Asymmetrical split-plot designs with clear effects

Abstract

The fractional factorial split-plot (FFSP) design is an important experimental design both in theory and in practice. There is extensive literature on the two-level FFSP design and its various variants. However, there is little work on the s-level FFSP design and its variants in the asymmetrical (i.e., mixed-level) case, where s is any prime or prime power. Such designs are commonly used e.g. in agriculture, medicine and chemistry. This paper provides the necessary and sufficient conditions for the existence of resolution III or IV regular $s^{(n_1+n_2)-(k_1+k_2)}(s^r)$ designs which contain clear main effects or two-factor interaction components. In particular, the sufficient conditions are proved through constructing the corresponding designs, and some examples are provided to illustrate the construction methods.

Appendix: Proofs of theorems

Proof of Theorem 1

Suppose that an $s_{\mathrm{III}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_s$ design D is determined by $C_s$. Note that $B_1 =\{c_1,\ldots ,c_{n_1}\}\subset H_a$. Then $n_1\le (s^{p_1}-1)/(s-1)$. If the $s^r$-level SP main effect $E=H(b_1, \ldots ,$$b_r)=\{d_1$, $\ldots$, $d_t\}$ is clear, then we have $d_ic_j^l\in H\backslash C_s, \mathrm{~for~} i =1,\ldots , t,\ j=1,\ldots , n_1+n_2,\ l=1,\ldots , s-1$. Note that the columns $d_ic_j^l$ are different from each other (otherwise $d_i$ is not clear) and $t=(s^{r}-1)/(s-1)$. We can conclude that

$$\begin{aligned} t(n_1+n_2)(s-1)\le \frac{s^{p}-1}{s-1}-n_1-n_2-t, ~~\text{ i.e., }~~ n_2\le \frac{s^{p-r}-1}{s-1}-n_1. \end{aligned}$$

When $n_1\le (s^{p_1}-1)/(s-1)$ and $n_2=(s^{p-r}-1)/(s-1)-n_1$, let $M_1\subset H_a$ such that $M_1$ is determined by $p_1$ independent columns with $|M_1|=n_1$, and

$$\begin{aligned} M_2= \left( \{b_r\}\otimes (H(a_1, \ldots ,a_{p_1},b_{r+1},\ldots ,b_{p_2})\backslash M_1 ) \right) \cup H_{b_r}. \end{aligned}$$

Hereafter, $|\cdot |$ denotes the cardinality of the set, and $T_1\otimes T_2=\{t_1t_2: t_1\in T_1,t_2\in T_2\}$ for any two sets $T_1$ and $T_2$. Then, $|M_2|=(s^{p-r}-1)/(s-1)+(s^{r}-1)/(s-1)-n_1$. Replacing $H_{b_r}$ with an $s^r$-level factor E in $M_2$, we can get an $s_{\mathrm{III}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_s$ design with $n_1\le (s^{p_1}-1)/(s-1)$ and $n_2=(s^{p-r}-1)/(s-1)-n_1$. Obviously, the $s^r$-level SP main effect E is clear. For $n_1\le (s^{p_1}-1)/(s-1)$ and $n_2 < (s^{p-r}-1)/(s-1)-n_1$, we can delete some columns from $M_2\backslash E$ to get the desired designs. The proof is completed. $\square$

Proof of Theorem 2

Suppose that an $s_{\mathrm{III}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_s$ design D is determined by $C_s$, and an s-level WP main effect, say $c_1$, is clear in D. Then we have $c_1c_i^l\in H_a\backslash B_1$ for $i =2,\ldots , n_1, l=1,\ldots , s-1$, $c_1c_j^l\in (H\backslash H_a)\backslash (B_2\cup H_{b_r})$ for $j =n_1+1,\ldots , n_1+n_2,l=1,\ldots , s-1$ and $c_1d_m^l\in (H\backslash H_a)\backslash (B_2\cup H_{b_r})$ for $m=1,\ldots , t, l=1,\ldots , s-1$. The columns above are different from each other, which implies that

$$\begin{aligned} (n_1-1)(s-1)\le & {} \frac{s^{p_1}-1}{s-1}-n_1, ~~\mathrm{i.e.},~~ n_1\le \frac{s^{p_1-1}-1}{s-1}+1; ~~\mathrm{and} \\ (n_2+t)(s-1)\le & {} \frac{s^{p}-1}{s-1}-\frac{s^{p_1}-1}{s-1}-n_2-t, ~~\mathrm{i.e.,}~~ n_2\le \frac{s^{p-1}-s^{p_1-1}}{s-1}-t. \end{aligned}$$

One can easily obtain the same result by similar arguments when D has a clear WP2FIC, which are omitted here.

When $n_1=(s^{p_1-1}-1)/(s-1)+1$ and $n_2=(s^{p-1}-s^{p_1-1})/({s-1})-t$, let $M_1=\{a_1\}\cup H(a_2, \ldots , a_{p_1})$ and $M_2=H(a_2,\ldots , a_{p_1}, b_1, \ldots , b_{p_2})\backslash H(a_2, \ldots , a_{p_1})$. Then $M=(M_1,M_2)$ is an $s_{\mathrm{III}}^{(n_1+n_{2}')-(k_1+k_{2}')}$ design with $n_1=|M_1|=(s^{p_1-1}-1)/(s-1)+1$, $n_{2}'=|M_2|=(s^{p-1}-s^{p_1-1})/(s-1)$, $k_1=n_1-p_1$ and $k_2'=n_2'-p_2-r$. Replacing $H_{b_r}$ with an $s^r$-level factor E in $M_2$, we can get an $s_{\mathrm{III}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_s$ design D with $n_1=(s^{p_1-1}-1)/(s-1)+1 \mathrm{~and~} n_2=(s^{p-1}-s^{p_1-1})/(s-1)-t$. It is obvious that the WP main effect $a_1$ and WP2FIC $a_1c$$(c\in M_1\backslash \{a_1\})$ are clear in D. When $n_1<(s^{p_1-1}-1)/(s-1)+1$ or (and) $n_2<(s^{p-1}-s^{p_1-1})/(s-1)-t$, we can get the designs with clear WP main effects or WP2FICs by deleting some columns from $M_1\backslash \{a_1\}$ or (and) $M_2\backslash H_{b_r}$. The proof is completed. $\square$

Proof of Theorem 3

For the case of (a), since the WP factors belong to $H_a$, we have $n_1\le (s^{p_1}-1)/(s-1)$. Without loss of generality, suppose that an $s_{\mathrm{III}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_s$ design D has a clear WS2FIC $c_1c_{n_1+1}$. Then we have $c_1c_{n_1+1}^l\in H\backslash C_s$ for $l=1,\ldots , s-1$, $c_1c_{n_1+1}c_i^l\in H\backslash C_s$ for $i=2,\ldots ,n_1,n_1+2,\ldots , n_1+n_2,l=1,\ldots , s-1$ and $c_1c_{n_1+1}d_m^l\in H\backslash C_s$ for $m=1,\ldots , t, l=1,\ldots , s-1$. The above columns are different from each other, which implies that

$$\begin{aligned}&(s-1)+(n_1+n_2-2)(s-1)+t(s-1)\le (s^{p}-1)/(s-1)-n_1-n_2-t,~\mathrm{i.e.,}\\&\quad n_2\le (s^{p-1}-1)/(s-1)-n_1-t+1. \end{aligned}$$

For the case of (b), similar to (a), we have $n_1\le (s^{p_1}-1)/(s-1)$. Suppose that an $s_{\mathrm{III}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_s$ design D has a clear s-level SP main effect $c_{n_1+1}$. Then we have $c_{n_1+1}c_i^l\in H\backslash C_s$ for $i=1,\ldots ,n_1,n_1+2,\ldots , n_1+n_2, l=1,\ldots , s-1$ and $c_{n_1+1}d_j^l\in H\backslash C_s$ for $j=1,\ldots , t, l=1,\ldots , s-1$. The above columns are different from each other, which implies that

$$\begin{aligned}&(n_1+n_2-1)(s-1)+t(s-1)\le (s^{p}-1)/(s-1)-n_1-n_2-t, ~\mathrm{i.e.,}\\&\quad n_2\le (s^{p-1}-1)/(s-1)-n_1-t+1. \end{aligned}$$

Without loss of generality, suppose that design D has a clear SP2FIC $c_{n_1+1}c_{n_1+2}$. Then we have $c_{n_1+1}c_{n_1+2}^l\in H\backslash C_s$ for $l=1,\ldots , s-1$, $c_{n_1+1}c_{n_1+2}c_i^l\in H\backslash C_s$ for $i=1,\ldots ,n_1,n_1+3,\ldots , n_1+n_2,l=1,\ldots , s-1$ and $c_{n_1+1}c_{n_1+2}d_m^l\in H\backslash C_s$ for $m=1,\ldots , t, l=1,\ldots , s-1$. The above columns are different from each other, which implies that

$$\begin{aligned}&(s-1)+(n_1+n_2-2)(s-1)+t(s-1)\le (s^{p}-1)/(s-1)-n_1-n_2-t,~\mathrm{i.e.,}\\&\quad n_2\le (s^{p-1}-1)/(s-1)-n_1-t+1. \end{aligned}$$

As for the case of (c), following similar arguments to cases of (a) and (b), we can get that $n_2\le (s^{p-1}-1)/(s-1)-n_1-t+1$. So, we only need to prove $n_1\le (s^{p_1}-1)/(s-1)-1$. If $n_1=(s^{p_1}-1)/(s-1)$, then every element in $H_a$ belongs to $B_1$, i.e., every element in $H_a$ is a WP factor. Recall that each column of $H_{b_r}$ is a component of the $s^r$-level SP factor. Note that when $p_2= r,$$H=H_a\cup H_{b_r} \cup (H_a\otimes H_{b_r})$, which implies that all the s-level SP factors come from $H_a\otimes H_{b_r}$, i.e., each s-level SP main effect (or SP2FIC) is aliased with the interaction of a WP factor and a component of the $s^r$-level factor. Obviously, the s-level SP main effect (or SP2FIC) is not clear. Thus, we get $n_1\le (s^{p_1}-1)/(s-1)-1$.

Now, it comes to prove the “if” parts. We need to construct $s_{\mathrm{III}}^{(n_1+n_2)-(k_1+k_2)}$$(s^r)_s$ designs containing clear s-level SP main effects, WS2FICs or SP2FICs, respectively.

For (a). Let $E=H_{b_r}=H(b_1,\ldots ,b_r)$, $M_1=H_a$. Without loss of generality, let $c_1=a_1$ and $d_1=b_1$. Then $c_1d_1=a_1b_1\in H\backslash H_a$. There are $s-1$ columns in H, $c_1d_1^l, l=1,\ldots , s-1$, which join $\{c_1,d_1\}$ to form $s-1$ distinct words of length three. The remaining $(s^{p}-1)/(s-1)-(s-1)-2$ columns in H can be divided into $((s^{p}-1)/(s-1)-(s-1)-2)/s=(s^{p-1}-1)/(s-1)-1$ different groups each with s columns such that any two columns in a group joining $\{c_1,d_1\}$ form a word of length four. Among them, $(s^{p_1}-1)/(s-1)-1$ groups have the form of $\{c_i,c_1d_1c_i, c_1d_1c_i^{2}, \ldots , c_1d_1c_i^{s-1}\}$, where $c_i\in H_a\backslash \{c_1\}$ and $c_1d_1c_i^{l} \in H\backslash H_a, i=2,\ldots , (s^{p_1}-1)/(s-1) , l=1, 2, \ldots , s-1$. The remaining $(s^{p-1}-s^{p_1})/(s-1)$ groups have the form of $\{f_{s_i}, c_1d_1f_{s_i}, c_1d_1f_{s_i}^{2}, \ldots , c_1d_1f_{s_i}^{s-1} \},i=1,\ldots ,(s^{p-1}-s^{p_1})/(s-1)$, where all of them are from $H\backslash H_a$. Note that the groups with the form of $\{d_i, c_1d_1d_i,$$c_1d_1d_i^{2}, \ldots ,$$c_1d_1d_i^{s-1}\},$$i=2,\ldots ,t,$ are from the latter $(s^{p-1}-s^{p_1})/(s-1)$ groups. From each latter group, we choose one column as an element of $M_2$, such that $H_{b_r}\backslash \{b_1\}\subset M_2$. Also adding $d_1$ into $M_2$, we have $|M_2|=(s^{p-1}-s^{p_1})/(s-1)+1$. Then replacing $H_{b_r}$ with an $s^r$-level factor, we can obtain an $s_{\mathrm{III}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_s$ design D with $n_1=(s^{p_1}-1)/(s-1)$ and $n_2=(s^{p-1}-s^{p_1})/(s-1)-t+1=(s^{p-1}-1)/(s-1)-n_1-t+1$. Obviously, the WS2FIC $c_1d_1$ is clear. When $n_1<(s^{p_1}-1)/(s-1)$ and $n_2=(s^{p-1}-1)/(s-1)-n_1-t+1$, we only need to delete some columns $c_i(i\ne 1)$ from $M_1$ and add $c_1d_1c_i$ into $M_2$ to get the desired design. Furthermore, when $n_2<(s^{p-1}-1)/(s-1)-n_1-t+1$, we can delete some columns from $M_2\backslash H_{b_r}$ to get the desired design.

For (b). Let $E=H(b_2,\ldots ,b_{r+1})$, $M_1$ be any $n_1$-subset of $H_a$ and

$$\begin{aligned} M_2=\{b_1\}\cup \left( \{b_1\}\otimes [H(a_1,\ldots ,a_{p_1},b_{2},\ldots ,b_{p_2})\backslash M_1 \backslash E \right) \cup E. \end{aligned}$$

Then $|M_1|=n_1\le (s^{p_1}-1)/(s-1)$ and $|M_2|=(s^{p-1}-1)/(s-1)-n_1+1$. Hence, $M=(M_1, M_2)$ is an $s_{\mathrm{III}}^{(n_1+n'_2)-(k_1+k'_2)}$ design D with $n_1\le (s^{p_1}-1)/(s-1)$ and $n_2=(s^{p-1}-1)/(s-1)-n_1+1$. Replacing E with an $s^r$-level factor, then we get an $s_{\mathrm{III}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_s$ design with $n_1\le (s^{p_1}-1)/(s-1)$ and $n_2=(s^{p-1}-1)/(s-1)-n_1-t+1$. Obviously, the s-level SP main effect $b_1$ is clear. For $n_2<(s^{p-1}-1)/(s-1)-n_1-t+1$, we only need to delete some columns from $M_2\backslash (E\cup \{b_1\})$ to get the required design.

Let $M_1=H_a=\{c_1,\ldots ,c_{(s^{p_1}-1)/(s-1)}\},$$E=H(b_1,\ldots ,b_r)=\{d_1,\ldots , d_t\}$ and $c_{n_1+1}=b_{r+1}$. Without loss of generality, let $d_1=b_1$. Then $c_{n_1+1}d_1=b_1b_{r+1}\in H\backslash H_a$. Similar to the above proof, there are $s-1$ columns in H, $c_{n_1+1}d_1^{l}$ for $l = 1,...,s-1,$ which join {$c_{n_1+1}, d_1$} to form $s-1$ distinct words of length three. The remaining $(s^{p}-1)/(s-1)-(s-1)-2$ columns in H can be divided into $(s^{p-1}-1)/(s-1)-1$ different groups each with s columns such that any two columns in a group join $\{c_1,d_1\}$ to form a word of length four. Among them, there are $(s^{p_1}-1)/(s-1)$ groups having the form of $\{c_i, c_{n_1+1}d_1c_i, c_{n_1+1}d_1c_i^{2}, \ldots , c_{n_1+1}d_1c_i^{s-1}\}$, where $c_i\in H_a$ and $c_{n_1+1}d_1c_i^{l} \in H\backslash H_a, i=1,\ldots , (s^{p_1}-1)/(s-1), l=1, 2, \ldots , s-1$. The remaining $(s^{p-1}-s^{p_1})/(s-1)-1$ groups have the form of $\{f_{s_i}, c_{n_1+1}d_1f_{s_i}, c_{n_1+1}d_1f_{s_i}^{2}, \ldots ,$$c_{n_1+1}d_1f_{s_i}^{s-1}\},i=1,\ldots ,(s^{p-1}-s^{p_1})/(s-1)-1,$ with all columns being from $H\backslash H_a$. Note that the groups with the form of $\{d_i, c_{n_1+1}d_1d_i, c_{n_1+1}d_1d_i^{2}, \ldots ,$$c_{n_1+1}d_1d_i^{s-1}\}$($i=2,\ldots ,t$) belong to the latter $(s^{p-1}-s^{p_1})/(s-1)-1$ groups. From each latter group, choose one column as an element of $M_2$, such that $\{d_2,\ldots ,d_t\}\subset M_2$. Adding $c_{n_1+1}$ and $d_1$ into $M_2$, we have $|M_2|=(s^{p-1}-s^{p_1})/(s-1)+1$. Replacing $E=H(b_1,\ldots ,b_r)$ with an $s^r$-level factor, we get the $s_{\mathrm{III}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_s$ design D with $n_1=(s^{p_1}-1)/(s-1)$ and $n_2=(s^{p-1}-1)/(s-1)-n_1-t+1$. From the construction of D, the SP2FIC $c_{n_1+1}d_1$ is clear. When $n_1<(s^{p_1}-1)/(s-1)$ and $n_2=(s^{p-1}-1)/(s-1)-n_1-t+1$, we only need to delete some columns $c_i$ from $M_1$ and add $c_{n_1+1}d_1c_i$ into $M_2$ to get the desired design. We can delete some columns from $M_2\backslash (E\cup \{c_{n_1+1}\})$ to get the design when $n_2<(s^{p-1}-1)/(s-1)-n_1-t+1$.

For (c). Let $H_{b_r}=H(b_1,\ldots ,b_r)$, $M_1=H_a\backslash \{a_1\}=\{c_1,\ldots ,c_{(s^{p_1}-1)/(s-1)-1}\},$$d_1=b_1$ and $c_{n_1+1}=a_1b_2$. Then we can get $c_{n_1+1}d_1=a_1b_1b_2\in H\backslash H_a$. There are $s-1$ columns in H, $c_{n_1+1}d_1^{l}, l = 1,...,s-1,$ which join {$c_{n_1+1}, d_1$} to form $s-1$ distinct words of length three. The remaining $(s^{p}-1)/(s-1)-(s-1)-2$ columns in H can be divided into $(s^{p-1}-1)/(s-1)-1$ different groups each with s columns such that any two columns in a group join $\{c_{n_1+1},d_1\}$ to form a word of length four. Among them, $(s^{p_1}-1)/(s-1)$ groups have the form of $\{c_i, c_{n_1+1}d_1c_i, c_{n_1+1}d_1c_i^{2}, \ldots , c_{n_1+1}d_1c_i^{s-1}\},$$i=1,\ldots , (s^{p_1}-1)/(s-1)$, where $c_i\in H_a$ and $c_{n_1+1}d_1c_i^{l} \in H\backslash H_a, l=1, 2, \ldots , s-1$. The remaining $(s^{p-1}-s^{p_1})/(s-1)-1$ groups with all columns in $H\backslash H_a$ are denoted as $\{f_{s_i}, c_{n_1+1}d_1f_{s_i}, c_{n_1+1}d_1f_{s_i}^{2}, \ldots ,$$c_{n_1+1}d_1f_{s_i}^{s-1}\},$$i=1,\ldots , (s^{p-1}-s^{p_1})/(s-1)-1$. Note that $\{b_1b_2, c_{n_1+1}d_1b_1b_2, \ldots , c_{n_1+1}d_1(b_1b_2)^{s-1}\}$ belongs to the former $(s^{p_1}-1)/(s-1)$ groups since $c_{n_1+1}d_1(b_1b_2)^{s-1}=a_1\in H_a$, and the columns of $H(b_1,\ldots ,b_r)\backslash$$\{b_1, b_1b_2\}$ belong to the later $(s^{p-1}-s^{p_1})/(s-1)-1$ groups. We choose one column from each latter group as an element of $M_2$ such that $H_{b_r}\backslash \{b_1, b_1b_2\}\subset M_2$. Then adding $\{c_{n_1+1}, b_1, b_1b_2\}$ into $M_2$, we have $|M_2|=(s^{p-1}-s^{p_1})/(s-1)+2$. Replacing $H_{b_r}$ with an $s^r$-level factor, we can obtain an $s_{\mathrm{III}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_s$ design D with $n_1=(s^{p_1}-1)/(s-1)-1$ and $n_2=(s^{p-1}-1)/(s-1)-n_1-t+1$. From the above construction of D, the SP2FIC $c_{n_1+1}d_1$ is clear. When $n_1<(s^{p_1}-1)/(s-1)-1$ and $n_2=(s^{p-1}-1)/(s-1)-n_1-t+1$, we only need to delete some columns $c_i$ from $M_1$ and add $c_{n_1+1}d_1c_i$ into $M_2$ to get the desired design. Furthermore, when $n_2<(s^{p-1}-1)/(s-1)-n_1-t+1$, we can delete some columns from $M_2\backslash (\{c_{n_1+1}\}\cup H_{b_r})$ to get the design with clear SP2FICs $c_{n_1+1}d_1$. Deleting $c_{n_1+1}$ from and adding $c_{n_1+1}d_1$ into the design constructed above, the SP main effect $c_{n_1+1}d_1$ is clear in the new design. The proof is completed. $\square$

Proof of Theorem 4

From Corollary 1 in Ai and Zhang (2004), if $n_1>(s^{p_1-1}-1)/(s-1)+1$, the $s^{n_1-k_1}$ designs have resolution at most III. Since the WP part of an $s_{\mathrm{IV}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_s$ design is an $s_{\mathrm{IV}}^{n_1-k_1}$ design, we have $n_1\le (s^{p_1-1}-1)/(s-1)+1$.

Next, for (a), we will show $n_2\le (s^{p}-1)/((s-1)(3s-2))+(5(s-1))/(3s-2)-n_1-t$. Suppose that an $s_{\mathrm{IV}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_s$ design is determined by $C_{s}$ and the WS2FIC $c_1c_{n_1+1}$ is clear. Then, we have

• $c_1c_{i_1}^l\in H\backslash C_s, ~ i_1=2, \ldots , n_1+n_2,~ l=1,\ldots ,s-1;$

• $c_{n_1+1}c_{i_2}^l\in H\backslash C_s, ~ i_2=2, \ldots ,n_1,n_1+2,\ldots , n_1+n_2,~ l=1,\ldots ,s-1;$

• $c_{1}d_{i_3}^l\in H\backslash C_s, c_{n_1+1}d_{i_3}^l\in H\backslash C_s, ~ i_3=1, \ldots , t,~ l=1,\ldots ,s-1;$

• $c_{1}c_{n_1+1}c_{i_4}^l\in H\backslash C_s, ~ i_4=2, \ldots , n_1, n_1+2, \ldots , n_1+n_2,~ l=1,\ldots ,s-1;$ and

• $c_{1}c_{n_1+1}d_{i_5}^l\in H\backslash C_s, ~ i_5=1, \ldots , t,~ l=1,\ldots ,s-1.$

Since the above columns are different from each other, we get

$$\begin{aligned}&{[}(n_1+n_2-1)+2(n_1+n_2-2)+3t](s-1)\le (s^p-1)/(s-1)-n_1-n_2-t, ~\hbox {i.e}.,\\&n_2\le (s^{p}-1)/((s-1)(3s-2))+(5(s-1))/(3s-2)-n_1-t. \end{aligned}$$

In the end, for (b), we show $n_2\le (s^{p}-s^{r+1})/((s-1)(3s-2))-n_1+1$. Similar to the above, suppose that an $s_{\mathrm{IV}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_s$ design is determined by $C_{s}$ and the WS2FIC $c_{1}d_1$ is clear. Then, we have

• $c_{1}c_{i_1}^l\in H\backslash C_s, ~ i_1=2,\ldots ,n_1+n_2, ~ l=1,\ldots ,s-1;$

• $c_{1}d_{i_2}^l\in H\backslash C_s, ~ i_2 =1,\ldots ,t, ~ l=1,\ldots ,s-1;$

• $d_{1}c_{i_3}^l\in H\backslash C_s, ~ i_3=2,\ldots ,n_1+n_2, ~ l=1,\ldots ,s-1;$ and

• $c_{1}d_1c_{i_4}^l\in H\backslash C_s, ~ i_4 =2,\ldots ,n_1+n_2, ~ l=1,\ldots ,s-1.$

Because these columns are different from each other, we get

$$\begin{aligned}&[3(n_1+n_2-1)+t](s-1)\le (s^{p}-1)/(s-1)-n_1-n_2-t, ~\hbox {i.e}.,\\&\quad n_2\le (s^{p}-s^{r+1})/((s-1)(3s-2))-n_1+1. \end{aligned}$$

For the conditions of an $s_{\mathrm{IV}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_s$ design containing the same type of clear SP2FICs, the proof is similar. This completes the proof. $\square$

Proof of Theorem 5

Without loss of generality, suppose that an $s_{\mathrm{IV}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_s$ design is determined by $C_{s}$ and the WP2FIC $c_{1}c_2$ is clear. Then we have

• $c_{1}c_{i_1}^l\in H_a\backslash B_1, ~ i_1 =2,\ldots , n_1, ~ l=1,\ldots , s-1;$

• $c_{2}c_{i_2}^l\in H_a\backslash B_1, ~ i_2 =3,\ldots , n_1, ~ l=1,\ldots , s-1;$ and

• $c_{1}c_2c_{i_3}^l\in H_a\backslash B_1, ~ i_3 =3,\ldots , n_1, ~ l=1,\ldots , s-1.$

Because the above columns are different from each other, we can get

$$\begin{aligned}&[(n_1-1)+2(n_1-2)](s-1)\le (s^{p_1}-1)/(s-1)-n_1,~ \hbox {i.e.},\\&\quad n_1\le (s^{p_1}-1)/((s-1)(3s-2))+(5(s-1))/(3s-2). \end{aligned}$$

Note that

• $c_{1}c_{i_4}^l, c_{2}c_{i_4}^l, c_{1}c_2c_{i_4}^l\in H\backslash H_a\backslash (B_2\cup H_{b_r}), i_4 =n_1+1,\ldots , n_1+n_2, l=1,\ldots ,s-1;$ and

• $c_{1}d_{i_5}^l, c_{2}d_{i_5}^l, c_{1}c_2d_{i_5}^l\in H\backslash H_a\backslash (B_2\cup H_{b_r}), i_5 =1,\ldots , t, ~ l=1,\ldots ,s-1.$

Since the above columns are different from each other, we have

$$\begin{aligned}&3(n_2+t)(s-1)\le (s^{p}-1)/(s-1)-(s^{p_1}-1)/(s-1)-n_2-t,~ \hbox {i.e}.,\\&\quad n_2\le (s^{p}-s^{p_1})/((s-1)(3s-2))-t. \end{aligned}$$

This completes the proof. $\square$

Proof of Theorem 6

Without loss of generality, suppose that an $s_{\mathrm{III}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_w$ design D is determined by $C_{w}$ and its s-level WP main effect $c_1$ is clear. Then we have

• $c_1c_i^l\in H_a\backslash B'_1, ~ i =2,\ldots , n_1, ~ l=1,\ldots ,s-1;$

• $c_1d_j^l\in H_a\backslash B'_1, ~ j =1,\ldots , t, ~ l=1,\ldots ,s-1;$ and

• $c_1c_m^l\in H\backslash H_a\backslash B'_2, ~ m =n_1+1,\ldots , n_1+n_2, ~ l=1,\ldots ,s-1$.

Since these columns are different from each other, we have

$$\begin{aligned}&(n_1+t-1)(s-1)\le (s^{p_1}-1)/(s-1)-n_1-t, \text{ i.e., } n_1\\&\quad \le (s^{p_1-1}-s^r)/(s-1)+1{, \text{ and }} \\&n_2(s-1)\le (s^{p}-1)/(s-1)-(s^{p_1}-1)/(s-1)-n_2, ~\mathrm{i.e.},\\&\quad n_2\le (s^{p-1}-s^{p_1-1})/(s-1). \end{aligned}$$

The proof of the necessity of the conditions for designs to contain clear WP2FICs is similar and thus omitted here.

Let

$$\begin{aligned}&W_1=\{a_1\}\cup H(a_2, \ldots , a_{p_1}), ~~ \mathrm{and}\\&W_2=H(a_2,\ldots , a_{p_1}, b_1, \ldots , b_{p_2})\backslash H(a_2, \ldots , a_{p_1}). \end{aligned}$$

Then $W=(W_1,W_2)$ corresponds to an $s_{\mathrm{III}}^{(n'_1+n_{2})-(k'_1+k_{2})}$ design with $n'_1=(s^{p_1-1}-1)/(s-1)+1$ and $n_{2}=(s^{p-1}-s^{p_1-1})/(s-1)$. Replacing $H(a_2,\ldots ,a_{r+1})$ with an $s^r$-level factor F, we can get an $s_{\mathrm{III}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_w$ design with $n_1=(s^{p_1-1}-1)/(s-1)-t+1=(s^{p_1-1}-s^r)/(s-1)+1$ and $n_2=(s^{p-1}-s^{p_1-1})/(s-1)$. From the construction of the design, it is obvious that the s-level WP main effect $a_1$ is clear. For any $c\in W_1\backslash \{a_1\}$, the WP2FIC $a_1c$ is clear. When $n_1<(s^{p_1-1}-s^r)/(s-1)+1$ and (or) $n_2<(s^{p-1}-s^{p_1-1})/(s-1)$, we only need to delete some columns from $W_1\backslash (\{a_1\}\cup F)$ and (or) some columns in $W_2$ to obtain the required designs. The proof is completed. $\square$

Proof of Theorem 7

Suppose that the design is determined by $C_w$ and the $s^r$-level WP main effect $F=H(a_1,\ldots ,a_r)$$=\{d_1,\ldots ,d_t\}$ is clear. Then we have $d_ic_{j_1}^l\in H_a\backslash B_1'$ for $i =1,\ldots , t, j_1=1,\ldots , n_1, l=1,\ldots ,s-1$, and $d_ic_{j_2}^l\in (H\backslash H_a)\backslash B_2'$ for $i =1,\ldots , t, j_2=n_1+1,\ldots , n_1+n_2, l=1,\ldots ,s-1$. Note that these columns are different from each other and $t=(s^r-1)/(s-1)$. We can conclude that

$$\begin{aligned}&tn_1(s-1)\le (s^{p_1}-1)/(s-1)-n_1-t, ~\mathrm{i.e.},~ n_1\le (s^{p_1-r}-1)/(s-1), ~~ \mathrm{and}\\&tn_2(s-1)\le (s^{p}-s^{p_1})/(s-1)-n_2, \text{ i.e., } n_2\le (s^{p-r}-s^{p_1-r})/(s-1). \end{aligned}$$

Let $F=H(a_1,\ldots ,a_r)$ be the $s^r$-level WP factor, $W_1\subset H(a_{r+1},\ldots ,a_{p_1})$ and

$$\begin{aligned} W_2\subset H(a_{r+1},\ldots ,a_{p_1},b_1,\ldots , b_{p_2})\backslash H(a_{r+1},\ldots ,a_{p_1}), \end{aligned}$$

such that $|W_1|=n_1\le (s^{p_1-r}-1)/(s-1)$ and $|W_2|=n_2\le (s^{p-r}-s^{p_1-r})/(s-1)$. Then we can get an $s_{\mathrm{III}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_w$ design with the $s^r$-level WP main effect F being clear. The proof is completed. $\square$

Proof of Theorem 8

Since the WP factors belong to $H_a$, we have

$$\begin{aligned} n_1+t\le (s^{p_1}-1)/(s-1), \text{ i.e., } n_1\le (s^{p_1}-1)/(s-1)-t=(s^{p_1}-s^r)/(s-1). \end{aligned}$$

Suppose the design is determined by $C_{w}$ and the SP main effect $c_{n_1+1}$ is clear. Then we have

$$\begin{aligned}&c_{n_1+1}c_i^l\in H\backslash C_{w}, ~~ i= 1,\ldots ,n_1,n_1+2,\ldots ,n_1+n_2, l=1,\ldots ,s-1; ~~ \mathrm{and}\\&c_{n_1+1}d_j^l\in H\backslash C_{w}, ~~ j =1,\ldots ,t, l=1,\ldots ,s-1. \end{aligned}$$

Since the above columns are different from each other, we can get

$$\begin{aligned}&(n_1+n_2-1+t)(s-1)\le (s^{p}-1)/(s-1)-n_1-n_2-t, ~ \mathrm{i.e.,}\\&n_2\le (s^{p-1}-1)/(s-1)-n_1-t+1=(s^{p-1}-s^r)/(s-1)-n_1+1. \end{aligned}$$

Let $W_1=H(a_1,\ldots ,a_{p_1})$ and

$$\begin{aligned} W_2=\{b_1\}\cup \left( \{b_1\}\otimes (H(a_1,\ldots ,a_{p_1},b_{2},\ldots ,b_{p_2})\backslash W_1)\right) . \end{aligned}$$

Then $W=(W_1,W_2)$ is an $s_{\mathrm{III}}^{(n'_1+n_{2})-(k'_1+k_{2})}$ design with $n'_1=(s^{p_1}-1)/(s-1)$ and $n_{2}=(s^{p-1}-s^{p_1})/(s-1)+1$. Replacing $H(a_1,\ldots ,a_{r})$ with an $s^r$-level factor F, we can get an $s_{\mathrm{III}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_w$ design with $n_1=(s^{p_1}-s^r)/(s-1)$ and $n_2=(s^{p-1}-s^r)/(s-1)-n_1+1$. From the construction of the design, it is obvious that the s-level SP main effect $b_1$ is clear. When $n_1<(s^{p_1}-s^r)/(s-1)$ and $n_2=(s^{p-1}-s^r)/(s-1)-n_1+1$, we only need to delete some columns $c_i$ from $W_1\backslash F$ and add $c_ib_1$ into $W_2$ to get the required designs. Furthermore, when $n_2<(s^{p-1}-s^r)/(s-1)-n_1+1$, we need to delete some columns from $W_2\backslash \{b_1\}$ to get the desired designs. This completes the proof. $\square$

Proof of Theorem 9

Suppose that an $s_{\mathrm{III}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_w$ design is determined by $C_{w}$ and the WS2FIC $c_1c_{n_1+1}$ is clear. Since the WP factors belong to $H_a$, we have

$$\begin{aligned} n_1+t\le (s^{p_1}-1)/(s-1), \text{ i.e., } n_1\le (s^{p_1}-1)/(s-1)-t=(s^{p_1}-s^r)/(s-1). \end{aligned}$$

Then we have

• $c_1 c_{n_1+1}^l\in H\backslash C_{w}, ~ l=1, \ldots , s-1;$

• $c_1c_{n_1+1}c_i^l\in H\backslash C_{w}, ~ i= 2, \ldots , n_1,n_1+2, \ldots , n_1+n_2, ~ l=1, \ldots , s-1;$ and

• $c_1c_{n_1+1}d_j^l\in H\backslash C_{w}, ~ j =1, \ldots , t, ~ l=1,\ldots , s-1.$

Since the above columns are different from each other, we can get

$$\begin{aligned}&(1+n_1+n_2-2+t)(s-1)\le (s^{p}-1)/(s-1)-n_1-n_2-t, ~ \mathrm{i.e.,}\\&n_2\le (s^{p-1}-1)/(s-1)-n_1-t+1=(s^{p-1}-s^r)/(s-1)-n_1+1. \end{aligned}$$

The necessary conditions for a design to contain clear SP2FICs can be proved similarly.

Without loss of generality, suppose $W_1=H_a=H(a_1,\ldots ,a_{p_1})$, $c_1=a_1$ and $c_{n_1+1}=b_1$. Then $c_1c_{n_1+1}=a_1b_1\in H\backslash H_a$. There are $s-1$ columns in H, $c_1c_{n_1+1}^{l}$ for $l=1,\ldots , s-1$, which join $\{c_1,c_{n_1+1}\}$ to form $s-1$ distinct words of length three. The remaining $(s^{p}-1)/(s-1)-(s-1)-2$ columns in H can be divided into $(s^{p-1}-1)/(s-1)-1$ different groups each with s columns such that any two columns in a group join $\{c_1,c_{n_1+1}\}$ to form a word of length four. Among them, $(s^{p_1}-1)/(s-1)-1$ groups have the form of $\{c_i, c_1c_{n_1+1}c_i, c_1c_{n_1+1}c_i^{2}, \ldots , c_1c_{n_1+1}c_i^{s-1}\}$ with $c_i\in H_a\backslash \{a_1\}$ and $c_1c_{n_1+1}c_i^{l} \in H\backslash H_a, i=2, \ldots , (s^{p_1}-1)/(s-1), l=1, \ldots , s-1$. The remaining $(s^{p-1}-s^{p_1})/(s-1)$ groups with all columns in $H\backslash H_a$ are denoted by $\{f_{s_i}, c_1c_{n_1+1}f_{s_i}, c_1c_{n_1+1}f_{s_i}^{2}, \ldots , c_1c_{n_1+1}f_{s_i}^{s-1}\},i=1, \ldots , (s^{p-1}-s^{p_1})/(s-1)$. Choose one column from each of the latter $(s^{p-1}-s^{p_1})/(s-1)$ groups as an element of $W_2$, and add $c_{n_1+1}$ into $W_2$. Then we have $|W_2|=(s^{p-1}-s^{p_1})/(s-1)+1$. Then replacing $H(a_1,\ldots ,a_{r})$ with an $s^r$-level factor F, we can obtain an $s_{\mathrm{III}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_w$ design with $n_1=(s^{p_1}-s^r)/(s-1)$ and $n_2=(s^{p-1}-s^r)/(s-1)-n_1+1$. It is obvious that the WS2FIC $c_1c_{n_1+1}$ is clear. For $n_1<(s^{p_1}-s^r)/(s-1)$ and $n_2=(s^{p-1}-s^r)/(s-1)-n_1+1$, we only need to delete some $c_i(i\ne 1)$ from $W_1\backslash F$ and add $c_1c_{n_1+1}c_i$ into $W_2$ to get the desired design. Furthermore, when $n_2<(s^{p-1}-s^r)/(s-1)-n_1+1$, we can delete some columns from $W_2\backslash \{c_{n_1+1}\}$ to get the design. The construction of designs with clear SP2FICs is similar. We omit the proof to save space. The proof is completed. $\square$

Proof of Theorem 10

Let $H(a_{r+1}, \ldots , a_{p_1}) \triangleq \{\alpha _1, \ldots , \alpha _{(s^{p_1-r}-1)/(s-1)}\},$$H(a_{r+1}, \ldots , a_{p_1}, b_1, \ldots , b_{p_2})\backslash H(a_{r+1}, \ldots ,$$a_{p_1})$$\triangleq \{\alpha _{(s^{p_1-r}-1)/(s-1)+1}, \ldots , \alpha _{(s^{p-r}-1)/(s-1)}\},$ and $S_i=\{\alpha _i\}\cup (\{\alpha _i, \alpha _i^{2}, \ldots , \alpha _i^{s-1}\}\otimes H(a_1,\ldots ,a_{r})),$$i=1,\ldots , (s^{p-r}-1)/(s-1).$ Then

$$\begin{aligned} H=H(a_1,\ldots ,a_{p_1},b_1,\ldots ,b_{p_2})=H(a_1,\ldots ,a_{r}) \cup \left( \cup _{i=1}^{(s^{p-r}-1)/(s-1)}S_i\right) . \end{aligned}$$

Clearly, the components of the WP $s^r$-level factor belong to $H(a_1,\ldots ,a_{r})$, the WP s-level factors belong to $S_i,i=1,\ldots , (s^{p_1-r}-1)/(s-1)$, and the SP factors belong to $S_i,$$i=(s^{p_1-r}-1)/(s-1)+1, \ldots , (s^{p-r}-1)/(s-1)$. Note that each $S_i$ contains at most one column of the design. Otherwise, the design would have resolution III. Thus, we have $n_1\le (s^{p_1-r}-1)/(s-1)$ and $n_2\le (s^{p-r}-s^{p_1-r})/(s-1)$.

When $n_1=(s^{p_1-r}-1)/(s-1)$ and $n_{2}=(s^{p-r}-s^{p_1-r})/(s-1)$, let

$$\begin{aligned} W_1= & {} H(a_1,\ldots ,a_r)\cup (\{a_r\}\otimes H(a_{r+1},\ldots ,a_{p_1})) ~~ \mathrm{and}\\ W_2= & {} \{a_r\}\otimes (H(a_{r+1},\ldots ,a_{p_1},b_{1},\ldots ,b_{p_2})\backslash H(a_{r+1},\ldots ,a_{p_1})). \end{aligned}$$

Then, $W=(W_1,W_2)$ is an $s_{\mathrm{IV}}^{(n'_1+n_{2})-(k'_1+k_{2})}$ design with $n'_1=|W_1|=(s^{p_1-r}-1)/(s-1)+t$ and $n_{2}=|W_2|=(s^{p-r}-1)/(s-1)-(s^{p_1-r}-1)/(s-1)=(s^{p-r}-s^{p_1-r})/(s-1)$. Replacing $H(a_1,\ldots ,a_r)$ with an $s^r$-level factor F, we get an $s_{\mathrm{IV}}^{(n_1+n_2)-(k_1+k_2)}(s^r)_w$ design with $n_1=(s^{p_1-r}-1)/(s-1)$ and $n_2=(s^{p-r}-s^{p_1-r})/(s-1)$. For any $c_1\in W_1\backslash F$ and $c_2\in W_2$, all of the WP2FICs $a_1c_1$ and WS2FICs $a_1c_2$ are clear. For $n_1< (s^{p_1-r}-1)/(s-1)$ and (or) $n_2<(s^{p-r}-s^{p_1-r})/(s-1)$, we delete some columns from $W_1\backslash F$ and (or) $W_2$ to get the required designs. This completes the proof. $\square$

Proof of Theorem 11

For (a). With a discussion similar to that of Theorem 10, we can get $n_1\le (s^{p_1-r}-1)/(s-1)$ and $n_2\le (s^{p-r}-s^{p_1-r})/(s-1)$. Next, we will prove that the equalities cannot hold at the same time. If $n_1=(s^{p_1-r}-1)/(s-1)$ and $n_2=(s^{p-r}-s^{p_1-r})/(s-1)$, then there is exactly one column of each of $S_i\ (i=1, \ldots , (s^{p-r}-1)/(s-1))$ belonging to the design. Suppose these columns are $e_i\in S_i\ (i=1,\ldots , (s^{p-r}-1)/(s-1))$. For any SP2FIC $e_ie_j^{\delta }\ ((s^{p_1-r}-1)/(s-1)< i, j\le (s^{p-r}-1)/(s-1), \delta =1, \ldots , s-1)$, if $e_ie_j^{\delta }\in S_k$ for some $k\ (1\le k \le (s^{p-r}-1)/(s-1))$, then there exists an $l\in \{1, \ldots , s-1\},$ such that $e_ie_j^{\delta }e_k^{l} \in H(a_1,\ldots ,a_{r})$, which implies that every $e_ie_j^{\delta }$ is not clear. Hence, the equalities cannot hold at the same time. The proof of (b) is similar to that of (a) and is omitted here. $\square$