FULL FACTORIAL ANALYSIS

Full Factorial Designs
RSM EVOP
Week 3
Knorr-Bremse Group
Content
• Analysis of experiments with two or more factorAnalysis of experiments with two or more factor
levels
• Use of diagnostic methods to assess the
usefulness of the modelusefulness of the model
• ExamplesExamples
• Introduction to Response Surface MethodologyIntroduction to Response Surface Methodology
(RSM) and to EVOP (Evolutionary Operation)
Knorr-Bremse Group 04 BB W3 Full, RSM & EVOP 08, D. Szemkus/H. Winkler Page 2/36

The Strategy of Experimentation
Collect information Fractional Factorial
Plackett-Burnam
Validate factors
Analyze behavior of
Plackett Burnam
Folding
2k Factorialy
important factors
Establish a model
Center Points
Blocking
Full Factorial
Determine optimal
settings
Full Factorial
Box-Behnken
g
RSM EVOPRSM
Taguchi
EVOP
Knowledge and complexity define the type of experiment
g p y yp p
The Model for 2 Factors with 2 Levels
InteractionsMean
errorxxbxbxbbY 211222110
++++=
Not explainable
i ti (N i )
Main effects
variation (Noise)
The null hypothesis:
All group means are equal or similar, we cannot state a difference
Which risk are we prepared to accept?
5 10% b bilit f / i ifi l l ( 0 05 0 1)5 - 10% probability of error / significance level (α = 0,05 – 0,1)
90% power of the test (1– β ; β = 0,1)

Experiments with 2 Factors
22 - factorial 32 - factorial22 – factorial with
center point
22 – factorial with center
and star points
DOE’ ith 3 f t l l lland star points
DOE’s with 3 or more factor levels allow
investigation of quadratic effects.
Tenable results are obtained from 9
experimental points. Often center points
li d l ti tare realized several times to ensure
correctness, estimate variation and gain
degrees of freedom
degrees of freedom.
Design of Experiment with 3 Factors
+ + +b x x b x x b x x12 1 2 13 1 3 23 2 3y b b x b x b x= + + +0 1 1 2 2 3 3
First order model Interaction model
Second order model
+ + +b x b x b x11 1
2
22 2
2
33 3
2
Second order model
+ + +b x b x b x11 1 22 2 33 3

Design of Experiment with 3 Factors
First order design Second order design
(23 factorial with center points)
g
(central composite)
Box Behnken DesignBox-Behnken Design
Example: 2 Factors and 3 Levels
Goal: Investigate the effects of pressure and temperature on the
chemical yield.chemical yield.
Output: Yield
Inputs: Temperature: 120 130 140 °CInputs: Temperature: 120, 130, 140 C
Pressure: 2, 3, 4 bar
D t
File: Full factorial 1.mtw
Data:
Temp
2 3 4
pressure
2 3 4
90.4 90.7 90.2
90.2 90.6 90.4
120
90.2 90.6 90.4
90.1 90.5 89.9
90.3 90.6 90.1
130
90.5 90.8 90.4
90.7 90.9 90.1
140

The Customization of a Design in Minitab
File: Full factorial 1.mtw
Stat
>DOE
>F t i l>Factorial
>Define Custom Factorial Design…
Temp Pressure Yield
120 2 90,4
120 3 90,7
120 4 90,2
130 2 90,1
130 3 90,5
130 4 89 9130 4 89,9
140 2 90,5
140 3 90,8
140 4 90 4140 4 90,4
120 2 90,2
120 3 90,6
120 4 90 4120 4 90,4
130 2 90,3
130 3 90,6
130 4 90,1130 4 90,1
140 2 90,7
140 3 90,9
140 4 90,1
How to Start Evaluation
Stat
>DOE
Enter the response variable and all terms
>Factorial
>Analyze Factorial Design…
p
(factors and interactions).

The Results as an ANOVA Table
General Linear Model: Yield versus Temp; Pressure
Factor Type Levels Values
Temp fixed 3 120; 130; 140
Pressure fixed 3 2; 3; 4
Analysis of Variance for Yield, using Adjusted SS for Tests
Source DF Seq SS Adj SS Adj MS F P
2 0 30111 0 30111 0 15056 8 47 0 009Temp 2 0,30111 0,30111 0,15056 8,47 0,009
Pressure 2 0,76778 0,76778 0,38389 21,59 0,000
Temp*Pressure 4 0,06889 0,06889 0,01722 0,97 0,470
Error 9 0,16000 0,16000 0,01778, , ,
Total 17 1,29778
S = 0,133333 R-Sq = 87,67% R-Sq(adj) = 76,71%
The error term is comparably
Proof the hypothesis:
Significance of temperature and
small, that means the factors
explain the variation properly.
Significance of temperature and
pressure.
Interaction not significant
Interaction not significant
Next: Reduce the Model and Analyze Residuals
Stat
>DOE
>Factorial
>Analyze Factorial Design…

Result: The Reduced Model
General Linear Model: Yield versus Temp; Pressure
23 % of the variation
Factor Type Levels Values
Temp fixed 3 120; 130; 140
Pressure fixed 3 2; 3; 4
is explained by temp and
59 % by pressure
Analysis of Variance for Yield, using Adjusted SS for Tests
Temp 2 0,30111 0,30111 0,15056 8,55 0,004
Pressure 2 0 76778 0 76778 0 38389 21 80 0 000Pressure 2 0,76778 0,76778 0,38389 21,80 0,000
Error 13 0,22889 0,22889 0,01761
Total 17 1,29778
S = 0,132691 R-Sq = 82,36% R-Sq(adj) = 76,94%
Effect plots coming next
Graphical Analysis
Stat
>DOE
Main effect plots or Multi-Vari Chart!
>Factorial
>Factorial Plots…

Graphical Presentation of the Effects
90,7
Temp Pressure
Main Effects Plot for Yield
Data Means
Stat
>Quality Tools
90,6
90,5
Mean
>Multi-Vari Chart…
90,4
90,3
90,2
140130120
90,2
432
Multi-Vari Chart for Yield by Pressure - Temp
90,9
90,8
90,7
2
3
4
Pressure
Multi-Vari Chart for Yield by Pressure - Temp
90,6
90,5
90,4
90,3
Yield
140130120
90,2
90,1
90,0
140130120
Temp
Residual Diagnostics
Residual Plots for Yield
99
90
t
N 18
AD 0,189
P-Value 0,888
0,2
0,1
al
Normal Probability Plot Versus Fits
50
10
1
Percen
0,0
-0,1
-0,2
Residua
0,300,150,00-0,15-0,30
1
Residual
90,890,690,490,290,0
Fitted Value
4 0,2
Histogram Versus Order
3
2
1
Frequency
0,2
0,1
0,0
-0,1
Residual
0,20,10,0-0,1-0,2
1
0
Residual
18161412108642
-0,2
Observation Order
Normal distributed – No alarming incident

Another Chemical Example
• Goal: The evaluation of a 2 factor design with interaction.
• Output variable: Yield
File:
Full factorial 2.mtw
• Input variable:
– Temperature: 75, 80, 85 °C
C t l t t ti 5 5 6 0 6 5 %– Catalyst concentration: 5,5, 6,0, 6,5 %
• Data: TemperatureCatalyst• Data:
75 80 85
76 55 52
82 56 63
TemperatureCatalyst
amount
5 5
Perform a complete
64 65 65
87 64 60
81 77 53
5,5
evaluation!
Present your results!
67 74 63
83 71 60
75 73 57
78 86 69
6
78 86 69
72 74 70
85 81 65
83 78 60
6,5
Example with 3 Factors
• Goal: Investigation of the effect of
crimp, process temperature and
moisture on the dye ability of nylon
Moist Crimp Temp Dye
2 2 2 38
3 2 2 36
2 1 2 34
moisture on the dye ability of nylon
fibers.
• Output:
Dye ability (higher values are better)
Zinc
2 1 2 34
3 1 1 28
2 2 2 36
3 1 2 35
Dye ability (higher values are better)
• Inputs:
– Crimp: low; high
3 1 2 35
1 2 2 36
3 1 1 27
3 1 3 26p ; g
– Temperature: low; medium; high
– Moisture: low; medium; high
N = 3 observations per treatment
3 2 1 29
2 1 3 33
2 2 3 31
1 1 3 32
• N = 3 observations per treatment 1 1 3 32
1 2 2 35
2 2 2 34
2 1 3 35
2 2 3 34
2 2 1 37
1 1 3 31
2 2 1 39
How are the optimal process settings?
2 2 1 39
Any further actions you would
suggest? File:
Full factorial 3 mtw
Full factorial 3.mtw

The Response Surface Method
Treatment of the quadratic model
Response surface graphs, wire
frame and contour plot are
additional graphs to interpret
resultsresults.
Th t ti f thThe computation of the
quadratic model is a special
application of multipleapplication of multiple
regression.
Example: Video Tape
• Goal: Find the settings for time and Starting DoE
temperature to optimize yield.
Time
g
• Actual settings:
Ti 75 i t
70 75 80
127.5 54.3 60.3
60 3
Time
– Time = 75 minutes
– Temperature = 130°
60.3
Temp 130.0 64.3
62.3
132.5 64.6 68.0
File:
• Starting points for the 2 factor design:
– Time: Lo = 70 minutes; Hi = 80 minutes
RSM1.mtw
Time: Lo 70 minutes; Hi 80 minutes
– Temperature: Lo = 127,5°; Hi = 132,5°

The RSM Evaluation
Stat
>DOE
>Response Surface
>Analyze Response Surface Design…
Check if all terms are
selected!selected!
Response Surface Regression: Yield versus Time; Temp
The RSM Evaluation
Response Surface Regression: Yield versus Time; Temp
The following terms cannot be estimated, and were removed.
Temp*Temp
The analysis was done using coded unitsThe analysis was done using coded units.
Estimated Regression Coefficients for Yield
Term Coef SE Coef T P
No significant interaction
and no quadratic effectsConstant 62,3000 1,155 53,953 0,000
Time 2,3500 1,000 2,350 0,143
Temp 4,5000 1,000 4,500 0,046
Time*Time -0,5000 1,528 -0,327 0,775
and no quadratic effects
Time*Temp -0,6500 1,000 -0,650 0,582
S = 2,00000 PRESS = *
R-Sq = 92,93% R-Sq(pred) = *% R-Sq(adj) = 78,80%
Analysis of Variance for Yield
i 4 105 209 105 209 26 3021 6 58 0 136Regression 4 105,209 105,209 26,3021 6,58 0,136
Linear 2 103,090 103,090 51,5450 12,89 0,072
Square 1 0,429 0,429 0,4286 0,11 0,775
Interaction 1 1,690 1,690 1,6900 0,42 0,582
Residual Error 2 8 000 8 000 4 0000Residual Error 2 8,000 8,000 4,0000
Pure Error 2 8,000 8,000 4,0000
Total 6 113,209
The model is represented b Y 62 3 + 2 35 time + 4 5 temp
The model is represented by: Y = 62,3 + 2,35 time + 4,5 temp

The Contour Plot
150
80
Yield
Contour Plot of Yield vs Temp; Time
Stat
>DOE
145
50
55
60
65
70
75
86,8>Response Surface
>Contour/Surface (Wireframe) Plot…
Temp
140
135
95
100
75
80
85
90
73,3
130
132,5
127,5
Time
100959085807570
125
Following the slope starting from the area under investigation and perform
further experiments on the way to the optimum.
This can be done using a graph but also using an equation.
The graph visualizes the way. This point lies far beyond of the experimental
th t li t d d
area, the contour lines are extended
Using the Regression Equation
The equation (first-order regression model) for the yield:
Y = 62,3 + 2,35 time + 4,5 temp (in coded units)
58,0
60,0
1
Contour Plot of Yield
62,3
64,0
66,0
0
mp
0
Te
10-1
-1
Time
The “easiest” way to improve the yield would be to follow the ascent which is
perpendicular the contour lines. The slope we have to follow in our contour plot is
4,50/2.35 = 1,91 based on our model. That means if we make a step for time of 1 (1∆4,50/2.35 1,91 based on our model. That means if we make a step for time of 1 (1∆
= 5 minutes) the step for temperature would be 1,91 (1∆ = 4,8°).
The ass mption the first order regression model is a good fit!
The assumption: the first-order regression model is a good fit!

Using the Regression Equation
We define now appropriate points for further experimentation. The first point
is where all x’s are 0, the mean or center of the experiment. Than we
choose the step size (∆) for a process variable lets say time which we callchoose the step size (∆) for a process variable, lets say time which we call
∆x1. The ∆x2 is calculated by 4,50/2,35 * ∆x1 which is 1,91. We also get now
experimental points for an area where the result does not increasep p
anymore.
Mean
Time Temperature
Time Temperature Trial YieldMean
coded coded
Time Temperature Trial Yield
Origin 0 0 75 130 6,6,7 62,3
∆ 1 1,91 5 4,8
5
Origin + 1∆ 1 1,91 80 134,8 8 73,3
Origin + 2∆ 2 3,82 85 139,6
Origin + 3∆ 3 5 73 90 144 5 9 86 8Origin + 3∆ 3 5,73 90 144,5 9 86,8
Origin + 4∆ 4 7,64 95 149,2
Origin + 5∆ 5 9,55 100 154 10 58,2
Th t i l 9 ill b f th t ti i ti t d i th t t
The area trial 9 will be further systematic investigated in the next step
The Next Design for the Optimized Yield
Experiments 1 - 6 have been conducted first in this example (2
levels with 2 center points) The results were confirmed by furtherlevels with 2 center points). The results were confirmed by further
runs with star points and 2 center points.
Code Zeit Code Temp Yield Time Temp
-1 -1 78,8 80 140
1 1 84 5 100 140
File:
RSM3 mtw
1 -1 84,5 100 140
-1 1 91,2 80 150
1 1 77,4 100 150
RSM3.mtw
0 0 89,7 90 145
0 0 86,8 90 145
1 414 0 83 3 76 145-1,414 0 83,3 76 145
1,414 0 81,2 104 145
0 -1,414 81,2 90 138
0 1,414 79,5 90 152
0 0 87 90 145
0 0 86 90 145
0 0 86 90 145

The Local Optimum has been Achieved
Stat
>DOE
>Response Surface
>Contour/Surface (Wireframe) Plot…
Surface Plot of Yield vs Temp; Time
90
Yield
Contour Plot of Yield vs Temp; Time
90
p
150,0
147,5
70
72
74
76
78
80
82
84
152
148
70
144
80
80 140
Yield
Temp
Temp
145,0
142,5
140 0
145
88
8
86
80 14090
100Time
Time
10095908580
140,0
The maximum from this experiment is about 89 for the yield. Open
questions: can we control the temperature in this area, does the yield
improvement justify the increase in energy consumption?
O ti i ti DOE 22 D i ith t i t
Optimization before Production Start
Optimization DOE: 22-Design with center point
• amount NaOH
amount DMS
• amount acid
concentr acid
Process Map
• Addition time
HOAc
• Temperature
• amount DMS
• Addition time
DMS
• Reaction time
• concentr. acid
• Addition time
acid
• hydrolysis time • amount MMS
p
HADS-
Formation Methylation Hydrolysis Distillation
Pareto Chart of the Effects
• amount MMS
• OHMA before
• DMHA before
• yield
• OHMA after
• DMHA after
• amount HADS
A
Pareto Chart of the Effects
(response is Umsatz, Alpha = ,20)
A: NaOH
B: DMS
StdOrder RunOrder CenterPt Blocks NaOH DMS Sales
B
1 1 1 1 6 60 39,01
2 2 1 1 30 60 79,84
3 3 1 1 6 72 40,15
4 4 1 1 30 72 98,35
AB
,
5 5 0 1 18 66 99,12
0 10 20 30 40 50

O ti i ti DOE 22 D i ith t i t
Optimization before Production Start
„robust process“
Optimization DOE: 22-Design with center point
setting
20 g NaOH
32 g DMS
Contour Plot of Sales
What does EVOP stand for ?
• A tool for process improvementp p
• A tool for capacity improvement
• Uses replicated 22 or 23 factorial DOE’s with center points• Uses replicated 22 or 23 factorial DOE s with center points
• Includes/empowers the operators
• Conducted during normal operation
• In a full scale plantp
Objective: Operate the plant to produce quality productsj p p p q y p
and gather information on how to improve products at the
same time. The participation of many persons requires a
good information flow. Small improvement steps need a
big sample size.

EVOP - Summary
Advantages
• A simple tool for process
Disadvantages
• Limited experimental possibilities• A simple tool for process
optimization for a team of
responsible operators.
• Limited experimental possibilities.
• Small number of factors for every
phase
• Tight experimental settings avoid
process disturbance or scrap.
phase.
• Per phase often several replicates
necessary
• Blocks can be used to explain
other effects.
necessary.
• Needs usually several phases and
th f ti
• Minor additional costs for the
experimentation.
therefore more time.
• One has to be prepared for a long
i t l h ith ll
• A meaningful method for
continuous improvement.
experimental phase with small
stepwise improvements.
With EVOP improvements will be achieved taking many small steps!
With EVOP improvements will be achieved taking many small steps!
The EVOP Concept
Yield for several variables (Temperature, Pressure)
Example of EVOPSurface
88.0
88.4
88.8
892
85.0
825 89.2
89.6
90.0
904
82.5
800
emp
90.480.0
77.5
Te
898825 89
89.25
88.75
75.0
88.5
8988.25
88 88.4
Phase1
89
89
Phase2
115.0112.5110.0107.5105.0
Pressure

The EVOP Concept
Phase I after 5 experimental points
Cost per Ton
9.0
tio
8.5
8 0
PurgeRat
86 91
8.0
7.5
Recycle:P
92 95
92
7.0
R
92 95
Phase I
7.57.06.56.05.5
Reflux Ratio
The EVOP Concept
Phase II after 5 experimental points
Cost per Ton
9.0
tio
8180
8.5
8 0
PurgeRat
82
8.0
7.5
Recycle:P
83
Phase II
84
7.0
R
Phase I
7.57.06.56.05.5
Reflux Ratio

The EVOP Concept
Phase III after 5 experimental points
Cost per Ton
9.0
tio
8586
80
Phase III
8.5
8 0
PurgeRat
8384
8.0
7.5
Recycle:P
Phase II
7.0
R
Phase I
7.57.06.56.05.5
Reflux Ratio
Summary
• Analysis of experiments with two or more factorAnalysis of experiments with two or more factor
levels
• Use of diagnostic methods for the estimation of
how close the model is to realityhow close the model is to reality
• ExamplesExamples
• Introduction to the Response Surface MethodIntroduction to the Response Surface Method
(RSM) and to the EVOP methodology
(Evolutionary Operation)(Evolutionary Operation)

FULL FACTORIAL ANALYSIS

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FULL FACTORIAL ANALYSIS