Ch 6 BB Improve

7/28/2019 Ch 6 BB Improve

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BB ProgramImprove Phase


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Five Broad Improvement ApproachesExit from Analyze Phase

Available?

Check Availability of

Useful Y = f (X)

A1

Yes

No

Effect Established?

Examine Nature of the Xs

Control Factors?

Examine Status of the Xs

Yes

Yes

No

No

Examine Feasibility

of ExperimentationFeasible?

Examine Nature of

the SolutionKnown?

A3

A4

A5

A2

Optimize using

graphical or quantitativeoptimization techniques Adopt Data BasedInnovative Approach

Conduct DOE

Obvious Solution

Adopt Innovation-Prioritization Approach

Yes

Yes

No

No


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Innovation-Prioritization Approach (A2) Most popular. Frequently used for improving service

processes

But the approach is risky, since we are dealing with a

poorly characterized process.

So, even if the problem is solved successfully, usually

It becomes difficult to hold on to the gains

Very little is learned about the process

Often requires sizeable capital investment or the solution

involves control procedures which are difficult to follow

on a routine basis


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Innovative (A4) and Obvious (A5)

Approaches In many cases it is found that the root cause of the problem is

a control factor and its optimal level is either known(specified standard) or obvious. Implementation of the knownor obvious solution will solve the problem. Note however that

even though the solution is obvious, identification of the rootcause may not be so.

But what to do when the root causes identified are found tobe within their specified operating tolerances or if theproblem remains unsolved even after maintaining the Xswithin their operating tolerances?

If the process is stable, then tightening operating tolerancesmay solve the problem. However, in case of an unstable

process, tightening of tolerances is not likely to be effective.So what do we do? The answer to this question is discussedin the next section.


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A Difficult Problem Consider a highly unstable continuous chemical process. The

team takes a 30000 ft view of the process, somehow estimatesthe approximate sigma level and proceeds to the analyze

phase. The control factors are identified easily. However the

team now finds it extremely difficult to establish the effectsof the control factors. The team also observes that the levelsof some of the control factors are adjusted quiet frequently inresponse to the unstable behavior of the process. Althoughthe standard operating ranges are available for the controlfactors, at times these limits are violated. More surprisingly,such violations are found to have no bearing on process

performance. In fact, in many cases the process performancedid not improve even after making the necessary processadjustments. Given this background, what should be theimprovement approach of the team?


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A Difficult Problem (Contd.) The above situation indicates presence of strong interactions

in the system.

So we have a situation where the root causes have been

identified but their effects are yet to be established. Also,experimentation is not feasible since we are dealing with acontinuous chemical process. The team needs to beinnovative enough to take care of the interactions.

Note that, there is no point in rejecting such problems on the

ground that these are primarily control problems (as identifiedduring the measure phase) and hence not ideal forbreakthrough. This is because, one finds it hard to controlsuch processes, particularly when the inputs vary a great dealdue to their natural or agricultural origin.


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Improve Phase Road Map

Find Optimal SolutionBrainstorm to Generate

Potential Solutions

Evaluate Potential Solutions

and Select the Optimum

Plan and Conduct DOE

Find Optimal Solution

Conduct Confirmatory Trials

Solution Confirmed?

Compare Expected Benefits

and Project Goals

A1 A2 A3

Go back to

appropriate stepof the Analyze orImprove phase

B

No

Yes

Find Optimal Solution

Formulate Innovative data

Collection and Analysis Plan

A4


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Improve Phase Road Map (Contd.)

Set Operating Tolerances

Assess Risks

Risk Acceptable?

Pilot Solutions

Results Satisfactory?

Develop Implementation Plan

Results Acceptable?

Go back to appropriatestep of the Analyze or

Improve phase

Go back to Definephase or close projectand document failure

B

Exit to Control Phase

Yes

Yes

Yes

No

No

No

A5


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Learning and Experimentation

Chinese Proverb

I hear and I forget

I see and I remember

I do and I understand


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Theory and Experiment Experiments complement, supplement and

stimulate theoretical research

Conceptualization

Experimentation

No sense of measurement

Thought experiment

Creation of instruments Experimental set up

(hardware and software)

Measurement

Theory

Propositions

Hypotheses

Observations

(Data)

Experiments

may betriggered by a

specific

societal need.

There may be

no concern foror relevance

to theory


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Role of Experimentation

Three Broad Objectives

Exploration/Discovery

Results are compared against the whole body ofknowledge

Investigation/Characterization/Optimization

Comparison of a set of treatment results

Verification/Confirmation/Demonstration

Results are compared against

standard/known/predicted values


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Black Box Model of Experimentation

PROCESSControlFactors

Responses

X1X2X3X4

Xp

.

.

Y1

Y2

Y3

Yr

.

.

. .Z1 Z2 Z3 Z4 Z5 Z6 Z7 Z8 Zq

Noise Factors

(Y1, Y2, ...., Yr) = f (X1, X2, ..., Xp; Z1, Z2, , Zq)


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How to ExperimentAn Example Consider a simple experiment to find the soaking temperature

that will maximize hardness of a steel component. Assumethat the experimenter selects three levels of soakingtemperature9500C, 10000C and 10500C forexperimentation.

Comparing with our Black Box model Control factor: Soaking temperature

Response: Hardness

Noise factors: Soaking time, Chemical composition, Heating curve,Ambient condition, Measurement and a host of other factors

Even in case of such a simple experiment, it is necessary toplan the experiment carefully


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Heat Treatment Experiment

Planning Questions

How many trials at each soaking temperature?

How many test pieces from each trial?

How many measurements on each test piece?

Should the number of trials / test pieces / measurements besame for each level of soaking temperature?

How to deal with the potential noise factor like chemicalcomposition?

How to deal with the potential noise factors like soaking time

and heating curve? How to deal with variation in ambient temperature?

Assuming three trials per level, what should be the order ofthe nine trials?

Answers are by no means trivial!


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Three Principles of Experimentation

Answers the planning questions

REPLICATION

RANDOMIZATION BLOCKING OR LOCAL CONTROL

Common to all experimentswhether single factoror multifactor

Multifactor experiments raise additional designquestions


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Replication

Replication means repetition of a trial

Why replicate?To obtain an estimate of

experimental error Why estimate experimental error?To test

statistical significance of the observed effects

Why test?To gain confidence that thepredicted effects will be realized in practice


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ReplicationPrimary, Secondary,

Soaking temperature 9500C

TP1 TP2M1 M2

TP3M3 M4 M6M5

Primary or Experimental Error

9500C

TP4 TP5M7 M8

TP6M9 M10 M12M11

Similar set up for 10000C and 10500C

Source df

Soaking temperature 2

Trial - Primary error (e1) 1 x 3 = 3

Test piece - Secondary error (e2) 4 x 3 = 12

Measurement- Tertiary error (e3) 6 x 3 = 18

TOTAL 35

Test pieceMeasurement

T

E

S

T

Genuine

repetition

of trials

Heat Treatment Example


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Randomization - 1Heat Treatment Example

Trial # Temp.

1 9502 950

3 1000

4 1000

5 1050

6 1050

TP2

TP1

TP3

M1

M2

M3

M4

M5

M6

Assume eighteen test pieces are cast

from the same heat

How should we allocate the test piecesto the trials?

Randomly! To protect against anyunforeseen bias resulting from the

condition of the test pieces

Having made the allocation, in what order should we conduct the trials?

Randomly! To protect against any unforeseen bias resulting from ahost of uncontrolled factors like furnace and environmental condition


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Randomization - 2Heat Treatment Example

Randomized trial order and

allocation of test pieces

Trial # Temp.

4 950

2 950

5 10001 1000

6 1050

3 1050

TP7

TP10

TP16

M1

M2

M3

M4

M5

M6

How to randomize?

Mechanical procedureDrawing

of numbered chips from a lot

Published random number table

MathematicalRAND function

in Excel

Avoid mental randomization!


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Randomization - 3

Gain

Protection from unforeseen bias Valid estimate of experimental errorif more than one replicate

is available

Higher the non-homogeneity of experimental material (e.g. onetest piece from each heat), the larger is the gain

Loss Simplicity of logistics

Protection from systematic bias only - Despite our besteffort, outliers may creep in!

Gain and loss from randomization


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Blocking Or Local Control - 1

How should we deal with the noise factors likesoaking time and heating curve?

Of course, as far as possible these are to be kept fixedat their standard operating level. Care should beexercised, so that they do not vary from trial to trial.

This is called local controlobjective is to block the

effect of the noise factors. Note that, these two can be control factors in other

experiments



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How should we deal with the noise factorchemicalcomposition?

We need 6 x 3 = 18 test pieces. How do we select them? All the eighteen test pieces from the same heat?

Local control of a noise factor should be avoided.Reproducibility of the results may be compromised

All the test pieces may not be obtainable from a single heat

One piece each from eighteen different heats? Results may vary too much on account of variation in

chemical composition itself. Effect of soaking temperature,the control factor of interest, may not be detected

Solution for both the above problems follows:

Blocking Or Local Control - 2Heat Treatment Example


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Blocking Or Local Control - 3

H1 H2 H3

950C TP12 TP24 TP36

1000C TP13 TP21 TP32

1050C TP15 TP22 TP35


Select three different heats (H1, H2, H3)covering as muchmaterial variation as possible

Select six test pieces randomly from each heat and allocate these(TP11, .., TP16, , TP31, , TP36) to the trials as follows

H1 H2 H3

950C TP14 TP23 TP31

1000C TP11 TP26 TP34

1050C TP16 TP25 TP33

Replicate 1 Replicate 2

Fair comparison of temperature levels is possible - WHY?

H1, H2 and H3 are called BLOCKS. Hence the name blocking


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Blocking Or Local Control - 3 How does blocking affect data analysis?

For simplicity, consider only one replicate and onemeasurement / test piece in our heat treatment example

H1 H2 H3

950C (TP12) Y11 (TP23) Y21 (TP33) Y31

1000C (TP13) Y12 (TP21) Y22 (TP32) Y32

1050C (TP11) Y13 (TP22) Y23 (TP31) Y33

Separate heat treatment ofeach test piece and conductingthe nine trials in random order

Two-way ANOVA with oneobservation per cell

Occasionally the block factor impose a restriction on randomizationof trial order

For example, if day is chosen as a block factor then randomizationof trial order is possible only within a day

ANOVA should reflect such restriction on randomization


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Types of Experiment Observational

Past QC records or observation of the factor levels and thecorresponding outcome without making any intervention

(1) Limited blocking/local control, if any (2) No randomization of

trial orderhence no valid estimate of experimental error (3)Approximate replicates

Correlation among the independent variables and autocorrelationof the dependent variables

Even if a factor is found significant, one must be careful to infer

cause and effect relationship Manipulative

Factor levels are changed as per plan and responses noted

Follows the three principles of experimentation

Should be preferred, wherever possible


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Types of Observational Studies Prospective

Sampling points are either predetermined randomly (Y yet to beobserved) or determined randomly from all the available records

Autocorrelation may be avoided by spacing out the sampling

points. Blocking of the timeline and drawing samples randomlyfrom each block is a good practice

Regression analysis

Retrospective

Y values are always available. Observations are classified

depending on the value of Y, say defectives and non-defectives. Random samples are drawn from each group and then the X

values of each group are compared to find the significant Xs, if any

Rare event or the X values are generated after sampling

Discriminant analysis


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Manipulative Experiment Henceforth by an experiment we shall mean a manipulative

experiment

Observational studies are usually conducted during theANALYZE phase

Usually, multifactor experiments play the most important roleduring the IMPROVE phase

By a multifactor experiment we shall mean an experimentinvolving more than one control factors

An experiment involving one control factor and one or moreblock factors is not a multifactor experiment

Rest of the material is devoted solely to multifactor experiments


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Multifactor ExperimentsAn ExampleAn experiment to increase hardness of a die cast engine component

Factor Code Factor Level 1 Level 2

A % Cu 0.10 0.20

B % Mg 0.05 0.07

C % Zn 0.03 0.06

D Water Cooling On Off

E Air Cooling On Off

Which factors affect hardness?

What is the rank of the factors withrespect to their impact on hardness?

What is the best factor level combination?

How much improvement can we expect atthe best factor level combination?

Response: Rockwellhardness (B scale)

Qualitative factor

How to designthe experiment?


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Traditional Approach

(One-factor-at-a-time)

Trial No. A B C D E Hardness Remark

1 1 1 1 1 1 56

2 2 1 1 1 1 63 A2 is better

3 2 2 1 1 1 68 B2 is better

4 2 2 2 1 1 69 C1 is better

5 2 2 2 2 1 72 D2 is better

6 2 2 2 2 2 75 E1 is better

OptimumCombinationA2B2C1D2E1

To what extent are the results of this experiment reproducible?

A2 is found better while the other factors are at B1C1D1E1. Is itreasonable to assume that A2 will be better even at B2C1D2E1, the

recommended optimum?


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Response GraphsA B AB

A1 A2

B1

B2

Avera

geresponse

A B AB

A1 A2

B1

B2Avera

geresponse

Averagerespon

se

A B AB

A1 A2

B1

B2

Averageresponse

A B AB

A1 A2

B1

B2

Averagerespon

se

A B AB

A1 A2

B1

B2

A B AB

A B AB

?

?

Three levelfactors ?


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Analysis of 2kExperiment

An Example

Single replicate of a 24

experiment

Response: Filtration rate of

a chemical produced in apressure vessel (to be

maximized)

Factors: Temperature (A),Pressure (B), Reactant

concentration (C), Stirringrate (D). Pilot plant

experiment

Level codes: Low (1), High(2)

Standard

order

A B C D Filtration

rate (Y)

(1) 1 1 1 1 45

a 2 1 1 1 71

b 1 2 1 1 48

ab 2 2 1 1 65c 1 1 2 1 68

ac 2 1 2 1 60

bc 1 2 2 1 80

abc 2 2 2 1 65

d 1 1 1 2 43

ad 2 1 1 2 100

bd 1 2 1 2 45

abd 2 2 1 2 104

cd 1 1 2 2 75

acd 2 1 2 2 86

bcd 1 2 2 2 70

abcd 2 2 2 2 96


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The 24 ExperimentANOVA Table

?15TOTAL

00Error

?1ABCD

1BCD

1ACD

1ABD?

1ABC

1CD

1BD

1AD

1CA

1BC

?

1AB

1D

1C

1B?

1A

SSdfSource

Eight Y valuescorrespondingto level 2 of A

Eight Y valuescorrespondingto level 1 of A

A2A1Form four

one way

tables Cell Total Cell Total

Form six

two way

tablesFour Y valuesFour Y values

B2

Four Y valuesFour Y valuesB1

A2A1

Cell Total

Cell Total

Cell Total

Cell Total

SSA =

(Cell Total)2 / 8 - CF

SSAB =

(Cell Total)2 / 4

CFSSA - SSB

Form four

three way

tables

SSABC =

(Cell Total)2 /2CFSSASSBSSC

SSABSSBC - SSAC

By Subtraction

TSS = Y2

- CF


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2kExperimentComputing SS

Computing interaction SS from the 2-way,3-way,..,

k-way tables is obviously very tedious

Yates method is very helpful for easy computationof the SS

We wont discuss Yates method, since MINITAB

is available

However, it will be instructive to compute the SS

from the effect contrasts using Excel


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The 24 Experiment - Computing

Factorial Effects and SS

Trial No. A B C D Y

1 - 1 - 1 - 1 - 1 45

2 +1 - 1 - 1 - 1 71

3 - 1 +1 - 1 - 1 484 +1 +1 - 1 - 1 65

5 - 1 - 1 +1 - 1 68

6 +1 - 1 +1 - 1 60

7 - 1 +1 +1 - 1 80

8 +1 +1 +1 - 1 65

9 - 1 - 1 - 1 +1 43

10 +1 - 1 - 1 +1 100

11 - 1 +1 - 1 +1 4512 +1 +1 - 1 +1 104

13 - 1 - 1 +1 +1 75

14 +1 - 1 +1 +1 86

15 - 1 +1 +1 +1 70

16 +1 +1 +1 +1 96

Y TotalL1 474 548 521 5021121

Y TotalL2 647 573 600 619

Recoded Design Matrix

1 -1, 2 +1 Let the factor-level totalsgiven at the last two rows ofthe table be denoted by F

i

Effect of A = EA = A2A1

= (547/8)(474/8) = 21.625

SS due to A = SSA= (A

2)2/8 + (A

1)2/8CF

= 4742/8 + 6472/811212/16

= 1870.563Effect and SS due to otherfactors are obtained similarly


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Effect Contrast Contrast for main effect of B

= Column B x Column Y= - Y1 - Y2 + Y3 + Y4 -Y5 - Y6 + Y7

+Y8 - Y9 - Y10 + Y11 + Y12 - Y13- Y14 + Y15 +Y16 = 573548 = 25

Esource = 2*Contrastsource/n

SSsource = (Contrastsource)2/n

n = Total number of observations

EB = (2*25)/16 = 3.125

SSB = (252

)/16 = 39.0625 We can verify thatSSB = (B1

2+B22)/8 - CF = 39.0625

(A12 +A2

2)/n(A1+ A2)2/(2*n)

= (A1A2)2/(2*n)

ContrastAB= ? . ContrastABCD= ?

Trial No. A B C D Y

1 - 1 - 1 - 1 - 1 45

2 +1 - 1 - 1 - 1 71

3 - 1 +1 - 1 - 1 48

4 +1 +1 - 1 - 1 65

5 - 1 - 1 +1 - 1 686 +1 - 1 +1 - 1 60

7 - 1 +1 +1 - 1 80

8 +1 +1 +1 - 1 65

9 - 1 - 1 - 1 +1 43

10 +1 - 1 - 1 +1 100

11 - 1 +1 - 1 +1 45

12 +1 +1 - 1 +1 10413 - 1 - 1 +1 +1 75

14 +1 - 1 +1 +1 86

15 - 1 +1 +1 +1 70

16 +1 +1 +1 +1 96

Y TotalL1 474 548 521 5021121

Y TotalL2 647 573 600 619


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Contrast and SS of InteractionsA B C D AB ABC ABCD Y

- 1 - 1 - 1 - 1 +1 - 1 +1 45

+1 - 1 - 1 - 1 - 1 +1 - 1 71

- 1 +1 - 1 - 1 - 1 +1 - 1 48

+1 +1 - 1 - 1 +1 - 1 +1 65

- 1 - 1 +1 - 1 +1 +1 - 1 68

+1 - 1 +1 - 1 - 1 - 1 +1 60

- 1 +1 +1 - 1 - 1 - 1 +1 80

+1 +1 +1 - 1 +1 +1 - 1 65

- 1 - 1 - 1 +1 +1 - 1 - 1 43

+1 - 1 - 1 +1 - 1 +1 +1 100

- 1 +1 - 1 +1 - 1 +1 +1 45

+1 +1 - 1 +1 +1 - 1 - 1 104

- 1 - 1 +1 +1 +1 +1 +1 75+1 - 1 +1 +1 - 1 - 1 - 1 86

- 1 +1 +1 +1 - 1 - 1 - 1 70

+1 +1 +1 +1 +1 +1 +1 96

474 548 521 502 560 553 5551121

647 573 600 619 561 568 566

ColAB = ColA x ColBColABC = ColAB x ColCColABCD = ColABC x ColD

Columns for other interactions are

obtained similarly ContrastAB = ColAB x ColY

= 561560 = 1

SSAB = 12/16 = 0.0625 (same as

obtained before from two way table)

SSABC = (568 - 553)2/16 = 14.0625

SSABCD = (566555)2/16 = 7.5625

SS for the other components can beobtained similarly


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The 24 ExperimentFinal ANOVASource df SS

MS F%

A 1 1870.5625 1870.562595.86** 32.3

AC 1 1314.0625 1314.062567.34** 22.6

AD 1 1105.5625 1105.562556.66** 19.0

D 1 855.5625 855.562543.85** 14.6

C 1 390.0625 390.0625

19.99** 06.5ABD 1 68.0625

68.0625B 1 39.0625

39.0625BCD 1 27.5625

27.5625BC 1 22.5625

Pooled

F.05,1,10 = 4.96F.01,1,10 = 10.04

% =SSSourcedfError x MSError

TSSx 100

% indicates

importance of the

component

Higher the F ratio,

the higher will be

%. But F ratio is

difficult to interpret


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The 24 ExperimentNPP

Effect

Percent

20100-10-20

99

95

90

80

7060504030

20

10

5

1

Effect Type

Not Significant

SignificantAD

AC

DC

A

Normal Probability Plot of the Effects(response is filtration rate, A lpha = .05)

Lenth's PSE = 2.625

MINITAB output. We can computethe effects from the contrasts asdescribed before

Insignificant effects are smaller inmagnitude and tend to be centered

around zerothe fitted line

Significant effects are larger inmagnitude and fall far away fromthe fitted line

The same five effects as found bypooling are found to be significant

NPP should not be viewed as an alternative to pooling, since we need an estimate oferror for constructing prediction band for the best combination. Lengths PSE(Pseudo Standard Error) in MINITAB output appears to be too conservative

NPP can be used for having better judgment on pooling

In many other cases, finding significant effects from NPP will not be as easy as above


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Practical Limitations of Full Factorials

Let us consider a situation where we wish to

investigate 13 factors, each at three levels

A FF experiment will require 313 trials

313 = 1594323

Even if it takes only 10 minutes to conduct a

trial, the complete experimental results will

be available only after 60 years!!

Industrial experiments involving ten or

more three level factors is common


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The Alternative Two practical alternatives

Fractional Factorials (2k-p, 3k-p designs)

Orthogonal Arrays (OA designsL4, L8, L16, L18, )

Above classification is conventional, but somewhatarbitraryFFs are orthogonal and OAs are saturated FFs

Here we shall discuss only OA designs developed by G.Taguchi because of their ease of construction and

flexibility Taguchis OA designs are available in MINITAB

3k-p and mixed level FFs are not available in MINITAB


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Orthogonal Array L8

1 2 3 4 5 6 7

1 1 1 1 1 1 1 1

2 1 1 1 2 2 2 2

3 1 2 2 1 1 2 2

4 1 2 2 2 2 1 1

5 2 1 2 1 2 1 2

6 2 1 2 2 1 2 1

7 2 2 1 1 2 2 18 2 2 1 2 1 1 2

One factor is assigned to one column -Maximum of seven two level factors

1s and 2s in each column - Two levelsof the factor assigned to the column

Each row constitute a trial

Assume four two level factors (A-D)are assigned to the first four columns

First trial: A1B1C1D1 Eighth trial: A2B2C1D2

Vacant columns are used for estimatingeither interactions or experimental error

Columns 1, 2 and 4 constitute the 23 design

1/16th fraction of the 27design or 27-4 design


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Orthogonal Array Series

Two level series: L4(23), L8(2

7), L16(215),

L16(215): 16 rows, 15 columns, each column consists of 1s

and 2s only. Similarly for other arrays

Three level series: L9(34), L27(313),

L9(34): 9 rows, 4 columns, each column consists of 1s, 2s

and 3s.

Mixed level series: L18(2

1

x 3

7

), L36(2

11

x 3

12

), . L18(2

1 x 37): 18 rows, one 2-level column and seven 3-

level column

In short, OAs are represented as L8, L16, L18,


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Meaning of Orthogonality

Orthogonal means balanced or seperabe or unbiased

The idea of balancing for clean separation of alternativesis used in many forms of experiments with which we are

familiar In football games, the changing of field and each team

getting a chance to kick-off is an act of balancing to avoidbias

Technically, orthogonality implies that in any pair ofcolumns the all possible factor-level combinationsappear with the same frequency

To explain, consider the array L9


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Designing a Simple OA Experiment

Step 1: Identify the control factors, noise factors and theresponse

Control factors are those factors whose optimal level can be fixedand monitored during experimentation as well as in actual practice

In case of field experiments, select about 5 -12 control factors.Reproducibility of results will be poor with less than five factors.Experimental error and cost of experimentation may be very largewith more than twelve factors.

The Cause and Effect Diagram can be very useful in identifyingimportant control and noise factors.

Identification of the noise factors is important for having properrandomization and blocking schemes.

Selection of the response is not trivial, although it may seem so.However, considerations for selecting proper response is beyond thescope of this course.


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Factor

Code

Factor Level 1 Level 2

A % Cu 0.10 0.20

B % Mg 0.05 0.07

C % Zn 0.03 0.06

D Water Cooling On Off

E Air Cooling On Off

Die Casting

Experiment

Response:

RockwellHardness(B scale)

Example: Factors, levels and response

Levels can be created using any scalecontinuous, discrete, ordinal or nominal

All the factors need not have the same number of levels. Such situations will be

discussed later


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Step 3: Decide on the interactions to be estimated

Experimentation keeping provisions for estimating too many

interactions (say more than three) is not a good practice.

Interactions are best dealt with through proper selection of factors and

their levels (Steps 1 and 2). Use supplementary measurements and

sliding level technique to deal with interactions. These techniques will

be explained later through examples and case studies.

Keep provisions for estimating only important interactions, which

cannot be tackled otherwise.


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Step 4: Draw the required linear graph

Required linear graph of the die casting experiment)

A BAxB

C D E

Poor Construction of Linear Graphs

Both the following representations are poor

A

B

A x BC

D

EA B

A x B

C D E


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Step 5: Choose an appropriate OA

Compute the total degrees of freedom (TDF) needed to estimate all

the main effects and the desired interactions. Compute Minimum Run

Size (MRS) and then the Desirable Run Size (DRS).

MRS = TDF + 1 DRS = MRS + 4 (error degrees of freedom)

Choose an OA from the appropriate series and of size >=DRS or of

size >=MRS and replicate. If the number of factors involved is ten or

more then use the MRS criteria.

For our die casting experiment, TDF = 5x1(five main effects) +1x1(assuming A x B is also desired) = 5 + 1 = 6. MRS = 6 + 1 = 7.

DRS = 7+4 = 11. Thus we have to choose either L16 or replicate L8.


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Choosing the right OAThe MRS criteria may fail

Die casting example: MRS= 7. So one more interaction can be

accommodated in L8. Can we accommodate one of AC, AD, BD etc?

YES. Can we accommodate one of CD, DE etc? NO.

Lesson: Two or more independent edges cannot be estimated usingL8. We have to use the array L16 in such cases.

Similarly, independent interactions involving two 3-level factors

cannot be estimated using L27 (see the standard LG of L27).

Another extension (2-level factors): Interactions forming two or

more independent triangles cannot be estimated using L16 . The

array L32 is necessary for this purpose (see standard LGs).


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Choosing the right OAWhen the DRS

criteria fail

Modify the required LG by dropping a few interactions so

that it becomes possible to use a smaller array.

Using a larger array for estimating one or two specific

interactions is likely to be unnecessary wastage of

resources.


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Step 6: Allocate the factors to the columns of the chosen

OA and construct the OA layout

If no interactions are present and complete randomization of the trials

is possible then any factor can be allocated to any of the columns.

When interactions are present, it will be convenient to proceed as

follows:

6.1: Choose a standard linear graph and modify it to match the required linear

graph

6.2: Allocate the factors to the nodes of the modified LG by comparing the

modified LG with the desired LG. 6.3: Construct the OA layout of the experiment

Alternatively, the interaction table of the chosen OA can also be used

for the purpose of allocation


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6.1: The required linear graph and its modification

A BAxB C D E

Required Linear Graph (Die casting experiment)

2 46

1

7

3 5

4 5 6 72

1

3

Standard Linear Graph (L8) Modified Linear Graph


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6.3: OA layout of the experiment

1 2 3 4 5 6 7

1 1 1 1 1 1 1 1

2 1 1 1 2 2 2 2

3 1 2 2 1 1 2 2

4 1 2 2 2 2 1 1

5 2 1 2 1 2 1 2

6 2 1 2 2 1 2 1

7 2 2 1 1 2 2 1

8 2 2 1 2 1 1 2

eEDAxB CBA

Column No.

Source

Preserve the

layout. It will

be useful

during data

analysis


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Step 7: Decide the number of replications and repetitions

Strictly speaking, to be decided based on expected experimental error

(process capability) and the least effect size that we want to detect

But rarely done in practice. Mostly decided based on cost of

experimentation and experimental budget

A block factor may be associated with the replicates

Recall that the experimental error is variation between trials and

repetition error is the variation within a trial

It is always advisable to have repetitions since repetitions are usuallyeasy to make more than one part/sample from each trial and more

than one measurement on each part

Measurement error is a common to both primary and secondary error.

So, it is extremely important to check for measurement adequacy


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Conducting the Experiment

Step 1: Construct the physical layout of the experiment

Delete the interaction and error columns from the OA layout

Substitute the actual levels of the control factors in place of the coded

levels of the OA layout

Trial

No

(1)

%Cu

(2)

%Mg

(4)

%Zn

(5)

WC

(6)

AC

1 .1 .05 .03 ON ON

2 .1 .05 .06 OFF OFF

3 .1 .07 .03 ON OFF

4 .1 .07 .06 OFF ON5 .2 .05 .03 OFF ON

6 .2 .05 .06 ON OFF

7 .2 .07 .03 OFF OFF

8 .2 .07 .06 ON ON

Physical layout

of our diecasting example


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Conducting the Experiment

Step 3: Design the experimental log sheet

The physical layout + the data column + the provisions for recording

special experimental conditions

Step 4: Conduct the trials and record results Exercise local control

Be careful in setting the levels of the control factors correctly. But no

special care should be exercised

Preserve the experimental units, input material etc., wherever feasible

Step 5: Repeat erroneous and missing trials, if possible Statistical treatment of missing values is difficult

Misplacement and breakage of parts before measurement may lead to

missing values


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Analysis of Experimental Data and

Confirmation of Results

Step 1: Identify significant effectsANOVA

Step 2: Find the best factor-level combination

Effect curves, production cost, operating cost

Step 3: Predict expected response at the best

combinationSimple prediction formulae or

regression analysis

Step 4: Confirm predictionsA small confirmationrun and comparison with the predictions


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Analysis of the Die Casting Experiment

Column # 1 2 3 4 5 6 7 Interaction AB

Source A B AB C D E e B1 B2 Effect

Level 1 549 573 579 555 537 591 567 A1 288 261 - 6.75

Level 2 576 552 546 570 588 534 558 A2 285 291 + 1.50

T = Total of the 16 observations = 1125

RSS = Raw sum of squares of the 16 observations = 712 + 732+ + 722 = 79685

CF = T2/16 = 11252 / 16 = 79101.56, TSS = RSSCF = 583.44 (df = 15)

A1 = Total of 8 observations at A1 = 549, A2 = Total of 8 observations at A2 = 576

SScol1 = SSA = (A1)2/8 + (A2)2/8CF = 45.56 (df = 1)

Similarly for the other columns / sources SSe1 = SScol7 = 567

2 /8+ 5582/8CF = 5.07 (df=1)

SS (pure error) = SSe2 = TSS - j Sscolj = 583.44(45.56+ +5.07) = 57.50 (df=8)

SSe2 can also be computed as ?

SSreplicate need not be computed since the 16 trials have been randomized completely

ANOVA table follows


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Analysis of the Die Casting Experiment

Source df SS MS F %A 1 45.6 45.6 6.55* 6.6

B 1 27.6 27.6 3.97 3.5

C 1 14.1 14.1 2.02 1.2D 1 162.6 162.6 23.39** 26.7

E 1 203.1 203.1 29.21** 33.6

AB 1 68.1 68.1 9.79* 10.5

e1 1 5.1 5.1 - -

e2 8 57.5 7.19 - -

(Poolederror)

(9) 62.6 6.95 - 17.9

Total 15 583.4 - - 100

ANOVA Table

% =SSsourcedfsource x MSerror

TSSx 100

Factors D and E are most

important. Factor A and

the interaction AB also

play significant role

* Significant at 95% level of confidence

** Significant at 99% level of confidence


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Prediction of expected response Main effect (significant): Expected response at A1 and A2 are

.)()( 2211 lyrespectiveTATAandTATA

If the factor A is insignificant then both are

zero and hence the expected response is just T-bar.

)()( 21 TAandTA

Interaction effect (significant): jijiji BABATBAofEffect )(

Assuming the effects of A, B and AB are additive, the expected

response at AiBj is then given by

E(Y) = (T-bar + Effect of Ai + Effect of Bj + Effect of AiBj)

Case 1: A, B and AB are significant: jijijiji BATBABATBTATYE )()()()(

Case 2: Only A and AB are significant: )()()()( jjijijii BBATTBABATATYE

Case 3: Only B and AB are significant: )()( iji ABATYE

Case 4: Only AB is significant:jiji BABATYE 2)(


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Die Casting Experiment: Prediction of

Mean Response at the Optimum

Optimum combination: A2B1C1D2E1

69.7831.7088.7350.7563.7125.71

)()()()()(

12112

121212212112

TEDBBA

TETDTBABATATYE EDCBA

Existing combination: A1B1C1D1E1

07.7131.7088.7313.6763.7100.72

)()( 1111111111

TEDBBAExistingYE EDCBA

Gain: (78.69 -71.07) RC = 7.6 RC

+ GREATER ROBUSTNESS


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i i i fid


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Die Casting Experiment : Confidence

Interval

From F table, F0.05, 1, 9 = 5.12

From ANOVA table, Ve = 6.95

ne = 16/(df of T + df of A + df of AB + df of D + df of E)

= 16/5 = 3.2 Thus 95% of the individual observations of the

confirmation run should lie within 78.69 6.82 RC.

Confirmation failure indicates that either the factorial

effects are not additive or important interactions havebeen ignored.


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Optimization: Complex Situations

Example 1: Consider an experiment involving two

responses Y1 and Y2. Assume A2 is better than A1

with respect to Y1 but A1 is better than A2 with

respect to Y2. How should we choose the best levelof A?

Example 2: Consider three factors A, B and C. The

combinations A1

B1

and B2

C1

are found to be the

best. How can we find the overall best combination

from the average response table?


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Analysis of 3-Level OA Experiments

Each main effect has two df and each interaction has 2 x 2 =

4 df.

SSA (for example) = (A12 + A2

2 + A32)/rCF, r = No. of

observations in the total Ai. If the two columns representing the interaction between two

3-level factors are available in the OA layout then the

interaction SS may be obtained by adding the SS of the two

columns. Alternatively, the interaction SS may be obtained

from the two-way table.

If needed, each SS can be partitioned further into components

of 1 df each.


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Analysis of 3-Level OA Experiments

For example, SSA (2df) = SSAL (1df) + SSAQ (1df)

SSAL = (A3 totalA1 total)2 / (2r), r = No. of observations in

Ai total.

SSAQ = (A1 total + A3 total2*A2 total)2

/ (6r) SSAB (4df) = SSALBL (1df) + SSALBQ (1df) + SSAQBL (1df) +

SSAQBQ (1df)

Partitioning as above will be useful when the available error

df is very limited Note that partitioning of SS into components of 1df is

meaningful only for quantitative factors


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Modification of OA: Mixed-Level Factors

Collapsing of Columns

To create a higher level column within a

lower level array Creating a 4-level column in a 2-level array

Creating a 8-level column in a 2-level array

Creating a 9-level column in a 3-level array

Creating a 6-level column in L18.


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Compound Factor Method

A1B1 (AB)1, A2B1 (AB)2, A1B2 (AB)3

Main effect of A = (AB)2(AB)1

Main effect of B = (AB)3(AB)1

SS(AB) = [{(AB)12 + (AB)22 + (AB)32}/(3r)]CF Assuming the assignment is to L9 and r is the no. of replicates

SSA = [(AB)1(AB)2]2/(6r)

SSB = {(AB)1(AB)3]2/(6r)

Note that SS(AB) SSA + SSB, since A and B are not orthogonal

If any of the two factors (say B) is found insignificant, then we may

consider (AB)1 = A1, (AB)2 = A2 and (AB)3 =A1 and recalculate SSA as

applicable for a dummy technique


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Effect Curves: 3 x 2 Experiment

111.67

129.33

118.33

110

122.33

132.33

105

110

115

120

125

130

135

A1 A2 A3

BHN

B1 B2

How can we avoid the interaction?


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Background

The bulk gas transmission and distribution network

consists of thirty two flow meters, of which two are

turbine meters and the rest are orifice meters.

The whole network can be partitioned into severaloverlapping segments. The gas balancing equation

for each such segment is given by

Unaccounted Gas (UAG) = Gas OutGas In

Gas In is the total flow measured by the input meters

Gas out is the total flow measured by the output meters

plus the Line Pack component


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Background

The company monitors daily UAG (as % of Gas

In) in all the segments. This project was undertaken

when the %UAG dropped to -0.6%.

This amounted to a loss of about Rs. five millionsper month.

It should however be mentioned that the real loss

may not be as high as above, since all the gas outpoints are not billing points


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Intermediate Studies

The study consisted of systematic investigation of the process

to identify the root causes for high variation of UAG. The

details of this investigation, consisting of many small studies,

will not be discussed here.

The main result that was obtained at the end of these studies

is that the UAG% could be kept near zero if the gas

temperatures at two particular stations could be controlled in

a particular fashion.

Such a result was surprising since the meters are supposed to

record flow in standard condition.

This led us to the flow validation study, discussed next.


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Flow Validation

Two main sources of errorError in on-line measurement of

gas composition, pressure, temperature and specific gravity

AND error in flow computation by the flow computers.

Detailed scrutiny of the calibration record eliminated the first

possibility.

Thus, although all the flow computers connected to the

SCADA (Supervisory Control And Data Acquisition) are

AGA -3 and AGA-7 compliant, it was decided to validate the

flow computations by the flow computers.


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Flow Validation Experiments

The array L25 was used to study the effect of six factors on

the error in flow computation.

The levels of the factors were chosen to cover the entire range

of operating conditions.

Five levels were selected for each factor since the operating

range was large and the theoretical flow computation

equations involved fourth order terms.

For each of the 25 experimental conditions, flow was

computed by each of the five computers and the same

compared with the corresponding standard value provided by

a standard laboratory.


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Factors and Levels

Factor UnitLevel

1 2 3 4 5

Tube Diameter (TD) mm 40 110 180 250 320

Diameter ratio (DR)* - 0.3 0.4 0.5 0.6 0.7

Absolute pressure (PR) KPa 500 2000 3500 5000 6500

Differential Pressure (DP) KPa 0.2 25 50 75 100

Temperature (TE) C 10 20 30 40 50

Specific Gravity (SG) - .50 .55 .60 .65 .70

* Orifice Diameter (OD) = TD*DR

Note that the factor OD has sliding levels. For example, the levels of OD

are 12, 16, 20, 24 and 28 when TD = 40 mm but the levels are 96, 128,

160, 192 and 224 when TD = 320 mm.


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Experimental Results

Trial

#

Flow (SCMH)

Bristol

Babcock

Flow Boss

600

Flow Boss

503ROC 809

Turbo

2500

Apex

(Standard)

1 41.5 37.7 42.261 42.049 42.258 42.248

. . . . . . .

. . . . . . .

18 24150.2 19323.07 21847.946 21867.01 21873.03 22054.52

. . . . . . .

25 243562.6 245153.01 247495.359 247428.3 247917.4 247924.7


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Analysis of Experimental Data Let Fi be the flow measured by the i

th computer and S be the

corresponding standard value. The linear model [log(Fi) = i +

*log(S)] was developed for the five computers. Ideally we should

have i = 0 and = 1. It was found that in all the five cases was

nearly unity. Accordingly, further analysis was carried out using the difference Z

= log(S)log(F). Performance summary is given below

Flow Computer Average Bias Mean Square Error Performance Rank

Bristol Babcock -0.01405 0.00072 3 (Worst)

Flow Boss 600 0.01735 0.00066 3

Flow Boss 503 0.00294 0.000063 2

ROC 809 0.00247 0.0000094 1

Turbo 2500 0.00217 0.0000089 1


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Corrective Action and Benefits

The matter was taken up with the supplier of the flow

computers (Bristol Babcock and Flow Boss)

Meanwhile these two computers were taken out of the

system and the flow meters were connected to the other

computers. Such a temporary corrective action resulted inmarked improvement in the variation of UAG.

Apart from reducing variation in UAG, the company now

had a methodology for identifying erratic meters (not

discussed here)

Further, it was realized that the present practice of flow

validation based on measurements made at a particular

condition is inadequate

Ch 6 BB Improve

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