Uniform augmented q-level designs

Abstract

In many practical applications, follow-up experimental designs are commonly used to explore the relationship between inputs and outputs steps by steps. As some additional resources could be available to the experimenter after the first step, some additional runs or factors may be added in the follow-up stage. The issue of uniform augmented q-level designs is investigated in this paper. Using the level permutation technique, the uniformity of augmented q-level designs is discussed under the average mixture discrepancy, and new lower bounds of average mixture discrepancy for augmented designs are obtained, which can be served as the benchmark for searching uniform augmented designs. Numerical examples show that the uniform augmented designs can be constructed with high efficiency.

Appendix

Proof of Proposition 1

The proof of Proposition 1 is similar to the Lemma 1 of both Zhou and Xu (2014) and Wang et al. (2020), based on the notations of $\varphi _{ij}$, one can combine the formula (1) with formula (2) and demonstrate that

$$\begin{aligned} \sum _{{{\varvec{d}}}'\in {{\mathcal {P}}}({{\varvec{d}}})}\sum _{i=1}^n\prod _{k=1}^sf_1(x_{ik})= & {} n\left[ (q-1)!{\sum _{j=0}^{q-1}}f_1\left( \frac{2j+1}{2q}\right) \right] ^{s} \end{aligned}$$

and

$$\begin{aligned} \sum _{{{\varvec{d}}}'\in {{\mathcal {P}}}({{\varvec{d}}})}\sum _{i=1}^n\sum _{j=1}^n\prod _{k=1}^sf(x_{ik},x_{jk})= & {} (q!)^s\sum _{i=1}^n\sum _{j=1}^n\left[ \frac{\alpha \beta }{q(q-1)}\right] ^{\varphi _{ij}} \left[ \frac{\beta }{q(q-1)}\right] ^{s-\varphi _{ij}} \\= & {} (q!)^s\left[ \frac{\beta }{q(q-1)}\right] ^{s} \sum _{i=1}^n\sum _{j=1}^n\alpha ^{\varphi _{ij}}, \end{aligned}$$

where

$$\begin{aligned} \beta = \left\{ \begin{array}{cc} \displaystyle \frac{(14q^2-4q+3)(q-1)}{24q},&{}~\text{ for } \text{ odd }~q,\\ \displaystyle \frac{(7q-2)(q-1)}{12}, &{} ~\text{ for } \text{ even }~q. \end{array}\right. \end{aligned}$$

Then, the formula (2) can be rewritten as

• (i) for odd q,

  $$\begin{aligned}&\left[ \overline{MD}({{\varvec{d}}})\right] ^2\nonumber \\&\quad = \frac{1}{(q!)^s}\sum _{{{\varvec{d}}}'\in {{\mathcal {P}}}({{\varvec{d}}})} \left[ \left( \frac{7}{12}\right) ^{s}-\frac{2}{n}\sum _{i=1}^n\prod _{k=1}^s f_1(x_{ik})+\frac{1}{n^2}\sum _{i=1}^n\sum _{j=1}^n\prod _{k=1}^sf(x_{ik},x_{jk})\right] \nonumber \\&\quad = \left( \frac{7}{12}\right) ^s-2\left( \frac{7q^2+1}{12q^2}\right) ^s +\frac{1}{n^2}\left( \frac{14q^2-4q+3}{24q^2}\right) ^s \sum _{i=1}^n\sum _{j=1}^n \alpha ^{\varphi _{ij}}\nonumber \\&\quad = \left( \frac{7}{12}\right) ^s-2\left( \frac{7q^2+1}{12q^2}\right) ^s +\frac{1}{n^2}\left( \frac{14q^2-4q+3}{24q^2}\right) ^s\nonumber \\&\quad \qquad \left[ \sum _{i=1}^n\sum _{j(=i)=1}^n\alpha ^{\varphi _{ij}} +\sum _{i=1}^n\sum _{j(\ne i)=1}^n\alpha ^{\varphi _{ij}}\right] \nonumber \\&\quad = \left( \frac{7}{12}\right) ^s-2\left( \frac{7q^2+1}{12q^2}\right) ^s +\frac{1}{n}\left( \frac{6q^2+1}{8q^2}\right) ^s\nonumber \\&\qquad +\frac{1}{n^2}\left( \frac{14q^2-4q+3}{24q^2}\right) ^s \sum _{i=1}^n\sum _{j(\ne i)=1}^n \alpha ^{\varphi _{ij}}, \end{aligned}$$

  (18)

• (ii) similarly, for even q,

  $$\begin{aligned}&\left[ \overline{MD}({{\varvec{d}}})\right] ^2\nonumber \\&\quad = \left( \frac{7}{12}\right) ^s -2\left( \frac{28q^2+1}{48q^2}\right) ^s +\frac{1}{n^2}\left( \frac{7q-2}{12q}\right) ^s \sum _{i=1}^n\sum _{j=1}^n \alpha ^{\varphi _{ij}}\nonumber \\&\quad = \left( \frac{7}{12}\right) ^s -2\left( \frac{28q^2+1}{48q^2}\right) ^s\nonumber \\&\qquad +\frac{1}{n^2}\left( \frac{7q-2}{12q}\right) ^s \left[ \sum _{i=1}^n\sum _{j(=i)=1}^n\alpha ^{\varphi _{ij}} +\sum _{i=1}^n\sum _{j(\ne i)=1}^n\alpha ^{\varphi _{ij}}\right] \nonumber \\&\quad = \left( \frac{7}{12}\right) ^s -2\left( \frac{28q^2+1}{48q^2}\right) ^s +\frac{1}{n}\left( \frac{3}{4}\right) ^s+\frac{1}{n^2}\left( \frac{7q-2}{12q}\right) ^s \sum _{i=1}^n\sum _{j(\ne i)=1}^n \alpha ^{\varphi _{ij}},\nonumber \\ \end{aligned}$$

  (19)

  which completes the proof. $\square$

Proof of Theorem 1

Based on Proposition 1 and Lemma 1, as for the portion of the fourth term in (3) and (4), one can obtain

$$\begin{aligned} \sum _{i=1}^n\sum _{j(\ne i)=1}^n\alpha ^{\varphi _{ij}}= & {} \sum _{i=1}^n\sum _{j(\ne i)=1}^n\text {e}^ {\eta \varphi _{ij}}\nonumber \\= & {} \sum _{i=1}^n\sum _{j(\ne i)=1}^n\sum _{t=0}^\infty \frac{\left( \eta \varphi _{ij}\right) ^t}{t!}\nonumber \\\ge & {} \displaystyle \sum _{t=0}^\infty \displaystyle \frac{p[z_{(v)}]^{t}+g[z_{(v+1)}]^{t}}{t!}\nonumber \\= & {} p\text {e}^{z_{(v)}}+g\text {e}^{z_{(v+1)}}, \end{aligned}$$

(20)

which completes the proof. $\square$

Proof of Theorem 2

According to Theorem 1 and Proposition 2, based on the notations of $\varphi _{ij}$, similar to (20), we can obtain that

$$\begin{aligned} \sum _{i=1}^{n}\sum _{j(\ne i)=1}^{n}\alpha ^{\varphi _{ij}}\ge & {} p_{1}\text {e}^{z_{1(v)}}+g_{1}\text {e}^{z_{2(v+1)}}, \end{aligned}$$

(21)

$$\begin{aligned} \sum _{i=n+1}^{n+{n_1}}\sum _{j(\ne i)=n+1}^{n+{n_1}}\alpha ^{\varphi _{ij}}\ge & {} p_{2}\text {e}^{z_{2(h)}}+g_{2}\text {e}^{z_{2(h+1)}}, \end{aligned}$$

(22)

$$\begin{aligned} \sum _{i=1}^n\sum _{j=n+1}^{n+{n_1}}\alpha ^{\varphi _{ij}}\ge & {} p_{3}\text {e}^{z_{3(f)}}+g_{3}\text {e}^{z_{3(f+1)}}. \end{aligned}$$

(23)

Based on the Eqs. (9) and (10) in Proposition 2 and inequalities (21)-(23), Theorem 2 is obviously, which completes the proof. $\square$

Proof of Proposition 3

Based on the formula (14), for odd q, we are able to easily get

$$\begin{aligned}&\left[ \overline{MD}({{\varvec{d}}}_{c})\right] ^2\nonumber \\&\quad = \left( \frac{7}{12}\right) ^{s+s_2} -2\left( \frac{7q^2+1}{12q^2}\right) ^{s+s_2} +\frac{1}{(n+n_1)^2}\left[ \sum _{i=1}^{n}\sum _{j=1}^{n} \prod _{k=1}^{s+s_2}f(x_{ik},x_{jk})\right. \nonumber \\&\qquad \left. +\sum _{i=n+1}^{n+n_1}\sum _{j=n+1}^{n+n_1} \prod _{k=1}^{s+s_2}f(x_{ik},x_{jk})+2\sum _{i=1}^{n}\sum _{j=n+1}^{n+n_1} \prod _{k=1}^{s+s_2}f(x_{ik},x_{jk})\right] \nonumber \\&\quad = \left( \frac{7}{12}\right) ^{s+s_2} -2\left( \frac{7q^2+1}{12q^2}\right) ^{s+s_2}\nonumber \\&\qquad +\frac{1}{(n+n_1)^2}\left( \frac{6q^2+1}{8q^2}\right) ^{s_2} \sum _{i=1}^{n}\sum _{j=1}^{n} \prod _{k=1}^{s}f(x_{ik},x_{jk})\nonumber \\&\qquad +\frac{1}{(n+n_1)^2}\left( \frac{14q^2-4q+3}{24q^2}\right) ^{s+s_2} \left[ \sum _{i=n+1}^{n+n_1}\sum _{j=n+1}^{n+n_1} \alpha ^{\varphi _{ij}}+2\sum _{i=1}^{n}\sum _{j=n+1}^{n+n_1} \alpha ^{\varphi _{ij}}\right] \nonumber \\&\quad = \left( \frac{7}{12}\right) ^{s+s_2} -2\left( \frac{7q^2+1}{12q^2}\right) ^{s+s_2}\nonumber \\&\qquad +\frac{1}{(n+n_1)^2}\left( \frac{6q^2+1}{8q^2}\right) ^{s_2} \sum _{i=1}^n\sum _{j=1}^n\prod _{k=1}^sf(x_{ik},x_{jk})\nonumber \\&\qquad +\frac{1}{(n+n_1)^2}\left( \frac{14q^2-4q+3}{24q^2}\right) ^{s+s_2} \left[ \sum _{i=n+1}^{n+n_1}\sum _{j(\ne i)=n+1}^{n+n_1} \alpha ^{\varphi _{ij}}+2\sum _{i=1}^{n}\sum _{j=n+1}^{n+n_1} \alpha ^{\varphi _{ij}}\right] \nonumber \\&\qquad +\frac{n_1}{(n+n_1)^2}\left( \frac{6q^2+1}{8q^2}\right) ^{s+s_2}. \end{aligned}$$

(24)

According to the formula (18), we have

$$\begin{aligned} \frac{1}{n^2}\sum _{i=1}^n\sum _{j=1}^n\prod _{k=1}^sf(x_{ik},x_{jk})= & {} [\overline{MD}({{\varvec{d}}}_0)]^2 -\left( \frac{7}{12}\right) ^s+2\left( \frac{7q^2+1}{12q^2}\right) ^s. \end{aligned}$$

(25)

Thus, the formula (15) can be easily obtained by substituting Eq. (25) into the formula (24). Similarly, the formula (16) can be obtained as follow, while q is even

$$\begin{aligned} \left[ \overline{MD}({{\varvec{d}}}_{c})\right] ^2= & {} \left( \frac{7}{12}\right) ^{s+s_2} -2\left( \frac{28q^2+1}{48q^2}\right) ^{s+s_2} +\frac{n_1}{(n+n_1)^2}\left( \frac{3}{4}\right) ^{s+s_2}\nonumber \\&+\frac{n^2}{(n+n_1)^2}\left( \frac{3}{4}\right) ^{s_2} \left\{ [\overline{MD}({{\varvec{d}}}_0)]^2 -\left( \frac{7}{12}\right) ^s+2\left( \frac{28q^2+1}{48q^2}\right) ^s\right\} \nonumber \\&+\frac{1}{(n+n_1)^2}\left( \frac{7q-2}{12q}\right) ^{s+s_2} \left[ \sum _{i=n+1}^{n+n_1}\sum _{j(\ne i)=n+1}^{n+n_1} \alpha ^{\varphi _{ij}}\right. \nonumber \\&\left. +2\sum _{i=1}^{n}\sum _{j=n+1}^{n+n_1} \alpha ^{\varphi _{ij}}\right] , \end{aligned}$$

(26)

which completes the proof. $\square$