minimum assignment problem

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:

How to Solve the Assignment Problem: A Complete Guide

Table of Contents

Assignment problem is a special type of linear programming problem that deals with assigning a number of resources to an equal number of tasks in the most efficient way. The goal is to minimize the total cost of assignments while ensuring that each task is assigned to only one resource and each resource is assigned to only one task. In this blog, we will discuss the solution of the assignment problem using the Hungarian method, which is a popular algorithm for solving the problem.

Understanding the Assignment Problem

Before we dive into the solution, it is important to understand the problem itself. In the assignment problem, we have a matrix of costs, where each row represents a resource and each column represents a task. The objective is to assign each resource to a task in such a way that the total cost of assignments is minimized. However, there are certain constraints that need to be satisfied – each resource can be assigned to only one task and each task can be assigned to only one resource.

Solving the Assignment Problem

There are various methods for solving the assignment problem, including the Hungarian method, the brute force method, and the auction algorithm. Here, we will focus on the steps involved in solving the assignment problem using the Hungarian method, which is the most commonly used and efficient method.

Step 1: Set up the cost matrix

The first step in solving the assignment problem is to set up the cost matrix, which represents the cost of assigning a task to an agent. The matrix should be square and have the same number of rows and columns as the number of tasks and agents, respectively.

Step 2: Subtract the smallest element from each row and column

To simplify the calculations, we need to reduce the size of the cost matrix by subtracting the smallest element from each row and column. This step is called matrix reduction.

Step 3: Cover all zeros with the minimum number of lines

The next step is to cover all zeros in the matrix with the minimum number of horizontal and vertical lines. This step is called matrix covering.

Step 4: Test for optimality and adjust the matrix

To test for optimality, we need to calculate the minimum number of lines required to cover all zeros in the matrix. If the number of lines equals the number of rows or columns, the solution is optimal. If not, we need to adjust the matrix and repeat steps 3 and 4 until we get an optimal solution.

Step 5: Assign the tasks to the agents

The final step is to assign the tasks to the agents based on the optimal solution obtained in step 4. This will give us the most cost-effective or profit-maximizing assignment.

Solution of the Assignment Problem using the Hungarian Method

The Hungarian method is an algorithm that uses a step-by-step approach to find the optimal assignment. The algorithm consists of the following steps:

Subtract the smallest entry in each row from all the entries of the row.
Subtract the smallest entry in each column from all the entries of the column.
Draw the minimum number of lines to cover all zeros in the matrix. If the number of lines drawn is equal to the number of rows, we have an optimal solution. If not, go to step 4.
Determine the smallest entry not covered by any line. Subtract it from all uncovered entries and add it to all entries covered by two lines. Go to step 3.

The above steps are repeated until an optimal solution is obtained. The optimal solution will have all zeros covered by the minimum number of lines. The assignments can be made by selecting the rows and columns with a single zero in the final matrix.

Applications of the Assignment Problem

The assignment problem has various applications in different fields, including computer science, economics, logistics, and management. In this section, we will provide some examples of how the assignment problem is used in real-life situations.

Applications in Computer Science

The assignment problem can be used in computer science to allocate resources to different tasks, such as allocating memory to processes or assigning threads to processors.

Applications in Economics

The assignment problem can be used in economics to allocate resources to different agents, such as allocating workers to jobs or assigning projects to contractors.

Applications in Logistics

The assignment problem can be used in logistics to allocate resources to different activities, such as allocating vehicles to routes or assigning warehouses to customers.

Applications in Management

The assignment problem can be used in management to allocate resources to different projects, such as allocating employees to tasks or assigning budgets to departments.

Let’s consider the following scenario: a manager needs to assign three employees to three different tasks. Each employee has different skills, and each task requires specific skills. The manager wants to minimize the total time it takes to complete all the tasks. The skills and the time required for each task are given in the table below:

The assignment problem is to determine which employee should be assigned to which task to minimize the total time required. To solve this problem, we can use the Hungarian method, which we discussed in the previous blog.

Using the Hungarian method, we first subtract the smallest entry in each row from all the entries of the row:

Next, we subtract the smallest entry in each column from all the entries of the column:

We draw the minimum number of lines to cover all the zeros in the matrix, which in this case is three:

Since the number of lines is equal to the number of rows, we have an optimal solution. The assignments can be made by selecting the rows and columns with a single zero in the final matrix. In this case, the optimal assignments are:

Emp 1 to Task 3
Emp 2 to Task 2
Emp 3 to Task 1

This assignment results in a total time of 9 units.

I hope this example helps you better understand the assignment problem and how to solve it using the Hungarian method.

Solving the assignment problem may seem daunting, but with the right approach, it can be a straightforward process. By following the steps outlined in this guide, you can confidently tackle any assignment problem that comes your way.

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Operations Research

1 Operations Research-An Overview

History of O.R.
Approach, Techniques and Tools
Phases and Processes of O.R. Study
Typical Applications of O.R
Limitations of Operations Research
Models in Operations Research
O.R. in real world

2 Linear Programming: Formulation and Graphical Method

General formulation of Linear Programming Problem
Optimisation Models
Basics of Graphic Method
Important steps to draw graph
Multiple, Unbounded Solution and Infeasible Problems
Solving Linear Programming Graphically Using Computer
Application of Linear Programming in Business and Industry

3 Linear Programming-Simplex Method

Principle of Simplex Method
Computational aspect of Simplex Method
Simplex Method with several Decision Variables
Two Phase and M-method
Multiple Solution, Unbounded Solution and Infeasible Problem
Sensitivity Analysis
Dual Linear Programming Problem

4 Transportation Problem

Basic Feasible Solution of a Transportation Problem
Modified Distribution Method
Stepping Stone Method
Unbalanced Transportation Problem
Degenerate Transportation Problem
Transhipment Problem
Maximisation in a Transportation Problem

5 Assignment Problem

Solution of the Assignment Problem
Unbalanced Assignment Problem
Problem with some Infeasible Assignments
Maximisation in an Assignment Problem
Crew Assignment Problem

6 Application of Excel Solver to Solve LPP

Building Excel model for solving LP: An Illustrative Example

7 Goal Programming

Concepts of goal programming
Goal programming model formulation
Graphical method of goal programming
The simplex method of goal programming
Using Excel Solver to Solve Goal Programming Models
Application areas of goal programming

8 Integer Programming

Some Integer Programming Formulation Techniques
Binary Representation of General Integer Variables
Unimodularity
Cutting Plane Method
Branch and Bound Method
Solver Solution

9 Dynamic Programming

Dynamic Programming Methodology: An Example
Definitions and Notations
Dynamic Programming Applications

10 Non-Linear Programming

Solution of a Non-linear Programming Problem
Convex and Concave Functions
Kuhn-Tucker Conditions for Constrained Optimisation
Quadratic Programming
Separable Programming
NLP Models with Solver

11 Introduction to game theory and its Applications

Important terms in Game Theory
Saddle points
Mixed strategies: Games without saddle points
2 x n games
Exploiting an opponent’s mistakes

12 Monte Carlo Simulation

Reasons for using simulation
Monte Carlo simulation
Limitations of simulation
Steps in the simulation process
Some practical applications of simulation
Two typical examples of hand-computed simulation
Computer simulation

13 Queueing Models

Characteristics of a queueing model
Notations and Symbols
Statistical methods in queueing
The M/M/I System
The M/M/C System
The M/Ek/I System
Decision problems in queueing

Hungarian Method

The Hungarian method is a computational optimization technique that addresses the assignment problem in polynomial time and foreshadows following primal-dual alternatives. In 1955, Harold Kuhn used the term “Hungarian method” to honour two Hungarian mathematicians, Dénes Kőnig and Jenő Egerváry. Let’s go through the steps of the Hungarian method with the help of a solved example.

Hungarian Method to Solve Assignment Problems

The Hungarian method is a simple way to solve assignment problems. Let us first discuss the assignment problems before moving on to learning the Hungarian method.

What is an Assignment Problem?

A transportation problem is a type of assignment problem. The goal is to allocate an equal amount of resources to the same number of activities. As a result, the overall cost of allocation is minimised or the total profit is maximised.

Because available resources such as workers, machines, and other resources have varying degrees of efficiency for executing different activities, and hence the cost, profit, or loss of conducting such activities varies.

Assume we have ‘n’ jobs to do on ‘m’ machines (i.e., one job to one machine). Our goal is to assign jobs to machines for the least amount of money possible (or maximum profit). Based on the notion that each machine can accomplish each task, but at variable levels of efficiency.

Hungarian Method Steps

Check to see if the number of rows and columns are equal; if they are, the assignment problem is considered to be balanced. Then go to step 1. If it is not balanced, it should be balanced before the algorithm is applied.

Step 1 – In the given cost matrix, subtract the least cost element of each row from all the entries in that row. Make sure that each row has at least one zero.

Step 2 – In the resultant cost matrix produced in step 1, subtract the least cost element in each column from all the components in that column, ensuring that each column contains at least one zero.

Step 3 – Assign zeros

Analyse the rows one by one until you find a row with precisely one unmarked zero. Encircle this lonely unmarked zero and assign it a task. All other zeros in the column of this circular zero should be crossed out because they will not be used in any future assignments. Continue in this manner until you’ve gone through all of the rows.
Examine the columns one by one until you find one with precisely one unmarked zero. Encircle this single unmarked zero and cross any other zero in its row to make an assignment to it. Continue until you’ve gone through all of the columns.

Step 4 – Perform the Optimal Test

The present assignment is optimal if each row and column has exactly one encircled zero.
The present assignment is not optimal if at least one row or column is missing an assignment (i.e., if at least one row or column is missing one encircled zero). Continue to step 5. Subtract the least cost element from all the entries in each column of the final cost matrix created in step 1 and ensure that each column has at least one zero.

Step 5 – Draw the least number of straight lines to cover all of the zeros as follows:

(a) Highlight the rows that aren’t assigned.

(b) Label the columns with zeros in marked rows (if they haven’t already been marked).

(d) Continue with (b) and (c) until no further marking is needed.

(f) Simply draw the lines through all rows and columns that are not marked. If the number of these lines equals the order of the matrix, then the solution is optimal; otherwise, it is not.

Step 6 – Find the lowest cost factor that is not covered by the straight lines. Subtract this least-cost component from all the uncovered elements and add it to all the elements that are at the intersection of these straight lines, but leave the rest of the elements alone.

Step 7 – Continue with steps 1 – 6 until you’ve found the highest suitable assignment.

Hungarian Method Example

Use the Hungarian method to solve the given assignment problem stated in the table. The entries in the matrix represent each man’s processing time in hours.

$\begin{array}{l}\begin{bmatrix} & I & II & III & IV & V \\1 & 20 & 15 & 18 & 20 & 25 \\2 & 18 & 20 & 12 & 14 & 15 \\3 & 21 & 23 & 25 & 27 & 25 \\4 & 17 & 18 & 21 & 23 & 20 \\5 & 18 & 18 & 16 & 19 & 20 \\\end{bmatrix}\end{array} $

With 5 jobs and 5 men, the stated problem is balanced.

$\begin{array}{l}A = \begin{bmatrix}20 & 15 & 18 & 20 & 25 \\18 & 20 & 12 & 14 & 15 \\21 & 23 & 25 & 27 & 25 \\17 & 18 & 21 & 23 & 20 \\18 & 18 & 16 & 19 & 20 \\\end{bmatrix}\end{array} $

Subtract the lowest cost element in each row from all of the elements in the given cost matrix’s row. Make sure that each row has at least one zero.

$\begin{array}{l}A = \begin{bmatrix}5 & 0 & 3 & 5 & 10 \\6 & 8 & 0 & 2 & 3 \\0 & 2 & 4 & 6 & 4 \\0 & 1 & 4 & 6 & 3 \\2 & 2 & 0 & 3 & 4 \\\end{bmatrix}\end{array} $

Subtract the least cost element in each Column from all of the components in the given cost matrix’s Column. Check to see if each column has at least one zero.

$\begin{array}{l}A = \begin{bmatrix}5 & 0 & 3 & 3 & 7 \\6 & 8 & 0 & 0 & 0 \\0 & 2 & 4 & 4 & 1 \\0 & 1 & 4 & 4 & 0 \\2 & 2 & 0 & 1 & 1 \\\end{bmatrix}\end{array} $

When the zeros are assigned, we get the following:

The present assignment is optimal because each row and column contain precisely one encircled zero.

Where 1 to II, 2 to IV, 3 to I, 4 to V, and 5 to III are the best assignments.

Hence, z = 15 + 14 + 21 + 20 + 16 = 86 hours is the optimal time.

Practice Question on Hungarian Method

Use the Hungarian method to solve the following assignment problem shown in table. The matrix entries represent the time it takes for each job to be processed by each machine in hours.

$\begin{array}{l}\begin{bmatrix}J/M & I & II & III & IV & V \\1 & 9 & 22 & 58 & 11 & 19 \\2 & 43 & 78 & 72 & 50 & 63 \\3 & 41 & 28 & 91 & 37 & 45 \\4 & 74 & 42 & 27 & 49 & 39 \\5 & 36 & 11 & 57 & 22 & 25 \\\end{bmatrix}\end{array} $

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Frequently Asked Questions on Hungarian Method

What is hungarian method.

The Hungarian method is defined as a combinatorial optimization technique that solves the assignment problems in polynomial time and foreshadowed subsequent primal–dual approaches.

What are the steps involved in Hungarian method?

The following is a quick overview of the Hungarian method: Step 1: Subtract the row minima. Step 2: Subtract the column minimums. Step 3: Use a limited number of lines to cover all zeros. Step 4: Add some more zeros to the equation.

What is the purpose of the Hungarian method?

When workers are assigned to certain activities based on cost, the Hungarian method is beneficial for identifying minimum costs.

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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 2023-01-02 UTC.

Solving assignment problem using min-cost-flow ¶

The assignment problem has two equivalent statements:

Given a square matrix $A[1..N, 1..N]$ , you need to select $N$ elements in it so that exactly one element is selected in each row and column, and the sum of the values of these elements is the smallest.
There are $N$ orders and $N$ machines. The cost of manufacturing on each machine is known for each order. Only one order can be performed on each machine. It is required to assign all orders to the machines so that the total cost is minimized.

Here we will consider the solution of the problem based on the algorithm for finding the minimum cost flow (min-cost-flow) , solving the assignment problem in $\mathcal{O}(N^3)$ .

Description ¶

Let's build a bipartite network: there is a source $S$ , a drain $T$ , in the first part there are $N$ vertices (corresponding to rows of the matrix, or orders), in the second there are also $N$ vertices (corresponding to the columns of the matrix, or machines). Between each vertex $i$ of the first set and each vertex $j$ of the second set, we draw an edge with bandwidth 1 and cost $A_{ij}$ . From the source $S$ we draw edges to all vertices $i$ of the first set with bandwidth 1 and cost 0. We draw an edge with bandwidth 1 and cost 0 from each vertex of the second set $j$ to the drain $T$ .

We find in the resulting network the maximum flow of the minimum cost. Obviously, the value of the flow will be $N$ . Further, for each vertex $i$ of the first segment there is exactly one vertex $j$ of the second segment, such that the flow $F_{ij}$ = 1. Finally, this is a one-to-one correspondence between the vertices of the first segment and the vertices of the second part, which is the solution to the problem (since the found flow has a minimal cost, then the sum of the costs of the selected edges will be the lowest possible, which is the optimality criterion).

The complexity of this solution of the assignment problem depends on the algorithm by which the search for the maximum flow of the minimum cost is performed. The complexity will be $\mathcal{O}(N^3)$ using Dijkstra or $\mathcal{O}(N^4)$ using Bellman-Ford . This is due to the fact that the flow is of size $O(N)$ and each iteration of Dijkstra algorithm can be performed in $O(N^2)$ , while it is $O(N^3)$ for Bellman-Ford.

Implementation ¶

The implementation given here is long, it can probably be significantly reduced. It uses the SPFA algorithm for finding shortest paths.

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Quadratic assignment problem

Author: Thomas Kueny, Eric Miller, Natasha Rice, Joseph Szczerba, David Wittmann (SysEn 5800 Fall 2020)

1 Introduction
2.1 Koopmans-Beckman Mathematical Formulation
2.2.1 Parameters
2.3.1 Optimization Problem
2.4 Computational Complexity
2.5 Algorithmic Discussions
2.6 Branch and Bound Procedures
2.7 Linearizations
3.1 QAP with 3 Facilities
4.1 Inter-plant Transportation Problem
4.2 The Backboard Wiring Problem
4.3 Hospital Layout
4.4 Exam Scheduling System
5 Conclusion
6 References

Introduction

The Quadratic Assignment Problem (QAP), discovered by Koopmans and Beckmann in 1957 [1] , is a mathematical optimization module created to describe the location of invisible economic activities. An NP-Complete problem, this model can be applied to many other optimization problems outside of the field of economics. It has been used to optimize backboards, inter-plant transportation, hospital transportation, exam scheduling, along with many other applications not described within this page.

Theory, Methodology, and/or Algorithmic Discussions

Koopmans-beckman mathematical formulation.

Economists Koopmans and Beckman began their investigation of the QAP to ascertain the optimal method of locating important economic resources in a given area. The Koopmans-Beckman formulation of the QAP aims to achieve the objective of assigning facilities to locations in order to minimize the overall cost. Below is the Koopmans-Beckman formulation of the QAP as described by neos-guide.org.

Quadratic Assignment Problem Formulation

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F = (F_{ij})} is an n x n matrix where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_{ij}} the required flow between facilities Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j} . Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D = (D_{ij})} is an n x n matrix where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D_{ij}} the distance between facilities Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j} [2] . Intuitively, the product of distance and flow represents cost, and the objective is to minimize this cost. Ideally, two facilities with lots of flow (more product or supplies) between the two should be placed closer together than two facilities that rarely interact (low flow).

Consider the set of numbers Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N = \big\{1, 2, ...n\big\} } . The group of bijections between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N } and itself is defined as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{n} = \{\phi: N \rightarrow N \}} . Each member Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi \in S_n} represents a particular arrangement of the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} elements. As an example, the permutation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi=(3 2 1)} represents facility 3 being placed at location 1, facility 1 at location 3 and facility 2 at location 2. Each permutation can be assigned a permutation matrix. The permutation matrix of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi } can be written as [3]

The operation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle X_{\phi}DX_{\phi}^{T} } permutes the values of the distance matrix Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D} so that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (X_{\phi}DX_{\phi}^{T})_{i,j} } represents the distance between facilities Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j} if the facilities are arranged according to the permutation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi} . The cost function between facilities Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j} , after being permuted to their location, is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_{i,j}} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (X_{\phi}DX_{\phi}^{T})_{i,j} } [3] . To find the total cost, sum over each Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j} for the facilities arranged according to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi} .

Inner Product

The inner product of two matrices Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A,B} is defined as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle A,B \rangle= \sum_{i=1}^{n}\sum_{j=1}^{n}a_{i,j}b_{i,j}} . The inner product

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle F,X_{\phi}DX_{\phi}^{T} \rangle = \sum_{i=1}^{n}\sum_{j=1}^{n}F_{i,j}(X_{\phi}DX_{\phi}^{T})_{i,j}}

thus describes the total cost of assigning facilities [3] to locations according to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi} . The objective function to minimize cost can be written as

Note: The true objective cost function only requires summing entries above the diagonal in the matrix comprised of elements

Since this matrix is symmetric with zeroes on the diagonal, dividing by 2 removes the double count of each element to give the correct cost value. See the Numerical Example section for an example of this note.

Optimization Problem

With all of this information, the QAP can be summarized as:

Computational Complexity

QAP belongs to the classification of problems known as NP-complete, thus being a computationally complex problem. QAP’s NP-completeness was proven by Sahni and Gonzalez in 1976, who states that of all combinatorial optimization problems, QAP is the “hardest of the hard”. [2]

Algorithmic Discussions

While an algorithm that can solve QAP in polynomial time is unlikely to exist, there are three primary methods for acquiring the optimal solution to a QAP problem:

Dynamic Program
Cutting Plane

Branch and Bound Procedures

The third method has been proven to be the most effective in solving QAP, although when n > 15, QAP begins to become virtually unsolvable.

The Branch and Bound method was first proposed by Ailsa Land and Alison Doig in 1960 and is the most commonly used tool for solving NP-hard optimization problems.

A branch-and-bound algorithm consists of a systematic enumeration of candidate solutions by means of state space search: the set of candidate solutions is thought of as forming a rooted tree with the full set at the root. The algorithm explores branches of this tree, which represent subsets of the solution set. Before one lists all of the candidate solutions of a branch, the branch is checked against upper and lower estimated bounds on the optimal solution, and the branch is eliminated if it cannot produce a better solution than the best one found so far by the algorithm.

Linearizations

The first attempts to solve the QAP eliminated the quadratic term in the objective function of

in order to transform the problem into a (mixed) 0-1 linear program. The objective function is usually linearized by introducing new variables and new linear (and binary) constraints. Then existing methods for (mixed) linear integer programming (MILP) can be applied. The very large number of new variables and constraints, however, usually poses an obstacle for efficiently solving the resulting linear integer programs. MILP formulations provide LP relaxations of the problem which can be used to compute lower bounds.

Numerical Example

Qap with 3 facilities.

Consider the QAP of placing 3 facilities in 3 locations. Let the distance and flow matrices be defined respectively as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D=\begin{bmatrix} 0 & 5 & 6 \\ 5 & 0 & 3.6 \\ 6 & 3.6 & 0 \end{bmatrix} } , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle F=\begin{bmatrix} 0 & 10 & 3 \\ 10 & 0 & 6.5\\ 3 & 6.5 & 0 \end{bmatrix} } .

Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P} be the set of permutation matrices for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_3 } . The problem can be stated as follows:

The table lists the resulting cost function to place the facilities according to each permutation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi \in S_3 } . Note that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi = (123) } is the identity mapping where the number of the facility matches the location number. Based on these results, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi_4=(321)} is the arrangement of facilities that minimizes the cost function with a value of 86.5. Facility 1 should be placed at location 3, Facility 2 remains at location 2 and Facility 3 is placed at location 1.

For the optimal solution Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi_4=(13)} , the calculation is as follows:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle XDX^T = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & 5 & 6 \\ 5 & 0 & 3.6 \\ 6 & 3.6 & 0 \end{bmatrix}\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 3.6 & 6 \\ 3.6 & 0 & 5 \\ 6 & 5 & 0 \end{bmatrix} }

Looking at the figure of the optimal solution, the flow Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \times} distance values are repeated in this matrix which is why the sum of all elements must be divided by 2. This gives Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle F,X_{\phi}DX_{\phi}^{T} \rangle = 173} so the minimum cost is 86.5.

Applications

Inter-plant transportation problem.

The QAP was first introduced by Koopmans and Beckmann to address how economic decisions could be made to optimize the transportation costs of goods between both manufacturing plants and locations. [1] Factoring in the location of each of the manufacturing plants as well as the volume of goods between locations to maximize revenue is what distinguishes this from other linear programming assignment problems like the Knapsack Problem.

The Backboard Wiring Problem

As the QAP is focused on minimizing the cost of traveling from one location to another, it is an ideal approach to determining the placement of components in many modern electronics. Leon Steinberg proposed a QAP solution to optimize the layout of elements on a blackboard by minimizing the total amount of wiring required. [4]

When defining the problem Steinberg states that we have a set of n elements

as well as a set of r points

In his paper he derives the below formula:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d_{s(i)s(j))}} is the length of wire needed to connect tow components if one is placed at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_{\alpha}} and the other is placed at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_{\beta}} . If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_{\alpha}} has coordinates Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x_{\alpha}, y_{\alpha}, z_{\alpha})} then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d_{s(i)s(j))}} can be calculated by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d_{s(i)s(j))} = \surd(\bigl(x_{\alpha} - x_{\beta}\bigr)^{2} + \bigl(y_{\alpha} - y_{\beta}\bigr)^{2} + \bigl(z_{\alpha} - z_{\beta}\bigr)^{2}) }

In his paper Steinberg a backboard with a 9 by 4 array, allowing for 36 potential positions for the 34 components that needed to be placed on the backboard. For the calculation, he selected a random initial placement of s1 and chose a random family of 25 unconnected sets.

The initial placement of components is shown below:

After the initial placement of elements, it took an additional 35 iterations to get us to our final optimized backboard layout. Leading to a total of 59 iterations and a final wire length of 4,969.440.

Hospital Layout

Building new hospitals was a common event in 1977 when Alealid N Elshafei wrote his paper on "Hospital Layouts as a Quadratic Assignment Problem". [5] With the high initial cost to construct the hospital and to staff it, it is important to ensure that it is operating as efficiently as possible. Elshafei's paper was commissioned to create an optimization formula to locate clinics within a building in such a way that minimizes the total distance that a patient travels within the hospital throughout the year. When doing a study of a major hospital in Cairo he determined that the Outpatient ward was acting as a bottleneck in the hospital and focused his efforts on optimizing the 17 departments there.

Elshafei identified the following QAP to determine where clinics should be placed:

Such that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{j \epsilon J }y_{ij} = 1 \forall i \epsilon I }

It can take a large amount of computational power to calculate the solution to a QAP problem, necessitating the use of a heuristic solution rather than using an optimizing algorithm. This approach is composed of two parts(Parts A and Part B). Part A develops the initial layout of the hospital, this is done by the following two strategies. Strategy 1 where we use a rank Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L_{i} } that designates the number of interactions facility Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i } has with with the other facilities. As well as ranking them in terms of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R_{j} } , which designates the sum of distances from j to all other locations. then we go through the list and find the facility with the greatest Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L_{i} } and the least Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R_{j} } . Strategy 2 says that at any step in the process we next place an unassigned facility that has the largest number of interactions with the most recently placed facility. Part B covers improving the initial solution developed in Part A, here at any step in the process we test whether the assignment is best for all of the facilities that are involved and make swaps where necessary until there are no more swaps to be made.

For the Cairo hospital with 17 clinics, and one receiving and recording room bringing us to a total of 18 facilities. By running the above optimization Elshafei was able to get the total distance per year down to 11,281,887 from a distance of 13,973,298 based on the original hospital layout.

Exam Scheduling System

The scheduling system uses matrices for Exams, Time Slots, and Rooms with the goal of reducing the rate of schedule conflicts. To accomplish this goal, the “examination with the highest cross faculty student is been prioritized in the schedule after which the examination with the highest number of cross-program is considered and finally with the highest number of repeating student, at each stage group with the highest number of student are prioritized.” [6]

Optimization of the QAP is easily stated but requires significant computational power. The number of potential facility/location combinations grows as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n!} . Thus for a QAP of just Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n=6 } facilities and location, there are Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 720} facility/location pairings. The QAP can be used in applications beyond just the location of economic activities that it was originally developed for by Koopmans and Beckmann. [1] The QAP has been used to optimize backboard wiring [4] , hospital layouts [5] , and to schedule exams [6] , along with other applications not defined on this page.

↑ 1.0 1.1 1.2 Koopmans, T., & Beckmann, M. (1957). Assignment Problems and the Location of Economic Activities. Econometrica, 25(1), 53-76. doi:10.2307/1907742
↑ 2.0 2.1 Quadratic Assignment Problem. (2020). Retrieved December 14, 2020, from https://neos-guide.org/content/quadratic-assignment-problem
↑ 3.0 3.1 3.2 Burkard, R. E., Çela, E., Pardalos, P. M., & Pitsoulis, L. S. (2013). The Quadratic Assignment Problem. https://www.opt.math.tugraz.at/~cela/papers/qap_bericht.pdf .
↑ 4.0 4.1 Leon Steinberg. The Backboard Wiring Problem: A Placement Algorithm. SIAM Review . 1961;3(1):37.
↑ 5.0 5.1 Alwalid N. Elshafei. Hospital Layout as a Quadratic Assignment Problem. Operational Research Quarterly (1970-1977) . 1977;28(1):167. doi:10.2307/300878
↑ 6.0 6.1 Muktar, D., & Ahmad, Z.M. (2014). Examination Scheduling System Based On Quadratic Assignment.
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Procedure, Example Solved Problem | Operations Research - Solution of assignment problems (Hungarian Method) | 12th Business Maths and Statistics : Chapter 10 : Operations Research

Chapter: 12th business maths and statistics : chapter 10 : operations research.

Solution of assignment problems (Hungarian Method)

First check whether the number of rows is equal to the numbers of columns, if it is so, the assignment problem is said to be balanced.

Step :1 Choose the least element in each row and subtract it from all the elements of that row.

Step :2 Choose the least element in each column and subtract it from all the elements of that column. Step 2 has to be performed from the table obtained in step 1.

Step:3 Check whether there is atleast one zero in each row and each column and make an assignment as follows.

Step :4 If each row and each column contains exactly one assignment, then the solution is optimal.

Example 10.7

Solve the following assignment problem. Cell values represent cost of assigning job A, B, C and D to the machines I, II, III and IV.

Here the number of rows and columns are equal.

∴ The given assignment problem is balanced. Now let us find the solution.

Step 1: Select a smallest element in each row and subtract this from all the elements in its row.

Look for atleast one zero in each row and each column.Otherwise go to step 2.

Step 2: Select the smallest element in each column and subtract this from all the elements in its column.

Since each row and column contains atleast one zero, assignments can be made.

Step 3 (Assignment):

Thus all the four assignments have been made. The optimal assignment schedule and total cost is

The optimal assignment (minimum) cost

Example 10.8

Consider the problem of assigning five jobs to five persons. The assignment costs are given as follows. Determine the optimum assignment schedule.

∴ The given assignment problem is balanced.

Now let us find the solution.

The cost matrix of the given assignment problem is

Column 3 contains no zero. Go to Step 2.

Thus all the five assignments have been made. The Optimal assignment schedule and total cost is

The optimal assignment (minimum) cost = ` 9

Example 10.9

Solve the following assignment problem.

Since the number of columns is less than the number of rows, given assignment problem is unbalanced one. To balance it , introduce a dummy column with all the entries zero. The revised assignment problem is

Here only 3 tasks can be assigned to 3 men.

Step 1: is not necessary, since each row contains zero entry. Go to Step 2.

Step 3 (Assignment) :

Since each row and each columncontains exactly one assignment,all the three men have been assigned a task. But task S is not assigned to any Man. The optimal assignment schedule and total cost is

The optimal assignment (minimum) cost = ₹ 35

Nash Balanced Assignment Problem

Conference paper
First Online: 21 November 2022
Cite this conference paper

Minh Hieu Nguyen 11 ,
Mourad Baiou 11 &
Viet Hung Nguyen 11

Part of the book series: Lecture Notes in Computer Science ((LNCS,volume 13526))

Included in the following conference series:

International Symposium on Combinatorial Optimization

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2 Citations

In this paper, we consider a variant of the classic Assignment Problem (AP), called the Balanced Assignment Problem (BAP) [ 2 ]. The BAP seeks to find an assignment solution which has the smallest value of max-min distance : the difference between the maximum assignment cost and the minimum one. However, by minimizing only the max-min distance, the total cost of the BAP solution is neglected and it may lead to an inefficient solution in terms of total cost. Hence, we propose a fair way based on Nash equilibrium [ 1 , 3 , 4 ] to inject the total cost into the objective function of the BAP for finding assignment solutions having a better trade-off between the two objectives: the first aims at minimizing the total cost and the second aims at minimizing the max-min distance. For this purpose, we introduce the concept of Nash Fairness (NF) solutions based on the definition of proportional-fair scheduling adapted in the context of the AP: a transfer of utilities between the total cost and the max-min distance is considered to be fair if the percentage increase in the total cost is smaller than the percentage decrease in the max-min distance and vice versa.

We first show the existence of a NF solution for the AP which is exactly the optimal solution minimizing the product of the total cost and the max-min distance. However, finding such a solution may be difficult as it requires to minimize a concave function. The main result of this paper is to show that finding all NF solutions can be done in polynomial time. For that, we propose a Newton-based iterative algorithm converging to NF solutions in polynomial time. It consists in optimizing a sequence of linear combinations of the two objective based on Weighted Sum Method [ 5 ]. Computational results on various instances of the AP are presented and commented.

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Minh Hieu Nguyen, Mourad Baiou & Viet Hung Nguyen

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ESSEC Business School of Paris, Cergy Pontoise Cedex, France

Ivana Ljubić

IBM TJ Watson Research Center, Yorktown Heights, NY, USA

Francisco Barahona

Georgia Institute of Technology, Atlanta, GA, USA

Santanu S. Dey

Université Paris-Dauphine, Paris, France

A. Ridha Mahjoub

Proposition 1 . There may be more than one NF solution for the AP.

Let us illustrate this by an instance of the AP having the following cost matrix

By verifying all feasible assignment solutions in this instance, we obtain easily three assignment solutions $(1-1, 2-2, 3-3), (1-2, 2-3, 3-1)$ , $(1-3, 2-2, 3-1)$ and $(1-3, 2-1, 3-2)$ corresponding to 4 NF solutions (280, 36), (320, 32), (340, 30) and (364, 28). Note that $i-j$ where $1 \le i,j \le 3$ represents the assignment between worker i and job j in the solution of this instance. $\square $

We recall below the proofs of some recent results that we have published in [ 10 ]. They are needed to prove the new results presented in this paper.

Theorem 2 [ 10 ] . $(P^{*},Q^{*}) = {{\,\mathrm{arg\,min}\,}}_{(P,Q) \in \mathcal {S}} PQ$ is a NF solution.

Obviously, there always exists a solution $(P^{*},Q^{*}) \in \mathcal {S}$ such that

Now $\forall (P',Q') \in \mathcal {S}$ we have $P'Q' \ge P^{*}Q^{*}$ . Then

The first inequality holds by the Cauchy-Schwarz inequality.

Hence, $(P^{*},Q^{*})$ is a NF solution. $\square $

Theorem 3 [ 10 ] . $(P^{*},Q^{*}) \in \mathcal {S}$ is a NF solution if and only if $(P^{*},Q^{*})$ is an optimal solution of $\mathcal {P(\alpha ^{*})}$ where $\alpha ^{*} = \frac{Q^{*}}{P^{*}}$ .

Firstly, let $(P^{*},Q^{*})$ be a NF solution and $\alpha ^{*} = \frac{Q^{*}}{P^{*}}$ . We will show that $(P^{*},Q^{*})$ is an optimal solution of $\mathcal {P(\alpha ^{*})}$ .

Since $(P^{*},Q^{*})$ is a NF solution, we have

Since $\alpha ^{*} = \frac{Q^{*}}{P^{*}}$ , we have $\alpha ^{*}P^{*}+Q^{*} = 2Q^{*}$ .

Dividing two sides of ( 6 ) by $P^{*} > 0$ we obtain

So we deduce from ( 7 )

Hence, $(P^{*},Q^{*})$ is an optimal solution of $\mathcal {P}(\alpha ^{*})$ .

Now suppose $\alpha ^{*} = \frac{Q^{*}}{P^{*}}$ and $(P^{*},Q^{*})$ is an optimal solution of $\mathcal {P}(\alpha ^{*})$ , we show that $(P^{*},Q^{*})$ is a NF solution.

If $(P^{*},Q^{*})$ is not a NF solution, there exists a solution $(P',Q') \in \mathcal {S}$ such that

We have then

which contradicts the optimality of $(P^{*},Q^{*})$ . $\square $

Lemma 3 [ 10 ] . Let $\alpha , \alpha ' \in \mathbb {R}_+$ and $(P_{\alpha }, Q_{\alpha })$ , $(P_{\alpha '}, Q_{\alpha '})$ be the optimal solutions of $\mathcal {P(\alpha )}$ and $\mathcal {P(\alpha ')}$ respectively, if $\alpha \le \alpha '$ then $P_{\alpha } \ge P_{\alpha '}$ and $Q_{\alpha } \le Q_{\alpha '}$ .

The optimality of $(P_{\alpha }, Q_{\alpha })$ and $(P_{\alpha '}, Q_{\alpha '})$ gives

By adding both sides of ( 8a ) and ( 8b ), we obtain $(\alpha - \alpha ') (P_{\alpha } - P_{\alpha '}) \le 0$ . Since $\alpha \le \alpha '$ , it follows that $P_{\alpha } \ge P_{\alpha '}$ .

On the other hand, inequality ( 8a ) implies $Q_{\alpha '} - Q_{\alpha } \ge \alpha (P_{\alpha } - P_{\alpha '}) \ge 0$ that leads to $Q_{\alpha } \le Q_{\alpha '}$ . $\square $

Lemma 4 [ 10 ] . During the execution of Procedure Find ( $\alpha _{0})$ in Algorithm 1 , $\alpha _{i} \in [0,1], \, \forall i \ge 1$ . Moreover, if $T_{0} \ge 0$ then the sequence $\{\alpha _i\}$ is non-increasing and $T_{i} \ge 0, \, \forall i \ge 0$ . Otherwise, if $T_{0} \le 0$ then the sequence $\{\alpha _i\}$ is non-decreasing and $T_{i} \le 0, \, \forall i \ge 0$ .

Since $P \ge Q \ge 0, \, \forall (P, Q) \in \mathcal {S}$ , it follows that $\alpha _{i+1} = \frac{Q_i}{P_i} \in [0,1], \, \forall i \ge 0$ .

We first consider $T_{0} \ge 0$ . We proof $\alpha _i \ge \alpha _{i+1}, \, \forall i \ge 0$ by induction on i . For $i = 0$ , we have $T_{0} = \alpha _{0} P_{0} - Q_{0} = P_{0}(\alpha _{0}-\alpha _{1}) \ge 0$ , it follows that $\alpha _{0} \ge \alpha _{1}$ . Suppose that our hypothesis is true until $i = k \ge 0$ , we will prove that it is also true with $i = k+1$ .

Indeed, we have

The inductive hypothesis gives $\alpha _k \ge \alpha _{k+1}$ that implies $P_{k+1} \ge P_k > 0$ and $Q_{k} \ge Q_{k+1} \ge 0$ according to Lemma 3 . It leads to $Q_{k}P_{k+1} - P_{k}Q_{k+1} \ge 0$ and then $\alpha _{k+1} - \alpha _{k+2} \ge 0$ .

Hence, we have $\alpha _{i} \ge \alpha _{i+1}, \, \forall i \ge 0$ .

Consequently, $T_{i} = \alpha _{i}P_{i} - Q_{i} = P_{i}(\alpha _{i}-\alpha _{i+1}) \ge 0, \, \forall i \ge 0$ .

Similarly, if $T_{0} \le 0$ we obtain that the sequence $\{\alpha _i\}$ is non-decreasing and $T_{i} \le 0, \, \forall i \ge 0$ . That concludes the proof. $\square $

Lemma 5 [ 10 ] . From each $\alpha _{0} \in [0,1]$ , Procedure Find $(\alpha _{0})$ converges to a coefficient $\alpha _{k} \in \mathcal {C}_{0}$ satisfying $\alpha _{k}$ is the unique element $\in \mathcal {C}_{0}$ between $\alpha _{0}$ and $\alpha _{k}$ .

As a consequence of Lemma 4 , Procedure $\textit{Find}(\alpha _{0})$ converges to a coefficient $\alpha _{k} \in [0,1], \forall \alpha _{0} \in [0,1]$ .

By the stopping criteria of Procedure Find $(\alpha _{0})$ , when $T_{k} = \alpha _{k} P_{k} - Q_{k} = 0$ we obtain $\alpha _{k} \in C_{0}$ and $(P_{k},Q_{k})$ is a NF solution. (Theorem 3 )

If $T_{0} = 0$ then obviously $\alpha _{k} = \alpha _{0}$ . We consider $T_{0} > 0$ and the sequence $\{\alpha _i\}$ is now non-negative, non-increasing. We will show that $[\alpha _{k},\alpha _{0}] \cap \mathcal {C}_{0} = \alpha _{k}$ .

Suppose that we have $\alpha \in (\alpha _{k},\alpha _{0}]$ and $\alpha \in \mathcal {C}_{0}$ corresponding to a NF solution ( P , Q ). Then there exists $1 \le i \le k$ such that $\alpha \in (\alpha _{i}, \alpha _{i-1}]$ . Since $\alpha \le \alpha _{i-1}$ , $P \ge P_{i-1}$ and $Q \le Q_{i-1}$ due to Lemma 3 . Thus, we get

By the definitions of $\alpha $ and $\alpha _{i}$ , inequality ( 9 ) is equivalent to $\alpha \le \alpha _{i}$ which leads to a contradiction.

By repeating the same argument for $T_{0} < 0$ , we also have a contradiction. $\square $

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Nguyen, M.H., Baiou, M., Nguyen, V.H. (2022). Nash Balanced Assignment Problem. In: Ljubić, I., Barahona, F., Dey, S.S., Mahjoub, A.R. (eds) Combinatorial Optimization. ISCO 2022. Lecture Notes in Computer Science, vol 13526. Springer, Cham. https://doi.org/10.1007/978-3-031-18530-4_13

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IBM Placement Paper | Quantitative Analysis Set - 5

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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