LECTURE 6 Response Surface Methodology II References: The basic paper: Box, G. E. P. and K. B. Wilson (1951). “On the experimental attainment of optimal conditions,” JRSS, Series B, 13 1–. Reviews: Hill, William J. and William G. Hunter (1966). “A review of rsponse surface methodology: a literature survey,” Technometrics, 571–591. Mead, R., and D. J. Pike (1975). “A review of response surface methodology from a biometric point of view,” Biometrics, 803–. Myers, Khori, and Carter (1989). “Response surface methodology 1966–88,” Technometrics, 66–88. There are several good books as well. 1. Steepest Ascent We know that to maximize the response, the movement of the design center must be in the direction of the directional derivatives of the response function, that is, in the direction of ∂φ ∂φ ∂φ = ,..., . ∂x ∂x1 ∂x p We then multiply by a constant A so that ∂φ x = A , ∂x where r A = . ∂φ 2 ∂x i 2 = 2 Thus xi r . For the first order model, ∂φ ˆ = βi , ∂xi = βˆ = / βˆ2 so xi A i and A r i . From this we see that the movement of xi up ˆ the path of steepest ascent is proportional to βi . Since this is the case it is easier c Steven Buyske and Richard Trout. 57 58 6. RESPONSE SURFACE METHODOLOGY II ˆ not to pick particular values of r butratherfixavalueofβi and make the other changes proportional to it. To see the procedure consider the following example: x1 x2 x3 x1 x2 x3 1 Base Level 0 0 0 225 4.25 91.5 (Center Point) 2 Unit Change 1 1 1 25 0.25 1.5 (divisor in codings) 3 βˆ from coded 7.5 -4.0 6.2 regression model 4 Uncoding of slope to 7.5 × 25 = 187.5 -1.0 9.3 original x scale −1.0 5 Change relative to one 1 -0.53 0.83 1 . =-0.0053 0.0496 ˆ 187 5 unit change in β1 = 1 6 Path of 1 -0.53 0.83 250 4.12 92.7 steepest ascent 2 -1.07 1.65 275 3.98 94.0 3 -1.6 2.48 300 3.85 95.2 Now discuss where the next design might be centered. Probably would choose center set ofpoints to center design. Where to center the next experiment? 1. Center the design on the path of steepest ascent. 2. Remember that points outside of ±1 are extrapolations from the model. 3. One possibility is to try a set of points along the path of steepest ascent, and then center the next experiment where the largest response was. 4. A more conservative approach would be to center the next experiment near the boundary of the first experiment. This procedure would be continued until is appears that a stationary point has been reached. One should look at lack of fit if at all possible as this might indicate aregionofhighcurvature. 2. Second Order Models Having reached the region the optimum we wish to explore this region more carefully, using a second order model. Orthogonality now becomes less important, while the prediction variance Var(yˆ) becomes more important. Again, there are two issues, one of design and the other of analysis. 2.1. Design. Since we are using second order models we must have at least three levels for each factor. For example, a 3k design would be possible. For example, if K = 2, 2. SECOND ORDER MODELS 59 x1 x2 -1 -1 0-1 +1 -1 -1 0 00 +1 0 -1 +1 0+1 +1 +1 We use a full second order model of the form ( ) = β + β + β + β + β ( 2 − ) + β ( 2 − ), E y 0 1x1 2x2 12x1x2 11 x1 c 22 x2 c = ¯2 = / K where c xi 2 3 for 3 designs. Discuss why we subtract off c. The design matrix is 2 − / 2 − / x0 x1 x2 x1x2 x1 2 3 x2 2 3 1 -1 -1 +1 1/3 1/3 1 0 -1 0 -2/3 1/3 1 +1 -1 -1 1/3 1/3 1 -1 0 0 1/3 -2/3 1 0 0 0 -2/3 -2/3 1 +1 0 0 1/3 -2/3 1 -1 +1 -1 1/3 1/3 10+10 -2/3 1/3 1+1+1+1 1/3 1/3 giving x x x x x x2 − cx2 − c 0 1 2 1 2 1 2 90 6 X X = 6 4 2 02 and Cov(β)ˆ = σ 2(X X)−1. We see that the estimates are independent of each other. The problem with 3k models is the large number of data points for relatively ˆ small k (e.g., k = 3givesn = 27, and k = 4givesn = 81). In addition,Var(βii) is relatively large; σ 2 Var(βˆ ) = for k = 2 ii 2 as opposed to σ 2 Var(βˆ ) = for k = 2. i 6 60 6. RESPONSE SURFACE METHODOLOGY II 2.2. Central Composite Designs. Another type of design, developed by Box and Wilson (1951), is called the Central Composite Design. These designs are first order designs (2k or a fraction of 2k, but almost always resolution V or better) augmented by center points and star, or axial, points. For example, for k = 2, 2k = 22 = 4 center point(s) = 1 star points = 4 giving, as the design matrix x1 x2 -1 -1 +1 -1 -1 +1 +1 +1 00 −α 0 +α 0 0 −α 0 +α Pictorially (assuming α>1) Draw picture Again using the same model as before we will let 2k + 2α2 ¯ c = = x2, n i where n equals the total number of points in the design. For the full second order model the design matrix will be, for k = 2, so c = (4 + 2α2)/9,andifweletα = 2, c = (4 + 8)/9 = 4/3, 2 − / 2 − / x0 x1 x2 x1x2 x1 4 3 x2 4 3 1 -1 -1 +1 -1/3 -1/3 1 +1 -1 -1 -1/3 -1/3 1 -1 +1 -1 -1/3 -1/3 1 +1 +1 +1 -1/3 -1/3 1 0 0 0 -4/3 -4/3 1 -2 0 0 8/3 -4/3 1 +2 0 0 8/3 -4/3 1 0 -2 0 -4/3 8/3 1 0 +2 0 -4/3 8/3 2. SECOND ORDER MODELS 61 giving x x x x x x2 x2 0 1 2 1 2 1 2 900 0 0 0 0120 0 0 0 X X = 0 0 12 0 0 0 . 000 4 0 0 000 0 20−12 000 0 −12 20 There are several points worth making 1. The linear and quadratic terms have greater precision than those of 3k de- signs (the interaction is slightly worse off). 2. The quadratic terms have a non-zero covariance. 3. For k = 2, the number of experimental runs is the same. In general, it is much less for the central composite. kCC3k 299 31527 42581 The question to discuss is that of selecting α. One of the criteria that can be used to select α is to make the estimates of the quadratic terms orthogonal, that is, make the X X matrix diagonal. k−p If we use a 2 design with 2k axial points and nc center points then 1 1 k−p 2 4 k−p 2 k−p 2 + 2k + nc − 2 2 2 α = , 4 will make X X diagonal. For example, if we let p = 0andnc = 1, k α 2 1.000 3 1.216 4 1.414 5 1.596 6 1.761 Now a logical question that might be asked is whether 3k or the orthogonal central composite design is better. An answer requires criteria to be developed to compare the designs. Later on in the semester we will have a discussion on optimality criteria. At that time we will develop procedures which will allow us to compare these designs. Another criteria that is used to select α is to make the design rotatable.A design is said to be rotatable when the variance of yˆ is a function only of the distance from the center of the design and not a function of the direction. Geometrically Draw picture 62 6. RESPONSE SURFACE METHODOLOGY II Points 1 and 2 are both distance ρ from center (0, 0) andsowouldhavethesame Var(yˆ). The concept of rotatable is not uniquely related to second order models or central composite designs. Myers, and Box and Hunter give a detailed description relating to rotatibility. Remember from regression that yˆ = x1×pβˆ p×1, where Var βˆ = σ 2(X X)−1. Thus Var(yˆ) = σ 2x1×p(X X)−1x, where x1×p is a particular row of the design matrix. For example, it turns out that, for first order designs, an orthogonal design is also a rotatable design. As an illustration consider a 22 factorial 1 −1 −1 1 +1 −1 X = 1 −1 +1 1 +1 +1 400 X X = 040 004 1/40 0 (X X)−1 = 01/40 001/4 If we take x to be row 1 of X, then 1/40 0 1 x(X X)−1x = 1 −1 −1 01/40 −1 001/4 −1 1 = 1/4 −1/4 −1/4 −1 = 3/4 −1 For row 2, 1 x(X X)−1x = 1/41/4 −1/4 1 = 3/4 −1 and so on for the remaining two rows. It must be emphasized that in general, for other than first order models, orthog- onal designs are not necessarily rotatable and vice versa.