

Hungarian Method
The Hungarian method is a computational optimization technique that addresses the assignment problem in polynomial time and foreshadows following primal-dual alternatives. In 1955, Harold Kuhn used the term “Hungarian method” to honour two Hungarian mathematicians, Dénes Kőnig and Jenő Egerváry. Let’s go through the steps of the Hungarian method with the help of a solved example.
Hungarian Method to Solve Assignment Problems
The Hungarian method is a simple way to solve assignment problems. Let us first discuss the assignment problems before moving on to learning the Hungarian method.
What is an Assignment Problem?
A transportation problem is a type of assignment problem. The goal is to allocate an equal amount of resources to the same number of activities. As a result, the overall cost of allocation is minimised or the total profit is maximised.
Because available resources such as workers, machines, and other resources have varying degrees of efficiency for executing different activities, and hence the cost, profit, or loss of conducting such activities varies.
Assume we have ‘n’ jobs to do on ‘m’ machines (i.e., one job to one machine). Our goal is to assign jobs to machines for the least amount of money possible (or maximum profit). Based on the notion that each machine can accomplish each task, but at variable levels of efficiency.
Hungarian Method Steps
Check to see if the number of rows and columns are equal; if they are, the assignment problem is considered to be balanced. Then go to step 1. If it is not balanced, it should be balanced before the algorithm is applied.
Step 1 – In the given cost matrix, subtract the least cost element of each row from all the entries in that row. Make sure that each row has at least one zero.
Step 2 – In the resultant cost matrix produced in step 1, subtract the least cost element in each column from all the components in that column, ensuring that each column contains at least one zero.
Step 3 – Assign zeros
- Analyse the rows one by one until you find a row with precisely one unmarked zero. Encircle this lonely unmarked zero and assign it a task. All other zeros in the column of this circular zero should be crossed out because they will not be used in any future assignments. Continue in this manner until you’ve gone through all of the rows.
- Examine the columns one by one until you find one with precisely one unmarked zero. Encircle this single unmarked zero and cross any other zero in its row to make an assignment to it. Continue until you’ve gone through all of the columns.
Step 4 – Perform the Optimal Test
- The present assignment is optimal if each row and column has exactly one encircled zero.
- The present assignment is not optimal if at least one row or column is missing an assignment (i.e., if at least one row or column is missing one encircled zero). Continue to step 5. Subtract the least cost element from all the entries in each column of the final cost matrix created in step 1 and ensure that each column has at least one zero.
Step 5 – Draw the least number of straight lines to cover all of the zeros as follows:
(a) Highlight the rows that aren’t assigned.
(b) Label the columns with zeros in marked rows (if they haven’t already been marked).
(c) Highlight the rows that have assignments in indicated columns (if they haven’t previously been marked).
(d) Continue with (b) and (c) until no further marking is needed.
(f) Simply draw the lines through all rows and columns that are not marked. If the number of these lines equals the order of the matrix, then the solution is optimal; otherwise, it is not.
Step 6 – Find the lowest cost factor that is not covered by the straight lines. Subtract this least-cost component from all the uncovered elements and add it to all the elements that are at the intersection of these straight lines, but leave the rest of the elements alone.
Step 7 – Continue with steps 1 – 6 until you’ve found the highest suitable assignment.
Hungarian Method Example
Use the Hungarian method to solve the given assignment problem stated in the table. The entries in the matrix represent each man’s processing time in hours.
\(\begin{array}{l}\begin{bmatrix} & I & II & III & IV & V \\1 & 20 & 15 & 18 & 20 & 25 \\2 & 18 & 20 & 12 & 14 & 15 \\3 & 21 & 23 & 25 & 27 & 25 \\4 & 17 & 18 & 21 & 23 & 20 \\5 & 18 & 18 & 16 & 19 & 20 \\\end{bmatrix}\end{array} \)
With 5 jobs and 5 men, the stated problem is balanced.
\(\begin{array}{l}A = \begin{bmatrix}20 & 15 & 18 & 20 & 25 \\18 & 20 & 12 & 14 & 15 \\21 & 23 & 25 & 27 & 25 \\17 & 18 & 21 & 23 & 20 \\18 & 18 & 16 & 19 & 20 \\\end{bmatrix}\end{array} \)
Subtract the lowest cost element in each row from all of the elements in the given cost matrix’s row. Make sure that each row has at least one zero.
\(\begin{array}{l}A = \begin{bmatrix}5 & 0 & 3 & 5 & 10 \\6 & 8 & 0 & 2 & 3 \\0 & 2 & 4 & 6 & 4 \\0 & 1 & 4 & 6 & 3 \\2 & 2 & 0 & 3 & 4 \\\end{bmatrix}\end{array} \)
Subtract the least cost element in each Column from all of the components in the given cost matrix’s Column. Check to see if each column has at least one zero.
\(\begin{array}{l}A = \begin{bmatrix}5 & 0 & 3 & 3 & 7 \\6 & 8 & 0 & 0 & 0 \\0 & 2 & 4 & 4 & 1 \\0 & 1 & 4 & 4 & 0 \\2 & 2 & 0 & 1 & 1 \\\end{bmatrix}\end{array} \)
When the zeros are assigned, we get the following:

The present assignment is optimal because each row and column contain precisely one encircled zero.
Where 1 to II, 2 to IV, 3 to I, 4 to V, and 5 to III are the best assignments.
Hence, z = 15 + 14 + 21 + 20 + 16 = 86 hours is the optimal time.
Practice Question on Hungarian Method
Use the Hungarian method to solve the following assignment problem shown in table. The matrix entries represent the time it takes for each job to be processed by each machine in hours.
\(\begin{array}{l}\begin{bmatrix}J/M & I & II & III & IV & V \\1 & 9 & 22 & 58 & 11 & 19 \\2 & 43 & 78 & 72 & 50 & 63 \\3 & 41 & 28 & 91 & 37 & 45 \\4 & 74 & 42 & 27 & 49 & 39 \\5 & 36 & 11 & 57 & 22 & 25 \\\end{bmatrix}\end{array} \)
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Frequently Asked Questions on Hungarian Method
What is hungarian method.
The Hungarian method is defined as a combinatorial optimization technique that solves the assignment problems in polynomial time and foreshadowed subsequent primal–dual approaches.
What are the steps involved in Hungarian method?
The following is a quick overview of the Hungarian method: Step 1: Subtract the row minima. Step 2: Subtract the column minimums. Step 3: Use a limited number of lines to cover all zeros. Step 4: Add some more zeros to the equation.
What is the purpose of the Hungarian method?
When workers are assigned to certain activities based on cost, the Hungarian method is beneficial for identifying minimum costs.
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Assignment problem: Hungarian method 3

Assignment problem: Hungarian Method Nui Ruppert (Mtk_Nr.: 373224) David Lenh (Mtk_Nr.: 368343) Amir Farshchi Tabrizi (Mtk-Nr.: 372894)
In this OR-Wiki entry we're going to explain the Hungarian method with 3 examples. In the first example you'll find the optimal solution after a few steps with the help of the reduced matrix. The second example illustrates a complex case where you need to proceed all the steps of the algorithm to get to an optimal solution. Finally in the third example we will show how to solve a maximization problem with the Hungarian method.
Inhaltsverzeichnis
- 1 Introduction
- 2 Example 1 – Minimization problem
- 3 Example 2 – Minimazation problem
- 4 Example 3 – Maximization problem
- 6 References
Introduction
The Hungarian method is a combinatorial optimization algorithm which was developed and published by Harold Kuhn in 1955. This method was originally invented for the best assignment of a set of persons to a set of jobs. It is a special case of the transportation problem. The algorithm finds an optimal assignment for a given “n x n” cost matrix. “Assignment problems deal with the question how to assign n items (e.g. jobs) to n machines (or workers) in the best possible way. […] Mathematically an assignment is nothing else than a bijective mapping of a finite set into itself […]” [1]
The assignment constraints are mathematically defined as:
To make clear how to solve an assignment problem with the Hungarian algorithm we will show you the different cases with several examples which can occur .
Example 1 – Minimization problem
In this example we have to assign 4 workers to 4 machines. Each worker causes different costs for the machines. Your goal is to minimize the total cost to the condition that each machine goes to exactly 1 person and each person works at exactly 1 machine. For comprehension: Worker 1 causes a cost of 6 for machine 1 and so on …
To solve the problem we have to perform the following steps:
Step 1 – Subtract the row minimum from each row.
Step 2 – Subtract the column minimum from each column from the reduced matrix.
The idea behind these 2 steps is to simplify the matrix since the solution of the reduced matrix will be exactly the same as of the original matrix.
Step 3 – Assign one “0” to each row & column.
Now that we have simplified the matrix we can assign each worker with the minimal cost to each machine which is represented by a “0”.
- In the first row we have one assignable “0” therefore we assign it to worker 3 .
- In the second row we also only have one assignable “0” therefore we assign it to worker 4 .
- In the third row we have two assignable “0”. We leave it as it is for now.
- In the fourth row we have one assignable “0” therefore we assign it. Consider that we can only assign each worker to each machine hence we can’t allocate any other “0” in the first column.
- Now we go back to the third row which now only has one assignable “0” for worker 2 .
As soon as we can assign each worker to one machine, we have the optimal solution . In this case there is no need to proceed any further steps. Remember also, if we decide on an arbitrary order in which we start allocating the “0”s then we may get into a situation where we have 3 assignments as against the possible 4. If we assign a “0” in the third row to worker 1 we wouldn’t be able to allocate any “0”s in column one and row two.
The rule to assign the “0”:
- If there is an assignable “0”, only 1 assignable “0” in any row or any column, assign it.
- If there are more than 1, leave it and proceed.
This rule would try to give us as many assignments as possible.
Now there are also cases where you won’t get an optimal solution for a reduced matrix after one iteration. The following example will explain it.
Example 2 – Minimazation problem
In this example we have the fastest taxi company that has to assign each taxi to each passenger as fast as possible. The numbers in the matrix represent the time to reach the passenger.
We proceed as in the first example.
Iteration 1:
Now we have to assign the “0”s for every row respectively to the rule that we described earlier in example 1.
- In the first row we have one assignable “0” therefore we assign it and no other allocation in column 2 is possible.
- In the second row we have one assignable “0” therefore we assign it.
- In the third row we have several assignable “0”s. We leave it as it is for now and proceed.
- In the fourth and fifth row we have no assignable “0”s.
Now we proceed with the allocations of the “0”s for each column .
- In the first column we have one assignable “0” therefore we assign it. No other “0”s in row 3 are assignable anymore.
Now we are unable to proceed because all the “0”s either been assigned or crossed. The crosses indicate that they are not fit for assignments because assignments are already made.
We realize that we have 3 assignments for this 5x5 matrix. In the earlier example we were able to get 4 assignments for a 4x4 matrix. Now we have to follow another procedure to get the remaining 2 assignments (“0”).
Step 4 – Tick all unassigned rows.
Step 5 – If a row is ticked and has a “0”, then tick the corresponding column (if the column is not yet ticked).
Step 6 – If a column is ticked and has an assignment, then tick the corresponding row (if the row is not yet ticked).
Step 7 - Repeat step 5 and 6 till no more ticking is possible.
In this case there is no more ticking possible and we proceed with the next step.
Step 8 – Draw lines through unticked rows and ticked columns. The number of lines represents the maximum number of assignments possible.
Step 9 – Find out the smallest number which does not have any line passing through it. We call it Theta. Subtract theta from all the numbers that do not have any lines passing through them and add theta to all those numbers that have two lines passing through them. Keep the rest of them the same.
(With this step we create a new “0”)
With the new assignment matrix we start to assign the “0”s after the explained rules. Nevertheless we have 4 assignments against the required 5 for an optimal solution. Therefore we have to repeat step 4 – 9.
Iteration 2:
Step 4 – Tick all unassigned row.
Note: The indices of the ticks show you the order we added them.
Iteration 3:
Iteration 4:
After the fourth iteration we assign the “0”s again and now we have an optimal solution with 5 assignments.
The solution:
- Taxi1 => Passenger1 - duration 12
- Taxi2 => Passenger4 - duration 11
- Taxi3 => Passenger2 - duration 8
- Taxi4 => Passenger3 - duration 14
- Taxi5 => Passenger5 - duration 11
If we define the needed duration as costs, the minimal cost for this problem is 56.
Example 3 – Maximization problem
Furthermore the Hungarian algorithm can also be used for a maximization problem in which case we first have to transform the matrix. For example a company wants to assign different workers to different machines. Each worker is more or less efficient with each machine. The efficiency can be defined as profit. The higher the number, the higher the profit.
As you can see, the maximal profit of the matrix is 13. The simple twist that we do is rather than try to maximize the profit, we’re going to try to minimize the profit that you don’t get. If every value is taken away from 13, then we can minimize the amount of profit lost. We receive the following matrix:
From now on we proceed as usual with the steps to get to an optimal solution.
With the determined optimal solution we can compute the maximal profit:
- Worker1 => Machine2 - 9
- Worker2 => Machine4 - 11
- Worker3 => Machine3 - 13
- Worker4 => Machine1 - 7
Maximal profit is 40.
The optimal solution is found if there is one assigned “0” for each row and each column.
[1] Linear Assignment Problems and Extensions, Rainer E. Burkard, Eranda Cela
[2] Operations Research Skript TU Kaiserslautern, Prof. Dr. Oliver Wendt
[3] The Hungarian method for the assignment problem, H. W. Kuhn, Bryn Mawr College
Fundamental of Operations Research, Lec. 16, Prof. G. Srinivasan
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- Discuss(20+)
- 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 This article is contributed by Yash Varyani . Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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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:

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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Assignment problem Hungarian method example
An assignment problem can be easily solved by applying Hungarian method which consists of two phases. In the first phase, row reductions and column reductions are carried out. In the second phase, the solution is optimized on iterative basis.
Step 0: Consider the given matrix. Step 1: In a given problem, if the number of rows is not equal to the number of columns and vice versa, then add a dummy row or a dummy column. The assignment costs for dummy cells are always assigned as zero. Step 2: Reduce the matrix by selecting the smallest element in each row and subtract with other elements in that row.
Step 3 : Reduce the new matrix column-wise using the same method as given in step 2. Step 4 : Draw minimum number of lines to cover all zeros. Step 5 : If Number of lines drawn = order of matrix, then optimally is reached, so proceed to step 7. If optimally is not reached, then go to step 6. Step 6: Select the smallest element of the whole matrix, which is NOT COVERED by lines. Subtract this smallest element with all other remaining elements that are NOT COVERED by lines and add the element at the intersection of lines. Leave the elements covered by single line as it is. Now go to step 4. Step 7: Take any row or column which has a single zero and assign by squaring it. Strike off the remaining zeros, if any, in that row and column (X). Repeat the process until all the assignments have been made. Step 8: Write down the assignment results and find the minimum cost/time.
Note: While assigning, if there is no single zero exists in the row or column, choose any one zero and assign it. Strike off the remaining zeros in that column or row, and repeat the same for other assignments also. If there is no single zero allocation, it means multiple numbers of solutions exist. But the cost will remain the same for different sets of allocations.
Example : Assign the four tasks to four operators. The assigning costs are given in Table.
Assignment Problem

Step 1: The given matrix is a square matrix and it is not necessary to add a dummy row/column Step 2: Reduce the matrix by selecting the smallest value in each row and subtracting from other values in that corresponding row. In row A, the smallest value is 13, row B is 15, row C is 17 and row D is 12. The row wise reduced matrix is shown in table below.
Row-wise Reduction

Step 3: Reduce the new matrix given in the following table by selecting the smallest value in each column and subtract from other values in that corresponding column. In column 1, the smallest value is 0, column 2 is 4, column 3 is 3 and column 4 is 0. The column-wise reduction matrix is shown in the following table.
Column-wise Reduction Matrix

Step 4: Draw minimum number of lines possible to cover all the zeros in the matrix given in Table
Matrix with all Zeros Covered

The first line is drawn crossing row C covering three zeros, second line is drawn crossing column 4 covering two zeros and third line is drawn crossing column 1 (or row B) covering a single zero. Step 5: Check whether number of lines drawn is equal to the order of the matrix, i.e., 3 ≠ 4. Therefore optimally is not reached. Go to step 6. Step 6: Take the smallest element of the matrix that is not covered by single line, which is 3. Subtract 3 from all other values that are not covered and add 3 at the intersection of lines. Leave the values which are covered by single line. The following table shows the details.
Subtracted or Added to Uncovered Values and Intersection Lines Respectively

Step 7: Now, draw minimum number of lines to cover all the zeros and check for optimality. Here in table minimum number of lines drawn is 4 which are equal to the order of matrix. Hence optimality is reached.
Optimality Matrix

Step 8: Assign the tasks to the operators. Select a row that has a single zero and assign by squaring it. Strike off remaining zeros if any in that row or column. Repeat the assignment for other tasks. The final assignment is shown in table below.
Final Assignment

Therefore, optimal assignment is:

Example : Solve the following assignment problem shown in Table using Hungarian method. The matrix entries are processing time of each man in hours.

Solution: The row-wise reductions are shown in Table
Row-wise Reduction Matrix

The column wise reductions are shown in Table.

Matrix with minimum number of lines drawn to cover all zeros is shown in Table.
Matrix will all Zeros Covered

The number of lines drawn is 5, which is equal to the order of matrix. Hence optimality is reached. The optimal assignments are shown in Table.
Optimal Assignment

Therefore, the optimal solution is:

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Assignment Method: Examples of How Resources Are Allocated
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What Is the Assignment Method?
The assignment method is a way of allocating organizational resources in which each resource is assigned to a particular task. The resource could be monetary, personnel , or technological.
Understanding the Assignment Method
The assignment method is used to determine what resources are assigned to which department, machine, or center of operation in the production process. The goal is to assign resources in such a way to enhance production efficiency, control costs, and maximize profits.
The assignment method has various applications in maximizing resources, including:
- Allocating the proper number of employees to a machine or task
- Allocating a machine or a manufacturing plant and the number of jobs that a given machine or factory can produce
- Assigning a number of salespersons to a given territory or territories
- Assigning new computers, laptops, and other expensive high-tech devices to the areas that need them the most while lower priority departments would get the older models
Companies can make budgeting decisions using the assignment method since it can help determine the amount of capital or money needed for each area of the company. Allocating money or resources can be done by analyzing the past performance of an employee, project, or department to determine the most efficient approach.
Regardless of the resource being allocated or the task to be accomplished, the goal is to assign resources to maximize the profit produced by the task or project.
Example of Assignment Method
A bank is allocating its sales force to grow its mortgage lending business. The bank has over 50 branches in New York but only ten in Chicago. Each branch has a staff that is used to bring in new clients.
The bank's management team decides to perform an analysis using the assignment method to determine where their newly-hired salespeople should be allocated. Given the past performance results in the Chicago area, the bank has produced fewer new clients than in New York. The fewer new clients are the result of having a small market presence in Chicago.
As a result, the management decides to allocate the new hires to the New York region, where it has a greater market share to maximize new client growth and, ultimately, revenue.
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MB0048 : State and discuss the methods for solving an assignment problem. How is Hungarian method better than other methods for solving an assignment problem?
Posted by: NIKHAT SHAHIN on: October 20, 2011
- In: Assignments | MB0048-Operations Research Assignments | MBA (SMU) | MBA 2nd Semester
MB0048 :State and discuss the methods for solving an assignment problem. How is Hungarian method better than other methods for solving an assignment problem? Answer : Assignment becomes a problem because each job requires different skills and the capacity or efficiency of each person with respect to these jobs can be different. This gives rise to cost differences. If each person is able to do all jobs with same efficiency then all costs will be the same and each job can be assigned to any person. When assignment is a problem it becomes a typical optimisation problem. Therefore, you can compare an assignment problem to a transportation problem.
Solution Methods
The assignment problem can be solved by the following four methods :
- Enumeration method
- Simplex method
Transportation method
Hungarian method
Enumeration method:
In this method, a list of all possible assignments among the given resources and activities is prepared. Then an assignment involving the minimum cost, time or distance or maximum profits is selected. If two or more assignments have the same minimum cost, time or distance, the problem has multiple optimal solutions. This method can be used only if the number of assignments is less. It becomes unsuitable for manual calculations if number of assignments is large
Simplex method:
The simplex method focuses on solving LPP of any enormity involving two or more decision variables.
The simplex algorithm is an iterative procedure for finding the optimal solution to a linear programming problem. The objective function controls the development and evaluation of each feasible solution to the problem. If a feasible solution exists, it is located at a corner point of the feasible region determined by the constraints of the system.
The simplex method simply selects the optimal solution amongst the set of feasible solutions of the problem. The efficiency of this algorithm is because it considers only those feasible solutions which are provided by the corner points, and that too not all of them. You can consider obtaining an optimal solution based on a minimum number of feasible solutions.
Transportation model is an important class of linear programs. For a given supply at each source and a given demand at each destination, the model studies the minimisation of the cost of transporting a commodity from a number of sources to several destinations.
As assignment is a special case of transportation problem it can also be solved using transportation model. But the degeneracy problem of solution makes the transportation method computationally inefficient for solving the assignment problem.
There are various ways to solve assignment problems. Certainly it can be formulated as a linear program (as we saw above), and the simplex method can be used to solve it. In addition, since it can be formulated as a network problem, the network simplex method may solve it quickly.
However, sometimes the simplex method is inefficient for assignment problems (particularly problems with a high degree of degeneracy). The Hungarian Algorithm developed by Kuhn has been used with a good deal of success on these problems and is summarized as follows.
Step 1. Determine the cost table from the given problem.
- If the no. of sources is equal to no. of destinations, go to step 3.
- If the no. of sources is not equal to the no. of destination, go to step2.
Step 2. Add a dummy source or dummy destination, so that the cost table becomes a square matrix. The cost entries of the dummy source/destinations are always zero.
Step 3. Locate the smallest element in each row of the given cost matrix and then subtract the same from each element of the row.
Step 4. In the reduced matrix obtained in the step 3, locate the smallest element of each column and then subtract the same from each element of that column. Each column and row now have at least one zero.
Step 5. In the modified matrix obtained in the step 4, search for the optimal assignment as follows:
(a) Examine the rows successively until a row with a single zero is found. Enrectangle this row ()and cross off (X) all other zeros in its column. Continue in this manner until all the rows have been taken care of.
(b) Repeat the procedure for each column of the reduced matrix.
(c) If a row and/or column has two or more zeros and one cannot be chosen by inspection then assign arbitrary any one of these zeros and cross off all other zeros of that row / column.
(d) Repeat (a) through (c) above successively until the chain of assigning () or cross (X) ends.
Step 6 . If the number of assignment () is equal to n (the order of the cost matrix), an optimum solution is reached.
If the number of assignment is less than n(the order of the matrix), go to the next step.
Step7. Draw the minimum number of horizontal and/or vertical lines to cover all the zeros of the reduced matrix.
Step 8. Develop the new revised cost matrix as follows:
(a)Find the smallest element of the reduced matrix not covered by any of the lines.
(b)Subtract this element from all uncovered elements and add the same to all the elements laying at the intersection of any two lines.
Step 9. Go to step 6 and repeat the procedure until an optimum solution is attained.
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1 response to "mb0048 : state and discuss the methods for solving an assignment problem. how is hungarian method better than other methods for solving an assignment problem".

Hey, do u have the codes of: Enumeration method, Simplex method or Transportation method

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