Design and Analysis of Multi-Factored Experiments. Fractional Factorial Designs. Design of Engineering Experiments – The 2 k-p Fractional Factorial Design.
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Fractional Factorial Designs
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
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?
4BC = 1 + a - b - ab -c - ac + bc + abc
(4A + 4BC)= 2(a - b - c + abc)
2(A + BC)= a - b - c + abc
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.
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.”
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
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)
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”
Consider a 2 some or all alias pairs one of the effects is KNOWN. This usually means “thought to be zero”4-1 with t.c.’s
1, ab, ac, bc, ad, bd, cd, abcd
Can estimate: A+BCD
- 8 t.c.’s
-Lose 1 effect
-Estimate other 14 in 7 alias pairs of 2
“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 (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:
The basic design; the design generator
Example: 2 inevitable in a many factor situation. It is generally better to study 5-1 : A, B, C, D, E
Step 1: In a 2k-p, we “lose” 2p-1.
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=BDE AB=DE ABC=CDE
B=ADE AC=4 BCD=ACE
C=ABCDE AD=BE BCE=ACD
- lose 1
- other 30 in 15 alias pairs of 2
- run 16 t.c.’s
See if they are (collectively) acceptable. inevitable in a many factor situation. It is generally better to study
Another option (among many others):
I = ABCDE
Next step: Find the 2 blocks (only inevitable in a many factor situation. It is generally better to study one of which will be run)
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
1 inevitable in a many factor situation. It is generally better to study
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:
There’s only one way to do it, i.e. either adding “d” or not adding “d”; one way will work, one won’t.
Example 2: or not adding “d”; one way will work, one won’t.
25-2 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
Choose the 3: Like in confounding schemes, 3rd or not adding “d”; one way will work, one won’t.
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:
Let’s find the 4 blocks: I =ABC = BCDE = ADE or not adding “d”; one way will work, one won’t.
Assume we run the Principal block (block 1)
Run the 8 t.c.’s and analyze: Since it’s a 2 or not adding “d”; one way will work, one won’t.5-2, 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
AC - +4-3
Good to know rule: or not adding “d”; one way will work, one won’t.
t.c. with even # letters in common with even-lettered effect is a + for that effect; t. c. with odd # letters in common with odd-lettered 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
# letters t.c. has in
Common with Effect
Interpretation of results often relies on making some assumptions
Confirmation experiments can be important
See the projection of this design into 3 factors
Every fractional factorial contains full factorials in fewer factors
The “flashlight” analogy
A one-half fraction will project into a full factorial in any k – 1 of the original factors
Complete defining relation: I = ABCE = BCDF = ADEF