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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

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

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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

[1] 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

 [1] 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

[1] 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.

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  • $\begingroup$ (+1): Thanks @neverknowsbest for your answer. Please my edits too. Thanks $\endgroup$ – MYaseen208 Apr 12 '14 at 16:24
  • $\begingroup$ 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. $\endgroup$ – neverKnowsBest Apr 12 '14 at 17:11
  • $\begingroup$ 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. $\endgroup$ – MYaseen208 Apr 12 '14 at 17:20
  • $\begingroup$ I guess they'll look at it after the end of the semester. $\endgroup$ – neverKnowsBest Apr 15 '14 at 17:22
  • $\begingroup$ That would be very late. $\endgroup$ – MYaseen208 Apr 15 '14 at 17:23

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