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Linear Programming Models: Graphical Methods 5/4/1435 (1-3 pm)noha hussein elkhidir
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Requirements of a Linear Programming ProblemRequirements of a Linear Programming Problem Formulating Linear Programming Problems. Formulating Linear Programming Problems. Graphical Solution to a Linear Programming ProblemGraphical Solution to a Linear Programming Problem -Graphical Representation of Constraints - Corner-Point Solution Method 5/4/1435 (1-3 pm)noha hussein elkhidir
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Linear Programming Applications Production. Diet Problem Example Labor Scheduling Example. 5/4/1435 (1-3 pm)noha hussein elkhidir
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When you complete this module you should be able to: - Formulate linear programming models, including an objective function and constraints -Graphically solve an LP problem with the corner-point method. - Construct and solve a minimization and maximization problem 5/4/1435 (1-3 pm)noha hussein elkhidir
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- Formulate production-mix, diet, and labor scheduling problems 5/4/1435 (1-3 pm)noha hussein elkhidir
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What about Linear Programming? - A mathematical technique to help plan and make decisions relative to the trade-offs necessary to allocate resources - Will find the minimum or maximum value of the objective - Guarantees the optimal solution to the model formulated 5/4/1435 (1-3 pm)noha hussein elkhidir
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Requirements of an LP Problem 1.LP problems seek to maximize or minimize some quantity (usually profit or cost) expressed as an objective function 2.The presence of restrictions, or constraints, limits the degree to which we can pursue our objective 5/4/1435 (1-3 pm)noha hussein elkhidir
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3.There must be alternative courses of action to choose from 4.The objective and constraints in linear programming problems must be expressed in terms of linear equations or inequalities 5/4/1435 (1-3 pm)noha hussein elkhidir
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Steps in Developing a Linear Programming (LP) Model 1)Formulation 2)Solution 5/4/1435 (1-3 pm)noha hussein elkhidir
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Properties of LP Models 1)Seek to minimize or maximize 2)Include “constraints” or limitations 3)There must be alternatives available 4)All equations are linear 5/4/1435 (1-3 pm)noha hussein elkhidir
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Example LP Model Formulation: The Product Mix Problem 1)Decision: How much to make of > 2 products? 2)Objective: Maximize profit 3)Constraints: Limited resources 5/4/1435 (1-3 pm)noha hussein elkhidir
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Example: Flair Furniture Co. Two products: Chairs and Tables Decision: How many of each to make this month? Objective: Maximize profit 5/4/1435 (1-3 pm)noha hussein elkhidir
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Flair Furniture Co. Data Tables (per table) Chairs (per chair) Hours Available Profit Contribution $7$5 Carpentry3 hrs4 hrs2400 Painting2 hrs1 hr1000 Other Limitations: Make no more than 450 chairs Make no more than 450 chairs Make at least 100 tables Make at least 100 tables 5/4/1435 (1-3 pm)noha hussein elkhidir
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Decision Variables: T = Num. of tables to make C = Num. of chairs to make Objective Function: Maximize Profit Maximize $7 T + $5 C 5/4/1435 (1-3 pm)noha hussein elkhidir
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Constraints: Have 2400 hours of carpentry time available 3 T + 4 C < 2400 (hours) Have 1000 hours of painting time available 2 T + 1 C < 1000 (hours) 5/4/1435 (1-3 pm)noha hussein elkhidir
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More Constraints: Make no more than 450 chairs C < 450 (num. chairs) Make at least 100 tables T > 100 (num. tables) Nonnegativity: Cannot make a negative number of chairs or tables T > 0 C > 0 5/4/1435 (1-3 pm)noha hussein elkhidir
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Model Summary Max 7T + 5C(profit) Subject to the constraints: 3T + 4C < 2400 (carpentry hrs) 2T + 1C < 1000 (painting hrs) C < 450 (max # chairs) T > 100 (min # tables) T, C > 0 (nonnegativity) 5/4/1435 (1-3 pm)noha hussein elkhidir
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Graphical Solution Graphing an LP model helps provide insight into LP models and their solutions. While this can only be done in two dimensions, the same properties apply to all LP models and solutions. 5/4/1435 (1-3 pm)noha hussein elkhidir
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Carpentry Constraint Line 3T + 4C = 2400 Intercepts (T = 0, C = 600) (T = 800, C = 0) 0 800 T C 600 0 Feasible < 2400 hrs Infeasible > 2400 hrs 3T + 4C = 2400 5/4/1435 (1-3 pm)noha hussein elkhidir
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Painting Constraint Line 2T + 1C = 1000 Intercepts (T = 0, C = 1000) (T = 500, C = 0) 0 500 800 T C 1000 600 0 2T + 1C = 1000 5/4/1435 (1-3 pm)noha hussein elkhidir
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0 100 500 800 T C 1000 600 450 0 Max Chair Line C = 450 Min Table Line T = 100 Feasible Region 5/4/1435 (1-3 pm)noha hussein elkhidir
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0 100 200 300 400 500 T C 500 400 300 200 100 0 Objective Function Line 7T + 5C = Profit 7T + 5C = $2,100 7T + 5C = $4,040 Optimal Point (T = 320, C = 360) 7T + 5C = $2,800 5/4/1435 (1-3 pm)noha hussein elkhidir
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