Robust Product and Process Designs - Quality …nutek-us.com/Robust_Design_Slides_0910.pdfDeming Juran Taguchi Ishikawa •TQM, SPC, Taguchi Methods, Fishbone Diagram …. aGenichi

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Robust Product and Process DesignsRobust Product and Process Designs

Instructor: Ranjit Roy◦ Mechanical engineer

IntroductionRef. Page N/A

◦ Industrial experience since 1973.◦ Independent consultant since 1987◦ Specializes in product and process design

improvement technique◦ Published books and developed technical software◦ Adjunct professor (Oakland University, Rochester,

MI since 1976)

Nutek, Inc. 3829 Quarton Road, Suite 102

Bloomfield Hills, MI 48302, USA.Ph: 248-540-4827, E-Mail: [email protected]

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What is Robustness?What is Robustness?

Robust Product & Process designs

Ref. Page 1-1

Sturdy, Rugged,Consistent, Dependable,..

What are examples of robust products and processes?

Built Strong to Last Long

Nutek, Inc. All Rights Reserved Robust Product and Process Designs www.Nutek-us.com Version 0804

Why design robust products?g g

It takes the licking, but keeps on ticking

Product Quality and RobustnessProduct Quality and Robustness

A desirable characteristic in a quality product is CONSISTENCY OF PERFORFORMANCE.

Ref. Page 1-2

What can deliver consistent performance?

• By definition, robust product s and processes deliver higher degree of consistency in performance.

Wh d t d i d b t it


• When products are designed robust, it performs consistently all the time, under all kinds of application environment.

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How do you design robust

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products and processes?

What is the approach?


Enemy of Consistency of PerformanceEnemy of Consistency of Performance

Performance is considered consistent when it suffers least variation.

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when it suffers least variation.

Variation in performance is common to all products, even among those that are machine-made with highest of precisions.

Variation may be due to many sources.


However, the primary cause of variation is the INFLUENCE of uncontrollable factors (called NOISEfactors)

Fish-Bone Diagram

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Variation Due to Noise FactorsVariation Due to Noise FactorsA direct way to reduce variation is to (1) eliminate the noise factors or (2) keep it from varying

Ref. Page N/A

from varying.

Unfortunately, neither is generally an option.

NOISE:- Cannot be eliminated- Cannot be controlled or held it fixed


Therefore, NOISE is expected to do its part and always tend to influence performance.

The Taguchi ApproachThe Taguchi Approach

“While NOISE often cannot be controlled or it is too expensive to control, perhaps, by some means,

Ref. Page N/A

its INFLUENCE to performance can be minimized.”

How do you control its influence without controlling the noise factor itself?

Dr. Taguchi suggested a revolutionary idea.


- Leave noise alone- Find proper levels of the controllable factors to minimize influence of noise

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Robust Design StrategyRobust Design Strategy

Reduce variation by minimizing the influence of uncontrollable factors.

Ref. Page N/A

Determine design parameters by selecting a combination of the CONTROL FACTORS such that the performance is insensitive (immune) to the influence of UNCONTROLLABLE FACTORS.


“ Reduce variability without actually removing the cause of variability.”

Desired Output Characteristic Desired Output Characteristic

ROBUST DESIGN

Ref. Page 1-3

Sensitive

OUTPUT

Less Sensitive


Input/Signal

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Robust Design Application MethodologiesRobust Design Application Methodologies

Methods: Undertake experimental studies with hardware or

Ref. Page 1-3

analytical simulation

When: The earlier the better. The return on investment is highest when applied in product design and development phases.

Specific Method:- Taguchi experimental design (PARAMETER DESIGN)- Outer array designs with NOISE factors (Process diagram & Ideal


Outer array designs with NOISE factors (Process diagram & Ideal function)

- Analysis ( Two step optimization, Signal-to-noise, Loss function, etc.)

(First part of this seminar will be dedicated to learning the experimental design techniques)

History of Quality ImprovementHistory of Quality Improvement

Ref. Page N/A

History of Quality Improvement History of Quality Improvement Activities and the Taguchi Activities and the Taguchi

Experimental Design TechniquesExperimental Design Techniques


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Quality & Cost Improvement Quality & Cost Improvement -- Lean Manufacturing History (Timeline )Lean Manufacturing History (Timeline )

Henry Ford Eiji Toyoda Taichi Ohno Shigeo Shingo

Ref. Page N/A

1850 1875 1900 1920 1940 1960 1980 2000

WW-I WW-IICivil War

Eli Whitney1765 – 1825, was an American inventor and manufacturer who is credited with creating the first cotton gin in 1793. • Interchangeable parts

•Assembly Line•Flow Lines•Manufacturing Strategy

•Just in Time and Toyota Production System

•Stockless Production

•World Class Manufacturing

Fredrick Taylor• Standardized work time study & work standards• Worker/management dichotomy

Frank & Lilian Gilbreth•Process Chart•Motion Study

Edward Joseph Genichi Kaoru Deming Juran Taguchi Ishikawa

•TQM, SPC, Taguchi Methods, Fishbone Diagram ….

Genichi Taguchi was born in Japan in 1924.W k d ith El t i C i ti L b t

Who is Taguchi?Ref. Page 1-4

Worked with Electronic Communication Laboratory (ECL) of Nippon Telephone and Telegraph Co.(1949 -61). Major contribution has been to standardize and simplify the use of the DESIGN OF EXPERIMENTS techniques. Published many books and papers on the subject.

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It all began with R. A. Fisher in England back in 1920’s.

What is the Design of Experiment Technique?Ref. Page 1-4

1920 s.Fisher wanted to find out how much rain, sunshine, fertilizer, and water produce the best crop.Design Of Experiments (DOE):

statistical technique studies effects of multiple variables simultaneouslydetermines the factor combination for optimum

lresult

Dr. Taguchi started his work in the early 1940’s Joined ECL to head the research department

Background of Genichi TaguchiRef. Page 1-5

Joined ECL to head the research departmentResearch focused primarily on combining engineering and statistical methods to improve cost and qualityIs Executive Director of American Supplier Institute in Dearborn, Michigan His method was introduced here in the U.S.A in 1980Most major manufacturing companies use it to i lit f d t d d iimprove quality of product and process designs

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DO IT UP-FRONT: Return on investment higher in design

What’s New? Philosophy !Ref. Page 1-5

g gThe best way is to build quality into the design

DO IT IN DESIGN. DESIGN QUALITY IN:Does not replace quality activities in productionMust not forget to do quality in design

Product Engineering Roadmap (Opportunities for Building Quality)

Ref. Page 1-6

Where do we do quality

* Design & Development

* Test & Validation

* ProductionReturn on Investment

improvement?* Design & Analysis

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Driving Questions For Quality ImprovementDriving Questions For Quality Improvement(Opportunities for Building Quality)(Opportunities for Building Quality)

DESIGN: New questions we may ask.


* Test & Validation

* Production

• Is the performance at its best or at optimum?

Will i f h ll h i d

DESIGN: New questions we may ask.

* Customer Requirements and Design Concepts (APQP)

• Will it perform the same way all the time, under all application environment?

• Is the design robust?

•Is the manufacturing process robust & adequate?

Leading Questions in Validation Test PlanningLeading Questions in Validation Test Planning(Opportunities for cost(Opportunities for cost--effective testing)effective testing)

TEST: New questions we may ask


* Test & Validation

* Production

• Will the product perform under extremes of application environment?

H t ff ti l t t d t

TEST: New questions we may ask.

* Customer Requirements and Design Concepts (APQP)

• How can we cost-effectively test products under all conditions before release?

• What is the worst of all possible application conditions?

•Can we produce the optimized products profitably?

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BRAINSTORMING: Plan experiments and follow through

What’s New? Discipline!

Ref. Page 1-6

through.TEAM WORK: Work as a team and not alone.CONSENSUS DECISIONS: Make decisions democratically as a team. Avoid expert based decisions.COMPLETE ALL EXPERIMENTS planned before making any conclusions.RUN CONFIRMATION EXPERIMENTS.

S om e th inking

Typical Old Approach (Series Process)

Ref. Page 1-7

S om e m ore thinking

T ry this


Try that

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Brainstorming

Application Phases (5-P’s)

(Experiment planning session)- What are we after?

Five-Phase Application ProcessRef. Page 1-7

Trial#1 Trial#2

P1. PLAN

P2. PRESCRIBE

P3. PERFORMConduct Experiments

- How do we evaluate performances?- Etc.


Predict Performance by Analysis of Results P4. PREDICT

P5. PROVEVerify and Prove (Confirmation Expts.)

Guiding Principle

“The secret is to work less as individual and more as a team. As a coach, I play not my eleven best, but my best eleven.”

- Knute Rockne


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CONSISTENCY OF PERFORMANCE: Quality may be viewed in terms of consistency of performance. To be consistent is to BE LIKE THE GOOD ONE’S ALL THE TIME. REDUCED VARIATION AROUND THE TARGET Q lit

What’s New? Definition of QualityRef. Page 1-8

REDUCED VARIATION AROUND THE TARGET: Quality of performance can be measured in terms of variations around the target.

This holds true also with performance of

any product or process.

Looks of ImprovementLooks of ImprovementFigure 1: Performance Before Experimental Study Figure 2: Performance After Study

(Y Y )

Improve Performance = Reduce σ and/or Reduce m

m = (Yavg -Yo )

Yavg. Yo

σσnew

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Poor Quality Not so Bad

Being on Target Most of the Time Ref. Page 1-9

Better Most Desirable

MEASURING COST OF QUALITY:

What’s New? Loss Function!Ref. Page 1-9

- Cost of quality extends far beyond rejection at the production

- Lack of quality causes a loss to the society. LOSS FUNCTION : A formula to quantify the amount of loss based on deviation from the target performance.

L = K ( y - y0 ) 2( y y0 )

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APPLICATION STEPS St f li ti l l

What’s New? Simpler and Standardized DOE.

Ref. Page 1-10

APPLICATION STEPS: Steps for applications are clearly defined.EXPERIMENT DESIGNS: Experiments are designed using special orthogonal arrays.ANALYSIS OF RESULTS: Analysis and conclusions follow standard guidelines.

Simpler and Standardized DOE Methodologies

Ref. N/A

“Things should be as simple as possible, but no simpler.”

- Albert Einstein

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PARAMETER DESIGN: Taguchi approach generally refers to the parameter design phase of the three quality engineering

DOE - the Taguchi Approach - Seminar ContentsRef. Page 1-11

the parameter design phase of the three quality engineering activities (SYSTEM DESIGN, PARAMETER DESIGN and TOLERANCE DESIGN) proposed by Taguchi.Off-line Quality Control Quality Loss FunctionSignal To Noise Ratio(s/n) For AnalysisReduced Variability, a Measure Of Quality

EXAMPLE APPLICATIONIt is an experimental technique that determines the solution

How Does DOE Technique Work?

Ref. Page 1-11

It is an experimental technique that determines the solution with minimum effort.In a POUND CAKE baking process with 5 ingredients, and with options to take HIGH and LOW values of each, it can determine the recipe with only 8 experiments. Full factorial calls for 32 experiments. Taguchi approach requires only 8

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Ingredients for Baking Pound CakeFactors Level-1 Level-2

A: EggA2A1

Ref. Page 1-12

gg

B: Butter

C: Milk

D: Flour

B1 B2

C1 C2

D2D1

FIVE factors at TWO levels each make 25 = 32 separate recipes (experimental condition) of the cake.

E: Sugar E1 E2

Experimental Trial Conditions by L-8 Orthogonal ArrayRef. Page N/A

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Experiment Design Using L-8 ArrayRef. Page N/A

3 2-L factors = 8 Vs. 4 Taguchi expts.7 ‘‘ ‘‘ = 128 Vs. 8 Taguchi expts.15 ‘‘ = over 32,000 Vs. 16 ‘‘

Orthogonal Array - a Fish FinderRef. Page 1-12

Fishing Net

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Standardized application and data analysis

Benefits of the Taguchi DOE?

Ref. Page 1-12

Higher probability of successOption to confirm predicted improvementImprovement quantified in terms of dollars

DC DOE-II

DOE/Taguchi Approach, Part I & Part IIRef. Page 1-15

• Interactions

•Loss Function •Problem solving

•Noise Factors, S/N, Analysis

•Robust Designs, ANOVADOE-I

•Experiment using Std. Orthogonal Arrays•Main effect studies and optimum condition

Interactions• Mixed level factors

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Module – 2Basic Concepts in Design of

Experiments


Things you should learn from discussions in this module:

Wh t F t ? [ A Ti B T t t ]

Factor and Level Characteristics Ref. Page 2-1

What are Factors? [ A:Time, B:Temperature, etc.]

What are Levels? [A1= 5 sec., A2= 10 sec. etc.]

How does continuous factors differ from discrete ones?

What are the considerations for determining the number of Levels of a Factor?

How does nonlinearity influence your decision about the number of levels?


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Nature of Influences of Factors at Different Levels

spon

se/

• Minimum TWO levels

Ref. Page 2-2

Res

ult/R

esQ

C

A1 A2 A3 A2

spon

se/

Res

pons

e

• Minimum TWO levels

• THREE levels desirable

• FOUR levels in rare cases

• Nonlinearity dictates levels for continuous factors only


Res

ult/R

esQ

C

A1 A2 A3R

esul

t/R/Q

C

A1 A2 A3 A4

Combination Possibilities – Full Factorial Combinations

ONE 2-level factor offer TWO test conditions (A1,A2).TWO 2-level factors create FOUR (22 = 4 )

NOTATIONS:A (A1,A2) or A represent 2-level factor

Cond.# A B C

Ref. Page 2-3

TWO 2 level factors create FOUR (2 4 ) test conditions A1B1 A1B2 A2B1 A2B2) .

THREE 2-level factors create

EIGHT (23 = 8) possibilities.

A1B1C1 A1B1C2

A1B2C1 A1B2C2

A2B1C1 A2B1C2

Simpler notations for all

possibilities or

Cond.# A B C

1 1 1 1

2 1 1 2

3 1 2 1

4 1 2 2

5 2 1 1

6 2 1 2


2 1 1 2 1 2

A2B2C1 A2B2C2

pfull factorial

6 2 1 2

7 2 2 1

8 2 2 2

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3 Factors at 2 le el 23 8

Full Factorial Experiments Based on Factors and Levels

Ref. Page 2-4

3 Factors at 2 level 23 = 84 Factors at 2 level 24 = 167 Factors at 2 level 27 = 12815 Factors at 2 level 215 = 32,768

What are Partial Factorial Experiments?Wh t O th l d h th d?


What are Orthogonal arrays and how are they used?

How are Orthogonal arrays used to design experiments?

What does the word “DESIGN” mean?

What are the common properties of Orthogonal Arrays?

Orthogonal Arrays– Experiment Design Tool

Ref. Page 2-4

2-Level Arrays

L4 (23 )L8 (27)L12 (211)L16 (215) . . . .

3-Level Arrays

L-4 Orthogonal ArrayTrial # 1 2 3

1 1 1 12 1 2 23 2 1 2 4 2 2 1


3 Level ArraysL9 (34), L18 (21 37) . . .

4-Level ArraysL16 (45) . . . .

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L-4 Orthogonal ArrayTrial #A B C1 1 1 12 1 2 23 2 1 2

Array Descriptions:1. Numbers represent factor levels2. Rows represents trial conditions3. Columns accommodate factors 3. Columns are balanced/orthogonal

Properties of Orthogonal ArraysRef. Page 2-5

Key observations: First row has all 1's. There is no row that has all 2's. All columns are balanced and maintains an order. Columns of the array are ORTHOGONAL or balanced. This means that there are equal number of levels in a column. The columns are also balanced between any two columns of the array which means that the level combinations exist in equal number. Within column 1, there are two 1's and two 2's. Between column 1 and 2 there is one each of 1 1 1 2 2 1 and 2 2

3 2 1 2 4 2 2 1

4. Each array is used for many experiments


Between column 1 and 2, there is one each of 1 1, 1 2, 2 1 and 2 2 combinations. Factors A, B And C all at 2-level produces 8 possible combinations (full factorial) Taguchi’s Orthogonal array selects 4 out of the 8.

How does One-Factor-at-a-time experiment differ from the one designed using an Orthogonal array?

Orthogonal Arrays for Common Experiment Designs

L (XY)nUse this array (L-4) to design

experiments with three 2-level factors

Ref. Page 2-6

L ( )n

No. of rows in the array

No. of levels in the columns.

No. of columns in the array.

e pe e ts t t ee e e acto s

1

1

2

2

1

2

2

1

1

2

1

2

1

2

4

3

xxxxxx

xxx

xxx

C A BTrial#Results

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(XY)Use this array (L-8) to design experiments with

seven 2-level factors

Ref. Page 2-6

L (XY)n




xx

xx

xx

xx

xx

xx

xx

1

1

1

1

2

2

2

1

2

4

3

5

6

7

1

2

1

2

2

1

1

1

1

2

2

2

2

1

1

2

2

1

1

2

1

1

2

2

1

2

1

2

1

2

1

2

1

2

2

1

1

2

2

1

1

2

Results

ETrial# A CB FD G

xx28 21 2 1 12



Use this array (L-9) to design experiments with four 3-level factors

Trial# A B C D Res lts

Ref. Page 2-7

L (XY)n



in the array.Trial# A B C D Results

1 1 1 1 1 xx

2 1 2 2 2 xx

3 1 3 3 3 xx

4 2 1 2 3 xx

5 2 2 3 1 xx

6 2 3 1 2 xx

7 3 1 3 2 xx

8 3 2 1 3 xx y8 3 2 1 3 xx

9 3 3 2 1 xx

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Steps in Experiment Design

Factors Level-1 Levl-2A:Time 2 Sec. 5 Sec.B:Material Grade 1 Grade 2

Step 1. Select the smallest orthogonal array

Ref. Page 2-7

B:Material Grade-1 Grade-2C:Pressure 200 psi 300 psi

L-4 Orthogonal ArrayTrial #A B C1 1 1 12 1 2 23 2 1 2 4 2 2 1

Step 2. Assign the factors to the columns (arbitrarily)

Step 3. Describe the trial conditions (individual experimental recipe)

Trial#1: A1B1C1 = 2 Sec. (Time), Grade-1 (Material), and 200 psi (Pressure)





Experiment Designs With Seven 2-Level FactorExperiments with seven 2-level factors are designed using L-8 arrays. An L-8 array has seven 2-level columns. The factors A, B, C, D, ... G can be assigned arbitrarily to the seven column as shown. The orthogonal arrays used in this manner to design experiments are called inner arrays.

Experiment Designs with More Factors?

Ref. Page 2-8

L8 Orthogonal Array

11

11

12

43

12

12

11

22

12

21

12

21

12

12

11

22

ETrl.# A CB FD G

Control Factors


22

22

56

87

21

21

22

11

12

21

21

12

12

12

11

22

Inner Array

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Ref. Page 2-9

C1 C2 C1 C2 C1 C2 C1 C2B1 B2

A1 A2B1 B2

G1 Trl#1G2G1G2 Trl#3

F1

F2E1

Full Factorial Arrangement with Seven 2-level Factors

G2 Trl#3G1G2 Trl#5G1 Opt. ? Trl#7G2G1G2 Trl#8G1 Trl#6

F1

F2

F1

F2

E2

E1

D1


G2G1 Trl#4G2G1G2 Trl#2

F1

F2E2

D2

Note: Trl#1, 2.. 8 are conditions based on L-8 array design. Opt. is optimum condition determined.

2-Level ArraysL (23)

Common Orthogonal Arrays

L (XY)

Ref. Page 2-10

L4 (23)L8 (27)L12 (211)L16 (215)

3-Level ArraysL9 (34)L (21 37)

L (XY)n





L18 (21 37)

4-Level ArraysL16 (45)

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PLANIdentify Project and Select Project Team

Define Project objectives Evaluation Criteria

Planning Before Designing ExperimentsRef. Page 2-10

Determine System Parameters (Control Factors, Noise Factors, Ideal Function, etc.)

DESIGN

Select Array and Assign Factors to the columns (inner and outer arrays)

CONDUCT EXPERIMENTS


CONDUCT EXPERIMENTS

ANALYZE RESULTS

Factor Effects, Optimum Condition, Predicted Performance, etc.

An ordinary kernel of corn, a little yellow seed, it just sits there. But add some oil, turn up the heat, and, pow. Withi d ti k

Popcorn Machine Performance Study (Example Experiment)

Ref. Page 2-11

Within a second, an aromatic snack sensation has come into being: a fat, fluffy popcorn.

Note: C. Cretors & Company in the U.S. was the first company to develop popcorn machines, about 100 years ago.


This example is used to demonstrate “cradle to grave”, mini planning, design, and analyses tasks involved in DOE.

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Project - Pop Corn Machine performance StudyObjective & Result - Determine best machine settingsQuality Characteristics - Measure unpopped kernels (Smaller is better)

Factors and Level Descriptions

Experiment Planning & DesignRef. Page 2-12

Factors and Level DescriptionsFactor Level I Level IIA: Hot Plate Stainless Steel Copper AlloyB: Type of Oil Coconut Oil Peanut OilC: Heat Setting Setting 1 Setting 2

1

Trial# C: Ht. Setting A: Hot plate B: Oil Type

C1: Setting 1 A1: Stainless B1: Coconut1C A B

1 11

Trial#Results


2

43

C1: Setting 1

C2: Setting 2

C2: Setting 2

A2: Copper

A1: Stainless

A2: Copper

B2: Peanut

B2: Peanut

B1: Coconut

1

22

2

21

2

12

2

43

Experiment Design & Results

1

Trial# C: Ht. Setting A: Hot plate B: Oil Type

C1: Setting 1 A1: Stainless B1: Coconut1C A B

1 11

Trial#Results

Ref. Page 2-12

12

43

g

C1: Setting 1

C2: Setting 2

C2: Setting 2

A2: Copper

A1: Stainless

A2: Copper

B2: Peanut

B2: Peanut

B1: Coconut

11

22

12

21

12

12

12

43

Design Layout (Recipes)

Expt.1: C1 A1 B1 or [Heat Setting 1, Stainless Plate, & Coconut Oil]Expt.2: C1 A2 B2 or [Heat Setting 1, Copper Plate, & Peanut Oil ]Expt.3: C2 A1 B2 or [Heat Setting 2, Stainless Plate, & Peanut Oil ]Expt.4: C2 A2 B1 or [Heat Setting 2, Copper Plate, & Coconut Oil ]


How to run experiments: Run experiments in random order when possible.

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Experimental Results and Analysis

1C A B

1 11

Trial#Results

5

Ref. Page 2-13

A1 =__

(5 + 7)/2 = 6.0

11

22

12

21

12

12

12

43

58

47

__T = (5 + 8 + 7 +4)/4 =

6

112

C A B121

122

123

Trial#Results

587 A =

__(8 + 4)/2 = 6 0


2 2 14 4A2 = (8 + 4)/2 = 6.0

Trend of Influence:

How do the factor behave?

Calculations: ( Min. seven, 3 x 2 + 1)

(5 + 8) / 2 = 6.5_C1 =

Analysis of Experimental ResultsRef. Page 2-14

How do the factor behave?♦ What influence do they have to the

variability of results?♦ How can we save cost?

Optimum Condition:♦ What condition is most desirable?

(7 + 4) / 2 = 5.5

(5 + 7) / 2 = 6.0

(8 + 4) / 2 = 6.0

(5 + 4) / 2 = 4.5

(8 + 7) / 2 = 7.5

_C2 =

1

_A2 =

_A1 =

_B1 = _B2 =

7

8

9UNPO

Main Effects(Average effects of factor

influence)


A1 Hot plate A2 B1 Oil B2 C1 Heat SettingC2

3

4

6

5

7PPED

KERNELS

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Nutek I

QC Plays a key roles in:◦ Understanding factor influence ◦ Determination of the most desirable condition

Role of Quality Characteristics (QC)Ref. Page 2-14

k, Inc.www.nutek

-us.co

mAll

Rights

Reserved

Design of

Determination of the most desirable condition.

Quality Characteristics

ExamplesNominal is Best: 5” dia. Shaft,12 volt battery, etc.Smaller is Better: noise, loss, rejects, surface roughness, etc.


Experiment

s Using the

Taguchi

Approach Sep

2005

Bigger is Better: strength, efficiency, S/N ratio, Income, etc.

Estimate of Performance at the Optimum Condition

7

8

9UNPOP

Main Effects(Also called: Factorial Effects or Column Effects)

Ref. Page 2-15

A1 Hot plate A2 B1 Oil B2 C1 Heat SettingC2

3

4

6

5

PED

KERNELS

A1

B1 C

2

Based on QC: Smaller is better

Optimum condition: ( Assuming A1 is less expensive than A2)A1

B1

C2


1 1 2

= 6.0 + ( 6 – 6 ) + (4.5 – 6.0 ) + ( 5.5 – 6.0 )= 4.0 (Assumption: Factor contributions are additive)

__Yopt = T +

__( A1 -

_( C2 -

__T )

__( B1 -

__T ) +

__T ) +

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Expected Performance:♦ What is the improved performance?♦ How can we verify it?♦ What is the boundary of expected performance?

(Confidence Interval, C.I.)Notes:

Interpretation of the Estimated Performance

Ref. Page 2-16

Notes:Generally, the optimum condition will not be one that has already been tested. Thus you will

need to run additional experiments to confirm the predicted performance.Confidence Interval (C.I.) on the expected performance can be calculated from ANOVA

calculation. These boundary values are used to confirm the performance.

Meaning: When a set of samples are tested at the optimum condition, the mean of the tested samples is expected to be close to the estimated performance.


3.5 Yavg. Yexp. = 4.0 4.5

Confidence level (C.L.), say 90%.

Confidence Interval, C.I. = +/- 0.50

(Calculation not shown)

Performance ImprovementImproved performance from DOE =

Estimated performance at the optimum condition (Yopt)Yopt = 4.0 (in this example)

Ref. Page 2-16

The estimated performance can be expressed in terms of a percent improvement, if the current performance is known.

Assuming that the current performance is the grand average of performance (YCurrent ) = 6.0

Improvement = x 100(Yopt - YCurrent )

YCurrent


= x 100 = - 33%(4 - 6 )

6

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Example 1: Plastic Molding ProcessFACTORS and LEVELSA: Injection Pressure A1 = 250 psi A2 = 350 psiB: Mold Temperature B1 = 150 deg. B2 = 200 deg.

An Example ExperimentRef. Page 2-16

C: Set Time C1 = 6 sec. C2 = 9 sec.

Where did these factors and levels come from?How do you determine:

Number of factors to include in the experimentNumber of levels for each factorThe values of the levels- for 2-level factors- for 3 or 4-level factors


- for 3 or 4-level factors

The first step in every experiment must be the experiment planning discussion with the project team.

Preparation for Meeting◦ Identify Project

One that gives “the biggest bang for the buck”.

◦ Form Team (3 – 12 people)

Planning –The Essential First StepRef. Page 2-17

( p p )People with first hand knowledgeInternal customersPeople responsible for implementation

◦ Schedule and convene an all-day experiment planning meeting with the team.

Inform and prepare all for a full day of meeting.

◦ As the project leader, invite all team members to attend the planning meeting.


p j , p g gSecure commitment to attend the meetingEncourage team members to bring all project information to the meeting, but discourage any formal research or documentations.Study subject project and bring information with respect to details of system breakdown (system into sub-systems, products into components) to the meeting.

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Process Diagram

Control Factors and Levels:* Sugar

Ref. Page 2-19

System/Process(Pound Cake Baking Process)

Mixing , Kneading, and allowing time for baking.

Input Output

Input to the Process:

Heat/Electricity

g* Butter* Flour* Milk

Result , Response, Quality Characteristic, or Overall Evaluation Criteria(OEC)- Evaluation(Readings)

- Observations


Noise Factors•Oven Type•Kitchen Temp•Humidity

Define SystemDefine SystemDefine SystemDefine SystemSystem may include all or part of the products or processes. It must System may include all or part of the products or processes. It must always include suspect areas (subalways include suspect areas (sub--processes or components) of the processes or components) of the

subject process under study. subject process under study.

Slide 66

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34

System definition utilizes knowledge about problem System definition utilizes knowledge about problem and probable causes.and probable causes.

Consider the process of baking cakes (shown next).)

Once the system is defined (boundaries), proceed to define:

- Outputs (Objectives)

Slide 67

- How outputs are measured- etc. (Planning discusions)

System View of Process(Cake Baking Process)

Ref. Page 2-19

Gather Ingredients Mix Ingredients Bake Cake in

O

Input Output

S

Ingredients (Make Batter) Oven

System may be defined one of the three different ways shown above (Dashed lined rectangles).


System

Note: OUTPUT of previous sub-process is INPUT to the next. For example: Batter is output of MIXING process, but input to the BAKING process.

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35

5 Steps Process Study Roadmap5 Steps Process Study Roadmap

43

5Run Tests to Confirm Solutions

Nutek & Team – on siteNutek off site

12Design and Describe Test

Analyze Test Results and Prescribe Solutions

3Carry Out Planned Tests and Collect Results

Nutek – off site

Project Team & Nutek

Slide 69

1Hold Experiment Planning Discussions

Recipes

Example Experiment Planning Meeting Topics

Project Title: Pound Cake Baking Process Optimization StudyObjective: Determine the recipe of the “overall best” cake.

Are there more than one objectives?H h bj i l d d d ifi d?

Ref. Page 2-20

How are the objectives evaluated, measured, and quantified?What are the criteria of evaluation?What are the relative weighting of these criteria?

Criteria Description Worst Reading

Best Reading

QC Relative Weight (Wt)

C1: Taste 0 8 B 60

C2: Moistness 25 – 70 gms 40 gms. N 25

Nutek, Inc. www.nutek-us.com All Rights Reserved Design of Experiments Using the Taguchi Approach Sep. 2005

g g

C3: Voids/Smoothness 6 0 S 15

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36

Example Experiment –Evaluation Criteria

Criteria Description Worst Reading

Best Reading

QC Relative Weight (Wt)

Sample ReadingsSample 1 Sample 2

Ref. Page 2-21

C1: Taste 0 8 B 60 5 6

C2: Moistness 25 – 70 gms 40 gms. N 25 46 35

C3: Voids/Smoothness 6 0 S 15 4 5

OEC = C1

C1rangex Wt1 +

( 1 - ) x Wt2 C2b – C2-nom

C2 – C2-nom + ( 1 - )x Wt2 C3rang

e

C3

OEC 1 (5/8) 60 [1 (46 40)/(70 40) ] 25 ( 1 4/6) 15

Calculation of OEC’s with Sample Readings:


OEC-1 = (5/8) x 60 + [1 – (46-40)/(70-40) ] x 25 - ( 1 – 4/6) x 15

= 37.5 + 20.0 + 5 = 62.5

OEC-2 = (6/8) x 60 + [1 – (40-35)/(70-40) ] x 25 - ( 1 – 5/6) x 15

= 45 + 20.83 + 2.5 = 68.33

Module - 3Experiment Designs to Study Interactions


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37

Things you should learn from discussions in this module:What is interaction?

Experiments to Study Interaction Between Factors

Ref. Pg. 3-1

Is interaction like a factor? Is it an input or an output?How many kinds of Interaction are there?Where does interactions show up?What can we do in our design to study interaction?How can you tell which interaction is stronger?


How can you tell which interaction is stronger?When Interactions are too many, what is a good way to design the experiment?

B1 = 0 glass of beer

A1B1

Plot of Presence of Interaction Between FactorsRef. Page 3 -2

A1 = 0 (Tabs of Aspirin) A2 = 2 Tabs

Hea

d P

ain


A2B1

A2B2

A1B2


• Interaction exits when the lines are non-parallel

• How are the two lines drawn?

• With FOUR data points A1B1, A1B2, etc.

• These points are calculated from the experimental results.

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Between TWO 2-level factors: AxB

Between TWO 3-level factors

Other Possible Types of Interactions

Between TWO factors: AxB, BxD, etc.Among THREE factors: AxBxCAmong FOUR factors: AxBxCxD

Ref. Page 3 -3

Between TWO 4-level factors

Between a 2-level and a 3-level factors, etc.

B1

B2

Among FOUR factors: AxBxCxD


A1 A2 A3

B3

Interaction effects can be theoretically determined from the array and the factor assignments. The task, however, is quite laborious. Fortunately, it has all been done by Taguchi.

Columns of Interaction Effects (AxB)Ref. Page 3 -5

Reference: Pages 208 -212 QUALITY ENGINEERING by Yuin Wu and Dr. Willie Hobbs Moore

Method: Interaction effect AxB (A in col 1, and B in col2 ) is the angle between the two lines which can be expressed in terms of results: Y, Y2, etc. This expression shows that it is the same for factor C in column 3.


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Number of Possible Factor Main Effects and Interaction EffectsAvg. Effect 2-factor 3-factor 4-factor 5-factor 6-factor 7-factor1 7 21 35 35 21 7 1

Obtainable Information from Full-Factorial, 27 = 128

Ref. Page 3 -5

(1 + 7 + 21 + 35 + 35 + 21 + 7 + 1) = 128 (27 = 128)

Calculation method: Two-Factor Interaction - two (say A and B) taken out of seven factors (The combination formula):

nCr = n!/[(n-r)! R! ] = 7 x 6 x 5! / [5!x 2 ] = 21


Ref: Page 374, STATISTICS FOR EXPERIMENTERS by Box, Hunter and Hunter

Scopes of Seminar: Learn how to study and make corrections for interactions between TWO 2-Level factors (AxB, BxC, etc)

For many 2-level arrays (L-12 is an exception), the interaction effect between two 2-level factors (AxB) is localized to a column. The location of the interaction effects depends on the location of the interacting factors itself. All possible interacting pars of factor locations have been calculated and are identified in the Triangular Table.

Columns of Localized InteractionsRef. Pg. 3-6

A1

B2

AxB3

AxB

D4Cols.

E5

What about


AxB1x2 => 3 BxE

2x5 => ?

at about

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Table for Determining the Interaction Location

Example: Interaction effects between two factors in column 2 and 4 will be mixed with factors (or interactions) in column 6.

Triangular Table for Two-Level Orthogonal Arrays

Ref. Pg. 3-6

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

(1) 23 5 74 6 89 11 1310 12 1415

(2) 61 7 54 10 811 9 1514 12 13

(3) 67 5 114 10 89 15 1314 12

(4) 21 3 1312 14 815 9 1110

(5) 23 13 1512 14 89 11 10

(6) 141 15 1312 10 811 9

(7) 1415 13 1112 10 89

(8) 21 3 54 6 7

2 x 4 => 6

5 x 10 => 15


(8) 21 3 54 6 7

(9) 23 5 74 6

(10) 61 7 54

Etc. xxxx xx

Linear Graphs for Interaction DesignLinear graphs are graphical representations of the interaction readings from the Triangular Table.

Linear Graphs – Selected Readings of the TTRef. Pg. 3-7

13 2 1

35

1

23

45

7

A BAxB


2466

7

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Facts:◦ Factors identified are generally more than what is possible to

study.◦ Interactions between two 2-level factors alone are always more

than the number of factors.

Selecting Interactions to Study – A Difficult Compromise

Ref. Pg. 3-8

◦ Most often the knowledge about interactions is absent or unavailable.

◦ Discussions about whether to study interactions or not, arises only after the factors to study and their levels are identified, late in the planning day (3 PM or later).

◦ When experimental scopes are firm, interaction studies are done at a cost of factors.

◦ A compromise between the number of factors and number interactions to be studied need to be determined by team consensus.


Conclusion:Number of interaction to study and selecting the ones to study among all possible pairs, are challenging tasks that the project team must accomplish by consensus (no data or science is applicable)

A: Egg (2-level) interaction: Egg x Milk (AxC)B: Butter (2-level) interaction: Butter x Milk (BxC)C: Milk (2-level)D: Flour (2-level)E: Sugar (2-level)

Example 2: Cake Baking ExperimentRef. Pg. 3-9

11

11

2

12

43

5

12

12

2

11

22

2

12

21

1

12

21

2

12

12

1

11

22

1

DTrial# A

AxCC

BxCB E Result

s66

75

54

62

52

Two Rules for Experiment Design

1. Treat the interacting factors first

2. Reserve columns for interactions as per the Triangular Table


2

22

6

87

1

21

2

11

2

21

1

12

2

12

1

22

82

52

78

Total of All Results = 521

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42

1. Select Array2 Assign factors to column and reserve columns for the

Experiment Design Steps for Interaction Study

Ref. Page N/A

2. Assign factors to column and reserve columns for the selected interaction

Treat interacting factors firstReserve columns for interaction based on TT

3. Describe experiments (IGNORE COLUMNS RESERVED FOR INTERACTION)


72

68

64

Plot of Factor Average Effects (Main Effects)

Plot of Factor and Interaction EffectsRef. Pg. 3-12

A1 A2 C1 C2 B1 B2 D1 D2 E1 E2

64

60

56

72

68

64

Interaction Column Effects


Because the QC = Bigger is better, the optimum condition is: A2 C1 B2 D1 E1

Based on the slope of the line, interaction AxC may be significant (statistical significance is determined by ANOVA)

(AxC)1 (AxC)2 (BxC)1(BxC)2

60

56

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43

Average effect of A1C1 is found byaveraging results which contain the

ff t f b th A & C

Test Data for Test of Presence of Interactions Plot

DTrial# A

AxCC

BxCB E Result

Ref. Pg. 3-13

effects of both A1 & C1._____(A1C1) = (66 + 75)/2 = 70.50

(The first two trial results only)Similarly_____(A1C2) = (54 + 62)/2 = 58.00___(A2C1) = (52 + 82)/2 = 67.00 and

11

11

22

22

12

43

56

87

12

12

21

21

11

22

22

11

12

21

12

21

12

21

21

12

12

12

12

12

11

22

11

22

C s66

75

54

62

52

82

52

78



_____(A2C2) = (52 + 78)/2 = 65.00


Pots for Test of Presence of Interaction


A1B1

Ref. Pg. 3-14

A1 = 0 (Tabs of Aspirin) A2 = 2 Tabs

Hea

d Pa

in


A2B1

A2B2

A1B2

• Interaction exits when the lines are non parallel


• Interaction exits when the lines are non-parallel• How are the two lines drawn?• With FOUR data points A1B1, A1B2, etc.• These points are calculated from the experimental results.

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Optimum Condition and the Expected Performance - 1

72

68

64

Plot of Factor Average Effects (Main Effects)

Ref. Pg. 3-16

A1 A2 C1 C2 B1 B2 D1 D2 E1 E2

60

56

Optimum Condition: A2 C1 B2 D1 E1 (Without Interaction)__ __ __ __ __ __ __ __ __ __ __

YOPT. = T + (A2 - T) + (C1 - T) + (B2 - T) + (D1 - T) + (E1 - T) __ __ __ __ __ __

= T + (66 - T) + (68.75 - T) + (74.25 - T) + (70 - T) + (65.5 - T)= 65.125 + 0.875 + 3.625 + 9.125 + 4.875 + 0.375= 65.125 + 18.875 = 84.00

Plots of Test of Presence of Interaction (shown earlier)

Ref. Page 3-14

Which interaction is stronger? What are the levels of the interacting factors?

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45

Optimum Condition and the Expected Performance-2

72

68


Ref. Pg. 3- 16 &17

(AxC)1 (AxC)2 (BxC)1(BxC)2

64

60

56

Optimum Condition: A1 C1 B2 D1 E1 (With Interaction)_ _ _ _ _ ___ _ _ _ _ _ _ _

YOPT = T + (A1-T) + (C1 -T) + ([AxC]1 -T) + (B2 -T) + (D1-T) + (E1 - T)


OPT. ( 1 ) ( 1 ) ([ ]1 ) ( 2 ) ( 1 ) ( 1 )

= 65.125 + (- 0.875) +3.625 +2.625 + 9.125 +4.875 +0.375

= 84.875

Expected Performance (alternative expression)

78.5

70.0

80

esul

t

A1C1 = 70.5

Ref. Pg. 3-17

C1 C2

C1 C2

A1

A2

B159.0

53.0

Aver

age

Effe

ct/R

e

A1C2 = 58.0

A2C1 = 67.0A2C2 = 65.0

60

70

50

Optimum Condition: A1 C1 B2 D1 E1 (With Interaction)

( ) ( ) ( ) ( )TTTTT −+−+−+−+= 5.657025.745.70

375.0875.4125.9375.5125.65 ++++=

( ) ( ) ( ) ( )TETDTBTCATY OPT −+−+−+−+= 11211.

= 84.875

p ( )

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46

M d l 4Module - 4Experiment Designs with Mixed-Level Factors



Standard orthogonal arrays are modified to use it for many

Designing Experiments with Mixed Factor Levels

Ref. Pg. 4-1

Standard orthogonal arrays are modified to use it for many experiment designs with mixed level factors.What is degrees of freedom (DOF)?How to determine requirements for the experiment in terms of DOF?How to determine which array is most suitable for modification?How to upgrade columns?H t d d l ?


How to downgrade columns?What is a combination design?

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47

Example Experiment (How to design?)1 Factor(X) at 4-Levels and 4 Factors (A B C and D) at 2 Levels

Example Experiments with Mixed Factor LevelsRef. Pg. 4-2

4 Factors (A, B, C and D) at 2-Levels

How do you determine which Array is most suitable?What are Degrees of Freedoms(DOF)?

Definitions of DOFDOF of a factor = number of level - 1DOF of a column = number of level - 1DOF f l b f l DOF


DOF of an array = total number of column DOFDOF of an experiment = total number of results - 1

DOF is used to determine the experimental needs and identify the orthogonal array suitable for the same.

Assume conversion rules:1 1 = > 1 1 2 = > 22 1 = > 3 2 2 = > 4

Columns Upgrading Method (2-level to 4-level)

Steps:

• Select a set of interacting group of columns

Ref. Pg. 4-2

L-8 ArrayCOL >>

EXPT # 1 2 3 4 5 6 71 1 1>1 1 1 1 1 12 1 1>1 1 2 2 2 23 1 2>2 2 1 1 2 24 1 2>2 2 2 2 1 15 2 1 3 2 1 2 1 2

New Column

Discarded Column

• Discard any one of the three columns

• Combine the remaining columns into a new columns following the conversion rules


5 2 1>3 2 1 2 1 26 2 1>3 2 2 1 2 17 2 2>4 1 1 2 2 18 2 2>4 1 2 1 1 2

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Experiment Design with the Modified Array

112

123

122

000

121

121

122

000

BTrial# X CA D Results50

62

70

Modified Array with Factor Assignments

Ref. Pg. 4-3

Col# Factor Description

Level - 1 Level - 2 Level - 3 Level - 4 (L2 – L1)

Main Effect (Also called Column Effect or Factorial Effect)

22

33

44

43

56

87

12

21

21

00

00

00

21

12

21

21

21

12

12

12

12

00

00

00

70

75

68

65

65

74


Description

7 Stator Structure 63.75 68.50 0 0 4.75

6 Contact Brush 66.75 65.50 0 0 1.25

5 Impregnation 64.75 67.50 0 0 2.75

4 Air Gap 63.25 69.00 0 0 5.75

1 Casement Structr. 56.00 72.50 66.50 69.50 16.50

Quality Characteristic: Smaller is better

ANOVA and Optimum Condition

Col# Factor Description

f S V F S’ P(%)

ANOVA Table

5 Impregnation 1 15.125 15.125 4.84 12.00 2.73

4 Air Gap 1 66.125 66.125 21.160 63.00 14.35

1 Casement Structr. 3 309.38 103.125 33.00 300.00 68.36

Ref. Pg. 4-5

Optimum Table: Estimated Performance at Optimum Condition

1 Casement Structr.

Col# Factor Description Level Description Level# Contributions

Present 1 -10.125

7 Stator Structure 1 45.125 45.125 14.44 42.00 9.57

All Other/Error 1 3.125 3.125 4.98

6 Contact Brush (1) (3.125) POOLED

p g

TOTALS: 7 438.875 100.00%


7 Stator Structure

5 Impregnation

4 Air Gap

Total Contributions from all factors and interactions - -16.750

Std. Design 1 -2.375

Present 1 -2.875

Softer 1 -1.375

Expected performance at the optimum condition -: 49.375

Current Grand Average of performance - 66.125

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How to design the experiment with Three 3-level factors (B, C & D) and One 2-level factor (A)?

L9 Array

Columns Downgrading (3-Level to 2-level)Ref. Pg. 4-6

9 yB C A D

Expt# 1 2 3 41 1 1 1 12 1 2 2 23 1 3 1' 34 2 1 2 35 2 2 1' 16 2 3 1 27 3 1 1' 28 3 2 1 3

( ' )indicates new modified levels 1 instead of 3.

Any column can be selected for Dummy Treatment.


Any column can be selected for DUMMY TREATMENT.Highest level can be replaced by any of the other levels unless dictated by knowledge of instability at a level.

9 3 3 2 18 3 2 1 39 3 3 2 1

Example 4: Lost Foam Casting ProcessProcess factors: 6 factors at 2 levels, 2 factors at 3 levels, & 1 factor at 4 levels

Minimum DOF requirement:DOF for 6 2-level factors 6 x (2 - 1) = 6DOF for 2 3-level factors 2 x (3 - 1) = 4

Modification of L-16 Array for Example 4 Ref. Pg. 4-7

DOF for 1 4-level factors 1 x (4 - 1) = 3Total DOF = 13

1 2 3 NEW 1 4 8 12 NEW 4 7 9 14 NEW 7 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 2 2 2 1 2 2 21 1 1 1 2 1 2 3 2 1 2 31 1 1 1 2 2 1 4=1' 2 2 1 41 2 2 2 1 1 1 1 2 1 2 31 2 2 2 1 2 1 2 2 2 1 41 2 2 2 2 1 1 3 1 1 1 11 2 2 2 2 2 1 4=1' 1 2 2 22 1 2 3 1 1 1 1 2 2 1 4


2 1 2 3 1 1 1 1 2 2 1 42 1 2 3 1 2 1 2 2 1 2 32 1 2 3 2 1 1 3 1 2 2 22 1 2 3 2 2 1 4=1' 1 1 1 12 2 2 4=1' 1 1 1 1 1 2 2 22 2 2 4=1' 1 2 1 2 1 1 1 12 2 2 4=1' 2 1 1 3 2 2 1 42 2 2 4=1' 2 2 1 4=1' 2 1 2 3Column conversion rules: 1 1 > 1 1 2 > 2 2 1 > 3 2 2 > 4

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Modified Array Ref. Page 4-8

Factor Assignments for Example 4 (L-16)Experiment Design Table Col.# FACTORS Level-1 Level-2 Level-3 Level-4

1 I:Coating Type Type 1 Type 2 Type 3 2 Unused/Upgraded M/U 3 Unused/Upgraded M/U

Ref. Pg. 4-9

3 Unused/Upgraded M/U 4 G:Sand Comp Plant X Plant Y Plant Z 5 A:Metal Head Low high 6 B:Sand Supplier Supplier 1 Supplier 2 7 H:Gating Type Plant X Plant Y Plant Z Plant W 8 Unused/Upgraded M/U 9 Unused/Upgraded M/U

10 C:Sand Perm 200 Perm 300 Perm 11 D:Metal Tem. 1430 F 1460 F 12 Un sed/Upgraded M/U


12 Unused/Upgraded M/U 13 E:Quench Type 450 F 725 F 14 Unused/Upgraded M/U 15 F:Gas Level None High

Characteristic: The Bigger The Better

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51

E = EngineT = Transmission

If E and T are two factors among many factors included

Factor Levels Compatibility Requirements

It is always possible to makeone 4-level factors bycombining two 2-level factors.

PP = Power Plant (E&T)

Ref. Page 4-10

among many factors included in a study, then

E1T1, E1T2, E2T1, and E2T2are all present in the experiment designed using an orthogonal array.

What if E2T2 combination does not exist?

PP = Power Plant (E&T)

PP (PP1,PP2,PP3,&PP4) OR

PP (E1T1, E1T2, E2T1,& E2T2)

Since E2T2 is absent,

PP (E1T1, E1T2, & E2T1)


Now PP is a three level factor.

What is Combination design?Consider: 3 3-Level Factors (A,B,C) and 2 2-Level Factors (X,Y)

Combination Design – a Special CaseRef. Pg. 4-11

Factors A B C (XY)EXPT.# 1 2 3 4

1 1 1 1 12 1 2 2 23 1 3 3 34 2 1 2 35 2 2 3 16 2 3 1 2


7 3 1 3 28 3 2 1 39 3 3 2 1

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___ ___MAIN EFFECT OF X = (XY)2 - (XY)1

Main Effects of the Combined Factors

nse/

Main Effect of factor X

Ref. Pg. 4-12

Assumption: There is no interaction between X and Y. This assumption is necessary since the

____ ____= X2Y1 - X1Y1 AT FIXED Y = Y1

___ ___MAIN EFFECT OF Y = (XY)3 - (XY)1

____ ____= X1Y2 - X1Y1 AT FIXED X = X1

Res

ult/R

espo

nQ

C

X1 A3 X2

pons

e/

Y1

X1

Main Effect of factor Y


assumption is necessary since the conclusions about the main effects are drawn based on a fixed value of the other factor. The main effect of X above is only valid when Y = Y1. R

esul

t/Res

pQ

C

Y1 A3 Y2

1

Ref. Page N/A

Module – 5Robust Design


g

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53

Changing Focus of Experimental Studies

# OFControl Factors

N i F

Ref. Pg. 5-3

FACTORS

R&D Adv Engg Design&Dev Test&Valid Mnf &Prod

Noise Factors


R&D Adv.Engg Design&Dev. Test&Valid. Mnf.&Prod.

Ref. Pg. N/A

C t l F t dControl Factor and Noise Interaction

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54

Plot of Interaction (Factor along x-axis)

A1N1= 50Interaction Between Alcohol and Consumed Food

Ref. Pg. 5-4

N1 = Hard liquorIntoxication

Level

A = Light snack A = Steak

N2 = Light beerA1N2 = 25

A2N2 = 20

A2N1 = 30


A1 = Light snack A2 = Steak Dinner

A: Type of Food N: Alcohol Strength

A1N1 = 50 A2N1 = 30

A1N2 = 25 A2N2 = 20

Plot of Interaction (Noise along x-axis)

A1N1 = 50 Interaction Between Alcohol and Food Consumed

Ref. Pg. 5-6

A1 = Light snack

A2N2 = 20

A2N1= 30

A2 = Steak dinnerA1N2 = 25

Intoxication Level

N = Hard Liquor N = Light


N1 = Hard Liquor N2 = Light Beer

A: Type of Food N: Alcohol Strength

A1N1 = 50 A2N1 = 30

A1N2 = 25 A2N2 = 20

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ANOTHER EXAMPLE OF NOISE AND CONTROL FACTOR INTERACTION

Example 5B:Number of average absent days by employees in a school district were identified to be influenced by control and noise factors and the results are as shown:

Control Factor - A:Prentive Measure (A1:Over the counter drugs, A2:Flue Shot)Noise Factor - N:Exposure to germs (N1:Administrative work, N2:Working with Kids)

Ref. Pg. 5-7

8.0

7.0

6.0

5.0

10.0

9.0

N2

N1

8.0

7.0

6.0

5.0

10.0

9.0 A1

Results: A1N1 = 6 days, A2N1 = 2 days, A1N2 = 10 days, and A2N2 = 3 days.

Plot of data reveals the robust factor level (A2:Flue Shot).

4.0

3.0

2.0

1.0

A2A1 Factor

N14.0

3.0

2.0

1.0

N2N1 Noise

A2

We will see how this interaction between noise and control factor is used to determine ROBUST

factor levels soon

Ref. Pg. N/A

factor levels soon.

Now, let’s learn about how we can smartly analyze MULTIPLE SAMPLE RESULTS for

DOE.

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56

11 1 1 1 1 1 1

ETr A B - C D F yz z z z z z Sd S/N

12.810 11 9 9 10 11 1.45 -22.2

Results (y)__

Average

10

Multiple Sample Results of Planned Experiments

Ref. Pg. N/A

22 1 1 1 2 2 2

23 1 2 2 1 1 2

14 1 2 2 2 2 1

15 2 1 2 1 2 2

26 2 1 2 2 1 1

27 2 2 1 1 2 1

18 2 2 1 2 1 2

9.310 12 8 8 10 12 0.85 9.3

12.911.7 11.8 11.5 14.3 14.4 14.1 1.43 12.9

14.112.7 12.7 12.6 15.6 15.6 15.4 1.57 14.1

13.213.8 13.5 13.8 13.3 12.8 12.4 0.56 13.2

14.913.2 13.5 13.4 16.2 16.6 16.4 1.67 14.9

13.912.6 12.9 12.1 15.4 15.8 14.8 1.59 13.9

13.312.3 11.7 12 15.1 14.3 14.2 1.43 13.3

Grand Averages => 13 1 1 32 13 1

-

10

Grand Averages => 13.1 1.32 13.1

Is the result of trial#1 better than that of trial#2?It is difficult to objectively answer using AVERAGE of results only.

Definition of Mean Squared Deviation (MSD)

Avg. Target

Ref. Pg. 5-12

_yi Y Yo

(Yi – Yo )2 = Squared of the deviation


( i o ) q

MSD = Σ(Yi - Yo)2/nAlso,

MSD = (STD. DEV.)2 + (Yavg. - Yo )2

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57

Why AVERAGE of results is inadequate?Why do we need a new YARDSTICK?

Yardstick for Comparison of Population Performance

Ref. Pg. 5-13

Why do we need a new YARDSTICK?

Nominal: MSD = [(Y1 -Yo) 2 +.(Y2 -Yo) 2 + (Y3 -Yo) 2 +. .]/n

Smaller: MSD = [( Y12 + .Y 2

2 +Y 32 + . .)]/n

Bigger: MSD = [(1 / Y12 + 1 / Y2

2 + 1/Y32 + ...... )]/n


S/N = - 10 LOG10 (MSD)Why - (Minus) Why 10

Logarithmic Measure of Sound Volume(Common household audio surround sound system)

Ref. Pg. N/A

Lower Volume => Lower Volume => -- 60 dB Higher volume=> 60 dB Higher volume=> -- 12 dB 12 dB (Range of Scale 0 to (Range of Scale 0 to –– 100 dB)100 dB)

Note: Lower magnitude of negative number is HIGHER as in S/N RatioNote: Lower magnitude of negative number is HIGHER as in S/N Ratio

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The magnitude of most earthquakes is measured on the Richter scale, invented by Charles F. Richter in 1934. The Richter magnitude is calculated from the amplitude of the largest seismic wave recorded for the earthquake, no matter what type of wave was the strongest.

The Richter magnitudes are based on a logarithmic scale (base 10). What this means is that for each whole number you go up on the Richter scale, the amplitude of the ground motion recorded by a seismograph goes up ten times.U i thi l it d 5 th k ld lt i t ti th l l f d

Ref. Pg. N/A

Using this scale, a magnitude 5 earthquake would result in ten times the level of ground shaking as a magnitude 4 earthquake (and 32 times as much energy would be released). To give you an idea how these numbers can add up, think of it in terms of the energy released by explosives: a magnitude 1 seismic wave releases as much energy as blowing up 6 ounces of TNT. A magnitude 8 earthquake releases as much energy as detonating 6 million tons of TNT. Pretty impressive, huh? Fortunately, most of the earthquakes that occur each year are magnitude 2.5 or less, too small to be felt by most people.

Body-wave magnitude formula:

l (A/T) Q(D h)


mb = log(A/T) + Q(D,h)

where A is the ground motion (in microns)T is the wave's period (in seconds), and Q(D,h) is a correction factor that depends on distance to the quake's epicenter D (in degrees) and focal depth h (in kilometers).

How are S/N Ratios compared? (QUALITEK-4 software output, SEM-EX06.Q4W)

How can we transform S/N Ratios back into

Benefits of Transforming MSD to Log Scale (S/N)

Ref. Pg. 5-14

expected results?

EXPT. A B C R1 R2 R3 S/N AVG.1 1 1 1 2 3 4 -9.86 32 1 2 2 4 5 3 -12.2 43 2 1 2 4 5 6 -14.1 54 2 2 1 3 5 7 -14.4 5

An experiment with three2-level factors (A, B and C)and 3 samples per trialyielded the followingresults (QC = Smaller isBetter)


Based on averages, condition results of trials 3 & 4 are equal.Based on S/N ratios, condition 3 is better, since comparing

-14.1 > -14.4 ( -14.1 is bigger than -14.4)

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Plot of Factor Average Effects

Optimum Condition Based on Analysis Using S/N Ratio

Ref. Pg. 5-14

A1 A2 B1 B2 C1 C2

T

-15

-14

-13

-11

-12


Highest values selected regardless of the Quality Characteristic.Does that mean the Quality Characteristic has no effect?When does it come into play?

Example 7: Engine Idle Stability Study (QUALITEK‐4 software output, SEM‐EX07.Q4W)

3 - Factors at 3-levels each3 - Repetitions ( normal operating noise)

S/N Analysis Example Application

Ref. Pg. 5-16

DESIGN FACTORS AND THEIR LEVELSCol. # FACTORS LEVEL-1 LEVEL-2 LEVEL-3

1 INDEXING -30 DEG O DEG +50 DEG2 OVERLAP AREA -30% 0 % +30 %3 SPARK ADVANCE 20 DEG 30 DEG 40 DEG4 UNUSED/UPGRADED M/U

CHARACTERISTIC: THE SMALLER THE BETTER

L9(34)

COND 1 2 3 4

TRIAL RESULTS Trial# R(1) R(2) R(3) S/N Ratios 1 20 25 26 -27.5358


1 1 1 1 12 1 2 2 23 1 3 3 34 2 1 2 35 2 2 3 1 6 2 3 1 27 3 1 3 28 3 2 1 39 3 3 2 1

0 5 6 53582 34 36 26 -30.1815 3 45 34 26 -31.0913 4 13 23 22 -25.9550 5 36 45 35 -31.8051 6 23 25 34 -28.86497 35 45 53 -33.0528 8 56 46 75 -35.5939 9 35 46 53 -33.1175

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Example Analysis of ResultsMAIN EFFECTSCOL#/FACTORS LEVEL 1 LEVEL 2 LEVEL 3 LEVEL 4 (L2-L1)1 INDEXING -29.6028 -28.8749 -33.9214 0.0000 0.72782 OVERLAP AREA -28.8478 -32.5268 -31.0245 0.0000 -3.67893 SPARK ADVANCE -30.6648 -29.7513 -31.9830 0.0000 0.9135

Ref. Pg. 5-16

QUALITY CHARACTERISTIC: THE SMALLER THE BETTERDATA TYPE : S/N RATIO

OPTIMUM TABLE

A N O V A TABLE DATA TYPE: S/N RATIO COL.#/FACTOR f S V F S' P(%)1 INDEXING 2 44.6445 22.3222 846.5925 44.5917 61.27152 OVERLAP 2 20.5283 10.2641 389.2777 20.4755 28.13453 SPARK ADV. 2 7.5517 3.7758 143.2037 7.4990 10.3040

OTHER/ERROR 2 0.0527 0.026 0.2898TOTAL: 8 72.7773 100.00%


OPTIMUM TABLECOL #/FACTORS LEVEL DESC. LEVEL# CONTRIBUTION

1 INDEXING O DEG 2 1.92472 OVERLAP AREA -30% 1 1.95183 SPARK ADVANCE 30 DEG 2 1.0484

TOTAL CONTRIBUTION FROM ALL FACTORS.... 4.9251CURRENT GRAND AVERAGE OF PERFORMANCE -30.7998EXPECTED RESULT AT OPTIMUM CONDITION... -25.8747

Multiple Factor Levels for Robustness

Ref. Pg. N/A

Multiple Factor Levels for Robustness against single noise

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Steps for Robust Design with Multiple Factors

Ref. Pg. N/A

Two-Step Optimization Strategy

In this approach product and process designs are achieved by

Ref. Pg. 5-12

In this approach product and process designs are achieved byadjusting factor levels to reduce variability and bring the meanperformance closer to the target. It follows two distinct steps with theassumption that reduction of variability is more important than beingon the target:

1.Reduce variability by adjusting levels of factors determined to beinfluential

2. Adjust performance mean to target by adjusting those factorswith less influence on variability

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Notation Factor Description Level 1 Level 2

A Injection Pressure 1,800 psi 2,250 psi

B Mold Closing Speed Low(Not revealed)

Moderate

C Mold Pressure 600 psi (4.1 950 psi

Example 7A: (QUALITEK-4, SEM-EX7a.Q4W)

Interaction: Interaction between factors A & B (AxB) was selected for the study.

Noise factors: Among the factors identified, th li t id d t ll bl

Ref. Pg. 5-13

11 1 1 1 1 1 1

22 1 1 1 2 2 2

ETr

A B - C D F Noise N1 Noise N2Test Results

For each trial condition:

MPA)D Backpressure 950 psi 1,075 psi

E Screw Speed 50 Sec. 65 Sec.

F Spear Temperature 325 Deg. C 380 Deg. C

Notation Factor Description Level 1 Level 2

the e list were considered uncontrollable, noise factor: Coolant type (Water N1 and Oil N2)

23 1 2 2 1 1 2

14 1 2 2 2 2 1

15 2 1 2 1 2 2

26 2 1 2 2 1 1

27 2 2 1 1 2 1

18 2 2 1 2 1 2

3 samples results exposed to noise condition N1.

3 samples results exposed to noise condition N2.

11 1 1 1 1 1 1

22 1 1 1 2 2 2

23 1 2 2 1 1 2

14 1 2 2 2 2 1

ETr A B - C D F N1z z z z z z N2 y

11.511.5 11.8 11.3 14.1 14.5 13.8 14.3 12.8

8.79.2 8.7 8.2 9.3 10.7 9.6 9.9 9.3

11.711.7 11.8 11.5 14.3 14.4 14.1 14.3 12.9

12 712 7 12 7 12 6 15 6 15 6 15 4 15 5 14 1

N1 N2

Results (y) ______

(Two-Step Optimization – Noise x Control factor InteractionsRef. Pg. 5-16

14 1 2 2 2 2 1

15 2 1 2 1 2 2

26 2 1 2 2 1 1

27 2 2 1 1 2 1

18 2 2 1 2 1 2

12.712.7 12.7 12.6 15.6 15.6 15.4 15.5 14.1

13.713.8 13.5 13.8 13.3 12.8 12.4 12.8 13.3

13.413.2 13.5 13.4 16.2 16.6 16.4 16.4 14.9

12.512.6 12.9 12.1 15.4 15.8 14.8 15.3 13.9

12.012.3 11.7 12 15.1 14.3 14.2 14.7 13.3


Average effect of A1N1 = (11.5 + 8.70 + 11.7 + 12.70 ) / 4 = 11.15 (Highlighted data)

Noise and Control Factor Interaction Effects

Noise N1

Noise N2

11.15 12.90 11.82 12.21 12.36 11.68 12.14 11.90 12.48 11.57 12.52 11.52

13.49 14.82 13.31 14.97 14.15 14.13 14.88 13.39 14.30 13.98 15.35 12.93

A1 A2 B1 B2 C1 C2 D1 D2 E1 E2 F1 F2

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11 1 1 1 1 1 1

22 1 1 1 2 2 2

23 1 2 2 1 1 2


11.511.5 11.8 11.3 14.1 14.5 13.8 14.3 12.8

8.79.2 8.7 8.2 9.3 10.7 9.6 9.9 9.3

11.711.7 11.8 11.5 14.3 14.4 14.1 14.3 12.9

N1 N2

Results (y) ______

(Two-Step Optimization – Noise x Control factor Interactions

Ref. Pg. 5-16

14 1 2 2 2 2 1

15 2 1 2 1 2 2

26 2 1 2 2 1 1

27 2 2 1 1 2 1

18 2 2 1 2 1 2

12.712.7 12.7 12.6 15.6 15.6 15.4 15.5 14.1

13.713.8 13.5 13.8 13.3 12.8 12.4 12.8 13.3

13.413.2 13.5 13.4 16.2 16.6 16.4 16.4 14.9

12.512.6 12.9 12.1 15.4 15.8 14.8 15.3 13.9

12.012.3 11.7 12 15.1 14.3 14.2 14.7 13.3



Average effect of A2N1 = ( 13.7 + 13.4 + 12.5 + 12.0)/4 = 12.90


Noise N1

Noise N2

11.15 12.90 11.82 12.21 12.36 11.68 12.14 11.90 12.48 11.57 12.52 11.52

13.49 14.82 13.31 14.97 14.15 14.13 14.88 13.39 14.30 13.98 15.35 12.93

A1 A2 B1 B2 C1 C2 D1 D2 E1 E2 F1 F2

11 1 1 1 1 1 1

22 1 1 1 2 2 2

23 1 2 2 1 1 2


11.511.5 11.8 11.3 14.1 14.5 13.8 14.3 12.8

8.79.2 8.7 8.2 9.3 10.7 9.6 9.9 9.3

11 711 7 11 8 11 5 14 3 14 4 14 1 14 3 12 9

N1 N2

Results (y) ______

(Two-Step Optimization – Noise x Control factor Interactions

Ref. Pg. 5-16

23 1 2 2 1 1 2

14 1 2 2 2 2 1

15 2 1 2 1 2 2

26 2 1 2 2 1 1

27 2 2 1 1 2 1

18 2 2 1 2 1 2

11.711.7 11.8 11.5 14.3 14.4 14.1 14.3 12.9

12.712.7 12.7 12.6 15.6 15.6 15.4 15.5 14.1

13.713.8 13.5 13.8 13.3 12.8 12.4 12.8 13.3

13.413.2 13.5 13.4 16.2 16.6 16.4 16.4 14.9

12.512.6 12.9 12.1 15.4 15.8 14.8 15.3 13.9

12.012.3 11.7 12 15.1 14.3 14.2 14.7 13.3

Grand Averages => 12.0 14.1 13.1Noise and Control Factor Interaction Effects

A1 A2 B1 B2 C1 C2 D1 D2 E1 E2 F1 F2

Noise N1

Noise N2

11.15 12.90 11.82 12.21 12.36 11.68 12.14 11.90 12.48 11.57 12.52 11.52

13.49 14.82 13.31 14.97 14.15 14.13 14.88 13.39 14.30 13.98 15.35 12.93

Average effect of A2N2 = ( 12.8+ 16.4 + 15.3 + 14.7)/4 = 14.82

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Noise N1

Noise N2

11.15 12.90 11.82 12.21 12.36 11.68 12.14 11.90 12.48 11.57 12.52 11.52

13.49 14.82 13.31 14.97 14.15 14.13 14.88 13.39 14.30 13.98 15.35 12.93

A1 A2 B1 B2 C1 C2 D1 D2 E1 E2 F1 F2

14.50

14.00

15.50

15.00

A2 B2

A2N1 – A2N2 = 12.90 –14.82

Ref. Pg. 5-16

13.50

13.00

12.50xx12.00

11.50

11.0

N2N1 Noise N2N1 Noise N2N1 Noise

B1

A1

C1

B2

C2

14.50

14 00

15.50

15.00

E1F1

A1N1 – A1N2 = 11,15 – 13.49

Factor levels selected based on strong

14.00

13.50

13.00

12.50xx12.00

11.50

11.0

N2N1 Noise N2N1 Noise N2N1 Noise

D1

E1

D2

E2

F2

Fig. 5D-1 Plot of Noise and Control factor interactions (NxA, NxB, .. NxF)

ginteractions:

B1, C1, D2 and F2

13.60

13.40

13.20

14.00

13.80

Ref. Pg. 5-17

13.00

12.80

12.60

12.40

12.20

A2A1 Factor B2B1 Factor C2C1 Factor

13.60

13 40

14.00

13.80

Factor levels selected based on main effect:

A1 and E2

13.40

13.20

13.00

12.80

12.60

12.40

12.20

D2D1 Factor E2E1 Factor F2F1 Factor

Fig. 5D-2 Plot of Average Effects of Factors (Main Effects of A, B, C, D, E, and F)

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(a) Two-Step Optimization (Noise Interaction) - Summary

1. Reduce Variability by identifying factors which interact with noise.

•Factors with strong interaction: B, C, D, and F (A and E are found to

Ref. Pg. 5-18

have less interaction with N, See interaction Plots above)

•Robust factor levels: B1, C1, D2 and F2

2. Adjust mean by selecting factors with least interaction with noise.

•Factors with lesser interaction: A and E

•Levels for mean closer to target: A1 and E2 (Smaller is better QC, see Plots of Main effect above)

11 1 1 1 1 1 1

22 1 1 1 2 2 2

23 1 2 2 1 1 2

14 1 2 2 2 2 1

ETr A B - C D F yz z z z z z Sd

12.811.5 11.8 11.3 14.1 14.5 13.8 1.45

9.39.2 8.7 8.2 9.3 10.7 9.6 0.85

12.911.7 11.8 11.5 14.3 14.4 14.1 1.43

14 112 7 12 7 12 6 15 6 15 6 15 4 1 57

N1 N2

Results (y) __

(b) Two-Step Optimization–Average and Standard Deviation AnalysesRef. Pg. 5-18

14 1 2 2 2 2 1

15 2 1 2 1 2 2

26 2 1 2 2 1 1

27 2 2 1 1 2 1

18 2 2 1 2 1 2

14.112.7 12.7 12.6 15.6 15.6 15.4 1.57

13.213.8 13.5 13.8 13.3 12.8 12.4 0.56

14.913.2 13.5 13.4 16.2 16.6 16.4 1.67

13.912.6 12.9 12.1 15.4 15.8 14.8 1.59

13.312.3 11.7 12 15.1 14.3 14.2 1.43

Grand Averages => 13.1 1.32

Factor Effects on Standard Deviation of Results

Average effect of A2 = ( 1.4 + 0.80 + 1.4 + 1.5)/4 = 1.33 on St.Dev. (data rounded)

A B - C D E F

1.33 1.13 - 1.26 1.50 1.26 1.57Level 1

1.31 1.51 - 1.38 1.14 1.38 1.07Level 2

-.01 .37 - .12 .35 .13 -.50Diff. L2 - L1

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1.40

1.35

1.30

1.25

1.20

1.15

1.10

1.50

1.45

Ref. Pg. 5-19

1.05


1.40

1.35

1.30

1.25

1.50

1.45

Factor levels selected based on Std. Deviation:

B1, D2 and F2

1.20

1.15

1.10

1.05


Fig. 5D-3 Plot of Average factor Effects on Standard Deviation of Results

(b) Two-Step Optimization (Mean & std. Deviation) - Summary

1. Reduce Variability by identifying factors with significant effects of standard deviation of results

Ref. Pg. 5-20

standard deviation of results.

•Significant Factors: B, D, and F (A and E are found to have less interaction with N, See interaction Plots above)

•Levels for least variability: B1, D2 and F2

2. Adjust mean by selecting factors with less interaction with noise.

•Factors with lesser effects on St.Dev: A, C and E

•Factor Levels: A1, C2 and E2 (Smaller is better QC, see Plots of Main effect above)

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11 1 1 1 1 1 1

22 1 1 1 2 2 2

23 1 2 2 1 1 2

14 1 2 2 2 2 1

ETr A B - C D F yz z z z z z Sd S/N

12.811.5 11.8 11.3 14.1 14.5 13.8 1.45 -22.2

9.39.2 8.7 8.2 9.3 10.7 9.6 0.85 9.3

12.911.7 11.8 11.5 14.3 14.4 14.1 1.43 12.9

14 112 7 12 7 12 6 15 6 15 6 15 4 1 57 14 1

N1 N2

Results (y) __S/N Ratios

-22.30

23 03

-22.21

-19.38

(c) Equivalent Two-Step Optimization – S/N Ratio AnalysisRef. Pg. 5-21

14 1 2 2 2 2 1

15 2 1 2 1 2 2

26 2 1 2 2 1 1

27 2 2 1 1 2 1

18 2 2 1 2 1 2

14.112.7 12.7 12.6 15.6 15.6 15.4 1.57 14.1

13.213.8 13.5 13.8 13.3 12.8 12.4 0.56 13.2

14.913.2 13.5 13.4 16.2 16.6 16.4 1.67 14.9

13.912.6 12.9 12.1 15.4 15.8 14.8 1.59 13.9

13.312.3 11.7 12 15.1 14.3 14.2 1.43 13.3

Grand Averages => 13.1 1.32 13.1-22.29

-23.03

-22.93

-22.50

-22.46

-23.50

Factor Average Effects Based on S/N Ratios

Average effect of A2 = ( -22.46 – 23.50 – 22.93 -22.50)/4 = -22.847 on S/N(data rounded)

A B - C D E F

-21.73 -21.89 - -22.48 -22.63 -22.55 -22.92Level 1

-22.847 -22.69 - -22.10 -21.95 -22.03 -21.66Level 2

-1.12 -.80 - .37 .68 .52 1.26Diff. L2 - L1

-21.60

-21.80

-22.00

-22.20

-22.40

-22.60

22 80

-21.20

-21.40

Factor levels selected

Ref. Pg. 5-22

-22.80

-23.00


-21.60

-21.80

-22.00

-22.20

-21.20

-21.40

se ectedbased on S/N:

A1, B1, D2 and F2

Note: In case of S/N analysis, higher level value is always selected no -22.40

-22.60

-22.80

-23.00


Fig. 5D-4 Plot of Average Effects of Factors (S/N Effects of A, B, C, D, E, and F)

selected no matter the original QC. The QC changes the way MSD is defined.

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1 1A:Injection Press. 2.073 18.966.0312.4862.486

2 1B:Mold Closing Sp. .865 7.913.0991.2771.277

3 1Int. AxB 1.865 17.055.5252.2772.277

4 (1)C:Mold Pressure Pooled(.278)

5 1D B k P 503 4 602 222915915

ANOVA Statistics (S/N of Results)

# DOFFactor & Int. S’ P(%)FVSS

Ref. Pg. 5-23

2Other /Error 26.38.412.824

7Total: 100%10.937

5 1D:Back Pressure .503 4.602.222.915.915

6 (1)E:Screw Speed Pooled(.545)

7 1F:Spear Temp. 2.743 25.097.6573.1563.156

Optimum condition: A1 B1 D2 F2 (Factors C & E are pooled)

Yopt = 22 29 + ( 21 73 22 29 ) + ( 21 89 22 29 ) + ( 21 95 22 29 ) + ( 21 66 22 29 )Yopt = - 22.29 + (-21.73 - - 22.29 ) + (-21.89 - - 22.29 ) + (-21.95 - - 22.29 ) + (-21.66 - - 22.29 )= - 22.29 +(0.56 + 0.4 + 0.34 + 0.63)= - 22.29 + 1.93= - 20.36 S/N (which translates to 10.4 in the original unite sof results)

(Pooled factors and interaction column effects are not included in the estimate of the optimum performance)

( c) Two-Step Optimization (S/N Analysis) - Summary

1. Reduce Variability by identifying factors with significant effects on

Ref. Pg. 5-24

y y y g gS/N.

• Significant Factors: A, B, D, and F (C and E are found to have less interaction with N, See interaction Plots above)

• Factor Levels: A1, B1, D2 and F2 (Larger value of S/N always preferred regardless of the QC)

2. Adjust mean by selecting factors with lesser effect on S/N.

• Factors with lesser effects on S/N: C and E

• Factor Levels: C2 and E2 (Smaller is better QC, see Plots of Main effect above)

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ANOTHER EXAMPLE OF SINGLE

Ref. Pg. N/A

NOISE AND MULTIPLE CONTROL FACTOR ROBUSTNESS

Example 7B: Results with Six Samples in Each Trial

Ref. Pg. 5-25

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Results Involved in Interaction Calculations

Ref. Pg. 5-26

Interaction Effects

Ref. Pg. 5-26

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Control to Noise Factors Interactions

Ref. Pg. 5-27

Five Separate Analyses for Factor Level Selection1. Analysis based on Robustness – This approach allows factor level

selection based on interactions between control and noise factor. Of course, when there are multiple noise influences, conflict in factor level selection may arise

Ref. Pg. N/A

arise.

2. Analysis using average of trail results – This should be performed when improvement in mean performance is desire.

3. Analysis using the standard deviation of results – This is a direct way to determine factors more influential on variability and determine factor levels for reduced variability in performance.

4. Analysis using the S/N ratios of trial results – This is preferred option to determine optimum condition based on mean and variability It is a preferreddetermine optimum condition based on mean and variability. It is a preferred alternative to robust design approach which can be comparatively time consuming.

5. Analysis based on interactions between two factors – Consideration of interactions between two 2-level factors produce information that dictate factor level selections when its effect is present and found significant. Unfortunately, different pairs of interactions may put conflicting demand on selection of the factor level.

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So, which type of analysis should one follow?

Recommended Analysis Approach

Ref. Pg. N/A

So, c type o a a ys s s ou d o e o o

• Variability - Standard Deviation

• Robustness - Robust Design by Noise interaction

• General Case (recommended) - S/N ratio

Analysis Approaches (Overview)

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AN EXAMPLE OF MULTIPLE NOISE AND MULTIPLE CONTROL FACTOR

Ref. Pg. N/A

AND MULTIPLE CONTROL FACTOR ROBUSTNESS

Structure and Information in Designs with Noise factors Ref. Pg. 5-33

ETr A B - C D F

Outer Array (L-4)

NC

NB

NA

1

1

1

1

2

2

2

1

3

2

1

2

4

1

2

2Control Factors

Noise

NA:Oven Type, NB:Tempertaure, NC:Humidity

11 1 1 1 1 1 1

22 1 1 1 2 2 2

23 1 2 2 1 1 2

14 1 2 2 2 2 1

15 2 1 2 1 2 2

26 2 1 2 2 1 1

27 2 2 1 1 2 1

18 2 2 1 2 1 2

Inner Array (L-8)

R12

R22

R32

R42

R52

R62

R72

R82

R11

R21

R31

R41

R51

R61

R71

R81

R13

R23

R33

R43

R51

R63

R73

R83

R14

R24

R34

R44

R54

R64

R74

R84


Order of Sophistication in Experiment Design (From most to least desirable options)

1. Formal treatment of noise factors by outer array design

2. Repeating experiments with "RANDOM NOISE“

3. Run multiple samples per trial (Simply repeat)

4. Run one sample per trial (Poor man's experiment)

e ay ( 8)

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Ref. Pg. N/A

Example 7C: Piston Bearing Design Optimization Study (Piston1.q4w in Qualitek-4)

Ref. Pg. N/AS/N of Results

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Ref. Pg. N/AMain Effects

Ref. Pg. N/AANOVA Output

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Ref. Pg. N/AOptimum Condition and Performance

Ref. Pg. N/A

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Ref. Pg. N/A

Noise EffectsNoise Effects (3 factors, 8 samples in each trial)

Ref. Pg. N/ANoise Effects

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Example 7D: Engine Idle Stability Study3 - Factors at 3-levels each3 - Repetitions ( normal operating noise)

S/N Analysis Example Application

Ref. Pg. 5-16

DESIGN FACTORS AND THEIR LEVELSCol. # FACTORS LEVEL-1 LEVEL-2 LEVEL-3

1 INDEXING -30 DEG O DEG +50 DEG2 OVERLAP AREA -30% 0 % +30 %3 SPARK ADVANCE 20 DEG 30 DEG 40 DEG4 UNUSED/UPGRADED M/U

CHARACTERISTIC: THE SMALLER THE BETTER

L9(34)

COND 1 2 3 4

TRIAL RESULTS Trial# R(1) R(2) R(3) S/N Ratios 1 20 25 26 -27.5358


1 1 1 1 12 1 2 2 23 1 3 3 34 2 1 2 35 2 2 3 1 6 2 3 1 27 3 1 3 28 3 2 1 39 3 3 2 1

0 5 6 53582 34 36 26 -30.1815 3 45 34 26 -31.0913 4 13 23 22 -25.9550 5 36 45 35 -31.8051 6 23 25 34 -28.86497 35 45 53 -33.0528 8 56 46 75 -35.5939 9 35 46 53 -33.1175

Example Analysis of Results

MAIN EFFECTS (QUALITEK-4 software output)COL#/FACTORS LEVEL 1 LEVEL 2 LEVEL 3 LEVEL 4 (L2-L1)1 INDEXING -29.6028 -28.8749 -33.9214 0.0000 0.72782 OVERLAP AREA -28.8478 -32.5268 -31.0245 0.0000 -3.67893 SPARK ADVANCE -30.6648 -29.7513 -31.9830 0.0000 0.9135

QUALITY CHARACTERISTIC THE SMALLER THE BETTER

Ref. Pg. 5-16

QUALITY CHARACTERISTIC: THE SMALLER THE BETTERDATA TYPE : S/N RATIO

A N O V A TABLE DATA TYPE: S/N RATIO COL.#/FACTOR f S V F S' P(%)1 INDEXING 2 44.6445 22.3222 846.5925 44.5917 61.27152 OVERLAP 2 20.5283 10.2641 389.2777 20.4755 28.13453 SPARK ADV. 2 7.5517 3.7758 143.2037 7.4990 10.3040



OPTIMUM TABLECOL #/FACTORS LEVEL DESC. LEVEL# CONTRIBUTION

1 INDEXING O DEG 2 1.92472 OVERLAP AREA -30% 1 1.95183 SPARK ADVANCE 30 DEG 2 1.0484

TOTAL CONTRIBUTION FROM ALL FACTORS.... 4.9251CURRENT GRAND AVERAGE OF PERFORMANCE -30.7998EXPECTED RESULT AT OPTIMUM CONDITION... -25.8747

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Recommended Experiment Design Considerations

Ref. Pg. 5-36Experiment Design Roadmap

Designs using standard arrays

g

Mixed-level factors and interaction

designs

•Consider incorporating NOISE (Outer array)

•Decide on number repititions

Assign factors to columns arbitrarily

Modify columns and reserve interaction

columns prperly


Run experiments in random order when

possible

Module – 6ANOVA


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80

What steps are involved in

L-8 Orthogonal ArrayA C AxC B D BxC E

Trial 1 2 3 4 5 6 7 RESULTS(y)1 1 1 1 1 1 1 1 42.00 2 1 1 1 2 2 2 2 50.00

Calculations of ANOVA Statistics Ref. Pg. 6-3

studying the influence of Interactions?

8 2 2 1 2 1 1 2 54.00

3 1 2 2 1 1 2 2 36.00 4 1 2 2 2 2 1 1 45.00 5 2 1 2 1 2 1 2 35.00 6 2 1 2 2 1 2 1 55.00 7 2 2 1 1 2 2 1 30.00 8 2 2 1 2 1 1 2 54.00

_or A1 = 173/4 = 43.25

Level Totals and their Average Effects

A = (y + y + y + y ) = 42 + 50 + 36 + 45 = 173


1_

or A2 = 174/4 = 43.50_

or C1 = 182/4 = 45.50_

or C2 = 165/4 = 41.25

A1 = (y1 + y2 + y3 + y4) = 42 + 50 + 36 + 45 = 173

A2 = (y5 + y6 + y7 + y8) = 35 + 55 + 30 + 54 = 174.

C1 = (y1 + y2 + y5 + y6) = 42 + 50 + 35 + 55 = 182.

C1 = (y3 + y4 + y7 + y8) = 36 + 45 + 30 + 54 = 165

Plot of Factor Average Effects

Optimum Condition without Interactions

Ref. Pg. 6-5

A1 A2 D1 D2C1 C2 B1 B2 E1 E2

30

40

50

T= 43.375_


Optimum Condition: A1C2B1D2E1

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What about effects of Interaction?How do we test for PRESENCE OF INTERACTION?Special calculations needed.


Ref. Pg. 6-6

Special calculations needed.How to determine if the Interaction is SIGNIFICANT?(Done in ANOVA for reserved columns only.)

40

50

T= 43.375_

Plot of Interaction Average Effects


(AxC)1 (AxC)2 (BxC)1 (BxC)2

30

____A1C1 = y1 + y2 = 42 + 50 = 92 or A1C1 = 92/2 = 46.00

____A1C2 = y3 + y4 = 36 + 45 = 81 A1C2 = 81/2 = 40.50

Plots for Test of Presence of Interactions

Ref. Pg. 6-8

A1C2 y3 y4 36 45 81 A1C2 81/2 40.50____

A2C1 = y5 + y6 = 35 + 55 = 90 A2C1 = 90/2 = 45.00____

A2C2 = y7 + y8 = 30 + 54 = 84 A2C2 = 84/2 = 42.00

50

Plots of Interactions52.5

49.5B2

46.0 A1


C1 C2

30

40

C1 C2

38.5

33

B1

45.040.5

42.5A1A2

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Calculation of Sums of Squares for ANOVA

n __ST = ∑ [ Yi - Y ] 2 Which can be reduced to the following form

i 1

Ref. Pg. 6-10

Step 4 Factor sum of squares:

i=1

8ST = ∑ Yi

2 - C.F (Correction Factor = T2 / n )i=1

= ( 422 + 502 + 362+ ..... + 542) - 15051.125= 599.88


Step 4. Factor sum of squares:

SA = A12/NA1 + A22/NA2 - C.F

= 1732 /4 + 1742/4 - 15051.125 = 0.125SB= B12/NB1 + B12/NB2 - C.F

= 1432 /4 + 2042 /4 - 15051.125 = 465.125

A1 = 8 A2 = 8B1 =5 B2 = 11C1 = 9 C2 = 7

L-4 Orthogonal ArrayTrial #A B C Result1 1 1 1 3

An Example Data for ANOVA

Ref. Page N/A

C1 9 C2 7

Avg. = 4,ST = 10, CF = 64

SA = 64/2 + 64/ 2 – 64 =0SB = 25/2 + 121/2 – 64 = 9SC = 81/2 + 49/2 – 65 = 1

1 1 1 1 32 1 2 2 53 2 1 2 24 2 2 1 6


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Se = ST - ( SA + SB + SC + SD + SE +SAxC + SBxC )= 599.88 - 599.88 = 0

St 5 E i t D f F d

Calculations of Error Terms and & DOF

Ref. Pg. 6-12

Step 5. Experiment Degrees of FreedomfT = n - 1 = 8 - 1 = 7 (Total number of results - 1)

fe = fT - ( fA + fB+ fC + fD +fE +fAxC + fBxC ) = 0Step 6. factor degrees of freedom

fA = (# levels of factor A) - 1 = 2 - 1 = 1fB = (# levels of factor B) - 1 = 2 - 1 = 1

How is Variance calculated?VA = SA /fA = 0.125 / 1 = 0.125VB = SB /fB = 465.125 / 1 = 465.125


How is F-ratios calculated?How is PUR SUMS OF SQUARES calculated?How is percent INFLUENCE calculated?

PA = SA/ST = 0.125/ 599.88 = 0.02%PB = SB/ST = 465.125/ 599.88 = 77.54%

ANOVA Table with Calculated Statistics

Ref. Page 6-14


How are SIGNIFICANCE of Factors and Interactionsdetermined?

What is the procedure for POOLING Factorsand Interactions?

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Step 1. Sum of squares of error termSe = ST - (SB + SC + SD ) = 599.88 - (592.4) = 7.5(neglecting all terms pooled)Note: Se will now be equal to (SA + SE + SAxC + SBxC) i.e. Sum of all factors pooled.Step 2. Degree of freedom of error termfe = fT - (fB + fC + fD )= 7 - 3 = 4

Method for Revising ANOVA Statistics Ref. Page 6-16

e T ( B C D )

Step 3. Variance of error termVe = Se /fe = 7.5 / 4 = 1.875


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Is factor C significant? Should we pool factor C ?

To study the possibility of pooling, is to perform test of significance for factor C. The significance test is carried out by comparing the experiment F-ratio with the standard table

Performing Test of Significance for Pooling

Ref. Page 6-17

significance test is carried out by comparing the experiment F-ratio with the standard table value for a desired Confidence Level (subjective: 90, 95 or 99% commonly used).

From the ANOVA table FC = 19.267 (From Experiment)

From The F-TABLE, Find F Value At n1 = DOF OF Factor C = 1n2 = DOF Of Error Term = 4 At A Confidence Level (say 95%) .

F (table) = 7.7086 (From F-Table)

Since F Is Greater Than the F(table)


Since FC Is Greater Than the F(table),Factor C should not be pooled.

Recommendations: Create non-zero DOF for the error term by pooling the weaker factors. Test for significance first, then pool. Attempt to pool until the DOF for the error term is about half the experiment DOF.

The F-ratios are contained in F-Table developed by R. A Fisher and are available in most common books on statistical science.

Reading the F-Table

Ref. Page 6-19

In some text, the F-Tables are specified by level of significance instead of the confidence level. The level of significance (symbol: Greek letter Alpha) is complementary to the level of confidence. That is to say 95% confidence level is equivalent to 5% level of significance.

Sample Table Readings at 95% confidence level: F (3, 8 ) = 4.0661(3rd column, 8th row) F (2,5) = 5.7862 ( 2nd column ,5th row)

To perform test of significance at other confidence levels, the corresponding F-Table must be secured first. F-TABLE (95%)

n2\n1 1 2 3 4 5 6 7 ..etc.1 161.45 199.50 215.71 224.58 230.16 233.99 236.77 2 18.513 19.000 19.164 19.247 19.296 19.330 19.353 3 10.128 9.5521 9.2766 9.1172 9.0135 8.9406 8.8868

See enlarged view in the next slide


4 7.7086 6.9443 6.5914 6.3883 6.2560 6.1631 6.09425 6.6079 5.7862 5.4095 5.1922 4.3874 4.2839 4.2066 6 5.9874 5.1433 4.7571 4.5337 4.3874 4.2839 4.2066 7 5.5914 4.7374 4.3468 4.1203 3.9725 3.8660 3.7870 8 5.3277 4.3459 4.0661 3.8378 3.6875 3.5806 3.5005 9 5.1174 4.2565 3.7626 3.6331 3.4817 3.3738 3.2927 10 4.9646 4.1028 3.7083 3.4780 3.3258 3.2172 3.1355 11 4.8443 3.9823 3.5874 3.3567 3.2039 3.0946 3.0123 12 4.7472 3.8853 3.4903 3.2592 3.1059 2.9961 2.9134 13 4.6672 3.8056 3.4105 3.1791 3.0254 2.9153 2.8321

in the next slide

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86

F-TABLE (95%)n2\n1 1 2 3 4 5 6 7 ..etc.1 161.45 199.50 215.71 224.58 230.16 233.99 236.77

F-Table for 95% Confidence Level

Ref. Page 6-19

2 18.513 19.000 19.164 19.247 19.296 19.330 19.353 3 10.128 9.5521 9.2766 9.1172 9.0135 8.9406 8.8868 4 7.7086 6.9443 6.5914 6.3883 6.2560 6.1631 6.09425 6.6079 5.7862 5.4095 5.1922 4.3874 4.2839 4.2066 6 5.9874 5.1433 4.7571 4.5337 4.3874 4.2839 4.2066 7 5.5914 4.7374 4.3468 4.1203 3.9725 3.8660 3.7870 8 5.3277 434590 4.0661 3.8378 3.6875 3.5806 3.5005 9 5 1174 4 2565 3 7626 3 6331 3 4817 3 3738 3 2927


9 5.1174 4.2565 3.7626 3.6331 3.4817 3.3738 3.2927 10 4.9646 4.1028 3.7083 3.4780 3.3258 3.2172 3.1355 11 4.8443 3.9823 3.5874 3.3567 3.2039 3.0946 3.0123 12 4.7472 3.8853 3.4903 3.2592 3.1059 2.9961 2.9134 13 4.6672 3.8056 3.4105 3.1791 3.0254 2.9153 2.8321 etc. .. ..

The confidence interval (C.I.) = +/- \/ (F(1,n2) x Ve/Ne)

h F(1 ) F l f f t bl t d f 1 & t i d

Confidence Level Formula & Calculation

Ref. Page 6-20

For factor C at level C1n2 = 4N = 8/(1+1) = 4 x F(1 4) = 7 7086 AT 95% Confidence Level

where F(1,n2) = F-value from f table at dof 1 & n2 at a requiredconfidence level.

n2 = degrees of freedom of the error term.Ve = variance of error term (from ANOVA).

ne = effective # of replications

= DOF of mean (=1 always) + DOF of the factorTotal # of results ( or # of S/N ratios)


n2 = 4Ne = 8/(1+1) = 4 x F(1,4) = 7.7086 AT 95% Confidence Level

Which gives C.I. = +/- 1.9034 AT 95% Confidence Level.

i.e the main effect of factor C at level C1 will be 45.5 +/- 1.9034in 95 out of every 100 experiments conducted.

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87

At the OPTIMUM condition,n2 = 4 Ne = 8/(1+3) = 2 F(1,4) = 7.7086 at 95% confidence l l

Confidence Interval for Optimum Performance

Ref. Page 6-21

level

C.I. = +/- 2.2039 at 95% confidence level

n2 = 4 Ne = 8/(1+3) = 2 F(1,4) = 7.7086 at 95% confidencelevel

C.I. = +/- 2.2039 at 95% confidence level


__Grand average of performance: T = 347 / 8 = 43.375

_ _ _ _ _ _ _Yopt = T + (B1 - T) + (D2 - T) + (C2 - T)

= 43.375 + (41.25 - 43.375) + (40 - 43.375) + (33 - 43.375)= 30.25

Optimum Condition & Performance in Table Form

Ref. Page 6-21


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6.7 Example 8B: Standard analysis with multiple runsQC= Smaller is better (Same design as Ex. 8A)

Trial \ Results R1 R2 R3 Average

ANOVA for Multiple Samples Test Results

Ref. Page 6-25

g1 38.00 42.00 46.00 42.002 45.00 50.00 55.00 50.003 38.00 36.00 34.00 36.004 55.00 45.00 35.00 45.005 30.00 35.00 40.00 35.006 65.00 55.00 45.00 55.007 40.00 30.00 20.00 30.008 58.00 54.00 50.00 54.00


DOF = 8 x 3 - 1 = 23

DOF with Multiple Samples Test Results

Ref. Page 6-26

B1 = (38.0 + 42.0 + 46.0) + (38.0 + 36.0 + 34.0)+ (30.0 + 35.0 + 40.0) + (40.0 + 30.0 + 20.0) = 429

Note: B1 is obtained from results of trial conditions 1, 3, 5 & 7.


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89

ANOVA Screen After Pooling Insignificant Factors

Ref. Page 6-27


Trial \ Rep. R1 R2 R3 S/N1 38.00 42.00 46.00 - 11.672 45.00 50.00 55.00 -20.673 38 00 36 00 34 00 12 72

S/N Analysis for NOMINAL QC

Ref. Page 6-28

For trial # 1,

MSD = [ ( y1 - yo )2 + (y2 - yo )2 ...... ]/n

3 38.00 36.00 34.00 -12.724 55.00 45.00 35.00 -19.635 30.00 35.00 40.00 -16.216 65.00 55.00 45.00 -24.657 40.00 30.00 20.00 -22.228 58.00 54.00 50.00 -23.16


or MSD = [ ( 38 - 40 )2 + ( 42 - 40 )2 + ( 46 - 40 )2] /3

= 14.667and

S/N = - 10 LOG10 (MSD) = -10 LOG10 (14.667) = -11.67

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90

Main Effects_A = [ -11 67 + (-20 67) + (-12 72) + (-19 63)]/ 4 = -16 17

Calculation of Main Effects (S/N –Nominal)

Ref. Page 6-29

A1 = [ -11.67 + (-20.67) + (-12.72) + (-19.63)]/ 4 = -16.17_A2 = [ -16.21 + (-24.65) + (-22.22) + (-23.16)] / 4 = -21.55, etc.

Main EffectsCol.# Factor Level 1 Level 2 (L2 - L1)1 FACTOR A -16.17 -21.55 - 5.392 FACTOR C -18.29 -19.43 - 1.143 INT A C 19 43 18 29 1 13


3 INT. AxC -19.43 -18.29 - 1.134 FACTOR B -15.70 -22.02 - 6.335 FACTOR D -18.04 -19.68 - 1.646 INT. BxC -17.66 -20.06 - 2.417 FACTOR E -19.54 -18.18 1.35

A N O V A Table

ANOVA Statistics (S/N –Nominal)

Ref. Page 6-30

Cols:FACTORS f S V F S' P1 FACTOR A 1 58.063 58.063 16.44 54.53 33.302 FACTOR C (1) (2.56) POOLED3 INT. AxC (1) (2.56) POOLED4 FACTOR B 1 80.029 80.029 22.659 76.50 46.715 FACTOR D (1) (5.33) POOLED6 INT. BxC 1 11.548 11.548 3.270 8.02 4.877 FACTOR E (1) (3.680) POOLEDOTHER/ERROR 4 14 130 3 53 15 10



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91

Optimum Performance (S/N –Nominal)

Ref. Page 6-31

Estimate of Performance at the Optimum Condition

FACTOR DESCRIPTION LEVEL LEVEL# CONTRIBUTIONFACTOR A A1 1 2.6940 FACTOR B B1 1 3.1628 INTERACT 2 2 x 4 1 1.2014TOTAL CONTRIBUTION FROM ALL FACTORS ... 7.0583 CURRENT GRAND AVERAGE OF PERFORMANCE -18.8604


EXPECTED RESULT AT OPTIMUM CONDITION ... -11.8021

Note : 1. The Expected result at optimum condition is the S/N Ratio of

Transformation of S/N

Ref. Page 6-32

the results.

2. Since S/N = - 10 log10(MSD)

and MSD = 10- (S/N /10 ) = 15.142

i.e. The results at optimum condition are expected to produce________

Y = 40 +/- \/(15.142)


( )

= 40 +/- 3.891 [Ref. 6-28]

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6. 9 Example 9B: S/N Analysis - Smaller is better(Same experiment as example 8A)

S/N Analysis with Bigger and Smaller QC

Ref. Page 6-32

MSD = ( Y12 + Y2

2 + Y32 + ......) /n

For Trial # 1MSD = ( 382 + 422 + 462 ) / 3 = 1774.666S/N = -10 LOG10(MSD) = -32.50

6.10 Example 9C: S/N analysis - Bigger is better (Same example 8B)MSD = [ 1/y1

2+ 1/y22 + ........ ] / n


For Trial # 1MSD = ( 1/382 + 1/422 + 1/462 ) / 3 = 5.773 x 10

S/N = - 10 Log10(MSD) = 32.38

Expression for Confidence Interval

Ref. Page 6-38

Expected Range of Results (at 90% C I )


Expected Range of Results (at 90% C. I.)

Lower limit: S/N = 34.2975Result = 51.86Mean Value: S/N = 35.349 Result = 58.54Upper Limit: S/N = 36.4005 Result = 66.07

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The expected limits in the original units of measurement can be calculated as follows: S/N = 34.2975, 35.349, and 36.4005

Transformation of C.I. in S/N into original Units

Ref. Page 6-38

WhenS/N = 34.2975, Yexp. = Sqr. Root { 1/( 10 - [S/N)/10] )}

= Sqr. Root ( 1/ 10 -3.42975 ) = 51.86

S/N = 35.349, Yexp. = Sqr. Root { 1/( 10 - [S/N)/10] )}

= Sqr Root ( 1/ 10 -3 5349 ) = 58 54


= Sqr. Root ( 1/ 10 3.5349 ) = 58.54

S/N = 36.4005, Yexp. = Sqr. Root { 1/( 10 - [S/N)/10] )}

= Sqr. Root ( 1/ 10 -3.64005 ) = 66.07

6.11 Understanding and Interpreting Error Terms(Signature Plots)

Relationship between the Results and the Error Term

Ref. Page 6-39

Trials

Array

Runs


1. Is the experiment satisfactory?2. What if the confirmation is unsatisfactory?3. When should you consider repeating the experiment

from the start?

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Module – 7Loss Function


Why Study Loss?Difference in Loss makes a difference.

Ref. Page N/A

e

After I t

Before

$3M$ R

even

ue ImprovementImprovement

Saving = 5M


$8M

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95

Loss Function – How it Evolved

Ref. Page N/A

$Loss $Loss Plot

Not good.

Loss

Not good.

Loss

Good.

No LossGood.

No Loss


Yo USL LSL The loss plot (GREEN LINE) results when parts are made to specification. Under this strategy, everything that falls within the specification limit is good. The loss due to rejects is essentially zero. At the same time, as soon as the part exceeds the limits, the loss equals cost of production ( a fixed value).

Loss Function – How it Evolved

Ref. Page N/A

$Loss L = K (Y - Yo )2

Not good.

Loss

Not good.

Loss

Good.

No LossLinear Cubic


Yo USL LSL

Natural trends of loss suggest it to be a continuous function from NO LOSS to a FIXED VALUE outside the specification limits (symmetrical)

Quadratic

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96

L = K (y - yo)2 $ Per unit of product for single unitFor multiple product the loss equation becomes:

Computation of Savings from Loss Function

Ref. Page 7-3

p p qL = K (MSD) $/Unit productWhereL = loss/unit product in $k = a constantyo = target value y = performance measure

Monthly production = 2,000 unitsM thl j ti 50 it


Monthly rejection = 50 unitCost of rejection = $8 per unit

Therefore: Loss=(50 x $ 8 )/2000 = $ 0.50/unit

Input for Saving Computation from Loss FunctionRef. Page 7-4


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97

Find Monthly SavingsLoss Function L = K (Y Y )2 For a single unit

Loss Function Manipulation

Ref. Page 7-5

Loss Function, L = K (Y-Yo) For a single unit.Thus

20 = K (Yo +/- TOL - Yo)2

(since, when y = yo +/- TOL, all parts are rejected andthe loss equals cost of rejecting a single unit)

or 20 = K (TOL)2


( )or K = 20/(TOL)2 = 20/(.35)2 = 163.265Also, L = K (MSD) for the population of parts

L B f E i t/C t D i BEFORE

Expected Savings from the Improvement

Ref. Page 7-6

Loss Before Experiment/Current DesignL = K (MSD) = 163.265 x .0475) = $ 7.75

Loss After Experiment/New DesignL = K (MSD) = 163.265 x .0145) = $ 2.36

BEFORES/N = 13.23307

OR MSD = .0475

AFTER S/N = 18.38631

OR MSD = 0.0145

Savings Per Month = 7.75 - 2.36 = $ 5.39/PER PRODUCT


Total Savings Per Month = 15000 x 5.39 = $ 80,850

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Normal DistributionA l di t ib ti ti t id l d i th N l

Mathematical Expression for Normal Distribution

Ref. Page 7-7

Among several distribution equations, most widely used is the Normal Distribution which was discovered by Gauss, Laplace and Demoivre, independently. But it is generally identified as Gaussian distribution or the Gaussian error law .

1 σ 2 Π

f(x) = e( x - μ ) 2

2 σ2


Where, x is the random variable, f(x) probability of x ( -∞ <= x <= + ∞ ) and μ = average (Mean) of population, σ = standard deviation of population

When S/N is given, MSD can be calculated

Improvements Expressed in terms of Reduced Variation

Ref. Page 7-8

MSD = 10 -[(S/N)/10]

also, since MSD = σ2 + (a - yo )2 a = sample average and yo = Target value Standard deviation, σ can also be calculated when a and yo are known or assumed.


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Since MSD = 10 -[(S/N)/10]

and MSD = σ2 + (a - yo )2 = σ2 when average is on target

Savings from Known Improvement in S/N

Ref. Page 7-11

σ improved = σ current { (MSD)improved /(MSD)current}0.5

σ improved = σ current 10

(Above equation should apply to all three quality characteristics)

[(S/N)current - (S/N)imprved]20


99.73 %95.45 %68.27 %

1σ-1σ 2σ-2σ 3σ-3σ

Area under Normal Distribution

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Module – 8Application Steps


◦ Can you measure results (Performance)?

Should you do it? (DOE APPLICABILITY TEST)

◦ Is it possible for you to test (4 -9)?

◦ Do you suspect many factors to influence results?

If answer to above questions is YES, you have much to gain from t t d i t h DOE


structured experiments such as a DOE.

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101


Application Guidelines

HOW TO APPLY- Planning- Experiment Design- Analysis of results

HOW TO PREPARE & PRESENT REPORT (Appendix)


8.1 Description of Application Steps1. Project Selection: Look for a project to apply2. Planning : Arrange for the planning/brainstorming session.

Application Guidelines

Ref. Page 8-1

3. Designing : Design experiment & Describe trial conditions.4. Doing :Carry out experiments5. Analyzing: Analyze results6. Confirming: Run confirmation tests

8.2 Brainstorming for Design of Experiments (DOE)Purpose of brainstorming session Who should conduct?Who should host the session?Who should attend?How many should attend?Topics of Discussions


Topics of Discussions

1. Objective of the study (What are you after?)2. Design Factors and their Levels3. Interaction Studies (Which factors are likely to interact?)4. "Noise" Variables (How to create a robust design?)5. Tasks Descriptions and Assignments (Who will do what, how and when?)

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1. Know how to plan experiments and conduct brainstorming session. Understand basic principles ofconsensus decision making process while working as a team. Know how to:- select criteria and prepare evaluation table- prepare data collection tables- select factors for experiments

Application and Analysis Check List Ref. Page 8-7

p- decide on the number and values of the levels- compromise on number of factors and interactions- allocate the work assignments for the project (tasks)

2. Know how to design experiments when:- all factors are at the same level (2, 3 or 4 levels)- you wish to study interactions between two 2-level factors- factors are at mixed levels(2, 3 or 4 levels)

3. Can carry out computations for factor average effects and determine:- main effects- optimum condition- performance at the optimum condition

4 Understand the philosophy behind robust design


4. Understand the philosophy behind robust design5. 5. Be familiar with the transformation of results to MSD and S/N, and know how to interpret the

conclusions from the analysis.6. Understand the purpose and mechanics of ANOVA calculations:

- interpret the ERROR term & implications of the experimental results - why to pool factors and when to pool them- carry out test for significance and pool factors- calculate confidence interval and what it means

7. Know how to apply the principles of brainstorming problem solving

Taguchi Experimental Design TechniquesTaguchi Experimental Design Techniques

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Training & Workshop

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Ranjit Roy, Ph.D.,P.E.,PMPAuthor, Trainer, Consultant

Nutek, Inc.3 829 Quarton Road

Bloomfield Hills, MI 48302.Tel : 1‐248‐540‐4827 Info@Nutek‐us.com

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Th k YThank YouWe appreciate your taking the time to

view our seminar slides.

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Robust Product and Process Designs - Quality …nutek-us.com/Robust_Design_Slides_0910.pdfDeming Juran Taguchi Ishikawa •TQM, SPC, Taguchi Methods, Fishbone Diagram …. aGenichi