A class of mixed orthogonal arrays obtained from projection matrix inequalities

Abstract

By using the relationship between orthogonal arrays and decompositions of projection matrices and projection matrix inequalities, we present a method for constructing a class of new orthogonal arrays which have higher percent saturations. As an application of the method, many new mixed-level orthogonal arrays of run sizes 108 and 144 are constructed.

1 Introduction

An $n \times m$ matrix A, having $k_{i}$ columns with $p_{i}$ levels, $i=1,2,\ldots,r$, $m=\sum_{i=1}^{r}k_{i}$, $p_{i} \neq p_{j}$ for $i \neq j$, is called an orthogonal array (OA) of strength d and size n if each $n \times d$ submatrix of A contains all possible $1 \times d$ row vectors with the same frequency. Unless stated otherwise, we use the notation $L_{n}(p_{1}^{k_{1}} \cdots p_{r}^{k_{r}})$ for an OA of strength 2. An orthogonal array is said to have mixed levels if $r \geq2$. Orthogonal arrays have been used extensively in statistical design of experiments, computer science and cryptography. Constructions of OAs have been studied extensively in the literature; see Hedayet et al. [1, 2], Zhang et al. [3, 7], Pang et al. [5], Pang [6], Zhang et al. [7], Chen et al. [8], Du et al. [9], etc. Zhang et al. [3] present a method of construction of orthogonal arrays of strength two by using a relationship between orthogonal arrays and decompositions of projection matrices. In the construction of new mixed orthogonal arrays, two goals should be kept in mind, first, we want the orthogonal array to be as close to a saturated main-effect plan as possible so that there will be a large number of factors and second, we want the $p_{i}$, the number of levels, to be as large as possible so that the design has a high degree of flexibility (see Mandeli [10]). In this paper by using projection matrix inequalities and further exploring the relationship, matrix images of a class of orthogonal arrays can be found. Therefore we construct a class of new orthogonal arrays. If, as in Mandeli [10], we still define the percent saturation of an OA $L_{n}(p_{1}^{k_{1}} \cdots p_{r}^{k_{r}})$ to be $\sum_{i=1}^{r}k_{i}(p_{i}-1)/(n-1) \times100$%, then the orthogonal arrays constructed in this paper have higher percent saturations.

2 Basic concepts and main theorems

The following definitions, notations, and results are needed in the sequel.

Definition 1

A matrix A is said to be an orthogonal projection matrix if it is idempotent ($A^{2}=A$) and symmetric ($A^{T}=A$).

Definition 2

Suppose that an experiment is carried out according to an array $A=(a_{ij})_{n \times m}=(a_{1},\ldots,a_{m})$, and $Y=(Y_{1},\ldots,Y_{n})^{T}$ is the experimental data vector. In the analysis of variance $S_{j}^{2}$, the sum of squares of the jth factor, is defined as

$$ S_{j}^{2}=\sum_{i=0}^{p_{j}-1} \frac{1}{|I_{ij}|}\biggl(\sum_{s \in I_{ij}}Y_{s} \biggr)^{2}-\frac{1}{n}\Biggl(\sum_{s=1}^{n}Y_{s} \Biggr)^{2}, $$

(1)

where $I_{ij}=\{s:a_{sj}=i\}$ and $|I_{ij}|$ is the number of elements in $I_{ij}$. From (1), $S_{j}^{2}$ is a quadratic form in Y and there exists a unique symmetric matrix $A_{j}$ such that $S_{j}^{2}=Y^{T}A_{j}Y$. The matrix $A_{j}$ is called the matrix image (MI) of the jth column $a_{j}$ of A, denoted by $m(a_{j})=A_{j}$. The MI of a subarray of A is defined as the sum of the MIs of all its columns. In particular, we denote the MI of A by $m(A)$. Let $1_{r}$ be the $r \times1$ vector of 1’s. Then $m(1_{r})=P_{r}$ where $P_{r}={\frac{1}{r}}1_{r}1_{r}^{T}$.

Let $(r)=(0, \ldots, r-1)^{T}$, $e_{i}(r)=(0\cdots0\stackrel{i}{1}0\cdots 0)_{1\times r}^{T}$ and $I_{r}$ be the identity matrix of order r. Then $m((r))=\tau_{r}$ where $\tau_{r}=I_{r}-P_{r}$. The following permutation matrices are very useful:

$$N_{r}=e_{1}(r)e_{2}^{T}(r)+ \cdots+e_{r-1}(r)e_{r}^{T}(r)+e_{r}(r)e_{1}^{T}(r) $$

and

$$K(p,q)=\sum_{i=1}^{p} \sum _{j=1}^{q} e_{i}(p)e_{j}^{T}(q) \otimes e_{j}(q)e_{i}^{T}(p), $$

where ⊗ is the usual Kronecker product in the theory of matrices. Sometimes, it is necessary and easy to use the following properties of these two permutation matrices to obtain the orthogonal arrays needed:

$$\begin{aligned}& N_{r} (r)=1_{r}+(r), \quad \operatorname{mod} r, \\& K(p,q) \bigl(1_{q} \otimes(p)\bigr)=(p)\otimes1_{q},\qquad K(p,q) \bigl((q)\otimes(p)\bigr)=(p)\otimes(q) \end{aligned}$$

and

$$K(p,q) (P_{q} \otimes\tau_{p})K^{T}(p,q)= \tau_{p} \otimes P_{q},\qquad K(p,q) (\tau_{q} \otimes\tau_{p})K^{T}(p,q)=\tau_{p} \otimes \tau_{q}. $$

Lemma 1

For any permutation matrix S and any array L, $m(S(L \otimes1_{r}))=S(m(L) \otimes P_{r})S^{T}$, and $m(S(1_{r} \otimes L))=S(P_{r} \otimes m(L))S^{T}$.

Lemma 2

Let A be an OA of strength 1, i.e.,

$$A=(a_{1},\ldots,a_{m})=\bigl(S_{1} \bigl(1_{r_{1}} \otimes(p_{1})\bigr), \ldots, S_{m} \bigl(1_{r_{m}} \otimes(p_{m})\bigr)\bigr), $$

where $r_{j}p_{j}=n$ and $S_{i}$ is a permutation matrix, for $i=1,\ldots,m$. The following statements are equivalent.

• (1) A is an OA of strength 2.

• (2) $m(A)$ is a projection matrix.

• (3) $m(a_{i})m(a_{j})=0$ ($i \neq j$).

• (4) The projection matrix $\tau_{n}$ can be decomposed as $\tau_{n}=m(a_{1})+ \cdots+m(a_{m})+ \triangle$, where $\operatorname{rk}(\triangle)=n-1-\sum_{j=1}^{m}(p_{j}-1)$ is the rank of the matrix △.

Lemma 3

Suppose L and H are OAs satisfying $m(L)m(H)=0$. Then $K=(L,H)$ is also an OA.

Lemma 4

Let $(L,H)$ and K be OAs of run size n. Then $(K,H)$ is also an OA if $m(K) \leq m(L)$, where $A\le B$ means that the difference $B-A$ is nonnegative.

Lemmas 1, 2, 3, and 4 can be found in Zhang et al. [3].

Now we state the following theorems.

Theorem 1

If $L=[(r) \otimes1_{p}, H]$ is an OA, then $m(H)\le I_{r}\otimes \tau_{p}$. If $L=[1_{p} \otimes(r), K]$ is an OA, then $m(K)\le\tau_{p} \otimes I_{r}$.

Proof

From Lemma 2, we have $m(L)=\tau_{r}\otimes P_{p} +m(H)$. Since $m(L)\le\tau_{rp}$, $\tau_{rp}=\tau_{r}\otimes P_{p} +I_{r} \otimes \tau_{p}$ and $m((r) \otimes1_{p})=\tau_{r}\otimes P_{p}$, it follows that $m(H)\le I_{r}\otimes\tau_{p}$.

Similarly, we can prove that $m(K)\le\tau_{p} \otimes I_{r}$. □

Corollary

If p is a prime and $D(r,m;p)$ is a p-level difference matrix, then both $D(r,m;p)\oplus(p)$ and $(p)\oplus D(r,m;p)$ are OAs, and $m(D(r,m;p)\oplus(p))\le I_{r}\otimes \tau_{p}$ and $m((p)\oplus D(r,m;p)) \le\tau_{p} \otimes I_{r}$.

Proof

From Bose and Bush [11], we see that

$$L=\bigl[(r) \otimes1_{p} ,D(r,m;p)\oplus(p)\bigr] $$

is an OA. From Theorem 1, it follows that $m(D(r,m;p)\oplus(p)) \le I_{r} \otimes\tau_{p}$. Similarly, we can prove that $(p)\oplus D(r,m;p)$ is an OA and $m((p)\oplus D(r,m;p))\le\tau_{p}\otimes I_{r}$. □

Theorem 2

If p is a prime and $D(r,m;p)$ is a p-level difference matrix, then $D(r,m;p)\oplus0_{q}\oplus(p)$ is an OA, and $m(D(r,m;p)\oplus0_{q}\oplus(p))\le I_{r}\otimes P_{q}\otimes \tau_{p}$.

Proof

From Lemma 1 and Theorem 1, we have $m(D(r,m;p)\oplus (p)\oplus0_{q} )\le I_{r}\otimes\tau_{p}\otimes P_{q}$,

$$\begin{aligned}& m\bigl(D(r,m;p)\oplus0_{q}\oplus(p)\bigr) \\& \quad = m\bigl(\bigl(I_{r}\otimes K(q,p)\bigr) \bigl(D(r,m;p)\oplus(p) \oplus0_{q}\bigr)\bigr) \\& \quad \le \bigl(I_{r}\otimes K(q,p)\bigr) (I_{r}\otimes \tau_{p}\otimes P_{q}) \bigl(I_{r}\otimes K(q,p) \bigr)^{T} \\& \quad = I_{r}\otimes P_{q}\otimes\tau_{p}. \end{aligned}$$

 □

Theorem 3

If $\tau_{p}\otimes \tau_{q}=\sum_{i=1}^{k}S_{i}(\tau_{p}\otimes P_{q})S_{i}^{T}$ is an orthogonal decomposition of $\tau_{p}\otimes\tau_{q}$, then $I_{r}\otimes\tau_{p}\otimes \tau_{q}=\sum_{i=1}^{k}(I_{r}\otimes S_{i})(I_{r}\otimes\tau_{p}\otimes P_{q})(I_{r}\otimes S_{i})^{T}$ is an orthogonal decomposition of $I_{r}\otimes\tau_{p}\otimes \tau_{q}$. If there exists an OA H such that $m(H)\le I_{r} \otimes \tau_{p}$, then $L=[(I_{r}\otimes S_{1})(H\otimes1_{q}),\ldots, (I_{r}\otimes S_{k})(H\otimes1_{q})]$ is also an OA.

Proof

Since $\tau_{p}\otimes \tau_{q}=\sum_{i=1}^{k}S_{i}(\tau_{p}\otimes P_{q})S_{i}^{T}$, we have

$$\begin{aligned} I_{r}\otimes\tau_{p}\otimes \tau_{q} =&I_{r}\otimes\Biggl(\sum_{i=1}^{k}S_{i}( \tau_{p}\otimes P_{q})S_{i}^{T}\Biggr) \\ =& \sum_{i=1}^{k}(I_{r}I_{r}I_{r}) \otimes\bigl(S_{i}(\tau_{p}\otimes P_{q})S_{i}^{T} \bigr) . \end{aligned}$$

Using the property $(ABC)\otimes(DEF)=(A\otimes D)(B\otimes E)(C\otimes F)$, we obtain

$$I_{r}\otimes\tau_{p}\otimes \tau_{q}= \sum _{i=1}^{k}(I_{r}\otimes S_{i}) (I_{r}\otimes \tau_{p}\otimes P_{q}) (I_{r}\otimes S_{i})^{T}. $$

Because of the orthogonality in each step, the above decomposition of $I_{r}\otimes\tau_{p}\otimes \tau_{q}$ is orthogonal. By Lemma 1, we have

$$m\bigl((I_{r}\otimes S_{i}) (H\otimes1_{q}) \bigr)=(I_{r}\otimes S_{i}) (H\otimes1_{q}) (I_{r}\otimes S_{i})^{T}\le(I_{r}\otimes S_{i}) (I_{r}\otimes\tau_{p}\otimes P_{q}) (I_{r}\otimes S_{i})^{T}. $$

By Lemmas 3 and 4, we see that $(I_{r}\otimes S_{i})(H\otimes1_{q})$ and $(I_{r}\otimes S_{j})(H\otimes 1_{q})$ is orthogonal ($i\ne j$). It follows from Lemma 2 that L is an OA. □

3 Some examples

These matrix inequalities in Theorems 1, 2, and 3 are very useful for construction of orthogonal arrays. We illustrate their applications with some examples. We begin with OAs $L_{9}(3^{4})$ and $L_{16}(4^{5})$ and their properties:

$$L_{9}\bigl(3^{4}\bigr)=\bigl((3)\otimes1_{3},1_{3} \otimes(3),a,b\bigr)=\left [ \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} 0&0&0&1&1&1&2&2&2 \\ 0&1&2&0&1&2&0&1&2 \\ 0&1&2&1&2&0&2&0&1 \\ 0&1&2&2&0&1&1&2&0 \end{array}\displaystyle \right ]^{T}. $$

For this OA $L_{9}(3^{4})$, there exists a $9 \times9$ permutation $T_{2}$ (as follows) such that $((3)\otimes1_{3},1_{3}\otimes (3))=T_{2}(a,b)$. From Lemma 2, we have $m(L_{9}(3^{4}))=\tau_{9}$, $m(a,b)=\tau_{9}-\tau_{3}\otimes P_{3}-P_{3}\otimes\tau_{3}=\tau_{3}\otimes \tau_{3}$ and $m((3)\otimes1_{3},1_{3}\otimes(3))=T_{2}(\tau_{3}\otimes \tau_{3})T_{2}^{T}$. Hence

$$\tau_{9}=\sum_{i=1}^{2}T_{i}( \tau_{3}\otimes \tau_{3})T_{i}^{T}, $$

where $T_{1}=I_{9}$ and

$$T_{2}=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} 1&0&0&0&0&0&0&0&0 \\ 0&0&0&0&0&1&0&0&0 \\ 0&0&0&0&0&0&0&1&0 \\ 0&0&0&0&0&0&0&0&1 \\ 0&1&0&0&0&0&0&0&0 \\ 0&0&0&1&0&0&0&0&0 \\ 0&0&0&0&1&0&0&0&0 \\ 0&0&0&0&0&0&1&0&0 \\ 0&0&1&0&0&0&0&0&0 \end{array}\displaystyle \right ). $$

On the other hand, we have $\tau_{9}-\tau_{3}\otimes P_{3}-P_{3}\otimes \tau_{3}=\tau_{3}\otimes\tau_{3}$. Also, $\tau_{3}\otimes \tau_{3}=\sum_{i=1}^{2}Q_{i}(\tau_{3}\otimes P_{3})Q_{i}^{T}$, where $Q_{1}=\operatorname{diag}(I_{3},N_{3},N_{3}^{2})K(3,3)$ and $Q_{2}=\operatorname{diag}(I_{3},N_{3}^{2},N_{3})K(3,3)$.

Consider the orthogonal array $L_{16}(4^{5})$:

$$\begin{aligned} L_{16}\bigl(4^{5}\bigr) =&\bigl((4)\otimes1_{4},1_{4} \otimes(4),c,d,f\bigr) \\ =&\left [ \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} 0&0&0&0&1&1&1&1&2&2&2&2&3&3&3&3 \\ 0&1&2&3&0&1&2&3&0&1&2&3&0&1&2&3 \\ 0&1&2&3&1&0&3&2&2&3&0&1&3&2&1&0 \\ 0&1&2&3&2&3&0&1&3&2&1&0&1&0&3&2 \\ 0&1&2&3&3&2&1&0&1&0&3&2&2&3&0&1 \end{array}\displaystyle \right ]^{T}. \end{aligned}$$

From Lemma 2, we have $\tau_{16}=\sum_{i=1}^{2} S_{i}(\tau_{4} \otimes P_{4})S_{i}^{T}+ \tau_{4} \otimes\tau_{4}$, where $S_{1}=I_{16}$ and $S_{2}=K(4,4)$.

Moreover, we have $\tau_{4} \otimes \tau_{4}=\sum_{i=3}^{5}S_{i}(\tau_{4} \otimes P_{4})S_{i}^{T}$, where $S_{3}=\operatorname{diag}(Q_{1}',Q_{2}',Q_{3}',Q_{4}')K(4,4)$, $S_{4}=\operatorname{diag}(Q_{1}',Q_{3}',Q_{4}',Q_{2}')K(4,4)$, $S_{5}=\operatorname{diag}(Q_{1}',Q_{4}',Q_{2}',Q_{3}')K(4,4)$, $Q_{1}'=I_{4}$, $Q_{2}'=I_{2}\otimes N_{2}$, $Q_{3}'=N_{2}\otimes I_{2}$ and $Q_{4}'=N_{2}\otimes N_{2}$.

Example 1

(Construction of orthogonal arrays of run size 108)

Orthogonally decompose the projection matrix $\tau_{108}$ as follows:

$$\begin{aligned} \tau_{108} =&I_{12}\otimes\tau_{9}+ \tau_{12}\otimes P_{9} \\ =&I_{12}\otimes\tau_{3}\otimes\tau_{3}+I_{12} \otimes\tau_{3}\otimes P_{3}+I_{12}\otimes P_{3}\otimes\tau_{3} \\ &{}+(P_{6}\otimes \tau_{2}+P_{3}\otimes \tau_{2}\otimes\tau_{2}+ \tau_{3}\otimes I_{4}+P_{3}\otimes\tau_{2}\otimes P_{2}) \otimes P_{9} \\ =&I_{12}\otimes\tau_{3}\otimes\tau_{3} +I_{12}\otimes\tau_{3}\otimes P_{3}+(P_{6} \otimes\tau_{2}+P_{3}\otimes\tau_{2}\otimes \tau_{2})\otimes P_{9} \\ &{}+I_{12}\otimes P_{3}\otimes\tau_{3}+( \tau_{3}\otimes I_{4}+P_{3}\otimes \tau_{2}\otimes P_{2})\otimes P_{9} \\ =& I_{12}\otimes \tau_{3}\otimes\tau_{3} +\bigl(I_{12}\otimes\tau_{3}+(P_{6}\otimes \tau_{2}+P_{3}\otimes \tau_{2}\otimes \tau_{2})\otimes P_{3}\bigr)\otimes P_{3} \\ &{}+Q\bigl(\bigl(I_{12}\otimes\tau_{3}+( \tau_{3}\otimes I_{4}+P_{3}\otimes \tau_{2}\otimes P_{2})\otimes P_{3}\bigr)\otimes P_{3}\bigr)Q^{T}, \end{aligned}$$

where $Q=I_{12}\otimes K(3,3)$.

Now we want to find OAs $H_{1}$, $H_{2}$ and $H_{3}$ such that $m(H_{1}) \leq I_{12}\otimes\tau_{3}\otimes\tau_{3}$, and $m(H_{2}) \leq I_{12}\otimes\tau_{3}+(P_{6}\otimes\tau_{2}+P_{3}\otimes \tau_{2}\otimes\tau_{2})\otimes P_{3}$, and $m(H_{3}) \leq I_{12}\otimes\tau _{3}+(\tau_{3}\otimes I_{4}+P_{3}\otimes\tau_{2}\otimes P_{2})\otimes P_{3}$.

Let ⊕ be the Kronecker sum $(\operatorname{mod} 3)$ in ordinary matrix theory and $D(12,12,3)$ be a difference matrix as follows:

$$D(12,12,3)= \left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} 0&0&0&1&1&0&0&1&0&2&2&0 \\ 0&0&0&0&2&0&2&0&2&0&0&1 \\ 0&0&1&0&0&2&1&2&0&0&1&0 \\ 0&0&2&2&0&1&0&0&1&1&0&0 \\ 0&1&2&2&0&0&1&1&2&0&2&2 \\ 0&1&2&1&2&1&2&2&2&2&1&0 \\ 0&1&0&0&2&2&0&2&1&1&2&2 \\ 0&1&1&2&1&2&2&0&0&2&0&2 \\ 0&2&1&2&1&0&0&2&2&1&1&1 \\ 0&2&1&0&0&1&2&1&1&2&2&1 \\ 0&2&2&1&2&2&1&1&0&1&0&1 \\ 0&2&0&1&1&1&1&0&1&0&1&2 \end{array}\displaystyle \right ). $$

It follows from the corollary of Theorem 1 that $H=L_{36}(3^{12})=D(12,12,3)\oplus(3)$ is an OA and $m(H)\le I_{12}\otimes\tau_{3}$. From the property $\tau_{3}\otimes \tau_{3}=\sum_{i=1}^{2}Q_{i}(\tau_{3}\otimes P_{3})Q_{i}^{T}$ and Theorem 3, we have an orthogonal array $H_{1}$:

$$H_{1}=L_{108}\bigl(3^{24}\bigr)= \bigl[(I_{r}\otimes Q_{1}) (H\otimes1_{3}),(I_{r} \otimes Q_{2}) (H\otimes1_{3})\bigr] $$

and $m(H_{1}) \le I_{12}\otimes\tau_{3}\otimes\tau_{3}$.

From the orthogonal array $L_{36}(6^{2}3^{4}2^{9})$ in Zhang et al. [4], we can get an orthogonal array $L_{36}^{1}(6^{2}3^{4}2^{9})=K(12,3)L_{36}(6^{2}3^{4}2^{9})=[L_{12}(2^{9})\otimes1_{3}, L_{36}(6^{2}3^{4})]$ satisfying $m( L_{36}^{1}(6^{2}3^{4} 2^{9}) )\le\tau_{36}=I_{12}\otimes\tau_{3}+\tau _{12}\otimes P_{3}$. Deleting the orthogonal array $L_{12}(2^{9})\otimes1_{3}$ from $L_{36}^{1}(6^{2}3^{4}2^{9})$, we obtain an OA $H_{2}=L_{36}(6^{2}3^{4})$ whose MI satisfies $m(H_{2}) \leq I_{12}\otimes\tau_{3}+(P_{6}\otimes\tau_{2}+P_{3}\otimes \tau_{2}\otimes\tau_{2})\otimes P_{3}$.

By the definition of an OA, there exists a permutation matrix T such that $TL_{36}(6^{2}3^{4}2^{9})$ contains the two columns $1_{6}\otimes(2)\otimes1_{3}$ and $1_{3}\otimes((2)\oplus (2))\otimes1_{3}$. Deleting these two columns from $TL_{36}(6^{2}3^{4}2^{9})$, we obtain an OA $H_{3}=L_{36}(6^{2}3^{4}2^{7})$ satisfying $m(H_{3}) \leq I_{12}\otimes\tau_{3}+(\tau_{3}\otimes I_{4}+P_{3}\otimes\tau_{2}\otimes P_{2})\otimes P_{3}$.

Hence we can construct a new OA $L_{108}(6^{4}3^{32}2^{7})$ as follows:

$$ L_{108}\bigl(6^{4}3^{32}2^{7}\bigr)= \bigl[H_{1},H_{2}\otimes1_{3}, Q(H_{3} \otimes1_{3})\bigr]. $$

(2)

The percent saturation for this OA is 84.2%.

Also, similarly constructing of $H_{2}$ and by use of the orthogonal arrays $L_{36}(3^{2}2^{20})$, $L_{36}(3^{3}2^{13})$, $L_{36}(6^{1}3^{2}2^{11})$ and $L_{36}(3^{1}2^{27})$ in Zhang et al. [3], $L_{36}(6^{1}3^{1}2^{18})$ in Hedayet et al. [2], $L_{36} (3^{12}2^{11})$ and $L_{36}(6^{1}3^{8}2^{10})$ in Zhang et al. [4], we can obtain seven OAs $L_{36}(3^{2}2^{11})$, $L_{36}(3^{1}2^{18})$, $L_{36}(6^{1}3^{2}2^{2})$, $L_{36}(6^{1}3^{1}2^{9})$, $L_{36}(6^{2}3^{4})$, $L_{36}(3^{12}2^{2})$ and $L_{36}(6^{1}3^{8}2^{1})$ whose MIs are less than or equal to $I_{12}\otimes\tau_{3}+(P_{6}\otimes\tau_{2}+P_{3}\otimes\tau_{2}\otimes\tau _{2})\otimes P_{3}$.

On the other hand, by the definition of an OA, any OA of run size 36 with two factors having two levels can contain the two columns $1_{6}\otimes (2)\otimes1_{3}$ and $1_{3}\otimes((2)\oplus(2))\otimes1_{3}$ through row permutations. For example, there exists OA $L_{36}(9^{1}3^{16})$ in Example 2, $L_{36}(18^{1}2^{2})$, $L_{36}(3^{12}2^{11})$, $L_{36}(2^{35})$, $L_{36}(3^{1}2^{27})$, $L_{36}(6^{1}3^{12}2^{2})$, $L_{36}(3^{2}2^{20})$, $L_{36}(3^{3}2^{13})$, $L_{36}(6^{1}3^{2}2^{11})$ in Zhang et al. [3], $L_{36}(6^{1}3^{1}2^{18})$ in Hedayet et al. [2], $L_{36}(6^{1}3^{8}2^{10})$, $L_{36}(6^{2}3^{9}2^{3})$, $L_{36}(6^{2}3^{5}2^{2})$, $L_{36}(6^{3}2^{8})$ in Zhang et al. [4], and $L_{36}(6^{2}3^{1}2^{10})$, $L_{36}(6^{3}3^{1}2^{4})$, $L_{36}(6^{3}3^{2}2^{3})$ in Xu [12]. Deleting these two columns $1_{6}\otimes(2)\otimes1_{3}$ and $1_{3}\otimes((2)\oplus (2))\otimes1_{3}$ from these arrays, we can obtain 17 OAs $L_{36}(9^{1}3^{14})$, $L_{36}(18^{1})$, $L_{36}(3^{12}2^{9})$, $L_{36}(2^{33})$, $L_{36}(3^{1}2^{25})$, $L_{36}(6^{1}3^{12})$, $L_{36}(3^{2}2^{18})$, $L_{36}(3^{3}2^{11})$, $L_{36}(6^{1}3^{2}2^{9})$, $L_{36}(6^{1}3^{1}2^{16})$, $L_{36}(6^{1}3^{8}2^{8})$, $L_{36}(6^{2}3^{9}2^{1})$, $L_{36}(6^{2}3^{5})$, $L_{36}(6^{3}2^{6})$, $L_{36}(6^{2}3^{1}2^{8})$, $L_{36}$ $(6^{3}3^{1}2^{2})$, and $L_{36}(6^{3}3^{2}2^{1})$ whose MIs are less than or equal to $I_{12}\otimes\tau_{3}+(\tau_{3}\otimes I_{4}+P_{3}\otimes\tau_{2}\otimes P_{2})\otimes P_{3}$.

Replacing $H_{2}$ in (2) by those seven OAs and $H_{3}$ in (2) by these 17 OAs, respectively, we can construct $(7+1)\times(17+1)-1=143$ OAs such as $L_{108}(6^{5}3^{28}2^{6})$, etc. The orthogonal arrays constructed not only are new but also have higher percent saturations.

Example 2

(Construction of orthogonal arrays of run size 144)

From the above properties of OAs $L_{9}(3^{4})$ and $L_{16}(4^{5})$, we have

$$\tau_{9}=\sum_{i=1}^{2}T_{i}( \tau_{3}\otimes\tau_{3})T_{i}^{T}, $$

where $T_{1}=I_{9}$ and $T_{2}$ is the above permutation matrix and

$$\tau_{16}=\sum_{i=1}^{2} S_{i}(\tau_{4} \otimes P_{4})S_{i}^{T}+ \tau_{4} \otimes\tau_{4}. $$

By using the properties $I_{9}=I_{9}I_{9}I_{9}$, $T_{i}I_{9}T_{i}^{T}=I_{9}$, $S_{i}P_{16}S_{i}^{T}=P_{16}$ ($i=1,2$), $(ABC) \otimes(DEF)=(A \otimes D)(B \otimes E)(C \otimes F)$, we can orthogonally decompose $\tau_{144}$ as follows:

$$\begin{aligned} \tau_{144} = & I_{9} \otimes\tau_{16}+ \tau_{9} \otimes P_{16} \\ = & \sum_{i=1}^{2}(T_{i} \otimes S_{i}) \bigl((I_{9} \otimes\tau_{4} + \tau_{3} \otimes\tau_{3}\otimes P_{4}) \otimes P_{4}\bigr) (T_{i} \otimes S_{i})^{T} +I_{9} \otimes\tau_{4}\otimes\tau_{4}. \end{aligned}$$

From the $L_{16}(4^{5})$, we have

$$\tau_{4} \otimes\tau_{4}=\sum_{i=3}^{5}S_{i}( \tau_{4} \otimes P_{4})S_{i}^{T}. $$

Hence, $\tau_{144}$ can be further decomposed as

$$\begin{aligned} \tau_{144} =& \sum_{i=1}^{2}(T_{i} \otimes S_{i}) \bigl((I_{9} \otimes\tau_{4} + \tau_{3} \otimes\tau_{3}\otimes P_{4}) \otimes P_{4}\bigr) (T_{i} \otimes S_{i})^{T} \\ &{}+ \sum_{i=3}^{5} (I_{9} \otimes S_{i}) (I_{9} \otimes \tau_{4}\otimes P_{4}) (I_{9}\otimes S_{i})^{T}. \end{aligned}$$

Now we want to find OAs $H_{1}$ and $H_{2}$ such that $m(H_{1}) \leq I_{9} \otimes\tau_{4} +\tau_{3} \otimes\tau_{3} \otimes P_{4}$ and $m(H_{2}) \leq I_{9} \otimes\tau_{4}$.

By the definition of an OA and through row permutation, any OA with two 3-level columns can contain the two columns $(3)\otimes 1_{12}$, $1_{3} \otimes(3) \otimes1_{4}$. For example, from the result in Bose and Bush [11], we see that $K_{1}=[1_{3} \otimes L_{12}(12),(3) \oplus D(12,12,3)]$ is an OA, where $D(12,12,3)$ is a difference matrix in Example 1.

It is clear that $K_{1}$ contains two columns $(3 )\otimes1_{12}$ and $(3) \oplus(3)\otimes1_{4}$, and $(\operatorname{diag}(I_{3}, N_{3}^{2}, N_{3}) \otimes I_{4})[(3)\otimes1_{12}, (3)\oplus(3)\otimes1_{4}] =[(3)\otimes1_{12}, 1_{3}\otimes(3)\otimes1_{4}]$.

Set $K_{2} = (\operatorname{diag}(I_{3}, N_{3}^{2},N_{3}) \otimes I_{4})K_{1}$, we get an OA $K_{2} = L_{36}(12^{1} 3^{12})$, which contains two columns $(3)\otimes1_{12}$ and $1_{3}\otimes(3)\otimes1_{4}$. Deleting these two columns from $K_{2}$, we obtain an OA $H_{1}=L_{36}(12^{1} 3^{10})$, whose MI satisfies $m(H_{1}) =\tau_{36}-\tau_{3} \otimes P_{12}-P_{3} \otimes\tau_{3} \otimes P_{4}=I_{9} \otimes \tau_{4} +\tau_{3} \otimes\tau_{3}\otimes P_{4}$.

Now we need to find another OA $H_{2}$ such that $m(H_{2}) \leq I_{9} \otimes\tau_{4}$. By using row permutations of the orthogonal array $L_{36}(9^{1} 2^{16})$ (cf. [13]), we can obtain an OA $L_{36}(9^{1} 2^{16})$ as follows:

$$\begin{aligned}& L_{36}\bigl(9^{1}2^{16}\bigr) \\ & \quad = \left [ \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} 0 &x & x & x &-y & x & y & x &-y &-y &-z &-y & y & x &-x &-z & y\\ 1 &x & x & x & z & z & y &-x & x & z &-x & z & z &-x &-y & z & z\\ 2 &x &-y &-x & y & y & y &-x &-x & x &-y &-y & z & z &-x &-x & y\\ 3 &x &-z & y & x &-x & y & x &-y & z & y & y & x &-y & x &-x &-y\\ 4 &x & x &-x &-y &-y & y & z &-x & y &-y & y &-x & y & x & x & z\\ 5 &x &-x & z &-z &-x & y & x & y & y & y &-y &-y &-x &-x & x & y\\ 6 &x &-x &-x &-x & x & y & z & x &-x &-y & x &-z & z & y & y &-z\\ 7 &x & y &-y & z &-z & y &-z & y &-z & y &-x & y & z &-z & z &-y\\ 8 &x &-x & x & y &-z & y &-x & z &-y & x & y &-y & x & x &-y & y \end{array}\displaystyle \right ], \end{aligned}$$

where $i=(i,i,i,i)^{T}$, for $i=0,1,\ldots,8$, $x=(0,0,1,1)^{T}$, $-x=(1,1,0,0)^{T}$, $y=(0,1,0,1)^{T}$, $-y=(1,0,1,0)^{T}$, $z=(0,1,1,0)^{T}$ and $-z=(1,0,0,1)^{T}$.

Deleting the column $(9) \otimes1_{4}$ from the $L_{36}(9^{1} 2^{16})$, we obtain an OA $H_{2}=L_{36}( 2^{16})$. By using Theorem 1, we see that $m(H_{2}) \le I_{9}\otimes\tau_{4}$.

From Theorem 3 the decomposition of $\tau_{144}$, we construct a new OA $L_{144}(12^{2} 3^{20} 2^{48})$ as follows:

$$\begin{aligned}& L_{144}\bigl(12^{2} 3^{20} 2^{48}\bigr) \\& \quad =\bigl[(T_{1} \otimes S_{1}) (H_{1} \otimes1_{4}),(T_{2} \otimes S_{2}) (H_{1} \otimes1_{4}), (I_{9} \otimes S_{3}) (H_{2}\otimes1_{4}), \\& \qquad (I_{9} \otimes S_{4}) (H_{2}\otimes1_{4}), (I_{9} \otimes S_{5}) (H_{2}\otimes1_{4}) \bigr]. \end{aligned}$$

(3)

The percent saturation for this OA is 76.4%. In fact, the percent saturations for the arrays constructed by Wang and Wu [14] are between 49.7% and 53.8% and the highest percent saturation for the arrays with 144 runs constructed by Mandeli [10] is 62.2%.

Also, by the definition of an OA, any OA of run size 36 with two factors having three levels can contain the two columns $(3) \otimes 1_{12}$ and $1_{3} \otimes(3) \otimes1_{4}$ through row permutations. For example, there exist OAs $L_{36}(6^{2} 3^{8} 2^{1})$, $L_{36}(6^{1} 3^{2} 2^{11})$, $L_{36}( 3^{2} 2^{20})$ in Zhang et al. [3], $L_{36}(12^{1} 3^{12})$, $L_{36}(6^{1} 3^{12} 2^{2})$, $L_{36}( 3^{13} 2^{4})$, $L_{36}( 3^{12} 2^{11})$, $L_{36}(6^{2} 3^{5} 2^{2})$, $L_{36}(6^{1} 3^{8} 2^{10})$, $L_{36}(6^{2} 3^{4} 2^{9})$, $L_{36}(6^{1} 3^{9} 2^{3})$ in Zhang et al. [4], $L_{36}(6^{3} 3^{2} 2^{3})$, $L_{36}(6^{3} 3^{3} 2^{1})$ in Xu [12], $L_{36}(6^{3} 3^{7})$ in Finney [15], etc. having $(3) \otimes 1_{12}$ and $1_{3} \otimes(3) \otimes1_{4}$. Deleting these two columns, we can obtain 14 OAs $L_{36}(6^{3} 3^{5})$, $L_{36}(6^{2} 3^{6} 2^{1})$, $L_{36}(6^{1} 2^{11})$, $L_{36}(6^{1} 3^{10} 2^{2})$, and so on. Replacing $H_{1}$ in (3) by these OAs, respectively, we can construct 195 OAs such as $L_{144}(6^{6}3^{10} 2^{48})$, etc. The orthogonal arrays constructed not only are new but also have higher percent saturations.