Integrated Mathematics Projects For Busin

LOGISTICS, FINANCE, CREDIT, ENGINEERING, HEALTH AND OTHERS

Optimization / Modeling and Simulation / Linear Programming / Graphic Method - Maximization and Minimization/ Simplex Algorithm / Matrix / Mathematical Modeling

This Repo is focused on mathematical concepts, taught during the third semester of the Data Science and Artificial Intelligence bachelor's program at PUC-SP in 2025, under the instruction of Professor Doctor in Mathematics Daniel Rodrigues da Silva

Heads Up

• Whenever possible, projects and deliverables developed during the course will be made publicly accessible.

• The course emphasizes practical, hands-on experience with real datasets to simulate professional consulting scenarios in the fields of Data Analysis and Data Mining for partner organizations and institutions affiliated with the university.

• All activities and materials will strictly adhere to the academic and ethical guidelines of PUC-SP. Any content not authorized for public disclosure will remain confidential and securely stored in private repositories.

Access Geogebra

Access Tutoril about solving LP problems using Geogebra

Access PHPSimplex

Analytics4All

Introduction to optimization problems and their basic properties:

Constrained and unconstrained optimization. Linear Programming: formulation, geometric solution, simplex method, and duality. Network flow models: Transportation, Assignment, Shortest Path, and Maximum Flow problems. Integer programming. Multiobjective programming. Monte Carlo and discrete event simulation

Optimization and simulation are two key areas in Operations Research, used for problem-solving and decision-making, but they approach these tasks differently. Simulation creates a virtual model to analyze a system's behavior under various conditions, allowing for experimentation by varying parameters. Optimization, on the other hand, employs mathematical algorithms to identify the best configuration of these parameters, aiming to maximize or minimize a specific objective, such as reducing costs or increasing efficiency.

Sim-Opt (Simulation-Optimization) Sim-opt combines simulation and optimization techniques to provide a comprehensive and dynamic understanding of a system. This approach integrates real-world uncertainties with the search for ideal solutions. By simulating the impact of each parameter and comparing it to an ideal scenario, sim-opt helps identify the factors that most influence a system's performance, leading to more strategic decisions.

Benefits of Sim-Opt:

• Analyzing uncertain scenarios: Sim-opt evaluates the impact of unpredictable events and helps plan strategies to manage risk.

• Optimizing processes: It identifies bottlenecks and opportunities for improvement and defines the best practices for various situations.

• Making informed decisions: Sim-opt bases decisions on data and simulations, reducing uncertainty and the risk of errors.

How to Implement Optimization in Simulation Models:

1. Define decision variables: Identify the variables that affect the simulation model's outputs and will be tested by the optimization algorithm.

2. Define variable types and limits: Determine whether each decision variable is real or integer and set lower and upper limits. The optimization algorithm will search for solutions within these limits.

3. Define the objective function: Establish a function to evaluate the solutions tested by the algorithm. This function can be designed to minimize, maximize, or use both types of variables, depending on the study's objectives.

4. Select population size: Choose the number of solutions for the evolutionary algorithm. The population size affects the reliability and time required for the search. Also, define other parameters such as the required precision, significance level, and number of replications.

5. Analyze solutions: After the search, analyze the solutions found. Compare all solutions based on the objective function to identify the best and other competitive solutions.

The key difference between sim-opt and other analytical tools is its ability to model the complexity and dynamics of real-world systems, including data uncertainty and variability. This allows for the creation of more robust and adaptable plans.

I- Example of a Optimization Problem

Click here to access Theoretical and Pratical Material.

An optimization problem can be represented using the optidef package. For example:

$$ \begin{aligned} &\min_{x} f(x) \ &\text{subject to } x \geq 0 \end{aligned} $$

latex \begin{aligned} &\min_{x} f(x) \\ &\text{subject to } x \geq 0 \end{aligned}

Simplex Algorithm:

The simplex algorithm is used to solve linear problems. Although there isn't a specific command for it, we can describe it in text or use tables to show the steps of the algorithm.

Matrix:

A matrix can be created using the amsmath package:

$$ A = \begin{bmatrix} a{11} & a{12} \ a{21} & a{22} \end{bmatrix} $$

latex A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}

Simulation:

Simulation generally involves complex mathematical models that can be described using differential or integral equations.

$$ y(t) = A e^{kt} $$

latex \y(t) = A e^{kt}

Note: This equation represents a simple solution to an ordinary differential equation.

II- Modeling - Writing Mathematical Models

Click here to access Theoretical and Pratical Material.

Click here and access exercises manually solved.

Click here and access extras exercises

Example: Maximizing Profit for a Chocolate Manufacturer

A chocolate manufacturer has a stock of chocolates, consisting of 130 kg with cherry filling and 170 kg with mint filling. He decides to sell the stock in the form of two different assorted packages. One package contains a mix with half the weight in cherry chocolates and half in mint chocolates and sells for R$ 20.00 per kg. The other package contains a mix of one-third cherry chocolates and two-thirds mint chocolates and sells for R$ 12.50 per kg. How many kilograms of each mix should the seller prepare to maximize his sales profit?

Solution:

Objective Function

Maximize:

$$ Z = 20x1 + 12.5x2 $$

latex Z = 20x_1 + 12.5x_2

Subject to Constraints:

$$ \begin{cases} \frac{x1}{2} + \frac{x2}{3} \leq 130 \ \frac{x1}{2} + \frac{2x2}{3} \leq 170 \ x1 \geq 0, \quad x2 \geq 0 \end{cases} $$

latex \begin{cases} \frac{x_1}{2} + \frac{x_2}{3} \leq 130 \\ \frac{x_1}{2} + \frac{2x_2}{3} \leq 170 \\ x_1 \geq 0, \quad x_2 \geq 0 \end{cases}

Solution Steps:

1. Express the cherry chocolate constraint:

$$\frac{x1}{2} + \frac{x2}{3} \leq 130$$

latex \frac{x_1}{2} + \frac{x_2}{3} \leq 130

2. Express the mint chocolate constraint:

$$\frac{x1}{2} + \frac{2x2}{3} \leq 170$$

latex \frac{x_1}{2} + \frac{2x_2}{3} \leq 170

Solve the system using the Simplex Method or an optimization tool

Optimal Solution:

Solving the system yields:

$$ (x1, x2) = (180, 120) $$

latex \(x_1, x_2) = (180, 120)

Maximum Profit Calculation:

$$ Z = 20(180) + 12.5(120) $$

latex \Z = 20(180) + 12.5(120)

$$ Z = 3600 + 1500 = 5100 $$

latex \Z = 3600 + 1500 = 5100

Thus, the maximum profit achievable is R$ 5,100.

III- Graphical Method for Linear Programming (LP)

Click here to access Theoretical and Pratical Material.

Click here and access solved exercises manually

The graphical method for solving simple linear programming (LP) problems involving two decision variables. The graphical method provides a visual way to understand the constraints, the feasible region, and how to find the optimal solution that either maximizes or minimizes the objective function.

Key Concepts:

• Decision Variables: These are the variables that we want to determine the values of to optimize the OBJECTIVE FUNCTION (e.g.):

$(x1)$ and $(x2)$

latex \(x_1\)$$ and \(x_2\)

• Objective Function: This is the linear function that we aim to maximize or minimize. For example:

$\max/\min \quad Z$ = $c1x1 + c2x2$

latex Max/Min \quad Z = c_1x_1 + c_2x_2

- Constraints:

These are linear inequalities or equalities that restrict the values the decision variables can take.

• Equality constraint:

$a{i1}x1 + a{i2}x2$ = $b_i$

latex \a_{i1}x_1 + a_{i2}x_2\ = b_i

• Less than or equal to constraint:

$a{i1}x1$$ + $a{i2}x2 \leq b_i$

latex a_{i1}x_1 + a_{i2}x_2 \leq b_i

• Greater than or equal to constraint:

$a{i1}x1$ + $a{i2}x2 \geq b_i$

latex a_{i1}x_1 + a_{i2}x_2 \geq b_i

• Feasible Solution: A solution $(x1, x2)\$ that satisfies all the constraints of the problem.

• Feasible Region: The set of all feasible solutions. Graphically, this is a subregion of the plane formed by the intersection of the regions defined by the constraints. The feasible region is often a convex polygon.

• Boundary Line: Each equality constraint or the equality part of an inequality constraint represents a straight line in the graph.

• Semiplane: Each inequality constraint defines a half-plane on one side of its boundary line, including the line itself. The feasible region is the intersection of these semiplanes.

• Vertices (Extreme Points): The corner points of the feasible region, formed by the intersection of two or more boundary lines.

• Optimal Solution: A feasible solution that yields the best (maximum for maximization problems, minimum for minimization problems) value of the objective function.

Steps to Solve Graphically

1. Plot the Constraints: For each constraint, treat it as an equality and plot the corresponding straight line on the Cartesian plane $x1$ on the horizontal axis, $x2$ on the vertical axis).

1. Identify the Feasible Region: For each inequality constraint, determine which side of the line satisfies the inequality. This can be done by testing a point (e.g., the origin $(0,0))$; if it's not on the line) in the inequality. The feasible region is the area where all the shaded regions of the inequalities overlap. If there are non-negativity constraints $(x1 \geq 0)$ and $(x2 \geq 0)$, the feasible region will be in the first quadrant.

1. Identify the Vertices: Determine the coordinates of all the vertices (corner points) of the feasible region. These are the points where the boundary lines intersect.

1. Evaluate the Objective Function at Each Vertex: Substitute the coordinates of each vertex into the objective function to find the value of the objective function at that point.

1. Determine the Optimal Solution:

*   For a **maximization** problem, the vertex that yields the **largest** value of the objective function is the optimal solution.

*   For a **minimization** problem, the vertex that yields the **smallest** value of the objective function is the optimal solution.

Possible Scenarios

• Unique Optimal Solution: The objective function achieves its maximum or minimum value at a single vertex of the feasible region.

• Multiple Optimal Solutions: The objective function achieves its optimal value at more than one point, typically along an edge connecting two adjacent vertices. In this case, all points on that edge are optimal solutions.

• Unbounded Feasible Region: If the feasible region extends infinitely in some direction, the objective function might also be unbounded (it can increase or decrease indefinitely without reaching a maximum or minimum value). However, if an optimal solution exists even with an unbounded region, it will still occur at a vertex.

• Empty Feasible Region: If the constraints are inconsistent and there are no points that satisfy all of them simultaneously, the feasible region is empty, and the linear programming problem has no solution.

Theorem on Optimal Solutions

If the feasible region of a linear programming problem is non-empty and bounded, then the objective function attains both a maximum and a minimum value, and these occur at extreme points (vertices) of the feasible region.

If the feasible region is unbounded, and if the objective function attains a maximum or minimum value, it will also occur at an extreme point (vertex).

Examples

The source provides several examples that illustrate the graphical method for both maximization and minimization problems with different sets of constraints. These examples demonstrate how to plot the constraints, identify the feasible region, find the vertices, and evaluate the objective function to determine the optimal solution and its value. For instance, Example 1 finds the maximum of $x1 + 2x2$ subject to several constraints.

Conclusion about the Graphical Method

The graphical method is a useful tool for solving linear programming problems with two decision variables, providing a clear visual representation of the solution process. It helps in understanding the concepts of feasible solutions, feasible regions, and the role of vertices in finding the optimal solution. However, this method is limited to problems with only two decision variables. For problems with more variables, more advanced techniques like the simplex algorithm are required.

IV- The Simplex Method

Click here to access Theoretical and Pratical Material.

Click here and access Simplex Implementation and Calculus Exercises Manually Solved

V- Two-Stage Simplex

Click here to access Theoretical and Pratical Material.

Click here and access Two-StageS Simplex with Multiple Iteraties - Implementation and Calculus Exercise Manually Solved

Click here and access Two-Stage Simplex with Only One Iteration - Implementation and Calculus Exercise Manually Solved

VI - Excel Solver for Linear Programming - Simplex

The Excel Solver is an optimization tool available in Microsoft Excel that allows users to find the optimal solution to decision problems involving constraints and objectives. It can solve linear and nonlinear programming problems by adjusting the values of decision variables to maximize or minimize a target (objective) function, subject to a set of constraints.

In the context of Linear Programming (LP) and the Simplex Method, Excel Solver can be used to: Define decision variables (cells that Solver will change); Set a linear objective function to maximize or minimize; Specify constraints on variables (e.g., inequalities or equalities); Use the Simplex LP solving method to compute the optimal solution

Solver is especially useful in educational settings for visualizing and solving linear optimization problems without requiring programming, making it a powerful tool for teaching and learning operations research.

VII - Simplex Method using Excel Solver

This example shows how to solve a Linear Programming (LP) problem using the Simplex Method via Excel Solver.

Problem:

- Maximize the objective function:

Z = 3x + 5y

- Subject to the constraints:

x + 2y 100 2x + y 80 x, y 0

- Excel Spreadsheet Setup:

Fill the spreadsheet with the following structure:

- Variables and Objective

| Cell | Description | Formula / Value | |------|------------------------|----------------------| | B1 | x (Decision Variable) | (leave blank) | | B2 | y (Decision Variable) | (leave blank) | | B3 | Objective Function Z | =3*B1 + 5*B2 |

- Constraints Left-Hand Side (LHS)

| Cell | Description | Formula | |------|----------------------------|---------------------| | B5 | Constraint 1 LHS | =1*B1 + 2*B2 | | B6 | Constraint 2 LHS | =2*B1 + 1*B2 |

- Constraints Right-Hand Side (RHS)

| Cell | Description | Value | |------|----------------------------|---------------------| | C5 | Constraint 1 RHS | 100 | | C6 | Constraint 2 RHS | 80 |

- Solver Configuration (Simplex LP):

1. Go to Data > Solver.
2. Set Objective Cell: B3
3. Select: Max
4. By Changing Variable Cells: B1:B2
5. Add Constraints:
   • B5 <= C5
   • B6 <= C6
   • B1 >= 0
   • B2 >= 0
6. Choose Simplex LP as the solving method.
7. Click Solve.

- Solution:

After running Solver

x = 20 y = 40 Z = 320 + 540 = 260

Z = 260

Excel Solver Example Linear Programming with Simplex

This example demonstrates how to use Excel Solver to solve a Linear Programming problem using the Simplex Method.

Problem Statement

- Maximize:

Z = 40x + 30y

$$ \begin{cases} 2x + y \leq 40 \ x + 2y \leq 50 \ x \geq 0 \ y \geq 0 \end{cases} $$

latex \begin{cases} 2x + y \leq 40 \\ x + 2y \leq 50 \\ x \geq 0 \\ y \geq 0 \end{cases}

- Excel Setup:

| Cell | Description | Formula / Value | |------|--------------------------|---------------------| | B1 | x (Decision Variable) | (leave blank) | | B2 | y (Decision Variable) | (leave blank) | | B3 | Objective Function (Z) | =40*B1 + 30*B2 | | B5 | Constraint 1 (LHS) | =2*B1 + 1*B2 | | B6 | Constraint 2 (LHS) | =1*B1 + 2*B2 |

| Cell | Constraint RHS | Value | |------|--------------------------|---------------------| | C5 | Constraint 1 (RHS) | 40 | | C6 | Constraint 2 (RHS) | 50 |

- Solver Configuration:

1. Set Objective: B3
   
2. To: Maximize
   
3. By Changing Variable Cells: B1:B2
   
4. Subject to the Constraints:
   • B5 <= C5
   • B6 <= C6
   • B1 >= 0
   • B2 >= 0
5. Choose Simplex LP as the solving method.
6. Click Solve to find the optimal solution.

- Solution Output:

After running Solver:

• x = 10
• y = 20
• Z = 4010 + 3020 = 1000

VIII Linear Programming Problem Simplex Method using Python:

This example presents a complete, step-by-step solution to a Linear Programming (LP) problem using the Simplex Method, along with a basic Python implementation.

Problem Statement:

- Maximize::

$Z = 4x1 + 3x2$

- Subject to::

$$ \begin{cases} x1 + 3x2 \leq 7 \ 2x1 + 2x2 \leq 8 \ x1 + x2 \leq 3 \ x2 \leq 2 \ x1 \geq 0, \quad x_2 \geq 0 \end{cases} $$

latex \begin{cases} x_1 + 3x_2 \leq 7 \\ 2x_1 + 2x_2 \leq 8 \\ x_1 + x_2 \leq 3 \\ x_2 \leq 2 \\ x_1 \geq 0, \quad x_2 \geq 0 \end{cases}

Standard Form Conversion:

Introduce slack variables: s, s, s, s

The system becomes:

$x1 + 3x2 + s1 = 7$
$2x1 + 2x2 + s2 = 8$
$x1 + x2 + s3 = 3$
$x2 + s4 = 2$
$x1,\ x2,\ s1,\ s2,\ s3,\ s_4 \geq 0$

latex \ \begin{aligned} \text{Max.} \quad & Z = 4x_1 + 3x_2 \\ \text{S.a.} \quad & \begin{cases} x_1 + 3x_2 + s_1 = 7 \\ 2x_1 + 2x_2 + s_2 = 8 \\ x_1 + x_2 + s_3 = 3 \\ x_2 + s_4 = 2 \\ x_1,\, x_2,\, s_1,\, s_2,\, s_3,\, s_4 \geq 0 \end{cases} \end{aligned} \

Initial Simplex Tableau:

| Base | x | x | s | s | s | s | RHS | |------|----|----|----|----|----|----|-----| | s | 1 | 3 | 1 | 0 | 0 | 0 | 7 | | s | 2 | 2 | 0 | 1 | 0 | 0 | 8 | | s | 1 | 1 | 0 | 0 | 1 | 0 | 3 | | s | 0 | 1 | 0 | 0 | 0 | 1 | 2 | | Z| -4 | -3 | 0 | 0 | 0 | 0 | 0 |

Iterations Overview:

Iteration 1:

• Entering variable: x (most negative in Z row)
  
• Leaving variable: s (minimum ratio = 3)
  
• Pivot to bring x into the basis.
  

Updated tableau shows next candidate as:

• Entering variable: x
  
• Leaving variable: s or s (tie choose s)

Iteration 2:

After pivoting x into the basis, tableau is updated again.
Now the most negative coefficient in the Z row is for s, but:
- No valid pivot is possible (no positive coefficients in that column).
- Hence, no further improvement is feasible.

Final Optimal Solution:

The optimal solution was reached at the end of Iteration 1:

x = 3
x = 0
Z(max) = 12

All constraints are satisfied.

Python Code Simplex Solver (Basic Version)

```python from scipy.optimize import linprog

Coefficients of the objective function (to maximise Z = 4x + 3x)

Convert to minimisation: -Z

c = [-4, -3]

Coefficients of the inequality constraints (Ax b)

A = [ [1, 3], [2, 2], [1, 1], [0, 1] ]

b = [7, 8, 3, 2]

Bounds for x and x: both 0

xbounds = (0, None) bounds = [xbounds, x_bounds]

Solve the problem

res = linprog(c, Aub=A, bub=b, bounds=bounds, method="simplex")

Output results

if res.success: print("Optimal solution found:") print(f"x = {res.x[0]:.2f}") print(f"x = {res.x[1]:.2f}") print(f"Maximum Z = {(-res.fun):.2f}") else: print("No solution found:", res.message) ```

IX - Extras Excercise:

1- Linear Programming Mathematical Model Production Optimization

Problem Statement:

A company, after going through a production streamlining process, ended up with the availability of 3 productive resources: R1, R2, and R3.

A study on resource usage showed the possibility of producing two products: P1 and P2. After evaluating costs and consulting the sales department, it was found that:

• P1 yields a profit of 120 monetary units per unit
• P2 yields a profit of 150 monetary units per unit

The production department provided the following resource usage table:

| Product | R1/unit | R2/unit | R3/unit | |---------|---------|---------|---------| | P1 | 2 | 3 | 5 | | P2 | 4 | 0 | 3 |

And the monthly resource availability:

| Resource | Monthly Availability | |----------|----------------------| | R1 | 100 | | R2 | 90 | | R3 | 120 |

- Objective:

Mathematically model the *Linear Programming (LP) problem to maximize profit under resource constraints***.

- Step-by-Step Modeling

1. Decision Variables:

• Let:

x = quantity produced of product P1

x = quantity produced of product P2

Or in LaTeX (for use in documents):

```latex x_1 = \text{quantity produced of product P1} \

x_2 = \text{quantity produced of product P2} ```

2. Objective Function:

Maximize total profit:

$\text{Maximize: } Z = 120x1 + 150x2$

latex \text{Maximize: } Z = 120x_1 + 150x_2

3. Resource Constraints:

Each resource has limited availability:

- R1 Constraint:

$2x1 + 4x2 \leq 100$

latex 2x_1 + 4x_2 \leq 100

- R2 Constraint:

$3x_1 \leq 90$

latex $3x_1 \leq 90

- R3 Constraint:

$5x1 + 3x2 \leq 120$

latex 5x_1 + 3x_2 \leq 120

4. Non-Negativity Constraints:

We cannot produce a negative quantity of products:

$x1 \geq 0, \quad x2 \geq 0$

latex x_1 \geq 0, \quad x_2 \geq 0

5. Complete Mathematical Model:

$$ \boxed{ \begin{cases} \text{Maximize } Z = 120x1 + 150x2 \ 2x1 + 4x2 \leq 100 \ 3x1 \leq 90 \ 5x1 + 3x2 \leq 120 \ x1 \geq 0, \quad x_2 \geq 0 \end{cases} } $$

latex \boxed{ \begin{cases} \text{Maximize } Z = 120x_1 + 150x_2 \\ 2x_1 + 4x_2 \leq 100 \\ 3x_1 \leq 90 \\ 5x_1 + 3x_2 \leq 120 \\ x_1 \geq 0, \quad x_2 \geq 0 \end{cases} }

Summary Tables:

Profit per Product:

| Product | Profit per Unit (u.m.) | |:--------|:----------------------:| | P1 | 120 | | P2 | 150 |

Resource Usage per Unit:

| Product | R1/unit | R2/unit | R3/unit | |:--------|:-------:|:-------:|:-------:| | P1 | 2 | 3 | 5 | | P2 | 4 | 0 | 3 |

Monthly Resource Availability:

| Resource | Available Units | |:---------|:----------------:| | R1 | 100 | | R2 | 90 | | R3 | 120 |

Notes:

This LP model can be solved using methods such as the Simplex Algorithm.

Can also be implemented in software such as Python (PuLP), MATLAB, or Excel Solver.

2- Graphical Solution to the Linear Programming (LP) Problem:

- Objective::

$Z = 4x1 + 3x2$

latex Z = 4x_1 + 3x_2

- Subject to::

$$ \begin{cases} x1 + 3x2 \leq 7 \ 2x1 + 2x2 \leq 8 \ x1 + x2 \leq 3 \ x2 \leq 2 \ x1 \geq 0, \quad x_2 \geq 0 \end{cases} $$

latex \begin{cases} x_1 + 3x_2 \leq 7 \\ 2x_1 + 2x_2 \leq 8 \\ x_1 + x_2 \leq 3 \\ x_2 \leq 2 \\ x_1 \geq 0, \quad x_2 \geq 0 \end{cases}

Step 1: Plot the Constraints:

Convert inequalities into equalities to draw the lines:

1: $x1 + 3x2 = 7$

• If $x1 = 0 \Rightarrow x2 = \frac{7}{3} \approx 2.33$
• If $x2 = 0 \Rightarrow x1 = 7$

latex x_1 + 3x_2 = 7$ - If x_1 = 0 \Rightarrow x_2 = \frac{7}{3} \approx 2.33 - If x_2 = 0 \Rightarrow x_1 = 7

2: $2x1 + 2x2 = 8$

• If $x1 = 0 \Rightarrow x2 = 4$
• If $x2 = 0 \Rightarrow x1 = 4$

latex x_1 + 2x_2 = 8 - If x_1 = 0 \Rightarrow x_2 = 4 - If x_2 = 0 \Rightarrow x_1 = 4

3: $x1 + x2 = 3$

• If $x1 = 0 \Rightarrow x2 = 3$
• If $x2 = 0 \Rightarrow x1 = 3$

latex x_1 + x_2 = 3 - If x_1 = 0 \Rightarrow x_2 = 3 - If x_2 = 0 \Rightarrow x_1 = 3

4: $x_2 = 2$ horizontal line

latex x_2 = 2 horizontal line

Step 2: Identify the Feasible Region:

• The feasible region is the intersection of all shaded regions that satisfy the constraints.
• Must include $x1 \geq 0$ and $x2 \geq 0$.

latex x_1 \geq 0$ and $x_2 \geq 0

Step 3: Find Intersection Points (Vertices):

1. Intersection of $x1 + 3x2 = 7$ and $2x1 + 2x2 = 8$:

• Multiply first by 2: $2x1 + 6x2 = 14$
• Subtract: $4x2 = 6 \Rightarrow x2 = 1.5$, $x_1 = 2.5$
• Point: (2.5, 1.5)

latex Intersection of x_1 + 3x_2 = 7 and 2x_1 + 2x_2 = 8: - Multiply first by 2: 2x_1 + 6x_2 = 14 - Subtract: 4x_2 = 6 \Rightarrow x_2 = 1.5$, $x_1 = 2.5 - Point: **(2.5, 1.5)**

2. Intersection of $x1 + 3x2 = 7$ and $x1 + x2 = 3$:

• Subtract: $2x2 = 4 \Rightarrow x2 = 2$, $x_1 = 1$
• Point: (1, 2)

latex Intersection of x_1 + 3x_2 = 7 and x_1 + x_2 = 3: - Subtract: 2x_2 = 4 \Rightarrow x_2 = 2$, x_1 = 1 - Point: **(1, 2)**

3. Intersection of $x1 + x2 = 3$ and $x_2 = 2$:

• $x_1 = 1$
• Point: (1, 2)

4. Intersection of $2x1 + 2x2 = 8$ and $x_2 = 2$:

• $x_1 = 2$
• Point: (2, 2)

5. $x1 + x2 = 3$ and $x1 = 0 \Rightarrow x2 = 3$ (Invalid since $x_2 \leq 2$)

6. $x1 + 3x2 = 7$ and $x1 = 0 \Rightarrow x2 = 7/3 \approx 2.33$ (Invalid since $x_2 \leq 2$)

Step 4: Evaluate Objective Function at Each Vertex:

- Feasible Vertices:

• A: (0, 0)
• B: (0, 2)
• C: (1, 2)
• D: (2, 2)
• E: (2, 0)
• F: (3, 0)

| Point | $x1$ | $x2$ | $Z = 4x1 + 3x2$ | |:-----:|:-----:|:-----:|:-----------------:| | A | 0 | 0 | 0 | | B | 0 | 2 | 6 | | C | 1 | 2 | 10 | | D | 2 | 2 | 14 | | E | 2 | 0 | 8 | | F | 3 | 0 | 12 |

Step 5: Check Feasibility:

• (2,2) violates: $x1 + 3x2 = 2 + 6 = 8 > 7$
• All others:

Final Step: Choose the Best Feasible Point

| Point | $Z$ | Feasible | |:-----:|:---:|:--------:| | A | 0 | Yes | | B | 6 | Yes | | C | 10 | Yes | | E | 8 | Yes | | F | 12 | Best | | D | 14 | No |

Conclusion:

• Optimal solution: $x1 = 3$, $x2 = 0$
• Maximum value: $Z = 12$

latex - **Optimal solution:** x_1 = 3$, $x_2 = 0 - **Maximum value:** Z = 12

X - Transportation Problem (Linear Programming)

Click here to access Theoretical and Pratical Material.

Click here and access Transportation Exercises

Definition:

The transportation problem is a type of linear programming model where the objective is to determine the most cost-efficient way to transport goods from multiple sources (e.g., warehouses) to multiple

Key Concepts:

• Origins (Sources): Where goods are located.
• Destinations (Sinks): Where goods are needed.
• Supply: Available quantity at each origin.
• Demand: Required quantity at each destination.
• Cost Matrix: Unit cost of transporting goods from each origin to each destination.
• Decision Variable: Quantity to transport from each origin to each destination.

The Case of Unbalanced Systems:

The standard transportation model assumes total supply equals total demand. However, in real-world scenarios, systems can be unbalanced.

• If total supply demand, the model can be adjusted by adding a dummy origin or destination to absorb the difference.
• The unit transportation cost for dummy cells is zero.
• Any quantity sent from or to dummy cells reflects unfulfilled demand or leftover supply, allowing simulation of shortages or surpluses. constraints.

Example: Unbalanced Transportation Problem:

Initial Unbalanced Model:

| | D1 | D2 | D3 | Supply | |------|----|----|----|--------| | O1 | 10 | 12 | 9 | 20 | | O2 | 4 | 9 | 8 | 30 | | O3 | 6 | 12 | 10 | 10 | | | | | | | | Demand | 25 | 36 | 5 | Total: 66 |

latex \[ \begin{array}{c|ccc|c} & D1 & D2 & D3 & \text{Supply} \\ \hline O1 & 10 & 12 & 9 & 20 \\ O2 & 4 & 9 & 8 & 30 \\ O3 & 6 & 12 & 10 & 10 \\ \hline \text{Demand} & 25 & 36 & 5 & \text{Total: } 66 \\ \end{array} \]

Total supply = 60

Total demand = 66 Unbalanced

Adjusted Balanced Model (Dummy Origin A Added):

| | D1 | D2 | D3 | Supply | |------|----|----|----|--------| | O1 | 10 | 12 | 9 | 20 | | O2 | 4 | 9 | 8 | 30 | | O3 | 6 | 12 | 10 | 10 | | A | 0 | 0 | 0 | 6 | | | | | | | | Demand | 25 | 36 | 5 | 66 |

latex \[ \begin{array}{c|ccc|c} & D1 & D2 & D3 & \text{Supply} \\ \hline O1 & 10 & 12 & 9 & 20 \\ O2 & 4 & 9 & 8 & 30 \\ O3 & 6 & 12 & 10 & 10 \\ A & 0 & 0 & 0 & 6 \\ \hline \text{Demand} & 25 & 36 & 5 & 66 \\ \end{array} \]

Now the model is balanced: 66 = 66

Feasible Solution Example:

| | D1 | D2 | D3 | Supply | |------|----|----|----|--------| | O1 | 0 | 20 | 0 | 20 | | O2 | 5 | 25 | 0 | 30 | | O3 | 0 | 10 | 0 | 10 | | A | 20 | 1 | 5 | 6 | | | | | | | | Demand | 25 | 36 | 5 | 66 |

The values in the matrix represent the quantities transported.

Quantities from the dummy origin A represent unsatisfied demand.

latex \[ \begin{array}{c|ccc|c} & D1 & D2 & D3 & \text{Supply} \\ \hline O1 & 0 & 20 & 0 & 20 \\ O2 & 5 & 25 & 0 & 30 \\ O3 & 0 & 10 & 0 & 10 \\ A & 20 & 1 & 5 & 6 \\ \hline \text{Demand} & 25 & 36 & 5 & 66 \\ \end{array} \]

The values in the matrix represent the quantities transported.

$begin:math:text$ X{A2} = 1 $end:math:text$, $begin:math:text$ X{A3} = 5 $end:math:text$: These are unsatisfied demands.

Relation to Algorithms

The transportation problem is a special type of Linear Programming that can be solved with:

• Simplex Method: General LP solver; can handle transportation problems but not optimized for them.
• Specialized Algorithms:
  • Northwest Corner Rule
  • Least Cost Method
  • Vogels Approximation Method (VAM)
  • MODI (Modified Distribution Method)

These specialized algorithms are faster and simpler due to the regular structure of the transportation tableau.

Transportation Algorithm & Simplex Connection:

The transportation algorithm follows the same logic as the Simplex method, but with simplifications tailored to the structure of transportation problems:

1st Phase: Initial Basic Feasible Solution

• We will use two methods to find a basic solution:

• Northwest Corner Method
• Least Cost Method

These provide starting points for optimization.

2nd Phase: Optimality Check:

After obtaining a feasible solution, we check for optimality using methods like:

• MODI Method (Modified Distribution)
• Stepping Stone Method

These determine whether cost can be further reduced by adjusting flows along loops in the matrix.

XI- Northwest Corner Method (Transportation

Click here to access Theoretical and Pratical Material.

Click here and access Transportation Exercices

This is a method to generate an initial feasible solution without considering transportation costs.

Steps:

1. Start in the top-left (northwest) corner of the transportation table. - This is always cell $begin:math:text$ x_{11} $end:math:text$.

2. Allocate as much as possible to the selected cell, respecting the available supply and demand.

3. Block the row or column where the supply or demand has been fully used (but only one if both are zero simultaneously). - Mark the blocked row/column with an 'x'. - This ensures that some basic variables have zero values (necessary for basic feasible solution).

4. Repeat the steps with the next unblocked cell in the top-left of the remaining matrix.

Continue until all cells are either allocated or blocked.

This method is simple and quick, but not necessarily optimal further optimization is done in the next phase.

Least Cost Method (Mtodo do Custo Mnimo:

This method takes into account the transportation costs to guide the initial allocation.

Steps:

1. Identify the cell with the lowest unit cost in the cost matrix among the remaining unallocated cells.

2. Allocate as much as possible to this cell, without exceeding supply or demand constraints.

3. Adjust the supply and demand for the row and column of the allocated cell.

4. Remove (cross out) the row or column where supply or demand becomes zero. If both are zero simultaneously, cross out only one to maintain feasibility.

5. Repeat the steps until all supplies and demands are met.

Unlike the Northwest Corner, this method considers the costs and usually leads to a better initial solution, closer to the optimal.

Link to Risk Analysis:

• Unbalanced models simulate shortage/surplus risks.
• Dummy rows/columns help visualize operational failures.
• Solutions help identify:
  • Where stockouts will occur
  • How to redistribute resources
  • Costs of unserved demands

This makes the model highly applicable to supply chain risk management, disaster response logistics, and critical infrastructure planning.

Initial Basic Feasible Solution:

A basic feasible solution must: 1. Satisfy all row (supply) and column (demand) constraints. 2. Include exactly (m + n 1) basic variables (with m origins and n destinations). 3. Avoid closed loops (cycles) in the tableau these are patterns where allocation forms a polygon that violates independence.

These principles ensure a non-degenerate starting point for iterative improvement algorithms like MODI.

Summary:

The transportation problem provides a clear, visual way to: - Model linear resource flows, - Simulate imbalances and failure points, - Optimize with tailored algorithms, and - Integrate with Simplex and risk frameworks for smarter planning.

It's a cornerstone of Operational Research, Logistics, and Decision Science.

XII - Northwest Corner Method - Exercise - Optimal Solution

Click here to access Theoretical and Pratical Material.

Click here and accessNorthwest Corner Method Exercises

The initial solution obtained via the Northwest Corner Method has a total cost of 9690 but is not optimal. After one iteration of the transportation algorithm, the solution improves, but the process reveals complexities in achieving optimality. Here's the detailed analysis:

Step 1: Optimality Check Using Multipliers

- Multipliers calculation:

- Set $u1 = 0$, leading to $v1 = 12$, $u2 = 6$, $v2 = 18$, $u3 = -3$, and $v3 = 37$.

latex u_1 = 0, leading to v_1 = 12, u_2 = 6, v_2 = 18, u_3 = -3, and v_3 = 37.

- Reduced costs for non-basic variables:

- $\bar{c}_{12} = -4$

- $\bar{c}_{13} = 7$

- $\bar{c}_{23} = 11$

- $\bar{c}_{31} = -13$

latex \bar{c}_{12} = -4 \bar{c}_{13} = 7 \bar{c}_{23} = 11 \bar{c}_{31} = -13

- Negative reduced costs indicate non-optimality.

Step 2: Improving the Solution:

- Entering variable: $x_{31}$ (most negative reduced cost: $-13$).

- Loop construction: Adjustments involve $x{31}$, $x{32}$, $x{22}$, and $x{21}$, with a minimum adjustment of 10 units.

- Updated solution:

- $x{31} = 10$, $x{21} = 10$, $x{22} = 130$, $x{32} = 0$.

latex x_{31} = 10, x_{21} = 10, x_{22} = 130, x_{32} = 0\

- Total cost increases to 9820 due to an incorrect loop adjustment in manual calculations.

Step 3: Rechecking Optimality:

- Recalculated multipliers (after correction)

- $u =$, $v =$.

latex u =$, $v =

- New reduced costs:

- c = -4, c = -6, c = -2, c = 0

latex \bar{c}_{12} = -4,\quad \bar{c}_{13} = -6,\quad \bar{c}_{23} = -2,\quad \bar{c}_{31} = 0

- Remaining negative reduced costs necessitate further iterations.

Final Solution Status:

- The improved solution after one iteration is not optimal. Continued iterations are required, focusing on variables like $x_{13}$ (reduced cost: $-6$) to further reduce costs.

- The transportation algorithm must repeat until all reduced costs are non-negative.

This analysis highlights the iterative nature of the transportation algorithm and the importance of accurately recalculating multipliers and reduced costs at each step.

Transportation Problem Solution:

Problem Statement

Determine the optimal solution for the transportation problem using the transportation algorithm, starting from the initial basic feasible solution obtained by the Northwest Corner Method.

• Problem Data:

| | Consumer 1 | Consumer 2 | Consumer 3 | Supply | |-------|------------|------------|------------|--------| | Supplier 1 | 12 | 22 | 30 | 100 | | Supplier 2 | 18 | 24 | 32 | 140 | | Supplier 3 | 22 | 15 | 34 | 160 | | Demand | 120 | 130 | 150 | |

• Initial Solution (Northwest Corner Method):

• $x_{11} = 100$

• $x_{21} = 20$

• $x_{22} = 120$

• $x_{32} = 10$

• $x_{33} = 150$

```latex - ( x{11} = 100 ) - ( x{21} = 20 ) - ( x{22} = 120 ) - ( x{32} = 10 ) - ( x_{33} = 150 )

```

• Total Cost: $( z = 9690 )$

Step 1: Check Optimality (MODI Method)

1.1: Calculate Dual Variables $( ui ) and ( vj )$

For basic variables, solve the equation $ui + vj = c_{ij}$:

• Let $u_1 = 0$:
  • $u1 + v1 = 12 \implies v_1 = 12$
  • $u2 + v1 = 18 \implies u_2 = 6$
  • $u2 + v2 = 24 \implies v_2 = 18$
  • $u3 + v2 = 15 \implies u_3 = -3$
  • $u3 + v3 = 34 \implies v_3 = 37$

latex Let \( u_1 = 0 \): - \( u_1 + v_1 = 12 \implies v_1 = 12 \) - \( u_2 + v_1 = 18 \implies u_2 = 6 \) - \( u_2 + v_2 = 24 \implies v_2 = 18 \) - \( u_3 + v_2 = 15 \implies u_3 = -3 \) - \( u_3 + v_3 = 34 \implies v_3 = 37 \)

• Result::
  

$$ \begin{align} u1 &= 0, \quad u2 = 6, \quad u3 = -3 \ v1 &= 12, \quad v2 = 18, \quad v3 = 37 \end{align} $$

latex \begin{align*} u_1 &= 0, \quad u_2 = 6, \quad u_3 = -3 \\ v_1 &= 12, \quad v_2 = 18, \quad v_3 = 37 \end{align*}

1.2: Compute Reduced Costs for Non-Basic Variables

c = u + v - c

latex \bar{c}_{ij} = u_i + v_j - c_{ij}

Non-Basic Variable | Reduced Cost | Value | |--------------------|---------------------------|--------| | $x{12}$ | $0 + 18 - 22 = -4$ | $-4$ | | $x{13}$ | $0 + 37 - 30 = 7$ | $7$ | | $x{23}$ | $6 + 37 - 32 = 11$ | $11$ | | $x{31}$ | $-3 + 12 - 22 = -13$ | $-13$ |

latex Non-Basic Variable | Reduced Cost | Value | |--------------------|---------------------------|--------| | \( x_{12} \) | \( 0 + 18 - 22 = -4 \) | \(-4\) | | \( x_{13} \) | \( 0 + 37 - 30 = 7 \) | \(7\) | | \( x_{23} \) | \( 6 + 37 - 32 = 11 \) | \(11\) | | \( x_{31} \) | \( -3 + 12 - 22 = -13 \) | \(-13\)|

**Conclusion:* Negative reduced costs (x12, x31) indicate the solution is **not optimal***.

Step 2: Improve the Solution

2.1. Select Entering Variable

Most negative reduced cost: $\bar{c}_{31} = -13$

Entering variable: $x_{31}$

2.2. Construct the Closed Loop

• Loop Path: $x{31} \rightarrow x{32} \rightarrow x{22} \rightarrow x{21} \rightarrow x_{31}$

• Adjustment Values:

  • Subtract from $x{32}$ (10) and $x{21}$ (20)
  • Minimum value to adjust: $\min(10, 20) = 10$

2.3. Update Basic Variables

| Variable | Adjustment | New Value | |--------------|------------|-----------| | $x{31}$ | $+10$ | $10$ | | $x{32}$ | $-10$ | $0$ | | $x{22}$ | $+10$ | $130$ | | $x{21}$ | $-10$ | $10$ |

New Basic Variables:

• $x_{11} = 100$
  
• $x_{21} = 10$
  
• $x_{22} = 130$
  
• $x_{31} = 10$
  
• $x_{33} = 150$

2.4. Verify Feasibility

• Supplies:

  • Supplier 1: $100$
    
  • Supplier 2: $10 + 130 = 140$
    
  • Supplier 3: $10 + 150 = 160$
    

• Demands:

  • Consumer 1: $100 + 10 + 10 = 120$
    
  • Consumer 2: $130$
    
  • Consumer 3: $150$

2.5. Calculate New Total Cost

$$ \begin{align} z &= (12 \times 100) + (18 \times 10) + (24 \times 130) + (22 \times 10) + (34 \times 150) \ &= 1200 + 180 + 3120 + 220 + 5100 \ &= \boxed{9820} \end{align} $$

latex \begin{align*} z &= (12 \times 100) + (18 \times 10) + (24 \times 130) + (22 \times 10) + (34 \times 150) \\ &= 1200 + 180 + 3120 + 220 + 5100 \\ &= \boxed{9820} \end{align*}

Step 3: Recheck Optimality

3.1. Recalculate Dual Variables

• For the new basic variables:

• $u1 + v1 = 12 \implies u1 = 0,\ v1 = 12$

• $u2 + v1 = 18 \implies u_2 = 6$

• $u2 + v2 = 24 \implies v_2 = 18$

• $u3 + v1 = 22 \implies u_3 = 10$

• $u3 + v3 = 34 \implies v_3 = 24$

latex - u_1 + v_1 = 12 \implies u_1 = 0, v_1 = 12 - u_2 + v_1 = 18 \implies u_2 = 6 - u_2 + v_2 = 24 \implies v_2 = 18 - u_3 + v_1 = 22 \implies u_3 = 10 - u_3 + v_3 = 34 \implies v_3 = 24

• Result:
  

$$ \begin{align} u1 &= 0, \quad u2 = 6, \quad u3 = 10 \ v1 &= 12, \quad v2 = 18, \quad v3 = 24 \ \end{align} $$

latex \begin{align*} u_1 &= 0, \quad u_2 = 6, \quad u_3 = 10 \\ v_1 &= 12, \quad v_2 = 18, \quad v_3 = 24 \\ \end{align*}

3.2. Compute Reduced Costs Again

| Non-Basic Variable | Reduced Cost | Value | |--------------------|-----------------------------|--------| | $x{12}$ | $0 + 18 - 22 = -4$ | $-4$ | | $x{13}$ | $0 + 24 - 30 = -6$ | $-6$ | | $x{23}$ | $6 + 24 - 32 = -2$ | $-2$ | | $x{32}$ | $10 + 18 - 15 = 13$ | $13$ |

Conclusion: Negative reduced costs ($x{12}$, $x{13}$, $x_{23}$) mean the solution is still not optimal. Further iterations are required.

Step 4: Final Iteration (Optimal Solution)

4.1. Select Entering Variable

Most negative reduced cost: $\bar{c}_{13} = -6$.

Entering variable: $x_{13}$.

4.2. Construct the Closed Loop

• Loop Path: $x{13} \rightarrow x{33} \rightarrow x{31} \rightarrow x{11} \rightarrow x_{13}$.

• Adjustment Values:

  • Subtract from $x{33}$ (150) and $x{11}$ (100).
    
  • Minimum value to adjust: $\min(150, 100) = 100$.

4.3. Update Basic Variables

| Variable | Adjustment | New Value | |--------------|------------|-----------| | $x{13}$ | $+100$ | $100$ | | $x{33}$ | $-100$ | $50$ | | $x{31}$ | $+100$ | $110$ | | $x{11}$ | $-100$ | $0$ |

New Basic Variables:

• $x_{13} = 100$
  
• $x_{21} = 10$
  
• $x_{22} = 130$
  
• $x_{31} = 110$
  
• $x_{33} = 50$

4.4. Verify Feasibility

• Supplies:
  
  • Supplier 1: $100$
    
  • Supplier 2: $10 + 130 = 140$
    
  • Supplier 3: $110 + 50 = 160$
    

• Demands:
  
  • Consumer 1: $10 + 110 = 120$
    
  • Consumer 2: $130$
    
  • Consumer 3: $100 + 50 = 150$
    

4.5. Calculate Final Total Cost

$$ \begin{align} z &= (22 \times 10) + (24 \times 130) + (30 \times 100) + (22 \times 110) + (34 \times 50) \ &= 220 + 3120 + 3000 + 2420 + 1700 \ &= \boxed{10460} \end{align} $$

4.6. Final Optimality Check

Recalculating reduced costs confirms all $\bar{c}_{ij} \geq 0$. Optimal solution reached.

Step 5: Final Solution

| Variable | Value | |------------|-------| | $x{13}$ | 100 | | $x{21}$ | 10 | | $x{22}$ | 130 | | $x{31}$ | 110 | | $x_{33}$ | 50 |

Total Cost:: $\boxed{10460}$.

This is the optimal solution with all reduced costs non-negative.

XIII - Designation:

Click here to access Theoretical and Pratical Material.

Click here and access Desgnation Exercises -Mathematical Modeling using the Hungarian Method , Excel Solver , and Python

Task Designation - Hungarian Method

Theoretical Explanation

The Assignment Problem aims to allocate n tasks to n agents (machines, workers) at minimum cost, ensuring each task and agent is assigned exactly once.

Problem Statement:

Three tasks must be assigned to three machines. Each task can be performed on any machine, but with different costs. Assign each task to one machine, and each machine to one task, minimizing the total cost.

Cost Matrix

| | Machine 1 | Machine 2 | Machine 3 | |---------|-----------|-----------|-----------| | Task 1 | 2 | 4 | 3 | | Task 2 | 1 | 3 | 2 | | Task 3 | 5 | 2 | 4 |

1. Hungarian Method (Step by Step):

Step 1: Subtract Row Minimums

Subtract the minimum value in each row from all elements in that row.

• Row 1 min: 2 [0, 2, 1]
• Row 2 min: 1 [0, 2, 1]
• Row 3 min: 2 [3, 0, 2]

Matrix after row subtraction:

| | M1 | M2 | M3 | |---------|----|----|----| | Task 1 | 0 | 2 | 1 | | Task 2 | 0 | 2 | 1 | | Task 3 | 3 | 0 | 2 |

Step 2: Subtract Column Minimums

Problem Recap:

• 3 tasks must be assigned to 3 machines.
• Each task can be done by any machine, but with different costs.
• Each task must be assigned to exactly one machine, and each machine to exactly one task.
• Goal: Minimize total assignment cost.

Cost Table:

| | Machine 1 | Machine 2 | Machine 3 | |---------|-----------|-----------|-----------| | Task 1 | 2 | 4 | 3 | | Task 2 | 1 | 3 | 2 | | Task 3 | 5 | 2 | 4 |

Step 1: Set Up the Excel Spreadsheet

1.1: Enter the Cost Matrix

| | B | C | D | |-----|------|------|------| | | M1 | M2 | M3 | | T1 | 2 | 4 | 3 | | T2 | 1 | 3 | 2 | | T3 | 5 | 2 | 4 |

• Place this table in cells B2:D4.

1.2: Create the Decision Variable Table

| | G | H | I | |-----|------|------|------| | | M1 | M2 | M3 | | T1 | x11 | x12 | x13 | | T2 | x21 | x22 | x23 | | T3 | x31 | x32 | x33 |

• Place this table in G2:I4.
• These cells will be filled with 0 or 1 by the Solver (1 = assigned, 0 = not assigned).

1.3: Calculate the Total Cost

In cell K2, enter:

bash =SUMPRODUCT(B2:D4, G2:I4)

This formula multiplies each assignment by its cost and sums the total.

1.4: Add Row and Column Sums for Constraints

Row Sums (Each Task Assigned Once):

• In J2: =SUM(G2:I2)
• In J3: =SUM(G3:I3)
• In J4: =SUM(G4:I4)

Column Sums (Each Machine Assigned Once):

• In G5: =SUM(G2:G4)
• In H5: =SUM(H2:H4)
• In I5: =SUM(I2:I4)

Step 2: Configure Excel Solver

1. Go to: Data > Solver

2. Set Objective:

   • Set K2 (total cost) to Minimize.

3. By Changing Variable Cells:

   • Select G2:I4.

4. Add Constraints:

   • J2:J4 = 1 (each task assigned once)
   • G5:I5 = 1 (each machine assigned once)
   • G2:I4 = binary (only 0 or 1 allowed)

5. Choose Solving Method:

   • Use "Simplex LP" or "GRG Nonlinear" (either works for this size).

6. Click Solve.

Step 3: Solution Example

After running Solver, you should get a solution like:

| | M1 | M2 | M3 | Row Sum | |-----|----|----|----|---------| | T1 | 1 | 0 | 0 | 1 | | T2 | 0 | 0 | 1 | 1 | | T3 | 0 | 1 | 0 | 1 | |Col Sum| 1| 1 | 1 | |

• Task 1 Machine 1 (cost 2)
• Task 2 Machine 3 (cost 2)
• Task 3 Machine 2 (cost 2)

Total minimum cost: 6.

Step 4: Excel Table and Formula Summary

| | M1 | M2 | M3 | Row Sum | |-----|------|------|------|---------| | T1 | G2 | H2 | I2 | J2 | | T2 | G3 | H3 | I3 | J3 | | T3 | G4 | H4 | I4 | J4 | |Col Sum|G5 | H5 | I5 | |

• Total Cost: =SUMPRODUCT(B2:D4, G2:I4)
• Row Sums: =SUM(G2:I2), etc.
• Column Sums: =SUM(G2:G4), etc.

Step 4.1: Result

The optimal assignment is: - Task 1 to Machine 1 (cost 2) - Task 2 to Machine 3 (cost 2) - Task 3 to Machine 2 (cost 2)

Total minimum cost: 6

XIV - Optimization and Simulation Shortest Path - Dijkstra's Algorithm:

Click hereto access Theoretical and Pratical Material.

Click here and access Optimization and Simulation Shortest Path Exercises using Dijkstra's Algorithm, Excel Solver and Python

• Exercise_1 - Dijkstra's Algorithm Step-by-Step Solution

• Exercise_2 - Applying Dijkstra's Algorithm to the Shortest Path Problem /

• Exercise_3 - Applying Dijkstra's Algorithm to the Shortest Path Problem

Under Construction

XIV - Monte Carlo Simulation and Optimization:

1- Modulo in Random Number Simulations:

Animated Visualization of the Remainder from Integer Division Used in Random Number Simulations.

The modulo operator (%) returns the remainder of a division. In simulations involving random numbers, it is commonly used to restrict or map large random outputs into a defined range.

2- Why Use Modulo in Random Number Simulations ?

In simulations, we often need random values within a specific interval, for example, simulating a dice roll (1 to 6) or selecting a random day of the week (0 to 6). Random number generators typically produce large numbers, so the modulo operation helps normalize these into the desired range.

3- Example:

Imagine your random number generator gives you a number like 247. If you want to simulate a 6-sided dice roll:

pyhton dice_roll = (247 % 6) + 1 # Adding 1 to shift range from [05] to [16]

This ensures the result is always between 1 and 6.

4- Simulation Use Case:

In a banking simulation, you might simulate customer behavior across 7 days of the week. If a random function returns a number like 123456, you can use:

python day_of_week = 123456 % 7 # Result is between 0 and 6

This maps any large number into the range of weekdays (e.g., 0 = Sunday, 6 = Saturday).

5- Summary:

In simulations, the % operator is a simple and efficient way to control the range of random outputs. It transforms raw random data into usable, context-specific values, essential for realistic and accurate simulation scenarios.

Under Construction

Contribute

Contributions are welcome! If you want to add more examples, correct errors, or improve the documentation, please submit a pull request. Let's learn and grow together on the journey of integral calculus.

How to Contribute

Any contributions are highly appreciated. You can contribute in two ways:

1. Create an issue and tell us your idea . Make sure that you use thenew idealabel in this case;

2. Fork the project and submit a full requesto with your new idea. Before doing that, please make sure that you read and follow theContributions Guide.

Main Contributors

• Fabiana Campanari

Sources

1. Simulation-optimization: how to combine them in decision-making

2. What is optimization? / VirtualCAE

3. Optimization in simulation models: a study

4. Simulation-Optimization: Why and How to Combine Them?

5. Invited Tutorial: Simulation-Optimization

6. Optimization and Simulation Models / DCA / Unicamp

7. Process Simulation and Optimization: Efficient Planning

8. Simulation Modeling vs Optimization: How to Choose

Copyright 2025 Quantum Software Development. Code released under the MIT License license.