- DOI: 10.1007/978-3-540-68279-0_2
- Corpus ID: 9426884

## The Hungarian method for the assignment problem

## 9,842 Citations

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## Hungarian Method

## Hungarian Method to Solve Assignment Problems

## What is an Assignment Problem?

## Hungarian Method Steps

- 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).

(d) Continue with (b) and (c) until no further marking is needed.

Step 7 – Continue with steps 1 – 6 until you’ve found the highest suitable assignment.

## Hungarian Method Example

With 5 jobs and 5 men, the stated problem is balanced.

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.

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## Hungarian Algorithm for Assignment Problem | Set 2 (Implementation)

Input: arr[][] = {{3, 5}, {10, 1}} Output: 4 Explanation: The optimal assignment is to assign job 1 to the 1st worker, job 2 to the 2nd worker. Hence, the optimal cost is 3 + 1 = 4. Input: arr[][] = {{2500, 4000, 3500}, {4000, 6000, 3500}, {2000, 4000, 2500}} Output: 4 Explanation: The optimal assignment is to assign job 2 to the 1st worker, job 3 to the 2nd worker and job 1 to the 3rd worker. Hence, the optimal cost is 4000 + 3500 + 2000 = 9500.

Different approaches to solve this problem are discussed in this article .

- For each row of the matrix, find the smallest element and subtract it from every element in its row.
- Repeat the step 1 for all columns.
- Cover all zeros in the matrix using the minimum number of horizontal and vertical lines.
- Test for Optimality : If the minimum number of covering lines is N , an optimal assignment is possible. Else if lines are lesser than N , an optimal assignment is not found 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.

Consider an example to understand the approach:

Let the 2D array be: 2500 4000 3500 4000 6000 3500 2000 4000 2500 Step 1: Subtract minimum of every row. 2500, 3500 and 2000 are subtracted from rows 1, 2 and 3 respectively. 0 1500 1000 500 2500 0 0 2000 500 Step 2: Subtract minimum of every column. 0, 1500 and 0 are subtracted from columns 1, 2 and 3 respectively. 0 0 1000 500 1000 0 0 500 500 Step 3: Cover all zeroes with minimum number of horizontal and vertical lines. Step 4: Since we need 3 lines to cover all zeroes, the optimal assignment is found. 2500 4000 3500 4000 6000 3500 2000 4000 2500 So the optimal cost is 4000 + 3500 + 2000 = 9500

Below is the implementation of the above approach:

Time Complexity: O(N 3 ) Auxiliary Space: O(N 2 )

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## Although the Hungarian method is an efficient methodfor solving an assignment problem, the branch-and-boundmethod can also be used to solve an assignment problem.Suppose a company has five factories and five warehouses.Each factory’s requirements must be met by a singlewarehouse, and each warehouse can be assigned to only onefactory. The costs of assigning a warehouse to meet afactory’s demand (in thousands) are shown in Table 77.Let xij 1 if warehouse i is assigned to factory j and 0otherwise. Begin by branching on the warehouse assigned tofactory 1. This creates the following five branches: x11 1,x21 1, x31 1, x41 1, and x51 1. How can we obtaina lower bound on the total cost associated with a branch?Examine the branch x21 1. If x21 1, no furtherassignments can come from row 2 or column 1 of the costmatrix. In determining the factory to which each of theunassigned warehouses (1, 3, 4, and 5) is assigned, we cannotdo better than assign each to the smallest cost in thewarehouse’s row (excluding the factory 1 column). Thus, theminimum-cost assignment having x21 1 must have a totalcost of at least 10 10 9 5 5 39.Similarly, in determining the warehouse to which eachof the unassigned factories (2, 3, 4, and 5) is assigned, wecannot do better than to assign each to the smallest cost inthe factory’s column (excluding the warehouse 2 row). Thus,the minimum-cost assignment having x21 1 must have atotal cost of at least 10 9 5 5 7 36. Thus, thetotal cost of any assignment having x21 1 must be at leastmax(36, 39) 39. So, if branching ever leads to a candidatesolution having a total cost of 39 or less, the x21 1 branchmay be eliminated from consideration. Use this idea to solvethe problem by branch-and-bound.

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## The Hungarian method for solving an assignment problem can also be...

## The Hungarian method for solving an assignment problem can also be used to solve.

B. A travelling salesman problem

## Solution(By Examveda Team)

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## Hungarian Maximum Matching Algorithm

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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.

## The Hungarian Algorithm Using an Adjacency Matrix

The hungarian algorithm using a graph.

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. \(_\square\)

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. \(_\square\)

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

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

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).

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

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

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 &amp; 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

## 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)

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

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

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.

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

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

The optimal assignment (minimum) cost

∴ The given assignment problem is balanced.

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

Solve the following assignment problem.

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.

The optimal assignment (minimum) cost = ₹ 35

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## Hungarian Algorithm for Assignment Problem

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The Hungarian method is a combinatorial optimization algorithm that solves the assignment problem in polynomial time and which anticipated later primal-dual methods.

The Hungarian Method: The following algorithm applies the above theorem to a given n × n cost matrix to ﬁnd an optimal assignment. Step 1. Subtract the smallest entry in each row from all the entries of its row. Step 2. Subtract the smallest entry in each column from all the entries of its column. Step 3.

THE HUNGARIAN METHOD FOR THE ASSIGNMENT PROBLEM' H. W. Kuhn Bryn Yaw College Assuming that numerical scores are available for the perform- ance of each of n persons on each of n jobs, the "assignment problem" is the quest for an assignment of persons to jobs so that sum n scores so obtained is as large as possible.

The Hungarian algorithm, aka Munkres assignment algorithm, utilizes the following theorem for polynomial runtime complexity ( worst case O (n3)) and guaranteed optimality: 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 …

The Hungarian algorithm for assignment problem to solve traveling salesman problem is applied and tree examples of application are included. 2 Highly Influenced PDF View 4 excerpts A Simulation of the Faculty-Assignment Problem: An Integer Programming Approach Chin W. Yang, P. Y. Kim Business 2015 TLDR

The Hungarian method for the assignment problem † H. W. Kuhn First published: March 1955 https://doi.org/10.1002/nav.3800020109 Citations: 6,025 † The preparation of this report was supported, in part, by the ONR Logistics Project, Department of Mathematics, Princeton University. PDF Tools Share Abstract

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.

This function is an implementation of the Hungarian algorithm (also known as the Kuhn-Munkres algorithm) which runs in O (N3) time. It solves the optimal assignment problem. Below is the implementation of the above approach: Python import dlib def minCost (arr): assignment = dlib.max_cost_assignment (arr)

The classical solution to the assignment problem is given by the Hungarian or Kuhn-Munkres algorithm, originally proposed by H. W. Kuhn in 1955 [3] and reﬁned by J. Munkres in 1957 [5]. The Hungarian algorithm solves the assignment problem in O(n3) time, where n is the size of one partition of the bipartite graph. This and other

The hungarian method for the assignment problem can help students to understand the material and improve their grades. Solve Now. Hungarian algorithm 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

The Assignment Problem (Using Hungarian Algorithm) | by Riya Tendulkar | Medium 500 Apologies, but something went wrong on our end. Refresh the page, check Medium 's site status, or find...

The classical method for solving this problem is Kuhn's Hungarian method [Kuh55]. This method is of major theoretical interest and is still used widely. It maintains a price for each object and an (incomplete) assignment of persons and objects. At each iteration, the method chooses an unassigned person and computes a

The Hungarian Method for the Assignment Problem Harold W. Kuhn Chapter First Online: 01 January 2009 8348 Accesses 163 Citations Abstract This paper has always been one of my favorite "children," combining as it does elements of the duality of linear programming and combinatorial tools from graph theory.

Use the Hungarian method to determine the optimal assignments. Solution In the given problem there are 5 operators and 5 Lathe. The problem can be formulated as 5 × 5 assignment problem with cij = weekly output (in pieces) from jth Lathe by ith operator. Let xij = {1, if jth Lathe is assigned to ith Operator; 0, otherwise.

In this lesson we learn what is an assignment problem and how we can solve it using the Hungarian method.

Although the assignment problem can be solved as an ordinary transportation problem or as Linear programming problem, its unique structure can be exploited, resulting in special purpose algorithm, is called Hungarian method. A numerical example is provided to illustrate the solution procedure developed in this paper.

The Primal Linear Program for Assignment Problem. Image by Author. An n×n matrix of elements rᵢⱼ (i, j = 1, 2, …, n) can be represented as a bipartite graph, G(U,V; E) with edge weights ...

Obviously, these edges will be the solution of the assignment problem. If we can't find perfect matching on the current step, then the Hungarian algorithm changes weights of the available edges in such a way that the new 0-weight edges appear and these changes do not influence the optimal solution.

The Hungarian method is a combinatorial optimization algorithm that solves the assignment problem in polynomial time and which anticipated later primal-dual Get math help online Improve your scholarly performance

Assignment problem Hungarian method example Phase 1. Step 0: Consider the given matrix. Phase 2: Step 3: Reduce the new matrix column-wise using the same Solve Now The Assignment Problem: An Example This section presents an example that shows how to solve an assignment problem using both the MIP solver and the CP-SAT solver. ...

Assignment problem methods - This paper is concerned with approximation methods for handling the classical assignment problem. These methods permit solution of ... Solution of assignment problems (Hungarian Method) Assignment model is a special application of Linear Programming Problem (LPP), in which the main objective is to assign the work or ...

The Online Stochastic Generalized Assignment Problem The solution of the transport problem by the potential method. Complete, detailed, step-by-step description of solutions. Hungarian method, dual simplex 834+ Math Tutors 88% Recurring customers 24771 Clients Get Homework Help

Although the Hungarian method is an efficient methodfor solving an assignment problem, the branch-and-boundmethod can also be used to solve an assignment problem.Suppose a company has five factories and five warehouses.Each factory's requirements must be met by a singlewarehouse, and each warehouse can be assigned to only onefactory.

The Hungarian method for solving an assignment problem can also be used to solve a travelling salesman problem. The Hungarian method is a combinatorial optimization algorithm that solves the assignment problem in polynomial time and which anticipated later primal-dual methods.

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 ...

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.

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 Hungarian Method THE HUNGARIAN METHOD FOR THE ASSIGNMENT PROBLEM'. H. W. Kuhn. Bryn Yaw College. Assuming that numerical scores are available for the perform-.

Here is the video about assignment problem - Hungarian method on Operations research, In this video we discussed what is assignment problem and how to solve ...

The method used for solving an assignment problem is called Hungarian method. The Hungarian method is a combinatorial optimization algorithm that solves the assignment problem in polynomial time and which anticipated later primal-dual methods.