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Andrew Johnson 3018836 IE5522 HW11 2008/04/17 17.1 (a) x2 pressure x3 cycle time (-1,+1,+1) y=24 7 8 (+1,+1,+1) y=28 (-1,+1,-1) y=24 3 4 (+1,+1,-1) y=17 x1 temperature (-1,-1,+1) y=24 5 6 (+1,-1,+1) y=16 1 2 (+1,-1,-1) y=15 (-1,-1,-1) y=12 (b) y=[12 15 24 17 24 16 24 28]; E1=0.25*[(y(2)-y(1))+(y(4)-y(3))+(y(6)-y(5))+(y(8)-y(7))] E2=0.25*[(y(3)-y(1))+(y(4)-y(2))+(y(7)-y(5))+(y(8)-y(6))] E3=0.25*[(y(5)-y(1))+(y(6)-y(2))+(y(7)-y(3))+(y(8)-y(4))] E1 = E2 = E3 = (c) -2 6.5000 6 //temperature main effect //pressure main effect //cycle time main effect x1 =[-1 +1 -1 +1 -1 +1 -1 +1]; x2 =[-1 -1 +1 +1 -1 -1 +1 +1]; x3 =[-1 -1 -1 -1 +1 +1 +1 +1]; x1x2=x1.*x2; x1x3=x1.*x3; x2x3=x2.*x3; x1x2x3=x1.*x2.*x3; E12=0.25*sum(x1x2.*y) E13=0.25*sum(x1x3.*y) E23=0.25*sum(x2x3.*y) E123=0.25*sum(x1x2x3.*y) E12 = E13 = E23 = E123 = 0.5000 0 -0.5000 5.5000 (d) b0=(1/8)*sum(y); b1=E1/2;b2=E2/2;b3=E3/2; b12=E12/2;b13=E13/2;b23=E23/2; b123=E123/2; b=[b0 b1 b2 b3 b12 b13 b23 b123] b = 20.0000 -1.0000 so 17.2 3.2500 3.0000 0.2500 2.7500 y b0 b1 x1 b2 x2 b3 x3 b12 x1 x 2 b13 x1 x3 b23 x2 x3 b123 x1 x2 x3 y 20 1x1 3.25 x 2 3x3 0.25 x1 x2 0.25 x 2 x3 2.75 x1 x2 x3 If the tests are done in the manner specified, changes between tests might be skewed or minimized by ambient conditions that change between morning and night. This will make the interpretations less meaningful because you do not get a good idea of changes due simply to variables, but also test time. To rectify this, randomize the testing order. 17.4 0 -0.2500 (a) y=[56 72 68 89 43 62 59 75]; x1 =[-1 +1 -1 +1 -1 +1 -1 +1]; x2 =[-1 -1 +1 +1 -1 -1 +1 +1]; x3 =[-1 -1 -1 -1 +1 +1 +1 +1]; x1x2=x1.*x2; x1x3=x1.*x3; x2x3=x2.*x3; x1x2x3=x1.*x2.*x3; E1=0.25*sum(x1.*y) E2=0.25*sum(x2.*y) E3=0.25*sum(x3.*y) E12=0.25*sum(x1x2.*y) E13=0.25*sum(x1x3.*y) E23=0.25*sum(x2x3.*y) E123=0.25*sum(x1x2x3.*y) E1 = 18 E2 = 14.5000 E3 = -11.5000 E12 = 0.5000 E13 = -0.5000 E23 = 0 E123 = -2 (b) x2 ounces of coffee x3 number of doughnuts 59 7 8 75 68 3 4 89 x1 hours of sleep 43 5 6 62 1 2 72 56 82 Ounces of Coffee 63.5 49.5 62 67 68.5 Number of doughnuts 51 Hours of sleep Hours of sleep 67 Ounces of Coffee 78.5 80.5 64 Number of Doughnuts 52.5 (c) The three variables all contribute to alertness levels. Each one by itself adds a certain alertness, but combining all three does not give you sum of all the alertness bonuses in a simple addition. The three-factor interaction effect of -2 means that all three variables combined will add 2% less alertness than a simple addition would suggest. 17.8 (a) x2 (-1, +1) y=39.2 3 4 (+1, +1) y=28.9 x1 (+1, +1) y=45 1 2 (+1, -1) y=30.5 y=[45 30.5 39.2 28.9]; (b) x1 =[-1 +1 -1 +1]; x2 =[-1 -1 +1 +1]; x1x2=x1.*x2; E1=0.5*sum(x1.*y) E2=0.5*sum(x2.*y) E12=0.5*sum(x1x2.*y) E1 = -12.4000 E2 = -3.7000 E12 = 2.1000 (c) y 45 x2(-) 40 14.5 35 30 10.3 x2(+) 25 x1 - + A change in y due to a change in x1 depends on x2, because the slopes are not the same. 17.18 (a) yA=[59 76.3 28.9 51.4];yB=[20.3 41.6 48.3 73.1]; x1 =[-1 +1 -1 +1]; x2 =[-1 -1 +1 +1]; x1x2=x1.*x2; E1A=0.5*sum(x1.*yA) E2A=0.5*sum(x2.*yA) E12A=0.5*sum(x1x2.*yA) E1B=0.5*sum(x1.*yB) E2B=0.5*sum(x2.*yB) E12B=0.5*sum(x1x2.*yB) E1A = 19.9000 E2A = -27.5000 E12A = 2.6000 E1B = 23.0500 E2B = 29.7500 E12B = 1.7500 (b) the fixture type main effect (E2A) shows a steep drop in productivity, so maybe the first crew got a bad type of fixtures (c) Eday=0.25*sum(x3.*[yA yB]) Eday = (d) -8.0750 E2day=0.25*sum([x2 x2].*x3.*[yA yB]) E2day = 28.6250 (e) Eday suggests one set of workers is much faster than the other set, but as it is seen from E2day, there is a huge correlation between fixture type and productivity. So the manner in which the test was done, with two crews, is not fair to the first crew who will get blamed for being slow but got stuck with the bad fixture types.