2.810 T.Gutowski 1

2.810 Quality and Variation

Part ToleranceProcess VariationTaguchi “Quality Loss Function” Random Variables and how variation grows with size and complexityQuality Control

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References;

Kalpakjian pp 982-991 (Control Charts)

“Robust Quality” by Genichi Taguchi and Don Clausing

A Brief Intro to Designed Experiments

Taken from Quality Engineering using Robust Design by Madhav S. Phadke, Prentice Hall, 1989

5 homeworks due Nov 13

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Interchangeable Parts;Go, No-Go; Part Tolerance

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Product specifications are given as upper and lower limits, for example the dimensional tolerance +0.005 in.

Upper Specification Limit

Lower Specification Limit

Target

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Process VariationProcess measurement reveals a distribution in output values.

0

5

10

15

20

25

30

35

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

Discrete probability distribution based upon measurements

Continuous “Normal” distribution

In general if the randomness is due to many different factors, the distribution will tend toward a “normal” distribution. (Central Limit Theorem)

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Tolerance is the specification given on the part drawing, and variation is the variability in the manufacturing process. This figure confuses the two by showing the process capabilities in terms of tolerance. Never the less, we can see that the general variability (expressed as tolerance over part dimension) one gets from conventional manufacturing processes is on the order of

to210000,1

10

410100

01.

Homework problem; can you come up with examples of products that have requirements that exceed these capabilities? If so then what?

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We can be much more specific about process capability by measuring the process variability and comparing it directly to the required tolerance. Common measures are called Process Capability Indices (PCI’s), such as,

6

LSLUSLC p

3

),min( LSLUSLC pk

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Case 1 In this case the out of specification parts are 4.2% + 0.4% = 4.6% What are the PCI’s?

Upper Specification Limit

Lower Specification Limit

Target

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Case 2 However, in general the mean and the target do not have to line up. What are the PCI’s? How many parts are out of spec?

Upper Specification Limit

Lower Specification Limit

Target

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Comparison

Case 1Cp = = 2/3

Cpk =

Min()=2/3

Out of Spec = 4.6%

Case 2Cp = = 2/3

Cpk =

Min()=1/3

Out of Spec = 16.1%

Note; the out of Spec percentages are off slightly due to round off errors

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Why the two different distributions at Sony?

20% Likelihood set will be returned

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Qu

alit

y

Loss

Deviation,

2

!2

)0()0()0()( f

ffQL

Goal Post Quality

Taguchi Quality Loss Function

QL = k 2

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Homework Problem

Estimate a reasonable factory tolerance if the Quality Loss ($) for a failure in the field is 100 times the cost of fixing a failure in the factory. Say the observed field tolerance level that leads to failure is field.

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Random variables and how variation grows with size and complexity

Random variable basics

Tolerance stack up

Product complexity

Mfg System complexity

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If the dimension “X” is a random variable, the mean is given by = E(X) (1)

and the variation is given by

Var(x) = E[(x - )2] (2)

both of these can be obtained from the probability density function p(x).

For a discrete pdf, the expectation operation is:

(3)

E(X) xii p(xi)

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Properties of the Expectation

1. If Y = aX + b; a, b are constants,

E(Y) = aE(X) + b (4)

2. If X1,…Xn are random variables,

E(X1 + … + Xn) = E(X1) +…+ E(Xn) (5)

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Properties of the Variance

1. For a and b constants

Var(aX + b) = a2Var(X) (6)

2. If X1,…..Xn are independent random variables

Var(X1+…+ Xn) = Var(X1)+ Var(X2)+ + Var(Xn) (7)

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If X1 and X2 are random variables and not necessarily independent, then

Var(X1 + X2) = Var(X1) + Var(X2) + 2Cov(X1Y) (8)

this can be written using the standard deviation “”, and the correlation “” as

(9)

where L = X1 + X2

L2 1

2 22 21 2

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If X1 and X2 are correlated ( = 1), then

(14)

for X1 = X2 = X0

(15)

for N (16)

or (17)

L2 1

2 22 21 2 (1 2 )2

L2 N 2 0

2

0 NL

20

2 4 L

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Now, if X1 and X2 are uncorrelated ( = 0) we get the result as in eq’n (7) or,

(10)

and for N (11)

If X1=X =Xo (12)

Or (13)

L2 1

2 22

2

1

2i

N

iL

L2 N 0

2

L N 0

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Complexity and Variation

As the number of variables grow so does the variation in the system;

This leads to; more complicated systems may be more likely to fail

L N 0

0 NL

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Homework; Consider the final dimension and variation of a stack of n blocks.

1, 2 …… nIf USL – LSL = ’, and Cp = 1a) How many parts are out of compliance?

b) Now USL-LSL=’, what is Cp? How many parts are out of spec?

c) Repeat a) with ’

Assume that target.

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Homework Problem: Experience shows that when composites are cured by autoclave processing on one sided tools the variation in thickness is about 7%. After careful measurements of the prepreg thickness it is determined that their variation is about 7%. What can you tell about the source of variation?

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Complexity and Reliabilityref. Augustine’s Laws

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Quality and System DesignData from D. Cochran

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Quality Control

Inputs “I”; Mat’l, Energy, Info

Operator inputs,”u”; initial settings, feedback, action?

Disturbances, “d”; temperature, humidity, vibrations, dust, sunlight

Outputs, “X”

Machine “M”

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Who controls what?

X = f (M, I, u, d)

Equipment Purchase

Q.C., Utilities, etc

Operator, Real Time Control

Physical Plant, etc

So who is in charge of quality?

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How do you know there is a quality problem?

1. Detection2. Measurement3. Source Identification4. Action5. Goal should be prevention

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Detection

Make problems obvious Poke yoke at the process level Clear flow paths and responsibility Andon board Simplify the system

Stop operations to attend to quality problems Stop line Direct attention to problem Involve Team

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Measurement

Statistical Process Control

Avera

ge v

alu

e x

Sampling period

Upper Control Limit

Lower Control Limit

Centerline

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Statistical Process Control Issues

Sampling Period

Establish Limits

Sensitivity to Change

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Source Identification; Ishikawa Cause and Effect Diagram

Man Machine

Material Method

Effect

Finding the cause of a disturbance is the most difficult part of quality control. There are only aids to help you with this problem solving exercise like the Ishikawa Diagrams which helps you cover all categories, and the “5 Whys” which helps you go to the root cause.

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Truck front suspension assembly

Problem; warranty rates excessive

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Setting the best initial parameters

Tables and Handbooks E.g. Feeds and speeds

Models E.g. Moldflow for injection molding

Designed Experiments E.g. Orthogonal Arrays

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Designed Experiments

1. Temp “T” (3 settings)2. Pressure “P” (3 settings)3. Time “t” (3 values)4. Cleaning Methods “K” (3 types)How Many Experiments?One at a time gives 34 = 81

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But what if we varied all of the factors at once?

Our strategy would be to measure one of the factors, say temperature, while “randomizing” the other factors. For example measure T2 with all combinations of the other factors e.g. (P,t,K) = (123), (231), (312).

Notice that all levels are obtained for each factor.

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“Orthogonal Array” for 4 factors at 3 levels. Only 9 experiments are needed

Exp temp pressure

time clean

1 1 1 1 1

2 1 2 2 2

3 1 3 3 3

4 2 1 2 3

5 2 2 3 1

6 2 3 1 2

7 3 1 3 2

8 3 2 1 3

9 3 3 2 1

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Homework

Can you design an orthogonal array for 3 factors at 2 levels?

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Summary – the best ways to reduce variation

Simplify design

Simplify the manufacturing system

Plan on variation and put in place a

system to address it

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Aircraft engine case study

LPC

HPC

Fan case

HPT

LPT

exhaustcase

diffuser

gearbox

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Engine Data

  engine A1

engine A2

engine B1

engine B2

engine C1

engine C2

number ofpart numbers

~2,000 ~2,000 ~1,400

~1,300

4,465 3,485

total numberof parts

~15,000

~19,000

~7,000

~7,000

26,073 23,580

weight [lb] 2.3k-3.5k

9k-10k 1.5k-1.6k

1.5k-1.6k

2.3k-3.5k

1.5k-1.6k

thrust [lb] unlessotherwise noted

14k-21k

40k-50k

4k-5k hp

7k-9k 14k-21k

7k-9k

by-pass ratio 0.36:1 4.9:1 - 5.15:1 0.34:1 6.2:1

  engine A1

engine A2

engine B1

engine B2

engine C1

engine C2

annualproduction

150 150 110 150 150 286

planned through-put time [days]

15 20 8 10 23 21

approx. takt time [shifts/engine]

7.30 7.30 6.64 4.87 4.87 2.55

Engine “complexity”

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Scheduled build times Vs part count

y = 0.001x + 3.995

R2 = 0.968

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10

15

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25

0 5,000 10,000 15,000 20,000 25,000 30,000

total number of parts

days

A1

A2

B1

B2

C2

C1

Sch

ed

ule

d b

uild

ti

mes

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Engine Delivery Late Times

A1 A2 B1 B2 C1 C20

10

20

30

40

50

60

avera

ge d

ays

late

A1 A2 B1 B2 C1 C2

engines

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Late times compared to scheduled times

0

20

40

60

80

100

120

140

0 5,000 10,000 15,000 20,000 25,000 30,000

total number of parts

days

planned throughput time

avg. actual throughput time

shortest actual throughput time

longest actual throughput time

Linear (planned throughput time)

A1A2

B1

B2 C1C2

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Reasons for delay at site A

2%

13%

38%

47%

0%

5%

10%

15%

20%

25%

30%

35%

40%

45%

50%

People shortage Quality Issues Part Shortages unknown

perc

enta

ge o

f occ

urr

ence

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Reasons for delay at site B (Guesses)

part lost at site3%

people shortage9%

build awaiting inspection

7%

Quality problem9%

part shortages67%

station not available1%

tools not available4%

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Reasons for delay at site A (data)

unknown rework paperwork Quality0%

10%

20%

30%

40%

50%

60%

70%

80%perc

enta

ge o

f occ

urr

ence

unknown rework paperwork Quality

Q

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Engines shipped over a 3 month period at aircraft engine factory “B”

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2

4

6

8

10

12

7-Jun 15-Jun 23-Jun 30-Jun 7-Jul 15-Jul 24-Jul 31-Jul 7-Aug 15-Aug 24-Aug 31-Aug

Weeks

en

gin

es s

hip

ped

per

week

month 1 month 2 month 3

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Engines shipped over a 3 month period at aircraft engine factory “C”

0

1

2

3

4

5

6

7

may june july august

weeks

en

gin

es

ship

ped