# Assignment Problem In Operations Research Questions

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: 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.

Job | ||||
---|---|---|---|---|

Person | 1 | 2 | 3 | 4 |

A | 20 | 25 | 22 | 28 |

B | 15 | 18 | 23 | 17 |

C | 19 | 17 | 21 | 24 |

D | 25 | 23 | 24 | 24 |

Solution.

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

#### Step 1

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

Table

On small screens, scroll horizontally to view full calculation

Job | ||||
---|---|---|---|---|

Person | 1 | 2 | 3 | 4 |

A | 0 | 5 | 2 | 8 |

B | 0 | 3 | 8 | 2 |

C | 2 | 0 | 4 | 7 |

D | 2 | 0 | 1 | 1 |

Step 2

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

Table

Job | ||||
---|---|---|---|---|

Person | 1 | 2 | 3 | 4 |

A | 0 | 5 | 1 | 7 |

B | 0 | 3 | 7 | 1 |

C | 2 | 0 | 3 | 6 |

D | 2 | 0 | 0 | 0 |

Step 3

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

- For each row or column with a single zero value cell that has not be assigned or eliminated, box that zero value as an assigned cell.
- 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.
- The above process may be continued until every zero cell is either assigned or crossed (X).

#### Step 4

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.

Table

Use Horizontal Scrollbar to View Full Table Calculation

#### Step 5

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.

Table

#### Step 6

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.

Table

Job | ||||
---|---|---|---|---|

Person | 1 | 2 | 3 | 4 |

A | 0 | 4 | 0 | 6 |

B | 0 | 2 | 6 | 0 |

C | 3 | 0 | 3 | 6 |

D | 3 | 0 | 0 | 0 |

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.

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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