Construction of orthogonal marginally coupled designs

Abstract

Marginally coupled designs (MCDs) were first introduced by Deng et al. (Stat Sin 25:1567–1581, 2015), as more economical designs than sliced space-filling designs which are the popular choices for computer experiments with both qualitative and quantitative factors. In an MCD, the design for qualitative factors is an orthogonal array, and the one for quantitative factors is a Latin hypercube design (LHD) with its rows corresponding to each level of any qualitative factor also forming a small LHD. As we know, orthogonality is a popular and important property for evaluating LHDs, but was not considered in existing results on MCDs. In this paper, we propose some approaches to constructing a new class of MCDs with orthogonality. In some cases, the designs for quantitative factors also satisfy the two dimensional space-filling property. Besides, the run sizes of the obtained designs are more flexible than the existing ones.

Appendices

Appendix A: Proofs

To prove the theoretical results of this paper, we first give the following two lemmas.

Lemma A.1

Let $\zeta$ be any column of $D_{1}$ which is an $OA(s^t,m,s,2)$ and $\xi$ be any column of $D_{2}$ which is an $L(s^{t},k)$. And $\xi$ can be represented as $\xi =\pm \eta _{1}\pm \eta _{2}s\pm \cdots \pm \eta _{t}s^{t-1}$, where $(\eta _{1},\eta _{2},\ldots ,\eta _{t})$ with entries from $\{-(s-1)/2,-(s-3)/2,\ldots ,(s-3)/2,(s-1)/2\}$ is an $s^{t}$ full factorial design. If $(\zeta ,\eta _{2},\ldots ,\eta _{t})$ is an $s^{t}$ full factorial design, then $(D_{1},D_{2})$ is an MCD.

Following the symbols in Algorithm 4, let $i_k,j_{k}=0,1,\ldots ,\lfloor s/2\rfloor -1$, $k=1,2,\ldots$. Let $W_{i_{1}j_{1}}^{(1)}=(a_{1}^{(1)},a_{2}^{(1)},b^{(1)})$, where $a_{1}^{(1)}$ and $a_{2}^{(1)}$ are any two columns in $\beta _{i_{1}}\oplus D^{(0)}$, and $b^{(1)}$ is any column in $\beta _{j_{1}}\oplus D^{(0)}$, $i_{1} \ne j_{1}$. Create $W_{i_{1}j_{1}i_{2}j_{2}}^{(2)}=(\beta _{i_{2}}\oplus W_{i_{1}j_{1}}^{(1)},b^{(2)})$, where $b^{(2)} \in \beta _{j_{2}}\oplus D^{(1)}$, $i_{2} \ne j_{2}$. In general, define $W_{i_{1}j_{1}i_{2}j_{2}\cdots i_{v}j_{v}}^{(v)}=(\beta _{i_{v}}\oplus W_{i_{1}j_{1}i_{2}j_{2}\cdots i_{v-1}j_{v-1}}^{(v-1)},b^{(v)})$, where $b^{(v)} \in \beta _{j_{v}}\oplus D^{(v-1)}$, $i_{v} \ne j_{v}$.

Lemma A.2

Suppose $D^{(0)}$, $W_{i_{1}j_{1}i_{2}j_{2}\cdots i_{v}j_{v}}^{(v)}$ and $e^{(v)}$ for $v=1,2,\ldots$ are as defined above, then

• (i) $W_{i_{1}j_{1}i_{2}j_{2}\cdots i_{v}j_{v}}^{(v)}$ is a full factorial design with $2+v$ factors;

• (ii) for any column $h \in e^{(1)}$, $(a_{1}^{(1)},a_{2}^{(1)},h)$ is a full factorial design with three factors; furthermore for any column $h \in e^{(v)}$ with $v\ge 2$, $(\beta _{i_{v}}\oplus W_{i_{1}j_{1}i_{2}j_{2}\cdots i_{v-1}j_{v-1}}^{(v-1)},h)$ is a full factorial design with $2+v$ factors.

Proof of Lemma A.1

To make sure $(D_{1},D_{2})$ is an MCD, it needs to prove that for each level of $\zeta$, the corresponding rows in $D_{2}$ form an $L(s^{t-1},k)$ after level-collapsing. Here, we collapse any level x of $\xi$ by $f(x)=\lfloor \frac{x+(s^{t}-1)/2}{s} \rfloor$. After level-collapsing, the levels of $\xi$ are collapsed to the $s^{t-1}$ levels $\{0,1,s^{t-1}-1\}$. Let $\lambda _{i}=\pm \eta _{i}+ ((s-1)/{2}) 1_{s^t}$, $i=1,\ldots ,t$, then the entries of $\lambda _{i}$ are all in $\{0,1,\ldots ,s-1\}$, and $\xi +((s^{t}-1)/2)1_{s^t}=\lambda _{1}+\lambda _{2}s+\cdots +\lambda _{t}s^{t-1}$. Thus $f(\xi )=\lambda _{2}+\lambda _{3}s+\cdots +\lambda _{t}s^{t-2}$. It is easy to see that for each level of $\zeta$, the corresponding rows in $(\lambda _{2},\ldots ,\lambda _{t})$ form an $s^{t-1}$ full factorial design, since $(\zeta ,\eta _{2},\ldots ,\eta _{t})$ is an $s^{t}$ full factorial design. Thus for each level of $\zeta$, the corresponding rows in $D_{2}$ form an $L(s^{t-1},k)$ after level-collapsing. This completes the proof. $\square$

Proof of Lemma A.2

(i) From the construction of $D^{(0)}$, it is easy to see that $D^{(0)}$ is an $OA(s^{2},s,s,2)$, and $(\alpha ,\ldots ,\alpha )$ is a row in $D^{(0)}$ for any $\alpha \in GF(s)$. Furthermore, the rows of $D^{(0)}$ form a linear space over GF(s). Then, for any $\alpha _{i},\alpha _{k} \in GF(s)$, $\alpha _{i}\alpha _{k} J_{s^{2} \times s}+D^{(0)}$ can be transformed into $D^{(0)}$ by row permutation. Thus if $\alpha _{j_{1}}=\alpha _{i_{1}}+\alpha _{t_{1}}$, $\alpha _{t_{1}}\ne 0$, then after row permutation, $\left( \beta _{i_{1}}\oplus D^{(0)},\beta _{j_{1}}\oplus D^{(0)}\right)$ can be transformed into

$$\begin{aligned} \left( \begin{array}{llll} (D^{(0)})^{T}&{}\quad (D^{(0)})^{T}&{}\quad \ldots &{}\quad (D^{(0)})^{T}\\ (\alpha _{t_{1}}\alpha _{0}J_{s^{2} \times s}+D^{(0)})^{T}&{}\quad (\alpha _{t_{1}}\alpha _{1}J_{s^{2} \times s}+D^{(0)})^{T}&{}\quad \ldots &{}\quad (\alpha _{t_{1}}\alpha _{s-1}J_{s^{2} \times s}+D^{(0)})^{T} \end{array} \right) ^{T}. \end{aligned}$$

So it is straightforward to obtain that $W_{i_{1}j_{1}}^{(1)}$ is a full factorial design with three factors. Furthermore, for $v=2,3,\ldots$, if $\alpha _{j_{v}}=\alpha _{i_{v}}+\alpha _{t_{v}}$, $\alpha _{t_{v}}\ne 0$, then after row permutation, $\left( \beta _{i_{v}}\oplus D^{(v-1)}, \beta _{j_{v}}\oplus D^{(v-1)}\right)$ can be transformed into

$$\begin{aligned} \left( \begin{array}{llll} (D^{(v-1)})^{T}&{}\quad (D^{(v-1)})^{T}&{}\quad \ldots &{}\quad (D^{(v-1)})^{T}\\ (\alpha _{t_{v}}\alpha _{0}J_{s^{v+1} \times s^{v}}+D^{(v-1)})^{T}&{}\quad (\alpha _{t_{v}}\alpha _{1}J_{s^{v+1} \times s^{v}}+D^{(v-1)})^{T}&{}\quad \ldots &{}\quad (\alpha _{t_{v}}\alpha _{s-1}J_{s^{v+1} \times s^{v}}+D^{(v-1)})^{T} \end{array} \right) ^{T}. \end{aligned}$$

Thus $W_{i_{1}j_{1}i_{2}j_{2}\cdots i_{v}j_{v}}^{(v)}$ is an s-level full factorial design with $2+v$ factors for $v=2,3,\ldots$.

(ii) In Algorithm 4, N is a full factorial design which can be written as

$$\begin{aligned} N = \left( \begin{array}{l@{\quad }l@{\quad }l@{\quad }l} \gamma ^{T} &{} \gamma ^{T} &{} \cdots &{}\gamma ^{T}\\ \alpha _{0}1_{s}^{T} &{} \alpha _{1}1_{s}^{T} &{} \cdots &{} \alpha _{s-1}1_{s}^{T}\ \end{array} \right) ^{T} \end{aligned}$$

and $G_{1} =(1_{s},\gamma )^{T}$, where $\gamma =(\alpha _{0},\cdots ,\alpha _{s-1})^{T}$. Then $(D^{(0)},e^{(0)})$ can be written as

$$\begin{aligned} (D^{(0)},e^{(0)}) = \left( \begin{array}{ll} \gamma 1_{s}^{T}+\alpha _{0}1_{s}\gamma ^{T}&{}\quad \alpha _{0}1_{s}\\ \gamma 1_{s}^{T}+\alpha _{1}1_{s}\gamma ^{T}&{}\quad \alpha _{1}1_{s}\\ \vdots &{}\quad \vdots \\ \gamma 1_{s}^{T}+\alpha _{s-1}1_{s}\gamma ^{T}&{}\quad \alpha _{s-1}1_{s} \end{array} \right) . \end{aligned}$$

From the definitions, it is obvious to obtain that

$$\begin{aligned}&(\beta _{i_{1}}\oplus D^{(0)},\beta _{j_{1}}\oplus e^{(0)})\\&\quad = \left( \begin{array}{ll} \alpha _{i_{1}}\alpha _{0}J_{s^{2}\times s}+D^{(0)}&{}\quad \alpha _{j_{1}}\alpha _{0}1_{s^{2}}+e^{(0)}\\ \vdots &{}\quad \vdots \\ \alpha _{i_{1}}\alpha _{s-1}J_{s^{2}\times s}+D^{(0)}&{}\quad \alpha _{j_{1}}\alpha _{s-1}1_{s^{2}}+e^{(0)} \end{array} \right) \\&\quad =\left( \begin{array}{ll} \alpha _{i_{1}}\alpha _{0}J_{s\times s}+\gamma 1_{s}^{T}+\alpha _{0}1_{s}\gamma ^{T}&{}\quad \alpha _{j_{1}}\alpha _{0}1_{s}+\alpha _{0}1_{s}\\ \alpha _{i_{1}}\alpha _{0}J_{s\times s}+\gamma 1_{s}^{T}+\alpha _{1}1_{s}\gamma ^{T}&{}\quad \alpha _{j_{1}}\alpha _{0}1_{s}+\alpha _{1}1_{s}\\ \vdots &{}\quad \vdots \\ \alpha _{i_{1}}\alpha _{0}J_{s\times s}+\gamma 1_{s}^{T}+\alpha _{s-1}1_{s}\gamma ^{T}&{}\quad \alpha _{j_{1}}\alpha _{0}1_{s}+\alpha _{s-1}1_{s}\\ \vdots &{}\quad \vdots \\ \vdots &{}\quad \vdots \\ \alpha _{i_{1}}\alpha _{s-1}J_{s\times s}+\gamma 1_{s}^{T}+\alpha _{0}1_{s}\gamma ^{T}&{}\quad \alpha _{j_{1}}\alpha _{s-1}1_{s}+\alpha _{0}1_{s}\\ \alpha _{i_{1}}\alpha _{s-1}J_{s\times s}+\gamma 1_{s}^{T}+\alpha _{1}1_{s}\gamma ^{T}&{}\quad \alpha _{j_{1}}\alpha _{s-1}1_{s}+\alpha _{1}1_{s}\\ \vdots &{}\quad \vdots \\ \alpha _{i_{1}}\alpha _{s-1}J_{s\times s}+\gamma 1_{s}^{T}+\alpha _{s-1}1_{s}\gamma ^{T}&{}\quad \alpha _{j_{1}}\alpha _{s-1}1_{s}+\alpha _{s-1}1_{s} \end{array} \right) . \end{aligned}$$

Note that $(\alpha _{i_{1}}\alpha _{k_1}J_{s\times s}+\gamma 1_{s}^{T}+\alpha _{k_2}1_{s}\gamma ^{T},\alpha _{j_{1}}\alpha _{k_1}1_{s}+\alpha _{k_2}1_{s})$ can be transformed into $(\gamma 1_{s}^{T}+\alpha _{k_2}1_{s}\gamma ^{T},\alpha _{j_{1}}\alpha _{k_1}1_{s}+\alpha _{k_2}1_{s})$ by permuting rows, for any $k_1,k_2=0,\ldots ,s-1$. Then $(\beta _{i_{1}}\oplus D^{(0)},\beta _{j_{1}}\oplus e^{(0)})$ can be transformed into

$$\begin{aligned} \left( \begin{array}{ll} D^{(0)},&{}\quad \alpha _{j_{1}}\alpha _{0}1_{s^{2}}+e^{(0)}\\ \vdots &{}\quad \vdots \\ D^{(0)},&{}\quad \alpha _{j_{1}}\alpha _{s-1}1_{s^{2}}+e^{(0)} \end{array} \right) . \end{aligned}$$

Thus $(a_{1}^{(1)},a_{2}^{(1)},h)$ is a full factorial design with three factors, where $\alpha _{j_{1}}\ne 0$. Furthermore, for $v\ge 2$, $(\beta _{i_{v}}\oplus D^{(v)},\beta _{j_{v}}\oplus e^{(v)})$ can be transformed into

$$\begin{aligned} \left( \begin{array}{ll} D^{(v-1)},&{}\quad \alpha _{j_{v}}\alpha _{0}1_{s^{2}}+e^{(v-1)}\\ \vdots &{}\quad \vdots \\ D^{(v-1)},&{}\quad \alpha _{j_{v}}\alpha _{s-1}1_{s^{2}}+e^{(v-1)} \end{array} \right) . \end{aligned}$$

Similarly, for any column $h \in e^{(v)}$ with $v\ge 2$, we can obtain that $(\beta _{i_{v}}\oplus W_{i_{1}j_{1}i_{2}j_{2}\cdots i_{v-1}j_{v-1}}^{(v-1)},h)$ is a full factorial design with $2+v$ factors, where $\alpha _{j_{v}}\ne 0$. This completes the proof. $\square$

Proof of Theorem 1

From the construction of Algorithm 1, it is easy to obtain that X is an LHD and $D_1$ is an OA. D being an MCD follows from Lemma A.1 with $t=2$. Any column of X denoted as $\xi$ can be represented as $\xi =\eta _1\pm \eta _2 s$, where $(\eta _1,\eta _2)$ is an $s^2$ full factorial design. From the construction, it is obvious that $(\zeta ,\eta _2)$ is an $s^2$ full factorial design, where $\zeta$ is any column of $D_{1}$. Then D is an MCD following Lemma A.1. From Lemma 1, if M is orthogonal, then X is orthogonal. $\square$

Proof of Corollary 1

For any two columns in $X_j$, denoted as $\xi _1$ and $\xi _2$, they can be expressed as $\xi _1=\eta _{11}\pm \eta _{12}s$ and $\xi _2=\eta _{21}\pm \eta _{22}s$, respectively. And $(\eta _{11},\eta _{12})$, $(\eta _{21},\eta _{22})$ and $(\eta _{12},\eta _{22})$ are $s^2$ full factorial designs. Collapse the level x of $X_j$ by $\lfloor ({x+(s^{2}-1)/2})/{s} \rfloor$ to $\{0,1,\ldots ,s-1\}$. Then $\xi _1$ and $\xi _2$ become $\eta _{12}+((s-1)/2 )1_{s^2}$ and $\eta _{22}+((s-1)/2) 1_{s^2}$, respectively. As $(\eta _{12}+((s-1)/2 )1_{s^2},\eta _{22}+((s-1)/2) 1_{s^2})$ is an $s^2$ full factorial design, then (i) is correct. The proofs of (ii) and (iii) are similar to that of (i) and thus omitted here. $\square$

Proof of Theorem 2

• (i) It is easy to see that $D_1$ is an OA. From the construction of Algorithm 2, following the idea of Lemma 2, each $O_i$ is an $OA(s^u,2(u-1),s,u)$. So each $O_i^{(l)}$ is an $OA(s^u,2(u-1),s,u)$ with levels $\{-(s-1)/2,-(s-3)/2,\ldots ,(s-1)/2\}$. From the properties of the rotation matrix $R_{a1}$, $Z_i^{(l)}$ is an OLHD. Then $D_2$ is an OLHD from Lemma 1.

• (ii) Combing the idea of (iv) in Lemma 2 and Lemma A.1, we can obtain that $(D_1,D_2)$ is an OMCD.

• (iii) Similarly to the proof of Theorem 1, after the levels of $(z_{i,j}^{(l)},z_{i,j'}^{(l)})$ are collapsed to $\{0,1,\ldots ,s-1\}$, they become $(\eta _j+\frac{s-1}{2}1_{s^u},\eta _{j'}+\frac{s-1}{2}1_{s^u})$, where $\eta _j$ and $\eta _{j'}$ are the two corresponding columns in $O_i^{(l)}$. Then (iii) can be obtained straightforwardly.

• (iv) For $1\le j\le u-1$ and $u\le j'\le 2u-2$, $z_{i,j}^{(l)}$ and $z_{i,j'}^{(l)}$ can be represented as $z_{i,j}^{(l)}=\varphi d_{2i-1}\pm d_{2i,j}$ and $z_{i,j'}^{(l)}=\varphi 'd_{2i}+d_{2i-1,j'}$, where $\varphi$ and $\varphi '$ are rows with entries from a signed permutation of $s,s^2,\ldots ,s^{u-1}$. After collapsing $z_{i,j}^{(l)}$ and $z_{i,j'}^{(l)}$ to s and $s^{u-1}$ separately, the corresponding columns are $d_{2i-1,\tau }$ and $d_{2i}$ respectively. As $(d_{2i-1,\tau },d_{2i})$ forms a full factorial design, $(z_{i,j}^{(l)},z_{i,j'}^{(l)})$ achieves stratification on $s\times s^{u-1}$. The proofs for the other cases are similar.

Proof of Theorem 3

We only need to prove that $D_2$ is mirror-symmetric as other results follow from Theorem 2. As $O_i$ for $i=1,\ldots ,k$ and L are mirror-symmetric designs, then each $O_i^{(l)}$ is a mirror-symmetric design. So if b is a row of $O_i^{(l)}$, $-b$ is also one of its rows. Then for $Z_i^{(l)}$, if $bR_{a1}$ is one of the rows, $-bR_{a1}$ is also one of its rows. Thus $Z_i^{(l)}$ is mirror-symmetric, furthermore, $D_2$ is mirror-symmetric. $\square$

Proof of Theorem 4

• (i) From the definitions of $D^{(0)}$ and $e^{(0)}$, it is easy to check that $D_1$ is an $OA(s^{2+v},(s-1)^{v}+(s-(2p+2))s^{v},s,2)$. From Theorem 1 of Sun and Tang (2017), we can obtain that $D_2$ is an orthogonal $L(s^{2+v}, \lfloor s/2\rfloor ^{v}(p+1)2^{1+v})$.

• (ii) Let $\zeta$ be any column of $D_{1}$, then $\zeta \in e^{(v)}$ or $\zeta \in \beta _{i}\oplus D^{(v-1)}$ for $i=2p+2,\ldots ,s-1$. Let $\xi$ be any column of $D_{2}$, then $\xi \in {\widetilde{H}}_{j_{1}j_{2}\ldots j_{v}}^{(i)}$ for some corresponding $j_1,j_2,\ldots ,j_v$. And there exist $W_{i_{1}j_{1}i_{2}j_{2}\cdots i_{v-1}j_{v-1}}^{(v-1)}$, $\beta _{i_{v}}$ and $\beta _{j_{v}}$, such that $\xi$ can be represented as $\xi =\pm \lambda _{1}\pm \lambda _{2}s\pm \cdots \pm \lambda _{2+v}s^{2+v-1}$ where $(\lambda _{2},\ldots ,\lambda _{2+v})=\beta _{i_{v}}\oplus W_{i_{1}j_{1}i_{2}j_{2}\cdots i_{v-1}j_{v-1}}^{(v-1)}$, and $\lambda _{1} \in \beta _{j_{v}}\oplus D^{(v-1)}$. From Lemma A.2, we can obtain that $(\zeta ,\lambda _{2},\ldots ,\lambda _{2+v})$ is a full factorial design. Then $(D_1,D_2)$ is an MCD from Lemma A.1.

• (iii) According to the construction of $D_2$, it is easy to see that $D_2$ achieves stratification on $s\times s$ grids in any two dimensions.

$\square$

Proof of Corollary 3

• (i) It is clear that Part (i) can be obtained from Theorem 4 and Lemma A.2.

• (ii) From Theorem 4, $(D_1, \varGamma )$ is an OMCD. We only need to prove that $(\varPsi , \varGamma )$ is an OMCD. From the construction in Algorithm 4, we can obtain that

  $$\begin{aligned} H_{00\ldots 0}^{(1)}=\left( \beta _{0}\oplus H_{00\ldots 0j_{v-1}|j_{v-1}=0}^{(1)}, \beta _{1}\oplus H_{00\ldots 0j_{v-1}|j_{v-1}=0}^{(1)} \right) . \end{aligned}$$

  Let $\xi$ be any column of $\varGamma$, then $\xi =\pm \eta _{1}\pm \eta _{2}s\pm \cdots \pm \eta _{2+v}s^{2+v-1}$, where $\eta _{1} \in \beta _{0}\oplus H_{00\ldots 0j_{v-1}|j_{v-1}=0}^{(1)}$, $\eta _{h} \in \beta _{1}\oplus H_{00\ldots 0j_{v-1}|j_{v-1}=0}^{(1)}$ for $h=2,\ldots ,2+v$ and $(\eta _1,\eta _2,\ldots ,\eta _{2+v})$ is a full factorial design. Let $\zeta$ be any column of $\varPsi$, from the definition of $\varPsi$, we can obtain that $(\zeta ,\eta _{2},\ldots ,\eta _{2+v})$ is a full factorial design from Lemma A.2. From Lemma A.1, $(\varPsi , \varGamma )$ is an OMCD. Now Part (ii) can be proved.

$\square$

Proof of Theorem 5

Let $l_{i}^{(r)}$ denote the ith column of $L_r$, $r=1,2$. Without loss of generality, we only consider the column $l_{1}^{(1)} \oplus (s l_{1}^{(2)})$, which is the first column of Y. Since $l_{1}^{(1)}$ is a permutation on $\{-(s-1)/2,-(s-3)/2,\ldots ,(s-3)/2,(s-1)/2\}$, and $l_{1}^{(2)}$ is a permutation on $\{-(n/s-1)/2,-(n/s-3)/2,\ldots ,(n/s-3)/2,(n/s-1)/2\}$, then $l_{1}^{(1)}\oplus (s l_{1}^{(2)})$ is a permutation on $\{-(n-1)/2,-(n-3)/2,\ldots ,(n-3)/2,(n-1)/2\}$. Thus Y is an $L(n,k_{2})$. It is clear that $\lfloor (Y_{j}+(n-1)/2)/s\rfloor =L_{2}+(n/s-1)/2$ for $j=1,2,\ldots ,s$, where $L_{2}+(n/s-1)/2$ is an $L(n/s,k_{2})$ with entries from $\{0,1,\ldots ,n/s-1\}$. So $D=(D_1,Y)$ can be transformed into $(D_1,1_s\otimes (L_{2}+(n/s-1)/2))$ after level-collapsing of Y. Since $(D_1,1_s\otimes (1,2,\ldots ,n/s)^T)$ is an $OA(n,s^m(n/s),2)$, $(D_1,1_s\otimes (l_{i}^{(2)}+(n/s-1)/2))$ is an $OA(n,s^m(n/s),2)$ too, $i=1,2,\ldots ,k_2$. Therefore, D is an MCD with m qualitative factors and $k_{2}$ quantitative factors. $\square$

Proof of Theorem 6

For two vectors $a=(a_1,\ldots ,a_n)^T$ and $b=(b_1,\ldots ,b_n)^T$, define $\odot$ operator as

$$\begin{aligned} a\odot b=\sum _{i=1}^n a_ib_i. \end{aligned}$$

Let $l_{v}^{(r)}$ denote the vth column of $L_r$ for $r=1,2$, $l_{uv}^{(r)}$ denote the uth element of $l_{v}^{(r)}$, and $d_{it}=l_{i}^{(1)}\oplus (s l_{t}^{(2)})$, where $i=1,2,\ldots ,k_{1}$ and $t=1,2,\ldots ,k_{2}$. For $1\le i_{1},i_{2}\le k_{1}$ and $1\le t_{1},t_{2}\le k_{2}$,

$$\begin{aligned} d_{i_1t_1}\odot d_{i_2t_2}= & {} \sum _{u=1}^s(l_{ui_1}^{(1)}\oplus (sl_{t_1}^{(2)}))\odot (l_{ui_2}^{(1)}\oplus (sl_{t_2}^{(2)}))\\= & {} (n/s)\sum _{u=1}^s l_{ui_1}^{(1)}l_{ui_2}^{(1)}+ s^2\sum _{u=1}^s l_{t_1}^{(2)}\odot l_{t_2}^{(2)}\\= & {} (n/s)l_{i_1}^{(1)}\odot l_{i_2}^{(1)}+s^3 l_{t_1}^{(2)}\odot l_{t_2}^{(2)}. \end{aligned}$$

From the construction procedure, it is easy to see that

$$\begin{aligned} \rho (l_{i_{1}}^{(1)},l_{i_{2}}^{(1)})=\frac{12}{s(s^{2}-1)}l_{i_{1}}^{(1)}\odot l_{i_{2}}^{(1)},\ \ \rho (l_{t_{1}}^{(2)},l_{t_{2}}^{(2)})=\frac{12s^{3}}{n(n^{2}-s^{2})}l_{t_{1}}^{(2)}\odot l_{t_{2}}^{(2)} \end{aligned}$$

and

$$\begin{aligned} \rho (d_{i_{1}t_{1}},d_{i_{2}t_{2}})=\frac{12}{n(n^{2}-1)}d_{i_{1}t_{1}}\odot d_{i_{2}t_{2}}. \end{aligned}$$

Then we can obtain that

$$\begin{aligned} \rho (d_{i_{1}t_{1}},d_{i_{2}t_{2}}) =\lambda \rho (l_{i_{1}}^{(1)},l_{i_{2}}^{(1)})+(1-\lambda )\rho (l_{t_{1}}^{(2)},l_{t_{2}}^{(2)}), \text{ where } \lambda =\frac{s^{2}-1}{n^{2}-1}. \end{aligned}$$

• (i) For $k_1=k_2$ and $1\le j_1,j_2\le k_2$, we can have that $\rho _{j_1j_2}(Y)=\lambda \rho _{j_1j_2}(L_1)+(1-\lambda )\rho _{j_{1}j_{2}}(L_2)$. From Corollary 2 of Huang et al. (2014), if $L_1$ and $L_2$ are both OLHDs, then Y is OLHD; furthermore, Y is second-order orthogonal if $L_1$ and $L_2$ are both second-order orthogonal.

• (ii) It is straightforward to see that (1) is true.

$\square$

Appendix B

Table 5 An OMCD of 81 runs in Example 4

Table 6 An OMCD of 81 runs in Example 4 (Table 5 continued)

Table 7 An OMCD of 32 runs in Example 5