1. 10.1 Chapter 10 Optimization Designs

2. 10.2 Optimization Designs CS RO R Focus: A Few Continuous Factors Output: Best Settings Reference: Box, Hunter & Hunter Chapter 15

3. 10.3 Optimization Designs CS RO R 2 k with Center Points Central Composite Des. Box-Behnken Des.

4. 10.4 Response Surface Methodology A Strategy of Experimental Design for finding optimum setting for factors. (Box and Wilson, 1951)

5. 10.5 Response Surface Methodology When you are a long way from the top of the mountain, a slope may be a good approximation You can probably use first–order designs that fit a linear approximation When you are close to an optimum you need quadratic models and second–order designs to model curvature

6. 10.6 Response Surface

7. 10.7 First Order Strategy

8. 10.8 Second Order Strategy

9. 10.9 Quadratic Models can only take certain forms MaximumMinimumSaddle PointStationary Ridge

10. 10.10 Sequential Assembly of Experimental Designs as Needed Fractional Factorial Linear model Full Factorial w/Center points Main effects Interactions Curvature check Central Composite Design Full quadratic model

11. 10.11 Example: Improving Yield of a Chemical Process Levels

12. 10.12 Strategy Fit a first order model Do a curvature check to determine next step

13. 10.13 Use a 2 2 Factorial Design With Center Points x1x21––2+–3–+4++500600700 x1x21––2+–3–+4++500600700

14. 10.14 Plot the Data 7080 125 135 130 Temperature (C) Time (min) 64.668.0 60.3 54.3 60.3 64.3 62.3

15. 10.15 Scaling Equations

16. 10.16 Results From First Factorial Design

17. 10.17 Curvature Check by Interactions Are there any large two factor interactions? If not, then there is probably not significant curvature. If there are many large interaction effects then we should not follow a path of steepest ascent because there is curvature. Here, time =4.7, temp =9.0, and time*temp =-1.3, so the main effects are larger and thus there is little curvature.

18. 10.18 Curvature Check with Center Points Compare the average of the factorial points,, with the average of the center points,. If they are close then there is probably no curvature? Statistical Test: If zero is in the confidence interval then there is no evidence of curvature.

19. 10.19 Curvature Check with Center Points No Evidence of Curvature!!

20. 10.20 Fit the First Order Model

21. 10.21 Path of Steepest Ascent: Move 4.50 Units in x 2 for Every 2.35 Units in x 1 Equivalently, for every one unit in x 1 we move 4.50/2.35=1.91 units in x 2 2.35 4.50 x 1 1.91 1.0 x 2

22. 10.22 The Path in the Original Factors 60708090 125 145 140 135 130 Temperature (C) Time (min) 64.668.0 60.3 54.3 60.3 64.3 62.3

23. 10.23 Scaling Equations

24. 10.24 Points on the Path of Steepest Ascent

25. 10.25 60708090100 110 125 145 140 135 130 150 155 Temperature (C) Time (min) 58.2 86.8 64.668.0 60.3 54.3 60.3 64.3 62.3 73.3 Exploring the Path of Steepest Ascent

26. 10.26 60708090100 110 125 145 140 135 130 150 155 Temperature (C) Time (min) 86.8 89.7 64.668.0 60.3 54.3 60.3 64.3 62.3 78.884.5 77.491.2 A Second Factorial

27. 10.27 Results of Second Factorial Design

28. 10.28 Curvature Check by Interactions Here, time =-4.05, temp =2.65, and time*temp =-9.75, so the interaction term is the largest, so there appears to be curvature.

29. 10.29 New Scaling Equations

30. 10.30 60708090100 110 125 145 140 135 130 150 155 Temperature (C) Time (min) 87.0 86.0 64.668.0 60.3 54.3 60.3 64.3 62.3 78.884.5 77.491.2 A Central Composite Design 83.381.2 79.5

31. 10.31 The Central Composite Design and Results

32. 10.32 Must Use Regression to find the Predictive Equation The Quadratic Model in Coded units

33. 10.33 Must Use Regression to find the Predictive Equation The Quadratic Model in Original units

34. 10.34 708090100 110 145 140 135 150 155 Temperature (C) Time (min) 87.0 86.0 78.884.5 77.491.2 The Fitted Surface in the Region of Interest 83.381.2 79.5

35. 10.35 Quadratic Models can only take certain forms MaximumMinimumSaddle PointStationary Ridge

36. 10.36 Canonical Analysis Enables us to analyze systems of maxima and minima in many dimensions and, in particular to identify complicated ridge systems, where direct geometric representation is not possible.

37. 10.37 Two Dimensional Example x 2      x 1  is a constant

38. 10.38 Quadratic Response Surfaces Any two quadratic surfaces with the same eigenvalues ( ’s) are just shifted and rotated versions of each other. The version which is located at the origin and oriented along the axes is easy to interpret without plots.

39. 10.39 The Importance of the Eigenvalues The shape of the surface is determined by the signs and magnitudes of the eigenvalues. Type of Surface Minimum Saddle Point Maximum Ridge Eigenvalues All eigenvalues positive Some eigenvalues positive and some negative All eigenvalues negative At least one eigenvalue zero

40. 10.40 Stationary Ridges in a Response Surface The existence of stationary ridges can often be exploited to maintain high quality while reducing cost or complexity.

41. 10.41 Response Surface Methodology Eliminate Inactive FactorsFractional Factorials (Res III) Goals Tools Find Path of Steepest Ascent Fractional Factorials (Higher Resolution from projection or additional runs) Follow PathSingle Experiments until Improvement stops Find Curvature Center Points (Curvature Check) Fractional Factorials (Res > III) (Interactions > Main Effects) Model Curvature Central Composite Designs Fit Full Quadratic Model Continued on Next Page Brainstorming Select Factors and Levels and Responses

42. 10.42 Understand the shape of the surface Goals Tools Canonical Analysis A-Form if Stationary Point is outside Experimental Region B-Form if Stationary Point is inside Experimental Region Find out what the optimum looks like Point, Line, Plane, etc. Reduce the Canonical Form with the DLR Method If there is a rising ridge, then follow it. Translate the reduced model back and optimum formula back to original coordinates If there is a stationary ridge, then find the cheapest place that is optimal.