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Slides by John Loucks St. Edward’s University

Slides by John Loucks St. Edward’s University. Chapter 8 Nonlinear Optimization Models. A Production Application Blending : The Pooling Problem Forecasting Adoption of a New Product. Introduction. Many business processes behave in a nonlinear manner.

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Slides by John Loucks St. Edward’s University

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  1. Slides by John Loucks St. Edward’s University

  2. Chapter 8Nonlinear Optimization Models • A Production Application • Blending: The Pooling Problem • Forecasting Adoption of a New Product

  3. Introduction • Many business processes behave in a nonlinear manner. • For instance, The quantity demanded for a product is often a nonlinear function of the price.

  4. A nonlinear optimization problem is any optimization problem in which at least one term in the objective function or a constraint is nonlinear. • Nonlinear terms include • The nonlinear optimization problems presented on the upcoming slides can be solved using computer software such as LINGO and Excel Solver. Introduction

  5. Example: Production Application • Par, Inc. • Par, Inc. manufactures golf bags • Two Models: Standard (S) and Deluxe (D) • Four Operations Required for each Bag • Cutting and Dyeing • Sewing • Finishing • Inspection and Packaging

  6. Example: Production Application • Assume that the demand for Standard (S) and Deluxe (D) golf bags are (projected sales – cost differential): S = 2250 – 15Ps D = 1500 – 5Pd where Ps= the price of a Standard bag Pd= the price of a Deluxe bag. We will need to isolate Ps and Pd: • 15Ps = 2250 – S 5Pd = 1500 – D • Ps = 2250/15 –S/15 Pd = 1500/5 – 1/5D • Ps = 150 – 1/15S Pd = 300 – 1/5D

  7. Example: Production Application • Profit Contribution as a Function of Demand • The profit contributions (revenue – cost) are: PsS – 70S (Standard bags) PdD – 150D (Deluxe bags) • Solving for Ps we get: • PsS – 70S • (150 – 1/15S)S – 70S • 80S – 1/15S^2 • Solving for Pd we get • PdD – 150D • (300 – 1/5D)D – 150D • 150D – 1/5D^2

  8. Example: Production Application • Total Profit Contribution • Total Profit Contrib. = 80S – 1/15S^2 + 150D – 1/5D^2 This function is an example of a quadratic function because the nonlinear terms have a power of 2.

  9. Par Inc, Unconstrained Solution • If we were to just solve the optimization equation, then we would find that it is: • S = 600, D = 375, Ps = 110, Pd = 225 • BUT WE HAVENT INCLUDED THE CONSTRAINTS! • 7/10s + 1d <= 630 (Cutting and Dyeing) • 1/2s + 5/6d <= 600 (Sewing) • 1s + 2/3d <= 708 (Finishing) • 1/10s + 1/4d <= 135 (Inspecting and Packing) • s, d >= 0

  10. Par Inc

  11. Par Inc

  12. Par Inc

  13. Dual Values • Recall that the dual value is the change in the value of the optimal solution per unit increase in the right-hand side of the constraint. • The interpretation of the dual value for nonlinear models is exactly the same as it is for LPs. • However, for nonlinear problems the allowable increase and decrease are not usually reported. • This is because for typical nonlinear problems the allowable increase and decrease are zero. • That is, if you change the right-hand side by even a small amount, the dual value changes.

  14. Blending: The Pooling Problem • Blending problems arise when a manager must decide how to blend two or more components (resources) to produce one or more products. • It is often the case that while transporting or storing the blending components, the components must share a pipeline or storage tank. • In this case, the components are called pooled components.

  15. Blending: The Pooling Problem • Two types of decisions arise: • What should the proportions be for the components that are to be pooled? • How much of the pooled and non-pooled components will be used to make each of the final products?

  16. Example: Blending - The Pooling Problem Grand Strand refinery wants to refine three petroleum components into regular and premium gasoline in order to maximize total profit contribution. Components 1 and 2 are pooled in a single storage tank. Component 3 has its own storage tank. The maximum number of gallons available for the three components are 5000, 10,000, and 10,000, respectively. The three components cost $2.50, $2.60, and $2.84, respectively. Regular gasoline sells for $2.90 and premium sells for $3.00. At least 10,000 gallons of regular gasoline must be produced. The product specifications for regular and premium gasoline are shown on the next slide.

  17. Grand Strand

  18. Example: Blending - The Pooling Problem • Product Specifications • Regular gasoline At most 30% component 1 • At least 40% component 2 • At most 20% component 3 • Premium gasoline At least 25% component 1 • At most 45% component 2 • At least 30% component 3

  19. Example: Blending - The Pooling Problem • Define the 6 Decision Variables y1 = gallons of component 1 in the pooling tank y2 = gallons of component 2 in the pooling tank xpr = gallons of pooled components 1 and 2 in regular gas xpp = gallons of pooled components 1 and 2 in premium gas x3r = gallons of component 3 in regular gasoline x3p = gallons of component 3 in premium gasoline

  20. Example: Blending - The Pooling Problem • Define the Objective Function Maximize the total contribution to profit (which is revenue from selling regular and premium gasolines minus cost of buying components 1, 2, and 3): Max 2.90(xpr + x3r) + 3.00(xpp + x3p) – 2.50y1 – 2.60y2 – 2.84(x3r + x3p) (Note: xpr + x3r = gallons of regular gasoline sold, xpp + x3p = gallons of premium gasoline sold, x3r + x3p = gallons of component 3 consumed)

  21. Example: Blending - The Pooling Problem • Define the 11 Constraints Components 1 and 2 consumed Similar to a “flow-in-flow-out” constraint: 1)y1 + y2 = xpr + xpp Component availability: 2)y1< 5,000 3)y2< 10,000 4) x3r + x3p< 10,000 Minimum production of regular gasoline: 5)xpr + x3r> 10,000

  22. Example: Blending - The Pooling Problem • Define the 11 Constraints (continued) Regular gasoline specifications: We have to measure the proportion of a component in the pool when calculating the specification constraints 6) 7) 8)

  23. Example: Blending - The Pooling Problem • Define the 11 Constraints (continued) Premium gasoline specifications: 9) 10) 11) Non-negativity: xpr , xpp , x3r , x3p , y1 , y2> 0

  24. Example: Blending - The Pooling Problem

  25. Example: Blending - The Pooling Problem • Solution (with pooling optimal sol. = $5831.43 • Without Pooling optimal sol. = $7100

  26. Forecasting Adoption of a New Product • Forecasting new adoptions (purchases) after a product introduction is an important marketing problem. • We introduce here a forecasting model developed by Frank Bass. • Nonlinear programming is used to estimate the parameters of the Bass forecasting model.

  27. Forecasting Adoption of a New Product • The Bass model has three parameters that must be estimated. • m is the number of people estimated to eventually adopt a new product • q is the coefficient of imitation which measures the likelihood of adoption due to a potential adopter influenced by someone who has already adopted the product • p is the coefficient of imitation which measures the likelihood of adoption assuming no influence from someone who has already adopted the product.

  28. Forecasting Adoption of a New Product • Developing the Forecasting Model • Ft , the forecast of the number of new adopters during time period t , is • Ft = (likelihood of a new adoption in time period t) • x (number of potential adopters remaining at • the end of time period t – 1)

  29. Forecasting Adoption of a New Product • Developing the Forecasting Model • Essentially, the likelihood of a new adoption is the likelihood of adoption due to innovation plus the likelihood of adoption due to imitation. • Let Ct - 1 denote the number of people who have adopted the product up to time t - 1. • Hence, Ct - 1 /m is the fraction of the number of people estimated to adopt the product by time t – 1. • The likelihood of adoption due to imitation is q(Ct - 1 /m). • The likelihood of adoption due to innovation and imitation is p + q(Ct - 1 /m).

  30. Forecasting Adoption of a New Product • Developing the Forecasting Model • The number of potential adopters remaining at the end of time period t – 1 is m-Ct - 1 . • Hence, the complete forecast model is given by • Ft = (p + q(Ct - 1 /m)) (m-Ct - 1)

  31. Forecasting Adoption of a New Product • Nonlinear Optimization Problem Formulation • Ft = (p + q(Ct - 1 /m)) (m-Ct - 1), t = 1, …., N • Et = Ft -St , t = 1, …., N • where N = number of time periods of data available • Et = forecast error for time period t • St = actual number of adopters (or a multiple of • that number such as sales) in time period t

  32. Example: Forecasting New-Product Adoption • Maid For You Maid For You is a residential cleaning service firm that has been quite successful developing a client base in the Chicago area. The firm plans to expand to other major metropolitan areas during the next few years. Maid For You would like to use its Chicago subscription data (on the next slide) to develop a model for forecasting service subscriptions in regions where it might expand. The first step is to estimate values for p (coefficient of innovation) and q (coefficient of imitation).

  33. Example: Forecasting New-Product Adoption • Subscribers and Cumulative Subscribers (1000s) Month Subscribers St Cum. Subscribers Ct 1 0.53 0.53 2 2.94 3.47 3 3.60 7.07 4 4.85 11.92 5 3.44 15.36 6 2.76 18.12 7 1.82 19.94 8 0.93 20.87 9 0.61 21.48

  34. Forecasting (General Form)

  35. Example: Forecasting New-Product Adoption • Define the Objective Function Minimize the sum of the squared forecast errors:

  36. Example: Forecasting New-Product Adoption • Define the Constraints Define the forecast for each time period: 1)F1 = pm 2)F2 = (p + q( 0.53/m)) (m – 0.53) 3)F3 = (p + q( 3.47/m)) (m – 3.47) 4)F4 = (p + q( 7.07/m)) (m – 7.07) 5)F5 = (p + q(11.92/m)) (m – 11.92) 6) F6 = (p + q(15.36/m)) (m – 15.36) 7)F7 = (p + q(18.12/m)) (m – 18.12) 8)F8 = (p + q(19.94/m)) (m – 19.94) 9)F9 = (p + q(20.87/m)) (m – 20.87)

  37. Example: Forecasting New-Product Adoption • Define the Constraints (continued) Define the forecast error for each time period: 10) E1 = F1 – 0.53 11) E2 = F2 – 2.94 12)E3 = F3 – 3.60 13)E4 = F4 – 4.85 14)E5 = F5 – 3.44 15) E6 = F6 – 2.76 16)E7 = F7 – 1.82 17)E8 = F8 – 0.93 18)E9 = F9 – 0.61

  38. Example: Forecasting New-Product Adoption • Optimal Forecast Parameter Values Parameter Value p 0.08 q 0.62 m 21.26 The value of the imitation parameter q = .62 is substantially larger than the value of the innovation parameter p = .08. Subscriptions gain momentum over time due mainly to very favorable word-of-mouth.

  39. Example: Forecasting New-Product Adoption • Optimal Solution MonthForecastSubscribersError 1 1.77 0.53 1.24 2 2.05 2.94 -0.89 3 3.29 3.60 -0.31 4 4.12 4.85 -0.73 5 4.03 3.44 0.59 6 3.14 2.76 0.38 7 1.93 1.82 0.11 8 0.88 0.93 -0.05 9 0.27 0.61 -0.34

  40. Example: Forecasting New-Product Adoption • Subscribers versus Forecasts Subscribers 5 4 3 2 1 Forecast Subscribers (1000s) Month 1 2 3 4 5 6 7 8 9

  41. End of Chapter 8

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