Fractional Factorial Designs Each replicate of a 2 k design requires - - PowerPoint PPT Presentation

ST 516 Experimental Statistics for Engineers II

Fractional Factorial Designs

Each replicate of a 2k design requires 2k runs. E.g. 64 runs for k = 6,

• r 1024 runs for k = 10.

When this is infeasible, we use a fraction of the runs. As a result, we can no longer estimate all effects, but with good choices we can still estimate main effects and usually two-factor interactions. Notation: 2k−p; e.g. 26−2 is a one-quarter (2−2) fraction of the full 26 factorial, with 24 = 16 out of the full 26 = 64 runs.

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ST 516 Experimental Statistics for Engineers II

Key Ideas

Sparsity of effects: With several factors, the system or process is likely to be driven primarily by some of the main effects and low-order interactions. Projection: Fractional factorials can be projected into stronger designs in the subset of active factors. Sequential experimentation: Runs of two (or more) fractional factorials can be combined into a larger design to estimate effects of interest.

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ST 516 Experimental Statistics for Engineers II

Example: a 23−1 design

The full 23 design, sorted by the ABC column: Treatment Effect Combination I A B AB C AC BC ABC a + +

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+ b +

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c +

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abc + + + + + + + + (1) +

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• ac

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Two-level Fractional Factorial Designs The One-Half Fraction

ST 516 Experimental Statistics for Engineers II

The 4 runs in the top half are a one-half fraction. They all have the same sign in the ABC column; ABC is the generator of this fraction. Within this fraction, we see that: I = ABC A = BC B = AC C = AB. I = ABC is the defining relation for this design; the other equations show various aliases; you can derive them from the defining relation by multiplying by A, B, and C, respectively.

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ST 516 Experimental Statistics for Engineers II

The top half is the principal fraction. (Note: it is also one block in a 2-block design with ABC confounded with blocks.) The bottom half is the alternate fraction, defined by I = −ABC. (It is also the principal block in that 2-block design.) In either fraction, we can calculate any main effect, but it really estimates the sum (or difference) of the effect and its alias. We write [A] → A + BC, etc.

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ST 516 Experimental Statistics for Engineers II

Design Resolution Because in these designs main effects are aliased with two-factor interactions but not with other main effects, they are called resolution III designs. Notation: 23−1

III .

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ST 516 Experimental Statistics for Engineers II

In general, a design has resolution r if: main effects are aliased only with interactions of at least r − 1 factors; more generally, k-factor interactions are aliased only with interactions of at least r − k factors. The resolution is the number of factors in the interaction that defines the design.

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ST 516 Experimental Statistics for Engineers II

Most important resolutions: Resolution III: Main effects are aliased with two-factor interactions, but not with each other. Resolution IV: Main effects are not aliased with two-factor interactions or with each other, but two-factor interactions are aliased with each other. Resolution V: No main effects or two-factor interactions are aliased with each other; two-factor interactions are aliased with three-factor interactions and main effects are aliased with four-factor interactions.

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ST 516 Experimental Statistics for Engineers II

Filtration rate example Suppose we had only the 8 runs in the 24−1 design generated by I = ABCD (resolution IV):

fraction <- with(filtration, filtration[A * B * C * D == 1,]) summary(lm(Rate ~ A * B * C * D, fraction))

Aliasing [A] → A + BCD [B] → B + ACD [C] → C + ABD [D] → D + ABC [AB] → AB + CD, and so on.

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ST 516 Experimental Statistics for Engineers II

Output

Call: lm(formula = Rate ~ A * B * C * D, data = fraction) Residuals: ALL 8 residuals are 0: no residual degrees of freedom! Coefficients: (8 not defined because of singularities) Estimate Std. Error t value Pr(>|t|) (Intercept) 70.75 NA NA NA A 9.50 NA NA NA B 0.75 NA NA NA C 7.00 NA NA NA D 8.25 NA NA NA A:B

• 0.50

NA NA NA A:C

• 9.25

NA NA NA B:C 9.50 NA NA NA

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ST 516 Experimental Statistics for Engineers II

Output, continued

A:D NA NA NA NA B:D NA NA NA NA C:D NA NA NA NA A:B:C NA NA NA NA A:B:D NA NA NA NA A:C:D NA NA NA NA B:C:D NA NA NA NA A:B:C:D NA NA NA NA Residual standard error: NaN on 0 degrees of freedom Multiple R-Squared: 1, Adjusted R-squared: NaN F-statistic: NaN on 7 and 0 DF, p-value: NA

All effects are listed; the effects with missing estimates are aliased with earlier effects.

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ST 516 Experimental Statistics for Engineers II

Alias structure

alias(lm(Rate ~ A * B * C * D, fraction))

Output

Model : Rate ~ A * B * C * D Complete : (Intercept) A B C D A:B A:C B:C A:D 0 0 0 0 0 1 B:D 0 0 0 0 0 1 C:D 0 0 0 0 1 A:B:C 0 0 0 1 0 A:B:D 0 0 1 0 0 A:C:D 0 1 0 0 0 B:C:D 1 0 0 0 0 A:B:C:D 1 0 0 0 0 0

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ST 516 Experimental Statistics for Engineers II

The alias table has a column for each effect that has an estimate in the output, and a row for each effect with a missing estimate. The “1” in a column shows which effect is aliased with the column label. Each estimate in the lm() output must be interpreted as the sum of the effects in its alias chain.

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ST 516 Experimental Statistics for Engineers II

Half-normal plot

library(gplots) qqnorm(aov(Rate ~ A * B * C * D, fraction), label = TRUE)

A, C, and D are all relatively large. Each is aliased with only a three-factor interaction, which we shall ignore, so we interpret them as main effects. The B main effect is aliased with ACD, so both of these seem small; in particular, B seems inactive.

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ST 516 Experimental Statistics for Engineers II

The AC estimate represents AC + BD; if B is inactive, it must represent AC. Likewise, the BC estimate represents BC + AD, and if B is inactive it must represent AD. Finally, the AB estimate represents AB + CD, so both seem small. Summary The plot suggests that the interesting effects are A, C, D, AC, and AD.

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ST 516 Experimental Statistics for Engineers II

Reduced Model

summary(lm(Rate ~ A * C + A * D, fraction))

Output

Call: lm(formula = Rate ~ A * C + A * D, data = fraction) Residuals: 1 4 6 7 10 11 13 16

• 1.25

0.25 -0.25 1.25 -0.25 1.25 -1.25 0.25 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 70.7500 0.6374 111.00 8.11e-05 *** A 9.5000 0.6374 14.90 0.00447 ** C 7.0000 0.6374 10.98 0.00819 ** D 8.2500 0.6374 12.94 0.00592 ** A:C

• 9.2500

0.6374

• 14.51

0.00471 ** A:D 9.5000 0.6374 14.90 0.00447 **

• Signif. codes:

0 *** 0.001 ** 0.01 * 0.05 . 0.1 1

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ST 516 Experimental Statistics for Engineers II

Output, continued

Residual standard error: 1.803 on 2 degrees of freedom Multiple R-squared: 0.9979, Adjusted R-squared: 0.9926 F-statistic: 188.6 on 5 and 2 DF, p-value: 0.005282

Note The 2 degrees of freedom for error come from the B and AB alias chains, which were left out of the model because they were small. So 1.803 is likely to be an under-estimate of σ. (The estimate from the full 16 runs is 4.417.) We should be very cautious in drawing conclusions about significance.

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ST 516 Experimental Statistics for Engineers II

Followup experiments Depending on what is seen in the analysis of the observed data, various follow-up experiments could be run. Running the other half fraction (where ABCD = −I) would provide confirmation of the selected model, and would yield more degrees of freedom for error. Because the runs would be randomized only within halves, the halves would have to be viewed as blocks. Alternatively, one or a few runs from the other half, or center points, could be chosen and used for validation: use the current model to predict the response, and compare with the actual result.

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