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1. The Problem 2. Tabulate Data 3. Translate the Constraints 4. The Objective Function 5. Linear Programming Problem 6. Production Schedule 7. No Waste.

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Presentation on theme: "1. The Problem 2. Tabulate Data 3. Translate the Constraints 4. The Objective Function 5. Linear Programming Problem 6. Production Schedule 7. No Waste."— Presentation transcript:

1 1. The Problem 2. Tabulate Data 3. Translate the Constraints 4. The Objective Function 5. Linear Programming Problem 6. Production Schedule 7. No Waste 8. Feasible Set 1

2 A furniture manufacturer makes two types of furniture - chairs and sofas. The manufacture of a chair requires 6 hours of carpentry, 1 hour of finishing, and 2 hours of upholstery. Manufacture of a sofa requires 3 hours of carpentry, 1 hour of finishing, and 6 hours of upholstery. Each day the factory has available 96 labor hours for carpentry, 18 labor-hours for finishing, and 72 labor-hours for upholstery. The profit per chair is $80 and per sofa is $70. How many chairs and sofas should be produced each day to maximize the profit? 2

3 It is helpful to tabulate data given in the problem. ChairSofaAvailable time Carpentry Finishing Upholstery Profit 6 hours3 hours96 labor-hours 1 hour 18 labor-hours 2 hours6 hours72 labor-hours $80$70 3

4 Translate each of the constraints (restrictions on labor-hours available) into mathematical language. Let x be the number of chairs and y be the number of sofas manufactured each day, respectively. 4

5 Carpentry: [number of labor-hours per day] = (number of hours required per chair) (number of chairs per day) + (number of hours required per sofa) (number of sofas per day) = 6 x + 3 y [number of labor-hours per day] < [maximum available] 6 x + 3 y < 96 5

6 Similarly, Finishing: x + y < 18 Upholstery: 2 x + 6 y < 72 Number of chairs and sofas cannot be negative: x > 0, y > 0 6

7 The objective of the problem is to optimize profit. Translate the profit ( objective function ) into mathematical language. [profit] = [profit from chairs] + [profit from sofas] = [profit per chair] [number of chairs] + [profit per sofa] [number of sofas] = 80 x + 70 y 7

8 The manufacturing problem can now be written as a mathematical problem. Find x and y for which 80 x + 70 y is as large as possible, and for which the following hold simultaneously: 8 This is called a linear programming problem.

9 In the manufacturing problem, each pair of numbers ( x,y ) that satisfies the system of inequalities is called a production schedule. 9

10 Which of the following is a production schedule for (11,6)?(6,11)? 10 YesNo

11 It seems clear that a factory will operate most efficiently when its labor is fully utilized (no waste). This would require x and y to satisfy the system 11

12 Solve 12 According to the graph of the three equations, there is no common intersection and therefore no solution.

13 The set of solutions to the system of inequalities is called the feasible set of the system. This represents all possible production schedules. 13

14 Find the feasible set for 14

15 Notice that (0,0) satisfies all the inequalities. Graph the boundaries: y < -2 x + 32 y < - x + 18 y < - x / x > 0, y > 0 15 Feasible Set

16 A linear programming problem asks us to find the point (or points) in the feasible set of a system of linear inequalities at which the value of a linear expression involving the variables, called the objective function, is either maximized or minimized. 16


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