The Fitted Response Surface Graph the fitted surface and its - PowerPoint PPT Presentation

ST 516 Experimental Statistics for Engineers II The Fitted Response Surface Graph the fitted surface and its standard error: response.R 1 / 17 Response Surface Methods Analyzing a Second-Order Response Surface

ST 516 Experimental Statistics for Engineers II Predicted Yield 182 78.5 5 . 5 7 79.5 180 75 178 79 80 Temperature 176 174 172 8 7 170 7 8 76 . 5 75 75.5 76 168 7 7 76.5 6 77.5 4 . 77 5 77 78 80 82 84 86 88 90 92 Time 2 / 17 Response Surface Methods Analyzing a Second-Order Response Surface

ST 516 Experimental Statistics for Engineers II 80 78 Yield 76 74 180 90 Temperature 175 85 Time 170 80 3 / 17 Response Surface Methods Analyzing a Second-Order Response Surface

ST 516 Experimental Statistics for Engineers II Standard Error 182 0.22 0.22 0.28 0.28 4 0 0.32 2 . 0.32 . 2 0 4 0.26 0.26 180 0 . 1 4 0.18 0.2 0.16 178 0.12 Temperature 176 174 172 170 0.32 0.32 0 6 . 2 2 6 0 . 168 0 0.24 0.24 3 . 0.22 0.22 . 3 0 78 80 82 84 86 88 90 92 Time 4 / 17 Response Surface Methods Analyzing a Second-Order Response Surface

ST 516 Experimental Statistics for Engineers II Standard Error 0.35 0.30 0.25 0.20 0.15 180 90 Temperature 175 85 Time 170 80 5 / 17 Response Surface Methods Analyzing a Second-Order Response Surface

ST 516 Experimental Statistics for Engineers II Finding the Optimum Write the full quadratic model as y = ˆ β 0 + x ′ b + x ′ Bx ˆ where ˆ ˆ 2 ˆ 2 ˆ      1 1  β 1 β 1 , 1 β 1 , 2 . . . β 1 , k x 1 ˆ 2 ˆ ˆ 2 ˆ 1 1 x 2 β 2 β 1 , 2 β 2 , 2 . . . β 2 , k       x =  , b =  , and B =  .  .   .   . . .  ... . . . . .       . . . . .    ˆ 2 ˆ 2 ˆ ˆ 1 1 x k β k β 1 , k β 2 , k . . . β k , k The the location of the stationary point (maximum, minimum, or saddle-point) is x s = − 1 2 B − 1 b . 6 / 17 Response Surface Methods Analyzing a Second-Order Response Surface

ST 516 Experimental Statistics for Engineers II In the example l <- lm(y ~ x1 + x2 + I(x1^2) + I(x2^2) + x1:x2, t11d6) lc <- coefficients(l) b <- lc[2:3] B <- matrix(c(lc[4], lc[6]/2, lc[6]/2, lc[5]), 2, 2) cat("Stationary point:", -solve(B, b) / 2, "\n") Output Stationary point: 0.3892304 0.3058466 In engineering units: ξ 1 = 85 + 5 ∗ 0 . 3892304 = 86 . 94615 , ξ 2 = 175 + 5 ∗ 0 . 3058466 = 176 . 5292 . 7 / 17 Response Surface Methods Analyzing a Second-Order Response Surface

ST 516 Experimental Statistics for Engineers II We want to round these, but how much? 87 and 177? 85 and 175? 8 / 17 Response Surface Methods Analyzing a Second-Order Response Surface

ST 516 Experimental Statistics for Engineers II We can test the hypothesis that the stationary point is x (0) : the general quadratic with a stationary point at x (0) is � 2 � 2 � � x 1 − x (0) x 2 − x (0) y = β ∗ 0 + β ∗ + β ∗ 1 , 1 1 2 , 2 2 � � � � x 1 − x (0) x 2 − x (0) + β ∗ + ǫ ; 1 , 2 1 2 fit this as a reduced model, and compare with the full model using the extra sum of squares ; reduced model has 4 parameters versus 6 in the full model, so test statistic is F 2 , n − p . Test specific values, or graph the statistic for a grid of values. 9 / 17 Response Surface Methods Analyzing a Second-Order Response Surface

ST 516 Experimental Statistics for Engineers II Test whether 87, 177 is optimal x10 <- (87 - 85) / 5 x20 <- (177 - 175) / 5 summary(aov(y ~ I((x1 - x10)^2) + I((x2 - x20)^2) + I((x1 - x10)*(x2 - x20)) + x1 + x2, t11d6)) Output Df Sum Sq Mean Sq F value Pr(>F) I((x1 - x10)^2) 1 18.901 18.901 266.553 7.88e-07 *** I((x2 - x20)^2) 1 8.681 8.681 122.423 1.09e-05 *** I((x1 - x10) * (x2 - x20)) 1 0.526 0.526 7.415 0.0296 * x1 1 0.004 0.004 0.057 0.8186 x2 1 0.134 0.134 1.895 0.2110 Residuals 7 0.496 0.071 --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 F = ((0 . 004 + 0 . 134) / 2) / 0 . 071 = 0 . 97, P ≈ . 5, do not reject H 0 . 10 / 17 Response Surface Methods Analyzing a Second-Order Response Surface

ST 516 Experimental Statistics for Engineers II We can test the same hypothesis without fitting the reduced model: the model is y ( x 1 , x 2 ) = β 0 + β 1 x 1 + β 2 x 2 + β 1 , 1 x 2 1 + β 1 , 2 x 1 x 2 + β 2 , 2 x 2 ˆ 2 ; so ∂ ˆ y = β 1 + 2 β 1 , 1 x 1 + β 1 , 2 x 2 , ∂ x 1 ∂ ˆ y = β 2 + 2 β 2 , 2 x 2 + β 1 , 2 x 1 . ∂ x 2 So the hypothesis is equivalent to β 1 + 2 β 1 , 1 x (0) + β 1 , 2 x (0) = 0 , 1 2 β 2 + 2 β 2 , 2 x (0) + β 1 , 2 x (0) = 0 . 2 1 11 / 17 Response Surface Methods Analyzing a Second-Order Response Surface

ST 516 Experimental Statistics for Engineers II x (0) � � That is, L β = 0 , where � � 2 x (0) x (0) 0 1 0 0 x (0) � � 1 2 L = x (0) 2 x (0) 0 0 1 0 1 2 and β = ( β 0 , β 1 , β 2 , β 1 , 1 , β 1 , 2 , β 2 , 2 ) ′ . Any hypothesis of this form can be tested as follows: L ( X ′ X ) − 1 L ′ � − 1 ( L ˆ SS L = ( L ˆ β ) ′ � β ) is the sum of squares associated with the hypothesis, MS L = SS L / 2 is the mean square, and F = MS L / MS E is the test statistic. 12 / 17 Response Surface Methods Analyzing a Second-Order Response Surface

ST 516 Experimental Statistics for Engineers II R commands betaHat <- coefficients(l) V <- summary(l)$cov.unscaled msE <- summary(l)$sigma^2 L <- rbind(c(0, 1, 0, 2 * x10, 0, x20), c(0, 0, 1, 0, 2 * x20, x10)) LbetaHat <- L %*% betaHat ssL <- sum(LbetaHat * solve(L %*% V %*% t(L), LbetaHat)) msL <- ssL / 2 F <- msL / msE We can calculate F for a grid of x (0) values and graph it. The region where F < F 2 , n − p ( α ) is a 100(1 − α )% confidence region for the optimal settings. 13 / 17 Response Surface Methods Analyzing a Second-Order Response Surface

ST 516 Experimental Statistics for Engineers II F−Statistic 182 180 178 4.737414 140 0 4 Temperature 176 20 174 60 172 120 170 100 168 0 8 78 80 82 84 86 88 90 92 Time 14 / 17 Response Surface Methods Analyzing a Second-Order Response Surface

ST 516 Experimental Statistics for Engineers II 150 F−Statistic 100 50 180 90 Temperature 175 85 Time 170 80 15 / 17 Response Surface Methods Analyzing a Second-Order Response Surface

ST 516 Experimental Statistics for Engineers II F−Statistic Detail 180 2 179 0 40 178 Temperature 4.737414 177 15 0 1 176 5 35 0 3 175 25 5 3 2 3 0 5 86 87 88 89 90 Time 16 / 17 Response Surface Methods Analyzing a Second-Order Response Surface

ST 516 Experimental Statistics for Engineers II We can round to 87 and either 176 or 177. We cannot round to 85 and 175. 17 / 17 Response Surface Methods Analyzing a Second-Order Response Surface