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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).

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?

4A -1 a - b ab - c ac - bc abc 4BC 1

a - b - ab - c - ac bc abc

Consider

(4A 4BC) 2(a - b - c abc) or 2(A

BC) a - b - c abc

overall

2(A BC) a - b - c abc 2(B AC) -a b - c

abc 2(C AB) -a - b c abc

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

Note 4ABC(abcabc)-(1abacbc) cannot be

estimated.

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 interactions0, then you can say either

(1) you get 3 factors for the price

of 2. (2) you get 3 factors at

1/2 price.

Suppose we run these 4 1, ab, c,

abc We would then estimate A B

C ABC AC BC

two main effects together usually less 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 CONFOUNDED members of the

pair are ALIASED

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)

Consider a 24-1 with t.c.s 1, ab, ac, bc, ad,

bd, cd, abcd

ABCD BACD CABD ABCD ACBD BCAD D ABC

Can estimate

- 8 t.c.s - Lose 1 effect - Estimate other 14 in

7 alias pairs of 2

Note

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.

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.

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 (238). Whatever else the real world

is, its Multi-factored.

The best way to learn how is to work (and

discuss) some examples

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 ABDE

(in the plus-minus table)

Step 2 Find the resulting alias pairs

ABDE ABDE ABCCDE BADE

AC4 BCDACE CABCDE ADBE

BCEACD DABE AEBD EABD

BC4 CD4

CE4

- lose 1 -other 30 in 15 alias pairs of 2 -run 16

t.c.s 15 estimates

AxABDEBDE

See if theyre (collectively) acceptable.

Another option (among many others) I

ABCDE (in the plus-minus table)

A4 AB3 B4 AC3

C4 AD3 D4

AE3 E4 BC3

BD3

BE3 CD3

CE3

DE3

4 4 way interaction 3 3 way interaction.

Assume we choose I ABDE

Next Step Find the 2 blocks using ABDE as the

confounded effect.

I II 1??????????c

a ac ab abc b

bc de cde ade acde abde

abcde bde bcde ad acd d

cd bd bcd abd abcd ae

ace e ce be bce

abe abce

Same process as a Confounding Scheme

- Which block to run if I ABDE (ABDE )?

Example 2

In a 25, there are 31 effects with 8 runs, there

are 7 df 7 estimates available

25-2 A, B, C, D, E

Must Lose 3 other 28 in 7 alias groups of 4

Choose the 3 Like in confounding schemes, 3rd

must be product of first 2

I ABC BCDE ADE A BC 5

DE B AC 3 4 C AB 3

4 D 4 3 AE E 4 3

AD BD 3 CE 3 BE 3 CD 3

Find alias groups

Assume we use this design.

Lets find the 4 blocks

using ABC , BCDE , ADE as confounded effects

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 abde bcde abcde cde

bce ace ce abce acde

- Which block should we run if IABCBCDEADE?

- 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.

Analysis 2k-p Design using MINITAB

- Create factorial design
- Statgtgt DOEgtgt Factorial
- gtgt Create Factorial Design
- Input data values.
- Analyze data
- Statgtgt DOEgtgt Factorial
- gtgt Analyze Factorial Design

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 IABCD.

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

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 AB -1.000 -0.500 AC

-18.500 -9.250 AD 19.000 9.500

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Alias StructureI ABCDA BCDB

ACDC ABDD ABCAB CDAC BDAD

BC

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 AC -18.500 -9.250

0.6374 -14.51 0.005 AD 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.

From S.A.S

1) 23 factorial (3 replicates for each of 8

cols)

A

L H Factor B

Factor B L H L H

8,10, 24,28, 16,16, 28,18, 18

20 19 23 19,16 27,16,

16,25, 30,23, 16 17 22

25

L H

C

Source DF Sum of Squares Mean

Square Fvalue PRgtF Model 7

432.0000000 61.71428571 3.38

0.0206 Error 16 292.0000000

18.25000000 Corr. Total 23

724.0000000 Source DF Type I SS

F Value PRgtF DF PRgtF A

1 73.50000000 4.03

0.0620 1 0.0620 B

1 253.50000000 13.89 0.0018 1

0.0018 C 1

24.00000000 1.32 0.2683 1

0.2683 AB 1 6.00000000

0.33 0.5744 1 0.5744 AC

1 13.50000000 0.74

0.4025 1 0.4025 BC 1

37.50000000 2.05 0.1710 1

0.1710 ABC 1 24.00000000

1.32 0.2683 1 0.2683

p-values

2) 24-1

Source DF Sum of Squares Mean

Square Fval Model 7

1296.0000000 185.14285714 Error

0 0.0000000

0.00000000 Corr. Total 7

1296.0000000 Source DF Type I

SS F Value PRgtF DF A BCD

1 220.50000000 . .

1 B ACD 1

760.50000000 . .

1 C ABD 1 72.00000000

. . 1 AB CD

1 18.00000000 . .

1 AC BD 1 112.50000000

. . 1 BC AD

1 40.50000000 . .

1 ABC D 1

72.00000000 . . 1

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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 gt 3fis 0, except EFG,

EFH, EGH, FGH

I BCD ABEFGH ACDEFGH

Did objectives get met?

A 4 5 6 E,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

An alternative

I ABCD ABEFGH CDEFGH

A 3 5 7 E,F,G,H 5 5 5 (XY)

BE 4 4 6 .... DE,CE 4 6 4

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

Minimum Detectable Effects in 2k-pWhen we test

for significance of an effect, we can also

determine the power of the test.H0 A 0Hl A

not 0

By looking at power tables, we can determine the

power of the test, by specifying s, and,

essentially, (what reduces to) D, 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) perspectiveGiven a value of a, and

for a given value of b (or power 1-b), along

with N and n, N r 2k-p of data

pointsn degrees of freedom for error term,

We can determine the MDE, the Minimum

Detectable Effect (i.e., the minimum detectable

D, so that the a and 1-b are achieved). The

results are expressed in s units, since we

dont know s. 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

2fis as possible are cleanly estimated. (All

3fis and higher are assumed zero).

Degrees of Freedom for Error Term

23 with 8 runs ? complete factorial ? 1 df for

error (ABC) 23 replicated twice ? 1 8 (for

repl.) ? 9 df for error25-1 replicated twice

? all 15 alias rows have mains or 2fis, ? only

16 df for error (replication) 25 no

replication ? there will be 16 3fis or higher

effects ? 16 df for error

TABLES ON NEXT PAGES ASSUME DF OF THIS PAGE

TABLES 11.33 11.36

Of course, these tables can be used either1)

find MDE, given a, 1-b, or2) find power, given

MDE required and a.

Ex 1 25-2, main effects and some 2fisa) a

.05, 1-b.90, b .10 ? MDE 3.5s (not good!!-

effect must be very large to be detectable)b)

Do a 25-1, 16 runs, with a .05, 1-b.90,

b.10 ? MDE 2.5s (not as bad!)Maybe settle

for 1-b.75, b.25 ? MDE 2.0s

Ex 2 27, unreplicated, 128 runsa .01,

1-b.99, b.01 ? MDE .88s (e.g., if s

estimate .3 microns, MDE .264 microns)Ex

3 27-2, unreplicated, 32 runs a .05,

1-b.95, b.05 ? MDE 1.5s

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