2-1 Linear Programming: Model Formulation and Graphical Solution Chapter 2 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

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2-1 Linear Programming: Model Formulation and Graphical Solution Chapter 2 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall

2-2 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Chapter Topics Model Formulation A Maximization Model Example Graphical Solutions of Linear Programming Models A Minimization Model Example Irregular Types of Linear Programming Models Characteristics of Linear Programming Problems

2-3 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Objectives of business decisions frequently involve maximizing profit or minimizing costs. Linear programming uses linear algebraic relationships to represent a firm’s decisions, given a business objective, and resource constraints. Steps in application: 1. Identify problem as solvable by linear programming. 2. Formulate a mathematical model of the unstructured problem. 3. Solve the model. 4. Implementation Linear Programming: An Overview

2-4 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Decision variables - mathematical symbols representing levels of activity of a firm. Objective function - a linear mathematical relationship describing an objective of the firm, in terms of decision variables - this function is to be maximized or minimized. Constraints – requirements or restrictions placed on the firm by the operating environment, stated in linear relationships of the decision variables. Parameters - numerical coefficients and constants used in the objective function and constraints. Model Components

2-5 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Summary of Model Formulation Steps Step 1 : Clearly define the decision variables Step 2 : Construct the objective function Step 3 : Formulate the constraints

2-6 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall LP Model Formulation A Maximization Example (1 of 4) Product mix problem - Beaver Creek Pottery Company How many bowls and mugs should be produced to maximize profits given labor and materials constraints? Product resource requirements and unit profit: Resource Requirements Product Labor (Hr./Unit) Clay (Lb./Unit) Profit ($/Unit) Bowl1440 Mug2350

2-7 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall LP Model Formulation A Maximization Example (2 of 4)

2-8 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall LP Model Formulation A Maximization Example (3 of 4) Resource 40 hrs of labor per day Availability:120 lbs of clay Decision x 1 = number of bowls to produce per day Variables: x 2 = number of mugs to produce per day Objective Maximize Z = $40x 1 + $50x 2 Function:Where Z = profit per day Resource 1x 1 + 2x 2  40 hours of labor Constraints:4x 1 + 3x 2  120 pounds of clay Non-Negativity x 1  0; x 2  0 Constraints:

2-9 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall LP Model Formulation A Maximization Example (4 of 4) Complete Linear Programming Model: MaximizeZ = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 2 + 3x 2  120 x 1, x 2  0

2-10 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall A feasible solution does not violate any of the constraints: Example:x 1 = 5 bowls x 2 = 10 mugs Z = $40x 1 + $50x 2 = $700 Labor constraint check:1(5) + 2(10) = 25 < 40 hours Clay constraint check:4(5) + 3(10) = 70 < 120 pounds Feasible Solutions

2-11 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall An infeasible solution violates at least one of the constraints: Example:x 1 = 10 bowls x 2 = 20 mugs Z = $40x 1 + $50x 2 = $1400 Labor constraint check:1(10) + 2(20) = 50 > 40 hours Infeasible Solutions

2-12 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Graphical solution is limited to linear programming models containing only two decision variables (can be used with three variables but only with great difficulty). Graphical methods provide visualization of how a solution for a linear programming problem is obtained. Graphical Solution of LP Models

2-13 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Coordinate Axes Graphical Solution of Maximization Model (1 of 12) Figure 2.2 Coordinates for Graphical Analysis Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 2 + 3x 2  120 x 1, x 2  0 X 1 is bowls X 2 is mugs

2-14 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Labor Constraint Graphical Solution of Maximization Model (2 of 12) Figure 2.3 Graph of Labor Constraint Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 2 + 3x 2  120 x 1, x 2  0

2-15 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Labor Constraint Area Graphical Solution of Maximization Model (3 of 12) Figure 2.4 Labor Constraint Area Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 2 + 3x 2  120 x 1, x 2  0

2-16 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Clay Constraint Area Graphical Solution of Maximization Model (4 of 12) Figure 2.5 Clay Constraint Area Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 2 + 3x 2  120 x 1, x 2  0

2-17 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Both Constraints Graphical Solution of Maximization Model (5 of 12) Figure 2.6 Graph of Both Model Constraints Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 2 + 3x 2  120 x 1, x 2  0

2-18 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Feasible Solution Area Graphical Solution of Maximization Model (6 of 12) Figure 2.7 Feasible Solution Area Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 2 + 3x 2  120 x 1, x 2  0

2-19 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Objective Function Solution = $800 Graphical Solution of Maximization Model (7 of 12) Figure 2.8 Objection Function Line for Z = $800 Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 2 + 3x 2  120 x 1, x 2  0

2-20 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Alternative Objective Function Solution Lines Graphical Solution of Maximization Model (8 of 12) Figure 2.9 Alternative Objective Function Lines Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 2 + 3x 2  120 x 1, x 2  0

2-21 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Optimal Solution Graphical Solution of Maximization Model (9 of 12) Figure 2.10 Identification of Optimal Solution Point Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 2 + 3x 2  120 x 1, x 2  0

2-22 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Optimal Solution Coordinates Graphical Solution of Maximization Model (10 of 12) Figure 2.11 Optimal Solution Coordinates Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 2 + 3x 2  120 x 1, x 2  0

2-23 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Extreme (Corner) Point Solutions Graphical Solution of Maximization Model (11 of 12) Figure 2.12 Solutions at All Corner Points Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 2 + 3x 2  120 x 1, x 2  0

2-24 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Optimal Solution for New Objective Function Graphical Solution of Maximization Model (12 of 12) Maximize Z = $70x 1 + $20x 2 subject to:1x 1 + 2x 2  40 4x 2 + 3x 2  120 x 1, x 2  0 Figure 2.13 Optimal Solution with Z = 70x x 2

2-25 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Standard form requires that all constraints be in the form of equations (equalities). A slack variable is added to a  constraint (weak inequality) to convert it to an equation (=). A slack variable typically represents an unused resource. A slack variable contributes nothing to the objective function value. Slack Variables

2-26 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Linear Programming Model: Standard Form Max Z = 40x x 2 + s 1 + s 2 subject to:1x 1 + 2x 2 + s 1 = 40 4x 2 + 3x 2 + s 2 = 120 x 1, x 2, s 1, s 2  0 Where: x 1 = number of bowls x 2 = number of mugs s 1, s 2 are slack variables Figure 2.14 Solution Points A, B, and C with Slack