A novel non-heuristic search technique for constructing uniform designs with a mixture of two- and four-level factors: a simple industrial applicable approach

Abstract

Uniformly scatter the design points over the experimental domain is one of the most widely used techniques to construct optimal designs (called, uniform designs) for real-world high-dimensional experiments with limited resources and without model pre-specification. Uniform designs are robust to the underlying model assumption and thus experimenters do not need to specify the models of their experiments in advance before conducting them. A uniform design affords a good design space coverage that yields more accurate approximations globally using fewer experimental trials. The construction of uniform designs is a significant challenge due to the computational complexity. The existing techniques are extremely time-consuming (heuristic search techniques), difficult for non-mathematicians experimenters, and optimal results are not guaranteed. This paper tries to help non-mathematicians experimenters by providing a simple non-heuristic search technique for constructing uniform designs for experiments with a mixture of two- and four-level factors. The efficiency of the new technique is investigated theoretically and numerically. A comparison study between the new technique and the existing techniques is given. Furthermore, the applicability of the new technique for real-world applications is discussed and demonstrated by two real industrial experiments. The results show that the new designs that are generated by the new technique are better than the existing recommended designs.

Appendix

Proof of Theorem 1

For any design $\Gamma =(\gamma _{ik})_{i=1,k=1}^{a,b}$, define the following distances among its runs $D^{\ne 0}_{ij}(\Gamma )=\sharp \{k:|\gamma _{ik}-\gamma _{jk}|\ne 0\}$ and $D^{=2}_{ij}(\Gamma )=\sharp \{k:|\gamma _{ik}-\gamma _{jk}|=2\}$, and let $i^{\star } =i-2^{t-1}n,$ $j^{\star } =j-2^{t-1}n$ and $C_a^b= \{a,a+1,\ldots ,b\}.$ From the construction procedures in Sect. 3 (cf. Step 4) with some algebra, we get the following relationships between the extended designs and their images

$$\begin{aligned}&D^{\ne 0}_{ij}\left( \mathbf{Z}^{(t)}_1\right) = \left\{ \begin{array}{ll} 2D^{\ne 0}_{ij}\left( E^{(t-1)}(\mathbf{X}_1)\right) ,\,i,j\in C_1^{2^{t-1}n}; \\ 2D^{\ne 0}_{i^{\star }j^{\star }}\left( E^{(t-1)}(\mathbf{X}_1)\right) ,\,i,j\in C_{2^{t-1}n+1}^{2^{t}n}; \\ 2^{t-1}s_1,\,ow. \\ \end{array} \right. \end{aligned}$$

(6)

$$\begin{aligned}&D^{\ne 0}_{ij}\left( \mathbf{Z}^{(t)}_2\right) = \left\{ \begin{array}{ll} D^{\ne 0}_{ij}\left( E^{(t-1)}(\mathbf{X}_2)\right) ,\,i,j\in C_1^{2^{t-1}n}; \\ D^{\ne 0}_{i^{\star }j^{\star }}\left( E^{(t-1)}(\mathbf{X}_2)\right) ,\,i,j\in C_{2^{t-1}n+1}^{2^{t}n}; \\ 2^{t-1}s_2,\,ow. \\ \end{array} \right. \end{aligned}$$

(7)

$$\begin{aligned}&D^{=2}_{ij}\left( \mathbf{Z}^{(t)}_2\right) =\left\{ \begin{array}{ll} D^{\ne 0}_{ij}\left( E^{(t-1)}(\mathbf{X}_2)\right) ,\,i,j\in C_1^{2^{t-1}n}; \\ D^{\ne 0}_{i^{\star }j^{\star }}\left( E^{(t-1)}(\mathbf{X}_2)\right) ,\,i,j\in C_{2^{t-1}n+1}^{2^{t}n}; \\ 0,\,ow.\\ \end{array} \right. \end{aligned}$$

(8)

For any t-image design $\mathbf{Z}^{(t)}=\left(\mathbf{Z}^{(t)}_1\,~\,\mathbf{Z}^{(t)}_2\right)=\left(z^{(t)}_{ik}\right)_{i=1,k=1}^{2^tn,2^{t}s_1+2^{t-1}s_2}\in U\left(2^tn,2^{2^{t}s_1}4^{2^{t-1}s_2}\right)$ with $2^tn$ runs and a mixture of $2^{t}s_1$ factors with two levels $z^{(t)}_{ik_1}\in \{-1,1\},\,1\le i\le 2^tn,\,1\le k_1\le 2^{t}s_1$ and $2^{t-1}s_2$ factors with four levels $z^{(t)}_{ik_2}\in \{1,2,3,4\},\,1\le i\le 2^tn,\,2^{t}s_1+1\le k_2\le 2^{t-1}s_2,$ let $z^{(t)}_{ijk_1}=z^{(t)}_{ik_1}- z^{(t)}_{jk_1}$ and $z^{(t)}_{ijk_2}=z^{(t)}_{ik_2}- z^{(t)}_{jk_2}.$ Then, the analytical expressions of the discrepancies WD and LD in (4) and (5) can be rewritten in the following formulas, respectively

$$\begin{aligned} \left[ WD\left( \mathbf{Z}^{(t)}\right) \right] ^2&=\frac{1}{2^{2t}n^{2}}\left( \frac{1}{4}\right) ^{2^ts_1+2^ts_2} \sum _{i=1}^{2^tn}\sum _{j=1}^{2^tn} \prod _{k_1=1}^{2^ts_1}\left( 6-{|z^{(t)}_{ijk_1}|}+\frac{1}{4}{|z^{(t)}_{ijk_1}|^2}\right) \nonumber \\&{}\quad \times \prod _{k_2=2^ts_1+1}^{2^ts_1+2^{t-1}s_2}\left( 24-4{|z^{(t)}_{ijk_2}|}+{|z^{(t)}_{ijk_2}|^2}\right) -\left( \frac{4}{3}\right) ^{2^ts_1+2^{t-1}s_2} \end{aligned}$$

(9)

and

$$\begin{aligned} \left[ LD\left( \mathbf{Z}^{(t)}\right) \right] ^2&=\frac{1}{2^{2t}n^{2}}\left( \frac{1}{2}\right) ^{2^ts_1+2^ts_2} \sum _{i=1}^{2^tn}\sum _{j=1}^{2^tn} \prod _{k_1=1}^{2^ts_1}\left( 2-\min \left\{ \frac{1}{2}{|z^{(t)}_{ijk_1}|},2-\frac{1}{2}{|z^{(t)}_{ijk_1}|}\right\} \right) \nonumber \\&{}\quad \times \prod _{k_2=2^ts_1+1}^{2^ts_1+2^{t-1}s_2}\left( 4-\min \left\{ {|z^{(t)}_{ijk_2}|},4-{|z^{(t)}_{ijk_2}|}\right\} \right) -\left( \frac{3}{4}\right) ^{2^ts_1+2^{t-1}s_2}.\end{aligned}$$

(10)

For any factor with four levels $z^{(t)}_{ik_2}\in \{1,2,3,4\}$ with some algebra, we get

$$\begin{aligned}&4-\min \left\{ {|z^{(t)}_{ijk_2}|},4-{|z^{(t)}_{ijk_2}|}\right\} =\left\{ \begin{array}{ll} 4,\,\,|z^{(t)}_{ijk_2}|=0,\\ 3,\,\,|z^{(t)}_{ijk_2}|=1,\\ 2,\,\,|z^{(t)}_{ijk_2}|=2,\\ 3,\,\,|z^{(t)}_{ijk_2}|=3.\\ \end{array} \right. \end{aligned}$$

(11)

$$\begin{aligned}&24-4{|z^{(t)}_{ijk_2}|}+{|z^{(t)}_{ijk_2}|^2}=\left\{ \begin{array}{ll} 24,\,\,|z^{(t)}_{ijk_2}|=0,\\ 21,\,\,|z^{(t)}_{ijk_2}|=1,\\ 20,\,\,|z^{(t)}_{ijk_2}|=2,\\ 21,\,\,|z^{(t)}_{ijk_2}|=3.\\ \end{array} \right. \end{aligned}$$

(12)

For any factor with two levels $z^{(t)}_{ik_1}\in \{-1,1\}$ with some algebra, we get

$$\begin{aligned}&2-\min \left\{ \frac{1}{2}{|z^{(t)}_{ijk_1}|},2-\frac{1}{2}{|z^{(t)}_{ijk_1}|}\right\} =\left\{ \begin{array}{ll} 2,\,\,|z^{(t)}_{ijk_1}|=0,\\ 1,\,\,|z^{(t)}_{ijk_2}|=2.\\ \end{array} \right. \end{aligned}$$

(13)

$$\begin{aligned}&6-{|z^{(t)}_{ijk_1}|}+\frac{1}{4}{|z^{(t)}_{ijk_1}|^2}=\left\{ \begin{array}{ll} 6,\,\,|z^{(t)}_{ijk_1}|=0,\\ 5,\,\,|z^{(t)}_{ijk_2}|=2.\\ \end{array} \right. \end{aligned}$$

(14)

From (11)–(14), the formulas (9) and (10) of the discrepancies WD and LD can be rewritten in the following efficient analytical expressions, respectively

$$\begin{aligned} \left[ {WD}\left( \mathbf{Z}^{(t)}\right) \right] ^2= & {} \frac{1}{4^{t}n^{2}}\left( \frac{3}{2}\right) ^{2^{t}s_1+2^{t-1}s_2}\sum _{i,j\in C_1^{2^{t}n}} \left( \frac{5}{6}\right) ^{D^{\ne 0}_{ij}\left( \mathbf{Z}^{(t)}_1\right) } \left( \frac{7}{8}\right) ^{D^{\ne 0}_{ij}\left( \mathbf{Z}^{(t)}_2\right) } \left( \frac{20}{21}\right) ^{D^{=2}_{ij}\left( \mathbf{Z}^{(t)}_2\right) } \nonumber \\&-\left( \frac{4}{3}\right) ^{2^{t}s_1+2^{t-1}s_2} \end{aligned}$$

(15)

and

$$\begin{aligned} \left[ {LD}\left( \mathbf{Z}^{(t)}\right) \right] ^2= & {} \frac{1}{4^{t}n^{2}}\sum _{i,j\in C_1^{2^{t}n}} \left( \frac{3}{4}\right) ^{D^{\ne 0}_{ij}\left( \mathbf{Z}^{(t)}_2\right) } \left( \frac{1}{2}\right) ^{D^{\ne 0}_{ij}\left( \mathbf{Z}^{(t)}_1\right) } \left( \frac{2}{3}\right) ^{D^{=2}_{ij}\left( \mathbf{Z}^{(t)}_2\right) }\nonumber \\&-\left( \frac{3}{4}\right) ^{2^{t}s_1+2^{t-1}s_2}. \end{aligned}$$

(16)

From (6)–(8), the sum term in (15) can be written as follows

$$\begin{aligned}&\sum _{i,j\in C_1^{2^{t}n}} \left( \frac{5}{6}\right) ^{D^{\ne 0}_{ij}\left( \mathbf{Z}^{(t)}_1\right) } \left( \frac{7}{8}\right) ^{D^{\ne 0}_{ij}\left( \mathbf{Z}^{(t)}_2\right) } \left( \frac{20}{21}\right) ^{D^{=2}_{ij}\left( \mathbf{Z}^{(t)}_2\right) } \nonumber \\&{}\quad =\sum _{i,j\in C_1^{2^{t-1}n}}\left( \frac{5}{6}\right) ^{2D^{\ne 0}_{ij}\left( E^{(t-1)}(\mathbf{X}_1)\right) } \left( \frac{7}{8}\right) ^{D^{\ne 0}_{ij}\left( E^{(t-1)}(\mathbf{X}_2)\right) } \left( \frac{20}{21}\right) ^{D^{\ne 0}_{ij}\left( E^{(t-1)}(\mathbf{X}_2)\right) } \nonumber \\&\qquad +\sum _{i\in C_{2^{t-1}n+1}^{2^tn}}\sum _{j\in C_1^{2^{t-1}n}}\left( \frac{5}{6}\right) ^{2^{t-1}s_1} \left( \frac{7}{8}\right) ^{2^{t-1}s_2} +\sum _{i\in C_1^{2^{t-1}n}} \sum _{j\in C_{2^{t-1}n+1}^{2^tn}} \left( \frac{5}{6}\right) ^{2^{t-1}s_1} \left( \frac{7}{8}\right) ^{2^{t-1}s_2} \nonumber \\&\qquad +\sum _{i,j\in C_{2^{t-1}n+1}^{2^{t}n}}\left( \frac{5}{6}\right) ^{2D^{\ne 0}_{i^{\star }j^{\star }}\left( E^{(t-1)}(\mathbf{X}_1)\right) } \left( \frac{7}{8}\right) ^{D^{\ne 0}_{i^{\star }j^{\star }}\left( E^{(t-1)}(\mathbf{X}_2)\right) } \left( \frac{20}{21}\right) ^{D^{\ne 0}_{i^{\star }j^{\star }}\left( E^{(t-1)}(\mathbf{X}_2)\right) } \nonumber \\&{}\quad = 2\left( \sum _{i,j\in C_1^{2^{t-1}n}}\left( \frac{5}{6}\right) ^{D^{\ne 0}_{ij}\left( E^{(t-1)}(\mathbf{X}_2)\right) +2D^{\ne 0}_{ij}\left( E^{(t-1)}(\mathbf{X}_1)\right) } +4^{t-1}n^2\left( \frac{5}{6}\right) ^{2^{t-1}s_1} \left( \frac{7}{8}\right) ^{2^{t-1}s_2}\right) . \end{aligned}$$

(17)

By the same technique, the sum term in (16) can be written as follows

$$\begin{aligned}&\sum _{i,j\in C_1^{2^{t}n}} \left( \frac{1}{2}\right) ^{D^{\ne 0}_{ij}\left( \mathbf{Z}^{(t)}_1\right) } \left( \frac{3}{4}\right) ^{D^{\ne 0}_{ij}\left( \mathbf{Z}^{(t)}_2\right) } \left( \frac{2}{3}\right) ^{D^{=2}_{ij}\left( \mathbf{Z}^{(t)}_2\right) } \nonumber \\&{}\quad = 2\left( \sum _{i,j\in C_1^{2^{t-1}n}}\left( \frac{1}{2}\right) ^{D^{\ne 0}_{ij}\left( E^{(t-1)}(\mathbf{X}_2)\right) +2D^{\ne 0}_{ij}\left( E^{(t-1)}(\mathbf{X}_1)\right) } +4^{t-1}n^2\left( \frac{1}{2}\right) ^{2^{t-1}s_1} \left( \frac{3}{4}\right) ^{2^{t-1}s_2}\right) . \end{aligned}$$

(18)

From Theorem 1 in Elsawah (2021b) with some algebra, we get the following relationship between the dissimilarity among the runs of any sub-design $\mathbf{X}_\sigma \in U(n,2^{s_\sigma }),\,\sigma =1,2$ and its corresponding extended design $E^{(t-1)}(\mathbf{X}_\sigma )\in U\left(2^{t-1}n,2^{2^{t-1}s_\sigma }\right)$

$$\begin{aligned} D^{\ne 0}_{ij}\left( E^{(t-1)}(\mathbf{X}_\sigma )\right) = \left\{ \begin{array}{ll} 2^{t-1}D^{\ne 0}_{(i-kn)(j-kn)}(\mathbf{X}_\sigma ),\,i,j\in C_{kn+1}^{kn+n},\,k\in C_ 0^{2^{t-1}-1}; \\ 2^{t-2}s_\sigma , \,ow. \\ \end{array} \right. \end{aligned}$$

(19)

From (19), the sum in (17) can be rewritten as follows

$$\begin{aligned}&\sum _{i,j\in C_1^{2^{t-1}n}}\left( \frac{5}{6}\right) ^{D^{\ne 0}_{ij}\left( E^{(t-1)}(\mathbf{X}_2)\right) +2D^{\ne 0}_{ij}\left( E^{(t-1)}(\mathbf{X}_1)\right) } \nonumber \\&{}\quad =\sum _{k\in C_0^{2^{t-1}-1}}\sum _{i,j\in C_{kn+1}^{(k+1)n}}\left( \frac{5}{6}\right) ^{2^{t-1}\left( 2D^{\ne 0}_{(i-kn)(j-kn)}(\mathbf{X}_1)+D^{\ne 0}_{(i-kn)(j-kn)}(\mathbf{X}_2)\right) } \nonumber \\&\qquad +\sum _{k,\ell (\ne \ell )\in C_0^{2^{t-1}-1}} \sum _{i\in C_{kn+1}^{(k+1)n}}\sum _{j\in C_{\ell n+1}^{(\ell +1)n}}\left( \frac{5}{6}\right) ^{2^{t-2}(2s_1+s_2)} \nonumber \\&{}\quad =2^{t-1}\sum _{i,j\in C_1^{n}}\left( \frac{5}{6}\right) ^{2^{t-1}\left( 2D^{\ne 0}_{ij}(\mathbf{X}_1)+D^{\ne 0}_{ij}(\mathbf{X}_2)\right) }+2^{t-1}(2^{t-1}-1)n^2\left( \frac{5}{6}\right) ^{2^{t-2}(2s_1+s_2)}. \end{aligned}$$

(20)

By the same technique, the sum term in (18) can be written as

$$\begin{aligned}&\sum _{i,j\in C_1^{2^{t-1}n}}\left( \frac{1}{2}\right) ^{D^{\ne 0}_{ij}\left( E^{(t-1)}(\mathbf{X}_2)\right) +2D^{\ne 0}_{ij}\left( E^{(t-1)}(\mathbf{X}_1)\right) } \nonumber \\&{}\quad =2^{t-1}\sum _{i,j\in C_1^{n}}\left( \frac{1}{2}\right) ^{2^{t-1}\left( D^{\ne 0}_{ij}(\mathbf{X}_2)+2D^{\ne 0}_{ij}(\mathbf{X}_1)\right) }+2^{t-1}(2^{t-1}-1)n^2\left( \frac{1}{2}\right) ^{2^{t-2}(s_2+2s_1)}. \end{aligned}$$

(21)

Combining (15)–(18) and (20)–(21) with some algebra, the proof can be completed. $\square$

Proof of Corollary 1

The proof is obvious from Theorem 1, where

$$\begin{aligned} \prod _{r=1}^{2} \Omega ^{2^{t-r+1}h_{ij}(\mathbf{X}_r)}=\Omega ^{2^{t-1}(h_{ij}(\mathbf{X})+h_{ij}(\mathbf{X}_1))}=\Omega ^{2^{t-1}h_{ij}(\mathbf{X}^{\star })}. \end{aligned}$$

(22)

$\square$

Proof of Corollary 2

Since the base design $\mathbf{X}$ is a saturated orthogonal design $\mathbf{X}\in SOD(n,2^{s}),$ then $h_{ij}(\mathbf{X})=\frac{n}{2}$ for any $i\ne j$ (cf. Mukerjee and Wu 1995). From Theorem 1 and (22), the proof can be completed. $\square$

Proof of Corollary 3

The proof is obvious from the proof of Corollary 2. $\square$

Proof of Theorem 2

From the definition of $H_{\mu _1,\mu _2}(\mathbf{X}),$ we get

$$\begin{aligned} \sum _{i=1}^{n}\sum _{j=1}^{n}\left( \prod _{r=1}^{2} \Omega ^{2^{t-r+1}h_{ij}(\mathbf{X}_r)}\right)= & {} \sum _{\mu _1=0}^{s_1}\sum _{\mu _2=0}^{s_2} \Omega ^{2^{t}\mu _1}\Omega ^{2^{t-1}\mu _2}\sharp \{(i,j):h_{ij}(\mathbf{X}_{1})=\mu _1,h_{ij}(\mathbf{X}_{2})=\mu _2\} \nonumber \\ {}= & {} n \sum _{\mu _1=0}^{s_1}\sum _{\mu _2=0}^{s_2} \Omega ^{2^{t}\mu _1}\Omega ^{2^{t-1}\mu _2} H_{\mu _1,\mu _2}(\mathbf{X}). \end{aligned}$$

(23)

From Theorem 1 and (23), the proof can be completed. $\square$

Proof of Corollary 4

The proof can be obtained from Corollary 1 by the same technique of (23). $\square$

Proof of Corollary 5

For any design without replicates $\mathbf{X}_1\in U(n,2^{s_1}),$ we get $h_{ij}(\mathbf{X}_1)=0$ for only $i=j$ and thus we get

$$\begin{aligned} \sum _{j\ne i=1}^{n} \Omega ^{2^{t-1}h_{ij}(\mathbf{X}_1)}= & {} \sum _{\mu =1}^{s_1}\Omega ^{2^{t-1}\mu }\sharp \{(i,j):h_{ij}(\mathbf{X}_{1})=\mu \} =n \sum _{\mu =1}^{s_1}\Omega ^{2^{t-1}\mu } H_{\mu }(\mathbf{X}_1). \end{aligned}$$

(24)

The proof can be completed from Corollary 2 and (24). $\square$

Proof of Corollary 6

The proof is obvious from Corollary 3 by the same way of (24). $\square$

Proof of Theorem 3

By the same technique of Theorem 2 in Elsawah (2020) with some algebra, the inverse of the MacWilliam transformations is given as follows

$$\begin{aligned} H_{\mu _1,\mu _2}(\mathbf{X})= & {} \frac{n}{2^{s}}\sum _{\delta _1=0}^{s_1}\sum _{\delta _2=0}^{s_2}\left( \prod _{r=1}^2P_{\mu _r}(\delta _r,2^{s_r})\right) W_{\delta _1,\delta _2}(\mathbf{X}). \end{aligned}$$

(25)

From (25), we get

$$\begin{aligned}&\sum _{\mu _1=0}^{s_1}\sum _{\mu _2=0}^{s_2}\left( \prod _{r=1}^{2}\Omega ^{2^{t-r+1}\mu _r}\right) H_{\mu _1,\mu _2}(\mathbf{X})\nonumber \\&{}\quad = \frac{n}{2^{s}}\sum _{\mu _1=0}^{s_1}\sum _{\mu _2=0}^{s_2}\left( \prod _{r=1}^{2}\Omega ^{2^{t-r+1}\mu _r}\right) \sum _{\delta _1=0}^{s_1}\sum _{\delta _2=0}^{s_2}\left( \prod _{r=1}^2P_{\mu _r}(\delta _r,2^{s_r})\right) W_{\delta _1,\delta _2}(\mathbf{X}) \nonumber \\&{}\quad =\frac{n}{2^{s}} \sum _{\delta _1=0}^{s_1}\sum _{\delta _2=0}^{s_2} W_{\delta _1,\delta _2}(\mathbf{X}) \prod _{r=1}^{2}\left( \sum _{\mu _r=0}^{s_r}\Omega ^{2^{t-r+1}\mu _r}P_{\mu _r}(\delta _r,2^{s_r})\right) . \end{aligned}$$

(26)

From the orthogonality property of the Krawtchouk polynomials, we get

$$\begin{aligned} \sum _{\mu _r=0}^{s_r}\Omega ^{2^{t-r+1}\mu _r}P_{\mu _r}(\delta _r,2^{s_r}) =\left( 1+\Omega ^{2^{t-r+1}}\right) ^{s_r}\left( \frac{1-\Omega ^{2^{t-r+1}}}{1+\Omega ^{2^{t-r+1}}}\right) ^{\delta _r}.\end{aligned}$$

(27)

Combining (26) and (27), we get

$$\begin{aligned}&\sum _{\mu _1=0}^{s_1}\sum _{\mu _2=0}^{s_2}\left( \prod _{r=1}^{2}\Omega ^{2^{t-r+1}\mu _r}\right) H_{\mu _1,\mu _2}(\mathbf{X})\nonumber \\&= \frac{n}{2^{s}} \left( \prod _{r=1}^{2}\left( 1+\Omega ^{2^{t-r+1}}\right) ^{s_r}\right) \sum _{\delta _1=0}^{s_1}\sum _{\delta _2=0}^{s_2} W_{\delta _1,\delta _2}(\mathbf{X}) \left( \prod _{r=1}^{2}\left( \frac{1-\Omega ^{2^{t-r+1}}}{1+\Omega ^{2^{t-r+1}}}\right) ^{\delta _r}\right) . \end{aligned}$$

(28)

Combining Theorem 2 and (28), the proof can be completed. $\square$

Proof of Corollary 7

The proof can be obtained from Corollary 4 by the same technique of the proof of Theorem 3. $\square$

Proof of Corollary 8

From the definition of the Krawtchouk polynomials, we get

$$\begin{aligned} \sum _{\mu =1}^{s_{1}} \Omega ^{2^{t-1}\mu }H_\mu (\mathbf{X}_{1})= & {} \frac{n}{2^{s_1}} \sum _{\delta =0}^{s_1} W_{\delta }(\mathbf{X}_1) \sum _{\mu =1}^{s_{1}} \Omega ^{2^{t-1}\mu } P_{\mu }(\delta ,2^{s_1}) \nonumber \\= & {} \frac{n}{2^{s_1}} \sum _{\delta =0}^{s_1} W_{\delta }(\mathbf{X}_1) \left[ \sum _{\mu =0}^{s_{1}} \Omega ^{2^{t-1}\mu } P_{\mu }(\delta ,2^{s_1})-1\right] . \end{aligned}$$

(29)

From (27) and (29), we get

$$\begin{aligned} \sum _{\mu =1}^{s_{1}} \Omega ^{2^{t-1}\mu }H_\mu (\mathbf{X}_{1})= & {} \frac{n}{2^{s_1}} \sum _{\delta =0}^{s_1} W_{\delta }(\mathbf{X}_1) \left[ \left( 1+\Omega ^{2^{t-1}}\right) ^{s_1}\left( \frac{1-\Omega ^{2^{t-1}}}{1+\Omega ^{2^{t-1}}}\right) ^{\delta }-1\right] . \end{aligned}$$

(30)

The proof can be obtained from Corollary 5 and (30). $\square$

Proof of Corollary 9

From Corollary 6 by the same technique of Corollary 8. $\square$

Proof of Theorem 4

From the definition of the MAP in (3), we get

$$\begin{aligned}&\sum _{i=1}^{n}\sum _{j=1}^{n}\Omega ^{2^{t-1}h_{ij}(\mathbf{X}^{\star })} =n+\Omega ^{2^{t-1}(s+s_1)}\sum _{i=1}^{n}\sum _{j(i\ne j)=1}^{n}\left( \left( \frac{1}{\Omega }\right) ^{2^{t-1}}\right) ^{\xi _{ij}(\mathbf{X}^{\star })} \nonumber \\&{}\quad =n+2\Omega ^{2^{t-1}(s+s_1)}\sum _{i=1}^{n}\sum _{j=i+1}^{n} \sum _{\ell =0}^{\infty }\frac{1}{\ell !}\left( \xi _{ij}(\mathbf{X}^{\star })\ln \left( \left( \frac{1}{\Omega }\right) ^{2^{t-1}}\right) \right) ^{\ell } \nonumber \\&{}\quad =n+n(n-1)\Omega ^{2^{t-1}(s+s_1)} \sum _{\ell =0}^{\infty }\frac{1}{\ell !}\left( \ln \left( \left( \frac{1}{\Omega }\right) ^{2^{t-1}}\right) \right) ^{\ell }M_{\ell }(\mathbf{X}^{\star }). \end{aligned}$$

(31)

Combining Corollary 1 and (31), the proof can be completed. $\square$

Proof of Corollary 10

From Corollary 2 by the same technique of Theorem 4. $\square$

Proof of Corollary 11

From Corollary 3 by the same technique of Theorem 4. $\square$

Proof of Theorem 5

From (15) and (16) with some algebra, we can get the following analytical framework for any of the above-mentioned discrepancies (WD and LD) of the design $\mathbf{Z}^{(t)}\in U\left(2^tn,2^{2^{t}s_1}4^{2^{t-1}s_2}\right)$

$$\begin{aligned} \left[ {Disc}\left( \mathbf{Z}^{(t)}\right) \right] ^2= & {} -\Delta ^{2^{t}s_1+2^{t-1}s_2}+\frac{1}{n}\left( \frac{1}{2}\right) ^t\Theta ^{2^{t}s_1+2^{t-1}s_2} +\frac{1}{n^{2}}\left( \frac{1}{2}\right) ^{2t}\left( \Theta \Omega \right) ^{2^{t}s_1+2^{t-1}s_2} \nonumber \\&\times \sum _{i,j,(i\ne j)\in C_1^{2^{t}n}} \left( \frac{1}{\Omega }\right) ^{\xi _{ij}\left( \mathbf{Z}^{(t)}\right) } \left( \frac{\Psi _2}{\Theta \Omega }\right) ^{\zeta _{ij}\left( \mathbf{Z}^{(t)}_2\right) }. \end{aligned}$$

(32)

From Lemmas 5 and 6 in Elsawah et al. (2021), for any two nonnegative sequences with $\sum _{i=1}^{c}A_{i} = a,\,\sum _{i=1}^{c}B_{i} = b$ and $\tau _i>1,\,i=1,2$ we have

$$\begin{aligned} \sum _{i=1}^{c}\tau _1^{A_i}\tau _2^{B_i}\ge \tau _1^{\lfloor \frac{a}{c}\rfloor }\tau _2^{\lfloor \frac{b}{c}\rfloor } \left\{ \begin{array}{ll} c(\tau _1+\tau _2-1)+\phi _1(1-\tau _1)+\phi _2(1-\tau _2),\, \phi _1+\phi _2>c;\\ c\tau _1\tau _2+\phi _1\tau _2(1-\tau _1)+\phi _2\tau _1(1-\tau _2),\, \phi _1+\phi _2\le c \end{array} \right. \end{aligned}$$

(33)

and

$$\begin{aligned} \sum _{i=1}^{c}\tau _1^{A_i}\tau _2^{B_i}\ge pe^{\vartheta _{(\omega )}}+(c-p)e^{\vartheta _{(\omega +1)}}, \end{aligned}$$

(34)

respectively, where $\phi _1=c\left( \lfloor \frac{a}{c}\rfloor -\frac{a}{c}+1\right) ,$ $\phi _2=c\left( \lfloor \frac{b}{c}\rfloor -\frac{b}{c}+1\right) ,$ $z_i = \ln \left( \tau _1^{A_i}\tau _2^{B_i}\right) ,$ $\omega$ is the largest integer such that $\vartheta _{(\omega )}\le \frac{a \ln \tau _1+b \ln \tau _2}{c}< \vartheta _{(\omega +1)}$ and $p=\frac{c}{\vartheta _{(\omega +1)}-\vartheta _{(\omega )}}\left( \vartheta _{(\omega +1)}-\frac{a \ln \tau _1+b \ln \tau _2}{c}\right) .$ From the definitions of $\xi _{ij}$ and $\zeta _{ij}$ with some algebra, we get

$$\begin{aligned}&\sum _{i(\ne j)=1}^{2^tn}\sum _{j=1}^{2^tn}\xi _{ij}\left( \mathbf{Z}^{(t)}\right) = 2^{3t-3}n^2(s+3s_1)-2^{2t-1}n(s+s_1). \end{aligned}$$

(35)

$$\begin{aligned}&\sum _{i(\ne j)=1}^{2^tn}\sum _{j=1}^{2^tn}\zeta _{ij}\left( \mathbf{Z}^{(t)}_2\right) =2^{3t-2}s_2n^2. \end{aligned}$$

(36)

Combining (32)–(36) with some algebra, the proof can be obtained. $\square$

Proof of Theorem 6

From Lemma 4 in Elsawah and Qin (2015), for any nonnegative sequence with $\sum _{i=1}^{c}V_{i} = v$ and $\varepsilon >1,$ we have

$$\begin{aligned} \sum _{i=1}^{c}\varepsilon ^{V_i}\ge \varepsilon ^{\lfloor \frac{v}{c}\rfloor } (\varpi +(c-\varpi )\varepsilon ),\end{aligned}$$

(37)

where $\varpi =c\left( \lfloor \frac{v}{c}\rfloor -\frac{v}{c}+1\right) .$ From the definitions of $\xi _{ij}$ with some algebra, we get

$$\begin{aligned} \sum _{i(\ne j)=1}^{n}\sum _{j=1}^{n} \xi _{ij}\left( \mathbf{X}^{\star }\right) = \frac{1}{2}n(s+s_1)(n-2). \end{aligned}$$

(38)

Combining Corollary 1, (34), (37) and (38) with some algebra, the proof can be obtained. $\square$

Proof of Corollary 12

The proof can be obtained from Corollary 8 and the fact that for any orthogonal array of strength $s_1$ $\mathbf{X}_1\in U(n,2^{s_1})$ we get $W_\delta (\mathbf{X}_1)=0,\,1\le \delta \le s_1.$ $\square$

Proof of Corollary 13

From Corollary 9 by the same technique of Corollary 12. $\square$

Proof of Corollary 14

From Theorem 1 by the same way of the proof of Corollary 2. $\square$

Proof of Corollary 15

From (4) and (5) with some algebra, we get the following formulas of the WD and LD of the extended design $E^{(t-1)}(\mathbf{X}_2)\in U(2^{t-1}n,2^{2^{t-1}s})$

$$\begin{aligned}&{[}{WD}\left( E^{(t-1)}(\mathbf{X}_2)\right) ]^2=-\left( \frac{4}{3}\right) ^{2^{t-1}s}+\frac{1}{4^{t-1}n^{2}}\left( \frac{3}{2}\right) ^{2^{t-1}s}\sum _{i=1}^{2^{t-1}n}\sum _{j=1}^{2^{t-1}n}\left( \frac{5}{6}\right) ^{h_{ij}\left( E^{(t-1)}(\mathbf{X}_2)\right) }. \end{aligned}$$

(39)

$$\begin{aligned}&{[}{LD}\left( E^{(t-1)}(\mathbf{X}_2)\right) ]^2=-\left( \frac{3}{4}\right) ^{2^{t-1}s}+\frac{1}{4^{t-1}n^{2}}\sum _{i=1}^{2^{t-1}n}\sum _{j=1}^{2^{t-1}n}\left( \frac{1}{2}\right) ^{h_{ij}\left( E^{(t-1)}(\mathbf{X}_2)\right) }. \end{aligned}$$

(40)

Combining (15)–(18), (39) and (40) with some algebra, the proof can be completed. $\square$

Proof of Corollary 16

The proof is obvious from Corollary 15 by taking $t=1$ and the fact that any full factorial design in $U(2^s,2^s)$ has $\hbox {LD-value}=0.$ $\square$

Proof of Theorem 7

From (32) with some algebra, we can get the following analytical framework for any of the above-mentioned discrepancies (WD and LD) of the design $\mathbf{Z}^{(t)}_{Flex}\in U\left(2^tn+n_1,2^{2^{t}s_1+m_1}4^{2^{t-1}s_2+m_2}\right)$

$$\begin{aligned} \left[ {Disc}\left( \mathbf{Z}^{(t)}_{Flex}\right) \right] ^2= & {} -\Delta ^{2^{t}s_1+2^{t-1}s_2+m_1+m_2} +\left( \frac{1}{2^tn+n_1}\right) ^2\left( \Theta \Omega \right) ^{2^{t}s_1+m_1+2^{t-1}s_2+m_2} \nonumber \\&\times \sum _{i,j \in C_1^{2^{t}n+n_1}} \left( \frac{1}{\Omega }\right) ^{\xi _{ij}\left( \mathbf{Z}^{(t)}_{Flex}\right) } \left( \frac{\Psi _2}{\Theta \Omega }\right) ^{\zeta _{ij}\left( \mathbf{Z}^{(t)}_{Flex}\right) }. \end{aligned}$$

(41)

The sum term in (41) can be rewritten as follows

$$\begin{aligned}&\sum _{i,j\in C_1^{2^{t}n+n_1}} \left( \frac{1}{\Omega }\right) ^{\xi _{ij}\left( \mathbf{Z}^{(t)}_{Flex}\right) } \left( \frac{\Psi _2}{\Theta \Omega }\right) ^{\zeta _{ij}\left( \mathbf{Z}^{(t)}_{Flex}\right) }=\sum _{i,j\in C_1^{2^{t}n}} \left( \frac{1}{\Omega }\right) ^{\xi _{ij}\left( \mathbf{Z}^{(t)}_{Flex}\right) } \left( \frac{\Psi _2}{\Theta \Omega }\right) ^{\zeta _{ij}\left( \mathbf{Z}^{(t)}_{Flex}\right) } \nonumber \\&\,\,\,\,\,\,\,\,\,\,\,\,+\sum _{i,j\in C_{2^{t}n+1}^{2^{t}n+n_1}} \left( \frac{1}{\Omega }\right) ^{\xi _{ij}\left( \mathbf{Z}^{(t)}_{Flex}\right) } \left( \frac{\Psi _2}{\Theta \Omega }\right) ^{\zeta _{ij}\left( \mathbf{Z}^{(t)}_{Flex}\right) } \nonumber \\&\,\,\,\,\,\,\,\,\,\,\,+2\sum _{i\in C_1^{2^{t}n},j\in C_{2^{t}n+1}^{2^{t}n+n_1}} \left( \frac{1}{\Omega }\right) ^{\xi _{ij}\left( \mathbf{Z}^{(t)}_{Flex}\right) } \left( \frac{\Psi _2}{\Theta \Omega }\right) ^{\zeta _{ij}\left( \mathbf{Z}^{(t)}_{Flex}\right) }. \end{aligned}$$

(42)

From (41) with some algebra, we get

$$\begin{aligned}&\left[ \left[ {Disc}\left( \mathbf{Z}^{(t)}_{Flex1}\right) \right] ^2+\Delta ^{2^{t}s_1+2^{t-1}s_2+m_1+m_2}\right] 4^tn^2 =\left( \Theta \Omega \right) ^{2^{t}s_1+m_1+2^{t-1}s_2+m_2} \nonumber \\&{}\quad \,\times \sum _{i,j \in C_1^{2^{t}n}} \left( \frac{1}{\Omega }\right) ^{\xi _{ij}\left( \mathbf{Z}^{(t)}_{Flex}\right) } \left( \frac{\Psi _2}{\Theta \Omega }\right) ^{\zeta _{ij}\left( \mathbf{Z}^{(t)}_{Flex}\right) }. \end{aligned}$$

(43)

$$\begin{aligned}&\left[ \left[ {Disc}\left( \mathbf{Z}^{(t)}_{Flex2}\right) \right] ^2+\Delta ^{2^{t}s_1+2^{t-1}s_2+m_1+m_2}\right] n_1^2 =\left( \Theta \Omega \right) ^{2^{t}s_1+m_1+2^{t-1}s_2+m_2} \nonumber \\&{}\quad \,\times \sum _{i,j \in C_{2^{t}n+1}^{2^{t}n+n_1}} \left( \frac{1}{\Omega }\right) ^{\xi _{ij}\left( \mathbf{Z}^{(t)}_{Flex}\right) } \left( \frac{\Psi _2}{\Theta \Omega }\right) ^{\zeta _{ij}\left( \mathbf{Z}^{(t)}_{Flex}\right) }. \end{aligned}$$

(44)

From (41)–(44) with some algebra, the proof can be completed. $\square$

Proof of Theorem 8

The proof can be obtained from Theorem 7 by the same technique of the proof of Theorem 5. $\square$