IE 361 Module 23Design and Analysis of Experiments: Part 4Prof.s Stephen B. Vardeman and Max D. MorrisReading: Section 7.1, Statistical Quality Assurance Methods for Engineers1Inthismodulewediscusswhatcanbedoneinamany-factorcontextwherethe number of possible combinations of levels of p factorsissolargethatdoinga complete factorial experiment is not practically possible, and one must makedo with information from only some fraction of the factorial. We limit ourdiscussion to fractions of 2pstudies, and consider first half fractions of these,and then other fractions that are powers of 1/2.Motivation for the Subject of 2p−qStudies and aQualitative IntroductionFor p factors, even 2pgets big fastfor p =10 2p=10242Example 23-1 (From an article by Hendrix in 1979 Chemtech) A 15-factorchemical experiment had factorsFactor Levels Factor LevelsA-Coating Roll Temp 115◦vs 125◦J-Feed Air to Dryer Preheat Yes vs NoB-Solvent Recycled vs Refined K-Dibutylfutile in Formula 12% vs 15%C-Polymer X-12 Preheat No vs Yes L-Surfactant in Formula .5% vs 1%D-WebType LX-14vsLB-17 M-DispersantinFormula .1%vs.2%E-Coating Roll Tension 30 vs 40 N-Wetting Agent in Formula 1.5% vs 2.5%F-Number of Chill Roll 1 vs 2 O-Time Lapse 10min vs 30minG-Drying Roll Temp 75◦vs 80◦P-Mixer Agitation Speed 100rpm vs 250rpmH-Humidity of Air F eed 75% vs 90%and response variabley = a measure of product cold crack resistance3A full factorial here would require at least 215=32, 768 runs of the process!An obvious "solution" is to collect data for only some (a fraction) of all possiblecombinations of levels of the factors.Qualitative points that ought to be “obvious”apriori if a fraction of a factorialis used as an experimental design are that• there must be some information loss (relative to the full factorial)• some ambiguity must inevitably follow because of the loss• careful planning and wise analysis are needed to hold these to a minimumExample 23-2 (hypothetical) 22−1... a h alf fraction of a 2 × 2 factorial.The full 22factorial structure is shown in Figure 1.4Figure 1: A 22Full Factorial Layout5Here, if combination (1) is used, combination ab must be employed or elseone learns nothing about the action of one of the factors. (If a is used, thenb must be employed or one learns nothing about the action of one of thefactors.) One the other hand, if there is a big difference in response betweenthe two combinations included in the half fraction, one doesn’t know whetherto attribute this to one or the other or to the interaction effect of the factorsAandB.Example 23-3 (hypothetical) a 23−1... Suppose that 23factorial effects areμ...=10,α2=3,β2=1,γ2=2,αβ22=2,αγ22=0,βγ22=0,αβγ222=0The corresponing means are then as pictured in Figure 2.6Figure 2: A 23Factorial With a "Good" Half Fraction Indicated With CircledCorners7Suppose further that one gets data adequate to essentially reveal the meanresponses for combinations a, b, c and abc (the 4 corners circled in Figure 2)but has no data on the other combinations. (This choice of 4 corners hasadmirable symmetry. If one collapses the cube in any direction one is left witha complete factorial arrangement in the two other factors. That means that if,in fact, one of the 3 factors is "inert" doing nothing to affect response, one hasa full factorial in the 2 factors that matter.) How then might one try to assessthe 23factorial effects if presented with information from the circled cornerson Figure 2?Remember that with the 23as pictured in Figure 2,α2=”right face average” − ”grand average”=38A "half-fraction version" of this might beα∗2=”available right face average” − ”available grand average”=13− 10=3!!! Here α∗2= α2!!! Is this something for nothing? Can we learn about theAmaineffectusingdatafromonly"4corners"?But note that a similar calculation for the C main effect givesγ∗2=”available back face average” − ”available grand average”=14− 10=4??? 4=γ*26= γ2=2 ???9The general story behind this situation is that for this 23−1fractional factorialα∗2= α2+ βγ22and γ∗2= γ2+ αβ22We were able to "recover" α2using only α∗2(that is based on only half of thecorners of the cube) because βγ22=0. We were unable to "recover" γ2using only γ∗2(that is based on only half of the corners of the cube) becauseαβ22=26=0. We can really only know α2+ βγ22(and not α2)andγ2+ αβ22(and not γ2) based on the half fraction. This is an exampleof confounding/aliasing/ambiguity that of necessity comes with use of only afractional factorial.Half Fractions of 2pFactorial StudiesThe issues to be addressed in order to use 2p−qfractional factorials are:10• how to rationally choose12qout of 2pcombinations for study• how to determine the corresponding aliasing/confounding pattern• howtododataanalysisWe proceed to consider these questions first in the context of half fractions(the q =1case) then for general 1/2qfractions.Choice of standard half fractions of 2pfactorialsTo choose what we will call a "standard" (a "good") half fraction of a 2pfactorial, we will11• write out signs for specifying levels for all possible combinations of levelsof the “first” p − 1 factors• then “multiply” these together for a given combination of the “first” factorsto arrive at a corresponding level to use fo r the “last” factorExample 23-4 (A 24−1fractional factorial) With 4 two-level factors A, B, Cand D, one proceeds as in the following table. (One multiplies − and + signsas if they were ±1’s and the last column gives the indicated combination offactor levels for the row, using the special 2pnaming convention introduced inModule 22.12A B C Product (used for D) Combination−−− − (1)+ −− + ad− + − + bd++−−ab−−++cd+ − + − ac− ++ − bc+++ + abcdExample 23-5 R. Snee in a 1985 ASQC Technical Supplement discussed a25−1chemical process study. The factors and their levels were as in thefollowing table.13Factor − +A-Solvent/Reactant low vs highB-Catalyst/Reactant .025 vs .035C-Temperature 150◦vs 160◦D-Reactant Purity 92% vs 96%E-pH of Reactant 8.0 vs 8.7In Snee’s study, the response va riable wasy = color indexSnee’s (unreplicated) data were14combination y combination ye −.63 d 6.79a 2.51 ade 6.47b −2.68 bde 3.45abe −1.66 abd 5.68c 2.06 cde 5.22ace 1.22 acd 9.38bce −2.09 bcd 4.30abc 1.93 abcde 4.05These are data from half of all 32 combinations of 2 levels