# Solve Assignment Problem Using Hungarian Method Of Assignment

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.

*Phase 1*

*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.

*Phase 2:*

*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*

*Solution:*

*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.

*Assignment Problem*

**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.

*Optimal Assignment*

Therefore, the optimal solution is:

The **assignment problem** is one of the fundamental combinatorial optimization problems in the branch of optimization or operations research in mathematics. It consists of finding a maximum weight matching (or minimum weight perfect matching) in a weightedbipartite graph.

In its most general form, the problem is as follows:

- The problem instance has a number of
*agents*and a number of*tasks*. Any agent can be assigned to perform any task, incurring some*cost*that may vary depending on the agent-task assignment. It is required to perform all tasks by assigning exactly one agent to each task and exactly one task to each agent in such a way that the*total cost*of the assignment is minimized.

If the numbers of agents and tasks are equal and the total cost of the assignment for all tasks is equal to the sum of the costs for each agent (or the sum of the costs for each task, which is the same thing in this case), then the problem is called the *linear assignment problem*. Commonly, when speaking of the *assignment problem* without any additional qualification, then the *linear assignment problem* is meant.

## Algorithms and generalizations[edit]

The Hungarian algorithm is one of many algorithms that have been devised that solve the linear assignment problem within time bounded by a polynomial expression of the number of agents. Other algorithms include adaptations of the primal simplex algorithm, and the auction algorithm.

The assignment problem is a special case of the transportation problem, which is a special case of the minimum cost flow problem, which in turn is a special case of a linear program. While it is possible to solve any of these problems using the simplex algorithm, each specialization has more efficient algorithms designed to take advantage of its special structure.

When a number of agents and tasks is very large, a parallel algorithm with randomization can be applied. The problem of finding minimum weight maximum matching can be converted to finding a minimum weight perfect matching. A bipartite graph can be extended to a complete bipartite graph by adding artificial edges with large weights. These weights should exceed the weights of all existing matchings to prevent appearance of artificial edges in the possible solution. As shown by Mulmuley, Vazirani & Vazirani (1987), the problem of minimum weight perfect matching is converted to finding minors in the adjacency matrix of a graph. Using the isolation lemma, a minimum weight perfect matching in a graph can be found with probability at least ½. For a graph with n vertices, it requires time.

## Example[edit]

Suppose that a taxi firm has three taxis (the agents) available, and three customers (the tasks) wishing to be picked up as soon as possible. The firm prides itself on speedy pickups, so for each taxi the "cost" of picking up a particular customer will depend on the time taken for the taxi to reach the pickup point. The solution to the assignment problem will be whichever combination of taxis and customers results in the least total cost.

However, the assignment problem can be made rather more flexible than it first appears. In the above example, suppose that there are four taxis available, but still only three customers. Then a fourth dummy task can be invented, perhaps called "sitting still doing nothing", with a cost of 0 for the taxi assigned to it. The assignment problem can then be solved in the usual way and still give the best solution to the problem.

Similar adjustments can be done in order to allow more tasks than agents, tasks to which multiple agents must be assigned (for instance, a group of more customers than will fit in one taxi), or maximizing profit rather than minimizing cost.

## Formal mathematical definition[edit]

The formal definition of the **assignment problem** (or **linear assignment problem**) is

- Given two sets,
*A*and*T*, of equal size, together with a weight function*C*:*A*×*T*→**R**. Find a bijection*f*:*A*→*T*such that the cost function:

is minimized.

Usually the weight function is viewed as a square real-valued matrix*C*, so that the cost function is written down as:

The problem is "linear" because the cost function to be optimized as well as all the constraints contain only linear terms.

The problem can be expressed as a standard linear program with the objective function

subject to the constraints

The variable represents the assignment of agent to task , taking value 1 if the assignment is done and 0 otherwise. This formulation allows also fractional variable values, but there is always an optimal solution where the variables take integer values. This is because the constraint matrix is totally unimodular. The first constraint requires that every agent is assigned to exactly one task, and the second constraint requires that every task is assigned exactly one agent.

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