A model is a simplified representation of a real-world situation. We use it to understand problems and test solutions before applying them in real life.
Deterministic Model
All inputs are known and fixed.
No uncertainty involved.
Example: Calculating the area of a room. You know the length and width for sure.
Stochastic (Probabilistic) Model
Some inputs are uncertain or random.
Involves probabilities.
Example: Forecasting next week’s sales. It might be affected by weather, holidays, or other random events.
Example 1: Austin Auto Auction
Let’s build a model to help an auctioneer set a starting bid for a used car.
The Goal: Set the starting bid (B) at 70% of the predicted final winning bid.
The Logic:
Start with the car’s original price (P).
Deduct for age (A): $800 per year.
Deduct for mileage (M): $0.025 per mile.
Example 1: Building the Model
First, let’s write an equation for the expected winning bid:
\[
\text{Expected Bid} = P - 800A - 0.025M
\]
Now, we can build the full model for the starting bid (B), which is 70% of that value:
\[
\begin{aligned}
B &= 0.7 \times (\text{Expected Winning Bid}) \\
B &= 0.7 \times (P - 800A - 0.025M) \\
B &= 0.7P - 560A - 0.0175M
\end{aligned}
\]
This is our final model! It’s a simple linear equation.
Example 1: Using the Model
Question: A car is 4 years old with 60,000 miles. Its original price was $12,500. What should the starting bid be?
Answer: We just plug the numbers into our model.
P = $12,500
A = 4 years
M = 60,000 miles
\[
\begin{aligned}
B &= 0.7(12,500) - 560(4) - 0.0175(60,000) \\
B &= 8750 - 2240 - 1050 \\
B &= \$5,460
\end{aligned}
\]
The auctioneer should set the starting bid at $5,460.
Example 1: What Are We Assuming?
Every model has assumptions. It’s important to know what they are!
What assumptions does this model make?
Only three factors matter: Original price, age, and mileage. It ignores things like the car’s condition, color, brand reputation, or if it’s a rare model.
Value loss is linear: It assumes a car loses exactly $800 in value every year and $0.025 every mile, forever. In reality, this isn’t true. A very old car could have a negative starting bid with this model, which doesn’t make sense!
Example 2: Ponderosa Development Corp.
This company builds and sells one style of house. Let’s model its costs, revenue, and profit to find its break-even point—the number of houses it needs to sell just to cover its costs.
Key Financials (per house, per month):
Selling Price: $115,000
Variable Costs (per house):
Land: $55,000
Materials: $28,000
Labor: $20,000
Sales Commission: $2,000
Total Variable Cost: $105,000
Fixed Costs (per month):
Office lease, utilities, etc.: $5,000
Employee Salaries: $35,000
Total Fixed Cost: $40,000
Monthly Salaries
Employee
Salary
President
$10,000
VP, Development
$6,000
VP, Marketing
$4,500
Project Manager
$5,500
Controller
$4,000
Office Manager
$3,000
Receptionist
$2,000
Total
$35,000
Example 2: The Profit Model
Let ‘x’ be the number of houses sold in a month.
1. Revenue Function r(x): How much money comes in? \[
r(x) = 115,000x
\]
2. Cost Function c(x): How much money goes out? \[
c(x) = \underbrace{105,000x}_{\text{Variable Costs}} + \underbrace{40,000}_{\text{Fixed Costs}}
\]
What is the break-even point? Break-even is when profit is zero, or when Revenue = Cost. \[
\begin{aligned}
115,000x &= 105,000x + 40,000 \\
10,000x &= 40,000 \\
x &= 4
\end{aligned}
\] Ponderosa needs to sell 4 houses per month to break even.
What is the profit if they sell 12 houses?\[
p(12) = 10,000(12) - 40,000 = \$80,000
\] They would make an $80,000 profit.
Example 2: Visualizing the Break-Even Point
A graph makes this crystal clear. Let’s plot the Cost and Revenue functions.
Introduction to Linear Programming (LP)
Linear Programming is a powerful mathematical tool used to find the best possible outcome (like maximum profit or minimum cost) in a given situation with certain limits or constraints.
Objective: The goal you want to maximize (e.g., profit) or minimize (e.g., cost).
Constraints: The rules or limitations you must follow (e.g., limited budget, time, or materials).
Feasible Solution: Any solution that satisfies all the constraints.
Optimal Solution: The feasible solution that gives the best possible value for the objective function.
The “Linear” in Linear Programming
So, what makes it “linear”?
Linear Functions: Each variable is simple (like x or y, not \(x^2\) or \(\sqrt{y}\)) and is multiplied by a constant.
Linear:3A + 5B
Not Linear:3A² + 5/B
Linear Constraints: These are linear functions restricted by inequalities or equalities.
Linear:2x + 3y ≤ 50
Linear:A = 100
This linearity makes the problems solvable with efficient methods.
LP Problem Formulation: The Blueprint
Every LP problem follows the same basic structure. Your goal is to translate a real-world problem into this mathematical format.
flowchart TD
A[Start: Understand the Business Problem] --> B{What are the Decisions to be Made?};
B --> C[Define Decision Variables <br> e.g., x₁, x₂];
C --> D{What is the Goal? <br> Maximize or Minimize?};
D --> E[Define the Objective Function <br> e.g., Profit = 10x₁ + 20x₂];
E --> F{What are the Limitations? <br> Resources, Demands?};
F --> G[Define the Constraints <br> e.g., 2x₁ + 3x₂ ≤ 100];
G --> H[State Non-Negativity <br> e.g., x₁, x₂ ≥ 0];
H --> I[Solve the Model for the <br> Optimal Solution];
LP Example 3: M&D Chemicals (Minimization)
Problem: M&D Chemicals needs to produce two products, A and B, at the lowest possible cost while meeting certain production demands.
Production Target: Total production of A and B must be at least 350 gallons.
Customer Order: Must produce at least 125 gallons of A.
Resource Limit: Only 600 hours of processing time are available.
Product A takes 2 hours/gallon.
Product B takes 1 hour/gallon.
Costs:
Product A costs $2/gallon.
Product B costs $3/gallon.
Objective: Minimize total production cost.
Example 3: M&D Chemicals Formulation
Let A = number of gallons of product A, and B = number of gallons of product B.
Minimize Cost:\[
\text{Min } 2A + 3B
\]
Subject to (s.t.):\[
\begin{array}{rll}
A &\ge 125 & \text{(Customer Demand)} \\
A + B &\ge 350 & \text{(Total Production)} \\
2A + B &\le 600 & \text{(Processing Time)} \\
A, B &\ge 0 & \text{(Non-negativity)}
\end{array}
\]
Optimal Solution: Produce 250 gallons of A and 100 gallons of B for a minimum cost of $800.
Example 3: Visualizing the M&D Solution
The valid production plans are in the shaded “Feasible Region”. The model finds the corner point in this region with the lowest cost.
LP Example 4: Iron Works, Inc. (Maximization)
Problem: Iron Works makes two products from steel. They want to maximize their profit for the month.
Resource Limit: 2000 pounds of steel available.
Product 1 needs 2 lbs of steel.
Product 2 needs 3 lbs of steel.
Contract: Must make at least 60 units of Product 1.
Production Limit: Can make at most 720 units of Product 2.
Profits:
Product 1: $100 per unit.
Product 2: $200 per unit.
Objective: Maximize total profit.
Example 4: Iron Works Formulation
Let \(x_1\) = number of units of Product 1, and \(x_2\) = number of units of Product 2.