Fractional factorial aliases and foldover — Examples

Choose $x_A,x_B\in \{-1,+1\}$ freely and set $x_C=x_A{x}_B$. This gives exactly the four runs listed in the Example, one for each of the four choices of $(x_A,x_B)$, and every run satisfies $x_A{x}_B{x}_C=+1$.

Here $H=\{I,ABC\}$. By [F2], effects have equal columns exactly when their symmetric difference is either $\mathrm{\varnothing }$ or $\{A,B,C\}$. Pairing each subset with its complement gives precisely $\{I,ABC\}$, $\{A,BC\}$, $\{B,AC\}$, and $\{C,AB\}$; these four pairs exhaust the eight subsets.

In each listed pair the columns are equal, so [F3] shows that the corresponding terms enter the fitted vector through ${\beta }_I+{\beta }_{ABC}$, ${\beta }_A+{\beta }_{BC}$, ${\beta }_B+{\beta }_{AC}$, or ${\beta }_C+{\beta }_{AB}$, respectively. The design by itself does not identify the two coefficients within any pair separately.

The equations give $x_A{x}_B{x}_D=+1$ and $x_C{x}_D{x}_E=+1$, so the defining words are $ABD$ and $CDE$. Their product is $ABCE$ because the repeated $D$ cancels in symmetric difference. The two incidence vectors are nonzero and distinct; over ${\mathbb{F}}_2$, two nonzero vectors are dependent only if equal, so these generators are independent. By [F2] these and the identity are exactly the four words in $H$. Their nonidentity lengths are $3,3,4$, hence the initial resolution is III.

For the complete mirror, [F3] changes the signs of the length-three relations ABD and CDE but preserves ABCE, giving $H_{[5]}=\{I,ABCE\}$. The two eight-run blocks are disjoint because ABD has opposite signs on them, and both lie in $Q=\{x:x_A{x}_B{x}_C{x}_E=+1\}$; choosing $x_A,x_B,x_C,x_D$ freely determines $x_E$, so $\mathrm{\mid }Q\mathrm{\mid }=16$ and their union is exactly this regular $2^{5-1}$ fraction. Its aliases are precisely $S$ with $S\mathrm{△}\{A,B,C,E\}$, so $A$ remains paired with $BCE$, while $C\mathrm{△}DE=CDE$ is absent and that pair separates. The shortest residual word has length $4$.

For the partial foldover on $J=\{A\}$, [F3] changes ABD and ABCE but preserves CDE, giving $H_J=\{I,CDE\}$. The two eight-run blocks are disjoint because ABD has opposite signs on them, and both lie in $Q^{\mathrm{\prime }}=\{x:x_C{x}_D{x}_E=+1\}$; choosing $x_A,x_B,x_C,x_D$ freely determines $x_E$, so their union is all 16 runs in this regular $2^{5-1}$ fraction. The complete residual pairs are precisely $S$ and $S\mathrm{△}\{C,D,E\}$: this includes $C=DE$ and excludes the original relation $A=BD$. The shortest residual word has length $3$.

The retained columns $c_C=c_{DE}$ in the partial design contribute as $({\beta }_C+{\beta }_{DE})c_C$. By [F4], only this combination enters the fitted vector for this pair, so the partial follow-up does not separately identify ${\beta }_C$ and ${\beta }_{DE}$ without a restriction or further runs.