Partial aliased effect number pattern and selection of optimal compromise designs

Abstract

Often, experimenters are only interested in estimating a few factor specified effects. In this paper, we broadly call a design which can reach this target a compromise design. First, for assessing and selecting this kind of designs we introduce a partial aliased effect number pattern (P-AENP), then we use this pattern to study class one two-level compromise designs. Some theoretical results are obtained and a number of class one clear, strongly clear and general optimal $2^{n-m}$ compromise designs with 8, 16, 32 and 64 runs are tabulated.

Appendix: Proofs of Lemma 1, Theorems 1–5 and Propositions 1–6 and Tables 3, 4

Proof of Lemma 1

Obviously, first we have $c_1={}^{\#}_{\,1}P_2^{(0)}$, since the design T has the R at least III. We only need to prove the second equation. Let S denote the set of the specified 2fi’s not aliased with any other 2fi, implying that S contains all the clear specified 2fi candidates and $\#\{S\}={}^{\#}_{\,2}P_2^{(0)}$. This also means that there are $\#\{S\}$ alias cosets each contains only one 2fi and the 2fi is specified. Moreover, the $\#\{S\}$ alias cosets can be separated into two parts, in one part every coset contains one main effect and in the other part every coset does not contain main effect. Obviously, the specified 2fi’s in the cosets of the first part are not clear, while the specified 2fi’s in the cosets of the second part are clear. By the definition of notation ${}^{\#}_{\,1}C_{2^*}^{(1)}$, the number of cosets in the first part is just ${}^{\#}_{\,1}C_{2^*}^{(1)}$. Thus, the second equation is obtained. $\square$

Proof of Theorem 1

Let T be a $2^{n-m}$ OCD with $\#\{G_1\}=f\le n$. By the definition of OCD, ${}_{1}^{\#}P^{(0)}_2(T)$ is maximum among all the compromise designs, so from Lemma 1 the number of clear specified main effects of T is maximum. Then we consider if the number of clear specified 2fi’s of T is maximum. First we note that by Lemma 1 when $n\ge N/2$ any compromise design must have ${}^{\#}_{2}P_2^{(0)}-{}^{\#}_{1}C_{2^*}^{(1)}=0$, since any $2^{n-m}$ design has no clear 2fi by Chen and Hedayat (1998). Thus, when $n\ge N/2$, ${}^{\#}_{2}P_2^{(0)}-{}^{\#}_{1}C_{2^*}^{(1)}$ is the number of clear specified 2fi’s of T and is maximum, but it is 0. When $n\le N/2$, for any $f\le n$ there exists a compromise $2^{n-m}$ design of R IV, such that ${}_{2}^{\#}P^{(0)}_1$ is maximum among all the P-AENPs of the compromise designs respect to the same $\Omega _1$. Since ${}_{2}^{\#}P^{(0)}_1(T)$ and ${}^{\#}_{2}P_2^{(0)}(T)$ is sequentially maximum, and by definition of ${}^{\#}_{1}C_{2^*}^{(1)}$, T has ${}^{\#}_{1}C_{2^*}^{(1)}(T)=0$. Hence the number ${}^{\#}_{2}P_2^{(0)}(T)-{}^{\#}_{1}C_{2^*}^{(1)}(T)$ is maximum. Again by Lemma 1, the theorem is proved. $\square$

Clearly, with a substitution of $\{G_1,G_1\times G_1\}$ by a general $\Omega _1$ to compute ${}^{\#}_{1}C_{2^*}^{(1)}$, Lemma 1 and Theorem 1 are also valid.

Proof of Theorem 2

The first equation of (11) is obviously valid, we only prove the second one.

First consider $n=M(q)+1$ with $q\ge 4$. By the definition of M(q), any $2^{n-m}$ design with $n=M(q)+1$ must only have a R of III or IV. If having the R III, then in the design there are at least three factors whose main effects and 2fi’s are not clear, i.e., its $f(q,M(q)+1)\le M(q)-2$, which leads to $f^*(q,M(q)+1)\le M(q)-2$. If having the R IV, then the design must have four factors in a defining word with length four and hence at least three of them can not be in $G_1$ because of the confounding between their 2fi’s, this leads to its $f(q,M(q)+1)\le M(q)-2$ and $f^*(q,M(q)+1)\le M(q)-2$.

Then, consider $n=M(q)+2$ with $q\ge 4$. Let $T=\{G_1:G_2\}$ be a class one $2^{n-m}$ CCD with $n=M(q)+2$ and $f=\#\{G_1\}=f^*(q,n)$. Because the $G_1$ has the R V, so $f^*(q,n)\le M(q)$. Furthermore, $f^*(q,n)\ne M(q)-1$, since if not, then we have $\#\{G_2\}=3$, $\alpha \in G_2$ and a $2^{n-m}$ CCD $T'=$$\{G_1:\{G_2{\setminus }\{\alpha \}\}\}$ with $\#\{T'\}=M(q)+1$ and $f=M(q)-1$. This conflicts with $f^*(q,M(q)+1)\le M(q)-2$. We can similarly prove $f^*(q,n)\ne M(q)$. Thus $f^*(q,n)\le M(q)-2$.

At last, consider $n>M(q)+2$ with $q\ge 4$. Since for $n\ge M(q)+1$ we have $f^*(q,n)\le f^*(q,n-1)$, for $n>M(q)+2$ have $f^*(q,n)\le M(q)-2$. Thus, the second equation of (11) is proved. $\square$

Proof of Theorem 3

If $T=\{G_1:G_2\}$ is a $2^{n-m}$ SCCD, then $G_1$ must have R VI. Consider $n\le M^{VI}(q)$ for any q. For $q\le 4$, since $M^{VI}(q)=q$, to make f be largest we can take $G_1=\{\mathbf{1,\ldots ,n}\}$, and for $q\ge 5$ and $5\le n\le M^{VI}(q)$, to make f be the largest we can take a subset of $M^{VI}_q$ as $G_1$ with $\#\{G_1\}=n$ and $G_2=\emptyset$. Then $\{G_1:G_2\}$ is a class one $2^{n-m}$ SCCD with $f=n$. Thus, we get the first equation of (16).

Then, we consider $n=M^{VI}(q)+1$ with $q\ge 5$. By the definition of $M^{VI}(q)$, any $2^{n-m}$ design with $n=M^{VI}(q)+1$ does not have the R VI, i.e., it must be of R III, IV or V. If it is of R III, then it has a defining word of length three, thus the main effects and 2fi’s of the three factors in the word are not strongly clear and hence its $f_s(q,M^{VI}(q)+1)\le M^{VI}(q)-2$, which leads $f^*_s(q,M^{VI}(q)+1)\le M^{VI}(q)-2$.

If the design has the R IV or V, then it has a defining word of length four or five, which causes that not all the main effects and 2fi’s of the four or five factors in the defining word are strongly clear, thus among the four or five factors at least four of them can not be in $G_1$. This leads to $f_s(q,M^{VI}(q)+1)\le M^{VI}(q)-3$ and $f^*_s(q,M^{VI}(q)+1)\le M^{VI}(q)-3$.

At last, with the similar arguments in the proof of Theorem 2, we can easily prove that the second and third equations of (16) for $n> M^{VI}(q)+1$ and $q\ge 5$ are also valid. The theorem is proved. $\square$

Proof of Theorem 4

First, from Proposition 3 we know that for $N/4+2\le n\le N/2$ and $f=2$ there not exist class one $2^{n-m}$ CCD but there exists GOCD. By P-AENP, for the parameters the GOCD must have the form $T=\{G_1:G_2\}=\{\mathbf{1,2}:A_1,\mathbf{1}A_2,\mathbf{2}A_3,\mathbf{12}A_4\}$ in isomorphism, where $\cup _{i=1}^4A_i\subseteq \overline{H}_2$ and $A_1\cap A_2=A_1\cap A_3=\emptyset$, such that for $\Omega _1=\{G_1,G_1\times G_1\}$ it has $({}_{\,1}^{\#}P^{}_2,{}_{\,2}^{\#}P^{}_1)(T;\Omega _1)=((2),(1))$, which sequentially maximizes the $({}_{\,1}^{\#}P^{}_2,{}_{\,2}^{\#}P^{}_1)$.

Thus, to prove Theorem 4 we only need to prove that if the T satisfies $(A_1\cup A_4)\cap (A_2\cup A_3)=\emptyset$ and $\#\{A_1\cap A_4\}+\#\{A_2\cap A_3\}=n-N/4-1$ then it also sequentially maximizes ${}_{\,2}^{\#}P_2$.

Note that it must be to have ${}_{\,2}^{\#}P_2(T;\Omega _1)=(0,\ldots ,0, {}_{\,2}^{\#}P_2^{(k)})$ for some k since there are only two cases which cause the 2fi $\mathbf{12}$ to be aliased with: (i) $\alpha _1\alpha _2=\mathbf{12}$ with $\alpha _1\in A_1$ and $\alpha _2\in \mathbf{12}A_4$ and (ii) $\beta _1\beta _2=\mathbf{12}$ with $\beta _1\in \mathbf{1}A_2$ and $\beta _2\in \mathbf{2}A_3$. From the structure of T and $(A_1\cup A_4)\cap (A_2\cup A_3)=\emptyset$, we know $A_1\cap A_2=A_1\cap A_3=\emptyset$ and $k=\#\{A_1\cap A_4\}+\#\{A_2\cap A_3\}$.

Since $\cup _{i=1}^4A_i\subseteq \overline{H}_2$ and $(A_1\cup A_4)\cap (A_2\cup A_3)=\emptyset$, we have $\sum _{i=1}^4\#\{A_i\}-\#\{A_1\cap A_4\} -\#\{A_2\cap A_3\}\le 2^{q-2}-1$. Thus, we have

$$\begin{aligned} k=\#\{A_1\cap A_4\}+\#\{A_2\cap A_3\} \ge n-2-(2^{q-2}-1)=n-N/4-1 \end{aligned}$$

(19)

and $n-N/4-1$ is the minimum value that k can reach, hence the T is a GOCD and has $({}_{\,1}^{\#}P^{}_2,{}_{\,2}^{\#}P^{}_1,{}_{\,2}^{\#}P^{}_2)(T;\Omega _1)=((2),(1),(0^{n-N/4-1},1))$. $\square$

Proof of Theorem 5

Let $\overline{T}$ denote the complementary set of the T, i.e., $\overline{T}=H_q\backslash T$, where $H_q=\{\mathbf{1,2,12}, \overline{H}_2,$$\mathbf{1}\overline{H}_2,\mathbf{2}\overline{H}_2,\mathbf{12}\overline{H}_2\}$. From Theorem 1(b) in Zhang and Mukerjee (2009) we know that for any $\gamma \in T$, $B_2(T,\gamma )=(1/2)(n-\#\{\overline{T}\}-1)+B_2(\overline{T},\gamma )$ and hence we have

$$\begin{aligned} B_2(T,\mathbf{1})=n-N/2+B_2(\overline{T},\mathbf{1}) \, \hbox { and }\, B_2(T,\mathbf{2})=n-N/2+B_2(\overline{T},\mathbf{2}). \end{aligned}$$

(20)

To prove T to be a GOCD, we first need to prove that the ${}_{\,1}^{\#}P^{}_2(T;\Omega _1)$ is sequentially maximized, i.e., to prove that the $B_2(T,\mathbf{1})+B_2(T,\mathbf{2})$ is minimized. By the equations (20), we only need to prove that both $B_2(\overline{T},\mathbf{1})$ and $B_2(\overline{T},\mathbf{2})$ are minimized. Since the $\overline{T}$ has the form $\{\mathbf{12},{A}_1,\mathbf{1}{A}_2,\mathbf{2}{A}_3,\mathbf{12}{A}_4\}$ and $(A_1\cup A_4)\cap (A_2\cup A_3)=\emptyset$, we have $B_2(\overline{T},\mathbf{1})=B_2(\overline{T},\mathbf{2})=0$ and by (20) and the definition of P-AENP, the ${}_{\,1}^{\#}P^{}_2(T;\Omega _1)$ is sequentially maximized and has ${}_{\,1}^{\#}P^{}_2(T;\Omega _1)=(0^{n-N/2},2).$

Furthermore, we note that $N/2\le n \le N-2$ and $\mathbf{12}\notin T$, so we have ${}_{\,2}^{\#}P^{}_1(T;\Omega _1)=(1)$, meaning that ${}_{\,2}^{\#}P^{}_1(T;\Omega _1)$ is also maximized.

Finally, consider ${}_{\,2}^{\#}P^{}_2(T;\Omega _1)$. By $(A_1\cup A_4)\cap (A_2\cup A_3)=\emptyset$ we have $B_2(\overline{T},\mathbf{12})=\#\{A_1\cap A_4\}+\#\{A_2\cap A_3\}$ and by Theorem 1(c) of Zhang and Mukerjee (2009) and $\mathbf{12} \notin T$, we have

$$\begin{aligned} B_2(T,\mathbf{12})=n-N/2+1+B_2(\overline{T},\mathbf{12}). \end{aligned}$$

Note that the ${}_{\,2}^{\#}P^{}_2(T;\Omega _1)$ being sequentially maximized is equivalent to the $B_2(T,\mathbf{12})$ being minimized. Now let’s see if $\#\{A_1\cap A_4\}+\#\{A_2\cap A_3\}$ is minimized.

There are only two cases (i) $A_1\cap A_4=A_2\cap A_3=\emptyset$ and (ii) $(A_1\cap A_4)\cup (A_2\cap A_3)\ne \emptyset$ in consideration. For Case (i), it implies that $A_i\cap A_j=\emptyset$ for any $i\ne j$ and hence $B_2(\overline{T},\mathbf{12})=0$. Thus we have $B_2(T,\mathbf{12})=n-N/2+1$. However, we note that for this case we have $\#\{\overline{T}\}=1+\sum _{i=1}^4\#\{A_i\}\le N/4$, thus by $n+\#\{\overline{T}\}=N-1$ we get ${}_{\,2}^{\#}P^{}_2(T;\Omega _1)=(0^{n-N/2},1)$ only for $n\ge 3N/4-1$.

Then consider Case (ii). Let $\overline{T}^*$ denote $\overline{T}\backslash \{\mathbf{12}\}\cup \{\mathbf{1,2}\}$, we have $\#\{\overline{T}^*\}=(N-1-n)+1$. Since $\cup _{i=1}^4A_i\subseteq \overline{H}_2$, $N/4+2\le \#\{\overline{T}^*\}\le N/2$. By applying (19) in the proof of Theorem 4 to $\overline{T}^*$, we have $B_2(\overline{T},\mathbf{12})=B_2(\overline{T}^*,\mathbf{12})=\#\{A_1\cap A_4\}+\#\{A_2\cap A_3\}=((N-1-n)+1)-N/4-1=3N/4-1-n$, which is minimum because of $\cup _{i=1}^4A_i\subseteq \overline{H}_2$ and $(A_1\cup A_4)\cap (A_2\cup A_3)=\emptyset$. Thus we have $n\le 3N/4-1$ and get ${}_{\,2}^{\#}P^{}_2(T;\Omega _1)=(0^{N/4-1},1)$, for $n\le 3N/4-1$.

Summing up the above reasoning, we conclude that the given design T is a class one GOCD and its P-AENP satisfies the Eq. (18), Theorem 5 being proved. $\square$

Proof of Proposition 1

For expression convenience, we denote $\mathbf{1,2,\ldots ,q}$ as $\mathbf{t_{\mathrm{1}},\ldots ,t_{\mathrm{q}}}$.

For (1), obviously the $T=M_q\cup \{\mathbf{t_{\mathrm{i}}t_{\mathrm{j}}}\}$ has R III and $f=M(q)-2$ and $n=M(q)+1$. If the T is not a CCD, then by $M_q$ having the R V, the only possibility is that there is an s different from i and j such that $\mathbf{t_{\mathrm{s}}(t_{\mathrm{i}}t_{\mathrm{j}})}$ is aliased with some 2fi $\mathbf{t_{\mathrm{u}}t_{\mathrm{v}}}$, This means that in the defining contrast subgroup of $M_q$ there is a length five word $\mathbf{t_{\mathrm{u}}t_{\mathrm{v}}}{} \mathbf{t_{\mathrm{s}}t_{\mathrm{i}}t_{\mathrm{j}}}$ containing the pair $\mathbf{t_{\mathrm{i}}t_{\mathrm{j}}}$, which conflicts with the assumption.

For (2), obviously the $T=M_q\cup \{\mathbf{t_{\mathrm{i}}t_{\mathrm{j}}t_{\mathrm{k}}}\}$ has the R IV and $f=M(q)-2$ and $n=M(q)+1$. If it is not a CCD, then, again by $M_q$ having the R V, there are only two possibilities: (i) there is a 2fi $\mathbf{t_{\mathrm{u}}t_{\mathrm{v}}}$ in $G_1\times G_1$ with $u\ne k$ and $v\ne k$ which is aliased with a 2fi $\mathbf{t_{\mathrm{s}}(t_{\mathrm{i}}t_{\mathrm{j}}t_{\mathrm{k}})}$, where the s is different from any one in $\{u,v,i,j,k\}$ and the $\mathbf{t_{\mathrm{i}}t_{\mathrm{j}}t_{\mathrm{k}}}$ is a column (factor) of T. Also the $\mathbf{t_{\mathrm{i}}t_{\mathrm{j}}t_{\mathrm{k}}}$ can be treated as a 3fi of the $M_q$. Thus, this means that for the $M_q$ there is a length six word $\mathbf{t_{\mathrm{u}}t_{\mathrm{v}}}{} \mathbf{t_{\mathrm{s}}t_{\mathrm{i}}t_{\mathrm{j}}t_{\mathrm{k}}}$ containing the triple $\mathbf{t_{\mathrm{i}}t_{\mathrm{j}}t_{\mathrm{k}}}$, which conflicts with the assumption; (ii) there is a 2fi $\mathbf{t_{\mathrm{u}}t_{\mathrm{v}}}$ in $G_1\times G_1$ with $u\ne k$ and $v\ne k$ aliased with the main effect $\mathbf{t_{\mathrm{i}}t_{\mathrm{j}}t_{\mathrm{k}}}$ of T. This means that for the $M_q$ there is a length five word $\mathbf{t_{\mathrm{u}}t_{\mathrm{v}}}{} \mathbf{t_{\mathrm{i}}t_{\mathrm{j}}t_{\mathrm{k}}}$ containing the triple $\mathbf{t_{\mathrm{i}}t_{\mathrm{j}}t_{\mathrm{k}}}$, which also conflicts with the assumption. $\square$

Proof of Proposition 2

For $q=7$, we have $M(7)=11$ and $M_7=\{\mathbf{1,2,3,4,5,}$$\mathbf{6,7,1234,3456,2367,}$$\mathbf{12357}\}$ is a largest design with R V. We note that the pair $\mathbf{17}$ does not appear in its all the six length 5 words $\{\mathbf{12348,34569,}$$\mathbf{2367t_{\mathrm{1}},4578t_{\mathrm{2}},289t_{\mathrm{1}}t_{\mathrm{2}}, 156t_{\mathrm{1}}t_{\mathrm{2}}}\}$, where $\mathbf{t_{\mathrm{1}}=10}$ and $\mathbf{t_{\mathrm{2}}=11}$. By Proposition 1, the design $\{\mathbf{2,3,4,}$$\mathbf{5,6,1234,3456,2367,}$$\mathbf{12357: 1,7,17}\}$ is a class one $2^{12-5}$ CCD of R III with $f=9$ and $n=12$. Also, we note that the triple $\mathbf{246}$ does not appear in all the six length 5 words and all the five length 6 words of the $M_7$. Then, by Proposition 1, the design $\{\mathbf{1,3,5,}$$\mathbf{6,7,1234,3456,2367,}$$\mathbf{12357: 2,4,246}\}$ is a class one $2^{12-5}$ CCD of R IV with $f=9$ and $n=12$.

For $q=5$, we have $M(5)=6$ and its $M_5$ is $\{\mathbf{1,2,3,4,5,2345}\}$. The pair $\mathbf{12}$ does not appear in its only length 5 word $\mathbf{23456}$ and hence $T_1=\{\mathbf{3,4,5,2345}: \mathbf{1,2,12}\}$ is a largest R III CCD with $f=4$ and $n=7$. Also, it is easy to know that $T_2=\{\mathbf{3,4,5,2345}: \mathbf{1,2,123}\}$ is a largest R IV CCD with $n=7$ and $f=4$. By the same way we can get the same result for the cases $q=4,6$. Proposition 2 is proved. $\square$

Proof of Proposition 3

The first equation of (12) is got by taking design $\{\mathbf{1}:\overline{H}_1\}$.

For the second and third equations we only need to consider the case $q>4$, since for $q=2,3,4$ we can directly verify them to be valid.

First, for any class one $2^{n-m}$ CCD $T=\{G_1:G_2\}$ with $f=2$ and $q>4$, the two columns in $G_1$ must be independent. So, we can assume that $G_1=\{\mathbf{1},\mathbf{2}\}$ and $G_2$ can only take columns from $\{\{\mathbf{I,1,2, 12}\}\overline{H}_2\}$. Thus, the only requirement for selecting $G_2$ is that any 2fi of $G_2$ is not allowed to alias with $\mathbf{1}$, $\mathbf{2}$ and $\mathbf{12}$. So, to make n be largest, the only choice of $G_2$ is $\{A_1,\mathbf{1}A_2,\mathbf{2}A_3,\mathbf{12}A_4\}$, where $A_i\subset \overline{H}_2$, $i=1,2,3,4$, $A_i\bigcap A_j=\emptyset$ for $i\ne j$ and $\bigcup _{i=1}^4A_i=\overline{H}_2$. It is easy to get $n^*(q,2)=2^{q-2}-1+2=N/4+1.$

Particularly, if take $G_2=\overline{H}_2$ then the design is of R III, and if take $G_2=\mathbf{1}\overline{H}_2$ then the T is of R IV.

Then, consider any class one $2^{n-m}$ CCD $T=\{G_1:G_2\}$ with $f=3$ and $q>4$. Without loss of generality, in isomorphism we can assume that $G_1=\{\mathbf{1},\mathbf{2},\mathbf{3}\}$, since $G_1$, as a sub-design, must have the R at least V. Thus $G_2$ can not contain any column of $H_3$, i.e., it can be taken only from $H_q{\setminus } H_3=\{\{\mathbf{I},H_3\}\overline{H}_3\}$ with the only limitation that any 2fi of $G_2\times G_2$ is not allowed to alias with any column of $H_3$ except for 123. So, to make n be largest the only choice of $G_2$ is $\{\{\mathbf{I,123}\}A_1, \{\mathbf{1,23}\}A_2, \{\mathbf{12,3}\}A_3, \{\mathbf{13,2}\}A_4\}$ with $A_i\subset \overline{H}_3, i=1,2,3,4$, $A_i\bigcap A_j=\emptyset$ for $i\ne j$ and $\bigcup _{i=1}^4A_i=\overline{H}_3$. Thus, we get $n^*(q,3)=2(2^{q-3}-1)+3=N/4+1.$

At last, we consider $f=4$ and $q>4$. With the same idea as above, for design $T=\{G_1:G_2\}$ we should take $G_1=\{\mathbf{1,2,3,4}\}$, and to make n be largest, the one of best choices of $G_2$ is $\{\mathbf{1234}\}\cup S_1\cup S_2$, where $S_1=\{\mathbf{1234}\}\overline{H}_4$ and $S_2$ is $\{\mathbf{1}A_1, \mathbf{2}A_2, \mathbf{3 }A_3, \mathbf{4 }A_4\}$ with $A_i\bigcap A_j=\emptyset$ for $i\ne j$ and $\bigcup _{i=1}^4A_i=\overline{H}_4$. By a calculating we get $n^*(q,4)=2(2^{q-4}-1)+5=N/8+3,$ Proposition 3 is proved. $\square$

Proof of Proposition 4

First we consider $f=5$. To insure a selected $T=\{G_1:G_2\}$ to be a CCD, for $G_1$ there are only the two possible choices in isomorphism: $\{\mathbf{1,2,3,4,1234}\}$ or $\{\mathbf{1,2,3,4,5}\}$. If $G_1=\{\mathbf{1,2,3,4,1234}\}$, then, to make $T=\{G_1:G_2\}$ be a largest CCD, the $G_2$ can and only can take one column from every $\{\beta \{\mathbf{I}, H_4\}\}$, where $\beta \in \overline{H}_4$. Thus the column number of the largest CCD T is $n=\#\{\overline{H}_4\}+5=N/16+4$.

If $G_1=\{\mathbf{1,2,3,4,5}\}$, then, to make $T=\{G_1:G_2\}$ be a largest CCD, from $H_5$ the $G_2$ can only take $\mathbf{1234}$ or $\mathbf{12345}$ and from every $\{\beta \{\mathbf{I},H_5\}\}$ can take $\{\mathbf{I,123,145,2345}\}\beta$, where $\beta \in \overline{H}_5$. For each $\beta$, in isomorphism the four columns $\{\mathbf{I,123,145,2345}\}\beta$ are the only largest choices from $\{\beta \{\mathbf{I},H_5\}\}$, since taking any more column from $\{\beta \{\mathbf{I},H_5\}\}$ must cause some 2fi of $G_2$ to alias with some 2fi of $G_1$. Thus, for this case we have $n=$$4\#\{\overline{H}_5\}+5+1=N/8+2$ and hence the choice $G_1=\{\mathbf{1,2,3,4,5}\}$ is the best and $n^*(q,5)=N/8+2$, the first equality being proved.

Then consider $f=6$. To insure a $\{G_1:G_2\}$ to be a CCD, the $G_1$ has one of the three selections: $\{\mathbf{1,2,3,4,5,6}\}$, $\{\mathbf{1,2,3,4,5,1234}\}$ and $\{\mathbf{1,2,3,4,5,12345}\}$ in isomorphism.

If we take $G_1=\{\mathbf{1,2,3,4,5,6}\}$, then, to make $\{G_1:G_2\}$ be a largest CCD, in isomorphism the $G_2$ and a largest selection $\{\mathbf{1234,1256,3456}\}$ from $H_6$ and has a largest choice $\{\mathbf{I,123,145,2345,246,}$$\mathbf{1346, 1256,356}\}\beta$ from every $\{\beta \{\mathbf{I},H_6\}\}$, where $\beta \in \overline{H}_6$, since choosing any one more column from some $\{\beta \{\mathbf{I},H_6\}\}$ will cause some 2fi of $G_1$ to be aliased with some 2fi of $G_2$. Thus the largest $n=8\#\{\overline{H}_6\}+6+3=N/8+1$.

If $G_1$ chooses $\{\mathbf{1,2,3,4,5,1234}\}$ or $\{\mathbf{1,2,3,4,5,12345}\}$, then by the similar argument the $G_2$ has an only largest choice $\{\mathbf{I,125}\}\beta$ from every $\{\beta \{\mathbf{I}, H_5\}\}$. Thus for this case we have the largest $n=2\#\{\overline{H}_5\}+6=N/16+4$.

Hence, the $\{G_1:G_2\}$ with $G_1=\{\mathbf{1,2,3,4,5,6}\}$ and its best corresponding $G_2$ is the largest class one CCD for $f=6$ and has $n^*(q,6)=N/8+1$. $\square$

Proof of Proposition 5

The first equation of (17) is obvious. Consider the second equation. For any $2^{n-m}$ class one SCCD $T=\{G_1:G_2\}$ with $f=2$, the two columns in $G_1$ must be independent. So, in isomorphism $G_1=\{\mathbf{1},\mathbf{2}\}$ and $G_2$ must be from $\{\{\mathbf{I,1,2, 12}\}\overline{H}_2\}$. Similar to the proof of the largest class one CCD when $f=2$, we know that the $G_2$ with the largest n has the only choice $\{A_1,\mathbf{1}A_2,\mathbf{2}A_3,$$\mathbf{12}A_4\}$, where $A_i\subset \overline{H}_2$, $i=1,2,3,4$, $A_i\bigcap A_j=\emptyset$ for $i\ne j$ and $\bigcup _{i=1}^4A_i=\overline{H}_2$. Then we have $n^*_s(q,2)=2^{q-2}-1+2=N/4+1$.

Then, consider the third equality of (17). Let $T=\{G_1:G_2\}$ be a class one $2^{n-m}$ SCCD with $f=3$. Without loss of generality we still assume that $G_1=\{\mathbf{1},\mathbf{2},\mathbf{3}\}$ and the SCCD $T=\{G_1:G_2\}$. Since T is a SCCD with $f=3$, any column in $H_3$ can not appear in $G_2$ and hence $G_2$ can only take columns from $\{\{\mathbf{I},H_3\}\overline{H}_3\}$. With the same argument as before, the $G_2$ has the only largest choice $\{A_1,\mathbf{1}A_2,\mathbf{2}A_3,\mathbf{12}A_4,\mathbf{3}A_5,\mathbf{13}A_6, \mathbf{23}A_7,$$\mathbf{123}A_8\}$, where $A_i\subset \overline{H}_3$, $i=1,2,\ldots ,8$, $A_i\bigcap A_j=\emptyset$ for $i\ne j$ and $\bigcup _{i=1}^8A_i=\overline{H}_3$. By a simple calculation we get $n^*_s(q,3)=2^{q-3}-1+3=N/8+2$.

For $f=4$, since $G_1$ must have at least R VI, we assume that $G_1=\{\mathbf{1,2,3,4}\}$ in isomorphism. Considering $T=\{G_1:G_2\}$ is a SCCD, any column in $H_4$ can not appear in $G_2$ and hence $G_2$ can only take columns from $\{\mathbf{I},H_4\}\overline{H}_4$ and the only largest selection is $\{\{\mathbf{I, 1234}\}A_1,\{\mathbf{1,234}\}A_2,$$\{\mathbf{2,134}\}A_3,\{\mathbf{12,34}\}A_4,\{\mathbf{3,}$$\mathbf{124}\}A_5,\{\mathbf{13,24}\}A_6, \{\mathbf{23,14}\}A_7,$$\{\mathbf{4,123}\}A_8\}$, where $A_i\subset \overline{H}_4$, $i=1,2,\ldots ,8$, $A_i\bigcap A_j=\emptyset$ for $i\ne j$ and $\bigcup _{i=1}^8A_i=\overline{H}_4$. By calculating, we get $n^*_s(q,4)=2(2^{q-4}-1)+4=N/8+2.$$\square$

Proof of Proposition 6

First we consider $f=5$. To insure a selected $\{G_1:G_2\}$ to be a SCCD, in isomorphism the $G_1$ has only one selection $\{\mathbf{1,2,3,4,5}\}$ and the $G_2$ can only select $\mathbf{12345}$ from $H_5$ and select two columns $\{\beta \{\mathbf{I,5}\}\}$ from every $\{\beta \{\mathbf{I},H_5\}\}$, where $\beta \in \overline{H}_5$. Thus by a calculation we get $n_s^*(q,5)=2\#\{\overline{H}_5\}+5+1=N/16+4$.

Then consider $f=6$. To insure a selected $\{G_1:G_2\}$ to be a SCCD, in isomorphism the $G_1$ has only two possible selections $\{\mathbf{1,2,3,4,5,12345}\}$ and $\{\mathbf{1,2,3,4,5,6}\}$. For $G_1=\{\mathbf{1,2,3,4,5,12345}\}$, the $G_2$ can and only can take the column $\mathbf{6}$ from $H_6$ and the two columns $\{\mathbf{I,6}\}\beta$ from every $\{\beta \{\mathbf{I},H_6\}\}$, where $\beta \in \overline{H}_6$. For $G_1=\{\mathbf{1,2,3,4,5,6}\}$, the $G_2$ can and only can take one column $\mathbf{12345}$ (or $\mathbf{123456}$) of $H_6$ and the two columns $\{\mathbf{I, 12345}\}\beta$ (or $\{\mathbf{I, 12345}\}\beta$) of every $\{\beta \{\mathbf{I},H_6\}\}$, where $\beta \in \overline{H}_6$. Thus, for both cases we have $n_s^*(q,6)=2\#\{\overline{H}_6\}+6+1=N/32+5$ and hence Proposition 6 is proved. $\square$

Table 3 Some class one $2^{n-m}$ CCDs, SCCDs and GOCDs for 8, 16 and 32 runs

Table 4 Some class one $2^{n-m}$ CCDs, SCCDs and GOCDs for 64 runs