Response surfaces

• We have a dependent variable y, independent variables x1, x2, ... ,xp

• The general form of the model

y = f(x1, x2, ... ,xp) +

Series1Series8

Series15Series22Series29

Series36

Series43

Series50

-10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 100.0 110.0 120.0 130.0 140.0 150.0

1471013161922252831343740434649

Surface Graph Contour Map

• The linear model

y = 0 + 1x1 + 2x2 +... + pxp + e

Surface Graph Contour Map

• The quadratic response model

Linear terms

Contour Map

225214

21322110 xxxxxxy

Surface Graph

Quadratic terms

00.250.5

0.751

1.251.5

1.752

2.252.5

2.753

3.253.5

The quadratic response model (3 variables)

Linear terms

3322110 xxxy

Quadratic terms

329318217236

225

214 xxxxxxxxx

To fit this model we would be given the data on y, x1, x2, x3. From that data we would compute:

329318217236

225

214 and ,,,, xxuxxuxxuxuxuxu

We then regress y on x1, x2, x3, u4, u5, u6 , u7, u8 and u9

Exploration of a response surfaceThe method of steepest ascent

Situation

• We have a dependent variable y, independent variables x1, x2, ... ,xp

The general form of the model

y = f(x1, x2, ... ,xp) + We want to find the values of x1, x2, ... ,xp to maximize (or minmize) y.

We will assume that the form of f(x1, x2, ... ,xp) is unknown. If it was known (e.g. A quadratic response model), we could estimate the parameters and determine the optimum values of x1, x2, ... ,xp using calculus

The method of steepest ascent:1. Choose a region in the domain of f(x1, x2, ... ,xp)

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 horizontal9. Consider fitting a quadratic response model in this region and

determining where it is optimal.

The method of steepest ascent:

domain of f(x1, x2, ... ,xp)

Initial region direction of

steepest ascent.

2nd region

Final region

Optimal

(x1, x2)

ExampleIn 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 x1 and x2 so that

•when V = 100 then x1 = 0 and when V = 700, x1 = 100.

• 100 to 700 are the feasible values of V.

Also •when D = 0.040 then x2 = 0 and when D = 0.100, x2 = 100.

• 0.040 to 0.100 are the feasible values of V.

–

6

040.010000 and

6

100 Thus 21

Dx

Vx

The domain for (x1, x2)

0

100

100

x2

x10

Initial Region (2k design)

0

20

40

60

80

100

0 20 40 60 80 100

x2

x1

x 1 x 2 y

16 74 22.3 16 74 14.9 16 74 22.4 16 74 20.9 16 74 33.9 24 74 30.9 24 74 48.4

24 74 33.9

24 74 43.0

24 74 29.6 16 86 19.7 16 86 10.6 16 86 16.0

16 86 16.6 16 86 11.4 24 86 19.9 24 86 13.7

24 86 20.0

24 86 26.5 24 86 11.9

Analysisx 1 x 2 y

16 74 22.3 16 74 14.9 16 74 22.4 16 74 20.9 16 74 33.9 24 74 30.9 24 74 48.4

24 74 33.9

24 74 43.0

24 74 29.6 16 86 19.7 16 86 10.6 16 86 16.0

16 86 16.6 16 86 11.4 24 86 19.9 24 86 13.7

24 86 20.0

24 86 26.5 24 86 11.9

intercept 90.3171 1.1142 -1.116

R 2 = 0.6197

Source S.S. d.f. M.S. F p-valueRegression 1293.4 2 646.7 13.849 0.0003

Error 793.84 17 46.696Total 2087.2 19

Direction of steepest ascent:(1, 2) = (1.114, -1.116)

Moving in the direction of steepest ascentDirection of steepest ascent:

(1, 2) = (1.114, -1.116)x 1 x 2 y0 20 80 24.221 21.67 78.33 28.432 23.34 76.65 33.903 25.01 74.98 28.134 26.68 73.31 46.775 28.35 71.63 56.396 30.02 69.96 54.987 31.69 68.28 63.788 33.37 66.61 82.679 35.04 64.94 83.19

10 36.71 63.26 75.8711 38.38 61.59 98.6812 40.05 59.92 108.3313 41.72 58.24 90.1414 43.39 56.57 108.3315 45.06 54.89 93.0216 46.73 53.22 96.4217 48.40 51.55 94.5618 50.07 49.87 78.2819 51.74 48.20 73.6020 53.41 46.53 61.76

0

20

40

60

80

100

0 20 40 60 80 100

0.00

20.00

40.00

60.00

80.00

100.00

120.00

0 5 10 15 20 25

Series3

Optimum(x1, x2) = (41.72, 58.24)

2nd Region (2k design)

0

20

40

60

80

100

0 20 40 60 80 100

x 1 x 2 y

37.718 52.241 97.6 37.718 52.241 96.0 37.718 52.241 100.9 37.718 52.241 97.5 37.718 52.241 99.7 45.718 52.241 93.5 45.718 52.241 104.8

45.718 52.241 93.4

45.718 52.241 94.7

45.718 52.241 95.5 37.718 64.241 94.9 37.718 64.241 90.7 37.718 64.241 97.6

37.718 64.241 92.1 37.718 64.241 96.2 45.718 64.241 92.5 45.718 64.241 92.3

45.718 64.241 95.4

45.718 64.241 98.6 45.718 64.241 96.1

Analysis

Direction of steepest ascent:(1, 2) = (-0.080, -0.227)

x 1 x 2 y

37.718 52.241 97.6 37.718 52.241 96.0 37.718 52.241 100.9 37.718 52.241 97.5 37.718 52.241 99.7 45.718 52.241 93.5 45.718 52.241 104.8

45.718 52.241 93.4

45.718 52.241 94.7

45.718 52.241 95.5 37.718 64.241 94.9 37.718 64.241 90.7 37.718 64.241 97.6

37.718 64.241 92.1 37.718 64.241 96.2 45.718 64.241 92.5 45.718 64.241 92.3

45.718 64.241 95.4

45.718 64.241 98.6 45.718 64.241 96.1

intercept 112.5391 -0.0802 -0.227

R 2 = 0.1815

Source S.S. d.f. M.S. F p-valueRegression 39.04 2 19.52 1.885 0.1822

Error 176.04 17 10.355Total 215.08 19

Moving in the direction of steepest ascentDirection of steepest ascent:

(1, 2) = (-0.080, -0.227)

0

20

40

60

80

100

0 20 40 60 80 100

90

100

110

0 5 10 15 20 25

x 1 x 2 y

0 41.72 58.24 95.231 41.64 58.01 109.732 41.56 57.79 96.983 41.48 57.56 109.974 41.40 57.33 105.295 41.32 57.11 107.406 41.24 56.88 99.157 41.16 56.65 101.278 41.08 56.43 102.989 41.00 56.20 95.57

10 40.92 55.97 92.2511 40.84 55.75 106.6912 40.76 55.52 108.3313 40.68 55.29 104.6414 40.60 55.07 96.1415 40.52 54.84 104.1116 40.44 54.61 100.1917 40.36 54.39 94.3918 40.28 54.16 104.6119 40.20 53.93 97.2420 40.12 53.71 99.02

To determine the precise optimum we will fit a quadratic response surface:

225214

21322110 xxxxxxy

The optimum then satisfies:

02 241311

xxx

y

02 251422

xxx

y

2453

231422

453

15241 4

2,

4

2

xx

which has solution:

The data35 40 45 50 55 60 65

25 55.3 49.0 46.7 54.9 58.6 56.0 54.8 25 47.7 46.5 51.3 53.4 64.4 58.1 57.7 30 76.3 73.4 82.8 83.3 81.1 77.7 67.4 30 70.8 73.1 71.8 74.9 80.9 68.7 73.8

35 76.2 80.3 79.7 97.6 90.2 90.4 83.9 35 72.7 79.9 96.2 104.9 101.3 94.8 82.9

40 71.3 86.8 92.6 104.0 97.2 107.8 93.8

40 72.3 92.3 97.3 110.4 103.6 102.3 91.5

45 64.9 84.2 87.6 102.6 99.3 94.7 103.7

45 64.6 84.1 87.8 86.9 106.3 92.8 85.7

50 57.2 60.0 71.6 90.8 88.8 82.4 81.4

50 48.1 74.3 84.4 90.5 87.9 93.8 79.3

55 44.3 40.5 46.8 57.4 57.3 86.1 73.5 55 34.2 46.8 48.5 56.6 67.3 72.6 68.2 60 24.2 30.8 38.2 40.8 41.9 55.4 52.6 60 18.7 28.1 43.8 52.7 45.2 51.3 51.5

x 2

x 1

Location of the data points

0

20

40

60

80

100

0 20 40 60 80 100

Fitting a quadratic response surface:

225214

21322110 xxxxxxy

intercept -244.0801 10.39662 4.53003 -0.14704 -0.05185 0.0316

Source S.S. d.f. M.S. F p-valueRegression 44846.92269 5 8969.38454 208.429563 0.0000

Error 4561.515879 106 43.0331687Total 49408.43857 111

2453

231422

453

15241 4

2,

4

2

xx

The optimum:

x 1 = 41.44, x 2 = 56.41

6

040.010000 and

6

100 Now 21

Dx

Vx

04.010000

6 and 1006 Hence 2

1 x

DxV

:are and of valuesoptimal theHence DV

0738.04.0

10000

56.116 D

and 6.34810044.416 V