Download - 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
Making a Quality Product
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
Making a Quality Product
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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Making a Quality Product
Sure! Why not?!
So start the process and make product.
Raw
Material
Process
Control
Processing
Cell
Process Control Chart
Product
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Making a Quality Product
The customer expects uniformity.
Does all the product behave the same and conform to the manufacturing specifications?
Raw
Material
Process
Control
Processing
Cell
Process Control Chart
Product
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Raw
Material
Product
Process
Control
Processing
Cell
Process Control Chart
Wait a second!
What’s this?
This product is different!
The customer won’t accept this part!
So this product gets trashed.
Making a Quality Product
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TRASH
Making a Quality Product
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
Process Control Chart
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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?
Making a Quality Product
Feedback $
$?
TRASH
Raw
Material
Process
Control
Processing
Cell
Process Control Chart
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.
Making a Quality Product
Raw
Material
Feedback
DataSPC Chart
Control
Charts
Product
Process
Control
Processing
Cell
Process Control Chart
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Raw
Material
Product
Design of
Experiments
. (DOE)
Process
Capability
Feedback DataSPC Chart
Feedback
Process
Control
Processing
Cell
Process Control Chart
Making a Quality Product
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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Making a Quality Product
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
Process Control Chart
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Raw
Material
Product
DataFeedbackFeedback
Process
Control
Processing
Cell
Process Control Chart
Making a Quality Product
Control Charts
ID out-of-control events
TYPES
Variable (measurable)
Attribute (yes/no, on/off)
SPC Chart
Design of Experiments
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
Making a Quality Product
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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Making a Quality Product
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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Making a Quality Product
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.
Making a Quality Product
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.
Making a Quality Product
We need process control to monitor the input variables
Projectile weight and diameter
Powder weight
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Making a Quality Product
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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Making a Quality Product
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
Making a Quality Product
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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.
Making a Quality Product
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.
Making a Quality Product
Variable Control Charts
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Making a Quality Product
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
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.
Making a Quality Product
980702 980703 980704
980705 980706 980707
lot variation location
• • • •• • • •• • •• • • •• ••
•
• • •• •• • • •• • •• • •• •
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123 29Big Bullet Final Test See Customer Spec
Kim Tester #7 Tester #7 inch 0.0
6.27.0
Making a Quality Product
01
Making a Quality Product
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
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.
2
4
6
8UCL
LCL
5
10
15
range
avera
ge •
•
lot variation location
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
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
2
4
6
8UCL
LCL
5
10
15
range
avera
ge
123 29Big Bullet Final Test See Customer Spec
Kim Tester #7 Tester #7 inch 0.0
6.27.0
•
•
Making a Quality Product
01
Making a Quality Product
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.
lot variation location
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
Copyright ISandR
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
2
4
6
8UCL
LCL
5
10
15
range
avera
ge
123 29Big Bullet Final Test See Customer Spec
Kim Tester #7 Tester #7 inch 0.0
6.27.0
•
•
Making a Quality Product
01
Making a Quality Product
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?
lot variation location
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
Copyright ISandR
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
2
4
6
8UCL
LCL
5
10
15
range
avera
ge
123 29Big Bullet Final Test See Customer Spec
Kim Tester #7 Tester #7 inch 0.0
6.27.0
•
•
Making a Quality Product
01
Making a Quality Product
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.
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
2
4
6
8UCL
LCL
5
10
15
range
avera
ge
123 29Big Bullet Final Test See Customer Spec
Kim Tester #7 Tester #7 inch 0.0
6.27.0
•
•
Making a Quality Product
01
Making a Quality Product
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
••
••
•
•
Copyright ISandR
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
2
4
6
8UCL
LCL
5
10
15
range
avera
ge
123 29Big Bullet Final Test See Customer Spec
Kim Tester #7 Tester #7 inch 0.0
6.27.0
•
•
Making a Quality Product
01
Making a Quality Product
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
Copyright ISandR
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
Copyright ISandR
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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Making a Quality Product
oo
oo o
o
o
oo
oo
o
This is precise but This is accurate but not accurate. not precise.
Copyright ISandR
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
The Characteristics of a Normal Distribution Curve
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.
Copyright ISandR
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
Copyright ISandR
0
The Characteristics of a Normal Distribution Curve
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
Copyright ISandR
SPLAT
MASH!
MATH!
The Characteristics of a Normal Distribution Curve
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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The Characteristics of a Normal Distribution Curve
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
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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.
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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
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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
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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
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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
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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
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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)
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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
Variable Control Charts
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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
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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
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Fishbone diagrams - to brainstorm process improvement
Effect
Machine Operator
Weather Raw Materials
PM
Temp
Shift
Season Storage
Preparation
Problem Solving
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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
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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
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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
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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
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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
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
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
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• 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)
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• 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)
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Design of Experiments
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
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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
Design of Experiments
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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
Design of Experiments
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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.
Design of 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
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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
Design of Experiments
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
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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.
Design of Experiments
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
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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)
Design of Experiments
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
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
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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
Design of Experiments
84
86
90
98
90
84
90
98
90
92
88
88
96
82
98
94
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
1
2
3
4
84
98
84
90
90
88
82
98
INPUT OUTPUT
Factor A Factor B Factor C Result 1 Result 2A1
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
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Design of Experiments
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
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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
Design of Experiments
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).
Design of Experiments
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.
Design of Experiments
Time
Start
Stop
L1
L2
W2
W1
A2
A1
Time
Start
Stop
t2
t1
A
B D
EC
F
G
Design of Experiments
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
Analysis of Variance
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
Analysis of Variance
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
Analysis of Variance
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
Analysis of Variance
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
Analysis of Variance
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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Analysis of Variance
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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