Response surfaces and optimal design — Examples

The matrix $B$ is symmetric and nonsingular, and its eigenvalues are $1$ and $-1$. The stationary point formula in [F2] gives $x_0=0$, which lies in the interior of $\mathcal{R}$.

Directly, $m(h,0)-m(0,0)=h^2>0$ and $m(0,h)-m(0,0)=-h^2<0$ for every $0<\mathrm{\mid }h\mathrm{\mid }<1$. Thus arbitrarily close points in the region give increments of both signs, verifying the saddle classification in [F2].

The information matrix is $M^{\ast }=\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right)$, with determinant $1$ and inverse $\left(\begin{array}{cc}2& -1\\ -1& 1\end{array}\right)$. Hence the leverage at $x$ is $u(x)=(1,x)M^{\ast -1}(1,x{)}^{\mathsf{T}}=(x-1{)}^2+1$, giving $u(0)=2$, $u(1)=1$, and $u(2)=2$. By [F3], all candidates satisfy $u(x)\le p=2$ and equality holds on the support $\{0,2\}$, so ${\xi }^{\ast }$ is D-optimal.

For an exact three-run design let $(n_0,n_1,n_2)$ be the nonnegative integer counts with sum $3$. Substitution into [F1] and the exact-count convention [F2] gives $\det M=(n_0{n}_1+4n_0{n}_2+n_1{n}_2)\mathrm{/}9$.

The ten count triples give determinants as follows: $(3,0,0),(0,3,0),(0,0,3)$ give $0$; $(2,1,0),(1,2,0),(0,2,1),(0,1,2)$ give $2\mathrm{/}9$; $(2,0,1),(1,0,2)$ give $8\mathrm{/}9$; and $(1,1,1)$ gives $6\mathrm{/}9$. Thus the exact maximum is $8\mathrm{/}9$, attained by either endpoint allocation with counts $2$ and $1$.

Since $8\mathrm{/}9<1=\det M({\xi }^{\ast })$, the exact three-run optimum and the approximate optimum differ. This finite calculation concerns only the stated candidate set and run count.