Design and Analysis of MultiFactored Experiments. Fractional Factorial Designs. Design of Engineering Experiments – The 2 kp Fractional Factorial Design.
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Design and Analysis ofMultiFactored Experiments
Fractional Factorial Designs
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For the principal fraction, notice that the contrast for estimating the main effect A is exactly the same as the contrast used for estimating the BC interaction.
This phenomena is called aliasing and it occurs in all fractional designs
Aliases can be found directly from the columns in the table of + and  signs
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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 + 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.
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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 those 4:
1, ab, c, abc;
We would then estimate
A + B
C + ABC
AC + BC
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
two main effects
together usually
less desirable
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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 231 designs)
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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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Consider a 241 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
 8 t.c.’s
Lose 1 effect
Estimate other 14 in 7 alias pairs of 2
Note:
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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 241 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 (252 = 8) than 3 factors completely (23 = 8). Whatever else the real world is, it’s Multifactored.
The best way to learn “how” is to work (and discuss) some examples:
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Design and Analysis ofMultiFactored Experiments
Aliasing Structure
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The basic design; the design generator
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Example: 251 : A, B, C, D, E
Step 1: In a 2kp, we “lose” 2p1.
Here we lose 1. Choose the effect to lose. Write it as a “Defining relation” or “Defining contrast.”
I = ABDE
Step 2: Find the resulting alias pairs:
*A=BDEAB=DEABC=CDE
B=ADEAC=4BCD=ACE
C=ABCDEAD=BEBCE=ACD
D=ABEAE=BD
E=ABDBC=4
CD=4
CE=4
 lose 1
 other 30 in 15 alias pairs of 2
 run 16 t.c.’s
15 estimates
*AxABDE=BDE
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See if they are (collectively) acceptable.
Another option (among many others):
I = ABCDE
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
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Next step: Find the 2 blocks (only one of which will be run)
I II
1caac
ababcbbc
decdeadeacde
abdeabcdebdebcde
adacddcd
bdbcdabdabcd
aeaceece
bebceabeabce
Same process
as a
Confounding
Scheme
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1
a
b
ab
c
ac
bc
abc
e
ae
be
abe
ce
ace
bce
abce
Next: Pick which block to run.
(say, block II)
Next: Go out and collect the data.
Next: Analyze it.
a.) find a proper Yates’ order.
i.) pick a letter and for a moment
call it “DEAD.” (assume we pick “d”)
ii.) use the remaining (“live”) letters
to form the STANDARD Yates’ order:
(see right column)
Now append the dead letter as needed to form
the block chosen to be run:
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There’s only one way to do it, i.e. either adding “d” or not adding “d”; one way will work, one won’t.
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Example 2:
252 A, B, C, D, E
Must “lose” 3; other 28
in 7 alias groups of 4
In a 25 , there are 31 effects; with 8 t.c., there are 7 df & 7 estimates available
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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 = 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
Assume we use this design.
Find alias groups:
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Let’s find the 4 blocks: I =ABC = BCDE = ADE
Assume we run the Principal block (block 1)
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Run the 8 t.c.’s and analyze: Since it’s a 252, we designate 2 letters as “DEAD” (say b, d), write a standard Yates’ order in the other (3) (live) letters, and append the dead letters to form the t.c.’s being run:
I = ABC = BCDE = ADE
t.c.
yield
(1)
(2)
(3)
Estimate
1
.

.
a (
bd)
2+52
A
.
c (b)
2+34
C
.
ac (d)
AC  +43
B
e (d)
.
4+32
E
D
.
ae (b)
AE 3+4
.
ce (
bd)
CE 3+23
.
ace
ACE 2+32
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Good to know rule:
t.c. with even # letters in common with evenlettered effect is a + for that effect; t. c. with odd # letters in common with oddlettered effect is a + for that effect; otherwise a  (minus)
abd in ABCDE:3 in common with 5
ODD with ODD +
abce in ABCDFG:3 in common with 6
ODD with EVEN 
# letters in effect
Even
Odd
# letters t.c. has in
Common with Effect
Even
Odd
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Interpretation of results often relies on making some assumptions
Ockham’srazor
Confirmation experiments can be important
See the projection of this design into 3 factors
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Every fractional factorial contains full factorials in fewer factors
The “flashlight” analogy
A onehalf fraction will project into a full factorial in any k – 1 of the original factors
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Complete defining relation: I = ABCE = BCDF = ADEF
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