Construction of uniform mixed-level designs through level permutations

Abstract

Uniform designs have been widely used in physical and computer experiments due to their robust performances. The level permutation method can efficiently construct uniform designs with both lower discrepancy and less aberration. However, the related existing literature has mostly discussed uniform fixed-level designs, the construction of uniform mixed-level designs has been quite few studied. In this paper, a novel level permutation method for constructing uniform mixed-level designs is proposed. Our main idea is to perform level permutations on a new class of designs, called minimum average discrepancy designs, rather than generalized minimum aberration designs as in the fixed-level case. Several theoretical results on the design optimality and construction are obtained. Numerical results suggest the good performance of the resulting designs under various popular discrepancies.

Appendix

Proof of Theorem 1

For any row ${\varvec{d}}_i=(d_{i,1},\ldots ,d_{i,n})$ of D, when all level permutations are considered, each n-tuple in $R_{s_1}^{n_1}\times \cdots \times R_{s_g}^{n_g}$ occurs $\prod _{k=1}^g[(s_k-1)!]^{n_k}$ times. Then

$$\begin{aligned}&\sum _{D'\in {\mathcal {P}}_D}\sum _{i=1}^N\prod _{k=1}^g\prod _{j=n_1+\cdots +n_{k-1}+1}^{n_1+\cdots +n_k}f_1\left( s_k^{-1}(d_{ij}+0.5)\right) \nonumber \\&\quad =\sum _{i=1}^N\sum _{D'\in {\mathcal {P}}_D}\prod _{k=1}^g\prod _{j=n_1+\cdots +n_{k-1}+1}^{n_1+\cdots +n_k}f_1\left( s_k^{-1}(d_{ij}+0.5)\right) \nonumber \\&\quad =N\prod _{k=1}^g ((s_k!)\alpha (s_k))^{n_k}, \end{aligned}$$

(10)

which is a constant. Similarly,

$$\begin{aligned}&\sum _{D'\in {\mathcal {P}}_D}\sum _{i=1}^N\prod _{k=1}^g\prod _{j=n_1+\cdots +n_{k-1}+1}^{n_1+\cdots +n_k}f\left( s_k^{-1}(d_{ij}+0.5),s_k^{-1}(d_{ij}+0.5)\right) \nonumber \\&\quad =N\prod _{k=1}^g ((s_k!)\beta (s_k)\gamma (s_k))^{n_k} \end{aligned}$$

(11)

is also a constant.

For any two distinct rows ${\varvec{d}}_{i_1}$ and ${\varvec{d}}_{i_2}$ of D, when all level permutations of D are considered, each identical pair (z, z) of the $n_k$ factors with $s_k$ levels occurs $(s_k-1)!$ times in the $\delta _{i_1,i_2,k}(D)$ positions of coincidence and each distinct pair $(z_1,z_2)$ occurs $(s_k-2)!$ times in the other $n_k-\delta _{i_1,i_2,k}(D)$ positions. Then

$$\begin{aligned}&\sum _{D'\in {\mathcal {P}}_D}\sum _{i_1<i_2}^N\prod _{k=1}^g\prod _{j=n_1+\cdots +n_{k-1}+1}^{n_1+\cdots +n_k} f\left( s_k^{-1}(d_{i_1j}+0.5),s_k^{-1}(d_{i_2j}+0.5)\right) \nonumber \\&\quad =\sum _{i_1<i_2}^N \sum _{D'\in {\mathcal {P}}_D}\prod _{k=1}^g\prod _{j=n_1+\cdots +n_{k-1}+1}^{n_1+\cdots +n_k} f\left( s_k^{-1}(d_{i_1j}+0.5),s_k^{-1}(d_{i_2j}+0.5)\right) \nonumber \\&\quad =\sum _{i_1<i_2}^N \prod _{k=1}^g\bigg [(s_k-1)!\sum _{z\in R_{s_k}}f\left( s_k^{-1}(z+0.5),s_k^{-1}(z+0.5)\right) \bigg ]^{\delta _{i_1,i_2,k}(D)} \nonumber \\&\qquad \times \bigg [(s_k-2)! \sum _{z_1\ne z_2\in R_{s_k}}f\left( s_k^{-1}(z_1+0.5),s_k^{-1}(z_2+0.5)\right) \bigg ]^{n_k-\delta _{i_1,i_2,k}(D)} \nonumber \\&\quad =\prod _{k=1}^g\big [(s_k!)\beta (s_k)\big ]^{n_k} \cdot \sum _{i_1<i_2}^N\prod _{k=1}^g(\gamma (s_k))^{\delta _{i_1,i_2,k}(D)}, \end{aligned}$$

(12)

The desired result follows by combining (3), (5), (10), (11) and (12). $\square$

Proof of Theorem 4

Let $V=R_{s_1}^{n_1}\times \cdots \times R_{s_g}^{n_g}$. For the $(N,s_1^{n_1}\cdots s_g^{n_g})$-design D without repeated points, let $m(l_1,\ldots ,l_n)$ denote the number of times that the point $(l_1,\ldots ,l_n)$ occurs in D. Then the design D can be uniquely determined by the following column vector of length $N_0=\prod _{k=1}^gs_k^{n_k}$, i.e.,

$$\begin{aligned} \mathbf{y}_D=\left( m(l_1,\ldots ,l_n)\right) _{(l_1,\ldots ,l_n)\in V}, \end{aligned}$$

(13)

where all points $(l_1,\ldots ,l_n)$ in V are arranged in the lexicographical order. The vector $\mathbf{y}_D$ is called the frequency vector of the design D. In particular, a full factorial $(N_0,s_1^{n_1}\cdots s_g^{n_g})$-design has $\mathbf{y}_D= {\varvec{1}}_{N_0}$, the $N_0$-vector of ones. By noticing

$$\begin{aligned} \sum _{i_1,i_2=1}^N\mathrm {e}^{\varDelta _{i_1,i_2}(D)}= & {} \sum _{(l_1,\ldots ,l_n),(l_1',\ldots ,l_n')\in V} m(l_1,\ldots ,l_n)\,m(l_1',\ldots ,l_n') \\&\times \prod _{k=1}^g\prod _{j={n_1}+\cdots +n_{k-1}+1}^{{n_1}+\cdots +n_k} (\gamma (s_k))^{\delta (l_j,l_j')}, \end{aligned}$$

the quadratic form of ${\bar{\phi }}(D)$ in terms of $\mathbf{y}_D$ is obtained as

$$\begin{aligned} {\bar{\phi }}(D) = c_0^n-2\prod _{k=1}^g(\alpha (s_k))^{n_k}+\frac{1}{N^2}\prod _{k=1}^g(\beta (s_k))^{n_k}\mathbf{y}_D^{{ \mathrm {\scriptscriptstyle T} }}{} \mathbf{A}{} \mathbf{y}_D, \end{aligned}$$

(14)

where for $i,j=1,\ldots ,s_k$ and $k=1,\ldots ,g$,

$$\begin{aligned} \mathbf{A}=\overbrace{\mathbf{A}_1\otimes \cdots \otimes \mathbf{A}_1}^{n_1}\otimes \overbrace{\mathbf{A}_2\otimes \cdots \otimes \mathbf{A}_2}^{n_2}\otimes \cdots \otimes \overbrace{\mathbf{A}_g\otimes \cdots \otimes \mathbf{A}_g}^{n_g}, \end{aligned}$$

$\mathbf{A}_k=(a_{i,j}^{(k)})_{s_k\times s_k}$, $a_{i,j}^{(k)}$ equals $\gamma (s_k)$ if $i=j$ and 1 otherwise, $c_0$, $\gamma (\cdot )$, $\alpha (\cdot )$ and $\beta (\cdot )$ are defined in Theorem 1, $\mathbf{A}\otimes \mathbf{B}$ denotes the Kronecker product of two matrices $\mathbf{A}$ and $\mathbf{B}$. Similarly,

$$\begin{aligned} {\bar{\phi }}(D^c) = c_0^n-2\prod _{k=1}^g(\alpha (s_k))^{n_k}+\frac{1}{(N_0-N)^2}\prod _{k=1}^g(\beta (s_k))^{n_k}\mathbf{y}_{D^c}^{{ \mathrm {\scriptscriptstyle T} }}{} \mathbf{A}{} \mathbf{y}_{D^c}, \end{aligned}$$

(15)

where $\mathbf{y}_{D^c}$ is the frequency vector of the design $D^c$. Since $D^c$ is the complementary design of D, we have

$$\begin{aligned} \mathbf{y}_D+\mathbf{y}_{D^c}=\mathbf{1}_{N_0}\quad \text {and}\quad \mathbf{1}_{N_0}^{{ \mathrm {\scriptscriptstyle T} }}{} \mathbf{y}_{D^c}=N_0-N. \end{aligned}$$

(16)

By noting that $\mathbf{A}_k\mathbf{1}_{s_k}=\gamma (s_k)+s_k-1$, we also obtain

$$\begin{aligned} \mathbf{A}{} \mathbf{1}_{N_0}=\prod _{k=1}^g(\gamma (s_k)+s_k-1)^{n_k}. \end{aligned}$$

(17)

Combining (14), (15), (16) and (17), the desired result follows. $\square$