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Introduction to Management Science 8th Edition by Bernard W. Taylor III Chapter 6 Linear Programming: Model Formulation and Graphical Solution Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 1 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 Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 2 Linear Programming An Overview Objectives of business firms frequently include maximizing profit or minimizing costs. Linear programming is an analysis technique in which linear algebraic relationships represent a firm’s decisions given a business objective and resource constraints. Steps in application: Identify problem as solvable by linear programming. Formulate a mathematical model of the unstructured problem. Solve the model. Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 3 Model Components and Formulation 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, that is maximized or minimized Constraints - 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 constraint equations. Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 4 Problem Definition A Maximization Model Example (1 of 2) 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: Product Resource Requirements Labor Clay Profit (hr/unit) (lb/unit) ($/unit) Bowl 1 4 40 Mug 2 3 50 Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 5 Problem Definition A Maximization Model Example (2 of 3) Resource Availability: 40 hrs of labor per day 120 lbs of clay Decision Variables: x1 = number of bowls to produce per day x2 = number of mugs to produce per day Objective Function: Maximize Z = $40x1 + $50x2 Where Z = profit per day Resource Constraints: 1x1 + 2x2 40 hours of labor 4x1 + 3x2 120 pounds of clay Non-Negativity Constraints: x1 0; x2 0 Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 6 Problem Definition A Maximization Model Example (3 of 3) Complete Linear Programming Model: Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2 40 4x1 + 3x2 120 x1, x2 0 Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 7 Feasible Solutions A feasible solution does not violate any of the constraints: Example x1 = 5 bowls x2 = 10 mugs Z = $40x1 + $50x2 = $700 Labor constraint check: 1(5) + 2(10) = 25 < 40 hours, within constraint Clay constraint check: 4(5) + 3(10) = 50 < 120 pounds, within constraint Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 8 Infeasible Solutions An infeasible solution violates at least one of the constraints: Example x1 = 10 bowls x2 = 20 mugs Z = $1400 Labor constraint check: 1(10) + 2(20) = 50 > 40 hours, violates constraint Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 9 Graphical Solution of Linear Programming Models 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. Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 10 Coordinate Axes Graphical Solution of Maximization Model (1 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2 40 4x1 + 3x2 120 x1, x2 0 Figure 6.1 Coordinates for Graphical Analysis Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 11 Labor Constraint Graphical Solution of Maximization Model (2 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2 40 4x1 + 3x2 120 x1, x2 0 Find the boundary first, where 1x1 + 2x2 = 40 Figure 6.1 Graph of Labor Constraint Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 12 Labor Constraint Area Graphical Solution of Maximization Model (3 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2 40 4x1 + 3x2 120 x1, x2 0 Determine which side is allowed, by checking if the easiest choice, x1 = 0 = x2 satisfies the inequality. Figure 6.3 Labor Constraint Area Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 13 Clay Constraint Area Graphical Solution of Maximization Model (4 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2 40 4x1 + 3x2 120 x1, x2 0 Figure 6.4 Clay Constraint Area Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 14 Both Constraints Graphical Solution of Maximization Model (5 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2 40 4x1 + 3x2 120 x1, x2 0 Figure 6.5 Graph of Both Model Constraints Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 15 Feasible Solution Area Graphical Solution of Maximization Model (6 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2 40 4x1 + 3x2 120 x1, x2 0 T: violates both constraints; S: violates the 1st constraint; R: feasible. Figure 6.6 Feasible Solution Area Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 16 Objective Solution = $800 Graphical Solution of Maximization Model (7 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2 40 4x1 + 3x2 120 x1, x2 0 Plot the profit function, Z, for A sample value, Z = $800. Figure 6.7 Objection Function Line for Z = $800 Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 17 Alternative Objective Function Solution Lines Graphical Solution of Maximization Model (8 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2 40 4x1 + 3x2 120 x1, x2 0 Plot the profit function for a few more sample values. Z increases Figure 6.8 Alternative Objective Function Lines Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 18 Optimal Solution Graphical Solution of Maximization Model (9 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2 40 4x1 + 3x2 120 x1, x2 0 Figure 6.9 Identification of Optimal Solution Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 19 Optimal Solution Coordinates Graphical Solution of Maximization Model (10 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2 40 4x1 + 3x2 120 x1, x2 0 The point B is determined as the common solution to 4x1 + 3x2 = 120 x1 + 2x2 = 40 Solve this system of equations… Figure 6.10 Optimal Solution Coordinates Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 20 Corner Point Solutions Graphical Solution of Maximization Model (11 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2 40 4x1 + 3x2 120 x1, x2 0 Figure 6.11 Profit at Each Corner Point Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 21 Optimal Solution for New Objective Function Graphical Solution of Maximization Model (12 of 12) Maximize Z = $70x1 + $20x2 subject to: 1x1 + 2x2 40 4x1 + 3x2 120 x1, x2 0 Figure 6.12 Optimal Solution with Z’ = 70x1 + 20x2 Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 22 Slack Variables Standard form requires that all constraints be in the form of equations. A slack variable is added to an inequality (≤ or ≥) constraint to convert it to an equation (=). A slack variable represents unused resources. A slack variable contributes nothing to the objective function (total profit, cost, etc.) value. Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 23 Linear Programming Model Standard Form Max Z = 40x1 + 50x2 subject to:1x1 + 2x2 + s1 = 40 4x1 + 3x2 + s2 = 120 x1, x2, s1, s2 0 Where: s2 x1 = number of bowls x2 = number of mugs s1, s2 are slack variables Figure 6.13 Solution Points A, B, and C with Slack Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 24 Problem Definition A Minimization Model Example (1 of 7) Two brands of fertilizer available - Super-Gro, Crop-Quick. Field requires at least 16 pounds of nitrogen and 24 pounds of phosphate. Super-Gro costs $6 per bag, Crop-Quick $3 per bag. Problem: How much of each brand to purchase to minimize total cost of fertilizer given following data ? Chemical Contribution Nitrogen (lb/bag) Phosphate (lb/bag) Super-gro 2 4 Crop-quick 4 3 Brand Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 25 Problem Definition A Minimization Model Example (2 of 7) Decision Variables: x1 = bags of Super-Gro x2 = bags of Crop-Quick The Objective Function: Minimize Z = $6x1 + 3x2 Where: $6x1 = cost of bags of Super-Gro $3x2 = cost of bags of Crop-Quick Model Constraints: 2x1 + 4x2 16 lb (nitrogen constraint) 4x1 + 3x2 24 lb (phosphate constraint) x1, x2 0 (non-negativity constraint) Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 26 Model Formulation and Constraint Graph A Minimization Model Example (3 of 7) Minimize Z = $6x1 + $3x2 subject to: 2x1 + 4x2 16 4x1 + 3x2 24 x1, x2 0 Figure 6.14 Graph of Both Model Constraints Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 27 Feasible Solution Area A Minimization Model Example (4 of 7) Minimize Z = $6x1 + $3x2 subject to: 2x1 + 4x2 16 4x1 + 3x2 24 x1, x2 0 Figure 6.15 Feasible Solution Area Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 28 Optimal Solution Point A Minimization Model Example (5 of 7) Minimize Z = $6x1 + $3x2 subject to: 2x1 + 4x2 16 4x1 + 3x2 24 x1, x2 0 Figure 6.16 Optimum Solution Point Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 29 Surplus Variables A Minimization Model Example (6 of 7) A surplus variable is subtracted from an inequality (≥ or ≤) constraint to convert it to an equation (=). A surplus variable represents an excess above a constraint requirement level. Surplus variables contribute nothing to the calculated value of the objective function. Subtracting slack variables in the farmer problem constraints: 2x1 + 4x2 - s1 = 16 (nitrogen) 4x1 + 3x2 - s2 = 24 (phosphate) Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 30 Graphical Solutions A Minimization Model Example (7 of 7) Minimize Z = $6x1 + $3x2 + 0s1 + 0s2 subject to: 2x1 + 4x2 – s1 = 16 4x1 + 3x2 – s2 = 24 x1, x2, s1, s2 0 Figure 6.17 Graph of Fertilizer Example Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 31 Irregular Types of Linear Programming Problems For some linear programming models, the general rules do not apply. Special types of problems include those with: Multiple optimal solutions Infeasible solutions Unbounded solutions Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 32 Multiple Optimal Solutions Beaver Creek Pottery Example Objective function is parallel to a constraint line. Maximize Z=$40x1 + 30x2 subject to: 1x1 + 2x2 40 4x1 + 3x2 120 x1, x2 0 Where: x1 = number of bowls x2 = number of mugs Figure 6.18 Example with Multiple Optimal Solutions Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 33 An Infeasible Problem Every possible solution violates at least one constraint: Maximize Z = 5x1 + 3x2 subject to: 4x1 + 2x2 8 x1 4 x2 6 x1, x2 0 Figure 6.19 Graph of an Infeasible Problem Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 34 An Unbounded Problem Value of objective function increases indefinitely: Maximize Z = 4x1 + 2x2 subject to: x1 4 x2 2 x1, x2 0 Figure 6.20 Graph of an Unbounded Problem Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 35 Characteristics of Linear Programming Problems A linear programming problem requires a decision—a choice amongst alternative courses of action. The decision is represented in the model by decision variables, of which any mix represents a choice. The problem encompasses a goal, expressed as an objective function, that the decision maker wants to achieve (optimize = minimize or maximize, as appropriate). Constraints exist that limit the extent of achievement of the objective. The objective and constraints must be definable by linear mathematical functional relationships. Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 36 Properties of Linear Programming Models Proportionality - The rate of change (slope) of the objective function and constraint equations is constant. Additivity - Terms in the objective function and constraint equations must be additive. Divisibility -Decision variables can take on any fractional value and are therefore continuous as opposed to integer in nature. Certainty - Values of all the model parameters are assumed to be known with certainty (non-probabilistic). Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 37 Problem Statement Example Problem No. 1 (1 of 4) Hot dog mixture in 1000-pound batches. Two ingredients, chicken ($3/lb) and beef ($5/lb). Recipe requirements: at least 500 pounds of chicken at least 200 pounds of beef Ratio of chicken to beef must be at least 2 to 1. Determine optimal mixture of ingredients that will minimize costs. Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 38 Solution Example Problem No. 1 (2 of 4) Step 1: Identify decision variables. x1 = lb of chicken x2 = lb of beef Step 2: Formulate the objective function. Minimize Z = $3x1 + $5x2 where Z = cost per 1,000-lb batch $3x1 = cost of chicken $5x2 = cost of beef Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 39 Solution Example Problem No. 1 (3 of 4) Step 3: Establish Model Constraints x1 + x2 = 1,000 lb x1 500 lb of chicken x2 200 lb of beef x1/x2 2/1, or x1 2 x2, or x1 - 2x2 0 x1, x2 0 The Model: Minimize Z = $3x1 + 5x2 subject to: x1 + x2 = 1,000 lb x1 500 x2 200 x1 - 2x2 0 x1,x2 0 Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 40 Solution Example Problem No. 1 (3 of 4) Step 4: Determine the corners: x2 200 lb of beef A x1 + x2 = 1,000 lb B x1 - 2x2 0 A: x2=200 & x1+x2=1,000 (x1,x2)=(800,200) Z(A) = $3x1+$5x2=$3,400 B: x1=2x2 & x1+x2=1,000 (x1,x2)=(667,333) Z(B) = $3x1+$5x2=$3,667 B . .A B is the optimal choice! Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 41 Example Problem No. 2 (1 of 3) Solve the following model graphically: Maximize Z = 4x1 + 5x2 subject to: x1 + 2x2 10 6x1 + 6x2 36 x1 4 x1, x2 0 Step 1: Plot the constraints as equations Figure 6.21 Constraint Equations Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 42 Example Problem No. 2 (2 of 3) Maximize Z = 4x1 + 5x2 subject to: x1 + 2x2 10 6x1 + 6x2 36 x1 4 x1, x2 0 Step 2: Determine the feasible solution space Figure 6.22 Feasible Solution Space and Extreme Points Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 43 Example Problem No. 2 (3 of 3) Maximize Z = 4x1 + 5x2 subject to: x1 + 2x2 10 6x1 + 6x2 36 x1 4 x1, x2 0 Step 3: Determine the corners; Step 4: Evaluate the objective function and identify the optimal solution. Figure 6.22 Optimal Solution Point Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 44 Chapter 6 - Linear Programming: Model Formulation and Graphical Solution 45