Mixed two- and four-level split-plot designs with combined minimum aberration

Abstract

When the levels of some factors in an experiment are difficult to be changed or controlled, fractional factorial split-plot (FFSP) designs are commonly used in which the factors are classified as the whole plot (WP) and sub-plot (SP) factors. Mixed-level designs are used in practice when the levels of the factors are not equal to each other. This paper considers the mixed-level FFSP designs with the WP factors being more important than the SP factors. It proposes the combined minimum aberration criterion of type WP (WP-MA$^{c}$) for mixed-level FFSP designs. Some optimal mixed-level FFSP designs are constructed under the WP-MA$^{c}$ criterion.

Appendices

Appendix A: Proofs of lemmas and theorems

Proof of Lemma 1

We prove that a design with some two-level factor or component of the four-level factor not involved in its defining contrast subgroup cannot be a WP-MA$^{c}$ design.

(1) Suppose that $D_1$ is a $2^{(n_{1}+n_{2})-(k_{1}+k_{2})}4^{1}_{w}$ design and the two-level factor $b_{1}$ is not involved in its defining contrast subgroup. Denote the independent defining words of design $D_{1}$ by

$$\begin{aligned} a_{1}z_{1},a_{2}z_{2},z_{3},\ldots ,z_{k_{1}},w_{1},w_{2},\ldots ,w_{k_{2}}, \end{aligned}$$

where $a_{1}z_{1},a_{2}z_{2},\ldots ,z_{k_{1}}$ are WP defining words, $w_1,w_2,\ldots ,w_{k_{2}}$ are SP defining words. The letters $a_{1},a_{2}$ and $b_{1}$ are not included in $z_{1},z_{2},\ldots ,z_{k_{1}},w_1,w_2,\ldots , w_{k_{2}}$. Then the defining contrast subgroup of design $D_{1}$ is

$$\begin{aligned} G(D_{1})=G_{1}\cup (a_{1}z_{1}G_{1})\cup (a_{2}z_{2}G_{1})\cup (a_{3}z_{1}z_{2}G_{1}), \end{aligned}$$

(6)

where $G_{1}$ is a subgroup generated by $z_{3},\ldots ,z_{k_{1}},w_1,w_2,\ldots ,w_{k_{2}}$ and $\gamma G_{1}=\{\gamma g: g\in G_{1}\}$ with $\gamma =a_{1}z_{1}, a_{2}z_{2}, a_{3}z_{1}z_{2}$. Consider the $2^{(n_{1}+n_{2})-(k_{1}+k_{2})}4^{1}_{w}$ design $D'_{1}$ determined by the independent defining words

$$\begin{aligned} a_{1}z_{1},a_{2}b_{1}z_{2},z_{3},\ldots ,z_{k_{1}},w_{1},w_{2},\ldots ,w_{k_{2}}. \end{aligned}$$

The defining contrast subgroup of $D'_{1}$ is

$$\begin{aligned} G(D'_{1})=G_{1}\cup (a_{1}z_{1}G_{1})\cup (a_{2}b_{1}z_{2}G_{1})\cup (a_{3}b_{1}z_{1}z_{2}G_{1}). \end{aligned}$$

Note that $D_{1}$ and $D'_{1}$ have the same defining words in $G_{1}\cup (a_{1}z_{1}G_{1})$, and each defining word $w\in (a_{2}z_{2}G_{1})\cup (a_{3}z_{1}z_{2}G_{1})$ has one less letter than the corresponding defining word $b_1w\in (a_{2}b_{1}z_{2}G_{1})\cup (a_{3}b_{1}z_{1}z_{2}G_{1})$. Then $D'_{1}$ has less combined aberration of type WP than $D_{1}$.

(2) The second case can be proved similarly. Now suppose that $D_2$ is a $2^{(n_{1}+n_{2})-(k_{1}+k_{2})}4^{1}_{w}$ design and the component $a_3$ of the four-level factor is not involved in its defining contrast subgroup. Note that any two $a_{i}$ that appear in a defining word can be replaced by the third $a_{i}$ because of $I=a_{1}a_{2}a_{3}$. Thus, at most one of $a_1$ and $a_2$ is involved in $G(D_{2})$. Without loss of generality, denote the independent defining words of design $D_{2}$ by

$$\begin{aligned} a_{1}z_{1},z_{2},\ldots ,z_{k_{1}},w_{1},w_{2},\ldots ,w_{k_{2}}, \end{aligned}$$

where $z_{1},z_{2},\ldots ,z_{k_{1}},w_1,w_2,\ldots ,w_{k_{2}}$ do not contain $a_{1}, a_{2}, a_{3}$. The defining contrast subgroup of design $D_{2}$ is

$$\begin{aligned} G(D_{2})=G_{1}\cup (a_{1}z_{1}G_{1})\cup (z_{2}G_{1})\cup (a_{1}z_{1}z_{2}G_{1}), \end{aligned}$$

where the meaning of the symbols are the same as in (6). Now, consider the $2^{(n_{1}+n_{2})-(k_{1}+k_{2})}4^{1}_{w}$ design $D'_{2}$ determined by

$$\begin{aligned} a_{1}z_{1},a_{2}z_{2},z_{3},\ldots ,z_{k_{1}},w_{1},w_{2},\ldots ,w_{k_{2}}. \end{aligned}$$

The defining contrast subgroup of $D'_{2}$ is

$$\begin{aligned} G(D'_{2})=G_{1}\cup (a_{1}z_{1}G_{1})\cup (a_{2}z_{2}G_{1})\cup (a_{3}z_{1}z_{2}G_{1}). \end{aligned}$$

Note that $D_{2}$ and $D'_{2}$ have the same defining words in $G_{1}\cup (a_{1}z_{1}G_{1})$, and each defining word $a_1w\in a_{1}z_{1}z_{2}G_{1}$ has the same length as the corresponding defining word $a_3w\in a_{3}z_{1}z_{2}G_{1}$, and each defining word $w\in z_{2}G_{1}$ has one less letter than the corresponding defining word $a_2w\in a_{2}z_{2}G_{1}$. Then $D'_{2}$ has less combined aberration of type WP than $D_{2}$. This completes the proof. $\square$

Proof of Lemma 2

(1) This is obvious from the fact that there are $2^{k}-1$ defining words in the defining contrast subgroup of D.

(2) Suppose that D is a WP-MA$^{c}$ $2^{(n_{1}+n_{2})-(k_{1}+k_{2})}4^{1}_{w}$ design determined by C and the independent defining words

$$\begin{aligned} a_{1}z_{1},a_{2}z_{2},z_{3},\ldots ,z_{k_{1}},w_{1},w_{2},\ldots ,w_{k_{2}}, \end{aligned}$$

where $z_{1},z_{2},\ldots ,z_{k_{1}},w_{1},w_{2},\ldots ,w_{k_{2}}$ contain all the two-level factors. The defining contrast subgroup of D is

$$\begin{aligned} G=G_{1}\cup (a_{1}z_{1}G_{1})\cup (a_{2}z_{2}G_{1})\cup (a_{3}z_{1}z_{2}G_{1}), \end{aligned}$$

where $G_{1}$ is the same as in (6). Removing the column $a_{3}$ in C we get a $2^{(n_{1}+n_{2}+2)-(k_{1}+k_{2})}$ design ${\widetilde{D}}$. Then the defining contrast subgroup of ${\widetilde{D}}$ is

$$\begin{aligned} {\widetilde{G}}=G_{1}\cup (a_{1}z_{1}G_{1})\cup (a_{2}z_{2}G_{1})\cup (a_{1}a_{2}z_{1}z_{2}G_{1}). \end{aligned}$$

From Lemma 2(2) of Wang et al. (2019), we can get that the total number of letters in all the defining words of ${\widetilde{D}}$ is $(n_{1}+n_{2}+2)\times 2^{k_{1}+k_{2}-1}=(n+2)\times 2^{k-1}.$ By comparing G and ${\widetilde{G}}$, the designs D and ${\widetilde{D}}$ have the same defining words in $G_{1}\cup (a_{1}z_{1}G_{1})\cup (a_{2}z_{2}G_{1})$, and each defining word $a_3z_{1}z_{2}w\in a_{3}z_{1}z_{2}G_{1}$ has one less letter than the corresponding defining word $a_1a_2z_{1}z_{2}w\in a_{1}a_2z_{1}z_{2}G_{1}$. Therefore, the total number of all letters in G is

$$\begin{aligned} \sum _{i=1}^{n+1}iB^{w}_{i}(D)+\sum _{i=1}^{n+1}iB^{s}_{i}(D)=(n+2)\times 2^{k-1}-2^{k-2}=n\times 2^{k-1}+3\times 2^{k-2}. \end{aligned}$$

(3) Removing the columns $a_{1},a_{2},a_{3}$ in C, we get a $2^{(n_{1}+n_{2})-(k_{1}+k_{2})}$ design $D_{0}$ with defining contrast subgroup

$$\begin{aligned} G_0=G_{1}\cup (z_{1}G_{1})\cup (z_{2}G_{1})\cup (z_{1}z_{2}G_{1}). \end{aligned}$$

From Lemma 2(3) of Wang et al. (2019), either all the defining words of $G_{0}$ have even lengths or $2^{k-1}$ of them have odd lengths. Since $G_{1}$ is a subgroup generated by $z_{3},\ldots ,z_{k_{1}},w_1, w_2,\ldots ,w_{k_{2}}$ and it can be regarded as the defining contrast subgroup of a two-level design, either all the defining words in $G_{1}$ have even lengths or $2^{k-3}$ of them have odd lengths.

By comparing $G_{0}$ and G, the designs $D_0$ and D have the same defining words in $G_{1}$, and each defining word $w\in (z_1G_{1})\cup (z_2G_{1})\cup (z_1z_2G_{1})$ has one less letter than the corresponding defining word $a_iw\in (a_1z_1G_{1})\cup (a_2z_2G_{1})\cup (a_3z_1z_2G_{1})$, $i=1,2,3$.

Let $M=(z_{1}G_{1})\cup (z_{2}G_{1})\cup (z_{1}z_{2}G_{1})$. Note that there are $2^{k-2}$ defining words in $z_{1}G_{1}$, $z_{2}G_{1}$ and $z_{1}z_{2}G_{1}$, respectively. There are three cases for the wordlength of defining words in $G_{0}$ and $G_{1}$.

(i) If the lengths of defining words in $G_{0}$ and $G_{1}$ are all even, then the lengths of defining words in M are all even and there are $3\times 2^{k-2}$ defining words with odd-length in G.

(ii) If $2^{k-1}$ defining words in $G_{0}$ have odd-length and all the defining words in $G_{1}$ have even-length, then there are $2^{k-1}$ odd-length and $2^{k-2}$ even-length defining words in M. Thus there are $2^{k-2}$ defining words with odd-length in G.

(iii) If $2^{k-1}$ defining words have odd-length in $G_{0}$ and $2^{k-3}$ ones have odd-length in $G_{1}$, then there are $3\times 2^{k-3}$ defining words with odd-length and $3\times 2^{k-3}$ defining words with even-length in M. So, there are $3\times 2^{k-3}+2^{k-3}=2^{k-1}$ defining words with odd-length in G. This completes the proof. $\square$

Proof of Theorem 1

Note that for a $2^{(n_{1}+n_{2})-(0+k_{2})}4^{1}_{w}$ design D, it has only the SP wordlength pattern $W_{2}$. The result follows immediately if it is regarded as a $2^{(n_{1}+n_{2})-k_{2}}4^{1}_{w}$ design. $\square$

Proof of Theorem 2

Without SP defining words in design D, the WP part of design D corresponds to a $2^{n_{1}-k_{1}}4^{1}$ design. Then, the $2^{(n_{1}+n_{2})-(k_{1}+0)}4^{1}_{w}$ design is a WP-MA$^{c}$ design if and only if the $2^{n_{1}-k_{1}}4^{1}$ design is an MA$^c$ design. $\square$

Proof of Theorem 3

It is obviously that the WP part of D constitutes an MA$^c$ $2^{n_{1}-1}4^{1}$ design. Hence $W^{c}_{1}(D)$ is sequentially minimized. We consider $W^{c}_{2}(D)$ in what follows. When $n_1=2m+1$, the SP wordlength pattern is $W^{c}_{2}=(0,...,0,1,1,0,...,0)$, where $B^{s}_{m+1+n_{2}}=1$, $B^{s}_{m+2+n_{2}}=1$. Hence we consider the $2^{(n_{1}+n_{2})-(1+1)}4^{1}_{w}$ designs satisfying $B^{s}_{i}=0, i\le m+n_{2}$. From Lemma 2, $\sum _{i=1}^{n+1} B_{i}^{s}=2$ and $\sum _{i=1}^{n+1} iB_{i}^{s}=2n+3-(n_{1}+1)=2m+2n_{2}+3$. Then,

$$\begin{aligned} B^{s}_{m+1+n_{2}}+B^{s}_{m+2+n_{2}}+\cdots&=2, \\ (m+1+n_{2})B^{s}_{m+1+n_{2}}+(m+2+n_{2})B^{s}_{m+2+n_{2}}+\cdots&=2m+2n_{2}+3. \end{aligned}$$

Subtracting $(m+1+n_{2})$ times the first equation from the second equation yields $B^{s}_{m+2+n_{2}}=1$, then $B^{s}_{m+1+n_{2}}=1$, $B^{s}_{m+2+n_{2}}=1$ is the unique solution. Thus D is a WP-MA$^{c}$ design.

When $n_{1}=2m$, the SP wordlength pattern is $W^{c}_{2}=(0,...,0,2,0,...,0)$, where $B^{s}_{m+1+n_{2}}=2$. Hence we consider only those $2^{(n_{1}+n_{2})-(1+1)}4^{1}_{w}$ designs satisfying $B_{i}^{s}=0, i\le m+n_{2}$. From Lemma 2, $\sum _{i=1}^{n+1} B_{i}^{s}=2$ and $\sum _{i=1}^{n+1} iB_{i}^{s}=2n+3-(n_{1}+1)=2m+2n_{2}+2$. Then,

$$\begin{aligned} B^{s}_{m+1+n_{2}}+B^{s}_{m+2+n_{2}}+\cdots&=2, \\ (m+1+n_{2})B^{s}_{m+1+n_{2}}+(m+2+n_{2})B^{s}_{m+2+n_{2}}+\cdots&=2m+2n_{2}+2. \end{aligned}$$

The unique solution of the two equations is $B^{s}_{m+1+n_{2}}=2$. Thus D is a WP-MA$^{c}$ design. $\square$

Proof of Theorem 4

By Section 4 in Zhang and Shao (2001), the WP part of D constitutes an MA$^c$ $2^{n_{1}-2}4^{1}$ design. Hence $W^{c}_{1}(D)$ is sequentially minimized. We consider $W^{c}_{2}(D)$ in what follows. The proof of WP-MA$^{c}$ property of D when $n_{1}=3$ and $n_{1}=4$ are first presented, and then the WP-MA$^{c}$ property of D when $r=0$, $r=1$ and $r=2$ are proved.

(1) When $n_{1}=3$, by (1), $W_{1}^{c}=(0,0,3,0,\ldots ,0)$, where $B^{w}_{3}=3$; $W_{2}^{c}=(0,\ldots ,0,3,1,0,\ldots ,0)$, where $B^{s}_{n_{2}+2}=3$ and $B^{s}_{n_{2}+3}=1$. We consider only the designs satisfying $B^{s}_{i}=0,i\le n_{2}+1$. From Lemma 2,

$$\begin{aligned} B^{w}_{3}+B^{s}_{n_{2}+2}+B^{s}_{n_{2}+3}+\cdots&=7,\\ 3B^{w}_{3}+(n_{2}+2)B^{s}_{n_{2}+2}+(n_{2}+3)B^{s}_{n_{2}+3}+\cdots&=4n+6. \end{aligned}$$

The unique solution of the two equations is $B^{s}_{n_{2}+2}=3$, $B^{s}_{n_{2}+3}=1$. Thus D is a WP-MA$^{c}$ design.

(2) When $n_{1}=4$, by (2), $W_{1}^{c}=(0,0,1,2,0,\ldots ,0)$, where $B^{w}_{3}=1$, $B^{w}_{4}=2$; $W_{2}^{c}=(0,\ldots ,0,1,3,0,\ldots ,0)$, where $B^{s}_{n_{2}+2}=1$ and $B^{s}_{n_{2}+3}=3$. We consider only the designs satisfying $B^{s}_{i}=0,i\le n_{2}+1$. From Lemma 2,

$$\begin{aligned} B^{w}_{3}+B^{w}_{4}+B^{s}_{n_{2}+2}+B^{s}_{n_{2}+3}+\cdots&=7,\\ 3B^{w}_{3}+4B^{w}_{4}+(n_{2}+2)B^{s}_{n_{2}+2}+(n_{2}+3)B^{s}_{n_{2}+3}+\cdots&=4n+6. \end{aligned}$$

The unique solution of the two equations is $B^{s}_{n_{2}+2}=1$, $B^{s}_{n_{2}+3}=3$. Thus D is a WP-MA$^{c}$ design.

(3) When $m>1$ and $r=0$, by (3), we get

$$\begin{aligned} z_{1}z_{2}&=a_{3}b_{1}\ldots b_{m}b_{(2m+1)}\ldots b_{(3m)}b_{(n_{1}-1)}b_{n_{1}},\\ z_{1}w&=a_{1}b_{(t+1)}\ldots b_{m}b_{(m+1)}\ldots b_{(m+t)}b_{(2m+1)}\ldots b_{(2m+t)}b_{n_{1}}c_{1}c_{2}\ldots c_{n_{2}},\\ z_{2}w&=a_{2}b_{1}\ldots b_{t}b_{(m+1)}\cdots b_{(m+t)}b_{(2m+t+1)}\cdots b_{(3m)}b_{(n_{1}-1)}c_{1}c_{2}\cdots c_{n_{2}},\\ z_{1}z_{2}w&=a_{3}b_{(t+1)}\cdots b_{m}b_{(m+t+1)}\cdots b_{(2m)}b_{(2m+t+1)}\cdots b_{(3m)}c_{1}c_{2}\cdots c_{n_{2}}. \end{aligned}$$

The WP wordlength pattern is $W^{c}_{1}=(0,\ldots ,0,2,1,0,\ldots ,0)$, where $B^{w}_{2m+2}=2$, $B^{w}_{2m+3}=1$.

When $l=0$, the SP wordlength pattern is $W^{c}_{2}=(0,\ldots ,0,1,3,0,\ldots ,0)$, where $B^{s}_{3t+1+n_{2}}=1$ and $B^{s}_{3t+2+n_{2}}=3$. Hence we consider the $2^{(n_{1}+n_{2})-(2+1)}4^{1}_{w}$ designs satisfying $B^{s}_{i}=0, i\le 3t+n_{2}$. By Lemma 2, there are four SP defining words and the sum of the wordlength of them should be $(4n+6)-(6m+7)=6m+7+4n_{2}$. Then,

$$\begin{aligned} B^{s}_{3t+1+n_{2}}+B^{s}_{3t+2+n_{2}}+\cdots&=4, \end{aligned}$$

(7)

$$\begin{aligned} (3t+1+n_{2})B^{s}_{3t+1+n_{2}}+(3t+2+n_{2})B^{s}_{3t+2+n_{2}}+\cdots&=6m+7+4n_{2}. \end{aligned}$$

(8)

Subtracting $(3t+1+n_{2})$ times (7) from (8) yields

$$\begin{aligned} B^{s}_{3t+2+n_{2}}+2B^{s}_{3t+3+n_{2}}+3B^{s}_{3t+4+n_{2}}+\cdots&=3, \end{aligned}$$

(9)

which forces $B^{s}_{3t+5+n_{2}}=B^{s}_{3t+6+n_{2}}=\cdots =0$. From (7) and (9), we can get three solutions, which are $B^{s}_{3t+1+n_{2}}=3$, $B^{s}_{3t+4+n_{2}}=1$; $B^{s}_{3t+1+n_{2}}=2$, $B^{s}_{3t+2+n_{2}}=1$, $B^{s}_{3t+3+n_{2}}=1$ and $B^{s}_{3t+1+n_{2}}=1$, $B^{s}_{3t+2+n_{2}}=3$. It shows that there does not exist any design having less combined aberration of type WP than D. Thus D is a WP-MA$^{c}$ design.

When $l=1$, the sub-plot wordlength pattern is $W^{c}_{2}=(0,\ldots ,0,3,1,0,\ldots ,0)$, where $B^{s}_{3t+3+n_{2}}=3$ and $B^{s}_{3t+4+n_{2}}=1$. Hence we consider the $2^{(n_{1}+n_{2})-(2+1)}4^{1}_{w}$ designs satisfying $B^{s}_{i}=0, i\le 3t+2+n_{2}$. From Lemma 2,

$$\begin{aligned} B^{s}_{3t+3+n_{2}}+B^{s}_{3t+4+n_{2}}+\cdots&=4, \end{aligned}$$

(10)

$$\begin{aligned} (3t+3+n_{2})B^{s}_{3t+3+n_{2}}+(3t+4+n_{2})B^{s}_{3t+4+n_{2}}+\cdots&=6m+7+4n_{2}. \end{aligned}$$

(11)

Subtracting $(3t+3+n_{2})$ times (10) from (11) yields

$$\begin{aligned} B^{s}_{3t+4+n_{2}}+2B^{s}_{3t+5+n_{2}}+3B^{s}_{3t+6+n_{2}}+\cdots&=1, \end{aligned}$$

(12)

which forces $B^{s}_{3t+5+n_{2}}=B^{s}_{3t+6+n_{2}}=\cdots =0$. From (10) and (12), we can get the unique solution $B^{s}_{3t+3+n_{2}}=3$, $B^{s}_{3t+4+n_{2}}=1$, thus proving the WP-MA$^{c}$ property of D.

When $m=1$, $r=0$, $n_{1}=5$ and $l=1$, from (3) we get

$$\begin{aligned} z_{1}=a_{1}b_{1}b_{2}b_{4}, z_{2}=a_{2}b_{2}b_{3}b_{5}, w=b_{1}b_{2}b_{5}c_{1}c_{2}\cdots c_{n_{2}}. \end{aligned}$$

Then

$$\begin{aligned} z_{1}z_{2}&=a_{3}b_{1}b_{3}b_{4}b_{5},\\ z_{1}w&=a_{1}b_{4}b_{5}c_{1}c_{2}\cdots c_{n_{2}},\\ z_{2}w&=a_{2}b_{1}b_{3}c_{1}c_{2}\cdots c_{n_{2}},\\ z_{1}z_{2}w&=a_{3}b_{2}b_{3}b_{4}c_{1}c_{2}\cdots c_{n_{2}}. \end{aligned}$$

It can be seen that the wordlength patterns of this design are

$$\begin{aligned} W^{c}_{1}&=(0,0,0,2,1,0,\ldots ,0),\\ W^{c}_{2}&=(0,\ldots ,0,3,1,0,\ldots ,0), \end{aligned}$$

where $B^{w}_{4}=2$, $B^{w}_{5}=1$, $B^{s}_{n_{2}+3}=3$, and $B^{s}_{n_{2}+4}=1$. We consider only the designs satisfying $B^{s}_{i}=0,i\le n_{2}+2$. From Lemma 2,

$$\begin{aligned} B^{s}_{n_{2}+3}+B^{s}_{n_{2}+4}+\cdots&=4,\\ (n_{2}+3)B^{s}_{n_{2}+3}+(n_{2}+4)B^{s}_{n_{2}+4}+\cdots&=4n_{2}+13. \end{aligned}$$

The unique solution of the two equations is $B_{n_{2}+3}=3$, $B_{n_{2}+4}=1$. Thus D is a WP-MA$^{c}$ design.

When $m>1$ and $r=1,2$, the proofs are similar to that of $m>1$ and $r=0$ and hence are omitted to save space. This completes the proof of Theorem 4. $\square$

Appendix B: Tables of the WP-MA$^c$ $2^{(n_1+n_2)-(k_1+k_2)}4^{1}_{w}$ designs with 8, 16, 32 and 64 runs

See Tables 1, 2, 3 and 4.

Table 1 WP-MA$^c$ $2^{(n_1+n_2)-(k_1+k_2)}4^{1}_{w}$ designs with 8 runs

Table 2 WP-MA$^c$ $2^{(n_1+n_2)-(k_1+k_2)}4^{1}_{w}$ designs with 16 runs

Table 3 WP-MA$^c$ $2^{(n_1+n_2)-(k_1+k_2)}4^{1}_{w}$ designs with 32 runs

Table 4 WP-MA$^c$ $2^{(n_1+n_2)-(k_1+k_2)}4^{1}_{w}$ designs with 64 runs