J. Response Surface Methodology

Transcription

1 J. Response Surface Methodology Response Surface Methodology. Chemical Example (Box, Hunter & Hunter) Objective: Find settings of R, Reaction Time and T, Temperature that produced maximum yield subject to () color index > 8, (2) viscosity of the product be at least 70 units and no more than units. Current settings: R = minutes and T = 30 C. Known standard deviation: from the past experience, σ =.5 and it is unlikely that this value changes in the experiment to be performed. Focus on maximizing the yield for now:. First stage of experimentation: Background: the experimenter believed, but was not certain, that he was some distance away from the maximum (some way down the smooth hillside that represents the true response surface). predominant local characteristics of the surface were its gradients local surface could be roughly represented by the planar model having slope β in the x direction and slope β 2 in the x 2 direction. y = β 0 + β x + β 2 x 2 + ε first-order model (planar model), where x and x 2 are coded variables for time and temperature, respectively. Design chosen: a first-order design 2 2 with 3 replicates of centerpoint. actual coded yield run run factors variables (grams) number order R T x x 2 y range of R: 70 to minutes range of T: 27.5 to 32.5 C coded variables: R minutes x = 5 minutes x 2 = T 30 C 2.5 C J.

2 Analysis: Least squares fit: Fitted equation: ˆβ = 2 (main effect R) = 4.70 = ˆβ 2 = 2 (main effect T) = 9.00 = std.err(effect) = σ = 2.5 =.5 nf 4 std.err(coefficient) = 2 std.err(effect) =.5 = 0. 2 ˆβ 0 = y = 62.0 std.err( ˆβ 0 ) = σ 7 =.5 7 = 0.57 ŷ = x x 2 Diagnostic checks for the adequacy of the planar model interaction check insignificant interaction expected. Note: interaction model is y = β 0 + β x + β 2 x 2 + β 2 x x 2 + ε. ˆβ 2 = 2 RT interaction effect = (.30) = which is less than 2 std.err(effect) in magnitude and consequently the interaction is insignificant. curvature check insignificant curvature expected. Note: second-order model is y = β 0 +β x +β 2 x 2 +β x 2 +β 2 x x 2 +β 22 x 2 2+ε. The column x 2 = x 2 2 provides a contrast for estimating β + β 22. β + β 22 = y f y c = 0.50 Var( β + β 22 ) = Var(y f ) + Var(y c ) = σ 2 ( + ) n f n c std.err(curvature) = σ + =.5. n f n c The curvature effect is insignificant since its magnitude is less than twice the standard error of the curvature. lack-of-fit check no lack-of-fit expected. Pure error SS amd MS: SS E = ( ) ( ) 2 /3 = 8.0 which is computed from the center points. MS E = s 2 = SS E /2 = 4.0. SS R = n f /4 R 2 = 4/ = J.2

3 SS T = n f /4 T 2 = 4/ = Residual SS: SS r = SS total SS R SS T = 0.9. Lack-of-fit SS: SS l = SS r SS E = 2.9. Source SS DF MS F P residual lack-of-fit pure error Estimation of experimental error: So, there is no evidence of lack of fit. s 2 = MS E = 4.0 and hence s = 2.0 agreeing fairly well with σ =.5. Fitted equation, contour diagram, and the path of steepest ascent. The contour line equation obtained from the fitted planar model is ( ŷ ) x x 2 = 0. Using the contour equation, calculate straight line equations at several values of ŷ: 56, 60, 64, 68, say. x 0 temperature ( C) path of steepest ascent y^ = 64 y^ = 60 y^ = 68 0 x 2 Steepest ascent direction: starting at the center of the first design, the path is followed by simultaneously moving ˆβ 2 = 4.50 units in x 2 for every ˆβ = 2.35 units moved in x ; or equivalently, 4.50/2.35=.9 units in x 2 direction for every unit in x direction. y^ = time (min.) 2. Probing for better settings along the path of steepest ascent: x x 2 R T run yield center ,6,7 62.3(average) path of steepest ascent Second stage experimentation: second design will be centered at R = 90 min. and T = 45 C. background: moving up the surface experimental error increases need to broden J.3

4 the factor space of the design by a factor of, say, two to increase the absolute magnitude of the effects in relation to the error. actual coded yield coded variables: run run factors variables (grams) R 90 minutes number order R T x x 2 y x = 0 minutes x 2 = T 45 C 5 C Analysis of the second first-order design (runs to 6): Start with a first order design for planar model (runs to 6), the results will most likely be inadequate; if the planar model fails, then augment to second order design to fit the second order model (runs 7 to 22). β 2 = 4.88 ± 0. and β + β 22 = 5.28 ±.30 are both significant. Consequently the planar model is inadequate. s 2 2 = 2 [ ( ) 2 /2] = 4.2 s 2 = 2. so that the pooled estimate of σ 2 from the two first-order designs is given by s 2 = 2 s2 + s = 4.07 s = 2.02 again agreeing reasonably well with σ =.5. The second first-order design is inadequate for fitting a second-order model: y = β 0 + β x + β 2 x 2 + β x 2 + β 22 x β 2 x x 2 + ε which prompts us to augment the second first-order design with four axial points and 2 center points. The combined design is called a central composite design (or a secondorder design) as given above. Analysis of the second-order design: The regression line is given by ŷ = x x x x x x 2. The response surface can be visualized by the following figures: J.4

5 x temperature ( C) x 2 yhat time (min.) x x2 Diagnostic check for model adequacy exactly quadratic checks in x and x 2 directions no descrepancy from exact quadratic expected x line 2 line Exact quadratic in x direction: β β 22 (coefficients of x 3 and of x x 2 2, respectively, in a third-order model) measures descrepancy from exact quadratic which is given by β β 22 = slope line slope line 2 of which a value of zero is expected if we have exact quadratic in x direction (i.e., a parabola and the two lines are parallel). β β 22 = y 8 y 7 2 y 2 + y 4 y y which has a value of.28 and Var( β β 22 ) = (σ 2 + σ 2 )/(2 2) σ2 = 2 σ2. Therefore, std.error( β β 22 )=.06 which indicates that the exact quadratic in x direction is justified. The exact quadratic in the x 2 direction can be obtained in a similar fashion:.93 ±.06, a borderline situation. J.5

6 Estimating experimental error: s 2 3 = 2 [ ( ) 2 /2] = 0.5 and the pooled estimate of σ 2 is [2 2++ s2 + s s 2 3] = 3.8 so that s =.78 which is close to σ =.5. Block effect: estimated by y y 2 =.70, the difference in the mean responses in the two blocks. The standard error of the block effect is then σ + = There 6 6 is some evidence that somewhat lower yields were obtained in the six augmented runs. Nature of the fitted surface: the average variance of fitted values is Var(ŷ) = pσ2 n = =.25, where p is the number of parameters in the model and n is the number of observations. Hence, the average standard error of fitted values is.25 =.06. The range of predicted values is about from 77 to 90 which is approximately 2 times the average standard error of the fitted values ( ). This implies that there.06 is no substantial lack of fit and it is worthwhile to interpret the fitted surface. Now, if we fit the color index and the viscosity (actual data not available), their fitted equations can be computed. It turned out that they are both planar. The constraints are indicated at the beginning. The response contours of the three responses were superimposed with the feasible region unshaded: x 0 temperature ( C) w=8 y=88 y=76 y=84 z= y= y= w=9 y=86 z= z=65 z=60 y=72 z= z=70 w=0 y=84 0 x 2 From the plot, we observe that the best settings are about R = 82.6 minutes and T = 47.5 C with yield of at least 88 grams time (min.) J.6

7 .2 Characterization of the second-order response surface. Express the fitted equation in canonical form. 2. Perform the canonical analysis on the canonical form to find the best test condition. Canonical forms: The second order fitted equation in x and x 2 can be written as where ŷ = ˆβ 0 + ˆβ x + ˆβ 2 x 2 + ˆβ x 2 + ˆβ 22 x ˆβ 2 x x 2 x = [ x x 2 ŷ = ˆβ 0 + x T b + x T Bx, ] [ ] [ ] ˆβ ˆβ ˆβ2 /2, b =, B = ˆβ 2 ˆβ 2 /2 ˆβ. 22 Note: if there are three factors with coded variables x, x 2, and x 3 then the second order fitted equation has canonical form as in equation (J.) with x = x x 2 x 3, b = ˆβ ˆβ 2, B = ˆβ 3 This can be generalized to general number of factors k. stationary point: x S = 2 B b. ˆβ ˆβ2 /2 ˆβ 3 /2 ˆβ 2 /2 ˆβ 22 ˆβ23 /2 ˆβ 3 /2 ˆβ 23 /2 ˆβ 33. (J.) response at x S : ŷ S = ˆβ xt b. Let the eigenvalues of B be λ,..., λ k and their corresponding eigenvectors be m,..., m k. Denote M = [m,..., m k ], and Λ = diag(λ,..., λ k ). Then the spectral decomposition of B is given by B = MΛM T. Let X = [X,..., X k ] T = M T (x x S ). Equation (J.) can be expressed as the following canonical form (known as canonical form B) ŷ = ŷ S + X T ΛX = ŷ S + λ X λ k X 2 k. (J.2) For k = 2, the surface of ŷ = ŷ S + λ X 2 + λ 2 X 2 2 takes one of the following forms: J.7

8 sign of other surface example λ λ 2 properties type λ > λ 2 simple maximum Figure (a) (hill) + + λ > λ 2 simple minimum response decrease toward S (valley) + λ > λ 2 saddle Figure (b) + saddle no λ 2 = 0 stationary Figure (c) center not ridge of unique; a line maximum of centers on the X 2 axis + no stationary ridge of minimum no λ 2 = 0 rising Figure (d) center at ridge infinity + no falling ridge.3 Rising (falling) ridge analysis : Suppose Then x S = the nearest point on the ridge to the center of the design Z = M T x S. λ 2 λ, Z = x S = M ( Z 0 ( Z Linear coefficient B 2 = 2λ 2 Z 2. ŷ S = ŷ S + λ 2 Z 2 2. Then the fitted equation in approximated canonical form (known as canonical form A): and the equation for the ridge is X = 0. Z 2 ). ŷ ŷ S = λ X 2 + B 2 X 2, ). J.8

9 yhat yhat yhat yhat The figures below give four types of response surfaces for 2-factor problems: simple maximum/minimum minimax (saddle) stationary ridge rising/falling ridge y^ = x x + 7.2x x 2 3.7x x x 2 = X X 2 y^ = x x x x x x x 2 = X X 2 y^ = x x x x x x x 2 = X X 2 y^ = x x + 8.9x x x x x 2 = X X 2 x3 x4 x4 x3 x3 x4 x3 x4 2 Multiple responses case. If there are few responses and only two to three important input variables, the problem can be solved by overlaying and examining the contour plots for the responses over a few pairs of input variables as seen in the chemical example. Otherwise,a more formal method is required. 2. If, among the responses, one is of primary importance, it can be formulated as a constrained optimization problem by optimizing this response subject to constraints imposed on the other responses. Typically one can employ standard optimization packages, such as linear or nonlinear programming, to solve the constrained optimization problem. 3. The previous method is not appropriate when the goal of the investigation is to find an op- J.9

10 timal balance among several response characteristics. For instance, in a development of a tire tread compound, the following responses are all important: (i) an abrasion index to be maximized, (ii) elongation at break to be within specified limits, and (iii) hardness to be within specified limits. An alternative to provide a trade-off among several responses is to transform each predicted response to a desirability value d, where 0 d. The value of the desirability function d decreases as the desirability of the corresponding response decreases. Derringer and Suich (9) proposed a class of desirability functions for three types of problems: α d = ŷ U a U, a ŷ U, d = 0 for ŷ > U for ST B α d = ŷ L U L, L ŷ U, d = 0 for ŷ < L and d = for ŷ > U for LT B { ŷ L d = α t L ŷ U α 2 t U, L ŷ t,, a ŷ U, d = 0 for ŷ < L or ŷ > U for NT B; where t is the target value, L the lower bound below which the product is considered to be unacceptable, U the upper bound above which the product is considered to be unacceptable under NTB and STB and to be nearly perfect under LTB, while a the smallest possible value for the response. The choices of α, α, and α 2 are more subjective. In some practical situations, the L and U values cannot be properly chosen. A different desirability function could then be defined over the wole real line or half line. For example, use of the exponential function appears to be ideal for this purpose: d = exp{ c ŷ a α }, a ŷ < for ST B; d = exp{ cŷ α }/ exp{ cl α }, L ŷ <, d = 0 for ŷ < L for LT B; { exp{ c ŷ t d = α }, < ŷ t, exp{ c 2 ŷ t α 2 ; }, t ŷ <, where the constants c and α can be used to fine-tune the scale and shape of the desirability function. A smaller α value makes the desirability function drop more slowly from its peak. A smaller c value increases the spread of the desirability function by lowering the whole exponential curve between 0 and. An overall desirability function d can be defined as the geometric mean of the desirability functions d i s for the responses: d = [d d m ] m. A more general desirability function would be d = m i= d w i i J.0

11 to reflect possible difference in the importance of the different responses, where the weights w i satisfy 0 < w i < and w + + w m =. Any setting for the input factors that maximizes the d value is chosen to be one which achieves an optimal balance over the m responses. Running a confirmation experiment at the identified settings will provide assurance that all responses are in a desirable region. 2. Chemical Example II (Montgomery) A chemical engineer is interested in determining the operating conditions that maximize the yield of a process. Two process variables influence process yield: R, (reaction time), and T, (temperature). Two constraints are imposed on the maximization of the yield: () the viscosity of the product is required to be at least 62 and at most 68 units; (2) the molecular weight is to be at most 3400 units; (3) the yield is required to be at least 78.5 units. With a first-order design centered on the current settings of R = 35 min. and T = 55 F, a planar was fitted nicely and yielded to the new design center of R = min. and T = F after some subsequent exploratory runs along the path of steepest ascent direction implied by the first-order design. The coded variables for the second stage of the experimentation are x R = R min. 5 min. and x T = T F. 5 F A first-order design was performed around these new settings and later augmented to the following CCD (with data): original coded responses factors variables yield viscosity molecular weight R T x R x T y y 2 y J.

12 The fitted equations for the three responses are ŷ = x R x T.38x 2 R.00x 2 T x R x T, ŷ 2 = x R 0.95x T 0.69x 2 R 6.69x 2 T, 25x R x T, and ŷ 3 = x R x T. The overlaid contour plot of the three responses with the imposed constraints is given below x y 3 = 3400 y 2 = 62 reaction time (min.) 90 y = 78.5 y 2 = 62 y 2 = temperature The two unshaded areas within the design ranges gives the feasible regions for setting the two variables. The experimenter would be most interested in the larger of the two feasible regions. A more formal method such as nonlinear programming method may be called for to find the solutions. The two solutions are R = 83.5 min., T = 77. F with ŷ = 79.5; and R = 86.6 min., T = F with ŷ = x The first solution is in the smaller region while the second is in the larger one. J.2