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.
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.
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.
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.
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.
Column-wise Reduction Matrix
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.
Therefore, the optimal solution is:
The Hungarian method is a combinatorial optimizationalgorithm that solves the assignment problem in polynomial time and which anticipated later primal-dual methods. It was developed and published in 1955 by Harold Kuhn, who gave the name "Hungarian method" because the algorithm was largely based on the earlier works of two Hungarian mathematicians: Dénes Kőnig and Jenő Egerváry.
James Munkres reviewed the algorithm in 1957 and observed that it is (strongly) polynomial. Since then the algorithm has been known also as the Kuhn–Munkres algorithm or Munkres assignment algorithm. The time complexity of the original algorithm was , however Edmonds and Karp, and independently Tomizawa noticed that it can be modified to achieve an running time. Ford and Fulkerson extended the method to general transportation problems. In 2006, it was discovered that Carl Gustav Jacobi had solved the assignment problem in the 19th century, and the solution had been published posthumously in 1890 in Latin.
Simple explanation of the assignment problem
In this simple example there are three workers: Armond, Francine, and Herbert. One of them has to clean the bathroom, another sweep the floors and the third washes the windows, but they each demand different pay for the various tasks. The problem is to find the lowest-cost way to assign the jobs. The problem can be represented in a matrix of the costs of the workers doing the jobs. For example:
|Clean bathroom||Sweep floors||Wash windows|
The Hungarian method, when applied to the above table, would give the minimum cost: this is $6, achieved by having Armond clean the bathroom, Francine sweep the floors, and Herbert wash the windows.
We are given a nonnegative n×nmatrix, where the element in the i-th row and j-th column represents the cost of assigning the j-th job to the i-th worker. We have to find an assignment of the jobs to the workers that has minimum cost. If the goal is to find the assignment that yields the maximum cost, the problem can be altered to fit the setting by replacing each cost with the maximum cost subtracted by the cost.
The algorithm is easier to describe if we formulate the problem using a bipartite graph. We have a complete bipartite graph with worker vertices () and job vertices (), and each edge has a nonnegative cost . We want to find a perfect matching with minimum cost.
Let us call a function a potential if for each . The value of potential is . It can be seen that the cost of each perfect matching is at least the value of each potential. The Hungarian method finds a perfect matching and a potential with equal cost/value which proves the optimality of both. In fact it finds a perfect matching of tight edges: an edge is called tight for a potential if . Let us denote the subgraph of tight edges by . The cost of a perfect matching in (if there is one) equals the value of .
The algorithm in terms of bipartite graphs
During the algorithm we maintain a potential y and an orientation of (denoted by ) which has the property that the edges oriented from T to S form a matching M. Initially, y is 0 everywhere, and all edges are oriented from S to T (so M is empty). In each step, either we modify y so that its value increases, or modify the orientation to obtain a matching with more edges. We maintain the invariant that all the edges of M are tight. We are done if M is a perfect matching.
In a general step, let and be the vertices not covered by M (so consists of the vertices in S with no incoming edge and consists of the vertices in T with no outgoing edge). Let be the set of vertices reachable in from by a directed path only following edges that are tight. This can be computed by breadth-first search.
If is nonempty, then reverse the orientation of a directed path in from to . Thus the size of the corresponding matching increases by 1.
If is empty, then let . is positive because there are no tight edges between and . Increase y by on the vertices of and decrease y by on the vertices of . The resulting y is still a potential. The graph changes, but it still contains M. We orient the new edges from S to T. By the definition of the set Z of vertices reachable from increases (note that the number of tight edges does not necessarily increase).
We repeat these steps until M is a perfect matching, in which case it gives a minimum cost assignment. The running time of this version of the method is : M is augmented n times, and in a phase where M is unchanged, there are at most n potential changes (since Z increases every time). The time needed for a potential change is .
Given workers and tasks, and an n×n matrix containing the cost of assigning each worker to a task, find the cost minimizing assignment.
First the problem is written in the form of a matrix as given below
a1 a2 a3 a4 b1 b2 b3 b4 c1 c2 c3 c4 d1 d2 d3 d4
where a, b, c and d are the workers who have to perform tasks 1, 2, 3 and 4. a1, a2, a3, a4 denote the penalties incurred when worker "a" does task 1, 2, 3, 4 respectively. The same holds true for the other symbols as well. The matrix is square, so each worker can perform only one task.
Then we perform row operations on the matrix. To do this, the lowest of all ai (i belonging to 1-4) is taken and is subtracted from each element in that row. This will lead to at least one zero in that row (We get multiple zeros when there are two equal elements which also happen to be the lowest in that row). This procedure is repeated for all rows. We now have a matrix with at least one zero per row. Now we try to assign tasks to agents such that each agent is doing only one task and the penalty incurred in each case is zero. This is illustrated below.
0 a2' a3' a4' b1' b2' b3' 0 c1' 0 c3' c4' d1' d2' 0 d4'
The zeros that are indicated as 0' are the assigned tasks.
Sometimes it may turn out that the matrix at this stage cannot be used for assigning, as is the case for the matrix below.
0 a2' a3' a4' b1' b2' b3' 0 0 c2' c3' c4' d1' 0 d3' d4'
In the above case, no assignment can be made. Note that task 1 is done efficiently by both agent a and c. Both can't be assigned the same task. Also note that no one does task 3 efficiently. To overcome this, we repeat the above procedure for all columns (i.e. the minimum element in each column is subtracted from all the elements in that column) and then check if an assignment is possible.
In most situations this will give the result, but if it is still not possible then we need to keep going.
All zeros in the matrix must be covered by marking as few rows and/or columns as possible. The following procedure is one way to accomplish this:
First, assign as many tasks as possible.
- Row 1 has one zero, so it is assigned. The 0 in row 3 is crossed out because it is in the same column.
- Row 2 has one zero, so it is assigned.
- Row 3's only zero has been crossed out, so nothing is assigned.
- Row 4 has two uncrossed zeros. Either one can be assigned (both are optimum), and the other zero would be crossed out.
Alternatively, the 0 in row 3 may be assigned, causing the 0 in row 1 to be crossed instead.
0' a2' a3' a4' b1' b2' b3' 0' 0 c2' c3' c4' d1' 0' 0 d4'
Now to the drawing part.
- Mark all rows having no assignments (row 3).
- Mark all (unmarked) columns having zeros in newly marked row(s) (column 1).
- Mark all rows having assignments in newly marked columns (row 1).
- Repeat for all non-assigned rows.
× 0' a2' a3' a4' × b1' b2' b3' 0' 0 c2' c3' c4' × d1' 0' 0 d4'
Now draw lines through all marked columns and unmarked rows.
× 0' a2' a3' a4' × b1' b2' b3' 0' 0 c2' c3' c4' × d1' 0' 0 d4'
The aforementioned detailed description is just one way to draw the minimum number of lines to cover all the 0s. Other methods work as well.
From the elements that are left, find the lowest value. Subtract this from every unmarked element and add it to every element covered by two lines.
Repeat steps 3–4 until an assignment is possible; this is when the minimum number of lines used to cover all the 0s is equal to max(number of people, number of assignments), assuming dummy variables (usually the max cost) are used to fill in when the number of people is greater than the number of assignments.
Basically you find the second minimum cost among the remaining choices. The procedure is repeated until you are able to distinguish among the workers in terms of least cost.
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- Mordecai J. Golin, Bipartite Matching and the Hungarian Method, Course Notes, Hong Kong University of Science and Technology.
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- Mike Dawes, The Optimal Assignment Problem, Course notes, University of Western Ontario.
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- Lecture: Fundamentals of Operations Research - Assignment Problem - Hungarian Algorithm, Prof. G. Srinivasan, Department of Management Studies, IIT Madras.
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- Solve any Assignment Problem online, provides a step by step explanation of the Hungarian Algorithm.
- ^Harold W. Kuhn, "The Hungarian Method for the assignment problem", Naval Research Logistics Quarterly, 2: 83–97, 1955. Kuhn's original publication.
- ^Harold W. Kuhn, "Variants of the Hungarian method for assignment problems", Naval Research Logistics Quarterly, 3: 253–258, 1956.
- ^J. Munkres, "Algorithms for the Assignment and Transportation Problems", Journal of the Society for Industrial and Applied Mathematics, 5(1):32–38, 1957 March.