assignment problem example

Procedure, Example Solved Problem | Operations Research - Solution of assignment problems (Hungarian Method) | 12th Business Maths and Statistics : Chapter 10 : Operations Research

Chapter: 12th business maths and statistics : chapter 10 : operations research.

Solution of assignment problems (Hungarian Method)

First check whether the number of rows is equal to the numbers of columns, if it is so, the assignment problem is said to be balanced.

Step :1 Choose the least element in each row and subtract it from all the elements of that row.

Step :2 Choose the least element in each column and subtract it from all the elements of that column. Step 2 has to be performed from the table obtained in step 1.

Step:3 Check whether there is atleast one zero in each row and each column and make an assignment as follows.

Step :4 If each row and each column contains exactly one assignment, then the solution is optimal.

Example 10.7

Solve the following assignment problem. Cell values represent cost of assigning job A, B, C and D to the machines I, II, III and IV.

Here the number of rows and columns are equal.

∴ The given assignment problem is balanced. Now let us find the solution.

Step 1: Select a smallest element in each row and subtract this from all the elements in its row.

Look for atleast one zero in each row and each column.Otherwise go to step 2.

Step 2: Select the smallest element in each column and subtract this from all the elements in its column.

Since each row and column contains atleast one zero, assignments can be made.

Step 3 (Assignment):

Thus all the four assignments have been made. The optimal assignment schedule and total cost is

The optimal assignment (minimum) cost

Example 10.8

Consider the problem of assigning five jobs to five persons. The assignment costs are given as follows. Determine the optimum assignment schedule.

∴ The given assignment problem is balanced.

Now let us find the solution.

The cost matrix of the given assignment problem is

Column 3 contains no zero. Go to Step 2.

Thus all the five assignments have been made. The Optimal assignment schedule and total cost is

The optimal assignment (minimum) cost = ` 9

Example 10.9

Solve the following assignment problem.

Since the number of columns is less than the number of rows, given assignment problem is unbalanced one. To balance it , introduce a dummy column with all the entries zero. The revised assignment problem is

Here only 3 tasks can be assigned to 3 men.

Step 1: is not necessary, since each row contains zero entry. Go to Step 2.

Step 3 (Assignment) :

Since each row and each columncontains exactly one assignment,all the three men have been assigned a task. But task S is not assigned to any Man. The optimal assignment schedule and total cost is

The optimal assignment (minimum) cost = ₹ 35

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Hungarian Method Examples

Now we will examine a few highly simplified illustrations of Hungarian Method for solving an assignment problem .

Later in the chapter, you will find more practical versions of assignment models like Crew assignment problem , Travelling salesman problem , etc.

Example-1, Example-2

Example 1: Hungarian Method

The Funny Toys Company has four men available for work on four separate jobs. Only one man can work on any one job. The cost of assigning each man to each job is given in the following table. The objective is to assign men to jobs in such a way that the total cost of assignment is minimum.

This is a minimization example of assignment problem . We will use the Hungarian Algorithm to solve this problem.

Identify the minimum element in each row and subtract it from every element of that row. The result is shown in the following table.

"A man has one hundred dollars and you leave him with two dollars, that's subtraction." -Mae West

On small screens, scroll horizontally to view full calculation

Identify the minimum element in each column and subtract it from every element of that column.

Make the assignments for the reduced matrix obtained from steps 1 and 2 in the following way:

• For every zero that becomes assigned, cross out (X) all other zeros in the same row and the same column.
• If for a row and a column, there are two or more zeros and one cannot be chosen by inspection, choose the cell arbitrarily for assignment.

An optimal assignment is found, if the number of assigned cells equals the number of rows (and columns). In case you have chosen a zero cell arbitrarily, there may be alternate optimal solutions. If no optimal solution is found, go to step 5.

Use Horizontal Scrollbar to View Full Table Calculation

Draw the minimum number of vertical and horizontal lines necessary to cover all the zeros in the reduced matrix obtained from step 3 by adopting the following procedure:

• Mark all the rows that do not have assignments.
• Mark all the columns (not already marked) which have zeros in the marked rows.
• Mark all the rows (not already marked) that have assignments in marked columns.
• Repeat steps 5 (ii) and (iii) until no more rows or columns can be marked.
• Draw straight lines through all unmarked rows and marked columns.

You can also draw the minimum number of lines by inspection.

Select the smallest element (i.e., 1) from all the uncovered elements. Subtract this smallest 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.

Now again make the assignments for the reduced matrix.

Final Table: Hungarian Method

Since the number of assignments is equal to the number of rows (& columns), this is the optimal solution.

The total cost of assignment = A1 + B4 + C2 + D3

Substituting values from original table: 20 + 17 + 17 + 24 = Rs. 78.

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Solving an Assignment Problem

This section presents an example that shows how to solve an assignment problem using both the MIP solver and the CP-SAT solver.

In the example there are five workers (numbered 0-4) and four tasks (numbered 0-3). Note that there is one more worker than in the example in the Overview .

The costs of assigning workers to tasks are shown in the following table.

The problem is to assign each worker to at most one task, with no two workers performing the same task, while minimizing the total cost. Since there are more workers than tasks, one worker will not be assigned a task.

MIP solution

The following sections describe how to solve the problem using the MPSolver wrapper .

Import the libraries

The following code imports the required libraries.

Create the data

The following code creates the data for the problem.

The costs array corresponds to the table of costs for assigning workers to tasks, shown above.

Declare the MIP solver

The following code declares the MIP solver.

Create the variables

The following code creates binary integer variables for the problem.

Create the constraints

Create the objective function.

The following code creates the objective function for the problem.

The value of the objective function is the total cost over all variables that are assigned the value 1 by the solver.

Invoke the solver

The following code invokes the solver.

Print the solution

The following code prints the solution to the problem.

Here is the output of the program.

Complete programs

Here are the complete programs for the MIP solution.

CP SAT solution

The following sections describe how to solve the problem using the CP-SAT solver.

Declare the model

The following code declares the CP-SAT model.

The following code sets up the data for the problem.

The following code creates the constraints for the problem.

Here are the complete programs for the CP-SAT solution.

Except as otherwise noted, the content of this page is licensed under the Creative Commons Attribution 4.0 License , and code samples are licensed under the Apache 2.0 License . For details, see the Google Developers Site Policies . Java is a registered trademark of Oracle and/or its affiliates.

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Hungarian Maximum Matching Algorithm

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The Hungarian matching algorithm , also called the Kuhn-Munkres algorithm, is a \(O\big(|V|^3\big)\) algorithm that can be used to find maximum-weight matchings in bipartite graphs , which is sometimes called the assignment problem . A bipartite graph can easily be represented by an adjacency matrix , where the weights of edges are the entries. Thinking about the graph in terms of an adjacency matrix is useful for the Hungarian algorithm.

A matching corresponds to a choice of 1s in the adjacency matrix, with at most one 1 in each row and in each column.

The Hungarian algorithm solves the following problem:

In a complete bipartite graph \(G\), find the maximum-weight matching. (Recall that a maximum-weight matching is also a perfect matching.)

This can also be adapted to find the minimum-weight matching.

Say you are having a party and you want a musician to perform, a chef to prepare food, and a cleaning service to help clean up after the party. There are three companies that provide each of these three services, but one company can only provide one service at a time (i.e. Company B cannot provide both the cleaners and the chef). You are deciding which company you should purchase each service from in order to minimize the cost of the party. You realize that is an example of the assignment problem, and set out to make a graph out of the following information: \(\quad\) Company\(\quad\) \(\quad\) Cost for Musician\(\quad\) \(\quad\) Cost for Chef\(\quad\) \(\quad\) Cost for Cleaners\(\quad\) \(\quad\) Company A\(\quad\) \(\quad\) $108\(\quad\) \(\quad\) $125\(\quad\) \(\quad\) $150\(\quad\) \(\quad\) Company B\(\quad\) \(\quad\) $150\(\quad\) \(\quad\) $135\(\quad\) \(\quad\) $175\(\quad\) \(\quad\) Company C\(\quad\) \(\quad\) $122\(\quad\) \(\quad\) $148\(\quad\) \(\quad\) $250\(\quad\) Can you model this table as a graph? What are the nodes? What are the edges? Show Answer The nodes are the companies and the services. The edges are weighted by the price.

What are some ways to solve the problem above? Since the table above can be thought of as a \(3 \times 3\) matrix, one could certainly solve this problem using brute force, checking every combination and seeing what yields the lowest price. However, there are \(n!\) combinations to check, and for large \(n\), this method becomes very inefficient very quickly.

The Hungarian Algorithm Using an Adjacency Matrix

The hungarian algorithm using a graph.

With the cost matrix from the example above in mind, the Hungarian algorithm operates on this key idea: if a number is added to or subtracted from all of the entries of any one row or column of a cost matrix, then an optimal assignment for the resulting cost matrix is also an optimal assignment for the original cost matrix.

The Hungarian Method [1] Subtract the smallest entry in each row from all the other entries in the row. This will make the smallest entry in the row now equal to 0. Subtract the smallest entry in each column from all the other entries in the column. This will make the smallest entry in the column now equal to 0. Draw lines through the row and columns that have the 0 entries such that the fewest lines possible are drawn. If there are \(n\) lines drawn, an optimal assignment of zeros is possible and the algorithm is finished. If the number of lines is less than \(n\), then the optimal number of zeroes is not yet reached. Go to the next step. Find the smallest entry not covered by any line. Subtract this entry from each row that isn’t crossed out, and then add it to each column that is crossed out. Then, go back to Step 3.

Solve for the optimal solution for the example in the introduction using the Hungarian algorithm described above. Here is the initial adjacency matrix: Subtract the smallest value in each row from the other values in the row: Now, subtract the smallest value in each column from all other values in the column: Draw lines through the row and columns that have the 0 entries such that the fewest possible lines are drawn: There are 2 lines drawn, and 2 is less than 3, so there is not yet the optimal number of zeroes. Find the smallest entry not covered by any line. Subtract this entry from each row that isn’t crossed out, and then add it to each column that is crossed out. Then, go back to Step 3. 2 is the smallest entry. First, subtract from the uncovered rows: Now add to the covered columns: Now go back to step 3, drawing lines through the rows and columns that have 0 entries: There are 3 lines (which is \(n\)), so we are done. The assignment will be where the 0's are in the matrix such that only one 0 per row and column is part of the assignment. Replace the original values: The Hungarian algorithm tells us that it is cheapest to go with the musician from company C, the chef from company B, and the cleaners from company A. We can verify this by brute force. 108 + 135 + 250 = 493 108 + 148 + 175 = 431 150 + 125 + 250 = 525 150 + 148 + 150 = 448 122 + 125 + 175 = 422 122 + 135 + 150 = 407. We can see that 407 is the lowest price and matches the assignment the Hungarian algorithm determined. \(_\square\)

The Hungarian algorithm can also be executed by manipulating the weights of the bipartite graph in order to find a stable, maximum (or minimum) weight matching. This can be done by finding a feasible labeling of a graph that is perfectly matched, where a perfect matching is denoted as every vertex having exactly one edge of the matching.

How do we know that this creates a maximum-weight matching?

A feasible labeling on a perfect match returns a maximum-weighted matching. Suppose each edge \(e\) in the graph \(G\) connects two vertices, and every vertex \(v\) is covered exactly once. With this, we have the following inequality: \[w(M’) = \sum_{e\ \epsilon\ E} w(e) \leq \sum_{e\ \epsilon\ E } \big(l(e_x) + l(e_y)\big) = \sum_{v\ \epsilon\ V} l(v),\] where \(M’\) is any perfect matching in \(G\) created by a random assignment of vertices, and \(l(x)\) is a numeric label to node \(x\). This means that \(\sum_{v\ \epsilon\ V}\ l(v)\) is an upper bound on the cost of any perfect matching. Now let \(M\) be a perfect match in \(G\), then \[w(M) = \sum_{e\ \epsilon\ E} w(e) = \sum_{v\ \epsilon\ V}\ l(v).\] So \(w(M’) \leq w(M)\) and \(M\) is optimal. \(_\square\)

Start the algorithm by assigning any weight to each individual node in order to form a feasible labeling of the graph \(G\). This labeling will be improved upon by finding augmenting paths for the assignment until the optimal one is found.

A feasible labeling is a labeling such that

\(l(x) + l(y) \geq w(x,y)\ \forall x \in X, y \in Y\), where \(X\) is the set of nodes on one side of the bipartite graph, \(Y\) is the other set of nodes, \(l(x)\) is the label of \(x\), etc., and \(w(x,y)\) is the weight of the edge between \(x\) and \(y\).

A simple feasible labeling is just to label a node with the number of the largest weight from an edge going into the node. This is certain to be a feasible labeling because if \(A\) is a node connected to \(B\), the label of \(A\) plus the label of \(B\) is greater than or equal to the weight \(w(x,y)\) for all \(y\) and \(x\).

A feasible labeling of nodes, where labels are in red [2] .

Imagine there are four soccer players and each can play a few positions in the field. The team manager has quantified their skill level playing each position to make assignments easier.

How can players be assigned to positions in order to maximize the amount of skill points they provide?

The algorithm starts by labeling all nodes on one side of the graph with the maximum weight. This can be done by finding the maximum-weighted edge and labeling the adjacent node with it. Additionally, match the graph with those edges. If a node has two maximum edges, don’t connect them.

Although Eva is the best suited to play defense, she can't play defense and mid at the same time!

If the matching is perfect, the algorithm is done as there is a perfect matching of maximum weights. Otherwise, there will be two nodes that are not connected to any other node, like Tom and Defense. If this is the case, begin iterating.

Improve the labeling by finding the non-zero label vertex without a match, and try to find the best assignment for it. Formally, the Hungarian matching algorithm can be executed as defined below:

The Hungarian Algorithm for Graphs [3] Given: the labeling \(l\), an equality graph \(G_l = (V, E_l)\), an initial matching \(M\) in \(G_l\), and an unmatched vertex \(u \in V\) and \(u \notin M\) Augmenting the matching A path is augmenting for \(M\) in \(G_l\) if it alternates between edges in the matching and edges not in the matching, and the first and last vertices are free vertices , or unmatched, in \(M\). We will keep track of a candidate augmenting path starting at the vertex \(u\). If the algorithm finds an unmatched vertex \(v\), add on to the existing augmenting path \(p\) by adding the \(u\) to \(v\) segment. Flip the matching by replacing the edges in \(M\) with the edges in the augmenting path that are not in \(M\) \((\)in other words, the edges in \(E_l - M).\) Improving the labeling \(S \subseteq X\) and \(T \subseteq Y,\) where \(S\) and \(T\) represent the candidate augmenting alternating path between the matching and the edges not in the matching. Let \(N_l(S)\) be the neighbors to each node that is in \(S\) along edges in \(E_l\) such that \(N_l(S) = \{v|\forall u \in S: (u,v) \in E_l\}\). If \(N_l(S) = T\), then we cannot increase the size of the alternating path (and therefore can't further augment), so we need to improve the labeling. Let \(\delta_l\) be the minimum of \(l(u) + l(v) - w(u,v)\) over all of the \(u \in S\) and \(v \notin T\). Improve the labeling \(l\) to \(l'\): If \(r \in S,\) then \(l'(r) = l(r) - \delta_l,\) If \(r \in T,\) then \(l'(r) = l(r) + \delta_l.\) If \(r \notin S\) and \(r \notin T,\) then \(l'(r) = l(r).\) \(l'\) is a valid labeling and \(E_l \subset E_{l'}.\) Putting it all together: The Hungarian Algorithm Start with some matching \(M\), a valid labeling \(l\), where \(l\) is defined as the labelling \(\forall x \in X, y \in Y| l(y) = 0, l(x) = \text{ max}_{y \in Y}(w\big(x, y)\big)\). Do these steps until a perfect matching is found \((\)when \(M\) is perfect\():\) (a) Look for an augmenting path in \(M.\) (b) If an augmenting path does not exist, improve the labeling and then go back to step (a).

Each step will increase the size of the matching \(M\) or it will increase the size of the set of labeled edges, \(E_l\). This means that the process will eventually terminate since there are only so many edges in the graph \(G\). [4]

When the process terminates, \(M\) will be a perfect matching. By the Kuhn-Munkres theorem , this means that the matching is a maximum-weight matching.

The algorithm defined above can be implemented in the soccer scenario. First, the conflicting node is identified, implying that there is an alternating tree that must be reconfigured.

There is an alternating path between defense, Eva, mid, and Tom.

To find the best appropriate node, find the minimum \(\delta_l\), as defined in step 4 above, where \(l_u\) is the label for player \(u,\) \(l_v\) is the label for position \(v,\) and \(w_{u, v}\) is the weight on that edge.

The \(\delta_l\) of each unmatched node is computed, where the minimum is found to be a value of 2, between Tom playing mid \((8 + 0 – 6 = 2).\)

The labels are then augmented and the new edges are graphed in the example. Notice that defense and mid went down by 2 points, whereas Eva’s skillset got back two points. However, this is expected as Eva can't play in both positions at once.

Augmenting path leads to relabeling of nodes, which gives rise to the maximum-weighted path.

These new edges complete the perfect matching of the graph, which implies that a maximum-weighted graph has been found and the algorithm can terminate.

The complexity of the algorithm will be analyzed using the graph-based technique as a reference, yet the result is the same as for the matrix-based one.

Algorithm analysis [3] At each \(a\) or \(b\) step, the algorithm adds one edge to the matching and this happens \(O\big(|V|\big)\) times. It takes \(O\big(|V|\big)\) time to find the right vertex for the augmenting (if there is one at all), and it is \(O\big(|V|\big)\) time to flip the matching. Improving the labeling takes \(O\big(|V|\big)\) time to find \(\delta_l\) and to update the labelling accordingly. We might have to improve the labeling up to \(O\big(|V|\big)\) times if there is no augmenting path. This makes for a total of \(O\big(|V|^2\big)\) time. In all, there are \(O\big(|V|\big)\) iterations each taking \(O\big(|V|\big)\) work, leading to a total running time of \(O\big(|V|^3\big)\).

• Matching Algorithms
• Bruff, D. The Assignment Problem and the Hungarian Method . Retrieved June 26, 2016, from http://www.math.harvard.edu/archive/20_spring_05/handouts/assignment_overheads.pdf
• Golin, M. Bipartite Matching & the Hungarian Method . Retrieved Retrieved June 26th, 2016, from http://www.cse.ust.hk/~golin/COMP572/Notes/Matching.pdf
• Grinman, A. The Hungarian Algorithm for Weighted Bipartite Graphs . Retrieved June 26, 2016, from http://math.mit.edu/~rpeng/18434/hungarianAlgorithm.pdf
• Golin, M. Bipartite Matching & the Hungarian Method . Retrieved June 26, 2016, from http://www.cse.ust.hk/~golin/COMP572/Notes/Matching.pdf

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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

• History of O.R.
• Approach, Techniques and Tools
• Phases and Processes of O.R. Study
• Typical Applications of O.R
• Limitations of Operations Research
• Models in Operations Research
• O.R. in real world

2 Linear Programming: Formulation and Graphical Method

• General formulation of Linear Programming Problem
• Optimisation Models
• Basics of Graphic Method
• Important steps to draw graph
• Multiple, Unbounded Solution and Infeasible Problems
• Solving Linear Programming Graphically Using Computer
• Application of Linear Programming in Business and Industry

3 Linear Programming-Simplex Method

• Principle of Simplex Method
• Computational aspect of Simplex Method
• Simplex Method with several Decision Variables
• Two Phase and M-method
• Multiple Solution, Unbounded Solution and Infeasible Problem
• Sensitivity Analysis
• Dual Linear Programming Problem

4 Transportation Problem

• Basic Feasible Solution of a Transportation Problem
• Modified Distribution Method
• Stepping Stone Method
• Unbalanced Transportation Problem
• Degenerate Transportation Problem
• Transhipment Problem
• Maximisation in a Transportation Problem

5 Assignment Problem

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

6 Application of Excel Solver to Solve LPP

• Building Excel model for solving LP: An Illustrative Example

7 Goal Programming

• Concepts of goal programming
• Goal programming model formulation
• Graphical method of goal programming
• The simplex method of goal programming
• Using Excel Solver to Solve Goal Programming Models
• Application areas of goal programming

8 Integer Programming

• Some Integer Programming Formulation Techniques
• Binary Representation of General Integer Variables
• Unimodularity
• Cutting Plane Method
• Branch and Bound Method
• Solver Solution

9 Dynamic Programming

• Dynamic Programming Methodology: An Example
• Definitions and Notations
• Dynamic Programming Applications

10 Non-Linear Programming

• Solution of a Non-linear Programming Problem
• Convex and Concave Functions
• Kuhn-Tucker Conditions for Constrained Optimisation
• Quadratic Programming
• Separable Programming
• NLP Models with Solver

11 Introduction to game theory and its Applications

• Important terms in Game Theory
• Saddle points
• Mixed strategies: Games without saddle points
• 2 x n games
• Exploiting an opponent’s mistakes

12 Monte Carlo Simulation

• Reasons for using simulation
• Monte Carlo simulation
• Limitations of simulation
• Steps in the simulation process
• Some practical applications of simulation
• Two typical examples of hand-computed simulation
• Computer simulation

13 Queueing Models

• Characteristics of a queueing model
• Notations and Symbols
• Statistical methods in queueing
• The M/M/I System
• The M/M/C System
• The M/Ek/I System
• Decision problems in queueing

Index     Assignment problem     Hungarian algorithm     Solve online    

The assignment problem

The assignment problem deals with assigning machines to tasks, workers to jobs, soccer players to positions, and so on. The goal is to determine the optimum assignment that, for example, minimizes the total cost or maximizes the team effectiveness. The assignment problem is a fundamental problem in the area of combinatorial optimization.

Assume for example that we have four jobs that need to be executed by four workers. Because each worker has different skills, the time required to perform a job depends on the worker who is assigned to it.

The matrix below shows the time required (in minutes) for each combination of a worker and a job. The jobs are denoted by J1, J2, J3, and J4, the workers by W1, W2, W3, and W4.

Each worker should perform exactly one job and the objective is to minimize the total time required to perform all jobs.

It turns out to be optimal to assign worker 1 to job 3, worker 2 to job 2, worker 3 to job 1 and worker 4 to job 4. The total time required is then 69 + 37 + 11 + 23 = 140 minutes. All other assignments lead to a larger amount of time required.

The Hungarian algorithm can be used to find this optimal assignment. The steps of the Hungarian algorithm can be found here , and an explanation of the Hungarian algorithm based on the example above can be found here .

HungarianAlgorithm.com © 2013-2024

Algorithms: The Assignment Problem

One of the interesting things about studying optimization is that the techniques show up in a lot of different areas. The “assignment problem” is one that can be solved using simple techniques, at least for small problem sizes, and is easy to see how it could be applied to the real world.

Assignment Problem

Pretend for a moment that you are writing software for a famous ride sharing application. In a crowded environment, you might have multiple prospective customers that are requesting service at the same time, and nearby you have multiple drivers that can take them where they need to go. You want to assign the drivers to the customers in a way that minimizes customer wait time (so you keep the customers happy) and driver empty time (so you keep the drivers happy).

The assignment problem is designed for exactly this purpose. We start with m agents and n tasks. We make the rule that every agent has to be assigned to a task. For each agent-task pair, we figure out a cost associated to have that agent perform that task. We then figure out which assignment of agents to tasks minimizes the total cost.

Of course, it may be true that m != n , but that’s OK. If there are too many tasks, we can make up a “dummy” agent that is more expensive than any of the others. This will ensure that the least desirable task will be left to the dummy agent, and we can remove that from the solution. Or, if there are too many agents, we can make up a “dummy” task that is free for any agent. This will ensure that the agent with the highest true cost will get the dummy task, and will be idle.

If that last paragraph was a little dense, don’t worry; there’s an example coming that will help show how it works.

There are special algorithms for solving assignment problems, but one thing that’s nice about them is that a general-purpose solver can handle them too. Below is an example, but first it will help to cover a few concepts that we’ll be using.

Optimization Problems

Up above, we talked about making “rules” and minimizing costs. The usual name for this is optimization. An optimization problem is one where we have an “objective function” (which tells us what our goals are) and one or more “constraint functions” (which tell us what the rules are). The classic example is a factory that can make both “widgets” and “gadgets”. Each “widget” and “gadget” earns a certain amount of profit, but it also uses up raw material and time on the factory’s machines. The optimization problem is to determine exactly how many “widgets” and how many “gadgets” to make to maximize profit (the objective) while fitting within the material and time available (the constraints).

If we were to write this simple optimization problem out, it might look like this:

In this case, we have two variables: g for the number of gadgets we make and w for the number of widgets we make. We also have three constraints that we have to meet. Note that they are inequalities; we might not use all the available material or time in our optimal solution.

Just to unpack this a little: in English, the above is saying that we make 45 dollars / euros / quatloos per gadget we make. However, to make a gadget needs 120 lbs of raw material 1, 80 lbs of raw material 2, and 3.8 hours of machine time. So there is a limit on how many gadgets we can make, and it might be a better use of resources to balance gadgets with widgets.

Of course, real optimization problems have many more than two variables and many constraint functions, making them much harder to solve. The easiest kind of optimization problem to solve is linear, and fortunately, the assignment problem is linear.

Linear Programming

A linear program is a kind of optimization problem where both the objective function and the constraint functions are linear. (OK, that definition was a little self-referential.) We can have as many variables as we want, and as many constraint functions as we want, but none of the variables can have exponents in any of the functions. This limitation allows us to apply very efficient mathematical approaches to solve the problem, even for very large problems.

We can state the assignment problem as a linear programming problem. First, we choose to make “i” represent each of our agents (drivers) and “j” to represent each of our tasks (customers). Now, to write a problem like this, we need variables. The best approach is to use “indicator” variables, where xij = 1 means “driver i picks up customer j” and xij = 0 means “driver i does not pick up customer j”.

We wind up with:

This is a compact mathematical way to describe the problem, so again let me put it in English.

First, we need to figure out the cost of having each driver pick up each customer. Then, we can calculate the total cost for any scenario by just adding up the costs for the assignments we pick. For any assignment we don’t pick, xij will equal zero, so that term will just drop out of the sum.

Of course, the way we set up the objective function, the cheapest solution is for no drivers to pick up any customers. That’s not a very good business model. So we need a constraint to show that we want to have a driver assigned to every customer. At the same time, we can’t have a driver assigned to mutiple customers. So we need a constraint for that too. That leads us to the two constraints in the problem. The first just says, if you add up all the assignments for a given driver, you want the total number of assignments for that driver to be exactly one. The second constraint says, if you add up all the assignments to a given customer, you want the total number of drivers assigned to the customer to be one. If you have both of these, then each driver is assigned to exactly one customer, and the customers and drivers are happy. If you do it in a way that minimizes costs, then the business is happy too.

Solving with Octave and GLPK

The GNU Linear Programming Kit is a library that solves exactly these kinds of problems. It’s easy to set up the objective and constraints using GNU Octave and pass these over to GLPK for a solution.

Given some made-up sample data, the program looks like this:

Start with the definition of “c”, the cost information. For this example, I chose to have four drivers and three customers. There are sixteen numbers there; the first four are the cost of each driver to get the first customer, the next four are for the second customer, and the next four are for the third customer. Because we have an extra driver, we add a “dummy” customer at the end that is zero cost. This represents one of the drivers being idle.

The next definition is “b”, the right-hand side of our constraints. There are eight constraints, one for each of the drivers, and one for each of the customers (including the dummy). For each constraint, the right-hand side is 1.

The big block in the middle defines our constraint matrix “a”. This is the most challenging part of taking the mathematical definition and putting it into a form that is usable by GLPK; we have to expand out each constraint. Fortunately, in these kinds of cases, we tend to get pretty patterns that help us know we’re on the right track.

The first line in “a” says that the first customer needs a driver. To see why, remember that in our cost information, the first four numbers are the cost for each driver to get the first customer. With this constraint, we are requiring that one of those four costs be included and therefore that a driver is “selected” for the first customer. The other lines in “a” work similarly; the last four ensure that each driver has an assignment.

Note that the number of rows in “a” matches the number of items in “b”, and the number of columns in “a” matches the number of items in “c”. This is important; GLPK won’t run if this is not true (and our problem isn’t stated right in any case).

Compared to the above, the last few lines are easy.

• “lb” gives the lower bound for each variable.
• “ub” gives the upper bound.
• “ctype” tells GLPK that each constraint is an equality (“strict” as opposed to providing a lower or upper bound).
• “vartype” tells GLPK that these variables are all integers (can’t have half a driver showing up).
• “s” tells GLPK that we want to minimize our costs, not maximize them.

We push all that through a function call to GLPK, and what comes back are two values (along with some other stuff I’ll exclude for clarity):

The first item tells us that our best solution takes 27 minutes, or dollars, or whatever unit we used for cost. The second item tells us the assignments we got. (Note for pedants: I transposed this output to save space.)

This output tells us that customer 1 gets driver 2, customer 2 gets driver 3, customer 3 gets driver 4, and driver 1 is idle. If you look back at the cost data, you can see this makes sense, because driver 1 had some of the most expensive times to the three customers. You can also see that it managed to pick the least expensive pairing for each customer. (Of course, if I had done a better job making up cost data, it might not have picked the least expensive pairing in all cases, because a suboptimal individual pairing might still lead to an overall optimal solution. But this is a toy example.)

Of course, for a real application, we would have to take into consideration many other factors, such as the passage of time. Rather than knowing all of our customers and drivers up front, we would have customers and drivers continually showing up and being assigned. But I hope this simple example has revealed some of the concepts behind optimization and linear programming and the kinds of real-world problems that can be solved.

• Branch and Bound Tutorial
• Backtracking Vs Branch-N-Bound
• 0/1 Knapsack
• 8 Puzzle Problem
• Job Assignment Problem
• N-Queen Problem
• Travelling Salesman Problem

Job Assignment Problem using Branch And Bound

Let there be N workers and N jobs. Any worker can be assigned to perform any job, incurring some cost that may vary depending on the work-job assignment. It is required to perform all jobs by assigning exactly one worker to each job and exactly one job to each agent in such a way that the total cost of the assignment is minimized.

Let us explore all approaches for this problem.

Solution 1: Brute Force  

We generate n! possible job assignments and for each such assignment, we compute its total cost and return the less expensive assignment. Since the solution is a permutation of the n jobs, its complexity is O(n!).

Solution 2: Hungarian Algorithm  

The optimal assignment can be found using the Hungarian algorithm. The Hungarian algorithm has worst case run-time complexity of O(n^3).

Solution 3: DFS/BFS on state space tree  

A state space tree is a N-ary tree with property that any path from root to leaf node holds one of many solutions to given problem. We can perform depth-first search on state space tree and but successive moves can take us away from the goal rather than bringing closer. The search of state space tree follows leftmost path from the root regardless of initial state. An answer node may never be found in this approach. We can also perform a Breadth-first search on state space tree. But no matter what the initial state is, the algorithm attempts the same sequence of moves like DFS.

Solution 4: Finding Optimal Solution using Branch and Bound  

The selection rule for the next node in BFS and DFS is “blind”. i.e. the selection rule does not give any preference to a node that has a very good chance of getting the search to an answer node quickly. The search for an optimal solution can often be speeded by using an “intelligent” ranking function, also called an approximate cost function to avoid searching in sub-trees that do not contain an optimal solution. It is similar to BFS-like search but with one major optimization. Instead of following FIFO order, we choose a live node with least cost. We may not get optimal solution by following node with least promising cost, but it will provide very good chance of getting the search to an answer node quickly.

There are two approaches to calculate the cost function:  

• For each worker, we choose job with minimum cost from list of unassigned jobs (take minimum entry from each row).
• For each job, we choose a worker with lowest cost for that job from list of unassigned workers (take minimum entry from each column).

In this article, the first approach is followed.

Let’s take below example and try to calculate promising cost when Job 2 is assigned to worker A. 

Since Job 2 is assigned to worker A (marked in green), cost becomes 2 and Job 2 and worker A becomes unavailable (marked in red). 

Now we assign job 3 to worker B as it has minimum cost from list of unassigned jobs. Cost becomes 2 + 3 = 5 and Job 3 and worker B also becomes unavailable. 

Finally, job 1 gets assigned to worker C as it has minimum cost among unassigned jobs and job 4 gets assigned to worker D as it is only Job left. Total cost becomes 2 + 3 + 5 + 4 = 14. 

Below diagram shows complete search space diagram showing optimal solution path in green. 

Complete Algorithm:  

Below is the implementation of the above approach:

Time Complexity: O(M*N). This is because the algorithm uses a double for loop to iterate through the M x N matrix.  Auxiliary Space: O(M+N). This is because it uses two arrays of size M and N to track the applicants and jobs.

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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: