1. 2-1 Copyright © 2013 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 Linear Programming: Formulation and Applications

2. 2-2 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. Model formulation steps: Step 1 : Clearly define the problem: Objectives and decision to be made Step 2 : Construct the objective function Step 3 : Formulate the constraints Step 4: Solve the model Linear Programming: An Overview

3. LP Model Formulation Product mix problem - Beaver Creek Pottery Company How many bowls and mugs should be produced to maximize profits given labor and materials constraints? Resource Requirements Product Labor (Hr./Unit) Clay (Lb./Unit) Profit ($/Unit) Bowl1440 Mug2350

4. 2-4 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall LP Model Formulation Resource 40 hrs of labor per day Availability:120 lbs of clay Decision B = number of bowls to produce per day Variables: M = number of mugs to produce per day Objective Maximize Z = $40B + $50M Function:Where Z = profit per day Resource 1B + 2M  40 hours of labor Constraints:4B + 3M  120 pounds of clay Non-Negativity B  0; M  0 Constraints:

5. 2-5 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Complete Linear Programming Model: MaximizeZ = $40B + $50M subject to:1B + 2M  40 4B + 3M  120 B, M  0 LP Model Formulation Product mix problem - Beaver Creek Pottery Company

6. 2-6 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall A feasible solution does not violate any of the constraints: Example:B = 5 bowls M = 10 mugs Z = $40B + $50M= $700 Labor constraint check:1(5) + 2(10) = 25 ≤ 40 hours Clay constraint check:4(5) + 3(10) = 70 ≤ 120 pounds Feasible Solutions

7. 2-7 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall An infeasible solution violates at least one of the constraints: Example:B = 10 bowls M = 20 mugs Z = $40B + $50M= $1400 Labor constraint check:1(10) + 2(20) = 50 > 40 hours Infeasible Solutions

8. 2-8 Copyright © 2013 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

9. 2-9 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Coordinate Axes Graphical Solution of Maximization Model (1 of 12) Figure 2.2 Coordinates for graphical analysis B is bowls M is mugs M B

10. 2-10 Copyright © 2013 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 40M+50B M B

11. Product mix problem - Beaver Creek Pottery Company Gather information needed to conduct model building and analysis (steps 2 and 3) –Available capacity in each activity/process –How much capacity in each activity needed by each product –Profitability of each product Given parameters (known data) BowlMug Unit Profit$40$50 Production Time (Hours ) Used Per Unit Produced Resources Available labor1240 clay43120

12. Complete Linear Programming Model: MaximizeZ = $40B + $50M subject to:1B + 2M  40 4B + 3M  120 B, M  0 LP Model Formulation Product mix problem - Beaver Creek Pottery Company

13. 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). Beaver Creek Pottery Company BowlMug Unit Profit4050 Resources Resources Used Per Unit ProducedAvailable Labor1240 Clay43120 2-13

14. 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). 2-14 Beaver Creek Pottery Company BowlMug Unit Profit4050 Resources Resources Used Per Unit ProducedAvailable Labor1240 Clay43120 Units produced

15. 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). 2-15 Beaver Creek Pottery Company BowlMug Unit Profit4050 Resources Resources Used Per Unit ProducedAvailable Labor1240 Clay43120 Units produced total profit

16. 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 consecutive cells. For example, if Quantity A <= Quantity B, put these three items (Quantity A, <=, Quantity B) in consecutive cells. 2-16 Beaver Creek Pottery Company BowlMug Unit Profit4050 Resources Resources Used Per Unit Producedresources usedAvailable Labor12<40 Clay43<120 BowlMug Units produced total profit

17. A Trial Solution 2-17 Let’s do this problem in Excel…. Pay attention to Model layout and organization Data accuracy Steps in the Solver Addin Max or Min non-negativity option Linear optimization

18. Identifying the Target Cell and Changing Cells (Excel 2010) Choose the “Solver” from the Data tab. 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. 2-18

19. Some Important Options (Excel 2007) 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. 2-19

20. The Complete Solver Dialogue Box (Excel 2007) 2-20

21. Template for Resource-Allocation Problems 3-21

22. 2-22 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall LP Model Formulation – Minimization (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 ?

23. 2-23 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Decision Variables: x 1 = bags of Super-gro x 2 = bags of Crop-quick The Objective Function: Minimize Z = $6x 1 + 3x 2 Where:$6x 1 = cost of bags of Super-Gro $3x 2 = cost of bags of Crop-Quick Model Constraints: 2x 1 + 4x 2  16 lb (nitrogen constraint) 4x 1 + 3x 2  24 lb (phosphate constraint) x 1, x 2  0 (non-negativity constraint) LP Model Formulation – Minimization (2 of 7)

24. 2-24 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Minimize Z = $6x 1 + $3x 2 subject to:2x 1 + 4x 2  16 4x 2 + 3x 2  24 x 1, x 2  0 Figure 2.16 Constraint lines for fertilizer model Constraint Graph – Minimization (3 of 7)

25. 2-25 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Figure 2.17 Feasible solution area Feasible Region– Minimization (4 of 7) Minimize Z = $6x 1 + $3x 2 subject to:2x 1 + 4x 2  16 4x 2 + 3x 2  24 x 1, x 2  0

26. 2-26 Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Figure 2.18 The optimal solution point Optimal Solution Point – Minimization (5 of 7) Minimize Z = $6x 1 + $3x 2 subject to:2x 1 + 4x 2  16 4x 2 + 3x 2  24 x 1, x 2  0 The optimal solution of a minimization problem is at the extreme point closest to the origin.

27. Template for Cost-Benefit Tradeoff Problems 3-27

28. Types of Functional Constraints TypeForm*Typical InterpretationMain Usage Resource constraintLHS ≤ RHS For some resource, Amount used ≤ Amount available Resource-allocation problems and mixed problems Benefit constraintLHS ≥ RHS For some benefit, Level achieved ≥ Minimum Acceptable Cost-benefit-trade-off problems and mixed problems Fixed-requirement constraint LHS = RHS For some quantity, Amount provided = Required amount Transportation problems and mixed problems * LHS = Left-hand side (a SUMPRODUCT function). RHS = Right-hand side (a constant). 3-28

29. Template for Mixed Problems 3-29