Homework

University of Missouri-Columbia
IMSE 8110 Engineering Experimentation
Homework: Chapter 8 and Handout Problems
Sergio García 909242
Melissa Rios 931546
Sumit Tayal 880581
November 15, 2004
Problems from photocopies of unknown book pages 168-177
5.2 A 25III2 design with generators I = 1234 = 135 is run
a) If it is known that factors 1, 2, and 3 all act independently of one another, and that
factors 3, 4, and 5 all act independently of one another, what (ignoring
interactions of three or more factors) can we estimate?
b) Can you suggest a better 25III2 design for these factors, or is it the given the best
available?
SOLUTION
a) The design of this problem is a subset of the following original 2 5 factorial design:
RUN
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
2
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
3
-1
-1
-1
-1
1
1
1
1
-1
-1
-1
-1
1
1
1
1
4
-1
-1
-1
-1
-1
-1
-1
-1
1
1
1
1
1
1
1
1
5 Treatment RUN 1 2 3 4 5 Treatment
-1
I
17 -1 -1 -1 -1 1
5
-1
1
18 1 -1 -1 -1 1
15
-1
2
19 -1 1 -1 -1 1
25
-1
12
20 1 1 -1 -1 1
125
-1
3
21 -1 -1 1 -1 1
35
-1
13
22 1 -1 1 -1 1
135
-1
23
23 -1 1 1 -1 1
235
-1
123
24 1 1 1 -1 1
1235
-1
4
25 -1 -1 -1 1 1
45
-1
14
26 1 -1 -1 1 1
145
-1
24
27 -1 1 -1 1 1
245
-1
124
28 1 1 -1 1 1
1245
-1
34
29 -1 -1 1 1 1
345
-1
123
30 1 -1 1 1 1
1345
-1
234
31 -1 1 1 1 1
2345
-1
1234
32 1 1 1 1 1
12345
If we highlight the treatment combinations for what the generator is true, we obtain the
following table:
1 5 32
2 
 8 treatment combinatio ns
4
4
I  1234 I  1 35
2 5III 2 
4  123
RUN
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
2
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
3
-1
-1
-1
-1
1
1
1
1
-1
-1
-1
-1
1
1
1
1
4
-1
-1
-1
-1
-1
-1
-1
-1
1
1
1
1
1
1
1
1
5  13
5 Treatment RUN 1 2 3 4 5 Treatment
-1
I
17 -1 -1 -1 -1 1
5
-1
1
18 1 -1 -1 -1 1
15
-1
2
19 -1 1 -1 -1 1
25
-1
12
20 1 1 -1 -1 1
125
-1
3
21 -1 -1 1 -1 1
35
-1
13
22 1 -1 1 -1 1
135
-1
23
23 -1 1 1 -1 1
235
-1
123
24 1 1 1 -1 1
1235
-1
4
25 -1 -1 -1 1 1
45
-1
14
26 1 -1 -1 1 1
145
-1
24
27 -1 1 -1 1 1
245
-1
124
28 1 1 -1 1 1
1245
-1
34
29 -1 -1 1 1 1
345
-1
134
30 1 -1 1 1 1
1345
-1
234
31 -1 1 1 1 1
2345
-1
1234
32 1 1 1 1 1
12345
We discard the treatment combinations that do not meet the required condition to obtain
the following eight runs
RUN
1
2
3
4
5
6
7
8
1
-1
1
-1
1
-1
1
-1
1
2
-1
-1
1
1
-1
-1
1
1
3 4 = 123 5 = 13 Treatment
-1
-1
1
5
-1
1
-1
14
-1
1
1
245
-1
-1
-1
12
1
1
-1
34
1
-1
1
135
1
-1
-1
23
1
1
1
12345
It is known that factors 1, 2, and 3 all act independently of one another, and that factors 3,
4, and 5 all act independently of one another
This means that the following treatment combinations are not necessary to carry out
because the factors are independent from each other:



12
34
23
The only runs that shall be carried out are shown in the following table
RUN
1
2
3
4
5
1
-1
1
-1
1
1
2
-1
-1
1
-1
1
3 4 = 123 5 = 13 Treatment
-1
-1
1
5
-1
1
-1
14
-1
1
1
245
1
-1
1
135
1
1
1
12345
b)
Eliminating those treatment combinations that are useless because of the
independence of 1, 2,and 3 and 3, 4, and 5 from each other respectively
we can choose a better design
2 5III 2
RUN
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
2
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
3
-1
-1
-1
-1
1
1
1
1
-1
-1
-1
-1
1
1
1
1
4
-1
-1
-1
-1
-1
-1
-1
-1
1
1
1
1
1
1
1
1
5 Treatment RUN 1 2 3 4 5 Treatment
-1
I
17 -1 -1 -1 -1 1
5
-1
1
18 1 -1 -1 -1 1
15
-1
2
19 -1 1 -1 -1 1
25
-1
12
20 1 1 -1 -1 1
125
-1
3
21 -1 -1 1 -1 1
35
-1
13
22 1 -1 1 -1 1
135
-1
23
23 -1 1 1 -1 1
235
-1
123
24 1 1 1 -1 1
1235
-1
4
25 -1 -1 -1 1 1
45
-1
14
26 1 -1 -1 1 1
145
-1
24
27 -1 1 -1 1 1
245
-1
124
28 1 1 -1 1 1
1245
-1
34
29 -1 -1 1 1 1
345
-1
134
30 1 -1 1 1 1
1345
-1
234
31 -1 1 1 1 1
2345
-1
1234
32 1 1 1 1 1
12345
5.15 Identify the design given in Table E5.15 and perform an appropriate analysis.
Assume V(yi) = σ2 = 2. What factors appear to influence the response? What features
need to be further elucidated? If you were allowed to perform more runs, what would
they be? Why?
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
x4
-1
-1
-1
1
-1
1
1
1
y
105
107
102
104
114
111
105
107
Solution
Since we have 4 factors the total number of treatments for the complete 2 4 factorial
design should be sixteen. We observe only eight in the fractional design. This means that
this is a one-half fraction factorial design: 2 41
Now let’s check which the confounding pair of factor X4 is.
We observe that X4 = - X1X2X3. This means that this is the alternate or complementary
fractional design of
2 4III1
We rearrange the treatment combinations and then we proceed to calculate the Effect
Estimates, and the Mean Sum Squares
x1
-1
1
-1
1
-1
1
-1
1
x2
-1
-1
1
1
-1
-1
1
1
x3 x4 = -x1x2x3
-1
1
-1
-1
-1
-1
-1
1
1
-1
1
1
1
1
1
-1
Factor Level
Combination
x4
x1
x2
x1x2x4
x3
x1x3x4
x2x3x4
x1x2x3
Coded
Response
104
105
107
111
102
105
107
114
1
 104  105  107  111  102  105  107  114  15  3.75
4
4
15 2
SS X 1 
 28.125
8
1
23
 X 2   104  105  107  111  102  105  107  114  
 5.75
4
4
SS X 2  66.125
 X1 
1
 104  105  107  111  102  105  107  114  1  0.25
4
4
 0.125
 X3 
SS X 3
1
104  105  107  111  102  105  107  114   1  0.25
4
4
 0.125
X4 
SS X 4
1
104  105  107  111  102  105  107  114  7  1.75
4
4
 6.125
 X 1X 2 
SS X 1 X 2
1
104  105  107  111  102  105  107  114  5  1.25
4
4
 3.125
 X 1X 3 
SS X 1 X 3
1
104  105  107  111  102  105  107  114  5  1.25
4
4
 3.125
 X 2X 3 
SS X 2 X 3
Factor Level
Combination
x1
x2
x3
x4
x1x2
x1x3
x2x3
Effect
Estimate
3.75
5.75
0.25
-0.25
1.75
1.25
1.25
SS
28.125
66.13
0.13
0.13
6.13
3.13
3.13
dof
1
1
1
1
1
1
1
MS error =
σ
MS i
F0
28.13
66.13
0.13
0.13
6.13
3.13
3.13
1.4142
19.89
46.76
0.09
0.09
4.33
2.21
2.21
F0.05,1, 7
5.59
5.59
5.59
5.59
5.59
5.59
5.59
Significant
Significant
The extra runs would definitely have to be involving main effects A, and B and
interaction AB
5.16 What design is given in Table E5.16? Perform an appropriate factorial analysis of
the data.
Would you like to have additional data to clarify what you have seen in this experiment?
If four more runs were authorized what would you suggest? Why? If eight more runs
were authorized what would you suggest? Why?
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
x4
-1
-1
-1
-1
1
1
1
1
x5
-1
1
1
-1
1
-1
-1
1
y1
18
16
18
15
17
31
18
30
y2
15
18
17
16
19
30
18
27
Solution
Since we have 5 factors the total number of treatments for the complete 2 5 factorial
design should be thirty two. We observe only eight in the fractional design. This means
that this is a one-fourth fraction factorial design: 2 52
Now let’s check which the confounding pair of factor X3 is.
We observe that X3 = X1X2. This means that this is the a fractional design of
25III2
We set the treatment combinations and then we proceed to calculate the Effect Estimates,
and the Mean Sum Squares
x1
-1
1
-1
1
-1
1
-1
1
x2 x3 = x1x2
-1
1
-1
-1
1
-1
1
1
-1
1
-1
-1
1
-1
1
1
x4 x5 = x1x2x4
-1
-1
-1
1
-1
1
-1
-1
1
1
1
-1
1
-1
1
1
y1
18
16
18
15
17
31
18
30
y2
15
18
17
16
19
30
18
27
Factor Level
Coded
Combination Response
x3
33.0
x1x5
34.0
x2x5
35.0
x1x2x3
31.0
x3x4x5
36.0
x1x4
61.0
x2x4
36.0
x1x2x3x4x5
57.0
We determine the interactions that are not confounded with any of the five main effects
I
x1
x2
x3
x4
x5
x1x2
x1x3
x1x4
x1x5
x2x3
x2x4
x2x5
x3x4
x3x5
x4x5
=
x1x2x3
=
x2x3
=
x1x3
=
x1x2
= x1x2x3x4
= x1x2x3x5
=
x3
=
x2
=
x2x3x4
=
x2x3x5
=
x1
=
x1x3x4
=
x1x3x5
=
x1x2x4
=
x1x2x5
= x1x2x3x4x5
= x1x2x4x5
=
x2x4x5
=
x1x4x5
= x1x2x3x4x5
=
x1x2x5
=
x1x2x4
=
x4x5
= x2x3x4x5
=
x2x5
=
x2x4
= x1x3x4x5
=
x1x5
=
x1x4
= x1x2x3x5
= x1x2x3x4
=
x1x2
I
x1
x2
x3
x4
x5
x1x2
x1x3
x1x4
x1x5
x2x3
x2x4
x2x5
x3x4
x3x5
x4x5
=
x1x2x3
=
x2x3
=
x1x3
=
x1x2
= x1x2x3x4
= x1x2x3x5
=
x3
=
x2
=
x2x3x4
=
x2x3x5
=
x1
=
x1x3x4
=
x1x3x5
=
x1x2x4
=
x1x2x5
= x1x2x3x4x5
= x1x2x4x5
=
x2x4x5
=
x1x4x5
= x1x2x3x4x5
=
x1x2x5
=
x1x2x4
=
x4x5
= x2x3x4x5
=
x2x5
=
x2x4
= x1x3x4x5
=
x1x5
=
x1x4
= x1x2x3x5
= x1x2x3x4
=
x1x2
X1X4 and X1X5 are not confounded with any of the five main factors. Next we calculate
the effect estimator
1
 33  34  35  31  18  30.5  18  28.5  43  5.375
8
8
2
43
SS X 1 
 115.56
16
1
5
 X 2   33  34  35  31  36  61  36  57  
 0.625
8
8
SS X 2  1.5624
 X1 
1
33  34  35  31  36  61  36  57    9  1.125
8
8
 5.0624
 X3 
SS X 3
1
 33  34  35  31  36  61  36  57   57  7.125
8
8
 203.0624
X4 
SS X 4
1
 33  34  35  31  36  61  36  57   1  0.125
8
8
 0.0625
 X5 
SS X 5
1
33  34  35  31  36  61  36  57   49  6.125
8
8
 150.04
 X 1X 4 
SS X 1 X 4
1
33  34  35  31  36  61  36  57    3  0.375
8
8
 1.125
 X 1X 5 
SS X 1 X 5
Factor Level
Combination
x1
x2
x3
x4
x5
x1x4
x1x5
k 3
n2
N  2 * 8  16
Effect
Estimate
2.6875
-0.3125
0.5625
3.5625
0.0625
3.0625
-0.1875


SS
dof
115.560
1.562
5.062
203.062
0.063
150.040
1.125
476.475
1
1
1
1
1
1
1
MS i
F0
115.56 66.21
1.56
0.90
5.06
2.90
203.06 116.35
0.06
0.04
150.04 85.97
1.13
0.64
F0.05,1, 7
5.59 Significant
5.59
5.59
5.59 Significant
5.59
5.59 Significant
5.59
y i  163  160  323
y i2  7011
( 323 ) 2
SST  7011 
 490 . 4375
16
SSE  490 . 4375  476 . 475  13 . 9625
MSE 
SSE
13 . 9625

 1 . 7453
( n  1) 2 k
8
From this result we are faced with the large significance of X1, X4, and X1X4. So we do
a full factorial design with only the two influential factors and their interaction with each
other.
The more runs allowed the better!

Problems from the book pages Chapter 8
8-4 Problem 6-21 describes a process improvement study in the manufacturing process of
an integrated circuit. Suppose that only eight runs could be made in this process. Set up
an appropriate 2 52 design and find the alias structure. Use the appropriate observations
from problem 6-21 as the observations in this design and estimate the factor effects. What
conclusions can you draw?
RUN
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
A
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
B
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
C
-1
-1
-1
-1
1
1
1
1
-1
-1
-1
-1
1
1
1
1
D
-1
-1
-1
-1
-1
-1
-1
-1
1
1
1
1
1
1
1
1
E Treatment
-1
I
-1
a
-1
b
-1
ab
-1
c
-1
ac
-1
bc
-1
abc
-1
d
-1
ad
-1
bd
-1
abd
-1
cd
-1
acd
-1
bcd
-1
abcd
YIELD
7
9
34
55
16
20
40
60
8
10
32
50
18
21
44
61
RUN
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
A
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
B
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
C
-1
-1
-1
-1
1
1
1
1
-1
-1
-1
-1
1
1
1
1
D
-1
-1
-1
-1
-1
-1
-1
-1
1
1
1
1
1
1
1
1
E Treatment
1
e
1
ae
1
be
1
abe
1
ce
1
ace
1
bce
1
abce
1
de
1
ade
1
bde
1
abde
1
cde
1
acde
1
bcde
1
abcde
YIELD
8
12
35
52
15
22
45
65
6
10
30
53
15
20
41
63
Design Generators
D = AB, and E = AC
Here you can see the eight treatment combinations for which the Design Generators are
true
RUN
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
A
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
B
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
C
-1
-1
-1
-1
1
1
1
1
-1
-1
-1
-1
1
1
1
1
D
-1
-1
-1
-1
-1
-1
-1
-1
1
1
1
1
1
1
1
1
D = AB E E = AC Treatment YIELD RUN A B C D D = AB E E = AC Treatment YIELD
1
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
-1
-1
1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
1
-1
1
-1
-1
1
-1
1
1
-1
1
-1
-1
1
-1
1
I
a
b
ab
c
ac
bc
abc
d
ad
bd
abd
cd
acd
bcd
abcd
7
9
34
55
16
20
40
60
8
10
32
50
18
21
44
61
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
-1
-1
-1
-1
1
1
1
1
-1
-1
-1
-1
1
1
1
1
-1
-1
-1
-1
-1
-1
-1
-1
1
1
1
1
1
1
1
1
1
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
-1
-1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
-1
1
-1
-1
1
-1
1
1
-1
1
-1
-1
1
-1
1
e
ae
be
abe
ce
ace
bce
abce
de
ade
bde
abde
cde
acde
bcde
abcde
8
12
35
52
15
22
45
65
6
10
30
53
15
20
41
63
RUN
2
7
12
13
19
22
25
32
A
1
-1
1
-1
-1
1
-1
1
B
-1
1
1
-1
1
-1
-1
1
C
-1
1
-1
1
-1
1
-1
1
Factor Level
D = AB E = AC Combination
-1
-1
a
-1
-1
bc
1
-1
abd
1
-1
cd
-1
1
be
-1
1
ace
1
1
de
1
1
abcde
Coded
Response
9
40
50
18
35
22
6
63
Since we don’t have the proper set up for using Yates Algorithm to calculate the Effect
Estimators we proceed to use the following formulas:
1
9  40  50  18  35  22  6  63  45  11.25
4
4
1
133
 B   9  40  50  18  35  22  6  63 
 33.25
4
4
1
43
 C   9  40  50  18  35  22  6  63  10.75
4
4
1
31
 D   9  40  50  18  35  22  6  63 
 7.75
4
4
1
9
 E   9  40  50  18  35  22  6  63   2.25
4
4
1
7
 BC  9  40  50  18  35  22  6  63 
 1.75
4
4
1
7
 BE  9  40  50  18  35  22  6  63   1.75
4
4
A 
1
45 2
2
SS A  9  40  50  18  35  22  6  63 
 253.125
8
8
1
1332
2
SS B   9  40  50  18  35  22  6  63 
 2211.125
8
8
1
432
2
SS C   9  40  50  18  35  22  6  63 
 231.125
8
8
1
312
2
SS D   9  40  50  18  35  22  6  63 
 120.125
8
8
1
92
2
SS E   9  40  50  18  35  22  6  63 
 10.125
8
8
2

1
 7
2
SS BC  9  40  50  18  35  22  6  63 
 6.125
8
8
1
72
2
SS BE  9  40  50  18  35  22  6  63 
 6.125
8
8
Factor Level
Combination
A
B
C
D
E
BC
BE
Effect
Estimate
11.25
33.25
10.75
7.75
2.25
-1.75
1.75
We Graph the Effect Estimators in a Probability Plot and determine that the main effects
A, B, C, and D are large, but we keep in mind, however that what we are really
estimating is:
A + BD + CE
B + AD
C + DE
D + AD
So there could be other interpretations because of the aliases
Factor Level
Combination
A
B
C
D
Effect
SS
Estimate
11.25 253.125
33.25 2211.130
10.75 231.125
7.75 120.125
dof
MS i
1
1
1
1
253.13
2211.13
231.13
120.13
F0
33.93
47.64
4.98
2.59
8-5 Suppose that you have made the eight runs in the 25-2 design in problem 8-4. What
additional runs would be required to identify the factor effects that are of interest? What
are the alias relationships in the combined design?
A fold over of the original design may be done by changing the sign on the generator. If
we assume that all three factors and the higher interactions are insignificant, this would
mean that all main effects can be separated from their two-factor interaction aliased in the
combined design
PROBLEM
EXAMPLE Thanksgiving Turkey
AGE IN WEEKS
WEIGHT (lbs)
ORIGIN
ei
28
13.31
Georgia
-0.4
20
8.9
Georgia
-1.4
32
15.1
Georgia
-0.2
22
10.4
Georgia
-0.8
AGE IN WEEKS
WEIGHT (lbs)
ORIGIN
ei
29
13.1
Virginia
-1
27
12.4
Virginia
-0.8
28
13.2
Virginia
-0.5
26
11.8
Virginia
-1
AGE IN WEEKS
WEIGHT (lbs)
ORIGIN
ei
Solution
Next page
21
27
29
23
25
11.5
14.2
15.4
13.1
13.8
Wisconsin Wisconsin Wisconsin Wisconsin Wisconsin
0.8
1
1.3
1.5
1.4
PROBLEM FROM HANDOUT PAGE 14-30
Since there are 6 factors and only 8 treatment combinations we can say that this is a 2 6III3 .
The Generator is:
I = ABD = -ACE = BCF
The effect estimators are as follows:
1
 165  55  35  105  95  105  25  215  160  40
4
4
 130
B 
 32.5
4
80
C 
 20
4
360
D 
 90
4
 240
E 
 60
4
120
F 
 30
4
 110
 AD 
 27.5
4
0
 CD   0
4
 120
 DE 
 30
4
25600
SS A 
 3200
8
16900
SS B 
 2112.5
8
6400
SS C 
 800
8
129600
SS D 
 16200
8
57600
SS E 
 7200
8
14400
SS F 
 30
8
SS AD  1512.5
A 
SS CD  0
SS DE  1800
Factor Level
Effect
SS
dof
MS i
Combination Estimate
A
40.00 3200.00 1
3200.00
B
-32.50 2112.50 1
2112.50
C
20.00
800.00 1
800.00
D
90.00 16200.00 1 16200.00
E
-60.00 7200.00 1
7200.00
F
30.00
30.00 1
30.00
AD
-27.50 1512.50 1
1512.50
CD
0
0.00 1
0.00
DE
-30 1800.00 1
1800.00