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3.3 Response surfaces

There still remains some art in both the design and the analysis of experiments, which can only be learned from experience. In addition, the role of engineering judgment should not be underestimated (from http://www.itl.nist.gov/div898/handbook/casestud.htm).

Definition 3.3.1   A response surface design is a DOE aimed at obtaining a simple non-linear function as model of the phenomenon.

Remark 3.3.2   Here again, the "asserting" phase is essential. When the reality of the phenomenon leads to a model $ z=Cte  x_{1}  x_{2}  x_{3}$, it is obviously better to seek for an affine model relating the logarithms.

Remark 3.3.3   Along the same lines, studies about capillarity are evaluating and using $ \cos\alpha$. To ask whether the physical reality is captured by $ \alpha$ or $ \cos\alpha$ is not useless : what is linear for one of the two parameters can not be linear for the other.

Proposition 3.3.4 (CC = Centered Cubic)   If we want to show that a linear model is the right one, a centered cubic distribution (i.e. a $ 2^{m}$ full design completed by repetitions of the center) is the best one. Using repeated points at the center allows both to study the repeatability and verify the absence of curvature.

Proposition 3.3.5 (CCF = Cubic Centered Faces)   When a full binary design has yet been ubdertaken (e.g. as a restriction of a fractional design to its most significant parameters), a good way to complete the design in order to adjust quadratic terms is using the centers of faces and repeat the central point.

Remark 3.3.6 (CCC = Circumscribed Centered Cubic)   When possible, centers of faces are to be taken not on the cube itself, but on a circumscribed surface. But this introduces five levels per factor, instead of three. Let us compare, for a three factors desing, several values of $ k$ : CCF is $ k=1$, while CCC is $ k>1$. With $ r=7$ replications of the center, we obtain the design $ A$ given TAB. 3.1 and matrix $ S=\raisebox{0.5 em}{\normalfont\textsf{t}}{A}.A$ given TAB. 3.2.


TAB. 3.1: CCF design
\begin{table}\begin{displaymath}A=\left(\begin{array}{cccccccccc}
1&-1&-1&-1&+1&...
...r
1&0&0&0&0&0&0&0&0&0
\end{array}\right)\end{displaymath}\par\par
\end{table}



TAB. 3.2: CCF design : matrix $ S$
\begin{table}\begin{displaymath}S=\left(\begin{array}{cccccccccc}
14+r&0&0&0&0&0...
...&0&0&0&0&0&8&8&8+2k^4
\end{array}\right)
\end{displaymath}\par\par
\end{table}


After some computations, it can be seen that eigenvalues of $ S\left(k, r\right)$ are $ \lambda=8$ (threefold), $ \lambda=8+2k^{2}$ (threefold), $ \lambda=2k^{4}$ (twofold) and other two $ \lambda_{1}\leq\lambda_{2}$ that depend on $ k$ and $ r$. When $ k=1$ and $ r\geq1$, the smallest eigenvalue is $ 2$. When $ k=1.2$ and $ r\geq5$, the smallest eigenvalue is $ 4$. When $ k>1.4$, all eigenvalues but one are greater than $ 8$, but $ r=7$ only enforces $ \lambda>5$.

Exercise 3.3.7   When adding columns like $ x^{2}-mean\left(x^{2}\right)$ instead of columns like $ x^{2}$ the eigenspaces relative to $ \lambda_{1}$ and $ \lambda_{2}$ are reorganized leading to new eigenvalues : $ 14+r$ and $ \lambda\left(k,r\right)$. Examine in details what happens.

Remark 3.3.8   It can be seen that $ k=\sqrt[4]{8}\approx1.68$ leds to an expression of $ \sigma_{Y}^{2}$ having a spherical symmetry, i.e. that depends only from $ \rho=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}$. For $ r=6$, majoration $ \sigma_{Y}^{2}<0.2 \sigma_{\nu}^{2}$ (resp. $ \sigma_{Y}^{2}<0.3 \sigma_{\nu}^{2}$) holds uniformly in the inscribed sphere $ \rho\leq1$ (resp the circumscribed sphere $ \rho\leq\sqrt{3}$).

Proposition 3.3.9 (Box-Behnken)   When there are no prior results, it is interesting to consider the centers of the edges of the cube (and repeat the central point).

Remark 3.3.10   A Box-Behnken design with three factors uses $ 12$ non-central points and $ r$ replications of the center. The eigenvalues of $ S$ are $ \lambda=4$ (order $ 5$), $ \lambda=8$ (threefold), the two others $ \lambda_{1}\leq\lambda_{2}$ being dependent on $ r$. Minoration $ \lambda_{1}\geq4$ requiers $ r\geq8$. Moreover, assuming $ r=3$ and $ \rho<\sqrt{2}$ only ensures $ \sigma_{Y}^{2}\leq0.5 \sigma_{\nu}^{2}$ (the standard discrépancy is strongly anisotropic).


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Previous: 3.2 Computing uncertainties Up: 3. Improving Next: A. Some computing tools   Contents


douillet@ensait.fr
2008-03-14