Download IENG 451 Lecture 04: Quality Matters

IENG 451 - Lecture 04
Quality Matters:
Cost of Quality, Yield and Variance
Reduction
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Quality is a multifaceted entity.

Traditional (OLD) definition of Quality:
• Fitness for Use
(i.e., products must meet requirements of
those who use them.)
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Two Aspects of “Fitness for Use”


Quality of Design –
•
all products intentionally made in various grades of
quality. (e.g., Autos differ with respect to size, options,
speed, etc.)
Quality of Conformance –
•
how well the product conforms to specifications. (e.g.,
If diameter of a drilled hole is within specifications then
it has good quality.)
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What's Wrong with "Fitness for
Use" Definition of Quality?

Unfortunately, quality as “Fitness for Use” has
become associated with the "conformance to
specifications" regardless of whether or not the
product is fit for use by customer.

Common Misconception:
•
Quality can be dealt with solely in manufacturing that is, by "gold plating" the product
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Cost of Quality Myth:
Higher Quality  Higher Cost
Total Cost
Failure
Cost
$
Quality Cost
Defect Rate
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Reduction of Variability



Modern Definition of Quality:
• Quality is inversely proportional to variability
If variability of product decreases  quality of
product increases
Quality Improvement –
•
Reduction of variability in processes and products
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Cost of (Poor) Quality:
Higher Quality  Lower Cost

Example: Manufacture of Copier Part
100 parts
Manufacturing Process
$20 / part
(10 scrap parts)
(75 good parts)
25% Non-conforming:
(25 parts)
Re-work Process
$4 / part
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75% Conform
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(15 good parts)
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Study finds excessive process
variability causes high defect rate

New process implemented
• NOW: manufacturing non-conformities =
5%
• SAVINGS: $22.89 – $20.53 = $2.36 / good part
• PRODUCTIVITY: 9% improvement
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Understanding Process
Variation
 Three
Aspects:
•Location
•Spread
•Shape
 Independence:
•changing Location does not impact Spread
•Frequently, the CLT lets us use Normal Curve
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Shape: Distributions


Distributions quantify the probability of an event
Events near the mean are most likely to occur, events
further away are less likely to be observed
35.0 
2.5
30.4
(-3)
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34.8
32.6
(-)
(-2)
37
()
39.2
(+)
43.6
41.4
(+3)
(+2)
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Standard Normal Distribution
• The Standard Normal Distribution has a mean () of 0 and
a standard deviation () of 1
• Total area under the curve, (z), from z = – to z =  is
exactly 1 ( -or- 100% of the observations)
• The curve is symmetric about the mean
• Half of the total area lays on either side, so:
(– z) = 1 – (z)
(z)

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z
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Standard Normal Distribution
• How likely is it that we would observe a data point
more than 2.57 standard deviations beyond the
mean?
• Area under the curve from – to z = 2.5  is found by
using the Standard Normal table, looking up the
cumulative area for z = 2.57, and then subtracting the
cumulative area from 1.
(z)

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z
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Standard Normal Distribution
• How likely is it that we would observe a data point
more than 2.57 standard deviations beyond the
mean?
• Area under the curve from – to z = 2.5  is found by
using the table on pp. 716-717, looking up the
cumulative area for z = 2.57, and then subtracting the
cumulative area from 1.
• Answer: 1 – .99492 = .00508, or about 5 times in 1000
(z)

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z
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What if the distribution isn’t a
Standard Normal Distribution?
 If
it is from any Normal Distribution, we can
express the difference from an observation to the
mean in units of the standard deviation, and this
converts it to a Standard Normal Distribution.
• Conversion formula is:
where:
z
x

x is the point in the interval,
 is the population mean, and
 is the population standard deviation.
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Example: Process Yield

Specifications are often set irrespective of process
distribution, but if we understand our process we can
estimate yield / defects.
• Assume a specification calls for a value of 35.0  2.5.
• Assume the process has a distribution that is Normally
distributed, with a mean of 37.0 and a standard deviation of
2.20.
• Estimate the proportion of the process output that will meet
specifications.
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Six Sigma - Motorola


Six Sigma* = 3.4 defects per million opportunities!
•
Every Motorola employee must show bottom line
results of quality project – finance, mail room,
manufacturing, etc.
•
•
•
•

± 6 standard deviations – after a 1.5 sigma shift!
identify problem;
develop measurement;
set goal;
close gap
Long term process – 5 years to fully implement
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Questions & Issues

There WILL be a lab tomorrow:
• It is a follow-on from last week, covering
process improvement
• Variation effects
• Cost of (poor) Quality
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