Entering Data in Minitab ... Solution. Stat → DOE → Define Custom Factorial
Design ... An experiment is run in a chemical process using a 32 factorial design.
Design of Experiment ANSY Tech. Inc. Muhammad Sohail Ahmed, Ph.D. Mubashir Siddiqui
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Day 7 Factorial Design – 3 Levels
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Factorial Designs § § § §
(Day 7)
2k Factorial with Center Point 32 Factorial Design 33 Factorial Design Confounding
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Methods to Obtain Curvature Estimate §
2k design with center points
§
3-Level Designs
§
Response Surface Method
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2k Factorial with Center Point k
First Order Model
y = β 0 + ∑ β jx j + j =1
∑∑ β i< j
x xj +ε
ij i
To model CURVATURE in Response Function k
k
j =1
j =1
y = β 0 + ∑ β jx j + ∑ β jjx 2j + ∑∑ β ij xi x j + ε i< j
where ßjj: pure second order or quadratic effects
B High
b A-B +
Let nc be the observations at center point (0,0) Let yF be the observations at corner points
ab
.
A+B +
(0,0)
Low
A+B -
A-B (1)
a
Low
High
A
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Problem Response: Yield of a chemical process Factors: Reaction Time and Reaction Temperature (a) Find significant factors, their effects. (b) Is there any evidence of curvature?
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Entering Data in Minitab
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Solution Stat à DOE à Define Custom Factorial Design
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Solution (cont’d) Stat à DOE à Analyze Factorial Design
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Solution (cont’d)
No evidence of second order curvature for the region tested © AnsyTech 2008
32 Design
Low level
Intermediate Level
High Level
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32 Design Model
Low level
Intermediate Level
High Level
Notice the quadratic nature of the model © AnsyTech 2008
Problem An experiment is run in a chemical process using a 32 factorial design. The design factors are temperature and pressure, and the response variable is yield. Data is shown in the table. (a) Using ANOVA, which factor has significant effects? (b) Analyze the residuals. (c) Verify that the second order model could be written, in natural variables, as y = -1335.63 + 18.56T + 8.59P – 0.072T2 – 0.0196P2 – 0.0384TP (d) Verify that if we let low, medium, and high levels be represented by -1, 0, and +1, then the second order model for yield is y = 86.81 + 10.4x1 + 8.42x2 – 7.17x12 – 7.86x22 – 7.69x1x2
Pressure (psig)
Temperature 0C
100
120
140
80
47.58, 48.77
64.97, 69.22
80.92, 72.60
90
51.86, 82.43
88.47, 84.23
93.95, 88.54
100
71.18, 92.77
96.57, 88.72
76.58, 83.04 © AnsyTech 2008
Solution Entering Data in Minitab P
T
Y
100
80
47.58
100
90
51.86
100
100
71.18
120
80
64.97
120
90
88.47
120
100
96.57
140
80
80.92
140
90
93.95
140
100
76.58
100
80
48.77
100
90
82.43
100
100
92.77
120
80
69.22
120
90
84.23
120
100
88.72
140
80
72.6
140
90
88.54
140
100
83.04
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Solution Solution (a) Using ANOVA, which factors are significant? Stat à ANOVA à Two-Way Select Y as response and Pressure and Temperature as Factors. Click on Graphs, Select Four in One for Residual Plots.
Pressure and Temperature are significant. Their interaction is not.
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Solution (cont’d) Solution (b) Analyze residuals. We will get residuals plot by selecting Four in One for Residual Plots. Residual Plots for Y Normal Probability Plot
Versus Fits
99 10 Residual
Percent
90 50 10
0 -10
1 -20
-10
0 Residual
10
20
50
60
Histogram
70 Fitted Value
80
90
Versus Order 10
4.5
Residual
Frequency
6.0
3.0 1.5 0.0
0 -10
-15
-10
-5
0 5 Residual
10
15
2
4
6 8 10 12 14 Observation Order
16
18
Residuals seem to be normally distributed in Normal Probability Plot. But the plot of residuals vs fits shows a converging nozzle. Hence there is non-constant variance.
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Residuals VS Fitted Values n n n n n
As the magnitude of observation increases, the variance increases. Error in measuring instrument. Data follows a non-normal or skewed § distribution. F-test is only slightly affected in balanced designs. In worse cases, data transformation is required.
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Solution (cont’d) Solution (c) Verify that the second order model could be written, in natural variables, as y = -1335.63 + 18.56T + 8.59P – 0.072T2 – 0.0196P2 – 0.0384TP
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Solution (cont’d) Stat à Regression à Regression Select Response and Predictors.
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Solution (cont’d) Solution (d) Verify that if we let low, medium, and high levels be represented by -1, 0, and +1, then the second order model for yield is y = 86.81 + 10.4x1 + 8.42x2 – 7.17x12 – 7.86x22 – 7.69x1x2
Enter data in Minitab using Coded Variables representing low, medium , and high levels as -1, 0, and +1 respectively. Then get square- and product-columns as shown in C24, C25, and C26.
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Solution (cont’d) Stat à Regression à Regression Select Response and Predictors.
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3-D view Of Response y=f(T,P)
http://www.livephysics.com/ptools/online- 3 d -f u n c t i o n - grapher.php?ymin=- 1 & x m i n = - 1 & z m i n = A u t o & y m a x = 1 & x m a x = 1 & z m a xAuto&f = =86.81%2B%2810.4*x%29%2B%288.42*y%29
-% 2 8 7 . 1 * x % 5 E 2 % 2 9 -% 287.86*y%5E2%29
-% 2 8 7 . 6 9 * x * y % 2 9
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Same Problem Using Factorial Design Stat à DOE à Define Custom Factorial Design
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Same Problem Using Factorial Design Stat à DOE à Analyze Factorial Design Select Response
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33 Design
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33 Design Three factors are being studied viz bottle type, shelf type, and workers to find their effects on the time it takes to stock 10 cases on shelves. Time data is as shown here. (a) Which of the factors are significant? (b) Specify appropriate levels of factors..
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Entering Data in Minitab
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Solution Solution (a) – Significant Factors Stat à DOE à Define Custom Factorial Design Select Factors. Check ‘General Factorial Design’
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Solution (cont’d) Stat à DOE à Analyze Factorial Design Select Response. Click Graphs. Select Four In One under Residuals Plot.
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Solution (cont’d) Residual Plots for Time Versus Fits 0.4
90
0.2 Residual
Percent
Normal Pro bability Plot 99
50 10 1 -0.50
-0.25
0.00 Residual
0.25
0.0 -0.2 -0.4
0.50
3
4
Histogram
5 Fitted Value
6
Versus Order 0.4
Residual
Frequency
8 6 4
0.0 -0.2
2 0
0.2
-0.32
-0.16
0.00 Residual
0.16
0.32
-0.4
1
5
10
15 20 25 30 35 40 Observation Order
45 50
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Solution (cont’d) Solution (b) Stat à DOE à Factorial Plots Main Effects Plot for Time Data Means W orker
5.6
Bot tle Type
5.2 4.8
Mean
4.4 4.0 1
2
3
1
2
3
Shelf Type
5.6 5.2 4.8 4.4 4.0 1
2
3
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Solution (cont’d) Interaction Plot for Time Data Means 1
2
3
1
2
3 6
5
Wor ker
Work er 1 2 3
4 6
5
Bo ttle T ype
Bottle Ty pe 1 2 3
4
She lf T ype
Shortest Time: Worker: 1 (or 3) Shelf Type:1
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Confounding
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Confounding n
n
n n
Block 1 (1) ab
Block 2 a b
Confounding is a design technique for arranging a complete factorial experiment in blocks, where the block size is smaller than the number of treatment combinations in one replicate. A confounding design is one where some treatment effects (main or interactions) are estimated by the same linear combination, of the experimental observations, as some blocking effects. In this case, the treatment effect and the blocking effect are said to be confounded. Thus confounding causes some information, usually about higher order interactions to be indistinguishable from blocks.
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Confounding Runs
A
B
Combination
1
-
-
(1)
2
+
-
a
3
-
+
b
4
+
+
ab
Assignment of four runs of a 22 Design in Two BLOCKS Block 1 (1) ab
Block 2 a b
Main effect of A = A = 21n [−(1) + a − b + ab ] Main effect of B = B = 21n [−(1) − a + b + ab ] Interaction Effect = AB = 21n [ab + (1) − a − b]
Which effects are confounded? See the SIGNS in Main and Interaction Effects. Here AB effect is confounded. © AnsyTech 2008
Problem Four factors are studied to find their effects on Filtration Rate. They are: Temperature(A), Pressure(B), Concentration of formaldehyde(C), and Stirring Rate(D). All combinations cannot be run using one batch of raw materials. So there will be two blocks. A B C D Comb Block 1 (1)=25 ab=45 ac=40 bc=60 ad=80 bd=25 cd=55 abcd=76 ABCD = +
Block 2 a=71 b=48 c=68 d=43 abc=65 bcd=70 acd=86 abd=104 ABCD = -
-1
-1
-1
-1
(1)
1
-1
-1
-1
a
-1
1
-1
-1
b
1
1
-1
-1
ab
-1
-1
1
-1
c
1
-1
1
-1
ac
-1
1
1
-1
bc
1
1
1
-1
abc
-1
-1
-1
1
d
1
-1
-1
1
ad
-1
1
-1
1
bd
1
1
-1
1
abd
-1
-1
1
1
cd
1
-1
1
1
acd
-1
1
1
1
bcd
1
1
1
1
abcd
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Entering Data in Minitab
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Solution Stat à DOE à Define Factorial Design Select Low/High and select Worksheet Data as Coded. Click on Designs, and specify the block column.
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Solution (cont’d)
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Solution (cont’d) Normal Plot of the Effects (response is Y, Alpha = .05) 99 A
95 90
Percent
F actor A B C D
AD D
80
C
70 60 50 40 30
Effect Typ e No t Sig nificant Sig nific ant Nam e Tem p Pre ss C onc Rate
20 10 5 1
AC
-20
-10
0 Effect
10
20
Lenth's PSE = 3.1875
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End of Day 7
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