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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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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

Assignment Problems

Definition of the assignment problem, mathematical formulation of the assignment problem, hungarian method for solving assignment problem, flow chart of hungarian method.

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.

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.

An assignment problem can be mathematically formulated as follows:

Minimise the total cost

\(x_{ij} =1\), if \(i^{th}\) person is assigned to the \(j^{th}\) job \(x_{ij}=0\), if \(i^{th}\) person is that assigned to the \(j^{th}\) job

subject to the constraints

i) \(\sum_{i=1}^n x_{ij} = 1, j=1, 2, \cdots n\)

which means that only one job is done by the \(i^{th}\) person, \(i= 1, 2, \cdots, n\)

ii) \(\sum_{i=1}^n x_{ij} = 1, j=1, 2, \cdots n\)

which means that only one person should be assigned to the \(j^{th}\) person, \(j= 1, 2, \cdots, n\)

The Hungarian method of assignment provides us with an efficient method of finding the optimal solution without having to make a-direct comparison of every solution. It works on the principle of reducing the given cost matrix to a matrix of opportunity costs.

Opportunity cost show the relative penalties associated with assigning resources to an activity as opposed to making the best or least cost assignment. If we can reduce the cost matrix to the extent of having at least one zero in each row and column, it will be possible to make optimal assignment.

The Hungarian method can be summarized in the following steps:

Step 1: Develop the Cost Table from the given Problem

If the no of rows are not equal to the no of columns and vice versa, a dummy row or dummy column must be added. The assignment cost for dummy cells are always zero.

Step 2: Find the Opportunity Cost Table

(a) Locate the smallest element in each row of the given cost table and then subtract that from each element of that row, and

(b) In the reduced matrix obtained from 2 (a) locate the smallest element in each column and then subtract that from each element. Each row and column now have at least one zero value.

Step 3: Make Assignment in the Opportunity Cost Matrix

The procedure of making assignment is as follows:

(a) Examine rows successively until a row with exactly one unmarked zero is obtained. Make an assignment single zero by making a square around it.

(b) For each zero value that becomes assigned, eliminate (Strike off) all other zeros in the same row and/ or column

(c) Repeat step 3 (a) and 3 (b) for each column also with exactly single zero value all that has not been assigned.

(d) If a row and/or column has two or more unmarked zeros and one cannot be chosen by inspection, then choose the assigned zero cell arbitrarily.

(e) Continue this process until all zeros in row column are either enclosed (Assigned) or struck off (x)

Step 4: Optimality Criterion

If the member of assigned cells is equal to the numbers of rows column then it is optimal solution. The total cost associated with this solution is obtained by adding original cost figures in the occupied cells.

If a zero cell was chosen arbitrarily in step (3), there exists an alternative optimal solution. But if no optimal solution is found, then go to step (5).

Step 5: Revise the Opportunity Cost Table

Draw a set of horizontal and vertical lines to cover all the zeros in the revised cost table obtained from step (3), by using the following procedure:

(a) For each row in which no assignment was made, mark a tick (√)

(b) Examine the marked rows. If any zero occurs in those columns, tick the respective rows that contain those assigned zeros.

(c) Repeat this process until no more rows or columns can be marked.

If a no of lines drawn is equal to the no of (or columns) the current solution is the optimal solution, otherwise go to step 6.

Step 6: Develop the New Revised Opportunity Cost Table

(a) From among the cells not covered by any line, choose the smallest element, call this value K

(b) Subtract K from every element in the cell not covered by line.

(c) Add K to very element in the cell covered by the two lines, i.e., intersection of two lines.

(d) Elements in cells covered by one line remain unchanged.

Step 7: Repeat Step 3 to 6 Unlit an Optimal Solution is Obtained

The flow chart of steps in the Hungarian method for solving an assignment problem is shown in following figures:

In a factory there are five operator \(O_1\), \(O_2\), \(O_3\), \(O_4\), \(O_5\) and five machine \(M_1\), \(M_2\), \(M_3\), \(M_4\), \(M_5\). The operating costs are given when the \(O_i\) operator operates the \(M_j\) machine \((i,j=1,2,..,5)\). But there is a restriction that \(O_3\) cannot be allowed to operate the third machine \(M_3\) and \(O_2\) cannot be allowed to operate the fifth machine \(M_5\). The cost matrix is given above. Find the optional assignment and the optimal assignment cost also.

[2018, 15M]

2) Solve the following assignment problem to maximize the sales: \(\begin{array}{|c|c|c|c|c|c|} \hline {} & {I} & {II} & {III} & {IV} & {V} \\ \hline {A} & {3} & {4} & {5} & {6} & {7} \\ \hline {B} & {4} & {15} & {13} & {7} & {6} \\ \hline {C} & {6} & {13} & {12} & {5} & {11} \\ \hline {D} & {7} & {12} & {15} & {8} & {5} \\ \hline {E} & {8} & {13} & {10} & {6} & {9} \\ \hline \end{array}\)

where \(I\), \(II\), \(III\), \(IV\) and \(V\) are Territories; \(A\), \(B\), \(C\), \(D\), \(E\) are Salesmen.

[2015, 10M]

3) Solve the minimum time assignment problem: \(\begin{array}{|c|c|c|c|c|} \hline { } & {I} & {II} & {III} & {IV} \\ \hline {A} & {3} & {12} & {5} & {4} \\ \hline {B} & {7} & {9} & {8} & {12} \\ \hline {C} & {5} & {11} & {10} & {12} \\ \hline {D} & {6} & {14} & {4} & {11} \\ \hline \end{array}\)

where \(I\), \(II\), \(III\) and \(IV\) are Machines; \(A\), \(B\), \(C\) and \(D\) are Jobs.

[2013, 15M]

4) A travelling salesman has to visit 5 cities. He wishes to start from a particular city, visit each city once and then return to his starting point. Cost of going from one city to another is given below:

You are required to find the least cost route.

[2004, 15M]

5) Find the optimal solution for the assignment problem with the following cost matrix: \(\begin{bmatrix}{6} & {1} & {9} & {11} & {12} \\ {2} & {8} & {17} & {2} & {5} \\ {11} & {8} & {3} & {3} & {3} \\ {4} & {10} & {8} & {6} & {11} \\ {8} & {10} & {11} & {5} & {13}\end{bmatrix}\)

Indicate clearly the rule you apply to arrive at the complete assignment.

[2003, 15M]

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Quadratic Assignment Problem (QAP)

The Quadratic Assignment Problem (QAP) is an optimization problem that deals with assigning a set of facilities to a set of locations, considering the pairwise distances and flows between them.

The problem is to find the assignment that minimizes the total cost or distance, taking into account both the distances and the flows.

The distance matrix and flow matrix, as well as restrictions to ensure each facility is assigned to exactly one location and each location is assigned to exactly one facility, can be used to formulate the QAP as a quadratic objective function.

The QAP is a well-known example of an NP-hard problem , which means that for larger cases, computing the best solution might be difficult. As a result, many algorithms and heuristics have been created to quickly identify approximations of answers.

There are various types of algorithms for different problem structures, such as:

• Precise algorithms
• Approximation algorithms
• Metaheuristics like genetic algorithms and simulated annealing
• Specialized algorithms

Example: Given four facilities (F1, F2, F3, F4) and four locations (L1, L2, L3, L4). We have a cost matrix that represents the pairwise distances or costs between facilities. Additionally, we have a flow matrix that represents the interaction or flow between locations. Find the assignment that minimizes the total cost based on the interactions between facilities and locations. Each facility must be assigned to exactly one location, and each location can only accommodate one facility.

Facilities cost matrix:

Flow matrix:

To solve the QAP, various optimization techniques can be used, such as mathematical programming, heuristics, or metaheuristics. These techniques aim to explore the search space and find the optimal or near-optimal solution.

The solution to the QAP will provide an assignment of facilities to locations that minimizes the overall cost.

The solution generates all possible permutations of the assignment and calculates the total cost for each assignment. The optimal assignment is the one that results in the minimum total cost.

To calculate the total cost, we look at each pair of facilities in (i, j) and their respective locations (location1, location2). We then multiply the cost of assigning facility1 to facility2 (facilities[facility1][facility2]) with the flow from location1 to location2 (locations[location1][location2]). This process is done for all pairs of facilities in the assignment, and the costs are summed up.

Overall, the output tells us that assigning facilities to locations as F1->L1, F3->L2, F2->L3, and F4->L4 results in the minimum total cost of 44. This means that Facility 1 is assigned to Location 1, Facility 3 is assigned to Location 2, Facility 2 is assigned to Location 3, and Facility 4 is assigned to Location 4, yielding the lowest cost based on the given cost and flow matrices.This example demonstrates the process of finding the optimal assignment by considering the costs and flows associated with each facility and location. The objective is to find the assignment that minimizes the total cost, taking into account the interactions between facilities and locations.

Applications of the QAP include facility location, logistics, scheduling, and network architecture, all of which require effective resource allocation and arrangement.

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Course info.

• Prof. Dimitris Bertsimas

Departments

• Electrical Engineering and Computer Science
• Sloan School of Management

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Introduction to mathematical programming, assignments.

The table below contains problem set assignments and due dates. Problems are from the course textbook, [BT] = Bertsimas, Dimitris, and John Tsitsiklis. Introduction to Linear Optimization . Belmont, MA: Athena Scientific, 1997. ISBN: 9781886529199.

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Quadratic Assignment Problem

The objective of the Quadratic Assignment Problem (QAP) is to assign \(n\) facilities to \(n\) locations in such a way as to minimize the assignment cost. The assignment cost is the sum, over all pairs, of the flow between a pair of facilities multiplied by the distance between their assigned locations.

Problem Statement

The quadratic assignment problem (QAP) was introduced by Koopmans and Beckman in 1957 in the context of locating “indivisible economic activities”. The objective of the problem is to assign a set of facilities to a set of locations in such a way as to minimize the total assignment cost. The assignment cost for a pair of facilities is a function of the flow between the facilities and the distance between the locations of the facilities.

Consider a facility location problem with four facilities (and four locations). One possible assignment is shown in the figure below: facility 2 is assigned to location 1, facility 1 is assigned to location 2, facility 4 is assigned to location 3, and facility 4 is assigned to location 3. This assignment can be written as the permutation \(p = \{2, 1, 4, 3\}\), which means that facility 2 is assigned to location 1, facility 1 is assigned to location 2, facility 4 is assigned to location 3, and facility 4 is assigned to location 3. In the figure, the line between a pair of facilities indicates that there is required flow between the facilities, and the thickness of the line increases with the value of the flow.

To calculate the assignment cost of the permutation, the required flows between facilities and the distances between locations are needed.

Then, the assignment cost of the permutation can be computed as \(f(1,2) \cdot d(2,1) + f(1,4) \cdot d(2,3) + f(2,4) \cdot d(1,3) + f(3,4) \cdot d(3,4)\) = \(3 \cdot 22 + 2 \cdot 40 + 1 \cdot 53 + 4 \cdot 55\) = 419. Note that this permutation is not the optimal solution.

Mathematical Formulation

Here we present the Koopmans-Beckmann formulation of the QAP. Given a set of facilities and locations along with the flows between facilities and the distances between locations, the objective of the Quadratic Assignment Problem is to assign each facility to a location in such a way as to minimize the total cost.

Sets \(N = \{1, 2, \cdots, n\}\) \(S_n = \phi: N \rightarrow N\) is the set of all permutations

Parameters \(F = (f_{ij})\) is an \(n \times n\) matrix where \(f_{ij}\) is the required flow between facilities \(i\) and \(j\) \(D = (d_{ij})\) is an \(n \times n\) matrix where \(d_{ij}\) is the distance between locations \(i\) and \(j\)

Optimization Problem \(\text{min}_{\phi \in S_n} \sum_{i=1}^n \sum_{j=1}^n f_{ij} \cdot d_{\phi(i) \phi(j)}\)

The assignment of facilities to locations is represented by a permutation \(\phi\), where \(\phi(i)\) is the location to which facility \(i\) is assigned. Each individual product \(f_{ij} \cdot d_{\phi(i) \phi(j)}\) is the cost of assigning facility \(i\) to location \(\phi(i)\) and facility \(j\) to location \(\phi(j)\).

Solve some QAPs!

Follow the links below to test your skill at finding good solutions to QAPs of various sizes. Notice that as the problem size increases, it becomes much more difficult to find an optimal solution. As \(n\) increases beyond a small number, it becomes impossible to enumerate and evaluate all possible assignment vectors. Instead, advanced solution algorithms are required to solve larger instances.

QAP of size 4

Qap of size 5, qap of size 6, qap of size 7, qap of size 8, qap of size 9.

• Anstreicher, K.M. 2003. Recent advances in the solution of quadratic assignment problems. Mathematical Programming Series B 97 , 27 - 42.
• Çela, E. 1998. The Quadratic Assignment Problem: Theory and Algorithms . Kluwer Academic Publishers, Dordrecht.
• Koopmans, T. C. and M. J. Beckmann. 1957. Assignment problems and the location of economic activities. Econometrica 25 , 53 - 76.
• QAPLIB Home Page

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Abstract: Linear programming is the seminal optimization problem that has spawned and grown into today's rich and diverse optimization modeling and algorithmic landscape. This article provides an overview of the recent development of first-order methods for solving large-scale linear programming.

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An approximation algorithm for the generalized assignment problem

• Published: February 1993
• Volume 62 , pages 461–474, ( 1993 )

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• David B. Shmoys 1 &
• Éva Tardos 1  

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The generalized assignment problem can be viewed as the following problem of scheduling parallel machines with costs. Each job is to be processed by exactly one machine; processing job j on machine i requires time p ij and incurs a cost of c ij ; each machine i is available for T i time units, and the objective is to minimize the total cost incurred. Our main result is as follows. There is a polynomial-time algorithm that, given a value C , either proves that no feasible schedule of cost C exists, or else finds a schedule of cost at most C where each machine i is used for at most 2 T i time units.

We also extend this result to a variant of the problem where, instead of a fixed processing time p ij , there is a range of possible processing times for each machine—job pair, and the cost linearly increases as the processing time decreases. We show that these results imply a polynomial-time 2-approximation algorithm to minimize a weighted sum of the cost and the makespan, i.e., the maximum job completion time. We also consider the objective of minimizing the mean job completion time. We show that there is a polynomial-time algorithm that, given values M and T , either proves that no schedule of mean job completion time M and makespan T exists, or else finds a schedule of mean job completion time at most M and makespan at most 2 T.

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Research partially supported by an NSF PYI award CCR-89-96272 with matching support from UPS, and Sun Microsystems, and by the National Science Foundation, the Air Force Office of Scientific Research, and the Office of Naval Research, through NSF grant DMS-8920550.

Research supported in part by a Packard Fellowship, a Sloan Fellowship, an NSF PYI award, and by the National Science Foundation, the Air Force Office of Scientific Research, and the Office of Naval Research, through NSF grant DMS-8920550.

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Shmoys, D.B., Tardos, É. An approximation algorithm for the generalized assignment problem. Mathematical Programming 62 , 461–474 (1993). https://doi.org/10.1007/BF01585178

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Received : 10 April 1991

Revised : 14 January 1993

Issue Date : February 1993

DOI : https://doi.org/10.1007/BF01585178

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