Emil's example turns out to have lots of generators for half and quarter fractions: I={any combinaton of A,B,C,D}. Only ABC, ABD, ACD, BCD, and ABCD can be used as resolution III generators. Unfortunately, no pair of them keeps main effects clear of other main effects. But the fractions do exist. They exist for ANY pair of tc's.
Look at the number of + (or -, it doesn't matter) factors in each tc (call them p1 and p2). If they differ by an even number (including 0), then I=ABCDE works. If they differ by an odd number, then there are 4-letter words that will work, consisting of deleting any factor in which the signs for the two tc's do not agree.
In Joe's original example, p1=2, p2=3, and signs disagree for A, D, and E. So BCDE, ABCE, and ABCD are 4-letter words that will word to create a half-fraction. In Emil's example, p1=5 and P2=4, and only E disagrees, so ABCD can generate a half-fraction.
For a quarter fraction of resolution at least 3, you need two independent generators (and a third implied by the binary product of the other two) such that all word lengths are at least three. This is not possible if you start with the 5-letter word, and in fact requires a 4-3-3 letter configuration, there thr three-letter generators have exactly one letter in common. Joe's example has ABCD and ABE, implying CDE, which works out fine. The only three-letter generators in Emil's example all omit E, so no resolution 3 design is possible.
Three letter words are found from two TCs by combining factors in combinations in which either (1) all three factors agree in sign, or (2) exactly one factor agrees in sign. In Joe's example, there are only two factors that agree in sign (B,C), so only possibility (2) exists. Any combination of 1 from {B,C} with two from {A,D,E} will form a 3-letter word. In Emil's example, only one factor disagrees, so we can only choose combinations in which all three signs agree. We are restricted to combinations of three letters from {A,B,C,D}
We can create TCs in which 0, 1, 2, 3, 4, or 5 signs agree. The 5 is eliminated for obvious reasons. The 0 is also eliminated, because we must have exactly 1 or 3 agreements in a 3-letter word. The 4 is eliminated based on generalization of Emil's example.
If there is only one agreement, then we can form any two 3-letter words for which the factor with the agreement is the common letter, and disjoint pairs of the other 4 factors fill out the 2 words. There are three such pairs (e.g. if E agrees, we can take ABE+CDE, ACE+BDE, or ADE+BCE).
If there are two agreements, then the two 3-letter words must each consist of a different factor in which agreement occurs, plus a pair of the remaining three with only one in common. Joe3's example os one of these, with B and C in agreement. Then solutions are ABD+ACE, ABD+CDE, ABE+ACD, ABE+CDE, BDE+ACD, and BDE+ACE
If there are three agreements, then there cannot be two words of type (2) [one agreement], because that would force the same two disagreeing factors into both words. Thus, one word MUST be the three agreeing factors. The other word can be any of the agreeing factors with the two disagreeing factors. If ABC agree, then we can use ABC+ADE, ACB+BDE, or ABC+CDE.
Hope this helps.
-Tom.
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Thomas Loughin
Simon Fraser University
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Original Message:
Sent: 04-04-2012 16:04
From: Joseph Voelkel
Subject: Pre-selecting tc's in fractional-factorial designs
Thanks, Emil.
Yes, you are of course correct--the solution depends on which tc's are selected.
This is, I hope, a more accurate way to phrase the question:
- For k factors in 2^m runs, if one selects a certain set of tc's (I selected 2 in the example), is there is clever way (that is, short of brute force) to determine which, if any geometric ff designs (greater than Res II) are consistent with that set of tc's?
Joe
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Joseph Voelkel
Rochester Institute of Technology
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