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Lecture 1

BU275 Lecture 1: BU-275 Lecture 1

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David Wheatley

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BU-275Lecture1 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+2y50 and 2x+4y28 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 a. A feasible solution is one that satisfies all constraints b. The feasible region is the set of all feasible solutions c. There is at most 1 feasible region, a problem with no feasible solutions is called
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