technical note 8
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©The McGraw-Hill Companies, Inc., 2006
McGraw-Hill/Irwin
Technical Note 8
Process Capability and Statistical Quality Control
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2
Process Variation Process Capability Process Control Procedures
– Variable data– Attribute data
Acceptance Sampling– Operating Characteristic Curve
OBJECTIVES
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Basic Forms of VariationAssignable variation
is caused by factors that can be clearly identified and possibly managed
Common variation is inherent in the production process
Example: A poorly trained employee that creates variation in finished product output.
Example: A poorly trained employee that creates variation in finished product output.
Example: A molding process that always leaves “burrs” or flaws on a molded item.
Example: A molding process that always leaves “burrs” or flaws on a molded item.
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Taguchi’s View of Variation
IncrementalCost of Variability
High
Zero
LowerSpec
TargetSpec
UpperSpec
Traditional View
IncrementalCost of Variability
High
Zero
LowerSpec
TargetSpec
UpperSpec
Taguchi’s View
Exhibits TN8.1 & TN8.2
Exhibits TN8.1 & TN8.2
Traditional view is that quality within the LS and US is good and that the cost of quality outside this range is constant, where Taguchi views costs as increasing as variability increases, so seek to achieve zero defects and that will truly minimize quality costs.
Traditional view is that quality within the LS and US is good and that the cost of quality outside this range is constant, where Taguchi views costs as increasing as variability increases, so seek to achieve zero defects and that will truly minimize quality costs.
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Process Capability Process limits
Specification limits
How do the limits relate to one another?
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Process Capability Index, Cpk
3
X-UTLor
3
LTLXmin=C pk
Shifts in Process Mean
Capability Index shows how well parts being produced fit into design limit specifications.
Capability Index shows how well parts being produced fit into design limit specifications.
As a production process produces items small shifts in equipment or systems can cause differences in production performance from differing samples.
As a production process produces items small shifts in equipment or systems can cause differences in production performance from differing samples.
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A simple ratio:
Specification Width_________________________________________________________
Actual “Process Width”
Generally, the bigger the better.
Process Capability – A Standard Process Capability – A Standard Measure of How Good a Process Is.Measure of How Good a Process Is.
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Process CapabilityProcess Capability
This is a “one-sided” Capability Index
Concentration on the side which is closest to the specification - closest to being “bad”
3
;3
XUTLLTLXMinC pk
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The Cereal Box Example
We are the maker of this cereal. Consumer Reports has just published an article that shows that we frequently have less than 15 ounces of cereal in a box.
Let’s assume that the government says that we must be within ± 5 percent of the weight advertised on the box.
Upper Tolerance Limit = 16 + .05(16) = 16.8 ounces Lower Tolerance Limit = 16 – .05(16) = 15.2 ounces We go out and buy 1,000 boxes of cereal and find that
they weight an average of 15.875 ounces with a standard deviation of .529 ounces.
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Cereal Box Process CapabilityCereal Box Process Capability Specification or Tolerance
Limits– Upper Spec = 16.8 oz– Lower Spec = 15.2 oz
Observed Weight– Mean = 15.875 oz– Std Dev = .529 oz
3
;3
XUTLLTLXMinC pk
)529(.3
875.158.16;
)529(.3
2.15875.15MinC pk
5829.;4253.MinC pk
4253.pkC
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What does a Cpk of .4253 mean?
An index that shows how well the units being produced fit within the specification limits.
This is a process that will produce a relatively high number of defects.
Many companies look for a Cpk of 1.3 or better… 6-Sigma company wants 2.0!
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Types of Statistical Sampling
Attribute (Go or no-go information)– Defectives refers to the acceptability of
product across a range of characteristics.– Defects refers to the number of defects per
unit which may be higher than the number of defectives.
– p-chart application
Variable (Continuous)– Usually measured by the mean and the
standard deviation.– X-bar and R chart applications
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Statistical Process Control (SPC) Charts
UCL
LCL
Samples over time
1 2 3 4 5 6
UCL
LCL
Samples over time
1 2 3 4 5 6
UCL
LCL
Samples over time
1 2 3 4 5 6
Normal BehaviorNormal Behavior
Possible problem, investigatePossible problem, investigate
Possible problem, investigatePossible problem, investigate
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Control Limits are based on the Normal Curve
x
0 1 2 3-3 -2 -1z
Standard deviation units or “z” units.
Standard deviation units or “z” units.
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Control Limits
We establish the Upper Control Limits (UCL) and the Lower Control Limits (LCL) with plus or minus 3 standard deviations from some x-bar or mean value. Based on this we can expect 99.7% of our sample observations to fall within these limits.
xLCL UCL
99.7%
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Example of Constructing a p-Chart: Required Data
1 100 42 100 23 100 54 100 35 100 66 100 47 100 38 100 79 100 1
10 100 211 100 312 100 213 100 214 100 815 100 3
Sample
No.
No. of
Samples
Number of defects found in each sample
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Statistical Process Control Formulas:Attribute Measurements (p-Chart)
p =Total Number of Defectives
Total Number of Observationsp =
Total Number of Defectives
Total Number of Observations
ns
)p-(1 p = p n
s)p-(1 p
= p
p
p
z - p = LCL
z + p = UCL
s
s
p
p
z - p = LCL
z + p = UCL
s
s
Given:
Compute control limits:
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1. Calculate the sample proportions, p (these are what can be plotted on the p-chart) for each sample
1. Calculate the sample proportions, p (these are what can be plotted on the p-chart) for each sample
Sample n Defectives p1 100 4 0.042 100 2 0.023 100 5 0.054 100 3 0.035 100 6 0.066 100 4 0.047 100 3 0.038 100 7 0.079 100 1 0.01
10 100 2 0.0211 100 3 0.0312 100 2 0.0213 100 2 0.0214 100 8 0.0815 100 3 0.03
Example of Constructing a p-chart: Step 1
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2. Calculate the average of the sample proportions2. Calculate the average of the sample proportions
0.036=1500
55 = p 0.036=1500
55 = p
3. Calculate the standard deviation of the sample proportion 3. Calculate the standard deviation of the sample proportion
.0188= 100
.036)-.036(1=
)p-(1 p = p n
s .0188= 100
.036)-.036(1=
)p-(1 p = p n
s
Example of Constructing a p-chart: Steps 2&3
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4. Calculate the control limits4. Calculate the control limits
3(.0188) .036 3(.0188) .036
UCL = 0.0924LCL = -0.0204 (or 0)UCL = 0.0924LCL = -0.0204 (or 0)
p
p
z - p = LCL
z + p = UCL
s
s
p
p
z - p = LCL
z + p = UCL
s
s
Example of Constructing a p-chart: Step 4
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21Example of Constructing a p-Chart: Step 5
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Observation
p
UCL
LCL
5. Plot the individual sample proportions, the average of the proportions, and the control limits
5. Plot the individual sample proportions, the average of the proportions, and the control limits
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22
Example of x-bar and R Charts: Required Data
Sample Obs 1 Obs 2 Obs 3 Obs 4 Obs 51 10.68 10.689 10.776 10.798 10.7142 10.79 10.86 10.601 10.746 10.7793 10.78 10.667 10.838 10.785 10.7234 10.59 10.727 10.812 10.775 10.735 10.69 10.708 10.79 10.758 10.6716 10.75 10.714 10.738 10.719 10.6067 10.79 10.713 10.689 10.877 10.6038 10.74 10.779 10.11 10.737 10.759 10.77 10.773 10.641 10.644 10.72510 10.72 10.671 10.708 10.85 10.71211 10.79 10.821 10.764 10.658 10.70812 10.62 10.802 10.818 10.872 10.72713 10.66 10.822 10.893 10.544 10.7514 10.81 10.749 10.859 10.801 10.70115 10.66 10.681 10.644 10.747 10.728
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Example of x-bar and R charts: Step 1. Calculate sample means, sample ranges,
mean of means, and mean of ranges.Sample Obs 1 Obs 2 Obs 3 Obs 4 Obs 5 Avg Range
1 10.68 10.689 10.776 10.798 10.714 10.732 0.1162 10.79 10.86 10.601 10.746 10.779 10.755 0.2593 10.78 10.667 10.838 10.785 10.723 10.759 0.1714 10.59 10.727 10.812 10.775 10.73 10.727 0.2215 10.69 10.708 10.79 10.758 10.671 10.724 0.1196 10.75 10.714 10.738 10.719 10.606 10.705 0.1437 10.79 10.713 10.689 10.877 10.603 10.735 0.2748 10.74 10.779 10.11 10.737 10.75 10.624 0.6699 10.77 10.773 10.641 10.644 10.725 10.710 0.13210 10.72 10.671 10.708 10.85 10.712 10.732 0.17911 10.79 10.821 10.764 10.658 10.708 10.748 0.16312 10.62 10.802 10.818 10.872 10.727 10.768 0.25013 10.66 10.822 10.893 10.544 10.75 10.733 0.34914 10.81 10.749 10.859 10.801 10.701 10.783 0.15815 10.66 10.681 10.644 10.747 10.728 10.692 0.103
Averages 10.728 0.220400
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Example of x-bar and R charts: Step 2. Determine Control Limit Formulas and
Necessary Tabled Values
x Chart Control Limits
UCL = x + A R
LCL = x - A R
2
2
x Chart Control Limits
UCL = x + A R
LCL = x - A R
2
2
R Chart Control Limits
UCL = D R
LCL = D R
4
3
R Chart Control Limits
UCL = D R
LCL = D R
4
3
From Exhibit TN8.7From Exhibit TN8.7
n A2 D3 D42 1.88 0 3.273 1.02 0 2.574 0.73 0 2.285 0.58 0 2.116 0.48 0 2.007 0.42 0.08 1.928 0.37 0.14 1.869 0.34 0.18 1.82
10 0.31 0.22 1.7811 0.29 0.26 1.74
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Example of x-bar and R charts: Steps 3&4. Calculate x-bar Chart and Plot Values
10.601
10.856
=).58(0.2204-10.728RA - x = LCL
=).58(0.2204-10.728RA + x = UCL
2
2
10.601
10.856
=).58(0.2204-10.728RA - x = LCL
=).58(0.2204-10.728RA + x = UCL
2
2
10.550
10.600
10.650
10.700
10.750
10.800
10.850
10.900
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Sample
Mea
ns
UCL
LCL
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Example of x-bar and R charts: Steps 5&6. Calculate R-chart and Plot Values
0
0.46504
)2204.0)(0(R D= LCL
)2204.0)(11.2(R D= UCL
3
4
0
0.46504
)2204.0)(0(R D= LCL
)2204.0)(11.2(R D= UCL
3
4
0.000
0.100
0.200
0.300
0.400
0.500
0.600
0.700
0.800
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Sample
RUCL
LCL
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Basic Forms of Statistical Sampling for Quality Control
Acceptance Sampling is sampling to accept or reject the immediate lot of product at hand
Statistical Process Control is sampling to determine if the process is within acceptable limits
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Acceptance Sampling
Purposes– Determine quality level– Ensure quality is within predetermined level
Advantages– Economy– Less handling damage– Fewer inspectors– Upgrading of the inspection job– Applicability to destructive testing– Entire lot rejection (motivation for
improvement)
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Acceptance Sampling (Continued)
Disadvantages– Risks of accepting “bad” lots and
rejecting “good” lots– Added planning and documentation– Sample provides less information than
100-percent inspection
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Acceptance Sampling: Single Sampling Plan
A simple goal
Determine (1) how many units, n, to sample from a lot, and (2) the maximum number of defective items, c, that can be found in the sample before the lot is rejected
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Risk Acceptable Quality Level (AQL)
– Max. acceptable percentage of defectives defined by producer
The(Producer’s risk)– The probability of rejecting a good lot
Lot Tolerance Percent Defective (LTPD)– Percentage of defectives that defines
consumer’s rejection point The (Consumer’s risk)
– The probability of accepting a bad lot
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Operating Characteristic Curve
n = 99c = 4
AQL LTPD
00.10.20.30.40.50.60.70.80.9
1
1 2 3 4 5 6 7 8 9 10 11 12
Percent defective
Pro
bab
ilit
y of
acc
epta
nce
=.10(consumer’s risk)
= .05 (producer’s risk)
The OCC brings the concepts of producer’s risk, consumer’s risk, sample size, and maximum defects allowed together
The OCC brings the concepts of producer’s risk, consumer’s risk, sample size, and maximum defects allowed together
The shape or slope of the curve is dependent on a particular combination of the four parameters
The shape or slope of the curve is dependent on a particular combination of the four parameters
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Example: Acceptance Sampling Problem
Zypercom, a manufacturer of video interfaces, purchases printed wiring boards from an outside vender, Procard. Procard has set an acceptable quality level of 1% and accepts a 5% risk of rejecting lots at or below this level. Zypercom considers lots with 3% defectives to be unacceptable and will assume a 10% risk of accepting a defective lot.
Develop a sampling plan for Zypercom and determine a rule to be followed by the receiving inspection personnel.
Zypercom, a manufacturer of video interfaces, purchases printed wiring boards from an outside vender, Procard. Procard has set an acceptable quality level of 1% and accepts a 5% risk of rejecting lots at or below this level. Zypercom considers lots with 3% defectives to be unacceptable and will assume a 10% risk of accepting a defective lot.
Develop a sampling plan for Zypercom and determine a rule to be followed by the receiving inspection personnel.
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Example: Step 1. What is given and
what is not? In this problem, AQL is given to be 0.01 and LTDP is given to be 0.03. We are also given an alpha of 0.05 and a beta of 0.10.
In this problem, AQL is given to be 0.01 and LTDP is given to be 0.03. We are also given an alpha of 0.05 and a beta of 0.10.
What you need to determine is your sampling plan is “c” and “n.”
What you need to determine is your sampling plan is “c” and “n.”
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Example: Step 2. Determine “c”
First divide LTPD by AQL.First divide LTPD by AQL.LTPD
AQL =
.03
.01 = 3
LTPD
AQL =
.03
.01 = 3
Then find the value for “c” by selecting the value in the TN8.10 “n(AQL)”column that is equal to or just greater than the ratio above.
Then find the value for “c” by selecting the value in the TN8.10 “n(AQL)”column that is equal to or just greater than the ratio above.
Exhibit TN 8.10Exhibit TN 8.10
c LTPD/AQL n AQL c LTPD/AQL n AQL0 44.890 0.052 5 3.549 2.6131 10.946 0.355 6 3.206 3.2862 6.509 0.818 7 2.957 3.9813 4.890 1.366 8 2.768 4.6954 4.057 1.970 9 2.618 5.426
So, c = 6.So, c = 6.
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Example: Step 3. Determine Sample Size
c = 6, from Tablen (AQL) = 3.286, from TableAQL = .01, given in problem
c = 6, from Tablen (AQL) = 3.286, from TableAQL = .01, given in problem
Sampling Plan:Take a random sample of 329 units from a lot. Reject the lot if more than 6 units are defective.
Sampling Plan:Take a random sample of 329 units from a lot. Reject the lot if more than 6 units are defective.
Now given the information below, compute the sample size in units to generate your sampling plan
Now given the information below, compute the sample size in units to generate your sampling plan
n(AQL/AQL) = 3.286/.01 = 328.6, or 329 (always round up)n(AQL/AQL) = 3.286/.01 = 328.6, or 329 (always round up)
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