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## Fractional Factorial Designs

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### 1, ab, ac, bc, ad, bd, cd, abcd. Can estimate: - 8 t.c.'s - Lose 1 effect ... 25 no replication there will be 16 '3fi's or higher' effects 16 df for error ... – PowerPoint PPT presentation

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Title: Fractional Factorial Designs

1
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).
2
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?
3
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.
4
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.
5
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
6
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)
7
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
8
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.
9
Clean estimates of the remaining member of the
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.
10
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
11
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
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
12
See if theyre (collectively) acceptable.
Another option (among many others) I
ABCDE (in the plus-minus table)
A4 AB3 B4 AC3
AE3 E4 BC3
BD3
BE3 CD3
CE3
DE3
4 4 way interaction 3 3 way interaction.
Assume we choose I ABDE
13
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 )?

14
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
15
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.
16
Lets find the 4 blocks
using ABC , BCDE , ADE as confounded effects
1 2 3 4 1
a b d abd bd
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?

17
• 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.
18
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

19
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.
20
Analysis of Variance for rate (coded
units) Source DF Seq SS
4 1663 1663 415.7
2-Way Interactions 3 1408
1408 469.5 Residual Error
0 0 0 0.0 Total
7 3071
21
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
22
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23
Alias Structure I ABCD A BCD B
ACD C ABD D ABC AB CD AC BD AD
BC
24
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
9.500 0.6374 14.90 0.004 Analysis of
Variance for rate (coded units) Source
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.
25
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
26
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
27
2) 24-1
28
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 40.50000000 . .
1 ABC D 1
72.00000000 . . 1
29
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35
- 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
36
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
37
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
38
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
39
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) perspective Given 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
points n degrees of freedom for error term,
40
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).
41
Degrees of Freedom for Error Term
42
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 2fis, ? only
16 df for error (replication) 25 no
replication ? there will be 16 3fis or higher
effects ? 16 df for error
43
TABLES 11.33 11.36
Of course, these tables can be used either 1)
find MDE, given a, 1-b, or 2) find power, given
MDE required and a.
44
Ex 1 25-2, main effects and some 2fis a) 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
45
Ex 2 27, unreplicated, 128 runs a .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