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    • #14677
      Andrea Pisano's avatarAndrea Pisano
      Participant

      Basic Black Belt Study Guide (Free)

      Hello, I’d really like to know if there is an easy method to determine design resolution (I, II, III…) on the basis of number of levels, factor and generators. For example: 2^8-2, 2^9-1, 2^27-1…
      I’m sure there’s a trick or short formula i can use without write down the entire table.
      Thank you!

    • #38015
      Ted Hessing's avatarTed Hessing
      Keymaster

      Six Sigma Study Guide Support

      Hi Andrea,

      Sorry for the very late reply. There is a shortcut, and you don’t need to write out the whole table.

      The rule: resolution is the length of the shortest word in the defining relation. If you’re given the generators, multiply them out (letters that appear twice cancel) and count the letters in the shortest word.

      Half fractions (2k-1): the best design uses all k factors in its one generator, so the resolution equals k.

      • 29-1: resolution IX
      • 227-1: resolution XXVII in theory, though nobody runs 67 million trials to get there

      Quarter fractions (2k-2): the best achievable resolution is 2k/3, rounded down.

      • 28-2: 16/3 rounds down to 5, so resolution V. For example, G = ABCD and H = ABEF give the words ABCDG, ABEFH and their product CDEFGH. The shortest has 5 letters.

      I checked both rules by brute force for k from 5 to 10. For smaller fractions (p of 3 or more), use a standard design table.

      One more memory aid: in a resolution R design, the orders of two aliased effects add up to at least R. In resolution III, main effects (1) alias with two-factor interactions (2), and 1 + 2 = 3. In resolution IV, main effects alias with three-factor interactions (1 + 3). In resolution V, main effects alias with four-factor interactions, and two-factor interactions alias with three-factor ones (2 + 3).

      There’s more in my Partial and Fractional Factorial Design article. Ask here if a specific design is giving you trouble.

      Ted

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