1. BUAD306 Chapter 19 – Linear Programming

2. Optimization QUESTION: Have you ever been limited to what you can get done because you don’t have enough ________? Examples of how we optimize in daily decisions? Why do we strive for optimization? Factors that limit optimization?

3. What is Linear Programming A model consisting of linear relationships that represent a firm’s objective and resource constraints Problems are typically referred to as constrained optimization problems

4. LP Objectives Maximize profits Maximize outputs Minimize costs Determine combinations of outputs to meet/surpass goals

5. Business Examples of LP Product Planning - Landscaping Portfolio Selection - Investments Menu Planning – Restaurants Route Planning / Pricing - Airlines

6. LP Assumptions Linearity - The impact of decision variables is linear in constraints and objective function Divisibility – non-integer values of decision variables are acceptable Certainty – values of parameters are known and constant Non-Negativity – negative values of decision variables are unacceptable

7. LP Concepts Objective - The goal of the LP model Decision Variables - Amounts of either inputs or outputs Constraints - Limitations of available alternatives: Equal to, Greater than, Less than Feasible Solution Space - The set of all feasible combinations of decision variables as defined by the constraints

8. Constraints Restrictions on the company’s resources Time Labor Energy Materials Money Restriction guidelines recipe for making food products engineering specifications

9. LP Steps Step 1: Set up LP model Step 2: Plot the constraints Step 3: Identify the feasible solution space Step 4: Plot the objective function Step 5: Determine the optimal solution

10. LP Model Example X1 = Quantity of Product 1 to produce X2 = Quantity of Product 2 to produce Decision Variables Maximize: 5x1 + 8x2 (profit) Objective Function Subject to: Labor: 2x1 + 4x2 <= 250 hours Material: 7x1 + 6X2 <= 100 pounds Constraints Product 1: X1 >= 10 units x2 >= 0 Non-negativity constraint

11. LP Example A pottery company employs artisans to produce clay bowls and mugs. The two primary resources used by the company are special pottery clay and skilled labor. Given these limited resources, the company would like to determine how many bowls and mugs to produce each day to maximize profit.

12. The two products have the following resource requirements for production and selling price per item produced: There are 40 hours of labor and 60 pounds of clay available each day. Formulate this problem as an LP model.

13. Plot Constraints Step 1: For each constraint, set x 1 = 0, get value for x 2 Step 2: For each constraint, set x 2 = 0, get value for x 1 Step 3: Plot these as coordinates and then draw constraint lines Labor Constraint:x 1 + 2x 2 <= 40 hrs Clay Constraint:3x 1 + 2x 2 <= 60 lbs.

14. Plot Constraints 0 5 10 15 20 25 30 35 40 45 01020304050 x1 x2 50 Clay Labor

15. Identify Feasible Solution Space This is the area that satisfies all of the LP model’s constraints simultaneously. Includes all of the points on the borders as well. 0 5 10 15 20 25 30 35 40 45 01020304050 x1 x2 50

16. Plot the Objective Function In order to locate the point in the feasible solution space that will maximize revenue, we now need to plot the Objective Function: Maximize Z = $25x 1 + 20x 2 We will select and arbitrary level of revenue, such as $300, in order to plot the objective function. 300 = $25x 1 + 20x 2 if x 1 = 0, then x 2 = 12(0, 15) if x 2 = 0, then x 1 = 12(12, 0)

17. Identify Feasible Solution Space Observation: Every point on this Line in the feasible solution area Will result in a revenue of $300. (Every combination of X1 and X2 on this line in this space will yield an $300 revenue total.) 0 5 10 15 20 25 30 35 40 45 01020304050 x1 x2 50 CAN WE DO BETTER THAN $300 GIVEN THE CONSTRAINTS???

18. Identify Feasible Solution Space 0 5 10 15 20 25 30 35 40 45 01020304050 x1 x2 50 Shift the objective function line to locate the optimal solution point.

19. Identify Feasible Solution Space 0 5 10 15 20 25 30 35 40 45 01020304050 x1 x2 50 Shift the objective function line to locate the optimal solution point.

20. Identify Feasible Solution Space 0 5 10 15 20 25 30 35 40 45 01020304050 x1 x2 50 Shift the objective function line to locate the optimal solution point.

21. Identify Feasible Solution Space 0 5 10 15 20 25 30 35 40 45 01020304050 x1 x2 50 Shift the objective function line to locate the optimal solution point. This is where the objective function line is TANGENT to the feasible solution space!

22. What if??? What if objective function was: 100x1 + 20x2?? How does that change the look of the objective function line? 0 5 10 15 20 25 30 35 40 45 01020304050 x1 x2 50 What if STEEP obj function slope? What if FLAT obj function slope?

23. Solving for the Optimal Solution Given our constraint equations: x 1 + 2x 2 = 40 (Labor) 3x 1 + 2x 2 = 60 (Clay) We set them equal to each other and solve for the optimal quantities of X 1 and X 2 … See board for answer!

24. Solving for the Optimal Solution Thus, the optimal solution is to produce: And to maximize the objective function: Maximize Z: 25x 1 + 20x 2 Z =

25. Testing the Optimal Solution You can test to make sure this is the optimal solution by solving for your revenue at the other extreme points (which in this case, intersect with the x 1 and x 2 axis.)

26. Confirm Answer 0 5 10 15 20 25 30 35 40 45 01020304050 x1 x2 50 (0,0) = (0,20) = (20,0) = (10,15) =

27. Delstate Jewelers manufacturers two types of semi-precious gems for sale to its wholesale customers, Gem 1 and Gem 2. Delstate realizes revenue of $4.00 for each Gem 1 sold and revenue of $3 per gem for each Gem 2 sold. To produce the gems, the company must cut and polish each gem. The company is limited to a certain number of minutes for each machine that is used in the cutting and polishing process as shown in the table below. Given this information, determine the optimal quantities of Gem 1 and Gem 2 that maximize revenue. ProductCuttingPolishing Gem 1147 Gem 2612 Total Minutes Available84 Example #1

28. Example #2 A leather shop makes custom, hand-tooled briefcases and luggage. The shop makes a $400 profit from each briefcase and $200 profit from each piece of luggage. The shop has a contract to provide a local store with exactly 30 items each month. A tannery supplies the shop with at least 80 square yards of leather per month. The shop must at least use this amount, but can use more if needed. Each briefcase requires 2 square yards of leather, each piece of luggage requires 8 square yards of leather. From past performance, the shop owner knows that they cannot make more than 20 briefcases per month. Determine the number of briefcases and luggage to produce each month in order to maximize revenue. Formulate an LP model and solve graphically. Determine the optimal quantities of briefcases and luggage and the maximized revenue amount. Also determine if there is any slack or surplus.

29. In Class Example max 5x1 + 6x2 st A2x1+3x2<=60 B4x1+x2<=80 Cx1+x2<=50 N/Nx1, x2 >=0 Will email the answers to you!!

30. NEW - In Class Example max 2x1 + 3x2 st Ax1+3x2<=60 B2x1+2x2<=80 Cx1+x2<=35 D3x1+2x2<=120 N/Nx1, x2 >=0 Will email the answers to you!!

31. Add’l LP Considerations Binding Slack/Surplus Redundant Constraints Multiple Optimal Solutions

32. Binding Constraint When a constraint forms the optimal corner point of the feasible solution space, it is BINDING. Moo!

33. Slack and Surplus A slack variable is a variable representing unused resources added to a <= constraint to make it an equality. A surplus variable is a variable representing an excess above a resource requirement that is subtracted from a >= constraint to make it an equality.

34. Redundant Constraint A constraint that does not form a unique boundary of the feasible solution space. Its removal would not alter the feasible solution space.

35. Multiple Optimal Solutions In some instances, you may get an objective function which is parallel to a constraint line This ends up NOT being tangent to the feasible solution space!

36. Sensitivity Analysis Assessing the impact of potential changes to the numerical values of an LP model Objective function coefficients Right-hand values of constraints Constraint coefficients Max 25x1 + 20x2 S.T. L: 1x1 + 2x2 <= 40 C: 3x1 + 2x2 <= 60 N/N: x1, x2 >=0

37. LINDO Computer-based modeling system for solving linear programming problems All PCs in the Purnell computer lab have LINDO installed

38. LINDO Steps Designate functions within LINDO Max/Min Constraints Solve with Sensitivity Analysis

39. LINDO Answers Notice that the results provide the same answers we obtained in our graphical approach.

40. Objective Function Ranges With this report, we can determine the range in which the objective coefficients can exist before the optimal solution changes: Revenue coefficient for Bowls can range between 10 and 30 without impacting solution Revenue coefficient for Mugs can range between 16.67 and 50 without impacting solution

41. RHS Ranges With this report, we can determine the range in which the RHS of the constraints can change while yielding the same savings/cost indicated by the Shadow (Dual) Price RHS for Labor constraint can range between 20 and 60 RHS for Clay constraint can range between 40 and 120

42. Shadow (Dual) Price The amount by which the value of the objective function would change if there is a one unit change in the RHS of that constraint

43. Shadow (Dual) Pricing If constraint is binding, shadow price = the amount objective function will change for each one unit change in the RHS If constraint has slack, shadow price = 0 and increasing or decreasing RHS has no effect - would only increase slack/surplus!

44. Reminder for Range Analysis: When you change the objective function values and test, you are verifying whether the optimal combination changes… When you change the constraint RHS of constraints, the optimal combination WILL change– what you are verifying is whether the shadow/dual price is in effect and if, thus, the change is worth it!

45. Review of LP Handout #3

46. Homework #5 – PIES! Max Z: 1.5x1 + 1.2X2 Subject to: Flour: 3x1 + 3x2 <= 2100 Sugar: 1.5x1 + 2x2 <= 1200 Time: 6x1 + 3x2 <= 3600 N/N: X1, X2 >= 0

47. Homework #5