Stat 5031 Quadratic Response Surface Methods (QRSM) Sanford Weisberg November 30, 2015

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1 Stat 5031 Quadratic Response Surface Methods (QRSM) Sanford Weisberg November 30, 2015 One Variable x = spacing of plants (either 4, 8 12 or 16 inches), and y = plant yield (bushels per acre). Each condition repeated 4 times, for a total of 12 observations. boxplot(y ~ x, main="crop Yield", xlab="spacing (inches)", ylab="yield (bu/acre)") Crop Yield Yield (bu/acre) Spacing (inches) # center and scale x m1 <- lm(y ~ factor(x), data) anova(m1) Analysis of Variance Table Response: y Df Sum Sq Mean Sq F value Pr(>F) 1

2 factor(x) e-08 *** Residuals Signif. codes: 0 '***' '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 tapply(data$y, data$x, mean) tapply(data$y, data$x, sd) View x.scaled as continuous to get a quadratic response surface: data$x.scaled <- 2 * ((data$x - min(data$x))/(max(data$x) - min(data$x))) -1 m2 <- lm(y ~ 1 + x.scaled + I(x.scaled^2), data) plot(y ~ x.scaled, data, xlab="scaled Spacing", ylab="yield") new.x <- seq(-1, 1, length=100) lines(new.x, predict(m2, data.frame(x.scaled=new.x)), lwd=3) Yield Scaled Spacing Where is this estmiated response curve maximized? 2

3 coef(m2) (Intercept) x.scaled I(x.scaled^2) What is the minimum number of distinct values of x required to fit a quadratic? Two variable QRSM y = β 0 + β 1 x 1 + β 2 x 2 + β 11 x β 22 x β 12 x 1 x 2 + ɛ If β 11, β 22 are negative and β β 11 β 22 flat spot is a maximum If β 11, β 22 are postive and β β 11 β 22 flat spot is a minimum Otherwise, flat spot is a saddle point, meaning there is a maximum for one of the xs and a minimum for the other. We hope the first two cases come up when we are trying to maximize/minimize y; if they don t then the optimum will be off in the wild blue yonder. Sometimes, the QRSM suggests that the place we want to be is a long way from where we gathered the data. In that case, we can use the QRSM to suggest a direction in which we should move to a new / region and go and gather some fresh and more relevant data. Special cases: β 12 = 0 β 11 = β 22 = 0 If β 11 = β 22 = 0 but β 12 0 model is bilinear. (This does not come up much) Estimation and design Estimate via OLS Possible design: 3 2. Requires 9 design points for a model with 1 intercept + 2 linear terms + 2 quadratics + 1 interaction, for 6 parameters Replication required for pure error estmiate of error, but one could only replicate the center point if wanted. Alternative is a central composite design. Also uses 9 design points; also requires replication often of just the center point. As elsewhere, we need to check constancy of variance as best we can, and normality, and may want to transform y. Example 14.2 from the Textbook (description: Example 13.8) Etch Rate infile <- " getdata <- read.table(infile, header=true) getdata 3

4 line ETCH UNIFORM etchfit <- lm(etch ~ + + I(^2) + I(^2) + *, getdata) summary(etchfit) Call: lm(formula = ETCH ~ + + I(^2) + I(^2) + *, data = getdata) Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) e-10 *** ** e-05 *** I(^2) I(^2) : ** --- Signif. codes: 0 '***' '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: on 6 degrees of freedom Multiple R-squared: , Adjusted R-squared: F-statistic: on 5 and 6 DF, p-value: Strange model: ETCH looks to be effectively bilinear, with no curvature. The fitted surface in 2D and can be examined in a coutour plot using the contour function. The contour function requires 3 arguments, a grid of n values for the x-axis (x1.new), a grid of m values for the y-axis (x2.new), and a matrix of values of the function to be plotted, so the value in the i-th row and j column of the matrix is the fitted value at (x1.new[i], x2.new[j]). Usually n = m and here I ve set n = m = 21: x1.new <- x2.new <- seq(-2, 2, length=21) # values for x and y axis The expand.grid function will create a data.frame with two columns of length mn that consists of all possible combinations of levels of x1.new and x2.new. 4

5 new.data <- expand.grid(=x1.new, =x2.new) head(new.data) Predictions are then computed at each of these nm values of the predictors using the predict function, and this becomes an n m matrix using the matrix function. fit <- matrix( predict(etchfit, new.data), nrow=21) par(mfrow=c(1, 2)) contour(x1.new, x2.new, fit, main="montgomery etch rate", xlab="", ylab="") persp(x1.new, x2.new, fit, main="montgomery etch rate", xlab="", ylab="", zlab="predicted") Montgomery etch rate Montgomery etch rate Predicted Example: Suppose the target ETCH rate were

6 Uniformity unifit <- update(etchfit, UNIFORM ~.) summary(unifit) Call: lm(formula = UNIFORM ~ + + I(^2) + I(^2) + :, data = getdata) Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) e-12 *** e-05 *** e-05 *** I(^2) *** I(^2) ** : e-05 *** --- Signif. codes: 0 '***' '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: on 6 degrees of freedom Multiple R-squared: , Adjusted R-squared: F-statistic: on 5 and 6 DF, p-value: 1.054e-05 fit <- matrix(predict(unifit, new.data), nrow=21) par(mfrow=c(1, 2)) contour(fit, main="montgomery uniformity", xlab="", ylab="") persp(x1.new, x2.new, fit, main="montgomery uniformity", xlab="", ylab="", zlab="predicted") 6

7 Montgomery uniformity Montgomery uniformity Predicted The contour for which ETCH 1100: contour(x1.new, x2.new, fit, xlab="", ylab="", main="montgomery uniformity with Etch rate contour") tarx1 <- ( *x2.new) / ( *x2.new) sele <- abs(tarx1) < 2 lines(tarx1[sele], x2.new[sele], lty=2, lwd=4) 7

8 Montgomery uniformity with Etch rate contour