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.