1 slides used in class may be different from slides in student pack chapter 9a process capability...
TRANSCRIPT
![Page 1: 1 Slides used in class may be different from slides in student pack Chapter 9A Process Capability and Statistical Quality Control Process Variation](https://reader036.vdocuments.us/reader036/viewer/2022062320/56649c9e5503460f9495de52/html5/thumbnails/1.jpg)
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Chapter 9A
Process Capability and Statistical Quality Control Process Variation Process Capability Process Control Procedures
– Variable data– Attribute data
Acceptance Sampling– Operating Characteristic Curve
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Basic Causes of Variation
Assignable causes are factors that can be clearly identified and possibly managed.
Common causes are inherent to the production process. In order to reduce variation due to common causes, the process must be changed.
Key: Determining which is which!
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Types of Control Charts Attribute (Go or no-go information)
– Defectives refers to the acceptability of product across a range of characteristics.
– p-chart application
Variable (Continuous)– Usually measured by the mean and the standard
deviation.– X-bar and R chart applications
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Types of Statistical Quality Control
StatisticalQuality Control
ProcessControl
AcceptanceSampling
VariablesCharts
AttributesCharts
Variables Attributes
StatisticalQuality Control
ProcessControl
AcceptanceSampling
VariablesCharts
AttributesCharts
Variables Attributes
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UCL
LCL
Samples over time
1 2 3 4 5 6
Normal Behavior
UCL
LCL
Samples over time
1 2 3 4 5 6
Possible problem, investigate
UCL
LCL
Samples over time
1 2 3 4 5 6
Possible problem, investigate
Statistical Process Control (SPC) Charts
Excellent review in exhibit TN8.5.
Look for trends!
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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. 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
SampleSample size
Number of defectives
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Statistical Process Control Formulas:Attribute Measurements (p-Chart)
p =Total Number of Defectives
Total Number of Observations
ns
)p-(1 p = p
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.
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 proportions.
0.037=1500
55 = p
3. Calculate the standard deviation of the sample proportion
.0188= 100
.037)-.037(1=
)p-(1 p = p n
s
Example of Constructing a p-chart: Steps 2&3
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4. Calculate the control limits.
3(.0188) .037
UCL = 0.0930LCL = -0.0197 (0)
p
p
z - p = LCL
z + p = UCL
s
s
Example of Constructing a p-chart: Step 4
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Example of Constructing a p-Chart: Step 55. Plot the individual sample proportions, the average
of the proportions, and the control limits
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Observation
p
p
UCL
LCL
p-bar
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R Chart
Type of variables control chart– Interval or ratio scaled numerical data
Shows sample ranges over time– Difference between smallest & largest values in
inspection sample
Monitors variability in process Example: Weigh samples of coffee &
compute ranges of samples; Plot
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R Chart Control Limits
k
RR
RDLCL
RDUCL
k
1ii
3R
4R
Sample Range Sample Range in samplein sample i i
# Samples# Samples
From Table From Table (function of sample (function of sample size)size)
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R Chart Example
You’re manager of a 500-room hotel. You want to analyze the time it takes to deliver luggage to the room. For 7 days, you collect data on 5 deliveries per day. Is the process in control?
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R Chart Hotel Data
SampleDay Delivery Time Mean Range
1 7.30 4.20 6.10 3.45 5.55 5.32
7.30 + 4.20 + 6.10 + 3.45 + 5.557.30 + 4.20 + 6.10 + 3.45 + 5.55 5 5
Sample Mean = Sample Mean =
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R Chart Hotel Data
SampleDay Delivery Time Mean Range
1 7.30 4.20 6.10 3.45 5.55 5.32 3.85
7.30 - 3.457.30 - 3.45Sample Range = Sample Range =
LargestLargest SmallestSmallest
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R Chart Hotel Data
SampleDay Delivery Time Mean Range
1 7.30 4.20 6.10 3.45 5.55 5.32 3.852 4.60 8.70 7.60 4.43 7.62 6.59 4.273 5.98 2.92 6.20 4.20 5.10 4.88 3.284 7.20 5.10 5.19 6.80 4.21 5.70 2.995 4.00 4.50 5.50 1.89 4.46 4.07 3.616 10.10 8.10 6.50 5.06 6.94 7.34 5.047 6.77 5.08 5.90 6.90 9.30 6.79 4.22
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R Chart Control Limits Solution
0894.303
216.8894.311.2
89437
224274853
4
1
RDRLCL
RDUCL
....
k
RR
R
k
ii
From Table From Table ((nn = 5) = 5)
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02468
1 2 3 4 5 6 7
R, Minutes
Day
R Chart Control Chart Solution
UCLUCL
R-barR-bar
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X Chart
Type of variables control chart– Interval or ratio scaled numerical data
Shows sample means over time Monitors process average Example: Weigh samples of coffee &
compute means of samples; Plot
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X Chart Control Limits
k
RR
k
XX
RAXLCL
RAXUCL
k
1ii
k
1ii
2X
2X
Range Range of of samplesample i i
# Samples# Samples
Mean of Mean of sample sample ii
From From TableTable
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X Chart Example
You’re manager of a 500-room hotel. You want to analyze the time it takes to deliver luggage to the room. For 7 days, you collect data on 5 deliveries per day. Is the process in control?
![Page 24: 1 Slides used in class may be different from slides in student pack Chapter 9A Process Capability and Statistical Quality Control Process Variation](https://reader036.vdocuments.us/reader036/viewer/2022062320/56649c9e5503460f9495de52/html5/thumbnails/24.jpg)
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X Chart Hotel Data
SampleDay Delivery Time Mean Range
1 7.30 4.20 6.10 3.45 5.55 5.32 3.852 4.60 8.70 7.60 4.43 7.62 6.59 4.273 5.98 2.92 6.20 4.20 5.10 4.88 3.284 7.20 5.10 5.19 6.80 4.21 5.70 2.995 4.00 4.50 5.50 1.89 4.46 4.07 3.616 10.10 8.10 6.50 5.06 6.94 7.34 5.047 6.77 5.08 5.90 6.90 9.30 6.79 4.22
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X Chart Control Limits Solution*
555.3894.358.0813.5
071.8894.358.0813.5
894.37
22.427.485.3
813.57
79.659.632.5
2
2
1
1
RAXLCL
RAXUCL
k
RR
k
XX
X
X
k
ii
k
ii
From TableFrom Table
((nn = 5) = 5)
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X ChartControl Chart Solution*
02468
1 2 3 4 5 6 7
X, Minutes
Day
UCLUCL
LCLLCL
X-barX-bar
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X AND R CHART EXAMPLEIN-CLASS EXERCISE
The following collection of data represents samples of the amount of force applied in a gluing process:
Determine if the process is in control
by calculating the appropriate upper and lower
control limits of the X-bar and R charts.
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X AND R CHART EXAMPLEIN-CLASS EXERCISE
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 Range1 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
R Chart Control Limits
UCL = D R
LCL = D R
4
3
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
60110220405872810
85610220405872810
2
2
.)=.(-..R - AxLCL =
.)=.(..R + AxUCL =
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 15Sample
Mea
ns
SamplemeanUCL
LCL
grandmean of x
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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
0.000
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
R
RangeUCLLCLR-bar
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SOLUTION:Example of x-bar and R charts:
1. Is the process in Control?
2. If not, what could be the cause for the process being out of control?
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Process Capability
Process limits - actual capabilities of process based on historical data
Tolerance limits - what process design calls for- desired performance of process
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Process Capability
How do the limits relate to one another?
You want: tolerance range > process range
Two methods of accomplishing this:
1. Make bigger 2. Make smaller
Bad idea Implies having greater control over processGood!
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Process Capability Measurement
Cp index = Tolerance range / Process range
What value(s) would you like for Cp?
Larger Cp indicates a more reliable and predictable process (less variability)
The Cp index is based on the assumption that the process mean is centered at the midpoint of the tolerance range
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-4
LTL UTL
6)( LTLUTL
C p
X
16
6
pC
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26
12
pC
-4
LTL UTL
X
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While the Cp index provides useful information on process variability, it does not give information on the process average relative to the tolerance limits. Note:
-4
9
UTLLTL
26
12
pC
X
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Cpk Index
3
X-UTLor
3
LTLXmin=Cpk
Together, these process capability Indices show how well parts being produced conform to design specifications.
X = process mean (Unknown but can be estimated with the grand mean)
= standard deviation (Unknown but can be estimated with the average range)
Refers to the LTL Refers to the UTL
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13
3,
3
9min
pkC
26
12
pC
-4
9
X
LTL UTLSince Cp and Cpk are different we can conclude that the process is not centered, however the Cp index tells us that the process variability is very low
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An example of the use of process capability indicesThe design specifications for a machined slot is 0.5± .003 inches. Samples have been taken and the process mean is estimated to be .501. The process standard deviation is estimated to be .001.
What can you say about the capability of this process to produce this dimension?
6
LTLUTLC p
1001.6
497.503.
pC
3
X-UTL ,
3
LTLXmin=Cpk
667..0013
.501-.503 ,
001.3
497.501.min=Cpk
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Process capability
0.497 inchesLTL
0.503 inchesUTL
Process mean0.501 inches
Machined slot(inches)
= 0.001inches
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Sampling Distributions(The Central Limit Theorem)
Regardless of the underlying distribution, if the sample is large enough (>30), the distribution of sample means will be normally distributed around the population mean with a standard deviation of :
nx
/
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Computing Process Capability Indexes Using Control Chart DataRecall the following info from our in class
exercise:
Since A2 is calculated on the assumption of three sigma limits:
728.10X
22.R
58.2 A
043.3)22(.58.
32 RA
x
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From the Central Limit Theorem:
So,
Therefore,
nx
5043.
096.5043.
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Suppose the Design Specs for the Gluing Process were 10.7 .2, Calculate the Cp and Cpk Indexes:
Answer:
694.096.6
5.109.10
6
LTLUTL
C p
3,
3min
XUTLLTLXC pk
597.597.,792.min096.3
728.109.10,
096.3
5.10728.10min
pkC
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-4
1.792.8
Note, multiplying each component of the Cpk calculation by 3 yields a Z value. You can use this to predict the % of items outside the tolerance limits:
From Appendix E we would expect:
.008 + .036 = .044 or 4.4%
non-conforming product from this process
.792 * 3 = 2.38 .597 * 3 = 1.79
.008 or .8% of the curve .036 or 3.6% of the curve
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Capability Index – In Class Exercise
You are a manufacturer of equipment. A drive shaft is purchased from a supplier close by. The blueprint for the shaft specs indicate a tolerance of 5.5 inches ± .003 inches. Your supplier is reporting a mean of 5.501 inches. And a standard deviation of .0015 inches.
What is the Cpk index for the supplier’s process?
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3,
3min
XUTLLTLXC pk
444.444.,888.min0015.3
501.5503.5,
0015.3
497.5501.5min
pkC
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Your engineering department is sent to the supplier’s site to help improve the capability on the shaft machining process. The result is that the process is now centered and the CP index is now .75. On a percentage basis, what is the improvement on the percentage of shafts which will be unusable (outside the tolerance limits)?
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To answer this question we must determine the percentage of defective shafts before and after the intervention from our engineering department
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Before:
-4
x.1.3x.88) =2.67
444.444.,888.min0015.3
501.5503.5,
0015.3
497.5501.5min
pkC
From Table
.089
From Table
.004
Total % outside
Tolerance = .089 + .004 = .093 or 9.3%
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AfterSince the process is centered then Cpk = Cp; Cp = UTL-LTL / 6so the tolerance limits are .75 x 6 = 4.5 apart each 2.25 from the mean
-4
2.22.25
From Table
.012
So % outside of
Tolerance =
.012(2) = .024
Or 2.4%
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So the percentage decrease in defective parts is 1 – (2.4/9.3) = 74%
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Additional Questions A circular bar is produced on a draw bench by drawing the bar through a die. The bar has an UTL of 3.06 inches and a LTL of 2.94 inches. The process is in control and normally distributed. The following capability indices have been computed:
5.03
XUTL
5.13
LTLX
What is the CP index? What is the process mean (X bar bar)? Suppose 1000 bars were produced with this process. How many bars would you expect to be within the specification limits?
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Basic Forms of Statistical Sampling for Quality Control Sampling to accept or reject the immediate
lot of product at hand (Acceptance Sampling).
Sampling to determine if the process is within acceptable limits (Statistical Process Control)
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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
Disadvantages– Risks of accepting “bad” lots and rejecting
“good” lots– Added planning and documentation– Sample provides less information than 100-
percent inspection – No information is obtained on the process. Just
sorting “good” parts from “bad” parts
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Risk
Acceptable Quality Level (AQL)– Max. acceptable percentage of defectives
defined by producer. (Producer’s risk)
– The probability of rejecting a good lot. Lot Tolerance Percent Defective (LTPD)
– Percentage of defectives that defines consumer’s rejection point.
(Consumer’s risk)– The probability of accepting a bad lot.