# Constructing one-ninth fraction of the $3^{5}$ design

I want to construct a $3^{5-2}$ design with $I=ABC$ and $I=CDE$ as generators. The complete defining relation for this design is: $I=ABC=CDE=ABC^{2}DE=ABD^{2}E^{2}$. This is a resolution III design with $x_4=2x_1+2x_2+x_3$ and $x_5=x_1+x_2+2x_4\:({\rm mod}\:3)$.

Question I wonder how to get $x_4=2x_1+2x_2+x_3$ and $x5=x_1+x_2+2x_4\:({\rm mod}\:3)$ in this specific problem.

Edited This is a question from Design and Analysis of Experiments by Douglas C. Montgomery.

Allow me to begin with the qualification that algebra is not my strong suit and in my area (industrial statistics) we use fractionated three-level factorials almost never. Is it possible there's an error in your question? I've generated the design given the $x_k$'s and it appears that the words you give are not actually words. I also worked out the columns given $I=ABC=CDE$ and got a different design.

Words Given the Design Construction

I went ahead and constructed your design from the $x_k$'s you give using R:

X<-as.data.frame(expand.grid(c(0:2),c(0:2),c(0:2)))
colnames(X)<-c("A","B","C")
X[,"D"] <- (2*X[,"A"] + 2*X[,"B"] + X[,"C"]) %% 3
X[,"E"] <- (X[,"A"] + X[,"B"] + 2*X[,"D"]) %% 3


When I did both

(X[,"A"]+X[,"B"]+X[,"C"]) %% 3


and

(X[,"C"]+X[,"D"]+X[,"E"]) %% 3


I get

 0 1 2 1 2 0 2 0 1 1 2 0 2 0 1 0 1 2 2 0 1 0 1 2 1 2 0


which suggests neither $ABC$ nor $CDE$ are words. Let $A=x_1$,...,$E=x_5$ and assume addition on $\mathbb{F}_3$. Then given \begin{align} D&=2A+2B+C,\\ E&=A+B+2D, \end{align} we have \begin{align} 2D &= A + B + 2C\\ E &= 2A + 2B + 2C\\ A+B+C+E &= 0\\ \end{align} so $I=ABCE$ should be a word. Indeed, in R

(X[,"A"]+X[,"B"]+X[,"C"]+X[,"E"]) %% 3


yields

  0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0


Similarly, \begin{align} A+B+D&=C \\ A+B+2C+D &= 0 \end{align} so I'd expect $I=ABC^2D$ to be a word, and in R we find

(X[,"A"]+X[,"B"]+2*X[,"C"]+X[,"D"]) %% 3


yields

 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0


It appears that the words are \begin{align} I=ABCE=ABC^2D=A^2B^2C^2E^2=A^2B^2CD^2=A^2B^2DE=ABD^2E^2=CDE^2=C^2D^2E. \end{align}

Design Construction Given the Words

Going the other way, let $A+B+C=0$ and $C+D+E=0$ then \begin{align} C&=2A+2B\\ 2C&=A+B\\ E&=2C+2D = A+B+2D. \end{align} It is more convenient (for me, at least) to treat $A$,$B$, and $D$ as coming from the $3^3$ and solve for $C$ and $E$. We can express $D$ as \begin{align} D &= 2C + 2E \\ &= A+B+2E \\ &= 2A+ 2B+C+2E \end{align} but since no word contains something that's not a multiple of $DE$ (or $D+E$ in my notation) it's not possible to solve for both independently of each other.

• (+1): Thanks @neverknowsbest for your answer. Please my edits too. Thanks – MYaseen208 Apr 12 '14 at 16:24
• He's my PhD advisor. Was the formula for D and E in the student solutions manual or the book? I can follow up with him if so. – neverKnowsBest Apr 12 '14 at 17:11
• Formula for D and E are given in the student solution manual. I believe these formulas are not correct. Might be typo in question too. – MYaseen208 Apr 12 '14 at 17:20
• I guess they'll look at it after the end of the semester. – neverKnowsBest Apr 15 '14 at 17:22
• That would be very late. – MYaseen208 Apr 15 '14 at 17:23