copyright © 2010, All rights reserved eStudy.us Linear Programming Hirschey Alternative Presentation Sweeney Alternative Presentation

copyright © 2010, All rights reserved eStudy.us Linear programming involves choosing a course of action when the mathematical model of the problem contains only linear functions. The maximization or minimization of some quantity is the objective in all linear programming problems. All LP problems have constraints that limit the degree to which the objective can be pursued. A feasible solution satisfies all the problem's constraints. An optimal solution is a feasible solution that results in the largest possible objective function value when maximizing (or smallest when minimizing). A graphical solution method can be used to solve a linear program with two variables.

copyright © 2010, All rights reserved eStudy.us If both the objective function and the constraints are linear, the problem is referred to as a linear programming problem. Linear functions are functions in which each variable appears in a separate term raised to the first power and is multiplied by a constant (which could be 0). Linear constraints are linear functions that are restricted to be "less than or equal to", "equal to", or "greater than or equal to" a constant.

copyright © 2010, All rights reserved eStudy.us Problem formulation or modeling is the process of translating a verbal statement of a problem into a mathematical statement. Formulating models is an art that can only be mastered with practice and experience. Every LP problems has some unique features, but most problems also have common features. General guidelines for LP model formulation are illustrated on the slides that follow.

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copyright © 2010, All rights reserved eStudy.us Small brewery produces ale and beer. Production limited by scarce resources: corn, hops, barley malt. Recipes for ale and beer require different proportions of resources. corn (lbs)hops (oz)malt (lbs)profit ($) available ale (1 barrel)5435$13 beer (1 barrel)15420$23 Brewer’s problem: choose product mix to maximize profits. Corn Subject to: 5A+15B≤480 Hops 4A+ 4B≤160 Malt35A+20B≤1190 A≥0 B≥0 Maximize13A+23BProfitObjective Function A=ale, B=Beer

copyright © 2010, All rights reserved eStudy.us A=ale, B=Beer Maximize Subject to: 13A+23BProfit 5A+15B≤480Corn 4A+ 4B≤160Hops 35A+20B≤1190Malt A≥0 B≥0 Malt Corn Hops

copyright © 2010, All rights reserved eStudy.us A=ale, B=Beer Maximize Subject to: 13A+23BProfit 5A+15B≤480Corn 4A+ 4B≤160Hops 35A+20B≤1190Malt A≥0 B≥0 MaltCornHops 34,0 0,32 12,28 26,14 Ale Beer 96,0 Corn Hops Malt 0,40 0, ,0 Corn – Hops

copyright © 2010, All rights reserved eStudy.us A=ale, B=Beer Maximize Subject to: 13A+23BProfit 5A+15B≤480Corn 4A+ 4B≤160Hops 35A+20B≤1190Malt A≥0 B≥0 Profit ($800) Beer 0,32 12,28 26,14 Corn Hops Malt Ale Profit ($1,000) Profit ($1,200) Hops and Corn or effective constraints By using all the Hops and Corn the Brewery will maximize profits at $800 With 210 lbs. of Malt remaining 35(12)+20(28)=980 13(26)+23(14)=$800 34,0

copyright © 2010, All rights reserved eStudy.us Maximize Subject to: 13A+23BProfit 5A+15B≤480Corn 4A+ 4B≤160Hops 35A+20B≤1190Malt A≥0 B≥0 Objective Function Maximize Subject to: 13A+23BProfit 5A+15B+S c =480Corn 4A+ 4B+S h = 160Hops 35A+20B++S m =1190Malt A, B, S c, S h, S m ≥ 0 Standard Form Add slack variable to turn inequalities into equalities

copyright © 2010, All rights reserved eStudy.us Maximize Subject to: 13A+23BProfit 5A+15B+S c =480Corn 4A+ 4B+S h = 160Hops 35A+20B++S m =1190Malt A, B, S c, S h, S m ≥ 0 Standard Form Add slack variable to turn inequalities into equalities Maximize Subject to: 13A+23BProfit 5A+15B+0=480Corn 4A+ 4B+0= 160Hops 35A+20B++210=1190Malt Standard Form at solution A=12 and B=28

copyright © 2010, All rights reserved eStudy.us Sofa Drink Company LouisvilleOwensboro Grape Soda Orange Soda1000 Blueberry Soda A soft drink company has two bottling plants, one located in Louisville and the other in Owensboro. Each plant produces three different soft drinks Grape, Orange, and Blueberry flavor. The capacities of the two plants in number of bottles per day, are as follows: A market survey indicates that during the month of May, there will be a demand for 24,000 bottles of Grape, 16,000 bottles of Orange and 48,000 bottles of Blueberry. The operating cost per day of running the plants in Louisville and Owensboro are respectively $6,000 and $4,000. How many days should the firm run each plant in the month of May so that the production cost is minimized while still meeting the market demand.

copyright © 2010, All rights reserved eStudy.us Sofa Drink Company Suppose that the firm runs the Louisville plant for x 1 days and the Owensboro plant for x 2 days in the month of May in order to meet the market demand. The per day operating cost of Louisville plant is $6000. Therefore, for x 1 days the operating cost will be $6,000x 1. The per day operating cost of the Owensboro plant is $4,000. Therefore, for x 2 days the operating cost will be $4,000x 2. Thus the total operating cost of two plants is given by:

copyright © 2010, All rights reserved eStudy.us

copyright © 2010, All rights reserved eStudy.us

copyright © 2010, All rights reserved eStudy.us

copyright © 2010, All rights reserved eStudy.us

copyright © 2010, All rights reserved eStudy.us Minimize Subject to: 6000x x 2 Cost 3000x x 2 -S g =24000Grape 3000x x 2 -S o = 16000Orange 2000x x 2 -S b = 48000Blueberry x 1, x 2, S g, S o, S b ≥ 0 Standard Form Add slack variable to turn inequalities into equalities Standard Form at solution x 1 =4 and x 2 =12 Minimize Subject to: 6000x x 2 Cost 3000x x 2 -0=24000Grape 3000x x 2 -0= 16000Orange 2000x x = 48000Blueberry x 1, x 2, S g, S o, S b ≥ 0