Transcript No Slide Title

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).
1
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?
2
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=(a+b+c+abc)-(1+ab+ac+bc)
3
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 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.”
4
Suppose we run these 4:
1, ab, c, abc;
We would then estimate
A+B
two main effects
C+ ABC
together usually
AC + BC
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
5
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)
6
Consider a 24-1 with t.c.’s
1, ab, ac, bc, ad, bd, cd, abcd
Can estimate: A+BCD
B+ACD
C+ABD
AB+CD
AC+BD
BC+AD
D +ABC
Note:
- 8 t.c.’s
- Lose 1 effect
- Estimate other 14 in 7 alias pairs of 2
7
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.”
8
“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.
9
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:
10
25-1: A, B, C, D, E
Example:
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:
- lose 1
-other 30 in
15 alias pairs
of 2
-run 16 t.c.’s
15 estimates
*A=BDE
B=ADE
C=ABCDE
D=ABE
E=ABD
{
*AxABDE=BDE
AB=DE
AC=4
AD=BE
AE=BD
BC=4
CD=4
CE=4
ABC=CDE
BCD=ACE
BCE=ACD
11
See if they’re (collectively) acceptable.
Another option (among many others):
I = ABCDE (in the plus-minus table)
A=4
B=4
C=4
D=4
E=4
AB=3
AC=3
AD=3
AE=3
BC=3
BD=3
BE=3
CD=3
CE=3
DE=3
4: 4 way interaction;
3: 3 way interaction.
Assume we choose
I = ABDE
12
Next Step: Find the 2 blocks using
ABDE as the confounded effect.
I
Same process
as a
Confounding
Scheme
{
1
c
ab abc
de cde
abde abcde
ad acd
bd bcd
ae
ace
be
bce
II
a
b
ade
bde
d
abd
e
abe
ac
bc
acde
bcde
cd
abcd
ce
abce
•Which block to run if I = ABDE (ABDE = “+”)?
13
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
14
Choose the 3: Like in confounding
schemes, 3rd must be product of first 2:
I
Find alias
groups:
= ABC = BCDE = ADE
A =
B =
C =
D =
E =
BD =
BE =
BC
AC
AB
4
4
3
3
= 5 = DE
= 3 = 4
= 3 = 4
= 3 = AE
= 3 = AD
= CE = 3
= CD = 3
Assume we use this design.
15
Let’s find the 4 blocks using ABC , BCDE , ADE
as confounded effects
1
2
3
1
a
b
abd
bd
ad
bc abc
c
acd
cd
abcd
de ade
bde
abe
be
ae
bcde abcde cde
ace
ce
abce
4
d
ab
bcd
ac
e
abde
bce
acde
•Which block should we run if I=ABC=BCDE=ADE?
16
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.
17
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
18
Example: 24-1 with defining relation I=ABCD.
A
-1
1
-1
1
-1
1
-1
1
B
-1
-1
1
1
-1
-1
1
1
C
-1
-1
-1
-1
1
1
1
1
D
-1
1
1
-1
1
-1
-1
1
rate
45
100
45
65
75
60
80
96
19
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
20
Fractional Factorial Fit: rate versus A, B, C, D
Estimated Effects and Coefficients
for rate (coded units)
Term
Effect
Constant
Coef
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
21
Normal Probability Plot of the Effects
(response is rate, Alpha = .99)
1.5
A:
B:
C:
D:
AD
1.0
A
B
C
D
Normal Score
A
0.5
D
0.0
C
B
-0.5
AB
-1.0
AC
-1.5
-30
-20
-10
0
10
20
30
Effect
22
Alias Structure
I + A*B*C*D
A + B*C*D
B + A*C*D
C + A*B*D
D + A*B*C
A*B + C*D
A*C + B*D
A*D + B*C
23
Fractional Factorial Fit: rate versus A, C, D
Estimated Effects and Coefficients for rate (coded units)
Term
Effect
Constant
{
Coef
SE Coef
T
P
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
Main Effects
3
1658.50
2-Way Interactions
2
Residual Error
Total
Adj MS
F
P
1658.50
552.833 170.10
0.006
1406.50
1406.50
703.250 216.38
0.005
2
6.50
6.50
7
3071.50
3.250
We should also include alias structure like A(+BCD) for all terms.
24
From S.A.S:
1) 23 factorial (3 replicates for each of 8 cols):
A
L
H
Factor B
L
H
L 8,10, 24,28,
C
18
19,16
H 16
20
27,16,
17
Factor B
L
H
16,16, 28,18,
19
23
16,25, 30,23,
22
25
25
Source
Model
Error
Corr. Total
DF Sum of Squares Mean Square Fvalue PR>F
7 432.0000000
61.71428571 3.38 0.0206
16 292.0000000
18.25000000
23 724.0000000
Source
DF
A
B
C
A*B
A*C
B*C
A*B*C
1
1
1
1
1
1
1
Type I SS
73.50000000
253.50000000
24.00000000
6.00000000
13.50000000
37.50000000
24.00000000
F Value
PR>F
4.03
13.89
1.32
0.33
0.74
2.05
1.32
0.0620
0.0018
0.2683
0.5744
0.4025
0.1710
0.2683
DF
1
1
1
1
1
1
1
PR>F
0.0620
0.0018
0.2683
0.5744
0.4025
0.1710
0.2683
“p-values”
26
2) 24-1
A
A Low
DL
CL DH
DL
CH DH
BL
36
51
A High
BH
BL
72
51
60
63
BH
69
78
AL
AH
BL BH BL BH
CL 36 72 51 69
CH 51 60 63 78
27
Source
Model
Error
Corr. Total
DF Sum of Squares Mean Square Fval
7 1296.0000000
185.14285714
0
0.0000000
0.00000000
7 1296.0000000
Source
A
+ BCD
B
+ ACD
C
+ ABD
A*B
+ CD
A*C
+ BD
B*C
+ AD
A*B*C + D
DF Type I SS
F Value PR>F
1 220.50000000
.
.
1 760.50000000
.
.
1
72.00000000
.
.
1
18.00000000
.
.
1 112.50000000
.
.
1
40.50000000
.
.
1
72.00000000
.
.
DF
1
1
1
1
1
1
1
28
TWO-LEVEL FRACTION AL FACTORIAL D ESIGN S
26-2
I = ABCD = ABEF = CDEF.
DESIGN -EASE
Using Design Ease, the first step is to tell the softw are that w e w ish to
cond uct a 26-2 d esign. The softw are then gave us the design! In other w ord s, the
softw are proposed its ow n 26-2 d esign, and provid ed a tem plate for d ata entry.
When asked for the d esign specifications (an available option), the follow ing w as
provid ed :
1/ 4 Replicate of 6 factors in 16 experim ents
6 Factors: A, B, C, D, E, F
Defining Contrast
I = ABCE = BCDF = ADEF
follow ed by the alias groups (apparently, om itting term s of four-w ay or higherord er interaction effects):
29
A LIA SES
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 w as not the design w e w ished to duplicate. Our defining
relation w as
I = ABCD = ABEF = CD EF,
w hile the softw are imposed on us:
I = ABCE = BCD F = AD EF.
30
We noticed that, if w e “sw itch” A w ith C, and D w ith E, the softw are’s d efining
relation becom es the sam e as ours. H ence, w e “m ad e believe” that the first
colum n represented factor C, the third colum n factor A, and the fourth colum n
factor E and fifth colum n factor D. This allow ed us to use the softw are’s tem plate
to “accom m od ate” the exact d esign w e w anted .
The last colum n are the values of the d epend ent variable that are typed in:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1.00
-1.00
-1.00
1.00
1.00
-1.00
1.00
1.00
-1.00
1.00
-1.00
-1.00
-1.00
1.00
-1.00
1.00
1.00
1.00
1.00
-1.00
-1.00
-1.00
1.00
-1.00
-1.00
-1.00
-1.00
-1.00
1.00
1.00
1.00
1.00
1.00
-1.00
1.00
1.00
-1.00
-1.00
-1.00
-1.00
-1.00
1.00
1.00
1.00
-1.00
-1.00
1.00
1.00
-1.00
1.00
1.00
-1.00
1.00
-1.00
-1.00
-1.00
1.00
1.00
-1.00
1.00
-1.00
1.00
-1.00
1.00
1.00
1.00
-1.00
-1.00
1.00
-1.00
-1.00
1.00
-1.00
-1.00
1.00
1.00
1.00
-1.00
-1.00
1.00
-1.00
-1.00
1.00
1.00
1.00
-1.00
1.00
-1.00
1.00
-1.00
1.00
-1.00
1.00
-1.00
-1.00
1.00
103.00
102.00
400.00
353.00
259.00
152.00
409.00
58.00
355.00
157.00
253.00
52.00
301.00
203.00
203.00
303.00
31
H ere is the ou tp u t w hen no effects w ere p roactively entered into the error term . The
AN OVA has no SSQ, nor d egrees of freed om , for RESIDUAL (see below ):
SOURCE
SUM OF
SQUARES
MODEL
RESIDUAL
COR TOTAL
210068.938
0.000
210068.938
IN TERCEPT
A
B
C
D
E
F
AB
AC
AD
AE
AF
BD
BF
ABD
ABF
DF
15
0
15
MEAN
SQUARE
F
VALUE PROB > F
14004.6
228.93750 1
1.68750 1
24.06250 1
-0.93750 1
-0.06250 1
-50.06250
1
100.18750 1
-0.18750
1
-0.68750
1
-0.06250
1
0.18750 1
0.18750 1
-0.93750
1
0.06250 1
-0.43750 1
1.06250 1
32
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:
SOURCE
MODEL
RESIDUAL
COR TOTAL
SUM OF
SQUARES
MEAN
DF SQUARE
210047.812 13
21.125
2
210068.938 15
16157.5
10.6
F
VALUE PROB > F
1529.71
0.0007
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.
33
Real World Example of 28-2:
Level
Factor
L
H
Product
Attributes
{
{
Managerial
Decision
Variables
A
B
C
D
E
F
G
H
Geography
Volume Cat.
Price Cat.
Seasonality
Shelf Space
Price
ADV
Loc. Q
E
W
L
H
L
H
NO YES
Normal Double
Normal 20% cut
None Normal(IF)
Normal Prime
34
- E, F, G, H
- B, C, D
important
main effects, but
not important
“less” important
very important
-A
- XY,X= B, C, D
Y= E, F, G, H
-EF,EG,EH,FG,FH,GH very important
- all > 3fi’s = 0, except EFG, EFH, EGH, FGH
35
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:
36
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
37
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
38
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,
39
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 don’t 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 2fi’s as possible are cleanly estimated. (All
3fi’s and higher are assumed zero).
40
Degrees of Freedom for Error Term
K
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
42 35
27
21
14
8
112 106 99
91
82
72
61
128 121 117
48
41
* 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
42
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.
43
Ex 1: 25-2, main effects and some 2fi’s
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
44
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
45