McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Table of Contents Chapter 2 (Linear Programming: Basic Concepts) Three Classic Applications of LP (Section 2.1)2.2 The Wyndor Glass Company Product Mix Problem (Section 2.2)2.3 Formulating the Wyndor Problem on a Spreadsheet (Section 2.3)2.4–2.8 The Algebraic Model for Wyndor (Section 2.4)2.9 The Graphical Method Applied to the Wyndor Problem (Section 2.5)2.10–2.20 Using the Excel Solver with the Wyndor Problem (Section 2.6)2.21–2.26 A Minimization Example—The Profit & Gambit Co. (Section 2.7)2.27–2.32 Introduction to Linear Programming (UW Lecture)2.33–2.48 These slides are based upon a lecture introducing the basic concepts of linear programming and the Solver to first-year MBA students at the University of Washington (as taught by one of the authors). The lecture is largely based upon a production problem using lego building blocks. The Graphical Method and Properties of LP Solutions (UW Lecture)2.49–2.57 These slides are based upon a lecture introducing the graphical method and other concepts about linear programming solutions to first-year MBA students at the University of Washington (as taught by one of the authors).

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Three Classic Applications of LP Product Mix at Ponderosa Industrial –Considered limited resources, and determined optimal mix of plywood products. –Increased overall profitability of company by 20%. Personnel Scheduling at United Airlines –Designed work schedules for all employees at a location to meet service requirements most efficiently. –Saved $6 million annually. Planning Supply, Distribution, and Marketing at Citgo Petroleum Corporation –The SDM system uses LP to coordinate the supply, distribution, and marketing of each of Citgo’s major products throughout the United States. –The resulting reduction in inventory added $14 million annually to Citgo’s profits.

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Wyndor Glass Co. Product Mix Problem Wyndor has developed the following new products: –An 8-foot glass door with aluminum framing. –A 4-foot by 6-foot double-hung, wood-framed window. The company has three plants –Plant 1 produces aluminum frames and hardware. –Plant 2 produces wood frames. –Plant 3 produces glass and assembles the windows and doors. Questions: 1.Should they go ahead with launching these two new products? 2.If so, what should be the product mix?

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Developing a Spreadsheet Model Step #1: Data Cells –Enter all of the data for the problem on the spreadsheet. –Make consistent use of rows and columns. –It is a good idea to color code these “data cells” (e.g., light blue).

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Developing a Spreadsheet Model Step #2: Changing Cells –Add a cell in the spreadsheet for every decision that needs to be made. –If you don’t have any particular initial values, just enter 0 in each. –It is a good idea to color code these “changing cells” (e.g., yellow with border).

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Developing a Spreadsheet Model Step #3: Target Cell –Develop an equation that defines the objective of the model. –Typically this equation involves the data cells and the changing cells in order to determine a quantity of interest (e.g., total profit or total cost). –It is a good idea to color code this cell (e.g., orange with heavy border).

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Developing a Spreadsheet Model Step #4: Constraints –For any resource that is restricted, calculate the amount of that resource used in a cell on the spreadsheet (an output cell). –Define the constraint in three consecutive cells. For example, if Quantity A ≤ Quantity B, put these three items (Quantity A, ≤, Quantity B) in consecutive cells.

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., A Trial Solution The spreadsheet for the Wyndor problem with a trial solution (4 doors and 3 windows) entered into the changing cells.

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Algebraic Model for Wyndor Glass Co. LetD = the number of doors to produce W = the number of windows to produce Maximize P = $300D + $500W subject to D ≤ 4 2W ≤ 12 3D + 2W ≤ 18 and D ≥ 0, W ≥ 0.

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Graphing the Product Mix

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Graph Showing Constraints: D ≥ 0 and W ≥ 0

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Nonnegative Solutions Permitted by D ≤ 4

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Nonnegative Solutions Permitted by 2W ≤ 12

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Boundary Line for Constraint 3D + 2W ≤ 18

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Changing Right-Hand Side Creates Parallel Constraint Boundary Lines

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Nonnegative Solutions Permitted by 3D + 2W ≤ 18

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Graph of Feasible Region

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Objective Function (P = 1,500)

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Finding the Optimal Solution

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Summary of the Graphical Method Draw the constraint boundary line for each constraint. Use the origin (or any point not on the line) to determine which side of the line is permitted by the constraint. Find the feasible region by determining where all constraints are satisfied simultaneously. Determine the slope of one objective function line. All other objective function lines will have the same slope. Move a straight edge with this slope through the feasible region in the direction of improving values of the objective function. Stop at the last instant that the straight edge still passes through a point in the feasible region. This line given by the straight edge is the optimal objective function line. A feasible point on the optimal objective function line is an optimal solution.

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Identifying the Target Cell and Changing Cells Choose the “Solver” from the Tools menu. Select the cell you wish to optimize in the “Set Target Cell” window. Choose “Max” or “Min” depending on whether you want to maximize or minimize the target cell. Enter all the changing cells in the “By Changing Cells” window.

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Adding Constraints To begin entering constraints, click the “Add” button to the right of the constraints window. Fill in the entries in the resulting Add Constraint dialogue box.

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., The Complete Solver Dialogue Box

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Some Important Options Click on the “Options” button, and click in both the “Assume Linear Model” and the “Assume Non-Negative” box. –“Assume Linear Model” tells the Solver that this is a linear programming model. –“Assume Non-Negative” adds nonnegativity constraints to all the changing cells.

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., The Solver Results Dialogue Box

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., The Optimal Solution

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., The Profit & Gambit Co. Management has decided to undertake a major advertising campaign that will focus on the following three key products: –A spray prewash stain remover. –A liquid laundry detergent. –A powder laundry detergent. The campaign will use both television and print media The general goal is to increase sales of these products. Management has set the following goals for the campaign: –Sales of the stain remover should increase by at least 3%. –Sales of the liquid detergent should increase by at least 18%. –Sales of the powder detergent should increase by at least 4%. Question: how much should they advertise in each medium to meet the sales goals at a minimum total cost?

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Profit & Gambit Co. Spreadsheet Model

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Algebraic Model for Profit & Gambit LetTV = the number of units of advertising on television PM = the number of units of advertising in the print media Minimize Cost = TV + 2PM (in millions of dollars) subject to Stain remover increased sales: PM ≥ 3 Liquid detergent increased sales:3TV + 2PM ≥ 18 Powder detergent increased sales:–TV + 4PM ≥ 4 and TV ≥ 0, PM ≥ 0.

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Applying the Graphical Method

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., The Optimal Solution

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Summary of the Graphical Method Draw the constraint boundary line for each constraint. Use the origin (or any point not on the line) to determine which side of the line is permitted by the constraint. Find the feasible region by determining where all constraints are satisfied simultaneously. Determine the slope of one objective function line. All other objective function lines will have the same slope. Move a straight edge with this slope through the feasible region in the direction of improving values of the objective function. Stop at the last instant that the straight edge still passes through a point in the feasible region. This line given by the straight edge is the optimal objective function line. A feasible point on the optimal objective function line is an optimal solution.

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Components of a Linear Program Data Cells Changing Cells (“Decision Variables”) Target Cell (“Objective Function”) Constraints

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Four Assumptions of Linear Programming Linearity Divisibility Certainty Nonnegativity

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., When is a Spreadsheet Model Linear? All equations (output cells) must be of the form = ax + by + cz + … where a, b, c are constants (data cells) and x, y, z are changing cells. Suppose C1:C6 are changing cells and D1:D6 are data cells. Which of the following can be part of an LP? –SUMPRODUCT(D1:D6, C1:C6) –SUM(C1:C6) –C1 * SUM(C4:C6) –SUMPRODUCT(C1:C3, C4:C6) –IF(C1 > 3, 2*C3 + C4, 3*C3 + C5) –IF(D1 > 3, C1, C2) –MIN(C1, C2) –MIN(D1, D2) * C1 –ROUND(C1)

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Why Use Linear Programming? Linear programs are easy (efficient) to solve The best (optimal) solution is guaranteed to be found (if it exists) Useful sensitivity analysis information is generated Many problems are essentially linear

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., The Graphical Method for Solving LP’s Formulate the problem as a linear program Plot the constraints Identify the feasible region Draw an imaginary line parallel to the objective function (Z = a) Find the optimal solution

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Properties of Linear Programming Solutions An optimal solution must lie on the boundary of the feasible region. There are exactly four possible outcomes of linear programming: –A unique optimal solution is found. –An infinite number of optimal solutions exist. –No feasible solutions exist. –The objective function is unbounded (there is no optimal solution). If an LP model has one optimal solution, it must be at a corner point. If an LP model has many optimal solutions, at least two of these optimal solutions are at corner points.

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Example #4 (Multiple Optimal Solutions) Minimize Z = 6x 1 + 4x 2 subject to x 1 ≤ 4 2x 2 ≤ 12 3x 1 + 2x 2 ≤ 18 and x 1 ≥ 0, x 2 ≥ 0.

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Example #5 (No Feasible Solution) Maximize Z = 3x 1 + 5x 2 subject to x 1 ≥ 5 x 2 ≥ 4 3x 1 + 2x 2 ≤ 18 and x 1 ≥ 0, x 2 ≥ 0.

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., Example #6 (Unbounded Solution) Maximize Z = 5x x 2 subject to x 1 ≤ 5 2x 1 –x 2 ≤ 2 and x 1 ≥ 0, x 2 ≥ 0.

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., The Simplex Method Algorithm 1.Start at a feasible corner point (often the origin). 2.Check if adjacent corner points improve the objective function: a)If so, move to adjacent corner and repeat step 2. b)If not, current corner point is optimal. Stop.