Chapter 10 Optimization Designs

1. Chapter 10Optimization Designs

2. C S R O R Optimization Designs Focus: A Few Continuous Factors Output: Best Settings Reference: Box, Hunter & Hunter Chapter 15

3. Optimization Designs C S R O 2k with Center Points Central Composite Des. Box-Behnken Des. R

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

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. Response Surface

7. First Order Strategy

8. Second Order Strategy

9. Quadratic Models can only take certain forms Maximum Minimum Saddle Point Stationary Ridge

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

11. Example: Improving Yield of a Chemical Process Levels

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

13. Use a 22 Factorial Design With Center Points x1x2 1 – – 2 + – 3 – + 4 + + 5 0 0 6 0 0 7 0 0

14. Plot the Data 135 64.6 68.0 60.3 64.3 62.3 130 Temperature (C) 54.3 60.3 125 70 80 Time (min)

15. Scaling Equations

16. Results From First Factorial Design

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

20. Fit the First Order Model

21. Path of Steepest Ascent: Move 4.50 Units in x2 for Every 2.35 Units in x1 Equivalently, for every one unit in x1 we move 4.50/2.35=1.91 units in x2 x 2 4.50 1.91 x 1.0 2.35 1

22. The Path in the Original Factors 145 140 135 Temperature (C) 64.6 68.0 60.3 64.3 62.3 130 54.3 60.3 125 60 70 80 90 Time (min)

23. Scaling Equations

24. Points on the Path of Steepest Ascent

25. Exploring the Path of Steepest Ascent 155 58.2 150 145 86.8 Temperature (C) 140 135 73.3 64.6 68.0 60.3 64.3 62.3 130 54.3 60.3 125 110 60 70 80 90 100 Time (min)

26. A Second Factorial 155 150 91.2 77.4 86.8 89.7 145 Temperature (C) 140 78.8 84.5 135 64.6 68.0 60.3 64.3 62.3 130 54.3 60.3 125 110 60 70 80 90 100 Time (min)

27. Results of Second Factorial Design

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. New Scaling Equations

30. A Central Composite Design 155 79.5 150 91.2 77.4 87.0 86.0 145 83.3 81.2 Temperature (C) 140 78.8 84.5 81.2 135 64.6 68.0 60.3 64.3 62.3 130 54.3 60.3 125 110 60 70 80 90 100 Time (min)

31. The Central Composite Design and Results

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

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

34. The Fitted Surface in the Region of Interest 155 79.5 150 91.2 77.4 87.0 86.0 145 83.3 81.2 Temperature (C) 140 78.8 84.5 81.2 135 110 70 80 90 100 Time (min)

35. Quadratic Models can only take certain forms Maximum Minimum Saddle Point Stationary Ridge

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

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

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

42. Tools Goals Canonical Analysis • A-Form if Stationary Point is outside Experimental Region • B-Form if Stationary Point is inside Experimental Region Understand the shape of the surface 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. Translate the reduced model back and optimum formula back to original coordinates