Finding the Alias Structure of a Fractional Factorial

The entire set of aliases in a fractional factorial design is called the alias structure of the design. There is a fairly easy method for writing down the alias structure of a fractional design. This method depends on some simple observations about multiplying columns of +1's and -1's:

1. The letter I denotes the column consisting entirely of +1's.
2. Note that any column multiplied by itself yields column I. For example, A·A = A² = I, B·B = B² = I, and so forth.
3. Multiplying column I by any other column does not change the column. For example, A·I = I·A = A.

Using these facts, we can obtain the alias structure of any fractional factorial as follows:

1. First, write the p assignments of additional factors in equation form. These p equations are called the design generators.
2. Multiply each generator from step 1 by its left side to put each generator into the form I = w, where w is a "word" composed of several letters representing particular experimental factors (e.g., D = ABC becomes I = ABCD). It is also possible tocreate words with "-" signs, such as D = -ABC. If this is done, the resulting design will use a different fraction of the runs from the full 2^k design.
3. Letting I = w₁, I = w₂, ..., I = w_p denote the p design generators from step 2, form all possible products of the words w_i (one at a time, two at a time, three at a time, etc.). Use the fact that squares of factors can be eliminated (e.g., A² = I and multiplying by I does not change anything). There will be a total of 2^p words formed. This collection is called the defining relation of the design.
4. Multiply each word in the defining relation by all 2^k - 1 effects based on k factors. Use the fact that squares of factors cancel out to simplify the products. The result is called the alias structure of the design.

Example:

Question: Determine the alias structure of the 2^5-2 design with the generators D = ABC and I = ACE.

Solution:

1. The design generators are D = ABC and E = AC.
2. The generators in standard form are I = ABCD and I = ACE
3. Only one product of the above words is possible, namely, (ABCD)(ACE) = A²BC²DE = BDE.
4. The defining relation is I = ACE = BDE = ABCD
5. Multiplying each of the 2⁵-1 effects by the defining relation gives
   • I = ACE = BDE = ABCD
   • A = CE = ABDE = BCD
   • B = DE = ACD = ABCE
   • C = AE = ABD = BCDE
   • D = BE = ABC = ACDE
   • E = AC = BD = ABCDE
   • AB = CD = ADE = BCE
   • AD = BC = ABE = CDE

Adapted from:

Devore, Jay L. and Farnum, Nicholas R., "Applied Statistics for Engineers and Scientists", Duxbury Press, 1999, pp.457-458.

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