A new method of finding component orthogonal arrays for order-of-addition experiments

Abstract

The order-of-addition experiments aim at determining the optimal order of m components such that the yields are optimized. The component orthogonal array (COA) allows to economically find out the optimal order by testing some carefully selected orders from all of the m! orders. This paper proposes a new method of finding COAs of broader run sizes. As an application of the new method, some COAs with 4, 5 and 6 components are tabulated. The D-efficiencies of the COAs found by the new method are investigated under the pair-wise ordering model.

Appendices

Appendix I: Proofs of theorems

Proof of Theorem

1. We first prove the “only if ” part. Suppose the fractional OofA design A is a COA(n, m, 2) which consists of the $i_1,i_2,\ldots ,i_n^{th}$ rows of $O_m$. Then the $i_1,i_2,\ldots ,i_n^{th}$ entries in X are 1’s with the remainder of the m! entries being 0’s. According to the definition of COA(n, m, 2), the order $c_ic_j$ appears totally $\lambda =\frac{n}{m(m-1)}$ times in the $i_1,i_2,\ldots ,i_n^{th}$ rows of each 2-column subdesign $(l_p,l_q)$ of $O_m$. This implies that there are totally $\lambda$ 1’s in the $i_1,i_2,\ldots ,i_n^{th}$ rows of each $e^{l_p,l_q}_{c_ic_j}$. Therefore, $e{^{l_p,l_q}_{c_ic_j}}'X=\lambda$ for any $p\ne q$ and $i\ne j$ with $1 \le p,q,i,j\le m$. Then $D_m'X=\lambda \mathbf{1}_{m^2(m-1)^2/2}$ follows from the structure of $D_m$ immediately.

Now, we prove the “if ” part. Suppose A consists of the $i_1,i_2,\ldots ,i_n^{th}$ rows of $O_m$ and X of A is the solution of Eq. (2). Clearly, $D_m'X=\lambda \mathbf{1}_{m^2(m-1)^2/2}$ means $e{^{l_p,l_q}_{c_ic_j}}'X=\lambda$. This implies that the order $c_ic_j$ appears totally $\lambda$ times in the $i_1,i_2,\ldots ,i_n^{th}$ rows of any 2-columns subdesign $(l_p,l_q)$ of $O_m$ for $p\ne q$ and $i\ne j$ with $1 \le p,q,i,j\le m$. Therefore A is a COA(n, m, 2). This completes the proof. $\square$

Proof of Theorem

2. Theorem  2 is a straightforward extension of Theorem  1. We omit the proof here. $\square$

Before proving Theorem  3, we first introduce some notation. For a fractional OofA design A, we use $A({c_{i_1},c_{i_2},c_{i_3},c_{i_4}})$ to denote the resulted design after dropping the other components from A, where $1\le {i_1}< {i_2}< {i_3}<{i_4}\le m$. Denote $n_{i}$ as the number of the $i^{th}$ row in $O_4$ which appears in $A({c_{i_1},c_{i_2},c_{i_3},c_{i_4}})$. Zhao et al. (2020) provided the necessary and sufficient conditions for constructing fractional OofA designs which have the same $D_{PWO}$-efficiency as the full OofA design. We introduce the results as Lemma  1 below.

Lemma 1

An n-row fractional OofA design A have the same $D_{PWO}$-efficiency as the full OofA design if and only if the following equations

$$\begin{aligned} \begin{array}{l} n_1+n_{24}= \frac{n}{12},\\ n_3+n_{22}= \frac{n}{12},\\ n_7+n_{18}= \frac{n}{12},\\ n_9+n_{16}= \frac{n}{12},\\ n_{10}+n_{15}= \frac{n}{12},\\ n_{13}+n_{12}= \frac{n}{12},\\ \end{array} \begin{array}{l} \ n_{1}+n_{2}+n_{3}+n_{9}= \frac{n}{6},\\ n_{1}+n_{3}+n_{5}+n_{7}= \frac{n}{6},\\ n_{1}+n_{7}+n_{13}+n_{19}= \frac{n}{6},\\ n_{3}+n_{9}+n_{15}+n_{21}= \frac{n}{6},\\ n_{7}+n_{8}+n_{13}+n_{15}= \frac{n}{6},\\ n_{9}+n_{11}+n_{13}+n_{15}= \frac{n}{6},\\ \end{array} \begin{array}{l} n_{4}-n_{3}-n_{9}-n_{14}= -\frac{n}{12},\\ n_{6}-n_{1}-n_{7}-n_{13}= -\frac{n}{12},\\ n_{14}-n_{9}-n_{13}-n_{15}= -\frac{n}{12},\\ n_{17}-n_{7}-n_{13}-n_{15}= -\frac{n}{12},\\ n_{20}-n_{1}-n_{3}-n_{7}= -\frac{n}{12},\\ n_{23}-n_{1}-n_{3}-n_{9}= -\frac{n}{12},\\ \end{array} \end{aligned}$$

hold for all $A(c_{i_1},c_{i_2},c_{i_3},c_{i_4})$’s with $1\le {i_1}< {i_2}< {i_3}<{i_4}\le m$.

Based on Lemma  1, we give the proof of Theorem  3 below.

Proof of Theorem

3. We first prove that the equations in Lemma  1 hold for COA(n, m, 4). By the definition of COA, each of the 24 orders of the 4 components $c_{i_1},c_{i_2},c_{i_3},c_{i_4}$ appears $\lambda =\frac{n}{m(m-1)(m-2)(m-3)}$ times in every 4-column sundesign of a COA(n, m, 4), which means that $A({c_{i_1},c_{i_2},c_{i_3},c_{i_4}})$ is a $\lambda \left( {\begin{array}{c}m\\ 4\end{array}}\right) =\frac{n}{24}$ replication of $O_4$. Clearly, for every 4-component combination $c_{i_1},c_{i_2},c_{i_3},c_{i_4}$ the equations in Lemma  1 are valid for $A({c_{i_1},c_{i_2},c_{i_3},c_{i_4}})$. As a result, a COA(n, m, 4) have the same $D_{PWO}$-efficiency as the full OofA design $O_m$. Of course, COA(n, m, t)’s with $4<t\le m$ have the same $D_{PWO}$-efficiencies as the full OofA design. This completes the proof. $\square$

Appendix II: Some useful design tables

Table 1 OofA design $O_3$ and its model matrices under the PWO and CP models

Table 2 Coefficient matrix $D'_4$ in Eq. (2)

Table 3 Equations obtained from Eq. (3)

Table 4 Rows and relative $D_{PWO}$-efficiencies of 2 COA(12, 4, 2)’s

Table 5 Rows and relative $D_{PWO}$-efficiencies of 36 COA(20, 5, 2)’s

Table 6 Rows and relative $D_{PWO}$-efficiencies of 2 COA(60, 5, 3)’s