Javier Garcia - Verdugo Sanchez - Six Sigma Training - W3 Full Factorial Designs RSM EVOP

Full Factorial DesignsRSM 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 FactorialPlackett-Burnam

Validate factors

Analyze behavior of

Plackett BurnamFolding

2k Factorialyimportant factors

Establish a model

Center PointsBlocking

Full Factorial

Determine optimal settings

Full FactorialBox-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

errorxxbxbxbbY211222110++++=

Not explainable i ti (N i )

Main effectsvariation (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 12

22 22

33 32

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:

Temp2 3 4

pressure2 3 4

90.4 90.7 90.290.2 90.6 90.4

12090.2 90.6 90.490.1 90.5 89.990.3 90.6 90.1

130

90.5 90.8 90.490.7 90.9 90.1

140


The Customization of a Design in Minitab

File: Full factorial 1.mtwStat

>DOE

>F t i l>Factorial

>Define Custom Factorial Design…Temp Pressure Yield120 2 90,4120 3 90,7120 4 90,2130 2 90,1130 3 90,5130 4 89 9130 4 89,9140 2 90,5140 3 90,8140 4 90 4140 4 90,4120 2 90,2120 3 90,6120 4 90 4120 4 90,4130 2 90,3130 3 90,6130 4 90,1130 4 90,1140 2 90,7140 3 90,9140 4 90,1


How to Start EvaluationStat

>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 ValuesTemp fixed 3 120; 130; 140Pressure fixed 3 2; 3; 4

Analysis of Variance for Yield, using Adjusted SS for Tests

Source DF Seq SS Adj SS Adj MS F P2 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 0,000Temp*Pressure 4 0,06889 0,06889 0,01722 0,97 0,470Error 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 andsmall, 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 variationFactor Type Levels ValuesTemp fixed 3 120; 130; 140Pressure fixed 3 2; 3; 4

is explained by temp and59 % by pressure

Analysis of Variance for Yield, using Adjusted SS for Tests

Source DF Seq SS Adj SS Adj MS F PTemp 2 0,30111 0,30111 0,15056 8,55 0,004Pressure 2 0 76778 0 76778 0 38389 21 80 0 000Pressure 2 0,76778 0,76778 0,38389 21,80 0,000Error 13 0,22889 0,22889 0,01761Total 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

90,7Temp Pressure

Main Effects Plot for YieldData Means

Stat

>Quality Tools

90,6

90,5

Me

an

>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

234

Pressure

Multi-Vari Chart for Yield by Pressure - Temp

90,6

90,5

90,4

90,3

Yie

ld

140130120

90,2

90,1

90,0


140130120Temp

Residual Diagnostics

Residual Plots for Yield

99

90

t

N 18AD 0,189P-Value 0,888

0,2

0,1

al

Normal Probability Plot Versus Fits

50

10

1

Per

cen

0,0

-0,1

-0,2

Res

idua

0,300,150,00-0,15-0,301

Residual90,890,690,490,290,0

Fitted Value

4 0,2

Histogram Versus Order

3

2

1

Freq

uenc

y

0,2

0,1

0,0

-0,1Res

idua

l

0,20,10,0-0,1-0,2

1

0

Residual18161412108642

-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 5282 56 63

TemperatureCatalyst amount

5 5

Perform a complete

64 65 6587 64 6081 77 53

5,5

evaluation!

Present your results!

67 74 6383 71 6075 73 5778 86 69

6

78 86 6972 74 7085 81 6583 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 Dye2 2 2 383 2 2 362 1 2 34moisture on the dye ability of nylon

fibers.• Output:

Dye ability (higher values are better)

Zinc2 1 2 343 1 1 282 2 2 363 1 2 35Dye ability (higher values are better)

• Inputs:– Crimp: low; high

3 1 2 351 2 2 363 1 1 273 1 3 26p ; g

– Temperature: low; medium; high– Moisture: low; medium; high

N = 3 observations per treatment

3 2 1 292 1 3 332 2 3 311 1 3 32• N = 3 observations per treatment 1 1 3 321 2 2 352 2 2 342 1 3 352 2 3 342 2 1 371 1 3 312 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 DoEtemperature to optimize yield.

Time

g

• Actual settings:

Ti 75 i t

70 75 80127.5 54.3 60.3

60 3

Time

– Time = 75 minutes

– Temperature = 130°

60.3Temp 130.0 64.3

62.3132.5 64.6 68.0

File:• Starting points for the 2 factor design:

– Time: Lo = 70 minutes; Hi = 80 minutesRSM1.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 EvaluationResponse Surface Regression: Yield versus Time; Temp

The following terms cannot be estimated, and were removed.Temp*TempThe 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 effects

Constant 62,3000 1,155 53,953 0,000Time 2,3500 1,000 2,350 0,143Temp 4,5000 1,000 4,500 0,046Time*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

Source DF Seq SS Adj SS Adj MS F Pi 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 0,072Square 1 0,429 0,429 0,4286 0,11 0,775Interaction 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 4,0000Pure 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

15080

Yield

Contour Plot of Yield vs Temp; TimeStat

>DOE

145

505560657075

86,8>Response Surface

>Contour/Surface (Wireframe) Plot…

Tem

p

140

13595

100

75808590

73,3

130

132,5

127,5

Time100959085807570

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 EquationThe 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

0mp

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 EquationWe 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 increase p panymore.

MeanTime Temperature

Time Temperature Trial YieldMeancoded 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.mtw0 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

90Yield

Contour Plot of Yield vs Temp; Time

90

p

150,0

147,5

7072747678808284

152148

70

144

80

80 140

Yield

Temp

Tem

p

145,0

142,5

140 0

14588

886

80 14090100Time

Time10095908580

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 StartOptimization DOE: 22-Design with center point

• amount NaOH amount DMS

• amount acidconcentr acid

Process Map• Addition timeHOAc

• Temperature

• amount DMS• Addition timeDMS

• Reaction time

• concentr. acid • Addition timeacid

• 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: NaOHB: 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

setting20 g NaOH32 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 products j p p p q y pand 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

88.088.488.8892

85.0

825 89.289.690.0904

82.5

800emp

90.480.0

77.5

Te

898825 89

89.2588.75

75.0

88.5

8988.25

88 88.4Phase1

89

89Phase2

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 0Pur

geR

at

86 918.0

7.5

Rec

ycle

:P

92 95

92

7.0

R 92 95Phase I

7.57.06.56.05.5

Reflux Ratio


The EVOP Concept

Phase II after 5 experimental points

Cost per Ton

9.0

tio 81808.5

8 0Pur

geR

at

82

8.0

7.5

Rec

ycle

:P 83Phase 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 0Pur

geR

at

8384

8.0

7.5

Rec

ycle

: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)


Engineering

Javier Garcia - Verdugo Sanchez - Six Sigma Training - W3 Full Factorial Designs RSM EVOP