Optimal two-level regular fractional factorial split-plot designs when the effects of subplot factors are more important

Abstract

Robust parameter design (RPD) is an engineering methodology that focuses on reducing the variation of a process by appropriately selecting the setting of its control factors so as to make it less sensitive to noise variation. Then control factors are crucial in achieving robustness. If the control factors and noise factors in such a design are treated as the sub-plot (SP) factors and whole plot factors, fractional factorial split-plot (FFSP) design can be used. A minimum aberration of type SP (SP-MA) criterion is proposed to construct two-level regular FFSP designs, which is based on the FFSP-RPD effect hierarchy principle. We also derive the construction results of SP-MA $2^{(n_1+n_2)-(k_1+k_2)}$ designs with $k_1+k_2\le 4$. Some of the SP-MA FFSP-RPDs are better than the MA FFSP-RPDs in Table 5 of Bingham and Sitter (Technometrics 45(1):80–89, 2003). Finally, the SP-MA $2^{(n_1+n_2)-(k_1+k_2)}$ designs will be searched and tabulated for 8-, 16-, 32- and 64-run designs.

Appendix: Proofs of lemmas and theorems

Proof of Lemma 2.3

Suppose d is an SP-MA $2^{(n_{1}+n_{2})-(k_{1}+k_{2})}$ design with $n_{1}$ WP factors $Q_{1},\ldots , Q_{n_{1}}$ and $n_{2}$ SP factors $R_{n_{1}+1},\ldots ,R_{n_{1}+n_{2}}$. Suppose the independent defining words of d are

$$\begin{aligned} U_{1},\ldots ,U_{k_{1}},V_{1},\ldots ,V_{k_{2}}, \end{aligned}$$

where $U_{i}$ ($i=1,\ldots ,k_{1}$) and $V_{j}$ ($j=1,\ldots ,k_{2}$) denote the WP and SP type defining words, respectively. Without loss of generality, suppose $R_{n_{1}+n_{2}}$ is not involved in any defining words of d. Suppose $d'$ is determined by the following independent defining words

$$\begin{aligned} U_{1},\ldots ,U_{k_{1}},V_{1}R_{n_{1}+n_{2}},V_{2},\ldots ,V_{k_{2}}. \end{aligned}$$

Clearly, $d'$ is also a $2^{(n_{1}+n_{2})-(k_{1}+k_{2})}$ design. We compare $W_{1}(d)$ with $W_{1}(d')$. For any SP type defining word of $d'$ which does not involve the letter $R_{n_{1}+n_{2}}$, it is also the SP type defining word of d. For the SP type defining word of $d'$ which involves the letter $R_{n_{1}+n_{2}}$, say $V_{1}R_{n_{1}+n_{2}}$, it has one more factor $R_{n_{1}+n_{2}}$ than the corresponding SP type defining word $V_{1}$ of d. Hence $d'$ has less aberration of SP type than d. This contradicts the assumption that d is an SP-MA $2^{(n_{1}+n_{2})-(k_{1}+k_{2})}$ design. This completes the proof. $\square$

Proof of Theorem 3.1

The design $d_{0}$ has only SP type defining words because the WP part of $d_{0}$ is a full $2^{n_1}$ design. Then $W_1(d_0)=W(d)$. By Definition 2.2, if the design d is an MA $2^{n-k_{2}}$ design, then the design $d_{0}$ is an SP-MA design. $\square$

Proof of Theorem 3.2

The WP part of design $d_{0}$ is a $2^{n_{1}-k_{1}}$ design d. Note that the design $d_{0}$ has only WP type defining words. Then $A_{i,0}(d_0)=A_i(d)$, for $i=1,\ldots ,n_1$. Thus the $2^{(n_{1}+n_{2})-(k_{1}+0)}$ design $d_{0}$ is an SP-MA design if and only if the design d is an MA $2^{n_{1}-k_{1}}$ design. $\square$

Proof of Theorem 3.3

The SP wordlength pattern of the design $d_{0}$ determined by (1) is

$$\begin{aligned} W_{1}(d_{0})=\left( 0,\ldots ,0,\ \underset{n_{1}+n_{2}-2}{1,}\ \underset{n_{1}+n_{2}-1}{1,}\ 0 \right) , \end{aligned}$$

where $A_{n_{1}+n_{2}-2,1}(d_{0})=1$, $A_{n_{1}+n_{2}-1,1}(d_{0})=1$. To prove that $d_0$ is an SP-MA design, we need only to consider the $2^{(n_{1}+n_{2})-(1+1)}$ designs d with $A_{i,1}(d)=0,i=1,\ldots , n-3$ and $A_{n-2,1}(d)\le 1$, where $n=n_{1}+n_{2}$. By Corollary 2.5(1) and (2), for any such design, we have

$$\begin{aligned}{} & {} \sum _{i=1}^n A_{i,0}(d)+A_{n-2,1}(d)+A_{n-1,1}(d)+A_{n,1}(d)=3, \end{aligned}$$

(5)

$$\begin{aligned}{} & {} \sum _{i=1}^n iA_{i,0}(d)+(n-2)A_{n-2,1}(d)+(n-1)A_{n-1,1}(d)+nA_{n,1}(d)=2n. \end{aligned}$$

(6)

Note that $\sum _{i=1}^n A_{i,0}(d)=1$ and only designs with $A_{1,0}(d)=A_{2,0}(d)=0$ are considered. Subtracting $n-2$ times the Eq. (5) from the Eq. (6) yields

$$\begin{aligned} A_{n-1,1}(d)+2A_{n,1}(d)=4-\sum _{i=1}^n iA_{i,0}(d)\le 1. \end{aligned}$$

(7)

By Corollary 2.5(3), the Eq. (7) has the unique solution $A_{n-2,1}(d)=1$, $A_{n-1,1}(d)=1$, $A_{3,0}(d)=1$. Hence the $2^{(n_{1}+n_{2})-(1+1)}$ design $d_{0}$ is an SP-MA design. $\square$

Proof of Theorem 3.5

The proof is divided into three parts according to $r_{1}+r_{2}=0,1,2$ modulo 3.

(1) For $r_{1}+r_{2}=0$ modulo 3, we first consider the case $r_{1}+r_{2}=0$, that is $r_{1}=0$ and $r_{2}=0$. By (2) and Table 1, we get

$$\begin{aligned} {\left\{ \begin{array}{ll} u=w_1w_{m+1}w_{2m+1},\\ v_{1}=B_{1}B_{2}C_{1}C_{2},\\ v_{2}=B_{2}B_{3}C_{2}C_{3}. \end{array}\right. } \end{aligned}$$

The SP wordlength pattern of the design $d_{0}$ is

$$\begin{aligned} W_{1}(d_{0})=\left( 0,\ldots ,0,\ \underset{2m+2l-1}{3,}\ \underset{2m+2l}{3,}\ 0,\ldots ,0\right) , \end{aligned}$$

where $A_{2m+2l-1,1}(d_{0})=3$, $A_{2m+2l,1}(d_{0})=3$. To prove that $d_0$ is an SP-MA design, we need only to consider the $2^{(n_{1}+n_{2})-(1+2)}$ designs d with $A_{i,1}(d)=0, i=1,\ldots , 2\,m+2\,l-2$ and $A_{2\,m+2\,l-1,1}(d)\le 3$. By Corollary 2.5(1) and (2), for any such design, we have

$$\begin{aligned}{} & {} \sum _{i=1}^n A_{i,0}(d)+A_{2m+2l-1,1}(d)+A_{2m+2l,1}(d)+\cdots =7, \end{aligned}$$

(8)

$$\begin{aligned}{} & {} \sum _{i=1}^n iA_{i,0}(d)+(2m+2l-1)A_{2m+2l-1,1}(d)+(2m+2l)A_{2m+2l,1}(d)+\cdots =4n.\nonumber \\ \end{aligned}$$

(9)

Note that $\sum _{i=1}^n A_{i,0}(d)=1$ and only designs with $A_{1,0}(d)=A_{2,0}(d)=0$ are considered. Subtracting $2m+2l-1$ times the Eq. (8) from the Eq. (9) yields

$$\begin{aligned} A_{2m+2l,1}(d)+2A_{2m+2l+1,1}(d)+\cdots =6-\sum _{i=1}^n iA_{i,0}(d)\le 3, \end{aligned}$$

(10)

where $n=n_1+n_2$. By Corollary 2.5(3), the Eq. (10) has the unique solution $A_{2m+2l-1,1}(d)=3$, $A_{2m+2l,1}(d)=3$, $A_{3,0}(d)=1$. Hence the $2^{(n_{1}+n_{2})-(1+2)}$ design $d_{0}$ determined by (2) is an SP-MA design.

Now, consider the case $r_{1}+r_{2}=3$, that is $r_{1}=1$ and $r_{2}=2$, or $r_{1}=2$ and $r_{2}=1$. From (2) and Table 1, we can get the three independent defining words as

$$\begin{aligned} {\left\{ \begin{array}{ll} u=w_1w_{m+1}w_{2m+1},\\ v_{1}=B_{1}B_{2}C_{1}C_{2}w_{n_1}s_{n_1+n_2-1},\\ v_{2}=B_{2}B_{3}C_{2}C_{3}s_{n_{1}+n_{2}-1}s_{n_{1}+n_{2}} \end{array}\right. } \end{aligned}$$

for $r_{1}=1$ and $r_{2}=2$, or

$$\begin{aligned} {\left\{ \begin{array}{ll} u=w_1w_{m+1}w_{2m+1},\\ v_{1}=B_{1}B_{2}C_{1}C_{2}w_{n_{1}-1}w_{n_{1}},\\ v_{2}=B_{2}B_{3}C_{2}C_{3}w_{n_{1}}s_{n_{1}+n_{2}} \end{array}\right. } \end{aligned}$$

for $r_{1}=2$ and $r_{2}=1$. The SP wordlength pattern of the design $d_{0}$ is

$$\begin{aligned} W_{1}(d_{0})=\left( 0,\ldots ,0,\underset{2m+2l+1}{3,}\ \underset{2m+2l+2}{3,}0,\ldots ,0 \right) , \end{aligned}$$

where $A_{2m+2l+1,1}(d_{0})=3$, $A_{2m+2l+2,1}(d_{0})=3$. The rest of the proof is similar to that for $r_{1}+r_{2}=0$ and is omitted.

(2) For $r_{1}+r_{2}=1$ modulo 3, that is $r_{1}=0$, $r_{2}=1$, or $r_{1}=1$, $r_{2}=0$, or $r_{1}=2$, $r_{2}=2$, the proofs are similar. We only prove the case of $r_{1}=0$, $r_{2}=1$. By (2) and Table 1, we get

$$\begin{aligned} {\left\{ \begin{array}{ll} u=w_1w_{m+1}w_{2m+1},\\ v_{1}=B_{1}B_{2}C_{1}C_{2}s_{n_{1}+n_{2}},\\ v_{2}=B_{2}B_{3}C_{2}C_{3}. \end{array}\right. } \end{aligned}$$

The SP wordlength pattern of the design $d_{0}$ is

$$\begin{aligned} W_{1}(d_{0})=\left( 0,\ldots ,0,\ \underset{2m+2l-1}{1,}\ \underset{2m+2l}{3,}\ \underset{2m+2l+1}{2,}\ 0,\ldots ,0 \right) , \end{aligned}$$

where $A_{2m+2l-1,1}(d_{0})=1$, $A_{2m+2l,1}(d_{0})=3$, $A_{2m+2l+1,1}(d_{0})=2$. To prove that $d_0$ is an SP-MA design, we need only to consider the $2^{(n_{1}+n_{2})-(1+2)}$ designs d with $A_{i,1}(d)=0$, $i=1, \ldots , 2\,m+2\,l-2$ and $A_{2\,m+2\,l-1,1}(d)\le 1$. By Corollary 2.5(1) and (2), for any such design, we have

$$\begin{aligned}{} & {} \sum _{i=1}^n A_{i,0}(d)+A_{2m+2l-1,1}(d)+A_{2m+2l,1}(d)+\cdots =7,\nonumber \\ \end{aligned}$$

(11)

$$\begin{aligned}{} & {} \sum _{i=1}^n iA_{i,0}(d)+(2m+2l-1)A_{2m+2l-1,1}(d)+(2m+2l)A_{2m+2l,1}(d)+\cdots =4n,\nonumber \\ \end{aligned}$$

(12)

where $n=n_1+n_2$. Note that $\sum _{i=1}^n A_{i,0}(d)=1$ and only designs with $A_{1,0}(d)=A_{2,0}(d)=0$ are considered. Subtracting $2m+2l-1$ times the Eq. (11) from the Eq. (12) yields

$$\begin{aligned} A_{2m+2l,1}(d)+2A_{2m+2l+1,1}(d)+\cdots =10-\sum _{i=1}^n iA_{i,0}(d)\le 7. \end{aligned}$$

(13)

By Corollary 2.5(3), two solutions of the Eq. (13) meet our requirements: $A_{2m+2l,1}(d)=6$, $A_{4,0}(d)=1$ and $A_{2\,m+2\,l-1,1}(d)=1$, $A_{2m+2l,1}(d)=3$, $A_{2m+2l+1,1}(d)=2$, $A_{3,0}(d)=1$. Suppose $d_1$ is a design with $A_{2m+2l,1}(d_1)=6$, $A_{4,0}(d_1)=1$. Let l be an SP factor in the SP defining words. By deleting all the words containing l, the remaining words define a $2^{(n_1+n_2-1)-(1+1)}$ design $d_2$ with $A_{2m+2l,1}(d_2)=2$, $A_{4,0}(d_2)=1$. From Corollary 2.5(2), we get $2(2m+2l)+4=2(3m+3l)$, then $m+l=2$, that is $m=1$ and $l=1$. Since $A_{4,0}(d)=1$, we have $r_1=1$ and $r_2=0$. Hence $n_1=4$, $n_2=3$. Without loss of generality, the WP factors are denoted as $w_1,w_2,w_3,w_4$ and the SP factors are denoted as $s_5,s_6,s_7$. Three independent defining words of the design $d_1$ is $u=w_1w_2w_3w_4, v_{1}=w_3w_4s_5s_6, v_{2}=w_1w_3s_5s_7$ and the wordlength pattern of the design $d_1$ is

$$\begin{aligned} {\left\{ \begin{array}{ll} ~W_{1}(d_1)&{}=(0,0,0,6,0,0,0),\\ ~W_{2}(d_1)&{}=(0,0,0,1,0,0,0), \end{array}\right. } \end{aligned}$$

By Definition 2.2, when $m+l=2$, the design $d_1$ is optimal; when $m+l>2$, the design $d_0$ is optimal.

(3) For $r_{1}+r_{2}=2$ modulo 3, that is $r_{1}=0$, $r_{2}=2$, or $r_{1}=1$, $r_{2}=1$, or $r_{1}=2$, $r_{2}=0$, the proofs are similar. We only prove the case of $r_{1}=0$, $r_{2}=2$. By (2) and Table 1, we get

$$\begin{aligned} {\left\{ \begin{array}{ll} u=w_1w_{m+1}w_{2m+1},\\ v_{1}=B_{1}B_{2}C_{1}C_{2}s_{n_{1}+n_{2}-1},\\ v_{2}=B_{2}B_{3}C_{2}C_{3}s_{n_{1}+n_{2}}. \end{array}\right. } \end{aligned}$$

The SP wordlength pattern of the design $d_{0}$ is

$$\begin{aligned} W_{1}(d_{0})=\left( 0,\ldots ,0,\ \underset{2m+2l}{2,}\ \underset{2m+2l+1}{3,}\ \underset{2m+2l+2}{1,}\ 0,\ldots ,0 \right) , \end{aligned}$$

where $A_{2m+2l,1}(d_{0})=2$, $A_{2m+2l+1,1}(d_{0})=3$, $A_{2m+2l+2,1}(d_{0})=1$. To prove that $d_0$ is an SP-MA design, we need only to consider the $2^{(n_{1}+n_{2})-(1+2)}$ designs d that have $A_{i,1}(d)=0$, $i=1, \ldots , 2\,m+2\,l-1$ and $A_{2\,m+2\,l,1}(d)\le 2$. By Corollary 2.5(1) and (2), for any such design, we have

$$\begin{aligned}{} & {} \sum _{i=1}^n A_{i,0}(d)+A_{2m+2l,1}(d)+A_{2m+2l+1,1}(d)+\cdots =7,\nonumber \\ \end{aligned}$$

(14)

$$\begin{aligned}{} & {} \sum _{i=1}^n iA_{i,0}(d)+(2m+2l)A_{2m+2l,1}(d)+(2m+2l+1)A_{2m+2l+1,1}(d)+\cdots =4n, \nonumber \\ \end{aligned}$$

(15)

where $n=n_1+n_2$. Note that $\sum _{i=1}^n A_{i,0}(d)=1$ and only designs with $A_{1,0}(d)=A_{2,0}(d)=0$ are considered. Subtracting $2m+2l$ times the Eq. (14) from the Eq. (15) yields

$$\begin{aligned} A_{2m+2l+1,1}(d)+2A_{2m+2l+2,1}(d)+\cdots =8-\sum _{i=1}^n iA_{i,0}(d)\le 5. \end{aligned}$$

(16)

By Corollary 2.5(3), two solutions of (16) meet our requirements: $A_{2m+2l,1}(d)=2$, $A_{2m+2l+1,1}(d)=3$, $A_{2m+2l+2,1}(d)=1$, $A_{3,0}(d)=1$ and $A_{2m+2l,1}(d)=2$, $A_{2m+2l+1,1}(d)=4$, $A_{4,0}(d)=1$. By Definition 2.2, the design d with $A_{2m+2l,1}(d)=2$, $A_{2m+2l+1,1}(d)=3$, $A_{2m+2l+2,1}(d)=1$, $A_{3,0}(d)=1$ is optimal. Hence the $2^{(n_{1}+n_{2})-(1+2)}$ design $d_{0}$ determined by (2) is an SP-MA design. This completes the proof of Theorem 3.5. $\square$

Proof of Theorem 3.7

The proof can be divided into seven parts according to $r_{1}+r_{2}=0,1,\ldots ,6$ modulo 7. All proofs of the seven cases are similar. We only give the proof of $r_{1}+r_{2}=0$ to save space.

When $r_{1}+r_{2}=0$, that is $r_{1}=0$ and $r_{2}=0$, by (3), we get

$$\begin{aligned} {\left\{ \begin{array}{ll} u=w_{4m+1}w_{5m+1}w_{6m+1},\\ v_{1}=B_{1}B_{2}B_{3}B_{4}C_{1}C_{2}C_{3}C_{4},\\ v_{2}=B_{3}B_{4}B_{5}B_{6}C_{3}C_{4}C_{5}C_{6},\\ v_{3}=B_{1}B_{3}B_{5}B_{7}C_{1}C_{3}C_{5}C_{7}. \end{array}\right. } \end{aligned}$$

The SP wordlength pattern of the design $d_{0}$ is

$$\begin{aligned} W_{1}(d_{0})=\left( 0,\ldots ,0,\ \underset{4m+4l-1}{6,}\ \underset{4m+4l}{7,}\ 0,0,\underset{4m+4l+3}{1,}\ 0,\ldots ,0 \right) , \end{aligned}$$

where $A_{4m+4l-1,1}(d_{0})=6$, $A_{4m+4l,1}(d_{0})=7$, $A_{4m+4l+3,1}(d_{0})=1$. To prove that $d_0$ is an SP-MA design, we only need to consider the $2^{(n_{1}+n_{2})-(1+3)}$ designs d that have $A_{i,1}(d)=0, i=1,\ldots , 4\,m+4\,l-2$ and $A_{4\,m+4\,l-1,1}(d)\le 6$. By Corollary 2.5(1) and (2), for any such design, we have

$$\begin{aligned}{} & {} \sum _{i=1}^n A_{i,0}(d)+A_{4m+4l-1,1}(d)+A_{4m+4l,1}(d)+\cdots =15, \nonumber \\ \end{aligned}$$

(17)

$$\begin{aligned}{} & {} \sum _{i=1}^n iA_{i,0}(d)+(4m+4l-1)A_{4m+4l-1,1}(d)+(4m+4l)A_{4m+4l,1}(d)+\cdots =8n,\nonumber \\ \end{aligned}$$

(18)

where $n=n_1+n_2$. Note that $\sum _{i=1}^n A_{i,0}(d)=1$ and only designs with $A_{1,0}(d)=A_{2,0}(d)=0$ are considered. Subtracting $4m+4l-1$ times the Eq. (17) from the Eq. (18) yields

$$\begin{aligned} A_{4m+4l,1}(d)+2A_{4m+4l+1,1}(d)+\cdots =14-\sum _{i=1}^n iA_{i,0}(d)\le 11. \end{aligned}$$

(19)

From (19), we obtain the wordlength patterns of possible designs having less aberration of type SP than $d_{0}$

$$\begin{aligned} \begin{aligned} \left\{ \begin{array}{ll} W_{1}(d)=\left( 0,\ldots ,0,\ \underset{4m+4l-1}{5,}\ \underset{4m+4l}{7,}\ \underset{4m+4l+1}{2,}\ 0,\ldots ,0 \right) ,&{}\\ W_{2}(d)=(0,0,1,0,\ldots ,0),&{} \end{array} \right. \end{aligned} \end{aligned}$$

(20)

$$\begin{aligned} \begin{aligned} \left\{ \begin{array}{ll} W_{1}(d)=\left( 0,\ldots ,0,\ \underset{4m+4l-1}{6,}\ \underset{4m+4l}{6,}\ \underset{4m+4l+1}{1,}\ \underset{4m+4l+2}{1,}\ 0,\ldots ,0 \right) ,&{}\\ W_{2}(d)=(0,0,1,0,\ldots ,0), \end{array} \right. \end{aligned} \end{aligned}$$

(21)

$$\begin{aligned} \begin{aligned} \left\{ \begin{array}{ll} W_{1}(d)=\left( 0,\ldots ,0,\ \underset{4m+4l-1}{6,}\ \underset{4m+4l}{6,}\ \underset{4m+4l+1}{2,}\ 0,\ldots ,0 \right) ,&{}\\ W_{2}(d)=(0,0,0,1,0,\ldots ,0). \end{array} \right. \end{aligned} \end{aligned}$$

(22)

In the following, we show that no design has the wordlength patterns (20–22). Suppose $d_1$ is a design with one of the possible wordlength patterns (20–22). Corollary 2.5(3) and the Eqs. (20–22) imply that $d_1$ has 8 odd-length defining words. By adding a new letter $n_{1}+n_{2}+1$ to each of the odd-length defining words, we have a $2^{(n_{1}+n_{2}+1)-(1+3)}$ design $d_2$. Clearly, every defining word of $d_2$ has even length and $d_2$ has the wordlength pattern as

$$\begin{aligned} \begin{aligned} \left\{ \begin{array}{ll} W_{1}(d_{2})=\left( 0,\ldots ,0,\ \underset{4m+4l}{12,}\ 0,\ \underset{4m+4l+2}{2,}\ 0,\ldots ,0 \right) ,&{}\\ W_{2}(d_{2})=(0,0,0,1,0,\ldots ,0). \end{array} \right. \end{aligned} \end{aligned}$$

(23)

Now we prove the impossibility of (23). Clearly, there exists a letter l appearing in one of the two longest words but not in the other. By deleting all the words containing l, the remaining words define a $2^{(n_{1}'+n_{2}')-(1+2)}$ design $d_{3}$ or a $2^{(n_{1}'+n_{2}')-(0+3)}$ design $d_{4}$ with $n_{1}'+n_{2}'\le n_{1}+n_{2}$. From Corollary 2.5(1), the wordlength patterns of designs $d_{3}$ and $d_{4}$ must be

$$\begin{aligned} \begin{aligned} \left\{ \begin{array}{ll} W_{1}(d_{3})=\left( 0,\ldots ,0,\ \underset{4m+4l}{5,}\ 0, \underset{4m+4l+2}{1,}\ 0,\ldots ,0 \right) ,&{}\\ W_{2}(d_{3})=(0,0,0,1,0,\ldots ,0), \end{array} \right. \end{aligned} \end{aligned}$$

or

$$\begin{aligned} W_{1}(d_{4})= \left( 0,\ldots ,0,\ \underset{4m+4l}{6,}\ 0,\ \underset{4m+4l+2}{1,}\ 0,\ldots ,0 \right) . \end{aligned}$$

Then, we have

$$\begin{aligned} \sum _{i=1}^n iA_{i,0}(d_3)+\sum _{i=1}^n iA_{i,1}(d_3)=24(m+l)+6\ne n2^{3-1}=4n \end{aligned}$$

and

$$\begin{aligned} \sum _{i=1}^n iA_{i,0}(d_4)+\sum _{i=1}^n iA_{i,1}(d_4)=28(m+l)+2\ne n2^{3-1}=4n. \end{aligned}$$

Both violates Corollary 2.5(2). Therefore such a design $d_2$ does not exist. Hence the $2^{(n_{1}+n_{2})-(1+3)}$ design $d_{0}$ determined by (3) is an SP-MA design. $\square$

The proof of Lemma 3.9 is similar to that of Lemma 2.3 and is omitted.

Proof of Theorem 3.10

The SP wordlength pattern of the design $d_{0}$ determined by (4) is

$$\begin{aligned} W_{1}(d_{0})=\left( 0,\ldots ,0,\ \underset{n_{1}+n_{2}-3}{2,}\ \underset{n_{1}+n_{2}-2}{2,}\ 0,0 \right) , \end{aligned}$$

where $A_{n_{1}+n_{2}-3,1}(d_{0})=2$, $A_{n_{1}+n_{2}-2,1}(d_{0})=2$. To prove that $d_0$ is an SP-MA design, we need only to consider the $2^{(n_{1}+n_{2})-(2+1)}$ designs d that have $A_{i,1}(d)=0,i=1, \ldots , n-4$ and $A_{n-3,1}(d)\le 2$, where $n=n_{1}+n_{2}$. By Corollary 2.5(1) and (2), for any such design, we have

$$\begin{aligned}{} & {} \sum _{i=1}^n A_{i,0}(d)+A_{n-3,1}(d)+A_{n-2,1}(d)+A_{n-1,1}(d)+A_{n,1}(d)=7, \end{aligned}$$

(24)

$$\begin{aligned}{} & {} \sum _{i=1}^n iA_{i,0}(d)+(n-3)A_{n-3,1}(d)+\cdots +nA_{n,1}(d)=4n, \end{aligned}$$

(25)

where $n=n_1+n_2$. Note that $\sum _{i=1}^n A_{i,0}(d)=3$ and only designs with $A_{1,0}(d)=A_{2,0}(d)=0$ are considered. Subtracting $n-3$ times the Eq. (24) from the Eq. (25) yields

$$\begin{aligned} A_{n-2,1}(d)+2A_{n-1,1}(d)+3A_{n,1}(d)=12-\sum _{i=1}^n iA_{i,0}(d)\le 3. \end{aligned}$$

(26)

By Corollary 2.5(3), three solutions of the Eq. (26) meet our requirements: $A_{n-3,1}(d)=2$, $A_{n-2,1}(d)=2$, $A_{3,0}(d)=2$, $A_{4,0}(d)=1$ and $A_{n-3,1}(d)=2$, $A_{n-2,1}(d)=1$, $A_{n-1,1}(d)=1$, $A_{3,0}(d)=3$, where n is odd number and $A_{n-3,1}(d)=1$, $A_{n-2,1}(d)=3$, $A_{3,0}(d)=3$,where n is even number. Note that the WP part of a $2^{(n_{1}+n_{2})-(2+1)}$ design d is a $2^{n_{1}-2}$ design $d^*$ which is determined by the WP type defining words of d. Thus, $A_{3,0}(d)=3$ in the second and third solution above implies that $A_{3}(d^*)=3$, i.e., all the three defining words of $d^*$ have length 3. This contradicts Lemma 2.4(3). Hence the $2^{(n_{1}+n_{2})-(2+1)}$ design $d_{0}$ determined by (4) is an SP-MA design. This completes the proof of Theorem 3.10. $\square$