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BU-75 Lecture

Intro to Linear Programming

• Motivating Example: Craft Brewery

• A constrained resource problem and a revenue maximization problem

In the brewery today, we have: 50 kg of barley, 28 ounces of Hops.

It takes 5kg of barley and 2 ounces of Hops to make 20L of Lager

It takes 2kg of barley and 4 ounces of Hops to make 20L of Ale

The 20L Lager sells for $80

The 20L Ale sells for $100

How many kegs of Lager and Ale should we make today to maximize revenue?

Solution Techniques:

Linear Programming Formulation:

1. Define Decision Variables: Let x represent the number of kegs of Lager. Let y represent the

number of kegs of Ale.

2. Objective Function: Maximize 80x+100y

a. The objective function calculates an objective value (ex. Revenue)

3. Constraints: Can be built using a resource table

Lager

Ale

Available

Barley

5

2

50

Hops

2

4

28

Barley Constraint: 5x+2y is less than or equal to 50

Hops Constraint: 2x+4y is less than or equal to 28

Left hand side computes the amount of resources used. Right hand side computes the resources

available.

Our Linear Program:

Max revenue: 80x+100y

Subject to: 5x+2y0 and 2+28

Non-negativity constraints x-y0

Solving Methods: Graphically using Corner Point Method

1. Graph the problem, by graphing the constraints as if they were lines with = signs

a. 5X+2Y=50 and 2X+4Y=28

2. Next, we identify the Feasible Region

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