what is an assignment problem give two applications

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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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International Symposium on Combinatorial Optimization

ISCO 2022: Combinatorial Optimization pp 172–186 Cite as

Nash Balanced Assignment Problem

• Minh Hieu Nguyen 11 ,
• Mourad Baiou 11 &
• Viet Hung Nguyen 11  
• Conference paper
• First Online: 21 November 2022

349 Accesses

2 Citations

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

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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Bertsimas, D., Farias, V.F., Trichakis, N.: The price of fairness. Oper. Res. January–February 59 (1), 17–31 (2011)

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INP Clermont Auvergne, Univ Clermont Auvergne, Mines Saint-Etienne, CNRS, UMR 6158 LIMOS, 1 Rue de la Chebarde, Aubiere Cedex, France

Minh Hieu Nguyen, Mourad Baiou & Viet Hung Nguyen

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Correspondence to Viet Hung Nguyen .

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Editors and affiliations.

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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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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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:

Unbalanced Assignment Problem: Definition, Formulation, and Solution Methods

Table of Contents

Are you familiar with the assignment problem in Operations Research (OR)? This problem deals with assigning tasks to workers in a way that minimizes the total cost or time needed to complete the tasks. But what if the number of tasks and workers is not equal? In this case, we face the Unbalanced Assignment Problem (UAP). This blog will help you understand what the UAP is, how to formulate it, and how to solve it.

What is the Unbalanced Assignment Problem?

The Unbalanced Assignment Problem is an extension of the Assignment Problem in OR, where the number of tasks and workers is not equal. In the UAP, some tasks may remain unassigned, while some workers may not be assigned any task. The objective is still to minimize the total cost or time required to complete the assigned tasks, but the UAP has additional constraints that make it more complex than the traditional assignment problem.

Formulation of the Unbalanced Assignment Problem

To formulate the UAP, we start with a matrix that represents the cost or time required to assign each task to each worker. If the matrix is square, we can use the Hungarian algorithm to solve the problem. But when the matrix is not square, we need to add dummy tasks or workers to balance the matrix. These dummy tasks or workers have zero costs and are used to make the matrix square.

Once we have a square matrix, we can apply the Hungarian algorithm to find the optimal assignment. However, we need to be careful in interpreting the results, as the assignment may include dummy tasks or workers that are not actually assigned to anything.

Solutions for the Unbalanced Assignment Problem

Besides the Hungarian algorithm, there are other methods to solve the UAP, such as the transportation algorithm and the auction algorithm. The transportation algorithm is based on transforming the UAP into a transportation problem, which can be solved with the transportation simplex method. The auction algorithm is an iterative method that simulates a bidding process between the tasks and workers to find the optimal assignment.

In summary, the Unbalanced Assignment Problem is a variant of the traditional Assignment Problem in OR that deals with assigning tasks to workers when the number of tasks and workers is not equal. To solve the UAP, we need to balance the matrix by adding dummy tasks or workers and then apply algorithms such as the Hungarian algorithm, the transportation algorithm, or the auction algorithm. Understanding the UAP can help businesses and organizations optimize their resource allocation and improve their operational efficiency.

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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

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12 Monte Carlo Simulation

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Quadratic Assignment Problem (QAP)

• Channel Assignment Problem
• Assignment Operators In C++
• Solidity - Assignment Operators
• Job Assignment Problem using Branch And Bound
• Range Minimum Query with Range Assignment
• Transportation Problem | Set 1 (Introduction)
• Transportation Problem | Set 2 (NorthWest Corner Method)
• Transportation Problem | Set 3 (Least Cost Cell Method)
• Transportation Problem | Set 4 (Vogel's Approximation Method)
• Assignment Operators in C
• QA - Placement Quizzes | Profit and Loss | Question 7
• QA - Placement Quizzes | Profit and Loss | Question 4
• QA - Placement Quizzes | Profit and Loss | Question 12
• QA - Placement Quizzes | Profit and Loss | Question 10
• QA - Placement Quizzes | Profit and Loss | Question 8
• QA - Placement Quizzes | Age | Question 4
• QA - Placement Quizzes | Profit and Loss | Question 9
• QA - Placement Quizzes | Profit and Loss | Question 11
• IBM Placement Paper | Quantitative Analysis Set - 5
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• Top 50 Array Coding Problems for Interviews
• Introduction to Recursion - Data Structure and Algorithm Tutorials
• Difference between BFS and DFS
• Must Do Coding Questions for Product Based Companies
• A* Search Algorithm
• Top 10 Algorithms in Interview Questions
• Sliding Window Technique
• How to write a Pseudo Code?
• Asymptotic Notation and Analysis (Based on input size) in Complexity Analysis of Algorithms

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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Table of Contents

Assignment Problem is a special type of linear programming problem where the objective is to minimize 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 optimized (Maximized or Minimized).”Several problems of management have a structure identical with the assignment problem.

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 person is given. His objective is to assign each person to one and only one job in such a way that the total cost of assignment is minimized.

APPLICATIONS OF ASSIGNMENT PROBLEM

Few applications of assignment problem are as follows:

• Assignment of employees to machines.
• Assignment of operators to jobs.
• Effectiveness of teachers and subjects.
• Allocation of machines for optimum utilization of space.
• Allocation of salesmen to different sales areas.
• Allocation of clerks to various counters.

In all the cases, the objective is to minimize the total time and cost or otherwise maximize the sales and returns.

HUNGARIAN METHOD

Hungarian method is named after Hungarian mathematician D Konig who developed the assignment problem.

The Hungarian method is also known as Flood’s Technique or the Reduced Matrix method. This method of assignment provides an efficient means of finding the optimal solutions without having to make a direct comparison of every option. It operates on a principle of matrix reduction.

STEPS IN HUNGARIAN METHOD

Hungarian method of assignment problem (minimization case) can be summarized in the following steps:

STEP 1: NOTE THE MATRIX

From the given problem, find out the cost table. Note that if the number of origins is not equal to the number of destinations then a dummy origin or destination must be added.

STEP 2: ROW REDUCTION

In each row of the table find out the smallest cost element, subtract this smallest cost element from each element in that row. So, that there will be at least one zero in each row of the new table. This new table is known as First Reduced Cost Table.

Step 3: COLUMN REDUCTION

In each column of the table find out the smallest cost element, subtract this smallest cost element from each element in that column. As a result of this, each row and column has at least one zero element. This new table is known as Second Reduced Cost Table.

STEP 4: MARK ASSIGNMENTS

Now determine an assignment as follows:

• For each row or column with a single zero element cell that has not be assigned or eliminated, box that zero element as an assigned cell.
• For every zero that becomes assigned, cross out all other zeros in the same row and for column.
• If for a row and for a column there are two or more zero and one can’t be chosen by inspection, choose the assigned zero cell arbitrarily.
• The above procedures may be repeated until every zero element cell is either assigned (boxed) or crossed out.

An optimum assignment is found, if the number of assigned cells is equal to the number of rows (and columns). In case a zero cell had chosen arbitrarily, there may be an alternate optimum. If no optimum solution is found i.e. some rows or columns without an assignment then go to Step 6.

STEP 5: MARKING ROWS AND COLUMNS

Draw a set of lines equal to the number of assignments which has been made in Step 4, covering all the zeros in the following manner:

• Mark check (√) to those rows where no assignment has been made.
• Examine the checked (√) rows. If any zero element cell occurs in those rows, check (√) the respective columns that contains those zeros.
• Examine the checked (√) columns. If any assigned zero element occurs in those columns, check (√) the respective rows that contain those assigned zeros.
• The process may be repeated until now more rows or column can be checked.
• Draw lines through all unchecked rows and through all checked columns.

STEP 7: REASSIGNMENT

Examine those elements that are not covered by a line. Choose the smallest of these elements and subtract this smallest from all the elements that do not have a line through them.

Add this smallest element to every element that lies at the intersection of two lines. Then the resulting matrix is a new revised cost table.

EXAMPLE OF HUNGARIAN METHOD

In a computer centre after studying carefully the three expert programmes, the head of computer centre, estimates the computer time in minutes required by the experts for the application programmes as follows:

Assign the programmers to the programmes in such a way that the total computer time is minimum.

The Hungarian method is used to obtain an optimal solution.

Step (1) & (2):

The minimum time element in row 1, 2 and 3 is 80, 80 and 110. respectively. Subtract these elements from all elements in this respective row.

The reduced time matrix is shown in following table (1) Table 1:

In reduced Table (1) the minimum time element in columns A, B, and C is 0,10 and 0 respectively; subtract these elements from all elements in this resp. column to get the reduced time matrix as shown in Table 2.

Step 3 (a):

Examine all the rows starting from first one and make the assignments at zero.

Assign this cell as shown in table 4.

(c) Since the number of Assignments (= 3) equal the no of rows (= 3), the optimal solution is obtained.

The pattern of assignment among programmers and programmes with their respective line (in minutes) is given below:

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