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Nonlinear Optimization Models: Applications and Examples

This presentation explores various applications of nonlinear optimization models, including a production application, constructing an index fund, Markowitz portfolio model, blending, and forecasting adoption of a new product introduction. The slides also provide insights into the behavior of nonlinear business processes and the impact of nonlinear factors on pricing and costs.

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Nonlinear Optimization Models: Applications and Examples

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

  2. Chapter 8: Nonlinear Optimization Models • A Production Application • Constructing an Index Fund • Markowitz Portfolio Model • Blending: The Pooling Problem • Forecasting Adoption of a New Product

  3. Introduction • Many business processes behave in a nonlinear manner. • The price of a bond is a nonlinear function of interest rates. • The price of a stock option is a nonlinear function of the price of the underlying stock. • The marginal cost of production often decreases with the quantity produced. • The quantity demanded for a product is often a nonlinear function of the price.

  4. Introduction • 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 x3, 1/x, 5x1x2, and log x. • The nonlinear optimization problems presented on the upcoming slides can be solved using computer software such as LINGO and Excel Solver.

  5. Example: Production Application • Armstrong Bike Co. Armstrong Bike Co. produces two new lightweight bicycle frames, the Flyer and the Razor, that are made from special aluminum and steel alloys. The cost to produce a Flyer frame is $100, and the cost to produce a Razor frame is $120. We can not assume that Armstrong will sell all the frames it can produce. As the selling price of each frame model – Flyer and Razor - increases, the quantity demanded for each model goes down.

  6. Example: Production Application • Assume that the demand for Flyer frames F and the demand for Razor frames R are given by: F = 750 – 5PF R = 400 – 2PR where PF = the price of a Flyer frame PR = the price of a Razor frame. • The profit contributions (revenue – cost) are: PF F - 100F for Flyer frames PR R - 120R for Razor frames

  7. Example: Production Application • Profit Contribution as a Function of Demand • Solving F = 750 - 5PF for PF we get: PF = 150 - 1/5 F Substituting 150 - 1/5 F for PF in PF F - 100F we get: PF F - 100F = F(150 - 1/5 F) - 100F = 50F - 1/5 F 2 • Solving R = 400 - 2PR for PR we get: PR = 200 - 1/2 R Substituting 200 - 1/2 R for PR in PR R - 120R we get: PR R - 120R = R(200 - 1/2 R) - 120R = 80R - 1/2 R2

  8. Example: Production Application • Total Profit Contribution Total Profit Contribution = 50F – 1/5 F2 + 80R – 1/2 R2 This function is an example of a quadratic function because the nonlinear terms have a power of 2.

  9. Example: Production Application A supplier can deliver a maximum of 500 pounds of the aluminum alloy and 420 pounds of the steel alloy weekly. The number of pounds of each alloy needed per frame is summarized below. Aluminum AlloySteel Alloy Flyer 2 3 Razor 4 2 How many Flyer and Razor frames should Armstrong produce each week?

  10. Example: Production Application • Problem Formulation Max 50F – 1/5 F2 + 80R – 1/2 R2(Total Weekly Profit) s.t. 2F + 4R< 500 (Aluminum Available) 3F + 2R< 420 (Steel Available) F, R> 0 (Non-negativity)

  11. Example: Production Application • Total Profit Contribution First, we will solve the unconstrained version of this nonlinear program to find the values of F and R that maximize the above total profit contribution function (with the production constraints ignored).

  12. Example: Production Application • Optimal Solution for Unconstrained Problem x2 250 200 150 100 50 3F + 2R< 420 Unconstrained Optimum (125, 80) Profit = $6,325.00 Feasible Region 2F + 4R< 500 x1 50 100 150 200 250 300

  13. Example: Production Application • Total Profit Contribution Now we will solve the constrained version of this nonlinear program to find the values of F and R that maximize the total profit contribution function with the production constraints enforced.

  14. Example: Production Application • Objective Function Contour Lines x2 250 200 150 100 50 $6,200.00 Contour $6,325.00 $5,500.00 Contour $6,075.47 Contour x1 50 100 150 200 250 300

  15. Example: Production Application • Optimal Solution for Constrained Problem x2 Constrained Optimum (92.45, 71.32) Profit = $6,075.47 250 200 150 100 50 $6,075.47 Contour x1 50 100 150 200 250 300

  16. Example: Production Application • Optimal Solution • Produce 92.45 Flyer frames per week. • Produce 71.32 Razor frames per week. • Profit per week is $6,075.47. • Use 470.2 pounds of aluminum alloy per week (of the 500 pounds available per week). • Use the entire 420 pounds of steel alloy available per week.

  17. Local and Global Optima • A feasible solution is a local optimum if there are no other feasible solutions with a better objective function value in the immediate neighborhood. • For a maximization problem the local optimum corresponds to a local maximum. • For a minimization problem the local optimum corresponds to a local minimum. • A feasible solution is a global optimum if there are no other feasible points with a better objective function value in the feasible region. • Obviously, a global optimum is also a local optimum.

  18. Multiple Local Optima • Nonlinear optimization problems can have multiple local optimal solutions, in which case we want to find the best local optimum. • Nonlinear problems with multiple local optima are difficult to solve and pose a serious challenge for optimization software. • In these cases, the software can get “stuck” and terminate at a local optimum. • There can be a severe penalty for finding a local optimum that is not a global optimum. • Developing algorithms capable of finding the global optimum is currently a very active research area.

  19. Multiple Local Optima • Consider the function • The shape of this function is shown on the next slide. • The hills and valleys in the graph show that this function has several local maximums and local minimums. • There are two local minimums, one of which is the the global minimum. • There are three local maximums, one of which is the global maximum.

  20. Multiple Local Optima

  21. Single Local Optimum • Consider the function f(X, Y) = X 2 - Y 2. • The shape of this function is shown on the next slide. • A function that is bowl-shaped down is called a concave function. • The maximum value for this particular function is 0 and the point (0, 0) gives the optimal value of 0. • Functions such as this one have a single local maximum that is also a global maximum. • This type of nonlinear problem is relatively easy to maximize.

  22. Single Local Optimum Concave Function f(X, Y) = X 2 - Y 2

  23. Single Local Optimum • Consider the function f(X, Y) = X 2 + Y 2. • The shape of this function is shown on the next slide. • A function that is bowl-shaped up is called a convex function. • The minimum value for this particular function is 0 and the point (0, 0) gives the optimal value of 0. • Functions such as this one have a single local minimum that is also a global minimum. • This type of nonlinear problem is relatively easy to minimize.

  24. 40 Z 20 4 4 2 2 0 0 -2 -2 -4 -4 X Y Single Local Optimum Convex Function f(X, Y) = X 2 + Y 2

  25. 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.

  26. Constructing an Index Fund • Index funds are a very popular investment vehicle in the mutual fund industry. • Vanguard 500 Index Fund is the largest mutual fund in the U.S. with over $70 billion in net assets in 2005. • An index fund is an example of passive asset management. • The key idea behind an index fund is to construct a portfolio of stocks, mutual funds, or other securities that closely matches the performance of a broad market index such as the S&P 500. • Behind the popularity of index funds is research that basically says “you can’t beat the market.”

  27. Example: Constructing an Index Fund • Lymann Brothers Investments Lymann Brothers has a substantial number of clients who wish to own a mutual fund portfolio that closely matches the performance of the S&P 500 stock index. A manager at Lymann Brothers has selected five mutual funds (shown on the next slide) that will be considered for inclusion in the portfolio. The manager must decide what percentage of the portfolio should be invested in each mutual fund.

  28. Example: Constructing an Index Fund • Mutual Fund Performance in 4 Selected Years Annual Returns (Planning Scenarios) Mutual Fund Year 1 Year 2 Year 3 Year 4 International Stock 25.64 27.62 5.80 -3.13 Large-Cap Blend 15.31 18.77 11.06 4.75 Mid-Cap Blend 18.74 18.43 6.28 -1.04 Small-Cap Blend 14.19 12.37 -1.92 7.32 Intermediate Bond 7.88 9.45 10.56 3.31 S&P 500 13.00 12.00 7.00 2.00

  29. Example: Constructing an Index Fund • Define the 9 Decision Variables IS = proportion of portfolio invested in international stock LC = proportion of portfolio invested in large-cap blend MC = proportion of portfolio invested in mid-cap blend SC = proportion of portfolio invested in small-cap blend IB = proportion of portfolio invested in intermediate bond R1 = portfolio return for scenario 1 (year 1) R2 = portfolio return for scenario 2 (year 2) R3 = portfolio return for scenario 3 (year 3) R4 = portfolio return for scenario 4 (year 4)

  30. Example: Constructing an Index Fund • Define the Objective Function Min (R1 – 13)2 + (R2 – 12)2 + (R3 – 7)2 + (R4 – 2)2 • Define the 6 Constraints (including non-negativity) 25.64IS + 15.31LC + 18.74MC + 14.19SC + 7.88IB = R1 27.62IS + 18.77LC + 18.43MC + 12.37SC + 9.45IB = R2 5.80IS + 11.06LC + 6.28MC - 1.92SC + 10.56IB = R3 - 3.13IS + 4.75LC - 1.04MC + 7.32SC + 3.31IB = R4 IS + LC + MC + SC + IB = 1 IS, LC, MC, SC, IB> 0

  31. Example: Constructing an Index Fund • Optimal Solution for Lymann Brothers Example • R1 = 12.51 (12.51% portfolio return for scenario 1) • R2 = 12.90 (12.90% portfolio return for scenario 2) • R3 = 7.13 ( 7.13% portfolio return for scenario 3) • R4 = 2.51 ( 2.51% portfolio return for scenario 4) • IS = 0 ( 0.0% of portfolio in international stock) • LC = 0 ( 0.0% of portfolio in large-cap blend) • MC = .332 (33.2% of portfolio in mid-cap blend) • SC = .161 (16.1% of portfolio in small-cap blend) • IB = .507 (50.7% of portfolio in intermediate bond) 100.0% of portfolio

  32. Example: Constructing an Index Fund • Lymann Brothers Portfolio Return vs. S&P 500 Return Scenario Portfolio Return S&P 500 Return 1 12.51 13.00 2 12.90 12.00 3 7.13 7.00 4 2.51 2.00

  33. Markowitz Portfolio Model • There is a key tradeoff in most portfolio optimization models between risk and return. • The index fund model (Lymann Brothers example) presented earlier managed the tradeoff passively. • The Markowitz mean-variance portfolio model provides a very convenient way for an investor to actively trade-off risk versus return. • We will now demonstrate the Markowitz portfolio model by extending the Lymann Brothers example.

  34. Example: Markowitz Portfolio Model • In the Lymann Brothers example there were four scenarios and the return under each scenario was defined by the variables R1, R2, R3, and R4. • If ps is the probability of scenario s, and there are n scenarios, then the expected return for the portfolio R is • If we assume that the four scenarios in the Lymann Brothers model are equally likely, then • or

  35. Example: Markowitz Portfolio Model • The measure of risk most often associated with the Markowitz model is the variance of the portfolio. • For our example, the portfolio variance is • For our example, the four planning scenarios are equally likely. Thus, • or

  36. Example: Markowitz Portfolio Model • The portfolio variance is the average of the sum of the squares of the deviations from the mean value under each scenario. • The larger the variance value, the more widely dispersed the scenario returns are about the average return value. • If the portfolio variance were equal to zero, then every scenario return Ri would be equal.

  37. Example: Markowitz Portfolio Model • There are two basic ways to formulate the Markowitz model: • (1) Minimize the variance of the portfolio subject to constraints on the expected return, and • (2) Maximize the expected return of the portfolio subject to a constraint on risk. • We will now demonstrate the first (1) formulation, assuming that Lymann Brothers’ client requires the expected portfolio return to be at least 9 percent.

  38. Example: Markowitz Portfolio Model • Define the Objective Function Minimize the portfolio variance: • Define the Constraints Define the return for each scenario: 25.64IS + 15.31LC + 18.74MC + 14.19SC + 7.88IB = R1 27.62IS + 18.77LC + 18.43MC + 12.37SC + 9.45IB = R2 5.80IS + 11.06LC + 6.28MC - 1.92SC + 10.56IB = R3 - 3.13IS + 4.75LC - 1.04MC + 7.32SC + 3.31IB = R4

  39. Example: Markowitz Portfolio Model • Define the Constraints (continued) All the money must be invested in the portfolio: IS + LC + MC + SC + IB = 1 Define the expected return for the portfolio: The portfolio return must be at least 9 percent: Non-negativity: IS, LC, MC, SC, IB> 0

  40. Example: Markowitz Portfolio Model • Optimal Solution • R1 = 10.63 (10.63% portfolio return for scenario 1) • R2 = 12.20 (12.20% portfolio return for scenario 2) • R3 = 8.93 ( 8.93% portfolio return for scenario 3) • R4 = 4.24 ( 4.24% portfolio return for scenario 4) • Rbar = 9.00 ( 9.00% expected portfolio return) • IS = 0 ( 0.0% of portfolio in international stock) • LC = .251 (25.1% of portfolio in large-cap blend) • MC = 0 ( 0.0% of portfolio in mid-cap blend) • SC = .141 (14.1% of portfolio in small-cap blend) • IB = .608 (60.8% of portfolio in intermediate bond) 100.0% of portfolio

  41. 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.

  42. 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?

  43. 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.

  44. 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

  45. 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

  46. 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)

  47. Example: Blending - The Pooling Problem • Define the 11 Constraints Conservation equation: 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

  48. 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 (y1/ (y1 + y2))xpp< .25(xpp + x3p) (y2/ (y1 + y2))xpp< .45(xpp + x3p) x3p> .3(xpp + x3p)

  49. Example: Blending - The Pooling Problem • Computer Output Objective Function Value = 1045000.000 VariableValueReduced Cost xpr 8000.000 0.000 x3r 2000.000 0.000 xpp 6000.000 0.000 x3p 2571.429 0.000 y1 5000.000 0.000 y2 9000.000 0.000

  50. Example: Blending - The Pooling Problem • Computer Output (continued) ConstraintSlack/SurplusDual Value 1 0.000 1.412 2 1000.000 0.000 3 5428.571 0.000 4 0.000 -3.061 5 142.857 0.000 6 1142.857 0.000 7 0.000 0.229 8 0.000 -2.197 9 0.000 0.865 10 0.000 0.000 11 0.000 -0.123

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