1. Lecture 18 Today: More Chapter 9 Next day: Finish Chapter 9

2. Example Chemical Engineer is interested in maximizing yield of a process 2 variables influence process yield: –reaction time (x 1 ) and reaction temperature (x 2 ) Current operating conditions have reaction time at 35 minutes and temperature at 155 o F, which give a yield of about 40% Best operating conditions may be far from current conditions

3. Example Engineer decide that reaction time should be investigated in the area of the operating conditions –time = 35 minutes (z 1 =30 or 40) –temperature = 155 o F (z 2 =150 or 160)

4. Example Fit regression line to data, include a quadratic term for curvature check

5. Method of Steepest Ascent (climbing the hill) Linear effects for first order model estimated by least squares: Take partial derivative with respect to each variable: Direction of steepest ascent:

6. Method of Steepest Ascent Several experiment trials are taken along the line from the center point of the design, in the direction of the steepest ascent until no further increase is observed The location where the maximum has occurred is the center point of the next first order design Design should have n c trials at the center point If curvature is detected, augment the design with additional trials so that the second order model can be estimated

7. Method of Steepest Ascent Trials are performed along the direction of steepest ascent…which trials? The steps are proportional to the estimated regression coefficients Equivalently, Actual step-size is determined by the experimenter based on expert knowledge and practical considerations

8. Example Origin is at ( x 1,x 2 )=(0,0)

9. Example Center for new first-order design is ( z 1,z 2 )=(85,175) Origin is at ( x 1,x 2 )=(0,0)

10. Example Fit first order model… Curvature indicates we are near the optimum Want to fit second order model

11. Analysis of Second Order Model Wish to find optimum of response surface Point will occur at: This point is called the stationary point Point could be a –maximum –minimum –saddle point

12. Analysis of Second Order Model Model Model in matrix form Solution for stationary point

13. Analysis of Second Order Model Characterizing the stationary point

14. Central Composite Design Cannot fit a second order model to the data we have thus far Central composite design is an efficient design for fitting a second order model Consists of: –first order design (factorial or fractional factorial) (coded -1 or +1) –center points (coded as 0) –axial points (coded ) Have n f, n c, and 2k trials for each type of design respectively

15. Central Composite Design The axial points chosen so that the design is rotatable Good choice of axial points Choice of number of center points

16. Example Center for new first-order design is ( z 1,z 2 )=(85,175) Origin is at ( x 1,x 2 )=(0,0)

17. Example

18. Finding the maximum