DOE Basics

LM Glasfiber R & D Green Belt- DMAIC CourseLM Glasfiber R & D Green Belt- DMAIC Course

LM Glasfiber Proprietary Information

Improve Phase Objectives

• The benefits of Design of Experiments (DOE’s)

• Key concepts and terms associated with DOE’s

• Performing a simple full factorial and fractional DOE’s and interpreting the results

• Awareness of screening designs and higher level response surface designs


DefineTollgate

DEFINEDEFINE

Step A: Identify Project CTQs

Step B: Develop Team Charter

Step C: Define Process Map

1

MEASUREMEASURE

MeasureTollgate

Step 1: Select CTQ Characteristics

Step 2: Define Performance Standards

Step 3: Measurement System Analysis (MSA)

2

ANALYZEANALYZE

AnalyzeTollgate

Step 4: Establish Process Capability

Step 5: Define Performance Objectives

Step 6: Identify Variation Sources)

3

IMPPROVEIMPPROVE

ImproveTollgate

Step 7: Screen Potential Causes

Step 8: Discover Variable Relationships

Step 9: Establish Operating Tolerances

4

CONTROLCONTROL

ControlTollgate

Step `10: Define and Validate the Measurement System on Xs

Step 11: Determine Process Capability

Step 12: Implement Process Control

5

Key Deliverables

Required• List of Project CTQs• Team Charter• High Level Process

Map (COPIS or SIPOC)

Tools That May Help• Project Risk

Assessment• Stakeholder Analysis• High Level Project Plan• In Frame/Out of Frame• Customer Survey

Methods (focus groups, interviews, etc.)

Required• List of Project CTQs• Team Charter• High Level Process

Map (COPIS or SIPOC)

Tools That May Help• Project Risk

Assessment• Stakeholder Analysis• High Level Project Plan• In Frame/Out of Frame• Customer Survey

Methods (focus groups, interviews, etc.)

Required•QFD/CTQ Tree•Operational definition, Specification limits, target, defect definition for Project Y(s)

•Measurement System Analysis

Tools That May Help•Data Collection Plan•Gage R&R•Detailed Process Map• FMEA•Pareto Analysis

Required•QFD/CTQ Tree•Operational definition, Specification limits, target, defect definition for Project Y(s)

•Measurement System Analysis

Tools That May Help•Data Collection Plan•Gage R&R•Detailed Process Map• FMEA•Pareto Analysis

Required•Baseline of Current Process Performance

•Normality Test •Statistical Goal Statement for Project

• List of Statistically Significant Xs

Tools That May Help•Benchmarking• Fishbone Diagram•Box Whisker Plots•Hypothesis Testing•Regression Analysis

Required•Baseline of Current Process Performance

•Normality Test •Statistical Goal Statement for Project

• List of Statistically Significant Xs

Tools That May Help•Benchmarking• Fishbone Diagram•Box Whisker Plots•Hypothesis Testing•Regression Analysis

Required• List of Vital Few Xs• Transfer Function(s)•Optimal Settings for Xs•Confirmation Runs/Results

• Tolerances on Vital Few Xs

Tools That May Help•Design of Experiments•New Process Maps• FMEA on new process•Process Modeling

Required• List of Vital Few Xs• Transfer Function(s)•Optimal Settings for Xs•Confirmation Runs/Results

• Tolerances on Vital Few Xs

Tools That May Help•Design of Experiments•New Process Maps• FMEA on new process•Process Modeling

Required•MSA Results on Xs•Post Improvement Capability

•Statistical Confirmation of Improvements

•Process Control Plan•Process Owner Signoff

Tools That May Help•Control Charts•Hypothesis Testing•CAP Plan

Required•MSA Results on Xs•Post Improvement Capability

•Statistical Confirmation of Improvements

•Process Control Plan•Process Owner Signoff

Tools That May Help•Control Charts•Hypothesis Testing•CAP Plan

Overall Project Completion Percentage

10% 20% 30% 40% 50% 60% 70% 80% 90% 100%

Key

Ste

ps:

The DMAIC Process


9. Establish Operating Tolerance

Deliverable:• Specify Tolerances

on the Vital Few X’s.

Tools:• Simulation

IMPROVE Phase Steps

DefineDefine MeasureMeasure AnalyzeAnalyze ControlControl

7. Screen Potential Causes

Deliverable:• Determine the Vital

Few X’s That Are Causing Changes in Y.

Tools:• Screening DOE

8. Discover Variable Relationships

Deliverable:• Establish Transfer

Function Between Y and Vital Few X’s.

• Determine Optimal Setting for the Vital Few X’s.

• Perform Confirmation Runs.

Tools:• Factorial Designs

ImproveImprove


What’s Improve Phase About. . .

• Develop an Improvement Strategy

• Determine which candidate x’s identified in the Analyze Phaseare truly “critical X’s”.

• If possible, determine a quantitative transfer function thatrelates your Y to these critical X’s

• Identify Improvement Actions• Determine optimal settings for the X’s

• Show the impact of the changes on meetingproject or business objectives.

• Validate the Improvement• Demonstrate the validity of your identified improvement

actions via additional experiments or a pilot study

• Develop a Plan to Implement the Change

Y = f(x)

It’s More than Just Designed ExperimentsIt’s More than Just Designed Experiments


Common “Improve” Tools

Basic Process Map Fishbone Box Plot Time Order Plots Hypothesis Tests Linear Regression Mistake Proofing

Intermediate DOE

Full Factorial Fractional

Factorial Intro to

Response Surface

Multivariate Regression

Advanced DOE

Response SurfaceTaguchi (Inner /

Outer Array)Simulation Models

Problem Sophistication• Complexity• Business Impact

• Risk• Data Availability

Match the Tool to the ProblemMatch the Tool to the Problem

Already Covered Covered in Improve Covered in DFSS Adv. Level III e.g. ProModel

LOWLOW HIGHHIGH


DOE - Terminology

Y = f (x1, x2, x3,……xn)

Response (Y)

• The measured outcome of an experiment

• The value observed for the CTQ being explored

Factors (x’s)

• The critical X’s which determine the response,Y

• They can be categorical or numerical

Levels

• In DOE’s we investigate the effect of each factor at more than one setting or value

DOE – Design of Experiments

Ranges

• The extreme values for each factor determines the range for that factor - the region of interest/investigation


65

75

85

95

M

axi

mu

m

-4.00

-2.00

0.00

2.00

4.00

-4.00

-2.00

0.00

2.00

4.00

A B

Y

x1x2

The dependence of Y on the x’s can be complex.And it is unknown!Where do we start?

DOE Challenge


Classical ApproachOFAT - One Factor at a Time

• Change one variable, X2,while holding all othersconstant.

• Find a maximum

• Hold X2 at the“maximum effect” level andrepeat the process for the other variables.

Benefits of DOEs

60

7080

90

Factor X1F

ac t

or

X2

100

OFAT• Requires more experiments than a DOE• Becomes unmanageable as the number of factors increases• Can be very expensive and time consuming – and may not work very well

DMAICSteps 7-8


DOE Approach

• Select factors and levels

• Select mathematical modeldesigned to obtain maximuminformation for the numberof factors/levels selected.

• In your experiments changethe factor levels in a systematicmanner so that all coefficients inthe model can be uniquely computed.(Orthogonality)

• Solve the resulting set of simultaneous equations to obtain the coefficients.

• Use statistical tests to determine if the coefficients are statistically significant, and if the resulting model (transfer function) is adequate.

• Use the results of your DOE to plan the next DOE (if needed).

Benefits of Design of Experiments

60

7080

90

Factor X1F

ac t

or

X2

100

DMAICSteps 7-8


2 Level Factorial Designs

For our example- 22 Full Factorial• 2 Factors, A and B• 2-levels (a High and a Low level for each factor)

Appropriate Mathematical Model (Minitab provides this)

Y = K + a*A + b*B + c*A*B (where K = a constant)

It is very convenient to work with the “Coded Values”


2 Level Factorial DesignsCoded Values

A B AB Response-1 -1 +1 115.8-1 +1 -1 116.8+1 -1 -1 106.7+1 +1 +1 124.3

Y = K + a*A + b*B + c*A*B

Y1 = 115.8 = K – a – b + cY2 = 116.8 = K – a + b – cY3 = 106.7 = K + a – b – cY4 = 124.3 = K + a + b + c

Solutions

response) average thejust is (K

115.94

124.3106.7116.8115.8K

0.44

116.8115.8106.7124.3a

4.654

115.8106.7116.8124.3b

4.154

106.7116.8115.8124.3c

Our Transfer Function

Y = 115.9 – 0.4*A + 4.54*B + 4.15*AB

115.8

116.8

106.7

124.3

(-1, -1) (+1, –1)

(-1, +1) (+1, +1)


-1

1

-1-1 1 1

108

113

118

123

B

A

Me

an

Interaction Plot (data means) for Response Y

A B

-1 1 -1 1

111.0

113.5

116.0

118.5

121.0

Res

pons

e Y

Main Effects Plot (data means) for Response Y

Each main effects plot shows the effect on the response when a factor is changed from it’s low level to its high level

In this interaction plot is shown:

• The effect on the response when A is held at it’s low level and B is changed from its low level to its high level.

• The effect on the response when A is held at it’s high level and B is changed from its low level to its high level.

• The two lines are not parallel. This indicates the presence of an interaction

Main & Interaction Effects Plots


Main & Interaction Effects Plots

Main Effect (e.g. DB = Main effect of Factor B in previous example)

This is a measure of the “effect” that a given factor has on the response when it is changed from its Low Level to its High Level.

DB= (avg. value of Y when B is High) – (avg. value of Y when B is Low)

9.32

106.7115.82

124.3116.82

YY

2

YYΔ 3142

B

Interaction Effects (e.g. DAB for previous example)

This is a measure of the “effect” that a given interaction term has on the response when it is changed from its Low Level to its High Level.

8.32

106.7116.82

124.3115.82

YY2

YYΔ 3241

AB

DMAICSteps 7-8


Randomization & ReplicationRandomization

• Whenever possible DOE runs should be executed in randomorder (Minitab will set up a random order for us).

• Randomization averages the effect of lurking variables overall factors in our experiments.

• A lurking variable is an unidentified variable (x) that influencesour response (Y).

Replication• Definition – Multiple execution of all aspects of an

experiment. To do a replicate means doing a run againentirely from the beginning.

• Replication is used to obtain an estimate of the errorassociated with the runs made in a DOE. This permits us touse hypothesis tests to determine which terms in our transferfunction are statistically significant


Statistical Significance

You can not test for the statistical significance of theterms in your DOE derived transfer function

IfThe number of terms in the final Transfer Function is

equal to or greater than the number ofexperimental runs

(Remember: The constant is a term)

Replication of experimental runs is the ideal approach to providingus with the required estimates of experimental error.

Let’s use Minitab to explore this furtherfor our 2-Factor Design example

Let’s use Minitab to explore this furtherfor our 2-Factor Design example

DMAICSteps 7-8


2-Factor Design with5 Replicates

Open Minitab

STATDOECreate Factorial Design

# Factors = 2Designs

# Center Points = 0# Replicates = 5# Blocks = 1

OKOK

To save the time of data entry open the Minitab worksheet2-Factor DOE Example.MTW

2-Factor DOE - Example


Transfer FunctionConstant and Coefficients Y = 115.93 – 0.41*A + 4.64*B + 4.14*A*B

2-Factor DOE - AnalysisMinitab Open worksheet STAT DOE Analyze Factorial Design

Select Response C7 OK

Fractional Factorial Fit Estimated Effects and Coefficients for Response (coded units) Term Effect Coef StDev Coef T PConstant 115.930 0.3471 333.98 0.000A -0.820 -0.410 0.3471 -1.18 0.255B 9.280 4.640 0.3471 13.37 0.000A*B 8.280 4.140 0.3471 11.93 0.000 Analysis of Variance for Response (coded units) Source DF Seq SS Adj SS Adj MS F PMain Effects 2 433.95 433.95 216.977 90.04 0.0002-Way Interactions 1 342.79 342.79 342.792 142.25 0.000Residual Error 16 38.56 38.56 2.410 Pure Error 16 38.56 38.56 2.410Total 19 815.30 Unusual Observations for Response Obs Response Fit StDev Fit Residual St Resid 9 112.600 115.840 0.694 -3.240 -2.33R 13 118.700 115.840 0.694 2.860 2.06R R denotes an observation with a large standardized residual

P-value for T-test

If P<0.05 then term (aka factor) is significant.

Rule: Even if P > 0.05 keep a main effect in the TF if it appears in an interaction term

Main and Interaction Effects

95% of the variation in the experimentsis accounted for by the model.

Only 5% is attributed to error. This modeldescribes the data very well.

DMAICSteps 7-8

0.95815.30

342.79433.95

0.05

815.30

38.56


Fractional Factorial Fit Estimated Effects and Coefficients for Response (coded units) Term Effect Coef StDev Coef T PConstant 115.930 0.3471 333.98 0.000A -0.820 -0.410 0.3471 -1.18 0.255B 9.280 4.640 0.3471 13.37 0.000A*B 8.280 4.140 0.3471 11.93 0.000 Analysis of Variance for Response (coded units) Source DF Seq SS Adj SS Adj MS F PMain Effects 2 433.95 433.95 216.977 90.04 0.0002-Way Interactions 1 342.79 342.79 342.792 142.25 0.000Residual Error 16 38.56 38.56 2.410 Pure Error 16 38.56 38.56 2.410Total 19 815.30 Unusual Observations for Response Obs Response Fit StDev Fit Residual St Resid 9 112.600 115.840 0.694 -3.240 -2.33R 13 118.700 115.840 0.694 2.860 2.06R R denotes an observation with a large standardized residual

DMAICSteps 7-82-Factor DOE - Analysis (cont’d)

P < 0.05 – Main Effects & Interactions Are Statistically Significant

DOEfactorialbalancedcompletelyaforonlyHolds

2

Effectt Coefficien

Why are the STDEV’s all the same?


Minitab2-Factor DOE Example.MTW

STAT DOE Factorial Plots Select & Set up Main Effects

Select & Set Up Interaction OK

-1

1

-1-1 1 1

108

113

118

123

B

A

Me

an

Interaction Plot (data means) for Response Y

A B

-1 1 -1 1

111.0

113.5

116.0

118.5

121.0

Res

pons

e Y

Main Effects Plot (data means) for Response Y

2-Factor DOE - Analysis (cont’d)


Residuals Analysis

What is a residual?

The difference between YExp , the response measured for a givenDOE run, and YPred , the response predicted by the transfer function.

R = (YExp – YPred) These are the prediction errors.

Example Replicate RunsA B YExp YPred R-1 -1 116.1 115.8 0.3-1 -1 116.9 115.8 1.1-1 -1 112.6 115.8 -3.2-1 -1 118.7 115.8 2.9-1 -1 114.9 115.8 -0.9

What do we expect for a welldesigned randomized DOE?

• The residuals should be normally distributedabout zero with s = the experimental errorstandard deviation (0.347 in our example).

• The values should be randomly distributedover the experimental runs.

-1 -0.5 0 0.5 1


Residuals Analysis

Minitab2-Factor DOE Example.MTW

STAT DOE Analyze Factorial Design Select C7 Response Graphs Select as shown OK


-3 -2 -1 0 1 2 3

-2

-1

0

1

2

Nor

mal

Sco

re

Residual

Normal Probability Plot of the Residuals(response is Response)

2 4 6 8 10 12 14 16 18 20

-3

-2

-1

0

1

2

3

Observation Order

Res

idua

l

Residuals Versus the Order of the Data(response is Response)

Residuals AnalysisNormality Plot of Residuals

We expect that all of our residuals willbelong to a normal distribution witha mean = 0.

We expect the normality plot to reflecta straight line.

Residuals vs. Order of Data

The residual for each replicate is plotted in the order that the replicate experiment was actually run.

If the errors are randomly spread across all the experiments, then we expect to see no evidence of a pattern in the plot.

The residuals should appear to berandomly scattered.

DMAICSteps 7-8


Residuals vs. Fitted values

For each experimental condition in theDOE the residuals of the replicates areplotted for the corresponding predictedvalue.

In our 2-Factor example we have fourexperimental conditions.

In a well designed DOE, one expects nostatistical difference in the spread of replicate values for the different experimental conditions

Residuals Analysis

105 115 125

-3

-2

-1

0

1

2

3

Fitted Value

Res

idua

l

Residuals Versus the Fitted Values(response is Response)

Response Y FITS1 RESI1

106.7 106.74 -0.04

107.5 106.74 0.76

105.9 106.74 -0.84

107.1 106.74 0.36

106.5 106.74 -0.24


116.1 115.84 0.26

116.9 115.84 1.06

112.6 115.84 -3.24

118.7 115.84 2.86

114.9 115.84 -0.94


116.5 116.84 -0.34

115.5 116.84 -1.34

119.2 116.84 2.36

114.7 116.84 -2.14

118.3 116.84 1.46


123.2 124.30 -1.10

125.1 124.30 0.80

124.5 124.30 0.20

124.0 124.30 -0.30

124.7 124.30 0.40

(+1, +1)

(-1, +1)(-1, -1)

(+1, -1)


Residuals Analysis

-1 0 1

-3

-2

-1

0

1

2

3

A

Res

idua

l

Residuals Versus A(response is Response)

-1 0 1

-3

-2

-1

0

1

2

3

B

Res

idua

l

Residuals Versus B(response is Response)

Residuals vs. Factors

For a given Factor the residuals are plotted for all experiments run with thatfactor at each of it’s levels.

In our example each factor has two levels.

Plotted at +1 are the residuals for allexperiments run with the factor at itshigh level.

Plotted at –1 are the residuals for allexperiments run with the factor at it’slow level.

For a well-controlled execution of a DOEone expects the spread in the residualsto be the same at each factor level.


Curvature

Why not simply spread theexperiments out as far aspossible over the design space?

The 2-level approach is based upon the assumptionthat the response of most natural processes variesin an approximately linear fashion over limitedregions of design space.

You would not want to span a region of the design space with too muchcurvature. There are other more sophisticated DOE methods to addresssuch situations (Response Surface Methods).

Response

Factor X

Low HighLow High

Response

Factor X

6070

80

90

Factor X1

Fa

c to

r X

2

100

We would like this We’d like to avoid this

DMAICSteps 7-8

See Notes Page


Center Points -Testing for Curvature

We can test for curvature in a 2-level design by addinga Center Point experiment to our design.

• The measured response for the “center point” experimentis compared to the predicted response.

- If the difference is statistically significant wrt theexperimental error, then evidence for curvature isfound. This will show up as a P-value < 0.05for “curvature” in the Minitab output.

• A true center point exists only if all factors are numerical,although multiple center points can be added if categoricalfactors are in the design

• Also, the replication of the center point experiment is anotherway to obtain an estimate of the experimental error in a DOE.


-1

10

20Response

Factor B-1

0Factor B

30

40

0 Factor A-1

1

1

Factor A

Center Points -Testing for Curvature

20Δ

Transfer Function(Coded Factors)

12B6A25Y

For Center Point: A = 0, B = 0

Predicted Y = 25Experimental Y = 45

Using the estimate ofexperimental error from theDOE, Minitab performs a t-testto determine if the differencebetween 20 and zero isstatistically significant.

If p < 0.05 in this t-test, we haveevidence for curvature.

202545 Δ


Testing for CurvatureLet’s add 5 replicates ofa center point experimentto our 2-Factor example

This is importantfor a

proper analysis(see notes page)

Minitab Open Worksheet 2-Factor Curv Example.MTW Note the (0,0) entries STAT DOE Analyze Factorial Design Select C7 Response Click “Terms” Include Center Point OK OK


Testing for Curvature

Estimated Effects and Coefficients for Response (coded units) Term Effect Coef StDev Coef T PConstant 115.930 0.3211 361.06 0.000A -0.820 -0.410 0.3211 -1.28 0.216B 9.280 4.640 0.3211 14.45 0.000A*B 8.280 4.140 0.3211 12.89 0.000Ct Pt 4.096 0.7180 5.71 0.000 Analysis of Variance for Response (coded units) Source DF Seq SS Adj SS Adj MS F PMain Effects 2 433.95 433.95 216.977 105.23 0.0002-Way Interactions 1 342.79 342.79 342.792 166.25 0.000Curvature 1 67.11 67.11 67.109 32.55 0.000Residual Error 20 41.24 41.24 2.062 Pure Error 20 41.24 41.24 2.062Total 24 885.09

P < 0.05 – The curvature is statistically significant

P = Predicted Center Point =115.930O = Mean of 5 experimental Center Points = 120.026O – P = 120.026 – 115.930 = 4.096

t-test results

F-testresults


23 – Full Factorial Example

Our Next Challenge

Optimize the Yield of our mostprofitable polymer forming reaction.

Where are we?

• We have good evidence from theAnalyze Phase and DOE screeningexperiments (more on this later)that reaction Temperature, theConcentration of the monomer, andthe source of the Catalyst arecritical x’s.

• Phase One of the Improve Plan is:

• Run a 2-level Full Factorial DOE• Obtain a transfer function• Predict the conditions for

optimum Yield.

We have three factors at two levels

Factor A = TemperatureFactor B = Concentration

We have one categorical factor

C = Catalyst vendor

The process development and plantengineers worked together to developthe DOE plan

Temperature: High=180 C Low=160 CConcentration: High=40% Low=20%Catalyst Vendor: High=Ed Low=Sally

Run a 23 Full factorial DOE with3 replicates and center points


23 – Full Factorial Example

Appropriate Mathematical Model (Minitab provides this)

Y = K + a*A + b*B + c*C + d*A*B + e*A*C + f*B*C + g*A*B*C

Coded Values For DOE RunsRun A B C AB AC BC ABC

1 -1 -1 -1 1 1 1 -12 1 -1 -1 -1 -1 1 13 -1 1 -1 -1 1 -1 14 1 1 -1 1 -1 -1 -15 -1 -1 1 1 -1 -1 16 1 -1 1 -1 1 -1 -17 -1 1 1 -1 -1 1 -18 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

A, B, C

With this full factorial we cansolve for all eight terms inthe model shown below


23 – Full Factorial – Exercise

Break into Teams

Help the Chemical Reaction Team interpret their DOE.

Minitab Worksheet ChemReaction.MTW

Using the previous 2-Factor example as a guideline, fully analyze the DOE data and prepare a 10 minute report to the class.

Report your prediction of the conditions for maximum Yield.

Note that the actual values for T, C, and the names of the vendors were entered for the factors (but the results are in coded variables). Explain in your report why this was necessary.

ReminderRemove insignificant terms fromthe model one at a time startingwith the highest P-value term.


23 – Full Factorial – Exercise

Estimated Effects and Coefficients for Yield (coded units) Term Effect Coef StDev Coef T PConstant 64.250 0.4061 158.21 0.000A 23.000 11.500 0.4061 28.32 0.000B -5.000 -2.500 0.4061 -6.16 0.000C 1.500 0.750 0.3632 2.06 0.050A*C 10.000 5.000 0.4061 12.31 0.000Ct Pt 0.200 0.9081 0.22 0.828 Analysis of Variance for Yield (coded units) Source DF Seq SS Adj SS Adj MS F PMain Effects 3 3340.87 3340.87 1113.62 281.34 0.0002-Way Interactions 1 600.00 600.00 600.00 151.58 0.000Curvature 1 0.19 0.19 0.19 0.05 0.828Residual Error 24 95.00 95.00 3.96 Lack of Fit 4 15.00 15.00 3.75 0.94 0.462 Pure Error 20 80.00 80.00 4.00Total 29 4036.07

Yield = 64.25 + 11.5*A - 2.50*B + 0.75*C + 5.00*A*C

Max Yield = 84% (A-High, B-Low, C-High (Ed)

No statistical evidencethat curvature is

significant


2- Level Fractional Factorials

2-Level Full Factorial Designs

Number Numberof Factors of Runs

1 22 43 84 165 326 647 1288 2569 512

10 1024• •• •

15 32,768• •• •

20 1,048,576

Fractional Factorial Designs

• The number of runs required for a2-level full factorial increases rapidlywith the number of factors (this getseven worse when you add replicates)

• With Fractional Factorial Designsit possible to obtain useful informationwith fewer runs.

• Of course there will be tradeoffs wrtthe information obtainable.



Y = K + a*A + b*B + c*C + d*A*B + e*A*C + f*B*C + g*A*B*C

Y1 = K – a – b – c + d + e + f – g Y2 = K + a – b – c – d – e + f + g Y3 = K – a + b – c – d + e – f + gY4 = K + a + b – c + d – e – f – gY5 = K – a – b + c + d – e – f + gY6 = K + a – b + c – d + e – f – gY7 = K – a + b + c – d – e + f – gY8 = K + a + b + c + d + e + f + g


1 -1 -1 -1 1 1 1 -12 1 -1 -1 -1 -1 1 13 -1 1 -1 -1 1 -1 14 1 1 -1 1 -1 -1 -15 -1 -1 1 1 -1 -1 16 1 -1 1 -1 1 -1 -17 -1 1 1 -1 -1 1 -18 1 1 1 1 1 1 1

What information can we obtain if we run only one halfthe number of runs in a 23 Full factorial Design?

Y5

Y8

Y3

Y2

+1,-1,-1

-1,+1,-1

+1,+1,+1

-1,-1,+1



What information can we obtain if we run only one halfthe number of runs in a 23 Full factorial Design?

With 4 equations we candetermine 4 terms

• A constant plus 3 others

• Let’s define some terms

[K] = K + A*B*C[A] = A + B*C[B] = B + A*C[C] = C + A*B


2 1 -1 -1 -1 -1 1 13 -1 1 -1 -1 1 -1 15 -1 -1 1 1 -1 -1 18 1 1 1 1 1 1 1

In this design the terms A and B*C are always set at the same level in each DOE run. Thus, when A is High, the term[A] = A + B*C will be High. Similar relationships exist for the other factors.

Coded Values For Defined TermsRun [A] [B] [C]

2 1 -1 -13 -1 1 -15 -1 -1 18 1 1 1

Y = [K] + [a]*[A] + [b]*[B] + [c]*[C]

Y2 = [K] + [a] – [b] – [c]Y3 = [K] – [a] + [b] – [c]Y5 = [K] – [a] – [b] + [c]Y8 = [K] + [a] + [b] + [c]

The coefficients maynot be the same asthose computed forthe full factorial model.


Aliasing (Confounding)

What information do we lose if we run only one halfthe number of runs in a 23 Full factorial Design?

Y = [K] + a*[A] + b*[B] + c*[C]

Y2 = [K] + [a] – [b] – [c]Y3 = [K] – [a] + [b] – [c]Y5 = [K] – [a] – [b] + [c]Y8 = [K] + [a] + [b] + [c]

[K] = K + A*B*C[A] = A + B*C[B] = B + A*C[C] = C + A*B

Aliasing Results in a Loss of Information

The effects of the terms that are coupled can not be evaluated independently. They are said to be aliased (or confounded) with each other.

• For example, if the term [A] is found to be significant, is it because A is important, B*C is important, or both?

• For example, A and B*C could both beimportant but act in opposite directions. In this case the term [A] might be insignificant.

Of course, if the interaction terms are all insignificant in our example, then you can resolve the main effects, A, B, and C.

Note

Our Half Fraction is Labeled 23-1

½ x 23 = 23-1


Aliasing (Confounding)

Let’s analyze a 23-1 DOE for our Chemical Reaction Problemand compare it to the previous 23 Full Factorial Analysis

Open > Fractional_ChemReaction.MTW

Estimated Effects and Coefficients for C8 (coded units) Term Effect Coef StDev Coef T PConstant 64.500 0.7217 89.37 0.000Temperat 23.000 11.500 0.7217 15.93 0.000Concentr 5.000 2.500 0.7217 3.46 0.009Vendor 3.000 1.500 0.7217 2.08 0.071 Analysis of Variance for C8 (coded units) Source DF Seq SS Adj SS Adj MS F PMain Effects 3 1689.00 1689.00 563.000 90.08 0.000Residual Error 8 50.00 50.00 6.250 Pure Error 8 50.00 50.00 6.250Total 11 1739.00 Estimated Coefficients for C8 using data in uncoded units

Transfer Function Comparison

23-1 DOE Analysis Y = 64.5 + 11.5*[A] + 2.5*[B] + 1.5*[C]

23 DOE Analysis Y = 64.3 + 11.5*A – 2.5*B + 0.75*C + 0.75*A*B + 5.0*A*C


Resolution and Aliasing

1. Hold up a number of fingers equal to the Design Resolution In this example:Resolution V = 5 Digits.

2. Use the other hand to grab a number of fingers equal to the Main or Interaction Effects you wish to investigate for aliasing. For example to determine aliasing for Main Effects, grab one finger.

3. The remaining number of fingers is the lowest level of interaction effects which are aliased. In this example, Main Effects are aliased with 4-way interactions.


Screening Designs

We have just learned about 2-level factorial designs in the order:

Full Factorial > Fractional Factorial

However, in practice we typically apply these designs in the order:

Fractional Factorial > Full Factorial (or RS Design)

Fractional Factorial Designs are often used to help us “Screen” a listof candidate factors that we believe may be critcal X’s and to do so

using a minimum number of experiments.

Typically, we screen for “Main Effects”


Screening DesignsEffects Hierarchy

• Main Effects occur frequently.

• 2-way interactions do occurand cannot be assumed tobe absent without goodreason.

• 3-way interactions seldomoccur in experimental DOE’s.

Fractional Factorial Designs arevery useful for screening for maineffects.

For example, during the Analyze Phasethe team finds evidence that 7 Factorsmay be critical x’s, but additionalconfirmation is required.

A 7-Factor Full Factorial Design wouldrequire 128 runs without replicates.

A 1/8 Fraction Factorial would require16 runs and have Resolution IV. Allmain effects are resolvable.

A 1/16 Fraction Factorial would requireonly 8 runs, but the main effects wouldbe aliased with the 2-way interactions.


Our Next Challenge

Optimize the Yield of our mostprofitable polymer forming reaction.

Where are we?

• We have good evidence from theAnalyze Phase and DOE screeningexperiments (more on this later)that reaction Temperature, theConcentration of the monomer, andthe source of the Catalyst arecritical x’s.

• Phase One of the Improve Plan is:

• Run a 2-level Full Factorial DOE• Obtain a transfer function• Predict the conditions for

optimum Yield.

We have three factors at two levels

Factor A = TemperatureFactor B = Concentration

We have one categorical factor

C = Catalyst vendor

The process development and plantengineers worked together to developthe DOE plan

Temperature: High=180 C Low=160 CConcentration: High=40% Low=20%Catalyst Vendor: High=Ed Low=Sally

Run a 23 Full factorial DOE with3 replicates and center points

Screening Designs

Remember this problem?How did the team developthis “Good Evidence”?


Screening Designs

The Team’s understanding atthe beginning of the project What Mother Nature Knows

The following 7 Factors may be critical X's

A. Temperature (160C - 180C)B. Monomer Concentration (20% - 40%)C. Catalyst Vendor (Sally - Ed)D. Stirring Speed ( 50 RPM - 100 RPM)E. Monomer Purity ( 90% - 98%)F. Pressure ( 100 PSI - 500 PSI)G. Acetone/Methanol Ratio - ( 0.25 - 0.50)

Yield = 64.25 +

11.50*A -

2.50*B +

0.75*C +

5.00*A*C

We need an efficient method for screening these"candidate critical X's"

so that we can identify the 'Vital Few"


Screening Designs

We have 7 candidate critical X's

A 1/8 fraction design will allow usto evaluate main effects.

This is a Resolution IV Design

Main effects are aliased with3-way interactions

2-way interactions are aliased with two way interactions

Let's evaluate the 7-Factor1/8 fraction screening DOE

7-Factor_Screening_DOE.MTW 7-Factors 3 Center Points for each

categorical factor (Sally/Ed) 22 experiments

If you have time - create this DOEin Minitab and inspect the aliasstructure


Screening Designs

Estimated Effects and Coefficients for Yield (coded units) Term Effect Coef StDev Coef T PConstant 62.435 0.7737 80.70 0.000Temperat 24.285 12.142 0.7737 15.69 0.000Concentr -4.460 -2.230 0.7737 -2.88 0.011Catalyst 2.354 1.177 0.6598 1.78 0.093Temperat*Catalyst 13.100 6.550 0.7737 8.47 0.000Ct Pt 0.073 1.4815 0.05 0.961 Analysis of Variance for Yield (coded units) Source DF Seq SS Adj SS Adj MS F PMain Effects 3 2469.08 2469.08 823.026 85.93 0.0002-Way Interactions 1 686.44 686.44 686.440 71.67 0.000Curvature 1 0.02 0.02 0.023 0.00 0.961Residual Error 16 153.24 153.24 9.578 Lack of Fit 4 16.99 16.99 4.249 0.37 0.823 Pure Error 12 136.25 136.25 11.354Total 21 3308.79

We find that only 3 of our 7candidate X's are significant.

Because only 3 factors are significantand no curvature is indicated, wecan fit our model completely. Noadditional experiments are required.

Looks like that 23 full factorial DOE that we did earlier was not needed.


Another Screening Design2-Level Plackett-Burman Designs are highly efficient Resolution III

designs for screening for main effects

A B C D E F G H J K L1 -1 1 -1 -1 -1 1 1 1 -1 11 1 -1 1 -1 -1 -1 1 1 1 -1

-1 1 1 -1 1 -1 -1 -1 1 1 11 -1 1 1 -1 1 -1 -1 -1 1 11 1 -1 1 1 -1 1 -1 -1 -1 11 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 11 -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

11 Factors

12 Runs

Each main effect is partially aliased with everytwo way interaction except for the two wayinteractions containing that effect.


Foldover Designs

Adding a second block of DOE runs with the signs of all the factors reversedwill convert any Resolution III factorial design to a Resolution IV designand break the aliases between the main effects and 2-way interactions.

This is especially popular in the use of Plackett-Burman Designs.

A B C D E F G H J K L1 1 -1 1 -1 -1 -1 1 1 1 -1 12 1 1 -1 1 -1 -1 -1 1 1 1 -13 -1 1 1 -1 1 -1 -1 -1 1 1 14 1 -1 1 1 -1 1 -1 -1 -1 1 15 1 1 -1 1 1 -1 1 -1 -1 -1 16 1 1 1 -1 1 1 -1 1 -1 -1 -17 -1 1 1 1 -1 1 1 -1 1 -1 -18 -1 -1 1 1 1 -1 1 1 -1 1 -19 -1 -1 -1 1 1 1 -1 1 1 -1 1

10 1 -1 -1 -1 1 1 1 -1 1 1 -111 -1 1 -1 -1 -1 1 1 1 -1 1 112 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -113 -1 1 -1 1 1 1 -1 -1 -1 1 -114 -1 -1 1 -1 1 1 1 -1 -1 -1 115 1 -1 -1 1 -1 1 1 1 -1 -1 -116 -1 1 -1 -1 1 -1 1 1 1 -1 -117 -1 -1 1 -1 -1 1 -1 1 1 1 -118 -1 -1 -1 1 -1 -1 1 -1 1 1 119 1 -1 -1 -1 1 -1 -1 1 -1 1 120 1 1 -1 -1 -1 1 -1 -1 1 -1 121 1 1 1 -1 -1 -1 1 -1 -1 1 -122 -1 1 1 1 -1 -1 -1 1 -1 -1 123 1 -1 1 1 1 -1 -1 -1 1 -1 -124 1 1 1 1 1 1 1 1 1 1 1

Block #1Resolution III

Block #2Resolution III

Block #1 + #2Resolution IV


BlockingTypical Blocking Situation

• We wish to run a DOE

• We know that there will be at least one source of variation that we cannot control over the all the experiments.

Examples:

• We do not have enough material from one batch to complete all the experiments.

• We will have to run the experiments in thefactory on two different days or two different shifts.

• We may have to run the experiments in different equipment

• What do we do?

• We divide the experiments into properly designed subsets of the total DOE. We call these subsets, “Blocks”.

Run A B C1 -1 -1 -12 +1 -1 -13 -1 +1 -14 +1 +1 -15 -1 -1 +16 +1 -1 +17 -1 +1 +18 +1 +1 +1

Block #1 Run A B C

1 -1 -1 -14 +1 +1 -16 +1 -1 +17 -1 +1 +1

Block #2Run A B C

2 +1 -1 -13 -1 +1 -15 -1 -1 +18 +1 +1 +1

+

Basic Assumption

The effect of the blocking variable is to shift each response Y by a common fixed amount, D.


Example: Compute the Coefficient “a”

Notice that the symmetry of the blocksis such that the “assumed” common shift effect, D , subtracts out for all terms except A*B*C which is aliased with the blockingvariable.

Blocking – How Does It Work?

-1,-1,-1 +1,-1,-1

-1,+1,-1

+1,+1,+1

+1,-1,+1

-1,-1,+1

-1,+1,+1

+1,+1,-1

Block #1 Run A B C AB AC BC ABC

1 -1 -1 -1 1 1 1 -14 +1 +1 -1 1 -1 -1 -16 +1 -1 +1 -1 1 -1 -17 -1 +1 +1 -1 -1 1 -1

Block #2Run A B C AB AC BC ABC

2 +1 -1 -1 -1 -1 1 13 -1 +1 -1 -1 1 -1 15 -1 -1 +1 1 -1 -1 18 +1 +1 +1 1 1 1 1

Y4 = K + a + b – c + d – e – f – gY6 = K + a – b + c – d + e – f – gY2 = K + a – b – c – d – e + f + g + DY8 = K + a + b + c + d + e + f + g + D

Y1 = K – a – b – c + d + e + f – gY7 = K – a + b + c – d – e + f – gY3 = K – a + b – c – d + e – f + g + DY5 = K – a – b + c + d – e – f + g + D

8

YYYYYYYYa 53718264


Response Surface AnalysisResponse surfaces with significant curvature cannot be accurately

described by the linear models obtained from 2-level designs. In such cases 3-level designs may be appropriate. These permit fitting the response to model equations containing quadratic or higher terms.

· Central Composite Designs: Most commonly used design (developed by Box and Wilson (1951)) for fitting quadratic response surfaces. These are first order designs augmented with center points and star (or axial) points

· Box-Behnken Designs: Box and Behnken (1960) developed a family of efficient three-level designs for fitting second-order response surfaces. Useful when corner points cannot be run due to physical limitations

· 3k Factorial Designs: A factorial arrangement with k factors each at three levels. Not very efficient design for large number of factors


Central Composite DesignsRun A B C

1 -1 -1 -12 -1 -1 +13 -1 +1 -14 -1 +1 +15 +1 -1 -16 +1 -1 +17 +1 +1 -18 +1 +1 +1

9 0 0 010 0 0 011 0 0 012 0 0 013 0 0 014 0 0 0

15 0 016 0 017 0 018 0 019 0 0

20 0 0

23 FactorialCorner Runs

Axial Runs

Center Runs

With the appropriate CCD one can start with a Factorial Designwith a few center runs. If curvature is indicated, then additionalcenter runs and axial runs can be added to complete the CCD.

With the appropriate CCD one can start with a Factorial Designwith a few center runs. If curvature is indicated, then additionalcenter runs and axial runs can be added to complete the CCD.

Axial RunsFor assessment ofquadratic terms

Assess linear and2-way interactions

Corner RunsPermit estimationof error and help toresolve curvature

Center Runs


Box-Behnken Designs

The Box-Behnken design doesn’t have any corners. It is suitable for the situation when corners are not feasible.

For example: if temperature and pressure are factors, it may not be possible to simultaneously set both at low or high levels.

Number of runs are equivalent (very close) to CCD

Exist only for number of factors = 3-7


Minitab uses regression methods to analyze response surface designsand provides R2 and R2-adjusted metrics which are indications ofhow well the model fits the data.

R2 is the fraction of the total variance thatis explained by the regression equation.

R2 is the fraction of the total variance thatis explained by the regression equation.

( )

( )

y y

y y

i ii

n

ii

n

2

1

2

1

1= --

-

=

=

å

å

^ Sum of squares of the residuals (or errors)

Total VarianceT

E

SS

SSR 12

R2 – Goodness of Fit

• A R2 close to 1 is an indication of a good fit.

• However, be wary of fits that equal 1 or are “too good to be true”• Example R2 = 1 if number of terms = number of runs (no error est.)

• The R2-adjusted metric was developed to help identify those situations


When the values of R2Adj and R2 are close,

you have greater confidence that your model has the right terms.

When the values of R2Adj and R2 are close,

you have greater confidence that your model has the right terms.

where: n = number of runs p = number of terms in the regression model (including the constant)

( )R nn p

RAdj2 21 1 1- -

-æèç

öø÷

-=

R2-Adjusted – Goodness of Fit

Any term (even if insignificant) added to the model will improve the R2 value.

• When # data points = # experiments, then R2 = 1 …….. A “Perfect Fit”

R2Adj is reduced when an insignificant term is added to the model.


Chemical Reactor Example

• The Critical X’s of Interest• Reaction Time• Temperature • Percent Catalyst

• Response Variable -CTQ• Percent conversion

CCD Exercise

Let’s First Run a 2-levelFull Factorial with3 Center Points

Minitab

Open Worksheet ChemReactorFactorial.MTW

How do things Look?

• Residuals?• Curvature?• Do we need to think about

continuing to build the CCDdesign?

Class Exercise


Estimated Effects and Coefficients for Yield (coded units) Term Effect Coef StDev Coef T PConstant 88.571 0.3293 268.97 0.000Time 6.172 3.086 0.3293 9.37 0.001Temperat 8.392 4.196 0.3293 12.74 0.000Concentr 12.298 6.149 0.3293 18.67 0.000Time*Concentr 4.278 2.139 0.3293 6.49 0.003Temperat*Concentr -6.813 -3.406 0.3293 -10.34 0.000Ct Pt -9.045 0.6306 -14.34 0.000 Analysis of Variance for Yield (coded units) Source DF Seq SS Adj SS Adj MS F PMain Effects 3 519.525 519.525 173.175 199.63 0.0002-Way Interactions 2 129.414 129.414 64.707 74.59 0.001Curvature 1 178.483 178.483 178.483 205.74 0.000Residual Error 4 3.470 3.470 0.867 Lack of Fit 2 1.347 1.347 0.674 0.63 0.612 Pure Error 2 2.123 2.123 1.061Total 10 830.891

CCD Exercise

There is evidence for curvature - The experimental center point average is ~ 10%lower than the predicted value. The team decides that a transfer function isneeded that better maps the curvature. Let's move on to a CCD design.


CCD Exercise

Chemical Reactor Example

• The Critical X’s of Interest- Reaction Time- Temperature - Percent Catalyst

• Response Variable -CTQ- Percent conversion

Let’s Set Up a RS Design

CCD - 6 center points

Open Minitab

StatDOECreate RS Design

Design Type: Central CompositeNumber of Factors: 3

Click on “Designs”• Full - 20 runs • 1 Block - 6 Center Points

OK

Click on “Factors”• Accept default values

OK

Click “OK” in RS Design Window

Class Exercise


CCD ExerciseData Analysis

File Open Worksheet ChemReactor-CCD.MTW

Stat DOE Analyze RS Design Responses: C7 Use Coded Units Graphs Select as appropriate OK OK Remove Terms from TF

based upon P-Value

Then go to RS Plots and look at contour and wire frame plots

21

0-2

60Concentration

70

80

-1

90

100

110

120

-10

1

Yield

-22Temperature

75

85

95

105

115

-1 0 1

-1

0

1

Temperature

Co

nce

ntra

tion

Contour Plot of Yield

Hold values: Time: 0.0


Estimated Regression Coefficients for Yield Term Coef StDev T PConstant 80.318 0.4819 166.664 0.000Time 3.829 0.3767 10.165 0.000Temperat 4.138 0.3767 10.985 0.000Concentr 6.321 0.3767 16.781 0.000Temperat*Temperat 10.427 0.3649 28.578 0.000Concentr*Concentr -2.145 0.3649 -5.880 0.000Time*Concentr 2.139 0.4921 4.346 0.001Temperat*Concentr -3.406 0.4921 -6.921 0.000 S = 1.392 R-Sq = 99.2% R-Sq(adj) = 98.7% Analysis of Variance for Yield Source DF Seq SS Adj SS Adj MS F PRegression 7 2831.37 2831.37 404.482 208.75 0.000 Linear 3 979.67 979.67 326.557 168.53 0.000 Square 2 1722.29 1722.29 861.144 444.43 0.000 Interaction 2 129.41 129.41 64.707 33.40 0.000Residual Error 12 23.25 23.25 1.938 Lack-of-Fit 7 13.70 13.70 1.957 1.02 0.508 Pure Error 5 9.55 9.55 1.910 Total 19 2854.62

CCD Exercise


DOE Reminders

Designing an Experiment• What Factors are likely to be important?• Over what ranges should the Factors be studied?• In what metrics should the Factors and Responses be considered?

- Linear - logarithmic - reciprocal scales ?

• Any multivariate transformations that should be made? - Example - perhaps the effects of Factors X1 and X2 can be more simply expressed as their

ratio or sum

• Validate your model with additional experimental runs!

Reality• One is least able to answer these questions at the outset of an experiment.

- Therefore, a sequential approach is often the best- Use a sequence of moderately sized designs- Reassess the situation after each design is analyzed

Use no more than 25% of your experimental budget on the first DOE.Use no more than 25% of your experimental budget on the first DOE.

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