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Lecture 18

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

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Lecture 18

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  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 (x1) and reaction temperature (x2) • Current operating conditions have reaction time at 35 minutes and temperature at 155 oF, 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 (z1=30 or 40) • temperature = 155 oF (z2=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 nctrials 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 ( x1,x2 )=(0,0)

  9. Example • Center for new first-order design is ( z1,z2 )=(85,175) • Origin is at ( x1,x2 )=(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 nf, nc, 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 ( z1,z2 )=(85,175) • Origin is at ( x1,x2 )=(0,0)

  17. Example

  18. Example • Finding the maximum

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