the smallest element and subtract it from every element in its row.
• Do the same (as step 1) for all columns.
• Cover all zeros in the matrix using minimum number of horizontal and vertical lines.
• Test for Optimality: If the minimum number of covering lines is n, an optimal assignment is possible and we are finished. Else if lines are lesser than n, we haven’t found the optimal assignment, and must proceed to step 5.
• Determine the smallest entry not covered by any line. Subtract this entry from each uncovered row, and then add it to each covered column. Return to step 3.

Try it before moving to see the solution

Explanation for above simple example:

  An example that doesn’t lead to optimal value in first attempt: In the above example, the first check for optimality did give us solution. What if we the number covering lines is less than n.

Time complexity : O(n^3), where n is the number of workers and jobs. This is because the algorithm implements the Hungarian algorithm, which is known to have a time complexity of O(n^3).

Space complexity :   O(n^2), where n is the number of workers and jobs. This is because the algorithm uses a 2D cost matrix of size n x n to store the costs of assigning each worker to a job, and additional arrays of size n to store the labels, matches, and auxiliary information needed for the algorithm.

In the next post, we will be discussing implementation of the above algorithm. The implementation requires more steps as we need to find minimum number of lines to cover all 0’s using a program. References: http://www.math.harvard.edu/archive/20_spring_05/handouts/assignment_overheads.pdf https://www.youtube.com/watch?v=dQDZNHwuuOY

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Unbalanced Assignment Problems

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.

Unbalanced Maximization Assignment problem example

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

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.

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.

– 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:

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Nash Balanced Assignment Problem

• Conference paper
• First Online: 21 November 2022
• Cite this conference paper

• Minh Hieu Nguyen 11 ,
• Mourad Baiou 11 &
• Viet Hung Nguyen 11  

Part of the book series: Lecture Notes in Computer Science ((LNCS,volume 13526))

Included in the following conference series:

• International Symposium on Combinatorial Optimization

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In this paper, we consider a variant of the classic Assignment Problem (AP), called the Balanced Assignment Problem (BAP) [ 2 ]. The BAP seeks to find an assignment solution which has the smallest value of max-min distance : the difference between the maximum assignment cost and the minimum one. However, by minimizing only the max-min distance, the total cost of the BAP solution is neglected and it may lead to an inefficient solution in terms of total cost. Hence, we propose a fair way based on Nash equilibrium [ 1 , 3 , 4 ] to inject the total cost into the objective function of the BAP for finding assignment solutions having a better trade-off between the two objectives: the first aims at minimizing the total cost and the second aims at minimizing the max-min distance. For this purpose, we introduce the concept of Nash Fairness (NF) solutions based on the definition of proportional-fair scheduling adapted in the context of the AP: a transfer of utilities between the total cost and the max-min distance is considered to be fair if the percentage increase in the total cost is smaller than the percentage decrease in the max-min distance and vice versa.

We first show the existence of a NF solution for the AP which is exactly the optimal solution minimizing the product of the total cost and the max-min distance. However, finding such a solution may be difficult as it requires to minimize a concave function. The main result of this paper is to show that finding all NF solutions can be done in polynomial time. For that, we propose a Newton-based iterative algorithm converging to NF solutions in polynomial time. It consists in optimizing a sequence of linear combinations of the two objective based on Weighted Sum Method [ 5 ]. Computational results on various instances of the AP are presented and commented.

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Minh Hieu Nguyen, Mourad Baiou & Viet Hung Nguyen

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ESSEC Business School of Paris, Cergy Pontoise Cedex, France

Ivana Ljubić

IBM TJ Watson Research Center, Yorktown Heights, NY, USA

Francisco Barahona

Georgia Institute of Technology, Atlanta, GA, USA

Santanu S. Dey

Université Paris-Dauphine, Paris, France

A. Ridha Mahjoub

Proposition 1 . There may be more than one NF solution for the AP.

Let us illustrate this by an instance of the AP having the following cost matrix

By verifying all feasible assignment solutions in this instance, we obtain easily three assignment solutions \((1-1, 2-2, 3-3), (1-2, 2-3, 3-1)\) , \((1-3, 2-2, 3-1)\) and \((1-3, 2-1, 3-2)\) corresponding to 4 NF solutions (280, 36), (320, 32), (340, 30) and (364, 28). Note that \(i-j\) where \(1 \le i,j \le 3\) represents the assignment between worker i and job j in the solution of this instance.     \(\square \)

We recall below the proofs of some recent results that we have published in [ 10 ]. They are needed to prove the new results presented in this paper.

Theorem 2 [ 10 ] . \((P^{*},Q^{*}) = {{\,\mathrm{arg\,min}\,}}_{(P,Q) \in \mathcal {S}} PQ\) is a NF solution.

Obviously, there always exists a solution \((P^{*},Q^{*}) \in \mathcal {S}\) such that

Now \(\forall (P',Q') \in \mathcal {S}\) we have \(P'Q' \ge P^{*}Q^{*}\) . Then

The first inequality holds by the Cauchy-Schwarz inequality.

Hence, \((P^{*},Q^{*})\) is a NF solution.     \(\square \)

Theorem 3 [ 10 ] . \((P^{*},Q^{*}) \in \mathcal {S}\) is a NF solution if and only if \((P^{*},Q^{*})\) is an optimal solution of \(\mathcal {P(\alpha ^{*})}\) where \(\alpha ^{*} = \frac{Q^{*}}{P^{*}}\) .

Firstly, let \((P^{*},Q^{*})\) be a NF solution and \(\alpha ^{*} = \frac{Q^{*}}{P^{*}}\) . We will show that \((P^{*},Q^{*})\) is an optimal solution of \(\mathcal {P(\alpha ^{*})}\) .

Since \((P^{*},Q^{*})\) is a NF solution, we have

Since \(\alpha ^{*} = \frac{Q^{*}}{P^{*}}\) , we have \(\alpha ^{*}P^{*}+Q^{*} = 2Q^{*}\) .

Dividing two sides of ( 6 ) by \(P^{*} > 0\) we obtain

So we deduce from ( 7 )

Hence, \((P^{*},Q^{*})\) is an optimal solution of \(\mathcal {P}(\alpha ^{*})\) .

Now suppose \(\alpha ^{*} = \frac{Q^{*}}{P^{*}}\) and \((P^{*},Q^{*})\) is an optimal solution of \(\mathcal {P}(\alpha ^{*})\) , we show that \((P^{*},Q^{*})\) is a NF solution.

If \((P^{*},Q^{*})\) is not a NF solution, there exists a solution \((P',Q') \in \mathcal {S}\) such that

We have then

which contradicts the optimality of \((P^{*},Q^{*})\) .     \(\square \)

Lemma 3 [ 10 ] . Let \(\alpha , \alpha ' \in \mathbb {R}_+\) and \((P_{\alpha }, Q_{\alpha })\) , \((P_{\alpha '}, Q_{\alpha '})\) be the optimal solutions of \(\mathcal {P(\alpha )}\) and \(\mathcal {P(\alpha ')}\) respectively, if \(\alpha \le \alpha '\) then \(P_{\alpha } \ge P_{\alpha '}\) and \(Q_{\alpha } \le Q_{\alpha '}\) .

The optimality of \((P_{\alpha }, Q_{\alpha })\) and \((P_{\alpha '}, Q_{\alpha '})\) gives

By adding both sides of ( 8a ) and ( 8b ), we obtain \((\alpha - \alpha ') (P_{\alpha } - P_{\alpha '}) \le 0\) . Since \(\alpha \le \alpha '\) , it follows that \(P_{\alpha } \ge P_{\alpha '}\) .

On the other hand, inequality ( 8a ) implies \(Q_{\alpha '} - Q_{\alpha } \ge \alpha (P_{\alpha } - P_{\alpha '}) \ge 0\) that leads to \(Q_{\alpha } \le Q_{\alpha '}\) .     \(\square \)

Lemma 4 [ 10 ] . During the execution of Procedure Find ( \(\alpha _{0})\) in Algorithm 1 , \(\alpha _{i} \in [0,1], \, \forall i \ge 1\) . Moreover, if \(T_{0} \ge 0\) then the sequence \(\{\alpha _i\}\) is non-increasing and \(T_{i} \ge 0, \, \forall i \ge 0\) . Otherwise, if \(T_{0} \le 0\) then the sequence \(\{\alpha _i\}\) is non-decreasing and \(T_{i} \le 0, \, \forall i \ge 0\) .

Since \(P \ge Q \ge 0, \, \forall (P, Q) \in \mathcal {S}\) , it follows that \(\alpha _{i+1} = \frac{Q_i}{P_i} \in [0,1], \, \forall i \ge 0\) .

We first consider \(T_{0} \ge 0\) . We proof \(\alpha _i \ge \alpha _{i+1}, \, \forall i \ge 0\) by induction on i . For \(i = 0\) , we have \(T_{0} = \alpha _{0} P_{0} - Q_{0} = P_{0}(\alpha _{0}-\alpha _{1}) \ge 0\) , it follows that \(\alpha _{0} \ge \alpha _{1}\) . Suppose that our hypothesis is true until \(i = k \ge 0\) , we will prove that it is also true with \(i = k+1\) .

Indeed, we have

The inductive hypothesis gives \(\alpha _k \ge \alpha _{k+1}\) that implies \(P_{k+1} \ge P_k > 0\) and \(Q_{k} \ge Q_{k+1} \ge 0\) according to Lemma 3 . It leads to \(Q_{k}P_{k+1} - P_{k}Q_{k+1} \ge 0\) and then \(\alpha _{k+1} - \alpha _{k+2} \ge 0\) .

Hence, we have \(\alpha _{i} \ge \alpha _{i+1}, \, \forall i \ge 0\) .

Consequently, \(T_{i} = \alpha _{i}P_{i} - Q_{i} = P_{i}(\alpha _{i}-\alpha _{i+1}) \ge 0, \, \forall i \ge 0\) .

Similarly, if \(T_{0} \le 0\) we obtain that the sequence \(\{\alpha _i\}\) is non-decreasing and \(T_{i} \le 0, \, \forall i \ge 0\) . That concludes the proof.     \(\square \)

Lemma 5 [ 10 ] . From each \(\alpha _{0} \in [0,1]\) , Procedure Find \((\alpha _{0})\) converges to a coefficient \(\alpha _{k} \in \mathcal {C}_{0}\) satisfying \(\alpha _{k}\) is the unique element \(\in \mathcal {C}_{0}\) between \(\alpha _{0}\) and \(\alpha _{k}\) .

As a consequence of Lemma 4 , Procedure \(\textit{Find}(\alpha _{0})\) converges to a coefficient \(\alpha _{k} \in [0,1], \forall \alpha _{0} \in [0,1]\) .

By the stopping criteria of Procedure Find \((\alpha _{0})\) , when \(T_{k} = \alpha _{k} P_{k} - Q_{k} = 0\) we obtain \(\alpha _{k} \in C_{0}\) and \((P_{k},Q_{k})\) is a NF solution. (Theorem 3 )

If \(T_{0} = 0\) then obviously \(\alpha _{k} = \alpha _{0}\) . We consider \(T_{0} > 0\) and the sequence \(\{\alpha _i\}\) is now non-negative, non-increasing. We will show that \([\alpha _{k},\alpha _{0}] \cap \mathcal {C}_{0} = \alpha _{k}\) .

Suppose that we have \(\alpha \in (\alpha _{k},\alpha _{0}]\) and \(\alpha \in \mathcal {C}_{0}\) corresponding to a NF solution ( P ,  Q ). Then there exists \(1 \le i \le k\) such that \(\alpha \in (\alpha _{i}, \alpha _{i-1}]\) . Since \(\alpha \le \alpha _{i-1}\) , \(P \ge P_{i-1}\) and \(Q \le Q_{i-1}\) due to Lemma 3 . Thus, we get

By the definitions of \(\alpha \) and \(\alpha _{i}\) , inequality ( 9 ) is equivalent to \(\alpha \le \alpha _{i}\) which leads to a contradiction.

By repeating the same argument for \(T_{0} < 0\) , we also have a contradiction.     \(\square \)

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Nguyen, M.H., Baiou, M., Nguyen, V.H. (2022). Nash Balanced Assignment Problem. In: Ljubić, I., Barahona, F., Dey, S.S., Mahjoub, A.R. (eds) Combinatorial Optimization. ISCO 2022. Lecture Notes in Computer Science, vol 13526. Springer, Cham. https://doi.org/10.1007/978-3-031-18530-4_13

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Balanced and Unbalanced Transportation Problems

The two categories of transportation problems are balanced and unbalanced transportation problems . As we all know, a transportation problem is a type of Linear Programming Problem (LPP) in which items are carried from a set of sources to a set of destinations based on the supply and demand of the sources and destinations, with the goal of minimizing the total transportation cost. It is also known as the Hitchcock problem.

Introduction to Balanced and Unbalanced Transportation Problems

Balanced transportation problem.

The problem is considered to be a balanced transportation problem when both supplies and demands are equal.

Unbalanced Transportation Problem

Unbalanced transportation problem is defined as a situation in which supply and demand are not equal. A dummy row or a dummy column is added to this type of problem, depending on the necessity, to make it a balanced problem. The problem can then be addressed in the same way as the balanced problem.

Methods of Solving Transportation Problems

There are three ways for determining the initial basic feasible solution. They are

1. NorthWest Corner Cell Method.

2. Vogel’s Approximation Method (VAM).

3. Least Call Cell Method.

The following is the basic framework of the balanced transportation problem:

The destinations D1, D2, D3, and D4 in the above table are where the products/goods will be transported from various sources O1, O2, O3, and O4. The supply from the source Oi is represented by S i . The demand for the destination Dj is d j . If a product is delivered from source Si to destination Dj, then the cost is called C ij .

Let us now explore the process of solving the balanced transportation problem using one of the ways known as the NorthWest Corner Method in this article.

Solving Balanced Transportation problem by Northwest Corner Method

Consider this scenario:

With three sources (O1, O2, and O3) and four destinations (D1, D2, D3, and D4), what is the best way to solve this problem? The supply for the sources O1, O2, and O3 are 300, 400, and 500, respectively. Demands for the destination D1, D2, D3, and D4 are 250, 350, 400, and 200, respectively.

The starting point for the North West Corner technique is (O1, D1), which is the table’s northwest corner. The cost of transportation is calculated for each value in the cell. As indicated in the diagram, compare the demand for column D1 with the supply from source O1 and assign a minimum of two to the cell (O1, D1).

Column D1’s demand has been met, hence the entire column will be canceled. The supply from the source O1 is still 300 – 250 = 50.

Analyze the northwest corner, i.e. (O1, D2), of the remaining table, excluding column D1, and assign the lowest among the supply for the appropriate column and rows. Because the supply from O1 is 50 and the demand for D2 is 350, allocate 50 to the cell (O1, D2).

Now, row O1 is canceled because the supply from row O1 has been completed. Hence, the demand for Column D2 has become 350 – 50 = 50.

The northwest corner cell in the remaining table is (O2, D2). The shortest supply from source O2 (400) and the demand for column D2 (300) is 300, thus putting 300 in the cell (O2, D2). Because the demand for column D2 has been met, the column can be deleted, and the remaining supply from source O2 is 400 – 300 = 100.

Again, find the northwest corner of the table, i.e. (O2, D3), and compare the O2 supply (i.e. 100) to the D2 demand (i.e. 400) and assign the smaller (i.e. 100) to the cell (O2, D2). Row O2 has been canceled because the supply from O2 has been completed. Column D3 has a leftover demand of 400 – 100 = 300.

Continuing in the same manner, the final cell values will be:

It should be observed that the demand for the relevant columns and rows is equal in the last remaining cell, which was cell (O3, D4). In this situation, the supply from O3 was 200, and the demand for D4 was 200, therefore this cell was assigned to it. Nothing was left for any row or column at the end.

To achieve the basic solution, multiply the allotted value by the respective cell value (i.e. the cost) and add them all together.

I.e., (250 × 3) + (50 × 1) + (300 × 6) + (100 × 5) + (300 × 3) + (200 × 2) = 4400.

Solving Unbalanced Transportation Problem

An unbalanced transportation problem is provided below. Because the sum of all the supplies, O1, O2, O3, and O4, does not equal the sum of all the demands, D1, D2, D3, D4, and D5, the situation is unbalanced.

The idea of a dummy row or dummy column will be applied in this type of scenario. Because the supply is more than the demand in this situation, a fake demand column will be inserted, with a demand of (total supply – total demand), i.e. 117 – 95 = 22, as seen in the image below. A fake supply row would have been introduced if demand was greater than supply.

Now this problem has been changed to a balanced transportation problem, and it can be addressed using any of the ways listed below to solve a balanced transportation problem, such as the northwest corner method mentioned earlier.

Frequently Asked Questions on Balanced and Unbalanced Transportation Problems

What is meant by balanced and unbalanced transportation problems.

The problem is referred to as a balanced transportation problem when both supplies and demands are equal. Unbalanced transportation is defined as a situation where supply and demand are not equal.

What is called a transportation problem?

The transportation problem is a type of Linear Programming Problem in which commodities are carried from a set of sources to a set of destinations while taking into account the supply and demand of the sources and destinations, respectively, in order to reduce the total cost of transportation.

What are the different methods to solve transportation problems?

The following are three approaches to solve the transportation issue:

• NorthWest Corner Cell Method.
• Least Call Cell Method.
• Vogel’s Approximation Method (VAM).

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What is Balanced or Unbalanced Assignment problem?

Operations Research

The Assignment problem can be Balanced or Unbalanced problem.

A Balanced problem means the no. of rows and no. of columns in the problem are equal. E. g. if the problem contains 4 workers and 4 jobs, then it is balanced.

Where as, an Unbalanced problem means the no. of rows and no. of columns are not equal. E. g. if the problem contains 4 workers and 3 jobs it is not balanced. Then first we need to balance the problem by taking a Dummy job (imaginary job).

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