Javier Garcia - Verdugo Sanchez - Six Sigma Training - W32 DOE Center Points and Blocks

2k factorial DOECenter Points

Blocks+1

Blocks+1

F aa c t o

C

o r B

-1 Factor A +1

-1 Week 3

Knorr-Bremse Group

About this Module

We know two ways to make 2 level designsWe know two ways to make 2 level designs more robust and more informative.

Center points to check for linearity

Incorporate and evaluate additional input as aIncorporate and evaluate additional input as a block factor

Knorr-Bremse Group 02 BB W3 center points & blocks 08, D. Szemkus/H. Winkler Page 2/27

Introduction of Center Points• We always are at the risk to overlook non linear relations

within the factor settings when using DOE’s with two factor levelslevels.

• The use of center points is an effective way to easily test for linearity (curvature)for linearity (curvature).

• An example:

– A chemical engineer wants to improve the yield. Two inputs are effecting the yield: reaction time and reaction temperaturetemperature.

– The chemical engineer decides to run an 2 x 2 design and adds center points in order to proof the linearity ofand adds center points in order to proof the linearity of this model.

Inputs:– Inputs:

• Reaction temperature: 150, 155 and 160 (°C)


• Reaction time: 30, 35 and 40 (minutes)

The DOE with 2 Factors and 3 Center Points

StdOrder CenterPt Temp. Time Yield 11 1 150 30 39,3

n = 4

mean = 40,425

1 1 150 30 39,32 1 160 30 403 1 150 40 40,94 1 160 40 41 5

,

n = 3

4 1 160 40 41,55 0 155 35 40,36 0 155 35 40,5

mean = 40,56 0 155 35 40,57 0 155 35 40,7

We want to know if there is a deviation between the actual center point and the theoretical value?

CenterPoints.mtw

( )2

centerpfactorcenterpfactor

Curvature nn

yynnSS

+−

=( )

345,40425,403*4

SS2

Curvature +−

=

centerpfactornn + 3

0096,0SSCurvature

=

L t thi l i Mi it b


Lets run this example in Minitab…

Create a DOE Stat>DOE>Factorial >Create factorial designs…

Open worksheet: center points mtw


Open worksheet: center points.mtw

Evaluation with Center PointsFactorial Fit: Yield 1 versus Temp.; Time

Estimated Effects and Coefficients for Yield 1 (coded units)

Stat>DOE>Factorial


Term Effect Coef SE Coef T PConstant 40,4250 0,1000 404,25 0,000

>Analyze Factorial Designs…

Temp. 0,6500 0,3250 0,1000 3,25 0,083Time 1,5500 0,7750 0,1000 7,75 0,016Temp.*Time -0,0500 -0,0250 0,1000 -0,25 0,826Ct Pt 0,0750 0,1528 0,49 0,672Ct Pt 0,0750 0,1528 0,49 0,672

S = 0,2 PRESS = *R-Sq = 97,26% R-Sq(pred) = *% R-Sq(adj) = 91,77%

Analysis of Variance for Yield 1 (coded units)

Source DF Seq SS Adj SS Adj MS F Pq j jMain Effects 2 2,82500 2,82500 1,41250 35,31 0,0282-Way Interactions 1 0,00250 0,00250 0,00250 0,06 0,826Curvature 1 0,00964 0,00964 0,00964 0,24 0,672

R id l E 2 0 08000 0 08000 0 04000Residual Error 2 0,08000 0,08000 0,04000Pure Error 2 0,08000 0,08000 0,04000

Total 6 2,91714

How do we decide?


How do we decide?

Main Effect Plot

Main Effects Plot for Yield 1Data Means

Stat>DOE>Factorial

41,25

41 00

Temp. TimeCornerCenter

Point Type

Data Means>Factorial Plots>Main Effect Plots…

41,00

40,75

n 40,50

40,25

Me

an

40,00

39,75

16015515039,50

403530

The actual center points don’t deviate from linearity significantly.

Therefore the null hypothesis is accepted


Therefore the null hypothesis is accepted.

Center Point Showing a Significant EffectRun the same experiment for the response column: yield 2.

Factorial Fit: Yield 2 versus Temp.; Time p ;


Term Effect Coef SE Coef T PConstant 40,4250 0,1000 404,25 0,000Temp. 0,6500 0,3250 0,1000 3,25 0,083Time 1 5500 0 7750 0 1000 7 75 0 016Time 1,5500 0,7750 0,1000 7,75 0,016Temp.*Time -0,0500 -0,0250 0,1000 -0,25 0,826Ct Pt 2,0750 0,1528 13,58 0,005

S = 0,2 PRESS = *R-Sq = 99,22% R-Sq(pred) = *% R-Sq(adj) = 97,67%

Analysis of Variance for Yield 2 (coded units)Analysis of Variance for Yield 2 (coded units)

Source DF Seq SS Adj SS Adj MS F PMain Effects 2 2,8250 2,82500 1,41250 35,31 0,028

i2-Way Interactions 1 0,0025 0,00250 0,00250 0,06 0,826Curvature 1 7,3811 7,38107 7,38107 184,53 0,005

Residual Error 2 0,0800 0,08000 0,04000Pure Error 2 0,0800 0,08000 0,04000


, , ,Total 6 10,2886

Main Effect Plots

42,5


Point Type

Main Effects Plot for Yield 2Data Means Stat

>DOEF t i l

42,0

41,5

an

Center >Factorial >Factorial Plots>Main Effect Plots…

41,0

40,5

40 0

Me

a

160155150

40,0

39,5403530

Temp Time P i t T

Main Effects Plot for Yield 1Data Means

41,25

41,00

40,75


Point Type

40,50

40,25

40 00

Me

an

160155150

40,00

39,75

39,50403530


Exercise with Real Data• Goal: Investigate the effects of concentration, ratio of B/A and

temperature on the yield

O t t ( ) Yi ld i %• Output (response): Yield in %

• Inputs:

C t ti l t hi h– Concentration → low; center; high

– Ratio B/A → low; center; high

T t l t hi h– Temperature → low; center; high

• Design: 2x2x2 factorial experiment with center points

P d• Procedure:

– Use the Minitab file: 3 fact. center points.mtw

A l f t i t ti d i ff t– Analyze for curvature, interaction and main effects

– Analyze the effects with the appropriate graphics

P f di ti– Perform diagnostics

– Calculate R² (How is the variation distributed)

P t lt d l i


– Present your results and conclusions

DOE Including BlocksThere are two ways to incorporate block factors!

1 The block factor is Advantage: fewer runs (effort)1. The block factor is confounded with the effect of the 3-way interaction

Advantage: fewer runs (effort)

Disadvantage: limited evaluation/information

A B C A*B*C Blocks-1 -1 -1 -1 11 -1 -1 1 2-1 1 -1 1 2

evaluation/information

1 1 -1 -1 1-1 -1 1 1 21 -1 1 -1 1-1 1 1 -1 11 1 1 1 2

A B C Blocks-1 -1 -1 11 -1 -1 1

1 1 1 1 2-1 1 -1 11 1 -1 1-1 -1 1 11 -1 1 11 1 1 1-1 1 1 11 1 1 1-1 -1 -1 21 -1 -1 21 1 1 22 Treat the block factor like -1 1 -1 21 1 -1 2-1 -1 1 21 -1 1 2-1 1 1 2

2. Treat the block factor like any independent factor.


1 1 1 21 1 1 2

Blocks Confounded with InteractionRun A B C A*B A*C B*C A*B*C Block

1 -1 -1 -1 1 1 1 -1 I2 1 -1 -1 -1 -1 1 1 II3 -1 1 -1 -1 1 -1 1 II3 -1 1 -1 -1 1 -1 1 II4 1 1 -1 1 -1 -1 -1 I5 -1 -1 1 1 -1 -1 1 II6 1 -1 1 -1 1 -1 -1 I7 -1 1 1 -1 -1 1 -1 I8 1 1 1 1 1 1 1 II

R A B C Bl kExample: 2 types of catalysts as block factor Run A B C Block1 -1 -1 -1 I4 1 1 -1 I6 1 -1 1 I

Example: 2 types of catalysts as block factorRule of thumb: High order interactions seldom contribute to the model.

7 -1 1 1 I

2 1 -1 -1 II

Therefore we overlay the 3-way interaction with an additional factor (catalyst)We assign catalyst A to the low level setting of

3 -1 1 -1 II5 -1 -1 1 II8 1 1 1 II

g y gthe 3-way interaction and catalyst B to the high level. We sacrificed the 3-way interaction to save 8 runs8 runs.

Be aware: in a real experiments we like to randomize the DOE within the blocks Minitab is taking care of that


blocks. Minitab is taking care of that.

Blocks Confounded with Interaction

1 Type I

212 Type II

12 12+1

12C

-11 2

-1 A +1


-1 A +1

Evaluation of an Example

• Goal: Analyze a DOE with 4 factors and 2 blocks

• Example: A chemical engineer wants to maximize the filtration rate of a p gchemical product which is produced in a pressure vessel. For the experiment he needs 16 runs. Only 8 runs per day are possible. The complete experiment requires 2 dayscomplete experiment requires 2 days.

• Output/response: Filtration rate in l/h

• Inputs:p

– Temperature

– Pressure

– Formaldehyde concentration

– Agitation speed

– Block variable day 1 vs. day 2

• Procedure:

U th Mi it b fil Bl k filt t DOE ith 4 f t 2 bl k– Use the Minitab file: Block filter.mtw, DOE with 4 factors, 2 blocks and 16 runs

– Analyze the data


a y e t e data

The Minitab Worksheet

Minitab file: Block filter.mtw

Blocks Temp Pressure F-Concentr. Agitator Filter rate1 1 -1 -1 -1 269 81 1 -1 -1 -1 269,81 -1 1 -1 -1 182,41 -1 -1 1 -1 258,41 1 1 1 -1 2471 1 1 1 1 2471 -1 -1 -1 1 163,41 1 1 -1 1 395,21 1 -1 1 1 326,81 -1 1 1 1 2662 -1 -1 -1 -1 1712 1 1 -1 -1 2472 1 -1 1 -1 2282 -1 1 1 -1 3042 1 -1 -1 1 3802 1 1 1 1 1712 -1 1 -1 1 1712 -1 -1 1 1 2852 1 1 1 1 364,8


The Evaluation

The factor pressure has A

31,82Factor Name

Pareto Chart of the Effects(response is Filtration rate, Alpha = ,05)

no significant effect.

We reduce the model

m B

ABD

C

D

AD

AC

A

A gitator

A TempB PressureC F -C oncentrationD

by this factor.Term

CD

ACD

ABC

BC

BCD

B

Effect

AB

BD

9080706050403020100

Lenth's PSE = 12,11252 26

Pareto Chart of the Standardized Effects(response is Filtration rate, Alpha = ,05)

AC

A

2,26Factor NameA TempC F -C oncentrationD A gitator

Term

D

AD

C

1086420


Standardized Effect

The EvaluationEstimated Effects and Coefficients for Filtration rate (coded units)

Term Effect Coef SE Coef T PConstant 266 24 4 337 61 39 0 000Constant 266,24 4,337 61,39 0,000Block -2,61 4,337 -0,60 0,562Temp 82,17 41,09 4,337 9,47 0,000F-Concentration 37,53 18,76 4,337 4,33 0,002Co ce t at o 3 ,53 8, 6 ,33 ,33 0,00Agitator 55,58 27,79 4,337 6,41 0,000Temp*F-Concentration -68,88 -34,44 4,337 -7,94 0,000Temp*Agitator 63,17 31,59 4,337 7,28 0,000

S = 17,3474 R-Sq = 96,73% R-Sq(adj) = 94,55%

Analysis of Variance for Filtration rate (coded units)

Source DF Seq SS Adj SS Adj MS F PBlocks 1 109 2 109 2 109 2 0 36 0 562Blocks 1 109,2 109,2 109,2 0,36 0,562Main Effects 3 44997,7 44997,7 14999,2 49,84 0,0002-Way Interactions 2 34939,4 34939,4 17469,7 58,05 0,000Residual Error 9 2708,4 2708,4 300,9Residual Error 9 2708,4 2708,4 300,9Total 15 82754,7

E l ti f ti l tti b f ll i h


Explanations for optimal settings by following graphs …

Interaction Plots

If the model shows significant effects of interactions we have to interpret these first.

Stat>DOE>Factorial

Interaction plots include all effects of the factors involved!

>Factorial Plots…>Interaction Plot

325 T

Interaction Plot for Filter rateData Means

380

Interaction Plot for Filter rateData Means

325

300

275

250

-11

Temp 380

360

340

320

-11

Temp

250

225

200

Me

an 300

280

260

240

Me

an

1-1

175

150

F-Concentr.1-1

220

200

Agitator

The optimal combination: High temperature and high agitation speed with a low formaldehyde concentration


with a low formaldehyde concentration

Multi Vari Chart

Stat>Quality Tools>Multi-Vari Chart>Multi Vari Chart…

Multi-Vari Chart for Filter rate by Temp - Agitator

400

1-1

-1 1-11

Temp

350

300ate

1

300

250Filt

er

ra

200

1-1F-Concentr.

Panel variable: Agitator


Residual DiagnosticsStat>DOE>Factorial >Analyze Fact…>Graph>Residual Plots 99

N 16 20

Normal Probability Plot Versus Fits

Residual Plots for Filter rate

>Four in one 90

50

10

Per

cent

N 16AD 0,369P-Value 0,384 10

0

-10Res

idua

l

40200-20-40

10

1

Residual400350300250200

-20

Fitted Value

Histogram Versus Order

4

3

2

requ

ency

20

10

0

Res

idua

l

20100-10-20

1

0

Residual

Fr

16151413121110987654321

-10

-20

Observation Order

R

The residuals don’t show any obvious pattern, the residuals are normal distributed Conclusion: The model is acceptable!


normal distributed. Conclusion: The model is acceptable!

Calculation of Components of VarianceStat> Anova> GLM…

Source DF Seq SS R2

Temp 1 27011 33%

F-concentr. 1 5633 7%

Agitator 1 12354 15%

Temp*F-Concentr 1 18975 23%Temp*F Concentr. 1 18975 23%

Temp*Agitator 1 15964 19%

E 10 2818 3%Error 10 2818 3%

Total 15 82755


Summary

In this module we have discussed how to use:

• Center points

Bl k• Blocks

This allows us to broaden the use of 2 factor levelThis allows us to broaden the use of 2 factor level designs. As a result we are more confident regarding our statementsstatements.


Appendix:Appendix:Evaluation of the Evaluation of the

examplep


The Graphical Evaluation with Minitab

4,303Factor NameA C t ti

Pareto Chart of the Standardized Effects(response is Yield, Alpha = ,05)

File: 3 fact. Center points.mtw

m

AC

AB

A A C oncentrationB Rel. B/AC Temp.

Term

BC

B

C

Standardized Effect

ABC

543210

2 447

Pareto Chart of the Standardized Effects(response is Yield, Alpha = ,05)

A

2,447Factor NameA C oncentrationB Rel. B/A

Step 1: Reduce the model from the overview to get the

Term AB

gbest model. B

6543210


Standardized Effect

The Reduced ModelF t i l Fit Yi ld C t ti R l B/AFactorial Fit: Yield versus Concentration; Rel. B/A

Estimated Effects and Coefficients for Yield (coded units)

Term Effect Coef SE Coef T PConstant 87,382 0,2573 339,65 0,000Concentration -3,500 -1,750 0,3017 -5,80 0,001Co ce t at o 3,500 , 50 0,30 5,80 0,00Rel. B/A 0,700 0,350 0,3017 1,16 0,284Concentration*Rel. B/A -0,800 -0,400 0,3017 -1,33 0,226

S = 0,853260 PRESS = 13,6690R-Sq = 84,00% R-Sq(pred) = 57,09% R-Sq(adj) = 77,15%

Analysis of Variance for Yield (coded units)

Source DF Seq SS Adj SS Adj MS F PSource DF Seq SS Adj SS Adj MS F PMain Effects 2 25,4800 25,4800 12,7400 17,50 0,0022-Way Interactions 1 1,2800 1,2800 1,2800 1,76 0,226Residual Error 7 5,0964 5,0964 0,7281Curvature 1 0,0297 0,0297 0,0297 0,04 0,857Pure Error 6 5,0667 5,0667 0,8444

Total 10 31,8564 File:


3 fact. Center points.mtw

Residual DiagnosticsStat>DOE>Factorial >Analyze Factorial Designs


>Analyze Factorial Designs>Graphs…>Residual Plots >Four in one Residual Plots for Yield

or

99

90

ent

N 11AD 0,465P-Value 0,203

1,0

0,5

ual

Normal Probability Plot Versus Fits

S

or

210-1-2

50

10

1

R id l

Per

ce

9089888786

0,0

-0,5

-1,0

Fitt d V l

Res

id

Stat>Regression>Regression…>Graphs… “Four in one”

Residual Fitted Value

6,0

4,5y

1,0

0 5

Histogram Versus Order

Graphs… Four in one

Store first the

4,5

3,0

1,5

0 0

Freq

uenc

y 0,5

0,0

-0,5

-1,0

Res

idua

l

Store first the residuals and fits

1,00,50,0-0,5-1,0-1,50,0

Residual1110987654321

Observation Order


Interpretation of ResultsStat>DOE>Factorial >F t i l Pl t


>Factorial Plots>Main Effect

Concentration Rel B/A P i t T

Main Effects Plot for YieldData Means

89

88

87

Concentration Rel. B/ACornerCenter

Point Type

10-1

87

86

10-1

Me

an

Temp.89

88

87

p

10-1

86

Only the concentration is significant. The position of the center points supports the assumption that the model is linear within the factor settings


supports the assumption that the model is linear within the factor settings.

Engineering

Javier Garcia - Verdugo Sanchez - Six Sigma Training - W32 DOE Center Points and Blocks