assignment problem form

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:

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Assignment problem

The problem of optimally assigning $ m $ individuals to $ m $ jobs. It can be formulated as a linear programming problem that is a special case of the transport problem :

maximize $ \sum _ {i,j } c _ {ij } x _ {ij } $

$$ \sum _ { j } x _ {ij } = a _ {i} , i = 1 \dots m $$

(origins or supply),

$$ \sum _ { i } x _ {ij } = b _ {j} , j = 1 \dots n $$

(destinations or demand), where $ x _ {ij } \geq 0 $ and $ \sum a _ {i} = \sum b _ {j} $, which is called the balance condition. The assignment problem arises when $ m = n $ and all $ a _ {i} $ and $ b _ {j} $ are $ 1 $.

If all $ a _ {i} $ and $ b _ {j} $ in the transposed problem are integers, then there is an optimal solution for which all $ x _ {ij } $ are integers (Dantzig's theorem on integral solutions of the transport problem).

In the assignment problem, for such a solution $ x _ {ij } $ is either zero or one; $ x _ {ij } = 1 $ means that person $ i $ is assigned to job $ j $; the weight $ c _ {ij } $ is the utility of person $ i $ assigned to job $ j $.

The special structure of the transport problem and the assignment problem makes it possible to use algorithms that are more efficient than the simplex method . Some of these use the Hungarian method (see, e.g., [a5] , [a1] , Chapt. 7), which is based on the König–Egervary theorem (see König theorem ), the method of potentials (see [a1] , [a2] ), the out-of-kilter algorithm (see, e.g., [a3] ) or the transportation simplex method.

In turn, the transportation problem is a special case of the network optimization problem.

A totally different assignment problem is the pole assignment problem in control theory.

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Algorithms: The Assignment Problem

One of the interesting things about studying optimization is that the techniques show up in a lot of different areas. The “assignment problem” is one that can be solved using simple techniques, at least for small problem sizes, and is easy to see how it could be applied to the real world.

Assignment Problem

Pretend for a moment that you are writing software for a famous ride sharing application. In a crowded environment, you might have multiple prospective customers that are requesting service at the same time, and nearby you have multiple drivers that can take them where they need to go. You want to assign the drivers to the customers in a way that minimizes customer wait time (so you keep the customers happy) and driver empty time (so you keep the drivers happy).

The assignment problem is designed for exactly this purpose. We start with m agents and n tasks. We make the rule that every agent has to be assigned to a task. For each agent-task pair, we figure out a cost associated to have that agent perform that task. We then figure out which assignment of agents to tasks minimizes the total cost.

Of course, it may be true that m != n , but that’s OK. If there are too many tasks, we can make up a “dummy” agent that is more expensive than any of the others. This will ensure that the least desirable task will be left to the dummy agent, and we can remove that from the solution. Or, if there are too many agents, we can make up a “dummy” task that is free for any agent. This will ensure that the agent with the highest true cost will get the dummy task, and will be idle.

If that last paragraph was a little dense, don’t worry; there’s an example coming that will help show how it works.

There are special algorithms for solving assignment problems, but one thing that’s nice about them is that a general-purpose solver can handle them too. Below is an example, but first it will help to cover a few concepts that we’ll be using.

Optimization Problems

Up above, we talked about making “rules” and minimizing costs. The usual name for this is optimization. An optimization problem is one where we have an “objective function” (which tells us what our goals are) and one or more “constraint functions” (which tell us what the rules are). The classic example is a factory that can make both “widgets” and “gadgets”. Each “widget” and “gadget” earns a certain amount of profit, but it also uses up raw material and time on the factory’s machines. The optimization problem is to determine exactly how many “widgets” and how many “gadgets” to make to maximize profit (the objective) while fitting within the material and time available (the constraints).

If we were to write this simple optimization problem out, it might look like this:

In this case, we have two variables: g for the number of gadgets we make and w for the number of widgets we make. We also have three constraints that we have to meet. Note that they are inequalities; we might not use all the available material or time in our optimal solution.

Just to unpack this a little: in English, the above is saying that we make 45 dollars / euros / quatloos per gadget we make. However, to make a gadget needs 120 lbs of raw material 1, 80 lbs of raw material 2, and 3.8 hours of machine time. So there is a limit on how many gadgets we can make, and it might be a better use of resources to balance gadgets with widgets.

Of course, real optimization problems have many more than two variables and many constraint functions, making them much harder to solve. The easiest kind of optimization problem to solve is linear, and fortunately, the assignment problem is linear.

Linear Programming

A linear program is a kind of optimization problem where both the objective function and the constraint functions are linear. (OK, that definition was a little self-referential.) We can have as many variables as we want, and as many constraint functions as we want, but none of the variables can have exponents in any of the functions. This limitation allows us to apply very efficient mathematical approaches to solve the problem, even for very large problems.

We can state the assignment problem as a linear programming problem. First, we choose to make “i” represent each of our agents (drivers) and “j” to represent each of our tasks (customers). Now, to write a problem like this, we need variables. The best approach is to use “indicator” variables, where xij = 1 means “driver i picks up customer j” and xij = 0 means “driver i does not pick up customer j”.

We wind up with:

This is a compact mathematical way to describe the problem, so again let me put it in English.

First, we need to figure out the cost of having each driver pick up each customer. Then, we can calculate the total cost for any scenario by just adding up the costs for the assignments we pick. For any assignment we don’t pick, xij will equal zero, so that term will just drop out of the sum.

Of course, the way we set up the objective function, the cheapest solution is for no drivers to pick up any customers. That’s not a very good business model. So we need a constraint to show that we want to have a driver assigned to every customer. At the same time, we can’t have a driver assigned to mutiple customers. So we need a constraint for that too. That leads us to the two constraints in the problem. The first just says, if you add up all the assignments for a given driver, you want the total number of assignments for that driver to be exactly one. The second constraint says, if you add up all the assignments to a given customer, you want the total number of drivers assigned to the customer to be one. If you have both of these, then each driver is assigned to exactly one customer, and the customers and drivers are happy. If you do it in a way that minimizes costs, then the business is happy too.

Solving with Octave and GLPK

The GNU Linear Programming Kit is a library that solves exactly these kinds of problems. It’s easy to set up the objective and constraints using GNU Octave and pass these over to GLPK for a solution.

Given some made-up sample data, the program looks like this:

Start with the definition of “c”, the cost information. For this example, I chose to have four drivers and three customers. There are sixteen numbers there; the first four are the cost of each driver to get the first customer, the next four are for the second customer, and the next four are for the third customer. Because we have an extra driver, we add a “dummy” customer at the end that is zero cost. This represents one of the drivers being idle.

The next definition is “b”, the right-hand side of our constraints. There are eight constraints, one for each of the drivers, and one for each of the customers (including the dummy). For each constraint, the right-hand side is 1.

The big block in the middle defines our constraint matrix “a”. This is the most challenging part of taking the mathematical definition and putting it into a form that is usable by GLPK; we have to expand out each constraint. Fortunately, in these kinds of cases, we tend to get pretty patterns that help us know we’re on the right track.

The first line in “a” says that the first customer needs a driver. To see why, remember that in our cost information, the first four numbers are the cost for each driver to get the first customer. With this constraint, we are requiring that one of those four costs be included and therefore that a driver is “selected” for the first customer. The other lines in “a” work similarly; the last four ensure that each driver has an assignment.

Note that the number of rows in “a” matches the number of items in “b”, and the number of columns in “a” matches the number of items in “c”. This is important; GLPK won’t run if this is not true (and our problem isn’t stated right in any case).

Compared to the above, the last few lines are easy.

• “lb” gives the lower bound for each variable.
• “ub” gives the upper bound.
• “ctype” tells GLPK that each constraint is an equality (“strict” as opposed to providing a lower or upper bound).
• “vartype” tells GLPK that these variables are all integers (can’t have half a driver showing up).
• “s” tells GLPK that we want to minimize our costs, not maximize them.

We push all that through a function call to GLPK, and what comes back are two values (along with some other stuff I’ll exclude for clarity):

The first item tells us that our best solution takes 27 minutes, or dollars, or whatever unit we used for cost. The second item tells us the assignments we got. (Note for pedants: I transposed this output to save space.)

This output tells us that customer 1 gets driver 2, customer 2 gets driver 3, customer 3 gets driver 4, and driver 1 is idle. If you look back at the cost data, you can see this makes sense, because driver 1 had some of the most expensive times to the three customers. You can also see that it managed to pick the least expensive pairing for each customer. (Of course, if I had done a better job making up cost data, it might not have picked the least expensive pairing in all cases, because a suboptimal individual pairing might still lead to an overall optimal solution. But this is a toy example.)

Of course, for a real application, we would have to take into consideration many other factors, such as the passage of time. Rather than knowing all of our customers and drivers up front, we would have customers and drivers continually showing up and being assigned. But I hope this simple example has revealed some of the concepts behind optimization and linear programming and the kinds of real-world problems that can be solved.

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Assignment problem

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Assignment Problem

• Feb 20, 2024

The assignment problem is a special case of transportation problem.

Suppose there are $n$ jobs to be performed and $n$ persons are available for these jobs. Assume that each person can do each job at a time, though with varying degree of efficiency.

Let $c_{ij}$ be the cost if the $i^{th}$ person is assigned the $j^{th}$ job. Then the problem is to find an assignment so that the total cost of performing all jobs is minimum. (i.e., which job should be assigned to which person with minimum cost). Such problem is called an Assignment Problem (AP).

The tabular form of the assignment problem is as follows

The above table is called the $n\times n$ cost-matrix, where $c_{ij}$ are real numbers.

Thus the objective in the Assignment Problem is to assign a number of jobs to the equal number of persons at a minimum cost or maximum profit. e.g. assigning men to offices, classes to rooms, drivers to trucks, problems to research teams etc.

Mathematical Formulation of AP

Mathematically, the AP can be stated as $$ \begin{equation}\label{eq2.4} \min z = \sum_{i=1}^n\sum_{j=1}^n x_{ij} c_{ij}. \end{equation} $$ subject to $$ \begin{equation}\label{eq2.5} \sum_{j=1}^n x_{ij} =1,; \text{ for } i=1,2,\ldots, n \end{equation} $$

$$ \begin{equation}\label{eq2.6} \sum_{i=1}^n x_{ij} =1,; \text{ for } j=1,2,\ldots,n \end{equation} $$ where $$ \begin{equation*} x_{ij}=\left{ \begin{array}{ll} 1, & \hbox{if $i^{th}$ person is assigned $j^{th}$ job;} \ 0, & \hbox{otherwise.} \end{array} \right. \end{equation*} $$ Constraint \eqref{eq2.5} indicate that one job is done by $i^{th}$ person $i=1,2,\ldots, n$ and constraint \eqref{eq2.6} indicate that one person should be assigned $j^{th}$ job $j=1,2,\ldots, n$.

It may be observed from the above formulation that AP is a special type of Linear programming problem.

Unbalanced Assignment Problem

If the cost matrix of an assignment problem is not a square matrix, the assignment problem is called an Unbalanced Assignment Problem . In such a case, add dummy row or dummy column with zero cost in the cost matrix so as to form a square matrix. Then apply the usual assignment method to find the optimal solution.

Maximal Assignment Problem

Sometimes, the assignment problem deals with the maximization of an objective function instead of minimization.. For example, it may be required to assign persons to jobs in such a way that the expected profit is maximum. In such a case, first convert the problem of maximization to minimization and apply the usual procedure of assignment.

The assignment problem of maximization type can be converted to minimization type by subtracting all the elements of the given profit matrix from the largest element.

Restrictions on Assignment

Sometimes due to technical or other difficulties do not permit the assignment of a particular facility to a particular job. In such a situation, the difficulty can be overcome by assigning a very high cost (say, infinity i.e. $\infty$) to the corresponding cell.

Because of assigning an infinite penalty to such a cell the activity will be automatically excluded from the optimal solution. Then apply the usual procedure to find the optimal assignment.

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Hungarian Algorithm for Assignment Problem | Set 1 (Introduction)

• 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

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Assignment problem in linear programming : introduction and assignment model.

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Assignment problem is a special type of linear programming problem which deals with the allocation of the various resources to the various activities on one to one basis. It does it in such a way that the cost or time involved in the process is minimum and profit or sale is maximum. Though there problems can be solved by simplex method or by transportation method but assignment model gives a simpler approach for these problems.

In a factory, a supervisor may have six workers available and six jobs to fire. He will have to take decision regarding which job should be given to which worker. Problem forms one to one basis. This is an assignment problem.

1. Assignment Model :

Suppose there are n facilitates and n jobs it is clear that in this case, there will be n assignments. Each facility or say worker can perform each job, one at a time. But there should be certain procedure by which assignment should be made so that the profit is maximized or the cost or time is minimized.

In the table, Co ij is defined as the cost when j th job is assigned to i th worker. It maybe noted here that this is a special case of transportation problem when the number of rows is equal to number of columns.

Mathematical Formulation:

Any basic feasible solution of an Assignment problem consists (2n – 1) variables of which the (n – 1) variables are zero, n is number of jobs or number of facilities. Due to this high degeneracy, if we solve the problem by usual transportation method, it will be a complex and time consuming work. Thus a separate technique is derived for it. Before going to the absolute method it is very important to formulate the problem.

Suppose x jj is a variable which is defined as

1 if the i th job is assigned to j th machine or facility

0 if the i th job is not assigned to j th machine or facility.

Now as the problem forms one to one basis or one job is to be assigned to one facility or machine.

The total assignment cost will be given by

The above definition can be developed into mathematical model as follows:

Determine x ij > 0 (i, j = 1,2, 3…n) in order to

Subjected to constraints

and x ij is either zero or one.

Method to solve Problem (Hungarian Technique):

Consider the objective function of minimization type. Following steps are involved in solving this Assignment problem,

1. Locate the smallest cost element in each row of the given cost table starting with the first row. Now, this smallest element is subtracted form each element of that row. So, we will be getting at least one zero in each row of this new table.

2. Having constructed the table (as by step-1) take the columns of the table. Starting from first column locate the smallest cost element in each column. Now subtract this smallest element from each element of that column. Having performed the step 1 and step 2, we will be getting at least one zero in each column in the reduced cost table.

3. Now, the assignments are made for the reduced table in following manner.

(i) Rows are examined successively, until the row with exactly single (one) zero is found. Assignment is made to this single zero by putting square □ around it and in the corresponding column, all other zeros are crossed out (x) because these will not be used to make any other assignment in this column. Step is conducted for each row.

(ii) Step 3 (i) in now performed on the columns as follow:- columns are examined successively till a column with exactly one zero is found. Now , assignment is made to this single zero by putting the square around it and at the same time, all other zeros in the corresponding rows are crossed out (x) step is conducted for each column.

(iii) Step 3, (i) and 3 (ii) are repeated till all the zeros are either marked or crossed out. Now, if the number of marked zeros or the assignments made are equal to number of rows or columns, optimum solution has been achieved. There will be exactly single assignment in each or columns without any assignment. In this case, we will go to step 4.

4. At this stage, draw the minimum number of lines (horizontal and vertical) necessary to cover all zeros in the matrix obtained in step 3, Following procedure is adopted:

(iii) Now tick mark all the rows that are not already marked and that have assignment in the marked columns.

(iv) All the steps i.e. (4(i), 4(ii), 4(iii) are repeated until no more rows or columns can be marked.

(v) Now draw straight lines which pass through all the un marked rows and marked columns. It can also be noticed that in an n x n matrix, always less than ‘n’ lines will cover all the zeros if there is no solution among them.

5. In step 4, if the number of lines drawn are equal to n or the number of rows, then it is the optimum solution if not, then go to step 6.

6. Select the smallest element among all the uncovered elements. Now, this element is subtracted from all the uncovered elements and added to the element which lies at the intersection of two lines. This is the matrix for fresh assignments.

7. Repeat the procedure from step (3) until the number of assignments becomes equal to the number of rows or number of columns.

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Quantitative Techniques: Theory and Problems by P. C. Tulsian, Vishal Pandey

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WHAT IS ASSIGNMENT PROBLEM

Assignment Problem is a special type of linear programming problem where the objective is to minimise the cost or time of completing a number of jobs by a number of persons.

The assignment problem in the general form can be stated as follows:

“Given n facilities, n jobs and the effectiveness of each facility for each job, the problem is to assign each facility to one and only one job in such a way that the measure of effectiveness is optimised (Maximised or Minimised).”

Several problems of management has a structure identical with the assignment problem.

Example I A manager has four persons (i.e. facilities) available for four separate jobs (i.e. jobs) and the cost of assigning (i.e. effectiveness) each job to each ...

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Solving an Assignment Problem

This section presents an example that shows how to solve an assignment problem using both the MIP solver and the CP-SAT solver.

In the example there are five workers (numbered 0-4) and four tasks (numbered 0-3). Note that there is one more worker than in the example in the Overview .

The costs of assigning workers to tasks are shown in the following table.

The problem is to assign each worker to at most one task, with no two workers performing the same task, while minimizing the total cost. Since there are more workers than tasks, one worker will not be assigned a task.

MIP solution

The following sections describe how to solve the problem using the MPSolver wrapper .

Import the libraries

The following code imports the required libraries.

Create the data

The following code creates the data for the problem.

The costs array corresponds to the table of costs for assigning workers to tasks, shown above.

Declare the MIP solver

The following code declares the MIP solver.

Create the variables

The following code creates binary integer variables for the problem.

Create the constraints

Create the objective function.

The following code creates the objective function for the problem.

The value of the objective function is the total cost over all variables that are assigned the value 1 by the solver.

Invoke the solver

The following code invokes the solver.

Print the solution

The following code prints the solution to the problem.

Here is the output of the program.

Complete programs

Here are the complete programs for the MIP solution.

CP SAT solution

The following sections describe how to solve the problem using the CP-SAT solver.

Declare the model

The following code declares the CP-SAT model.

The following code sets up the data for the problem.

The following code creates the constraints for the problem.

Here are the complete programs for the CP-SAT solution.

Except as otherwise noted, the content of this page is licensed under the Creative Commons Attribution 4.0 License , and code samples are licensed under the Apache 2.0 License . For details, see the Google Developers Site Policies . Java is a registered trademark of Oracle and/or its affiliates.

Last updated 2024-08-28 UTC.

Assignment Problem: Linear Programming

The assignment problem is a special type of transportation problem , where the objective is to minimize the cost or time of completing a number of jobs by a number of persons.

In other words, when the problem involves the allocation of n different facilities to n different tasks, it is often termed as an assignment problem.

The model's primary usefulness is for planning. The assignment problem also encompasses an important sub-class of so-called shortest- (or longest-) route models. The assignment model is useful in solving problems such as, assignment of machines to jobs, assignment of salesmen to sales territories, travelling salesman problem, etc.

It may be noted that with n facilities and n jobs, there are n! possible assignments. One way of finding an optimal assignment is to write all the n! possible arrangements, evaluate their total cost, and select the assignment with minimum cost. But, due to heavy computational burden this method is not suitable. This chapter concentrates on an efficient method for solving assignment problems that was developed by a Hungarian mathematician D.Konig.

"A mathematician is a device for turning coffee into theorems." -Paul Erdos

Formulation of an assignment problem

Suppose a company has n persons of different capacities available for performing each different job in the concern, and there are the same number of jobs of different types. One person can be given one and only one job. The objective of this assignment problem is to assign n persons to n jobs, so as to minimize the total assignment cost. The cost matrix for this problem is given below:

The structure of an assignment problem is identical to that of a transportation problem.

To formulate the assignment problem in mathematical programming terms , we define the activity variables as

for i = 1, 2, ..., n and j = 1, 2, ..., n

In the above table, c ij is the cost of performing jth job by ith worker.

Generalized Form of an Assignment Problem

The optimization model is

Minimize c 11 x 11 + c 12 x 12 + ------- + c nn x nn

subject to x i1 + x i2 +..........+ x in = 1          i = 1, 2,......., n x 1j + x 2j +..........+ x nj = 1          j = 1, 2,......., n

x ij = 0 or 1

In Σ Sigma notation

x ij = 0 or 1 for all i and j

An assignment problem can be solved by transportation methods, but due to high degree of degeneracy the usual computational techniques of a transportation problem become very inefficient. Therefore, a special method is available for solving such type of problems in a more efficient way.

Assumptions in Assignment Problem

• Number of jobs is equal to the number of machines or persons.
• Each man or machine is assigned only one job.
• Each man or machine is independently capable of handling any job to be done.
• Assigning criteria is clearly specified (minimizing cost or maximizing profit).

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Goal programming Linear programming Simplex Method Transportation Problem

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Assignment Problem

Calvin has a math assignment at school where he has to evaluate a lot of expressions. Calvin decides to not to waste much of his time. There are expressions overall. By looking at Susie’s answers, Calvin has figured that the answers to all questions form a non decreasing sequence.

He decides that all his answers are going to be between and (inclusive). He fills his answer sheet with a random non-decreasing sequence of length where each element is between and .

Here is the part where the real problem starts for Calvin. He does not want to choose a large value of , because, he will have a lot of options to choose from. Also, if he chooses a very small value of , a lot of answers would become equal and the teacher will become suspicious.

If , f(i) being the frequency or number of times occurs in the sequence of values he picked. Calvin wants to find out expected value of . Help him solve the problem.

For example, if & , the possible sequences are:

expected value of

Input Format The first line contains an integer which refers to the number of test cases. lines follow, each containing numbers, and for the corresponding test cases.

Constraints

Output Format Output lines, each containing answer to the corresponding test case. Error of upto is allowed.

Sample Input

Sample Output

Explanation For second testcase we have

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