ST3232: Design and Analysis of Experiments

ST3232: Design and Analysis of Experiments 2012/2013: Semester II Tutorial 8 1. A 2 5 2 design was used to investigate the effect of A = condensation temperature, B = amount of material 1, C = solvent volume, D = cdondensation time, and E = amount of material 2 on yield. The results obtained are as follows: e = 23.2 ad = 16.9 cd = 23.8 bde = 16.8 ab = 15.5 bc = 16.2 ace = 23.4 abcde = 18.1 (i) Verify that the design generators used were I = ACE and I = BDE. Arrange runs and write the signs of each factor as follows: e - - - - + ad + - - + - bde - + - + + ab + + - - - cd - - + - - ace + - + - + bc - + + - - abcde + + + + + It is identified that the signs of E are the products of the signs of B and D, and are also the products of the signs of A and C; that is, E = AC, E = BD. Hence the design generators are I = ACE and I = BDE. (ii) Write down the complete defining relation and the aliases for this design. The other generator is given by ACEBDE = ABCD. The complete defining relation is I = ACE = BDE = ABCD. The alias structure of the design is as follows: 1

(iii) Estimate the main effects. The scaled estimates of the effects are: A = CE = ABDE = BCD B = ABCE = DE = ACD AB = BCE = ADE = CD C = CE = BCDE = ABD AC = E = ABCDE = BD BC = ABE = CDE = AD ABC = BE = ACDE = D a b ab c ac bc abc -1.525-5.175 1.825 2.275 2.275-1.275-0.675 The estimated main effects are a, b, c, and d = abc. (iv) Prepare an ANOVA table. Verify that the AB and AD interactions are available to use as error. The ANOVA table: Df Sum Sq Mean Sq A 1 4.651 4.651 B 1 53.561 53.561 C 1 10.351 10.351 A:B 1 6.661 6.661 A:C 1 10.351 10.351 B:C 1 3.251 3.251 A:B:C 1 0.911 0.911 A reduced model without ABC interaction yields the following ANOVA table: Df Sum Sq Mean Sq F value Pr(>F) A 1 4.651 4.651 5.1043 0.26528 B 1 53.561 53.561 58.7778 0.08257. C 1 10.351 10.351 11.3594 0.18362 A:B 1 6.661 6.661 7.3100 0.22553 A:C 1 10.351 10.351 11.3594 0.18362 B:C 1 3.251 3.251 3.5679 0.30997 Residuals 1 0.911 0.911 2

It is identified that AB and BC=AD are not significant, and hence they can be used as error. (v) Plot the residuals versus the fitted values. Also construct a normal probability plot of the residuals. Comment on the results. 2. This question concerns with 2 6 3 designs. (i) Construct a 2 6 3 III design. A 2 6 3 III design can be constructed using the defining relation I = ABD = ACE = BCF. The design is given below: Run F 1 - - - + + + 2 + - - - - + 3 - + - - + - 4 + + - + - - 5 - - + + - - 6 + - + - + - 7 - + + - - + 8 + + + + + + (ii) Determine the effects that may be estimated if a second fraction of the design in (i) is run with all signs reversed. The total defining relation of the design in (i) is I = ABD = ACE = BCF = BCDE = ACDF = ABEF = D. This determines the alias structure of the design as follows: A = BD = CE = ABCF= ABCDE = CDF =BEF = AD B = AD = ABCE = CF = CDE = ABCDF = AEF = BD AB = D = BCE = ACF =ACDE = BCDF = EF = ABD C = ABCD = AE = ABF = BDE = ADF = ABCEF = CD AC = BCD = E = ABF = ABDE = DF = BCEF = ACD BC = ACD = ABE = F = DE = ABDF = ACEF = BCD ABC = CD = BE = AF = ADE = BDF = CEF = ABCD 3

The design with all signs reversed is as follows: Run F 1 + + + - - - 2 - + + + + - 3 + - + + - + 4 - - + - + + 5 + + - - + + 6 - + - + - + 7 + - - + + - 8 - - - - - - The defining relation of the design is I = -ABD = -ACE = - BCF. The complete defining relation is I = ABD = ACE = BCF = BCDE = ACDF = ABEF = D. The alias structure of the design is as follows: A = -BD = -CE = -ABCF= ABCDE = CDF =BEF = -AD B = -AD = -ABCE = -CF = CDE = ABCDF = AEF =-BD AB = -D = -BCE = -ACF =ACDE = BCDF = EF = -ABD C = -ABCD = -AE = -ABF = BDE = ADF = ABCEF = - CD AC = -BCD = -E = -ABF = ABDE = DF = BCEF = -ACD BC = -ACD = -ABE = -F = DE = ABDF = ACEF = -BCD ABC = -CD = -BE = -AF = ADE = BDF = CEF = -ABCD The first and second fraction together can estimate the following effects: A +ABCDE + CDF +BEF B +CDE + ABCDF +AEF AB + ACDE+BCDF + EF C + BDE+ ADF + ABCEF AC + ABDE + DF+ BCEF BC + DE + ABDF + ACEF ABC + ADE + BDF + CEF BD +CE +ABCF+AD AD +ABCE +CF +BD D +BCE +ACF +ABD ABCD +AE +ABF + CD BCD +E +ABF +ACD ACD +ABE +F +BCD CD +BE +AF +ABCD (iii) Determine the effects that may be estimated if a second fraction of the design in (i) is run with signs for factor A reversed. The design is given by 4

Run F 1 + - - + + + 2 - - - - - + 3 + + - - + - 4 - + - + - - 5 + - + + - - 6 - - + - + - 7 + + + - - + 8 - + + + + + The defining relation is I = -ABD = -ACE = BCF The total defining relation is I = ABD = ACE = BCF = BCDE = ACDF = ABEF = D. The estimable effects can be identified in the same way as in (ii). 3. An industrial engineer is conducting an experiment using a Monte Carlo simulation model of an inventory syster. The independent variables in her model are the order quantity (A), the reorder point (B), the setup cost (C), the backorder cost (D), and the carrying cost rate (E). The response variable is average annual cost. To conserve computer time, she decides to investigate these factors using a 2 5 2 III design with I = ABD and I = BCE. The results she obtains are: de=95 ae=134 b=158 abd=190 cd=92 ac=187 bce=155 abcde = 185 (i) Verify that the treatment combinations given are correct. Estimate the effects, assuming three-factor and higher interactions are negligible. The design is as follows: de - - - + + ae + - - - + b - + - - - abd + + - + - cd - - + + - ac + - + - - bce - + + - + abcde + + + + + 5

It is easy to verify that D = AB and E = BC; that is the defining relations are I = ABD and I = BCE. The scaled estimates of the effects are: a b ab c ac bc abc 49 45-18 10.5 13.5-14.5-14.5 From the complete defining relation I=ABD = BCE = ACDE, the alias structure are identified as follows: A = BD = ABCE = CDE B=AD = CE = ABCDE AB = D = ACE = BCDE C = ABCD = BE = ADE AC = BCD = ABE = DE BC = ADC = E = ABDE ABC = CD = AE =BDE Ignoring three-factor and higher interactions, the following effects are estimated: A + BD = 49 B+AD + CE = 45 AB + D = -18 C +BE = 10.5 AC + DE = 13.5 BC + E = -14.5 CD + AE =-14.5 (ii) Suppose the following second faction is added to the first: ade=136 e=93 ab=187 bd=153 acd=139 c=99 abce=191 bcde = 150 How is this second fraction obtained? Add this data to the original fraction and estimate the effects. 6

The design is as follows: e - - - - + ade + - - + + bd - + - + - ab + + - - - c - - + - - acd + - + + - bcde - + + + + abce + + + - + The design is generated by I = ABD = BCE. Putting the two fractions together, re-arrange the runs, we have the following design: e - - - - + ae + - - - + b - + - - - ab + + - - - c - - + - - ac + - + - - bce - + + - + abce + + + - + de - - - + + ade + - - + + bd - + - + - abd + + - + - cd - - + + - acd + - + + - bcde - + + + + abcde + + + + + The design is a 2 5 1 design generated by I = BCE. The scaled estimates of the effects are: a b ab c ac bc abc d ad bd abd cd acd bcd abcd 88.5 98.5-20 13 14.5-16.5-12 -16-8.5 9.5 12-17 -12.5 12.5 8 7

(iii) Suppose the following faction is added to the first: abc=189 ce=96 bcd=154 acde =135 abe=193 bde=152 ad=137 (1) = 98 How is this second fraction obtained? Add this data to the original fraction and estimate the effects. The design of this fraction is as follows: (1) - - - - - ad + - - + - bde - + - + + abe + + - - + ce - - + - + acde + - + + + bcd - + + + - abc + + + - - The design is obtained by I = ABD = BCE. Putting two fractions together: (1) - - - - - ae + - - - + b - + - - - abe + + - - + ce - - + - + ac + - + - - bce - + + - + abc + + + - - de - - - + + ad + - - + - bde - + - + + abd + + - + - cd - - + + - acde + - + + + bcd - + + + - abcde + + + + + 8

The above is a 2 5 1 generated by I = ACDE. The scaled estimates of the effects are: a b ab c ac bc abc d ad bd abd cd acd bcd abcd 87.5 100.5-18.5 9 12-14 -16-17.5-10.5 10.5 10.5-13 -15 15 12 4. Carbon anodes used in a smelting process are baked in a ring furnace. An experiment is run in the furnace to determine which factors influence the weight of packing material that is struck to the anodes after baking. Six variables are of interest, each at two levels: A = pitch/fines ratio (0.45, 0.55); B = packing material type (1, 2); C = packing material temperature ( ambient, 325 Celsius); D = flue location (inside, outside); E = pit temperature (ambient, 195 Celsius); and F = delay time before packing (zer0, 24 hours). A 2 6 3 design is run, and three replicates are obtained at each of the design points. The weight of packing material stuck to the anodes is measured in grams. The data in run order are as follows: abd = (984, 826, 936) abcdef = (1275, 976, 1457) be = (1217, 1201, 890) af = (1474, 1164, 1541) def = (1320, 1156, 913) cd = (765, 705, 821) ace = (1338, 1254, 1294) bcf = (1325, 1299, 1253) We wish to minimize the amount of stuck packing material. (i) Verify that the eight runs correspond to a 2 6 3 III design. Give the alias structure. The design is given by Run F Weights def - - - + + + (1320, 1156, 913) af + - - - - + (1474, 1164, 1541) be - + - - + - (1217, 1201, 890) abd + + - + - - (984, 826, 936) cd - - + + - - (765, 705, 821) ace + - + - + - (1338, 1254, 1294) bcf - + + - - + (1325, 1299, 1253) abcdef + + + + + + (1275, 976, 1457) The generators of the design are I = ABD = ACE = BCF. The resolution is therefore 3. The alias structure can then be easily obtained. (ii) Use the average weight as the response. What factors appear to be influential? See the R-codes (iii) Use the range of the weights as the response. What factors appear to be influential? See the R-codes (iv) What recommendation would you make to the process engineers? 9