Fractional Factorial Designs

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Fractional Factorial Designs

Consider a 2k, but with the idea of running fewer than 2k treatment combinations.

Example: (1) 23 design- run 4 t.c.’s written as 23-1 (1/2 of 23) (2) 25 design- run 8 t.c.’s written as 25-2 (1/4 of 25)

2k designs with fewer than 2k t.c.’s are called 2-level fractional factorial designs. (initiated by D.J. Finney in 1945).

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Example: Run 4 of the 8 t.c.’s in 23: a, b, c, abc

It is clear that from the(se) 4 t.c.’s, we cannot estimate the 7 effects (A, B, AB, C, AC, BC, ABC) present in any 23 design, since each estimate uses (all) 8 t.c.’s.

What can be estimated from these 4 t.c.’s?

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4A = -1 + a - b + ab - c +ac - bc + abc4BC= 1 + a - b - ab - c - ac + bc + abc

Consider(4A + 4BC)= 2(a - b - c + abc) or2(A + BC)= a - b - c + abc

overall:2(A + BC)= a - b - c + abc2(B + AC)= -a + b - c + abc2(C + AB)= -a - b + c + abc

In each case, the 4 t.c.’s NOT run cancel out.

Note: 4ABC=(a+b+c+abc)-(1+ab+ac+bc) cannot be estimated.

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Had we run the other 4 t.c.’s: 1, ab, ac, bc, We would be able to estimate A - BC B - AC C - AB(generally no better or worse than with + signs)NOTE: If you “know” (i.e., are willing to assume) that all interactions=0, then you can say either (1) you get 3 factors for

“the price” of 2. (2) you get 3 factors at

“1/2 price.”

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Suppose we run these 4: 1, ab, c, abc;We would then estimate A + B C+ ABC AC + BC }

two main effectstogether usuallyless desirable

In each case, we “Lose” 1 effect completely, and get the other 6 in 3 pairs of two effects.

{members of the pair are CONFOUNDEDmembers of the pair are ALIASED

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With 4 t.c.’s, one should expect to get only 3 “estimates” (or “alias pairs”) - NOT unrelated to “degrees of freedom being one fewer than # of data points” or “with c columns, we get (c-1) df.”

In any event, clearly, there are BETTER and WORSE sets of 4 t.c.’s out of a 23.(Better & worse 23-1 designs)

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Consider a 24-1 with t.c.’s1, ab, ac, bc, ad, bd, cd, abcd

Can estimate:

- 8 t.c.’s- Lose 1 effect- Estimate other 14 in 7 alias pairs of 2

Note:

A+BCDB+ACDC+ABDAB+CDAC+BDBC+ADD +ABC

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Prospect in fractional factorial designs is attractive if in some or all alias pairs one of the effects is KNOWN. This usually means “thought to be zero.”

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“Clean” estimates of the remaining member of the pair can then be made.

For those who believe, by conviction or via selected empirical evidence, that the world is relatively simple, 3 and higher order interactions (such as ABC, ABCD, etc.) may be announced as zero in advance of the inquiry. In this case, in the 24-1 above, all main effects are CLEAN. Without any such belief, fractional factorials are of uncertain value. After all, you could get A + BCD =0, yet A could be large +, BCD large -; or the reverse; or both zero.

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Despite these reservations fractional factorials are almost inevitable in a many factor situation. It is generally better to study 5 factors with a quarter replicate (25-2= 8) than 3 factors completely (23=8). Whatever else the real world is, it’s Multi-factored.

The best way to learn “how” is to work (and discuss) some examples:

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Example: 25-1: A, B, C, D, E

Step 1: In a 2k-p, we must “lose” 2p-1. Here we lose 1. Choose the effect to lose. Write it as a “Defining relation” or “Defining contrast.”

I = ABDEStep 2: Find the resulting alias pairs: *A=BDE AB=DE ABC=CDE B=ADE AC=4 BCD=ACE C=ABCDE AD=BE BCE=ACD D=ABE AE=BD E=ABD BC=4 CD=4 CE=4

{- lose 1-other 30 in15 alias pairsof 2-run 16 t.c.’s 15 estimates

*AxABDE=BDE

(in the plus-minus table)

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See if they’re (collectively) acceptable.

Another option (among many others): I = ABCDE (in the plus-minus table) A=4 AB=3 B=4 AC=3 C=4 AD=3 D=4 AE=3 E=4 BC=3 BD=3 BE=3 CD=3 CE=3 DE=3

Assume we choose I = ABDE

4: 4 way interaction;3: 3 way interaction.

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Next Step: Find the 2 blocks using ABDE as the confounded effect.

I II1c a acab abc b bcde cde ade acdeabde abcde bde bcdead acd d cdbd bcd abd abcdae ace e cebe bce abe abce

{Same processas a ConfoundingScheme

•Which block to run if I = ABDE (ABDE = “+”)?

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Example 2:

25-2 A, B, C, D, EIn a 25, there are 31 effects; with 8 runs, there are 7 df & 7 estimates available { Must “Lose” 3; other 28

in 7 alias groups of 4

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Choose the 3: Like in confounding schemes, 3rd must be product of first 2:

I = ABC = BCDE = ADE

A = BC = 5 = DEB = AC = 3 = 4C = AB = 3 = 4D = 4 = 3 = AEE = 4 = 3 = ADBD = 3 = CE = 3 BE = 3 = CD = 3

Assume we use this design.

Find aliasgroups:

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1 2 3 4 1 a b d abd bd ad ab bc abc c bcd acd cd abcd ac de ade bde e abe be ae abdebcde abcde cde bce ace ce abce acde

Let’s find the 4 blocks using ABC , BCDE , ADE as confounded effects

•Which block should we run if I=ABC=BCDE=ADE?

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Block 1: (in the plus-minus table)• the sign for ABC = “-”• the sign for BCDE = “+”• the sign for ADE = “-” }

Defining relation is:I = -ABC = BCDE = -ADE

Exercise: find the defining relations for the other blocks.

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Analysis 2k-p Design using MINITAB

1. Create factorial design:Stat>> DOE>> Factorial>> Create Factorial Design

2. Input data values.3. Analyze data:

Stat>> DOE>> Factorial >> Analyze Factorial Design

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A B C D rate-1 -1 -1 -1 45

1 -1 -1 1 100-1 1 -1 1 45 1 1 -1 -1 65-1 -1 1 1 75 1 -1 1 -1 60-1 1 1 -1 80 1 1 1 1 96

Example: 24-1 with defining relation I=ABCD.

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Analysis of Variance for rate (coded units)

Source DF Seq SS Adj SS Adj MS F

Main Effects 4 1663 1663 415.7 *

2-Way Interactions 3 1408 1408 469.5 *

Residual Error 0 0 0 0.0

Total 7 3071

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Fractional Factorial Fit: rate versus A, B, C, D

Estimated Effects and Coefficients for rate (coded units)

Term Effect Coef

Constant 70.750

A 19.000 9.500

B 1.500 0.750

C 14.000 7.000

D 16.500 8.250

A*B -1.000 -0.500

A*C -18.500 -9.250

A*D 19.000 9.500

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3020100-10-20-30

1.5

1.0

0.5

0.0

-0.5

-1.0

-1.5

Effect

No

rma

l Sco

re

AC

AB

B

C

D

A

AD

Normal Probability Plot of the Effects(response is rate, Alpha = .99)

A: AB: BC: CD: D

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Alias Structure

I + A*B*C*DA + B*C*DB + A*C*DC + A*B*DD + A*B*CA*B + C*DA*C + B*DA*D + B*C

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Fractional Factorial Fit: rate versus A, C, D

Estimated Effects and Coefficients for rate (coded units)

Term Effect Coef SE Coef T P

Constant 70.750 0.6374 111.00 0.000

A 19.000 9.500 0.6374 14.90 0.004

C 14.000 7.000 0.6374 10.98 0.008

D 16.500 8.250 0.6374 12.94 0.006

A*C -18.500 -9.250 0.6374 -14.51 0.005

A*D 19.000 9.500 0.6374 14.90 0.004

Analysis of Variance for rate (coded units)

Source DF Seq SS Adj SS Adj MS F P

Main Effects 3 1658.50 1658.50 552.833 170.10 0.006

2-Way Interactions 2 1406.50 1406.50 703.250 216.38 0.005

Residual Error 2 6.50 6.50 3.250

Total 7 3071.50

{

We should also include alias structure like A(+BCD) for all terms.

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From S.A.S:

1) 23 factorial (3 replicates for each of 8 cols):

L H Factor B Factor B L H L H 8,10, 24,28, 16,16, 28,18, 18 20 19 2319,16 27,16, 16,25, 30,23, 16 17 22 25

A

CL

H

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Source DF Sum of Squares Mean Square Fvalue PR>FModel 7 432.0000000 61.71428571 3.38 0.0206Error 16 292.0000000 18.25000000 Corr. Total 23 724.0000000

Source DF Type I SS F Value PR>F DF PR>F

A 1 73.50000000 4.03 0.0620 1 0.0620B 1 253.50000000 13.89 0.0018 1 0.0018C 1 24.00000000 1.32 0.2683 1 0.2683A*B 1 6.00000000 0.33 0.5744 1 0.5744A*C 1 13.50000000 0.74 0.4025 1 0.4025B*C 1 37.50000000 2.05 0.1710 1 0.1710A*B*C 1 24.00000000 1.32 0.2683 1 0.2683

“p-values”

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2) 24-1

A Low A High BL BH BL BH 36 69 72 51

60 63 51 78

A

DD

DL

DH

CL

CH

L

H

BL BH BL BH

36 72 51 69

51 60 63 78

AL AH

CL

CH

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Source DF Sum of Squares Mean Square Fval

Model 7 1296.0000000 185.14285714Error 0 0.0000000 0.00000000 Corr. Total 7 1296.0000000

Source DF Type I SS F Value PR>F DFA + BCD 1 220.50000000 . . 1 B + ACD 1 760.50000000 . . 1C + ABD 1 72.00000000 . . 1A*B + CD 1 18.00000000 . . 1A*C + BD 1 112.50000000 . . 1B*C + AD 1 40.50000000 . . 1A*B*C + D 1 72.00000000 . . 1

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TWO-LEVEL FRACTIONAL FACTORIAL DESIGNS

26-2

I = ABCD = ABEF = CDEF.

DESIGN-EASE

Using Design Ease, the first step is to tell the software that we wish to

conduct a 26-2 design. The software then gave us the design! In other words, the

software proposed its own 26-2 design, and provided a template for data entry.

When asked for the design specifications (an available option), the following was

provided:1/ 4 Replicate of 6 factors in 16 experiments

6 Factors: A, B, C, D, E, FDefining Contrast

I = ABCE = BCDF = ADEF

followed by the alias groups (apparently, omitting terms of four-way or higher-order interaction effects):

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A L I A S E S

A = BCE = DEF B = ACE = CDF C = ABE = BDF D = AEF = BCF E = ABC = ADF F = ADE = BCD

AB = CE AC = BE AD = EF

AE = BC = DF AF = DE BD = CF BF = CD

ABD = ACF = BEF = CDE ABF = ACD = BDE = CEF

We noted that this was not the design we wished to duplicate. Our defining

relation was

I = ABCD = ABEF = CDEF,

while the software imposed on us:

I = ABCE = BCDF = ADEF.

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We noticed that, if we “switch” A with C, and D with E, the software’s defining

relation becomes the same as ours. Hence, we “made believe” that the first

column represented factor C, the third column factor A, and the fourth column

factor E and fifth column factor D. This allowed us to use the software’s template

to “accommodate” the exact design we wanted.

The last column are the values of the dependent variable that are typed in:

1 1 1.00 1.00 1.00 -1.00 1.00 -1.00 103.002 1 -1.00 1.00 -1.00 1.00 1.00 -1.00 102.003 1 -1.00 1.00 1.00 1.00 -1.00 1.00 400.004 1 1.00 -1.00 1.00 -1.00 -1.00 1.00 353.005 1 1.00 -1.00 -1.00 1.00 1.00 1.00 259.006 1 -1.00 -1.00 -1.00 -1.00 -1.00 -1.00 152.007 1 1.00 1.00 -1.00 -1.00 -1.00 1.00 409.008 1 1.00 -1.00 -1.00 -1.00 1.00 -1.00 58.009 1 -1.00 -1.00 -1.00 1.00 -1.00 1.00 355.0010 1 1.00 -1.00 1.00 1.00 -1.00 -1.00 157.0011 1 -1.00 -1.00 1.00 -1.00 1.00 1.00 253.0012 1 -1.00 -1.00 1.00 1.00 1.00 -1.00 52.0013 1 -1.00 1.00 -1.00 -1.00 1.00 1.00 301.0014 1 1.00 1.00 -1.00 1.00 -1.00 -1.00 203.0015 1 -1.00 1.00 1.00 -1.00 -1.00 -1.00 203.0016 1 1.00 1.00 1.00 1.00 1.00 1.00 303.00

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Here is the output when no effects were proactively entered into the error term. The

ANOVA has no SSQ, nor degrees of freedom, for RESIDUAL (see below):

SUM OF MEAN FSOURCE SQUARES DF SQUARE VALUE PROB > F

MODEL 210068.938 15 14004.6RESIDUAL 0.000 0COR TOTAL 210068.938 15

INTERCEPT 228.93750 1A 1.68750 1B 24.06250 1C -0.93750 1D -0.06250 1E -50.06250 1F 100.18750 1AB -0.18750 1AC -0.68750 1AD -0.06250 1AE 0.18750 1AF 0.18750 1BD -0.93750 1BF 0.06250 1ABD -0.43750 1ABF 1.06250 1

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When we specified that the two alias rows without any main, nor two-factor

interaction, effects should be removed from the model (which, then, automatically

enters them into the error term (RESIDUAL to Design Ease), the following ANOVA

resulted:

SUM OF MEAN FSOURCE SQUARES DF SQUARE VALUE PROB > F

MODEL 210047.812 13 16157.5 1529.71 0.0007RESIDUAL 21.125 2 10.6COR TOTAL 210068.938 15

It is clear that Design Ease not only provides an analysis of the experiment,

but also provides aid in designing the experiment, in the first place, while providing, if

asked, the defining relation and alias groups.

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Real World Example of 28-2:

Factor L HLevel

A Geography E WB Volume Cat. L HC Price Cat. L HD Seasonality NO YESE Shelf Space Normal DoubleF Price Normal 20% cutG ADV None Normal(IF)H Loc. Q Normal Prime

{ProductAttributes

{ManagerialDecisionVariables

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- E, F, G, H important- B, C, D main effects, but

not important- A “less” important- XY,X= B, C, D very important Y= E, F, G, H-EF,EG,EH,FG,FH,GH very important- all > 3fi’s = 0, except EFG, EFH, EGH, FGH

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I = BCD = ABEFGH = ACDEFGH

Did objectives get met? A = 4 = 5 = 6E,F,G,H = 4 = 5 = 6

(XY) BE = 3 = 4 = 7 .... DE,CE = 3 = 6 = 5

... EF = 5 = 4 = 5 ... EFG = 6 = 3 = 4

Results:

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I = ABCD = ABEFGH = CDEFGH

A = 3 = 5 = 7E,F,G,H = 5 = 5 = 5

(XY) BE = 4 = 4 = 6 .... DE,CE = 4 = 6 = 4

... EF = 6 = 4 = 4 ... EFG = 7 = 3 = 3

An alternative:

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Minimum Detectable Effects in 2k-p

When we test for significance of an effect, we can also determine the

power of the test.H0: A = 0

Hl: A not = 0

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By looking at power tables, we can determine the power of the test, by specifying , and, essentially, (what reduces to) , the true value of the A effect

(since D = [true A - 0], = true A).

Here, we look at the issue from an opposite (of sorts) perspective:

Given a value of , and for a given value of (or power = 1-), along with N and ,

N = r • 2k-p = # of data points

= degrees of freedom for error term,

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We can determine the “MDE,” the Minimum Detectable Effect (i.e., the minimum detectable , so

that the and 1- are achieved). The results are expressed in “ units,” since we don’t know .

Note: if r = 1 (no replication), there may be few (or even no) df left for error. The tables that follow assume that there are at least 3 df for an error

estimate.

First, a table of df, assuming that all main effects and as many 2fi’s as possible are cleanly estimated. (All

3fi’s and higher are assumed zero).

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Degrees of Freedom for Error TermK

N 3 4 5 6 7 8 9 10 11

4 0

8 1* 0 0 0 0

16 9** 5 0 2 1 0 0 0 0

32 25 21 16*** 10 6 3 1 0 5

64 57 53 48 42 35 27 21 14 8

128 121 117 112 106 99 91 82 72 61

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* 23 with 8 runs complete factorial 1 df for error (ABC)** 23 replicated twice 1 + 8 (for repl.) 9 df for error***25-1 replicated twice all 15 alias rows have mains or

2fi’s, only 16 df for error (replication)*** 25 no replication there will be 16 “3fi’s or higher” effects 16 df for error

TABLES ON NEXT PAGES ASSUME DF OF THIS PAGE

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TABLES 11.33 – 11.36

Of course, these tables can be used either

1) find MDE, given , 1-, or2) find power, given MDE [required] and .

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Ex 1: 25-2, main effects and some 2fi’sa) = .05, 1-=.90, [ = .10] MDE = 3.5 (not good!!- effect must be very large to be detectable)

b) Do a 25-1, 16 runs, with = .05, 1-=.90, [=.10] MDE = 2.5 (not as bad!)

Maybe settle for 1-=.75, [=.25] MDE = 2.0

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Ex 2: 27, unreplicated, = 128 runs

= .01, 1-=.99, [=.01] MDE = .88 (e.g., if estimate = .3 microns, MDE = .264 microns)

Ex 3: 27-2, unreplicated, = 32 runs = .05, 1-=.95, [=.05] MDE = 1.5