describe the hungarian method of solving the assignment problem

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).

(c) Highlight the rows that have assignments in indicated columns (if they haven’t previously 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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Hungarian Algorithm for Assignment Problem | Set 2 (Implementation)

Given a 2D array , arr of size N*N where arr[i][j] denotes the cost to complete the j th job by the i th worker. Any worker can be assigned to perform any job. The task is to assign the jobs such that exactly one worker can perform exactly one job in such a way that the total cost of the assignment is minimized.

Input: arr[][] = {{3, 5}, {10, 1}} Output: 4 Explanation: The optimal assignment is to assign job 1 to the 1st worker, job 2 to the 2nd worker. Hence, the optimal cost is 3 + 1 = 4. Input: arr[][] = {{2500, 4000, 3500}, {4000, 6000, 3500}, {2000, 4000, 2500}} Output: 4 Explanation: The optimal assignment is to assign job 2 to the 1st worker, job 3 to the 2nd worker and job 1 to the 3rd worker. Hence, the optimal cost is 4000 + 3500 + 2000 = 9500.

Different approaches to solve this problem are discussed in this article .

Approach: The idea is to use the Hungarian Algorithm to solve this problem. The algorithm is as follows:

• For each row of the matrix, find the smallest element and subtract it from every element in its row.
• Repeat the step 1 for all columns.
• Cover all zeros in the matrix using the minimum number of horizontal and vertical lines.
• Test for Optimality : If the minimum number of covering lines is N , an optimal assignment is possible. Else if lines are lesser than N , an optimal assignment is not found and must proceed to step 5.
• Determine the smallest entry not covered by any line. Subtract this entry from each uncovered row, and then add it to each covered column. Return to step 3.

Consider an example to understand the approach:

Let the 2D array be: 2500 4000 3500 4000 6000 3500 2000 4000 2500 Step 1: Subtract minimum of every row. 2500, 3500 and 2000 are subtracted from rows 1, 2 and 3 respectively. 0   1500  1000 500  2500   0 0   2000  500 Step 2: Subtract minimum of every column. 0, 1500 and 0 are subtracted from columns 1, 2 and 3 respectively. 0    0   1000 500  1000   0 0   500  500 Step 3: Cover all zeroes with minimum number of horizontal and vertical lines. Step 4: Since we need 3 lines to cover all zeroes, the optimal assignment is found.   2500   4000  3500  4000  6000   3500   2000  4000  2500 So the optimal cost is 4000 + 3500 + 2000 = 9500

For implementing the above algorithm, the idea is to use the max_cost_assignment() function defined in the dlib library . This function is an implementation of the Hungarian algorithm (also known as the Kuhn-Munkres algorithm) which runs in O(N 3 ) time. It solves the optimal assignment problem. 

Below is the implementation of the above approach:

Time Complexity: O(N 3 ) Auxiliary Space: O(N 2 )

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This paper has always been one of my favorite “children,” combining as it does elements of the duality of linear programming and combinatorial tools from graph theory. It may be of some interest to tell the story of its origin.

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H.W. Kuhn, On the origin of the Hungarian Method , History of mathematical programming; a collection of personal reminiscences (J.K. Lenstra, A.H.G. Rinnooy Kan, and A. Schrijver, eds.), North Holland, Amsterdam, 1991, pp. 77–81.

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Kuhn, H.W. (2010). The Hungarian Method for the Assignment Problem. In: Jünger, M., et al. 50 Years of Integer Programming 1958-2008. Springer, Berlin, Heidelberg. https://doi.org/10.1007/978-3-540-68279-0_2

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Assignment problem: Hungarian method 3

Assignment problem: Hungarian Method Nui Ruppert (Mtk_Nr.: 373224) David Lenh (Mtk_Nr.: 368343) Amir Farshchi Tabrizi (Mtk-Nr.: 372894)

In this OR-Wiki entry we're going to explain the Hungarian method with 3 examples. In the first example you'll find the optimal solution after a few steps with the help of the reduced matrix. The second example illustrates a complex case where you need to proceed all the steps of the algorithm to get to an optimal solution. Finally in the third example we will show how to solve a maximization problem with the Hungarian method.

Inhaltsverzeichnis

• 1 Introduction
• 2 Example 1 – Minimization problem
• 3 Example 2 – Minimazation problem
• 4 Example 3 – Maximization problem
• 6 References

Introduction

The Hungarian method is a combinatorial optimization algorithm which was developed and published by Harold Kuhn in 1955. This method was originally invented for the best assignment of a set of persons to a set of jobs. It is a special case of the transportation problem. The algorithm finds an optimal assignment for a given “n x n” cost matrix. “Assignment problems deal with the question how to assign n items (e.g. jobs) to n machines (or workers) in the best possible way. […] Mathematically an assignment is nothing else than a bijective mapping of a finite set into itself […]” [1]

The assignment constraints are mathematically defined as:

To make clear how to solve an assignment problem with the Hungarian algorithm we will show you the different cases with several examples which can occur .

Example 1 – Minimization problem

In this example we have to assign 4 workers to 4 machines. Each worker causes different costs for the machines. Your goal is to minimize the total cost to the condition that each machine goes to exactly 1 person and each person works at exactly 1 machine. For comprehension: Worker 1 causes a cost of 6 for machine 1 and so on …

To solve the problem we have to perform the following steps:

Step 1 – Subtract the row minimum from each row.

Step 2 – Subtract the column minimum from each column from the reduced matrix.

The idea behind these 2 steps is to simplify the matrix since the solution of the reduced matrix will be exactly the same as of the original matrix.

Step 3 – Assign one “0” to each row & column.

Now that we have simplified the matrix we can assign each worker with the minimal cost to each machine which is represented by a “0”.

- In the first row we have one assignable “0” therefore we assign it to worker 3 .

- In the second row we also only have one assignable “0” therefore we assign it to worker 4 .

- In the third row we have two assignable “0”. We leave it as it is for now.

- In the fourth row we have one assignable “0” therefore we assign it. Consider that we can only assign each worker to each machine hence we can’t allocate any other “0” in the first column.

- Now we go back to the third row which now only has one assignable “0” for worker 2 .

As soon as we can assign each worker to one machine, we have the optimal solution . In this case there is no need to proceed any further steps. Remember also, if we decide on an arbitrary order in which we start allocating the “0”s then we may get into a situation where we have 3 assignments as against the possible 4. If we assign a “0” in the third row to worker 1 we wouldn’t be able to allocate any “0”s in column one and row two.

The rule to assign the “0”:

- If there is an assignable “0”, only 1 assignable “0” in any row or any column, assign it.

- If there are more than 1, leave it and proceed.

This rule would try to give us as many assignments as possible.

Now there are also cases where you won’t get an optimal solution for a reduced matrix after one iteration. The following example will explain it.

Example 2 – Minimazation problem

In this example we have the fastest taxi company that has to assign each taxi to each passenger as fast as possible. The numbers in the matrix represent the time to reach the passenger.

We proceed as in the first example.

Iteration 1:

Now we have to assign the “0”s for every row respectively to the rule that we described earlier in example 1.

- In the first row we have one assignable “0” therefore we assign it and no other allocation in column 2 is possible.

- In the second row we have one assignable “0” therefore we assign it.

- In the third row we have several assignable “0”s. We leave it as it is for now and proceed.

- In the fourth and fifth row we have no assignable “0”s.

Now we proceed with the allocations of the “0”s for each column .

- In the first column we have one assignable “0” therefore we assign it. No other “0”s in row 3 are assignable anymore.

Now we are unable to proceed because all the “0”s either been assigned or crossed. The crosses indicate that they are not fit for assignments because assignments are already made.

We realize that we have 3 assignments for this 5x5 matrix. In the earlier example we were able to get 4 assignments for a 4x4 matrix. Now we have to follow another procedure to get the remaining 2 assignments (“0”).

Step 4 – Tick all unassigned rows.

Step 5 – If a row is ticked and has a “0”, then tick the corresponding column (if the column is not yet ticked).

Step 6 – If a column is ticked and has an assignment, then tick the corresponding row (if the row is not yet ticked).

Step 7 - Repeat step 5 and 6 till no more ticking is possible.

In this case there is no more ticking possible and we proceed with the next step.

Step 8 – Draw lines through unticked rows and ticked columns. The number of lines represents the maximum number of assignments possible.

Step 9 – Find out the smallest number which does not have any line passing through it. We call it Theta. Subtract theta from all the numbers that do not have any lines passing through them and add theta to all those numbers that have two lines passing through them. Keep the rest of them the same.

(With this step we create a new “0”)

With the new assignment matrix we start to assign the “0”s after the explained rules. Nevertheless we have 4 assignments against the required 5 for an optimal solution. Therefore we have to repeat step 4 – 9.

Iteration 2:

Step 4 – Tick all unassigned row.

Note: The indices of the ticks show you the order we added them.

Iteration 3:

Iteration 4:

After the fourth iteration we assign the “0”s again and now we have an optimal solution with 5 assignments.

The solution:

- Taxi1 => Passenger1 - duration 12

- Taxi2 => Passenger4 - duration 11

- Taxi3 => Passenger2 - duration 8

- Taxi4 => Passenger3 - duration 14

- Taxi5 => Passenger5 - duration 11

If we define the needed duration as costs, the minimal cost for this problem is 56.

Example 3 – Maximization problem

Furthermore the Hungarian algorithm can also be used for a maximization problem in which case we first have to transform the matrix. For example a company wants to assign different workers to different machines. Each worker is more or less efficient with each machine. The efficiency can be defined as profit. The higher the number, the higher the profit.

As you can see, the maximal profit of the matrix is 13. The simple twist that we do is rather than try to maximize the profit, we’re going to try to minimize the profit that you don’t get. If every value is taken away from 13, then we can minimize the amount of profit lost. We receive the following matrix:

From now on we proceed as usual with the steps to get to an optimal solution.

With the determined optimal solution we can compute the maximal profit:

- Worker1 => Machine2 - 9

- Worker2 => Machine4 - 11

- Worker3 => Machine3 - 13

- Worker4 => Machine1 - 7

Maximal profit is 40.

The optimal solution is found if there is one assigned “0” for each row and each column.

[1] Linear Assignment Problems and Extensions, Rainer E. Burkard, Eranda Cela

[2] Operations Research Skript TU Kaiserslautern, Prof. Dr. Oliver Wendt

[3] The Hungarian method for the assignment problem, H. W. Kuhn, Bryn Mawr College

Fundamental of Operations Research, Lec. 16, Prof. G. Srinivasan

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Assignment Problem: Maximization

There are problems where certain facilities have to be assigned to a number of jobs, so as to maximize the overall performance of the assignment.

The Hungarian Method can also solve such assignment problems , as it is easy to obtain an equivalent minimization problem by converting every number in the matrix to an opportunity loss.

The conversion is accomplished by subtracting all the elements of the given matrix from the highest element. It turns out that minimizing opportunity loss produces the same assignment solution as the original maximization problem.

• Unbalanced Assignment Problem
• Multiple Optimal Solutions

Example: Maximization In An Assignment Problem

At the head office of www.universalteacherpublications.com there are five registration counters. Five persons are available for service.

How should the counters be assigned to persons so as to maximize the profit ?

Here, the highest value is 62. So we subtract each value from 62. The conversion is shown in the following table.

On small screens, scroll horizontally to view full calculation

Now the above problem can be easily solved by Hungarian method . After applying steps 1 to 3 of the Hungarian method, we get the following matrix.

Draw the minimum number of vertical and horizontal lines necessary to cover all the zeros in the reduced matrix.

Select the smallest element from all the uncovered elements, i.e., 4. Subtract this element from all the uncovered elements and add it to the elements, which lie at the intersection of two lines. Thus, we obtain another reduced matrix for fresh assignment. Repeating step 3, we obtain a solution which is shown in the following table.

Final Table: Maximization Problem

Use Horizontal Scrollbar to View Full Table Calculation

The total cost of assignment = 1C + 2E + 3A + 4D + 5B

Substituting values from original table: 40 + 36 + 40 + 36 + 62 = 214.

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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

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