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Transcript of 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
Heat Treatment Example
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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
Heat Treatment Example
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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Designing a Simple OA Experiment
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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Designing a Simple OA Experiment
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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Designing a Simple OA Experiment
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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Designing a Simple OA Experiment
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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Designing a Simple OA Experiment
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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Designing a Simple OA Experiment
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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Designing a Simple OA Experiment
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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Designing a Simple OA Experiment
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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Designing a Simple OA Experiment
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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Designing a Simple OA Experiment
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