1. Response surfaces

2. We have a dependent variable y, independent variables x 1, x 2,...,x p The general form of the model y = f(x 1, x 2,...,x p ) +  Surface Graph Contour Map

3. The linear model y =  0 +  1 x 1 +  2 x 2 +... +  p x p + e Surface Graph Contour Map

4. The quadratic response model Linear terms Contour Map Surface Graph Quadratic terms

5. The quadratic response model (3 variables) Linear terms Quadratic terms To fit this model we would be given the data on y, x 1, x 2, x 3. From that data we would compute: We then regress y on x 1, x 2, x 3, u 4, u 5, u 6, u 7, u 8 and u 9

6. Exploration of a response surface The method of steepest ascent

7. Situation We have a dependent variable y, independent variables x 1, x 2,...,x p The general form of the model y = f(x 1, x 2,...,x p ) +  We want to find the values of x 1, x 2,...,x p to maximize (or minmize) y. We will assume that the form of f(x 1, x 2,...,x p ) is unknown. If it was known (e.g. A quadratic response model), we could estimate the parameters and determine the optimum values of x 1, x 2,...,x p using calculus

8. The method of steepest ascent: 1.Choose a region in the domain of f(x 1, x 2,...,x p ) 2.Collect data in that region 3.Fit a linear model (plane) to that data. 4.Determine from that plane the direction of its steepest ascent. (direction (  1,  2,...,  p )) 5.Move off in the direction of steepest ascent collecting on y. 6.Continue moving in that direction as long as y is increasing and stop when y stops increasing. 7.Choose a region surrounding that point and return to step 2. 8.Continue until the plane fitted to the data is horizontal 9.Consider fitting a quadratic response model in this region and determining where it is optimal.

9. The method of steepest ascent: domain of f(x 1, x 2,...,x p ) Initial region direction of steepest ascent. 2 nd region Final region Optimal (x 1, x 2 )

10. Example In this example we are interested in how the life (y) of a lathe cutting tool depends on Lathe velocity (V) and Cutting depth (D). In particular we are interested in what settings of V and D will result in the maximum life (y) of the tool. The variables V and D have been recoded into x 1 and x 2 so that when V = 100 then x 1 = 0 and when V = 700, x 1 = 100. 100 to 700 are the feasible values of V. Also when D = 0.040 then x 2 = 0 and when D = 0.100, x 2 = 100. 0.040 to 0.100 are the feasible values of V. –

11. The domain for (x 1, x 2 ) 0 100 x2x2 x1x1 0

12. Initial Region (2 k design) x2x2 x1x1

13. Analysis Direction of steepest ascent: (  1,  2 ) = (1.114, -1.116)

14. Moving in the direction of steepest ascent Direction of steepest ascent: (  1,  2 ) = (1.114, -1.116) Optimum (x 1, x 2 ) = (41.72, 58.24)

15. 2 nd Region (2 k design)

16. Analysis Direction of steepest ascent: (  1,  2 ) = (-0.080, -0.227)

17. Moving in the direction of steepest ascent Direction of steepest ascent: (  1,  2 ) = (-0.080, -0.227)

18. To determine the precise optimum we will fit a quadratic response surface: The optimum then satisfies: which has solution:

19. The data

20. Location of the data points

21. Fitting a quadratic response surface: The optimum: