Making A Quality Product

Making a Quality Product

What is required to make a product?

A REVIEW / INTRODUCTION OF PROBLEM SOLVING TOOLS FOR ACHIEVING PROCESS CONTROL AND WASTE REDUCTION

Product

??????

please contact [email protected] for an animated PowerPoint presentation

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Raw

Material

Product

Processing

Cell


The process needs:

the raw materials ...

the equipment to produce the product ...

Is that all?

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Raw

Material

Product

Process

Control

Processing

Cell

Process Control Chart


The process also needs ... regulation or control using ...

A limited amount of the raw materials ...

- how much raw material can be processed at one time?

A limited range on the control factors ...

- temperature: how hot or cold?

- time: what duration?

Monitoring of materials and parameters ...

Is this enough to always make a good product?

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Sure! Why not?!

So start the process and make product.

Raw

Material

Process

Control

Processing

Cell


Product

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The customer expects uniformity.

Does all the product behave the same and conform to the manufacturing specifications?

Raw

Material

Process

Control

Processing

Cell


Product

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Raw

Material

Product

Process

Control

Processing

Cell


Wait a second!

What’s this?

This product is different!

The customer won’t accept this part!

So this product gets trashed.


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TRASH


And there is more trash,

and more ...

and more ...

Hey, this is getting expen$ive!!How can this be improved?

Product

$

$

Raw

Material

Process

Control

Processing

Cell


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Tell the operator when bad product is made and to watch the process better.

But the operator claims all process parameters are being maintained!

What else can be done?


Feedback $

$?

TRASH

Raw

Material

Process

Control

Processing

Cell


Product

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Find out what conditions produce very good or bad product.

Inspection establishes data on the normal output of all product.

It would be easiest to monitor all output and look at what conditions existed when a deviation from normal occurs.

Data is easily organized and interpreted with a Control Chart.


Raw

Material

Feedback

DataSPC Chart

Control

Charts

Product

Process

Control

Processing

Cell


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Raw

Material

Product

Design of

Experiments

. (DOE)

Process

Capability

Feedback DataSPC Chart

Feedback

Process

Control

Processing

Cell



Control

Charts

Data has several uses ...

Control Charts produce improvements by comparingtypical and unusual data

Efficient experiments produce data that results in an improved process yielding a better product

Data is used to estimate the ability of the process to produce conforming product

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Control Charts

ID out-of-control events

TYPES

Variable (measurable)

Attribute (yes/no, on/off)

SPC Chart

Raw

Material

Product

Design of Experiments

Optimize Output

Reduce Variation

Factorial Design

Conventional & Taguchi

Engineering Analysis

Process Capability

Cp > 2.0

Cpk > 1.5

DataFeedbackFeedback

Process

Control

Processing

Cell


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Raw

Material

Product

DataFeedbackFeedback

Process

Control

Processing

Cell



Control Charts

ID out-of-control events

TYPES

Variable (measurable)

Attribute (yes/no, on/off)

SPC Chart


Optimize Output

Reduce Variation

Factorial Design

Conventional & Taguchi

Engineering Analysis

Process Capability

Cp > 2.0

Cpk > 1.5

We will look at

How to identify an out-of-control process with statistical process

control (SPC).

How to predict the amount of non-conforming product from the

process data.

How to improve the process by conducting efficient experiments.

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SPC Chart


SPC and DOE Reduce Variation in a Process

Control Charts - reduce special (non-random) causes. They are used by the operator as a feedback mechanism to correct problems shown by the control chart.

Engineering Analysis - compares the process

capability to process tolerance. Scrap is reduced

when parts are processed through areas capable

of holding tolerance.

Design of Experiments - analyze the influence of factors

that cause variation. Factors are deliberately changed in an

controlled and organized fashion so that their effects can

be analyzed and then optimized to reduce output variation.

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Here is an example of making and testing bullets to illustrate:

control charts

design of experiments

engineering analysis

The test of a well made bullet is to hit the target bull’s eye

•

This is what the customer and manufacturer wants!

Is this always produced?

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Of course we can’t expect every bullet to be identical.

••

•••

•

So we will look the process of making a bullet and show:

process control -

How are factors controlled in the manufacturing?

control charts -

Why do weed need control charts?

Show the measure of good performance.

Show when the process has poor performance.

engineering analysis -

Predict the amount of scrap.

design of experiments -

Show how to improve process performance.Copyright ISandR

A heavier weight projectile is slower so it hits the target lower than a lighter and faster projectile, but too little weight and the wind affects the path.

The path of a smaller diameter projectile is erratic since the projectile wobbles, but too large and it doesn't fit the barrel.

More powder weight makes the projectile faster and less makes it slower.

No case factors influence bullet quality. Here, this was chosen for convenience, but acquiring from an approved vendor could reduce monitoring.


So what are the input factors to be controlled in the manufacture. Let’s assume only three factors require monitoring for process control.

Projectile

Powder

case

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The projectile has manufacturing limitations:

a maximum and minimum weight

a maximum and minimum diameter

The powder has manufacturing limitations:

a maximum and minimum weight

So let’s look at the process control or “rainbow” charts for several of the most recent lots of bullets.


We need process control to monitor the input variables

Projectile weight and diameter

Powder weight

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Process control monitors the input variables

DATE

TIME

INITIALS

NOTES

MAX

MIN

980701 980702 980703 980704 980705 980706 980707

Operation Characteristic: WEIGHT of PROJECTILE

DATE

TIME

INITIALS

NOTES

MAX

MIN

980701 980702 980703 980704 980705 980706 980707

Operation Characteristic: DIAMETER of PROJECTILE

DATE

TIME

INITIALS

NOTES

MAX

MIN

980701 980702 980703 980704 980705 980706 980707

Operation Characteristic: WEIGHT of POWDER

Projectile weight and diameter

Powder weight

Here are the “rainbow” charts for the lots 980701 through 980707

PROJECTILE WEIGHT OK

PROJECTILE DIAMETER OK

POWDER WEIGHT OK

Let’s look at the testing of these lots.

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Control Charts monitor the output variables

To measure the quality of the product, a few of the bullets from lot must be tested; this is called a sample. A sample is used because you can’t use the entire lot in testing or there would be nothing left to sell.

The quality of the lot is determined bythe spread of the hole pattern

andthe distance the center of the spread is to the center of the bull’s eye .

••

••

••

Here is the testing of lot 980701.

Let’s look closer at this pattern and put the results into a control chart.

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So we will look the process of making a bullet and show:

Let’s look atprocess control -

How are factors controlled in the manufacturing?

control charts -

Why do weed need control charts?

Show the measure of good performance.

Show when the process has poor performance.

DISCUSSION


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The diameter of the blue circle around the pattern is 7 inches in diameter. This circle represents the pattern spread and is a measure of variation.

This distance from the center of the pattern to the center of the bull’s eye is 6.5 inches. This is the a location measurement which compares the output to the desired or true value.

A proper evaluation requires a variation and a location measurement. Control charts plot both location and variation output measurements.


Control Charts monitor the output variables

•• •

• •

•

Test pattern of lot 980701

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Any change in people, equipment, materials, methods or environment to be noted on the reverse side; the notes will help to make corrections / improvements when indicated by the control chart.

Part Number Chart No.

Variable Control Chart (Average and Range)Part Name (Product) Operation (Process) Specification Limits

Operator Machine Gage Unit of Measure Zero Equals

DATE

TIME

1

2

3

4

5

SUM

AVERAGE

RANGE

NOTES1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

Control charts plot both location and variation outputmeasurements. On this control chart the location is called the average and the variation is called the range.

Control charts also have boundaries called UCL and LCL which stands for upper and lower control limits. These boundaries represent values that a stable process should not exceed. When the control boundaries are exceeded, the operator needs look for something that may be wrong with the process.

Let’s fill out the chart with the results from 980701.


Variable Control Charts

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DATE

TIME

1

2

3

4

5

SUM

AVERAGE

RANGE

NOTES1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

2

4

6

8UCL

LCL

5

10

15

range

avera

ge

UCL

The chart is provided with previously established process control limits. First fill in the information required into the header.

123 29Big Bullet Final Test See Customer Spec

Kim Tester #7 Tester #7 inch 0.001

6.5 this is the distance of the pattern from the bull’s eye - the location of the sample data7.0 this is the diameter of the pattern - the variation of the sample data

•

•Let’s look at the tests for the remaining lots.

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Making a Quality ProductOh good! We are just in time to see the tests of lots 980702 thruough 980707.

• • • •• • • •• • •• • • •• ••

that is 980702 to 980704

•• • •• •• • • •• • •• • •• •

that is 980705 to 980707

Record the patterns of location and variation from the targets and then plot them on the control chart.

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6.5, 2.5 7.0, 3.0 5.0, 5.5

7.0, 1.5 7.5, 1.0 13.5, 1.5

Fill in the table in with the variation and location results.

980702 6.5 2.5

980703 7.0 3.0

980704 5.0 5.5

980705 7.0 1.5

980706 7.5 1.0

980707 13.5 1.5

Use this table to fill in the control chart.


980702 980703 980704

980705 980706 980707

lot variation location

• • • •• • • •• • •• • • •• ••

•

• • •• •• • • •• • •• • •• •

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6.27.0


01





DATE

TIME

1

2

3

4

5

SUM

AVERAGE

RANGE

NOTES1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26


2

4

6

8UCL

LCL

5

10

15

range

avera

ge •

•


980702 6.5 2.5

980703 7.0 3.0

980704 5.0 5.5

980705 7.0 1.5

980706 7.5 1.0

980707 13.5 1.5UCL

Fill in the information for lots 980702 to 980704.

02 03 04

2.5 3.0 5.56.5 7.0 5.0 Now plot the points on the average and range graphs

•

••

• • •

Let’s look at this before finishing.Copyright ISandR





DATE

TIME

1

2

3

4

5

SUM

AVERAGE

RANGE

NOTES1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

2

4

6

8UCL

LCL

5

10

15

range

avera

ge



6.27.0

•

•


01


02 03 04

2.5 3.0 5.56.5 7.0 5.0

••

•

• • •

All of the location and variationdata looks normal so the process is behaving as expected.

None of the new values exceed the dotted lines which are the control limits that signal when to look for problems within the process.

Let’s continue.


980702 6.5 2.5

980703 7.0 3.0

980704 5.0 5.5

980705 7.0 1.5

980706 7.5 1.0

980707 13.5 1.5UCL

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DATE

TIME

1

2

3

4

5

SUM

AVERAGE

RANGE

NOTES1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

2

4

6

8UCL

LCL

5

10

15

range

avera

ge



6.27.0

•

•


01


02 03 04

2.5 3.0 5.07.5 7.0 5.0

••

•

• • •

Ok, you know there is somethingwrong with the remaining data.

05 06 07 Think about where the data

becomes unusual and what to do.

1.5 2.0 2.57.0 7.5 13

•••

•

••Do you see a problem?


980702 6.5 2.5

980703 7.0 3.0

980704 5.0 5.5

980705 7.0 1.5

980706 7.5 1.0

980707 13.5 1.5UCL

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DATE

TIME

1

2

3

4

5

SUM

AVERAGE

RANGE

NOTES1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

2

4

6

8UCL

LCL

5

10

15

range

avera

ge



6.27.0

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•


01


02 03 04 05 06 07

2.5 3.0 5.0 1.5 2.0 2.57.5 7.0 5.0 7.0 7.5 13

••

•

• • •

OK. There are some hints here!

• Did you think the red locationvalue was a problem?

• Did you think the blue variationvalue was a problem?

• Are both a problem?

• Maybe neither are a problem. Do both values have to exceed a limit at the same time to act?

What do you think and why?

Take a minute to think.UCL

••

••

•

•

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Thinking

UCL

LCL

5

10

15

ran

ge

Big BulletKim Tester

Any change in people, equipment, materials, methods or environment to be noted on the reverse

Variable Control Chart (Average and Range)Part Name (Product) Operation (Process)

Operator Machine

DATE

TIME

1

2

3

4

5

SUM

AVERAGE

RANGE

NOTES1 2 3 4 5 6 7 8 9 10 11

ave

rag

e

6.27.0

•

•

01

••

•

• • •

UCL

••

••

2

4

6

8

•

•

02 03 04 05 06 07

2.5 3.0 5.0 1.5 2.0 2.57.5 7.0 5.0 7.0 7.5 13

So what do you think?

Oh-Oh!? The red dot, the blue dot,...

Both, neither,...

Maybe it’s a trick and it’s all the above.

OK. Here’s the answer and why.





DATE

TIME

1

2

3

4

5

SUM

AVERAGE

RANGE

NOTES1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

2

4

6

8UCL

LCL

5

10

15

range

avera

ge



6.27.0

•

•


01


02 03 04 05 06 07

2.5 3.0 5.0 1.5 2.0 2.57.5 7.0 5.0 7.0 7.5 13

••

•

• • •

Certainly you would stop and look ifthe location upper control limit wasexceeded. That means the holepattern has shifted a large distanceaway from the bull’s eye and that isbad.

But the location has exceeded thelower control limit (LCL).

That would mean that the hole patternwas close to the bull’s eye and that’sgood. Why tell anyone if the processis better than what is expected?

Well if the process got better perhapswe can figure out why the process isbetter. So always look at what ishappening to the process when anycontrol limit is exceeded.

A special note. The variation limithas not been exceeded at the sametime as the location value. Thismeans that this may be a rareexception when a limit is exceededalthough the process is okay.

UCL

••

••

•

•

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DATE

TIME

1

2

3

4

5

SUM

AVERAGE

RANGE

NOTES1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

2

4

6

8UCL

LCL

5

10

15

range

avera

ge



6.27.0

•

•


01


02 03 04 05 06 07

2.5 3.0 5.0 1.5 2.0 2.57.5 7.0 5.0 7.0 7.5 13

••

•

• • •

Now if a variation and a locationcontrol limit are exceeded at thesame time there is usually a realproblem.

But the variation limit has beenexceeded by itself. Does this meanthere is a probelm?

YES!

A “well behaved” process will usuallyhave stable variation. When variationchanges there is a good chance thatsomething has definitely influencedthe process.

When any control limit is exceeded,assume there is a problem and lookfor a source that influences thevariation and/or the location value.

How are these problems identified?

UCL

••

••

•

•

Copyright ISandR

Come on snake

eyes!

It would be valuable to know when a process is producing parts that meet a desirable outcome (like high reliability or yield) and if it was not producing, why not?

Control charts are used to visually show when a process is producing parts within specification and when is it not producing parts within specification.

We want to build parts that would be identical, but we know all parts are not the same. The parts vary.

Probability relates the possibility of meeting and not meeting a desirable outcome.

The discussion of control charts requires some understanding of probability.

Control Charts and Probability

SPC Chart

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Come on snake

eyes!

Just as in gambling we cannot predict what will be the outcome of an event before it happens,

for instance rolling a two with a pair of dice,

we can know how frequently we should expect that event to occur.

When we make an item we can also predict how frequently the part should be out of some desirable range. When the frequency gets too high then we should look for the source that causes the part to vary too much so it is unacceptable.

We can pictorially represent the shape of how frequently events occur.

Control Charts and Probability

SPC Chart

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What is the probability of rolling a “one” with one die?

A = the number of ways an event can happen

B = the number of way an event fails to happen

A + B = the total number of all possibilities

Probability is calculated by dividing A by the sum of A and B

Probability = 16.6%

What is the probability of a “head” with a coin toss?

5 ways fail to get

a one

1 way to

get a one

Probability =A

A + B=

1

1 + 5

Come on snake

eyes!

Probability

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What is the probability of a “head” on a coin toss?

A = the number of ways an event can happen

B = the number of way an event fails to happen

A + B = the total number of all possibilities

Probability is calculated by dividing A by the sum of A and B

Probability = 50%

What is the probability of tossing two coins and both are “heads”?

1 way to fail to

get a head

1 way to

get a head

Probability =A

A + B=

1

1 + 1

Come on snake

eyes!

TAILS

Probability

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What is the probability of tossing two coins and

both are “heads”?

What are all the

combinations?

Probability = 25%

What is the probability of tossing coins five

consecutive times and getting “heads”?

3 ways to fail to

get two heads

1 way to get

two heads

Probability =A

A + B=

1

1 + 3

Come on snake

eyes!

HH HT

TH TT

Probability

HT

TH TT

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What is the probability of tossing coins five consecutive times and getting “heads”?

What are all the combinations?

31 ways to fail to get

five heads

1 way to get five heads

Probability = 3.125%

Note how some outcomes are more

likely and some are less likely and how this

influences the shape of the distribution.

What is the probability of rolling a

“two” with a pair of dice?

Come on snake

eyes!

HHHHHHHHHTHHHTHHHHTTHHTHHHHTHTHHTTHHHTTTHTHHHHTHHTHTHTHHTHTTHTTHHHTTHTHTTTHHTTTTTHHHHTHHHTTHHTHTHHTTTHTHHTHTHTTHTTHTHTTTTTHHHTTHHTTTHTHTTHTTTTTHHTTTHTTTTTHTTTTT

Probability

0 1 1

1 5 5

2 10 10

3 10 10

4 5 5

5 1 1

0

2

4

6

8

10

12

0 1 2 3 4 5

Probability =A

A + B=

1

1 + 31

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What is the probability of rolling a

“two” with a pair of dice??

What are all the outcomes from 2 dice?

36 total combinations

1 way to get a two

35 ways to fail to get a two

Probability = 2.78%

The developing shape is similar to the

“Normal Distribution Curve”.

Come on snake

eyes!

Probability =A

A + B=

1

1 + 35

1st die 2nd die

6 1,2,3,4,5,6

5 1,2,3,4,5,6

4 1,2,3,4,5,6

3 1,2,3,4,5,6

2 1,2,3,4,5,6

1 1,2,3,4,5,6

2 1 1 11

3 2 2 12

4 3 3 31

5 4 4 41

6 5 5 51

7 6 6 61

8 5 5 62

9 4 4 63

10 3 3 64

11 2 2 65

0

2

4

6

8

2 3 4 5 6 7 8 9 10 11 12

0

2

4

6

8

The graphical

presentation

Probability

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oo

oo o

o

o

oo

oo

o

This is precise but This is accurate but not accurate. not precise.

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The Characteristics of a Normal Distribution Curve

Probability

When we make an item the location

(mean/average) is not a zero value as

shown here. There is an actual length

or weight or whatever is important

enough to be measured.

All items do not have the same value; this

is the variation.

The shape of the curve results from the

fact that most items will have a value

at the peak of the curve and other

items will have other values, but these

will occur less frequently.

variation

0

mean

-15 -10 -5 0 5 10 15

maybe a histogram of parts being measured would help more

Copyright ISandR


variation

Sstandard deviation = +/- 5

0

-15 -10 -5 0 5 10 15

location

Xmean = 0

u o

Probability

The Normal Distribution Curve has a

location and a variation value which

describes the entire shape of the

curve.

Literally these are the essential variables of

the mathematical equation

The location value is called the mean.

The variation value is called the

standard deviation.

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95

0

-8

5

3

0

3

-8

3

5

ProbabilityNote how changes in location and variation affect the characteristics of a

Normal Distribution Curve

Horizontally the graphs show changes in variation

The standard deviation is, from left to right, 3, 5, and 9

As the standard

deviation gets

bigger, the curves

gets wider and

lower.

The change in location moves

the curve left and right

Vertically the graphs show changes in location

The mean is, from top to bottom, 0, -8, and 5

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0


100% of all possibilities are within the curve!

Probability

axis marked in units of std. dev.

+/- S INSIDE OUTSIDE

32.75%

4.56%

???%

1 1 1 68.25%

2 2 2 95.44%

3 3 3 99.73%

This describes the possibilities

of obtaining an outcome for any

process that is totally random

+/- S3 2 1 1 2 3

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SPLAT

MASH!

MATH!


How are the location (X) and variation (S) values determined?

0

S = ??

X = ??

Probability

Gather a sample from the group to be evaluated.

Measure the response (length, time, pressure, ...).

Calculate the mean, X, by adding all the measured values and divide by the number of measurements added together.

find X of 5 measurements: 2, 4, 5, 8, 9

(2+4+5+8+9) = 28

28 / 5 = 5.6

Calculate the standard deviation, S, by summing the square obtained from subtracting each measured value from the average, divide this sum by the number of measurements minus 1, and then take the square root of that number.

find S of same 5 measurements 2, 4, 5, 8, 9

(5.6-2)2+(5.6-4)2+(5.6-5)2+(5.6-8)2+(5.6-9)2 = 33.2

33.2 / (5-1) = 8.3

(8.3)1/2 = 2.88

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Probability

What is variation and of what is it composed?

Variation is composed of common and special sources.

Common cause of variation - is the stable random pattern caused by

natural or inherent conditions of a process. Performance is predictable

and is a state of statistical control. This is the type of variation handled

by probability and depicted with the Normal Distribution Curve.

Special cause of variation - is a source of variation that is intermittent,

unpredictable unstable; sometimes called assignable causes. This is

tool wear, a balance missing a weight, a misread gage.

Cpk = X - nearest limit3 s

Cpk = 1

says the manufacturing tolerance is equal to 6

sigma and is evenly centered about the process

capability

Process Capability

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Process Capability

You are a car salesperson.

You want to sell your customer a new SUV (Sports Utility Vehicle).

Assume the width of the car represents the capability of the process (that’s what

your selling) and the width of the garage door represents the customer’s

specifications (they are limited to what can be bought)

To get the SUV through the door is

easiest when the door is much wider

than the car.

It is easiest to meet requirements

when the customer’s specification is

big compared to what the process

delivers.

Customer Specification Process Capability

Comparison of Cp (Process Capability) and Cpk (where the Process

Capability is centered with respect to the specifications)

Customer SpecificationProcess Capability

Cp = 1 Cp > 1Cp < 1

Cpk < Cp Cpk < Cp

k

Process Capability

Cpk = Cp Cpk << Cp

Process

Disruption

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Process Capability

when Cp > or = 1 then it it starts to get easier to get the car

through the garage door

to get a calculation of process capability

remove all assignable causes - this is done with the control

chart

once all random events achieved in the process

get x bar and std dev

calculate Cp and Cpk

calculate process yield

Copyright ISandR

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.46410.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.42470.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 0.38590.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557 0.3520 0.34830.4 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3156 0.3121

0.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.27760.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483 0.24510.7 0.2420 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177 0.21480.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.18670.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.1660 0.1635 0.1611

1.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.13791.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.1230 0.1210 0.1190 0.11701.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.09851.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.08231.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0721 0.0708 0.0694 0.0681

1.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.05591.6 0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0.0475 0.0465 0.04551.7 0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 0.0375 0.03671.8 0.0359 0.0351 0.0344 0.0336 0.0329 0.0329 0.0314 0.0307 0.0301 0.02941.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233

2.0 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.01832.1 0.0179 0.0174 0.0170 0.0166 0.0162 0.0158 0.0154 0.0150 0.0146 0.01432.2 0.0139 0.0136 0.0132 0.0129 0.0125 0.0122 0.0119 0.0116 0.0113 0.01102.3 0.0107 0.0104 0.0102 0.0099 0.0096 0.0094 0.0091 0.0089 0.0087 0.00842.4 0.0082 0.0080 0.0078 0.0075 0.0073 0.0071 0.0069 0.0068 0.0066 0.0064

2.5 0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051 0.0049 0.00482.6 0.0047 0.0045 0.0044 0.0043 0.0041 0.0040 0.0039 0.0038 0.0037 0.00362.7 0.0035 0.0034 0.0033 0.0032 0.0031 0.0030 0.0029 0.0028 0.0027 0.00262.8 0.0026 0.0025 0.0024 0.0023 0.0023 0.0022 0.0021 0.0021 0.0020 0.00192.9 0.0019 0.0018 0.0018 0.0017 0.0016 0.0013 0.0015 0.0015 0.0014 0.0014

3.0 0.0013 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.0011 0.0010 0.00103.1 0.0010 0.0009 0.0009 0.0009 0.0008 0.0008 0.0008 0.0008 0.0007 0.00073.2 0.0007 0.0007 0.0006 0.0006 0.0006 0.0006 0.0006 0.0005 0.0005 0.00053.3 0.0005 0.0005 0.0005 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.00033.4 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0002

Process Capability

This is called a “z” table.

The table is used to find the

probability that events will

occur.

In the next slide we will look

up 1.81 because we want to

know what is the possibility

of an event occurring 1.81

standard deviations away

from the mean.

This illustrates how to look

up 1.81 and see that it

represents 0.0351 or 3.51%

probability an event will

occur.

Careful when using these; the table can be single or double tailed. This one is single tailed; the probability is

for one tail.

Copyright ISandR

1.81 S = 0.0351

or 3.51% of all

events within one tail

at 1.81 standard

deviations units and

beyond

Normalized Gaussian

0

20

40

60

80

100

120

-4 -3.6 -3.2 -2.8 -2.4 -2 -1.6 -1.2 -0.8 -0.4 0 0.4 0.8 1.2 1.6 2 2.4 2.8 3.2 3.6 4

std dev units

Process Capability

Using the Z Table(Using a single tailed table)

Z tables are used to determine the percent probability of an event in thetail of a distribution (a variation in an input or an output variable).

This is a look up table for the % probability between two events, themean (x bar) and another event, the distance between them given instandard deviation units.

Remember that these predictions work if the distribution is Normal / Gaussian. If the data is not

Normal then use control charts to find the assignable causes.

X = 1.7

S = 0.010

UCL = 1.715

LCL = 1.670

Calculate the control limits in standard deviation units

UCL = (1.715 - 1.7) / 0.010 = (0.015) / 0.010 = 1.50

LCL = (1.670 - 1.7) / 0.010 = (0.030) / 0.010 = 3.00

look up the “z” fraction beyond the points 1.50 and 3.00

Z1.50 = 0.0668 and Z3.00 = 0.00135 or

add together and make it a percent: 6.815% out-of-spec

z 0.00 0.01

0.0 0.5000 0.49600.1 0.4602 0.45620.2 0.4207 0.41680.3 0.3821 0.37830.4 0.3446 0.3409

0.5 0.3085 0.30500.6 0.2743 0.27090.7 0.2420 0.23890.8 0.2119 0.20900.9 0.1841 0.1814

1.0 0.1587 0.15621.1 0.1357 0.13351.2 0.1151 0.11311.3 0.0968 0.09511.4 0.0808 0.0793

1.5 0.0668 0.06551.6 0.0548 0.05371.7 0.0446 0.04361.8 0.0359 0.03511.9 0.0287 0.0281

2.0 0.0228 0.02222.1 0.0179 0.01742.2 0.0139 0.01362.3 0.0107 0.01042.4 0.0082 0.0080

2.5 0.0062 0.00602.6 0.0047 0.00452.7 0.0035 0.00342.8 0.0026 0.00252.9 0.0019 0.0018

3.0 0.0013 0.00133.1 0.0010 0.00093.2 0.0007 0.00073.3 0.0005 0.00053.4 0.0003 0.0003

2.9 0.0019

3.0 0.0013 0.3.1 0.0010

1.4 0.0808

1.5 0.0668 0.01.6 0.0548 0

Calculate the UCL & LCL in

units found in a standard

Normal Distribution table

the second curve is

centered on zero by

subtracting the

average value

the third curve has

the variation scaled

in whole units of

standard deviation

Calculate the out-of-spec parts for this process.

Process Capability

Copyright ISandR

Measurement adds variation

Adjust machines to get x bar in the center of UCL and LCL so Cpk

becomes as large as possible

What happen when Cpk produces yields of 0.98, 0.95, and 0.92?

(0.90) * (0.95) * (0.92) = 0.7866

When variation improves, get smaller, then yields improve.

Process Capability

Copyright ISandR

Assume a product tolerance of 1.00 +/- 0.01

1.00

1.02

1.00

0.99

1.01

0.99

1.00

1.01

1.00

1.01

x = 1.003

s = 0.009486

LCL = 0.99

UCL = 1.01

Cp = 0.02 / 0.0009486 = 0.35

Cpk = (1.10 - 1.003) / (3 * 0.009486)

= 0.007 / 0.028458

= 0.245

Z statistics

(1.003 - 0.99) / 0.009486 = 1.37

1.37 = 0.0853 or 8.53%

(1.003 - 1.01) / 0.009486 = 0.7379

0.7379 (about .74) = .2296 or 22.96%

total of 31.46% Out of Tolerance

0

Process Capability

Copyright ISandR

Assume a product tolerance of 2.5 + / - 0.05

2.49

2.50

2.54

2.50

2.47

2.49

2.51

2.52

2.54

2.53

x = 2.509

s = 0.02331

LCL = 2.45

UCL = 2.55

Cp = 0.10 / (6 * 0.02331) = 0.71

Cpk = (2.509 - 2.55) / (3 * 0.02331)

= 0.041 / 0.06993

= 0.586

Z statistics

(2.509 - 2.55) / 0.02331= 1.76

1.76 = 0.0392 or 3.92%

(2.509 - 2.45) / 0.02331= 2.53

2.53 = 0.0057 or 0.57%

total of 3.44% Out of Tolerance

0

Process Capability

Copyright ISandR

All Data

Yes / No, Good / Bad, Pass / Fail Measurable

VariableAttribute

Defects

UnlimitedDefectives

limitedX bar & R individual &

moving x bar

mixed sample

size

short run

production C u p np

fixed

sample

size

variable

sample

size

variable

sample

size

fixed

sample

size Best to use variable

Control Charts

Copyright ISandR

Control ChartsP CHART

When variable data cannot be obtained

When charting fraction rejected as non-conforming

When screening multiple characteristics for potential control charts

When tracking the quality level of a process before (how? By counting the number of defective items from a

sample and then plotting the percent defective)

Conditions

to be of help: there should be some rejects in each observed sample

the higher the quality level, the larger the sample size needs to be, since needs rejects. For example, 20% of a

product is rejectable......................................................................................................

needed. However, a sample of 1000 will give a ......................................................................................

sample if 0.1% of the product is rejectable

UCL - pbar + ( 3(pbar (1-pbar)/n)1/2 .... LCL

0

want a normal dist

UCL 7 LCL are calc

based on a small

sample size; LCL

usually = 0; if LCL = -3s

then part is bad <

0.0135%

calc pbar on n = 20 ( also

LCL and UCL)

Copyright ISandR





DATE

TIME

1

2

3

4

5

SUM

AVERAGE

RANGE

NOTES1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

Variable Control Charts

Copyright ISandR

Fac tors for Cont rol lim it s

n d2 D3 D. A2 A2

2 1 .1 28 0 3 .26 7 1.8 8 1.8 8

3 1 .6 93 0 2 .57 5 1.0 2 3 1.1 87

4 2 .0 59 0 2 .28 2 0.7 2 9 0.7 96

5 2 .3 26 0 2 .11 5 0.5 7 7 0.6 91

6 2 .5 34 0 2 .00 4 0.4 8 3 0.5 48

7 2 .7 04 0 .07 6 1 .92 4 0.4 1 9 0.5 08

8 2 .8 47 0 .13 6 1 .86 4 0.3 7 3 0.4 33

9 2 .9 7 0 .18 4 1 .81 6 0.3 3 7 0.4 12

1 0 3 .0 78 0 .22 3 1 .77 7 0.3 0 8 0.3 62

RANGES CHART

Subgroups Included: ,________________________

Sum of Ranges = S =

Number of Subgroups = K =

Subgroup Size = n =

Average Range = Rbar = S/K = ________

D3 factor =

LCLR = D3Rbar = ( )( ) = ________

D4 factor =

UCLR = D4Rbar = ( )( ) = ________

AVERAGES CHART

Subgroups Included: ,________________________

Sum of Averages = S =

Number of Subgroups = K =

Grand Average = X = S /K= ________

A2 factor =

UCLX= X + A2R = + ( )( ) = ________

LCLX= X - A2R = - ( )( ) = ________

Variable Control Charts Worksheet

Copyright ISandR

Control ChartsX bar and R Chart

works for even for non-normal distribution because of central limit

theorem. The examples are from the Boeing Manual for Supplier’s.

Xbar

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Pareto Charts

100%

looking for largest contributor;

20% 0f items cause 80% of the

problem

Other

Problem Solving

Copyright ISandR

Fishbone diagrams - to brainstorm process improvement

Effect

Machine Operator

Weather Raw Materials

PM

Temp

Shift

Season Storage

Preparation

Problem Solving

Copyright ISandR

Fishbone diagrams - to brainstorm process improvement

this is a way of analyzing problem is also known as Ishikowa Diagram

the effect is usually negative - a problem

the problem should be specific and clearly stated

the ideas are generated by using brainstorming

remind the group to ask themselves “what would cause the problem?”

keep attention on the effect of the problem

the goal is to find as many sources for variation as possible that cause the problem

Effect

Machine Operator

Environment Raw Materials

PM

Temp

Shift

Season Storage

Preparation

Measurement

s

Methods

Problem Solving

Copyright ISandR

RANK CAUSE

let each person rank the problems, tally all votes, rank causes

SCATTER PLOTS

CV - coefficient of Variation, of +1 and -1 for strong proportional and

inverse proportional

Problem Solving

Copyright ISandR

Example: If a box of capacitors has an mean (xbar) = 100 uF and

the standard deviation (s) = 2 uF, what is the probability when

pulling out 1 capacitor that ...

• 1 the cap is greater than 106 uF; (106-100)2 = 3; z(3) = 0.0013; 1.3%

• 98 uF < X < 102 uF; (100-98)/2 and (102-100)/2; z is 2(.1587); 100 - 31.74 = 68.26%

• 100 uF < X < 102 uF; (100-100)/2 and (102-100)/2; z is .50 + .1587 or 100-65.87 =34.13%

• 102 uF < X < 104 uF; (102-100)/2 and (104-100)/2; z is .1587 - .0228 or 13.59%

• <95 uF; (100-95)/2 = 2.5; z is 0.0062; .62%

0

Process Capability

Copyright ISandR

2 .50 00

2 .50 10

2 .49 90

2 .50 40

2 .50 20

2 .49 80

2 .50 10

2 .50 00

2 .50 50

2 .50 40

2 .50 10

2 .50 00

2 .50 30

2 .50 10

2 .50 30

2 .50 20

2 .50 20

2 .50 30

2 .49 90

2 .5 01 5

0 .00 18 96

These parts are tolerance as 2.500” +/- 0.005”.

What is Cp, Cpk and how many parts will be made out-of-spec?

xbar

std dev

Process Capability

Copyright ISandR

Construct a Control Chart for Individual and Moving Average

individ |MR|

1 1 0.0 00 0

2 9 .99 30 0.0 07 0

3 9 .99 90 0.0 06 0

4 1 0.0 07 0 0.0 08 0

5 1 0.0 09 0 0.0 02 0

6 1 0.0 01 0 0.0 08 0

7 1 0.0 00 0 0.0 01 0

8 1 0.0 11 0 0.0 11 0

9 1 0.0 00 0 0.0 11 0

1 0 1 0.0 03 0 0.0 03 0

1 1 9 .99 70 0.0 06 0

1 2 9 .99 80 0.0 01 0

1 3 9 .99 80 0.0 00 0

1 4 1 0.0 02 0 0.0 04 0

1 5 9 .98 90 0.0 13 0

1 6 1 0.0 01 0 0.0 12 0

1 7 9 .99 90 0.0 02 0

1 8 1 0.0 04 0 0.0 05 0

1 9 1 0.0 10 0 0.0 06 0

2 0 9 .99 00 0.0 20 0

a vg 1 0.0 00 6 0.0 06 6

s tdde v 0 .00 59

UCL 10 .0 18 2

LCL 9.9 82 9

UCL = Xbar + (2.66)(MRbar)

LCL = Xbar - (2.66)(MRbar)

Specification

10.000” +/- 0.010”

Control Charts





DATETIME

12345

SUMAVERAGERANGENOTES

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

Individual and Moving Range

Copyright ISandR

• Gages have an error distribution

• R&R are the two different ways of characterizing gage measurements

• The total variation is a function of the parts and the gage.

• Look at how much of the variation is due to the gage variation

– Rule of thumb - gage tolerance is a tenth of the minimum tolerance unit.

Gage Repeatability and Reproducibility (Gage R&R)

Copyright ISandR

• Repeatability of a gage - one person measures one part with the same instrument

• Reproducibility of a gage - different operators measure one part with the same instrument

“true reading”

NIST traceable

Repeatability

Accuracy

A

B

C

Reproducibility

SB = sC - sA

??????????

Gage Repeatability and Reproducibility (Gage R&R)

Copyright ISandR


Objectives of Design of Experiments (DOE)

DOE is a systematic approach to correlate the response between process inputs and outputs (independent and dependent variables).

tool pressure yield

temperature tolerance

The objective is to optimize for highest yield and best tolerance.

A good experiment doe not change only one variable since variables can interact (i.e.. temperature and wind or gasoline and air).

Consider the interaction of temperature and wind velocity in the wind chill factor; more wind makes it feel colder and an engine requires both gasoline and air.

The experiment that changes variables one-at-a-time (OAAT) requires more experiments and cannot evaluate interactions; it is inefficient at best.

PROCESS

Copyright ISandR

SPC vs. DOESPC - find special effects

Control Charts:

capability studies, sampling frequency, Cp, Cpk

Passive measure of the performance of a process and alerts when the process is out-of-control

Eliminate process variation by eliminating assignable or special causes

DOE - reduces the magnitude of randomness

Balanced orthogonal arrays, ANOVA, main effects, interacting variables

Actively manipulate factors in order to measure their effects on the process


Copyright ISandR

A Simple Experiment

Perform an experiment to determine if temperature affects yield.

Change temperature and follow the percent yield.

T1 has yields of 82 88 86 93 89

T2 has yields of 89 98 88 91 92

Find mean of each of the yields at each temperature

Plot mean and range then connect the means

Have to be careful about the conclusions. There could still be another factor not held constant thatcould influence the variation.

Duplicate the experiment to increase confidence in the result.

T1 T2


Copyright ISandR

If you want to know the affects from two factors (i.e. temperature and pressure),

then vary both T and P and record the results of the dependent variable (i.e. yield).

Don’t just change one factor at a time.

Run an experiment. Change the pressure between 50 and 100 psi and the temperature between 25 and 75 degrees F; record the yield results.

50 25 9450 75 92100 25 96100 75 88

Calculate the mean yield for each factor

T1 = 95; T2 = 91; P1 = 93; P2 = 93

Plot the response. From the main effects, pressure has little influence on the yield,

while temperature has a larger effect on the yield.

This experiment evaluates all possible combinations, but one cannot always run full factorial experiments, so use designed experiments.


INPUT RESULT

P T Yield

P1 T1

P1 T2

P2 T1

P2 T2

INPUT RESULT

P T Yield

A Two Factor Experiment

88

90

92

94

96

P1 P2 T1 T2

Copyright ISandR

This experiment looks at the affects of temperature and pressure on the yield of a process.

Determine the number of levels for each variable (we will choose only two).

Determine all combinations of the variables.

Two variables at two levels creates four combinations

Run the experiment in duplicate to confirm the results

Average the results of the runs at each variable level

Average at T1 = (90 + 92 + 84 + 88) / 4 = 88.5

Average at T2 = (94 + 96 + 88 + 90) / 4 = 92

Average at P1 = (90 + 92 + 94 + 96) / 4 = 93

Average at P2 = (84 + 88 + 88 + 90) / 4 = 87.5

This is a main effects experiment using a full factorial orthogonal balanced array


INPUT OUTPUT

Temp Press Result1 Result2

T1

T1

T2

T2

P1

P2

P1

P2

90

84

94

88

92

88

96

90

86

90

92

94

96

P1 P2 T1 T2

88

Copyright ISandR

This is a three factor experiment (A, B, C) and the effect on the process (yield,

tolerance, etc...).

Get engineering recommendations for each level of A1, A2, B1, B2, C1, and C2.

Determine the combination of factors.

The formula for all combinations

(number of levels) (factors) = (2)3 = (2)(2)(2) = 8

This is a full factorial array

Run the experiment, preferably obtaining more than one result.

Calculate the mean result for each factor at each level.

Plot the main effects.

Run the experiment again to confirm the results.


INPUT OUTPUT

Factor A Factor B Factor C Result 1 Result 2A1

A1

A1

A1

A2

A2

A2

A2

B1

B1

B2

B2

B1

B1

B2

B2

C1

C2

C1

C2

C1

C2

C1

C2

Three Factor Full Factorial Experiment

Copyright ISandR

Three Factor Full Factorial Experiment

Run the experiment and gather the results

Calculate the mean for each factor at each level

Plot the main effects

Run the experiment again to confirm the results

Next, compare a partial and full factorial array by using the same experimental data (we will compare the results in this full factorial to an array called an “L4” with Taguchi nomeclature)


84

86

88

90

92

94

A1 A2 B1 B2 C1 C2

A1 = (84 + 90 + 86 + 92 + 90 + 88 + 98 + 88) / 8 = 89.5

A2 = (90 + 96 + 84 + 82 + 90 + 98 + 98 + 94) / 8 = 91.5

B1 = (84 + 90 + 86 + 92 + 90 + 96 + 84 + 82) / 8 = 88.0

B2 = (90 + 88 + 98 + 88 + 90 + 98 + 98 + 94) / 8 = 93.0

C1 = (84 + 90 + 90 + 88 + 90 + 96 + 90 + 98) / 8 = 90.7

C2 = (86 + 92 + 98 + 88 + 84 + 82 + 98 + 94) / 8 = 90.2

84

86

90

98

90

84

90

98

90

92

88

88

96

82

98

94

INPUT OUTPUT


A1

A1

A1

A2

A2

A2

A2

B1

B1

B2

B2

B1

B1

B2

B2

C1

C2

C1

C2

C1

C2

C1

C2

Copyright ISandR

This is a comparison of a full factorial array to a partial factorial for the same experimental data.

Run the experiment.

Instead of running the experiment, compare the data from the identical rows in each array.

Calculate the mean for each factor at each level.

Plot the main effects.

Note the difference between a full (solid line) and a partial factorial (broken line) in the main effects responses; B is similar, but C has a larger influence, and A is less.

There is a price to pay for running a less trials; it is information loss.

The information, in either a full or partial factorial, could be influenced by an omitted variable and the response could be different than a straight line (only plotted 2 points). A partial factorial could be influenced by an interaction between factors.

Three Factor Full vs. Partial Factorial DOE


84

86

90

98

90

84

90

98

90

92

88

88

96

82

98

94

INPUT OUTPUT


A1

A1

A1

A2

A2

A2

A2

B1

B1

B2

B2

B1

B1

B2

B2

C1

C2

C1

C2

C1

C2

C1

C2

1

2

3

4

84

98

84

90

90

88

82

98

INPUT OUTPUT


A1

A2

A2

B1

B2

B1

B2

C1

C2

C2

C1

1

2

3

4

84

86

88

90

92

94

A1 A2 B1 B2 C1 C2

A1 = (84 + 90 + 98 + 88) / 4 = 90.7

A2 = (84 + 82 + 90 + 98) / 4 = 88.5

B1 = (84 + 90 + 84 + 82) / 4 = 85.0

B2 = (98 + 88 + 90 + 98) / 4 = 93.5

C1 = (84 + 90 + 90 + 98) / 4 = 94.0

C2 = (98 + 88 + 84 + 82) / 4 = 90.2

Copyright ISandR


1 3 2

A1

A2

B1

B2

(AxB)1

(AxB)2

INPUT OUTPUT

A

A1

A1

A2

A2

A

B1

B2

B1

B2

(AxB)

(AxB)1

(AxB)2

(AxB)2

(AxB)1

Result

This is a Taguchi L4 array and a linear diagram where the interaction (called confounding) between factors occur.

The actual array can be set up to acquire information on interactions. Here 3 represents the column where interaction, if any, between factors 1 and 2 occurs.

Suppose we ran an experiment to determine the result from changing the levels for each factor. The proof of interaction would be found from calculating the mean of the response and

plotting the results.

Plot the results and determine if there is an interaction.

= (90 + 85) / 2= (92 + 88) / 2

= (98 + 92) / 2 90

= (85 + 88) / 2 85

= (98 + 88) / 2 92= (85 + 92) / 2 88

Copyright ISandR

INPUT OUTPUT

A

A1

A1

A2

A2

A

B1

B2

B1

B2

(AxB)

(AxB)1

(AxB)2

(AxB)2

(AxB)1

Result

90

85

92

88

The upper left plot shows there is little interaction between factorsA and B because the lines representing (AxB) do not intersect. Themore non-parallel the lines, the greater is the interaction.

The upper right plot is the main effects. Here we see that A and Bboth contribute to the output result.

Design of ExperimentsINPUT OUTPUT

A B (AxB) R A1 A2

A1 B1 (AxB)1 90 90

A1 B2 (AxB)2 85 85

A2 B1 (AxB)2 92 92

A2 B2 (AxB)1 88 88

87.5 90

A1 A2

b1 90 92

b2 85 88A1 A2

87.5 90

Interaction of A and B

80

85

90

95

A1 A2

82

84

86

88

90

92

A1 A2 B1 B2

NO

INTERACTION80

85

90

95

A1 A2

Interaction of A and B

B1

B2

= (90 + 85) / 2 = 87.5

= (92 + 88) / 2 = 90

= (90 + 92) / 2 = 91

= (85 + 88) / 2 = 86.5

= (90 + 88) / 2 = 89

= (85 + 92) / 2 = 88.5

A1

A2

B1

B2

(AxB)1

(AxB)2

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This is an orthogonal Taguchi array called an L8. The symmetry or balance allows the evaluation of each factor independently of other factors varied at the same time. How does this pattern work?

Consider factor A. While factor A is “1”, factor B has 2 levels at “0” and “1”, as does C, D, E, F, and G.; and while factor A is “0” , factor B has 2 levels at “0” and “1”, as does C, D, E, F, and G.

The same symmetry is true for each of the other factors as they are tested at each level of “0” and “1”.

This is a “balanced array”. Each factor can be tested with a balanced influence from every other factor. Up to 7 factors can be tested, but confounding will occur.

DOE: Full or Partial Factorial Arrays


FACTOR

A B C D E F G

1 1 1 1 1 1 1 1

2 1 1 1 0 0 0 0

3 1 0 0 1 1 0 0

4 1 0 0 0 0 1 1

5 0 1 0 1 0 1 0

6 0 1 0 0 1 0 1

7 0 0 1 1 0 0 18 0 0 1 0 1 1 0

trial

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Here is the L8 array with the linear diagram(s) which provides the columns where the interactions/confounding occurs. Columns A, B, D, and G will be free of confounding, while C, E, and F will provide information on interactions or have information mixed with that from factors from the “ends” of each line.

Note that not all columns need be used in an experiment. Only the column patterns that minimize confounding might be used (you could chose only A, B, D, and G). However, all of the trials must be conducted (use all of the rows).


A

B D

EC

F

G

A

B D

EC

F

G

FACTOR

A B C D E F G

1 1 1 1 1 1 1 1

2 1 1 1 0 0 0 0

3 1 0 0 1 1 0 0

4 1 0 0 0 0 1 1

5 0 1 0 1 0 1 0

6 0 1 0 0 1 0 1

7 0 0 1 1 0 0 18 0 0 1 0 1 1 0

trial

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• This is where you do an experiment

• You will set up an experiment to determine what influences the time a pendulum takes to make one complete period of an arc swing.

• Consider weight, length of radius, and the start angleprior to beginning the arc swing.

• Label these independent variables W, L, A and determine if there is any interaction with W and L. The dependent variable, time, is labeled t.

• Use two levels for each factor.


Time

Start

Stop

L1

L2

W2

W1

A2

A1

Time

Start

Stop

t2

t1

A

B D

EC

F

G


FACTOR

A B C D E F G

1 1 1 1 1 1 1 1

2 1 1 1 0 0 0 0

3 1 0 0 1 1 0 0

4 1 0 0 0 0 1 1

5 0 1 0 1 0 1 0

6 0 1 0 0 1 0 1

7 0 0 1 1 0 0 18 0 0 1 0 1 1 0

trial

W L (WxL) A

This is one of the assignments of the factors weight (W), length (L), interaction of weight and length (WxL), and the angle (A). Other columns could be used as well to eliminate undesirable interaction (i.e. B, D, F, and G, respectively).

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ANOVA stands for

ANalysis Of VAriance

The technique assigns the percent contribution to the result from each factor, including the noise.

Analysis of Variance

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First, define the terms of an ANOVA table, then evaluate a sample experiment.

Column 1 - this is the degrees of freedom for a factor and is the number of levels minus 1. The degrees of freedom for the entire matrix, DOFTotal, is equal to

DOFTotal = (number of runs) x (number of repetitions) - 1

DOFError = (DOFTotal ) - (the sum of all factors’ DOF)

Column 2 - this is the sum of squares , SS (demonstrated in an example)

Column 3 - this is the mean SS, which equals SS divided by DOF for that row.

Column 4 - the F Ratio is the mean SS for the factor divided by the mean SS for the error.

Column 5 - the percent contribution is the SS for the factor divided by the SSTotal


Factor

A

B

C

Error

Total

DOF

(Degrees of

Freedom) 1

SS (Sum of

Squares) 1

SS (Mean

of SS) 3 F Ratio 4

Percent

Contribution 5

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Use this T and P experimental data for calculation and to fill out the ANOVA table

Column 1 - this is the degrees of freedom

The degrees of freedom for a factor is the number of levels minus 1. runs repetitions

DOFA = 2 - 1 = 1

DOFB = 2 - 1 = 1 ... there is no factor C

Total degrees of freedom equals the number of runs, times the number of repetitions, minus 1

DOFTotal = (4) x (2) - 1 = 7

The degrees of freedom for the error is DOFTotal minus DOF from each of the factors A and B

DOFError = 7 - 1 - 1 = 5

Fill in the table

11

57


DOF

(Degrees of

Freedom) 1

SS (Sum of

Squares) 1

SS (Mean

of SS) 3 F Ratio 4

Percent

Contribution 5

Factor

A

B

C

Error

Total

RUN

1

2

3

4

P

50

50

10

0

10

0

T

25

75

25

75

Result1

98

92

89

98

Result1

95

96

88

87

INPUT OUTPUT

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Column 2 - this is the sum of squares

A correction factor (CF) must be calculated

CF = (98 + 95 + 92 + 98 + 96 + 89 + 88 + 98 + 87)2 / 8 = 66866.1

SSA = [(98 + 95 + 92 + 96)2 /4] - [(89 + 88 + 98 + 87)2 / 4] - CF = 45.125

SSB = [(98 + 95 + 89 + 88)2 /4] - [(92 + 96 + 98 + 87)2 / 4] - CF = 1.125

SSTotal = (98)2 + (95)2 + (92)2 + (98)2 + (96)2 + (89)2 + (88)2 + (98)2 + (87)2 - CF = 72.875

Error makes up the difference between the total and the contribution from each factor

SSError = SSTotal - SSA - SSB = 72.875 - 45.125 - 1.125 = 26.575

Fill in the table

45.125

1.123

26.575

72.875


RUN

1

2

3

4

P

50

50

10

0

10

0

T

25

75

25

75

Result1

98

92

89

98

Result1

95

96

88

87

INPUT OUTPUT

DOF

(Degrees of

Freedom) 1

SS (Sum of

Squares) 1

SS (Mean

of SS) 3 F Ratio 4

Percent

Contribution 5

Factor

A

B

C

Error

Total

1

1

5

7

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Column 3 - this is the mean SSwhich equals SS divided by DOF

= 45.125 / 1 = 45.125

= 1.125 / 1 = 1.125

= 26.575 / 5 = 5.315

= 72.875 / 7 = 10.41

Enter these numbers

45.125

1.125

5.315

10.41


SSA

SSB

SSError

SSTotal

RUN

1

2

3

4

P

50

50

10

0

10

0

T

25

75

25

75

Result1

98

92

89

98

Result1

95

96

88

87

INPUT OUTPUT

45.125

1.123

26.575

72.875

DOF

(Degrees of

Freedom) 1

SS (Sum of

Squares) 1

SS (Mean

of SS) 3 F Ratio 4

Percent

Contribution 5

Factor

A

B

C

Error

Total

1

1

5

7

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Column 4 - the F Ratio is the mean SS for the factor divided by the mean SS for the error

FA = 46.126 / 5.315 = 8.49

FB = 1.126 / 5.315 = 0.211

Fill in the Table

8.490.211


RUN

1

2

3

4

P

50

50

10

0

10

0

T

25

75

25

75

Result1

98

92

89

98

Result1

95

96

88

87

INPUT OUTPUT

45.125

1.123

26.575

72.875

DOF

(Degrees of

Freedom) 1

SS (Sum of

Squares) 1

SS (Mean

of SS) 3 F Ratio 4

Percent

Contribution 5

Factor

A

B

C

Error

Total

1

1

5

7

45.125

1.125

5.315

10.41

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Column 5 - the percent contribution is the SS for the factor divided by the total SS

%A = (45.125 / 72.825) x 100 = 61.9%

%B = (1.125 / 72.825) x 100 = 1.5%

%Error = (26.575 / 72.825) x 100 = 36.5%

Fill in the Table

61.91.5

36.5

RUN

1

2

3

4

P

50

50

10

0

10

0

T

25

75

25

75

Result1

98

92

89

98

Result1

95

96

88

87

INPUT OUTPUT

45.125

1.123

26.575

72.875

DOF

(Degrees of

Freedom) 1

SS (Sum of

Squares) 1

SS (Mean

of SS) 3 F Ratio 4

Percent

Contribution 5

Factor

A

B

C

Error

Total

1

1

5

7

45.125

1.125

5.315

10.41

8.490

0.211

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• the end

• THE END

• THIS IS REALLY THE END

•THE END!

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