Optimum mechanism for breaking the confounding effects of mixed-level designs

Abstract

Fractional factorial designs (FFD’s) are no doubt the most widely used designs in the experimental investigations due to their efficient use of experimental runs to study many factors simultaneously. One consequence of using FFD’s is the aliasing of factorial effects. Follow-up experiments may be needed to break the confounding. A simple strategy is to add a foldover of the initial design, the new fraction is called a foldover design. Combining a foldover design with the original design converts a design of resolution r into a combined design of resolution $r+1$. In this paper, we take the centered $L_2$-discrepancy $({\mathcal {CD}})$ as the optimality measure to construct the optimal combined design and take asymmetrical factorials with mixed two and three levels, which are most commonly used in practice, as the original designs. New and efficient analytical expressions based on the row distance of the ${\mathcal {CD}}$ for combined designs are obtained. Based on these new formulations, we present new and efficient lower bounds of the ${\mathcal {CD}}$. Using the new formulations and lower bounds as the benchmarks, we may implement a new algorithm for constructing optimal mixed-level combined designs. By this search heuristic, we may obtain mixed-level combined designs with low discrepancy.

Appendix

Proof of Theorem 1

From (3.1), we can put the uniformity criterion measured by ${\mathcal {CD}}$ based on any mixed two and three levels combined design as follows

$$\begin{aligned}{}[{\mathcal {CD}}({{\mathfrak {P}}}_c)]^2= & {} \left( \frac{13}{12}\right) ^{m}-\frac{2}{2n}\sum _{i=1}^{2n}\prod _{k=1}^{m}\Xi _{ik}+\frac{1}{(2n)^{2}}\sum _{i=1}^{2n}\sum _{j=1}^{2n}\prod _{k=1}^{m}\Xi _{ijk}\nonumber \\= & {} \left( \frac{13}{12}\right) ^{m}-\frac{2}{2n}\left[ \sum _{i=1}^{n}+\sum _{i=n+1}^{2n}\right] \prod _{k=1}^{m}\Xi _{ik}\nonumber \\&+\frac{1}{4n^{2}}\left[ \sum _{i=1}^{n}\sum _{j=1}^{n}+\sum _{i=n+1}^{2n}\sum _{j=n+1}^{2n}+2\sum _{i=n+1}^{2n}\sum _{j=1}^{n}\right] \prod _{k=1}^{m}\Xi _{ijk}\nonumber \\= & {} \left( \frac{13}{12}\right) ^{m}-\frac{2}{2n}\sum _{i=1}^{n}\left[ \prod _{k=1}^{m}\Xi _{ik}({{\mathfrak {P}}})+\prod _{k=1}^{m}\Xi _{ik}({{\mathfrak {P}}}_f)\right] \nonumber \\&+\frac{1}{4n^{2}}\sum _{i=1}^{n}\sum _{j=1}^{n}\left[ \prod _{k=1}^{m}\Xi _{ijk}({{\mathfrak {P}}})+\prod _{k=1}^{m}\Xi _{ijk}({{\mathfrak {P}}}_f)+2\prod _{k=1}^{m}\Xi _{ijk}(\gamma _k)\right] , \end{aligned}$$

where $\Xi _{ijk}({{\mathfrak {P}}})=\Xi _{ijk},~\Xi _{ik}({{\mathfrak {P}}})=\Xi _{ik}$ and $\Xi _{ik}({{\mathfrak {P}}}_f)=1+\frac{1}{2}|u_{ik}(\gamma _k)-\frac{1}{2}|-\frac{1}{2}|u_{ik}(\gamma _k)-\frac{1}{2}|^{2}$ and $\Xi _{ijk}({{\mathfrak {P}}}_f)=1+\frac{1}{2}|u_{ik}(\gamma _k)-\frac{1}{2}|+\frac{1}{2}|u_{jk}(\gamma _k)-\frac{1}{2}|-\frac{1}{2}|u_{ik}(\gamma _k)-u_{jk}{(\gamma _k)}|.$ From the above definition of $u_{ik}(\gamma _k)$ it is easy to show that $\Xi _{ik}({{\mathfrak {P}}}_f)=\Xi _{ik}({{\mathfrak {P}}})$ and $\Xi _{ijk}({{\mathfrak {P}}}_f)=\Xi _{ijk}({{\mathfrak {P}}}),$ which completes the proof. $\square$

Proof of Theorem 2

Following (3.3), we noted that

$$\begin{aligned} u^{(\gamma _k)}_{ik} \text{ and } u_{ik}\in \left\{ \begin{array}{ll} \left\{ \frac{1}{4}, \frac{3}{4}\right\} ,\quad k=1,\ldots ,m_1,\\ \left\{ \frac{1}{6}, \frac{1}{2}, \frac{5}{6}\right\} ,\quad k=m_1+1,\ldots ,m. \end{array} \right. \end{aligned}$$

Then

$$\begin{aligned} \Xi _{ik}= & {} \left\{ \begin{array}{ll} \frac{35}{32},\quad k=1,\ldots ,m_1,\\ \frac{10}{9}, \quad k=m_1+1,\ldots ,m,~u_{ik} \ne \frac{1}{2}, \end{array} \right. \\ \Xi _{ijk}= & {} \left\{ \begin{array}{ll} \frac{5}{4}, \quad k=1,\ldots ,m_1,~~ u_{ik}=u_{jk},\\ \frac{4}{3},\quad k=m_1+1,\ldots ,m,~~ u_{ik}= u_{jk}\ne \frac{1}{2} \end{array} \right. \end{aligned}$$

and

$$\begin{aligned}\Xi _{ijk}(\gamma _k)=\left\{ \begin{array}{ll} \frac{5}{4}, ~~ k=1,\ldots ,m_1,~u_{ik}=u_{jk}{(\gamma _k)},\\ \frac{4}{3},~~k=m_1+1,\ldots ,m,~~u_{ik}= u_{jk}{(\gamma _k)}\ne \frac{1}{2}. \end{array} \right. \end{aligned}$$

Therefore, we have

Substituting (i) and (ii) into (3.3), then (3.4) holds, which completes the proof. $\square$

Proof of Lemma 3

Note that the above sum term can be put in the following form

$$\begin{aligned} \sum _{l=1}^{n}\mu _1^{x_l}\mu _2^{y_l}=\sum _{l=1}^{n}e^{a x_l+b y_l}=\sum _{l=1}^{n}\sum _{\theta =0}^{\infty }\frac{{(a x_l+b y_l)^\theta }}{\theta !}=\sum _{\theta =0}^{\infty }\frac{{1 }}{\theta !}\sum _{l=1}^{n}z_{l}^\theta , \end{aligned}$$

where $a=\ln \mu _1,~b=\ln \mu _2$ and $z_{l}=a x_l+b y_l.$ Now, from Lemma 4 in Elsawah and Qin (2014), for any positive integer $\theta ,$ we have

$$\begin{aligned} \sum _{l=1}^{n}z_{l}^{\theta }\ge \left\{ \begin{array}{ll} q_1w_1^{\theta }+q_2w_2^{\theta }+(p_1-q_2)w_3^{\theta },&{}\quad p_1>q_2, \\ p_2w_1^{\theta }+p_1w_2^{\theta }+(q_2-p_1)w_4^{\theta },&{}\quad p_1\le q_2. \end{array} \right. \end{aligned}$$

Then, we get

$$\begin{aligned} \sum _{l=1}^{n}\mu _1^{x_l}\mu _2^{y_l}\ge & {} \left\{ \begin{array}{ll} \sum _{\theta =0}^{\infty }\frac{q_1{w_1}^{\theta }+q_2{w_2}^{\theta }+(p_1-q_2){w_3}^{\theta }}{\theta !},&{}\quad p_1>q_2,\\ \sum _{\theta =0}^{\infty }\frac{p_2{w_1}^{\theta }+p_1{w_2}^{\theta }+(q_2-p_1){w_4}^{\theta }}{\theta !},&{}\quad p_1\le q_2, \end{array} \right. \nonumber \\\ge & {} \left\{ \begin{array}{ll}q_1e^{w_1}+q_2e^{w_2}+(p_1-q_2)e^{w_3},&{}\quad p_1>q_2,\\ p_2e^{w_1}+p_1e^{w_2}+(q_2-p_1)e^{w_4},&{} \quad p_1\le q_2. \end{array} \right. \end{aligned}$$

$\square$