tm 620: quality management session seven – 9 november 2010 control charts, part i –variables
TRANSCRIPT
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TM 620: Quality Management
Session Seven – 9 November 2010
• Control Charts, Part I– Variables
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Recall: What is Quality?
• Juran – Quality is fitness for use
=> we should be able to determine a set of measurable characteristics which define quality
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Recall: What is Quality?
• Taguchi – Loss from quality is proportional to the amount of variability in the system– Why?
• => if we reduce variation, we reduce loss from quality
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Quality Improvement
• The reduction of variability in processes and products
Equivalent definition:
• The reduction of waste
• Waste is any activity for which the customer will not pay
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Recall: Cost of Quality
Cost of Failure
Cost of Control
Total Cost
Traditional View
Quality Level
Costs
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Traditional Loss Function
x
LSL USL
T
TLSL USL
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Example (Sony, 1979)
Comparing cost of two Sony television plants in Japan and San Diego. All units in San Diego fell within specifications. Japanese plant had units outside of specifications.
Loss per unit (Japan) = $0.44Loss per unit (San Diego) = $1.33
How can this be?
Sullivan, “Reducing Variability: A New Approach to Quality,” Quality Progress, 17, no.7, 15-21, 1984.
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Example
x
T
U.S. Plant (2 = 8.33)
Japanese Plant (2 = 2.78)
LSL USL
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Taguchi Loss Function
x
T
x
T
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Taguchi Loss Function
T
L(x)
x
k(x - T)2
L(x) = k(x - T)2
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Estimating Loss Function
Suppose we desire to make pistons with diameter D = 10 cm. Too big and they create too much friction. Too little and the engine will have lower gas mileage. Suppose tolerances are set at D = 10 + .05 cm. Studies show that if D > 10.05, the engine will likely fail during the warranty period. Average cost of a warranty repair is $400.
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Estimating Loss Function
10
L(x)
10.05
400
400 = k(10.05 - 10.00)2
= k(.0025)
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Estimating Loss Function
10
L(x)
10.05
400
400 = k(10.05 - 10.00)2
= k(.0025)
k = 160,000
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Example 2
Suppose we have a 1 year warranty to a watch. Suppose also that the life of the watch is exponentially distributed with a mean of 1.5 years. The warranty costs to replace the watch if it fails within one year is $25. Estimate the loss function.
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Example 2
1.5
L(x)
25
1
f(x)
25 = k(1 - 1.5)2
k = 100
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Example 2
1.5
L(x)
25
1
f(x)
25 = k(1 - 1.5)2
k = 100
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Single Sided Loss Functions
Smaller is better
L(x) = kx2
Larger is better
L(x) = k(1/x2)
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Example 2
L(x)
25
1
f(x)
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Example 2
L(x)
25
1
f(x)
25 = k(1)2
k = 25
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Expected Loss
Expected Loss: Piston DiameterProbability Probability
Diameter Process A Process B9.925 0.000 0.0259.950 0.200 0.0759.975 0.200 0.20010.000 0.200 0.40010.025 0.200 0.20010.050 0.200 0.07510.075 0.000 0.025
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Expected Loss: Piston DiameterProbability Probability
Diameter Loss Process A Process B9.925 900 0.000 0.0259.950 400 0.200 0.0759.975 100 0.200 0.20010.000 0 0.200 0.40010.025 100 0.200 0.20010.050 400 0.200 0.07510.075 900 0.000 0.025
Expected Loss
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Expected Loss: Piston DiameterProbability Weighted Probability
Diameter Loss Process A Loss Process B9.925 900 0.000 0.0 0.0259.950 400 0.200 80.0 0.0759.975 100 0.200 20.0 0.20010.000 0 0.200 0.0 0.40010.025 100 0.200 20.0 0.20010.050 400 0.200 80.0 0.07510.075 900 0.000 0.0 0.025
Expected Loss
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Expected Loss: Piston DiameterProbability Weighted Probability Weighted
Diameter Loss Process A Loss Process B Loss9.925 900 0.000 0.0 0.025 22.59.950 400 0.200 80.0 0.075 30.09.975 100 0.200 20.0 0.200 20.010.000 0 0.200 0.0 0.400 0.010.025 100 0.200 20.0 0.200 20.010.050 400 0.200 80.0 0.075 30.010.075 900 0.000 0.0 0.025 22.5
Expected Loss
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Expected Loss
Expected Loss: Piston DiameterProbability Weighted Probability Weighted
Diameter Loss Process A Loss Process B Loss9.925 900 0.000 0.0 0.025 22.59.950 400 0.200 80.0 0.075 30.09.975 100 0.200 20.0 0.200 20.010.000 0 0.200 0.0 0.400 0.010.025 100 0.200 20.0 0.200 20.010.050 400 0.200 80.0 0.075 30.010.075 900 0.000 0.0 0.025 22.5
Exp. Loss = 200.0 145.0
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Expected Loss
Recall, X f(x) with finite mean and variance 2.
E[L(x)] = E[ k(x - T)2 ]
= k E[ x2 - 2xT + T2 ]
= k E[ x2 - 2xT + T2 - 2x + 2 + 2x - 2 ]
= k E[ (x2 - 2x+ 2) - 2 + 2x - 2xT + T2 ]
= k{ E[ (x - )2 ] + E[ - 2 + 2x - 2xT + T2 ] }
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Expected Loss
E[L(x)] = k{ E[ (x - )2 ] + E[ - 2 + 2x - 2xT + T2 ] }
Recall, Expectation is a linear operator and E[ (x - )2 ] = 2
E[L(x)] = k{2 - E[ 2 ] + E[ 2x - E[ 2xT ] + E[ T2 ] }
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Expected Loss
Recall,
E[ax +b] = aE[x] + b = a + b
E[L(x)] = k{2 - 2 + 2 E[ x - 2T E[ x ] + T2 }
=k {2 - 2 + 22 - 2T + T2 }
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Expected Loss
Recall, E[ax +b] = aE[x] + b = a + b
E[L(x)] = k{2 - 2 + 2 E[ x - 2T E[ x ] + T2 }
=k {2 - 2 + 22 - 2T + T2 }
=k {2 + ( - T)2 }
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Expected Loss
Recall, E[ax +b] = aE[x] + b = a + b
E[L(x)] = k{2 - 2 + 2 E[ x - 2T E[ x ] + T2 }
=k {2 - 2 + 22 - 2T + T2 }
=k {2 + ( - T)2 }
= k { 2 + ( x - T)2 } = k (2 +D2 )
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Since for our piston example, x = T,
D2 = (x - T)2 = 0
L(x) = k2
Example
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Example (Piston Diam.)
Expected Loss: Piston DiameterProbability Weighted Probability Weighted
Diameter (x-)2
Process A (x-)2
Process B (x-)2
9.925 0.0056 0.000 0.0000 0.025 0.00019.950 0.0025 0.200 0.0005 0.075 0.00029.975 0.0006 0.200 0.0001 0.200 0.000110.000 0.0000 0.200 0.0000 0.400 0.000010.025 0.0006 0.200 0.0001 0.200 0.000110.050 0.0025 0.200 0.0005 0.075 0.000210.075 0.0056 0.000 0.0000 0.025 0.0001
Var = 0.0013 0.0009E[LA(x)] = .0013*k E[LB(x)] = .0009*k
= 200 = 145
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Example (Sony)
x
T
U.S. Plant (2 = 8.33)
Japanese Plant (2 = 2.78)
LSL USL
E[LUS(x)] = 0.16 * 8.33 = $1.33
E[LJ(x)] = 0.16 * 2.78 = $0.44
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Tolerance (Pistons)
10
L(x)
10.05
400
400 = k(10.05 - 10.00)2
= k(.0025)
k = 160,000
Recall,
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Tolerance
Suppose repair for an engine which will fail during warranty can be made for only $200 10
L(x)
10.05
400
LSL USL
200
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Tolerance
Suppose repair for an engine which will fail during warranty can be made for only $200
200 =160,000(tolerance)2 10
L(x)
10.05
400
LSL USL
200
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Tolerance
Suppose repair for an engine which will fail during warranty can be made for only $200
200 = 160,000(tolerance)2
tolerance = (200/160,000)1/2
= .0354
10
L(x)
10.05
400
LSL USL
200
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Statistical Thinking
• All work occurs in a system of interconnected processes
• All process have variation
• Understanding variation and reducing variation are important keys to success
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Variability
• A certain amount of variability is inescapable
• Therefore, no two products are identical
• The larger the variability, the greater the probability that the customer will perceive its existence
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Sources of Variability
Include:
• Differences in materials
• Differences in the performance and operation of the manufacturing equipment
• Differences in the way the operators perform their tasks
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Variability and Statistics
• Variability is difference from the target• Characteristics of quality must be measurable
Therefore,• Variability is described in statistical terms• We will use statistical methods in our quality
improvement activities
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Breaking down workplace barriers
• Train people in the basics of the tools and methods
• Deal with those who look down on or fear numerical methods
• Encourage a data-driven culture of decision making
• Lose the jargon• No “lying with statistics”
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Recall: Types of Errors
• Type I error – Producers risk– Probability that a good product will be rejected
• Type II error– Consumers risk– Probability that a nonconforming product will be
available for sale
• Type III error– Asking the wrong question
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Data on Quality Characteristics
• Attribute data– Discrete– Often a count of some type
• Variable data– Continuous– Often a measurement, such as length,
voltage, or viscosity
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Terms
• Specifications
• Target (or Nominal) Value
• Upper Specification Limit
• Lower Specification Limit
• Random Variation
• Non-random Variation
• Process stability
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Terms
• Nonconforming: failure to meet one or more of the specifications
• Nonconformity: a specific type of failure
• Defect: a nonconformity serious enough to significantly affect the safe or effective use of the produce or completion of the service
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Nonconforming vs. Defective
• A nonconforming product is not necessarily unfit for use
• A nonconforming product is considered defective if if it has one or more defects
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Classroom Exercise
• For a product or service in your job:– Name a quality characteristic– Give an example of a nonconformity that is
not a defect– Give an example of a defect
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Types of Inspection
• Receiving
• In Process
• Final
• None
• One Hundred Percent
• Acceptance Sampling
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Quality Design & Process Variation
60 80 100 120 140
60 140
14060
Lower Spec Limit
Upper Spec Limit
AcceptanceSampling
Statistical ProcessControl
ExperimentalDesign
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Variation and Control
• A process that is operating with only common causes of variation is said to be in statistical control.
• A process operating in the presence of special or assignable cause is said to be out of control.
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Finding Trends and Special Causes
• Inspection does not tell you about a problem until it becomes a problem
• We need a mechanism to help us spot special causes when they occur
• We need mechanism to help us determine when we have a trend in the data
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Statistical Process Control
• Originally developed by Walter Shewhart in 1924 at the Bell Telephone Laboratories
• Late 1920s, Harold Dodge and Harry Romig developed statistically based acceptance sampling
• Not recognized by industry until after World War II
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Definition
• Statistical Process Control (SPC):– “a methodology for monitoring a process to
identify special causes of variation and signal the need to take corrective action when it is appropriate”
(Evans and Lindsay)
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Statistical Process Control Tools
• The magnificent seven
• The tool most often associated with Statistical Process Control is Control Charts
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Common Causes
Special Causes
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Histograms do not take into account changes over time.
Control charts can tell us when a process changes
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Control Chart Applications• Establish state of statistical control• Monitor a process and signal when it goes out of
control• Determine process capability
• Note: Control charts will only detect the presence of assignable causes. Management, operator, and engineering action is necessary to eliminate the assignable cause.
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Capability Versus Control
Control
Capability
Capable
Not Capable
In Control Out of Control
IDEAL
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Developing Control Charts
1. Prepare– Choose measurement– Determine how to collect data, sample
size, and frequency of sampling– Set up an initial control chart
2. Collect Data– Record data– Calculate appropriate statistics– Plot statistics on chart
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Types of Sampling
• Random Samples – Each piece has an equal chance of being selected for
inspection
• Systematic Samples– According to time or sequence
• Rational subgroups– A group of data that is logically homogeneous– Computing variation between subgroups
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Approaches to Rational Subgrouping
• Each sample consists of units that were produced at the same time (or as closely together as possible)– Used when the primary purpose of the control chart is
to detect process shifts
• Each sample consists of units of product that are representative of all units that have been produced since the last sample was taken– Used when the purpose of the control chart is to
make lot sentencing decisions
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Next Steps
3. Determine trial control limits– Center line (process average)
– Compute UCL, LCL
4. Analyze and interpret results– Determine if in control
– Eliminate out-of-control points
– Re-compute control limits as necessary
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Warning Limits on Control Charts
Some suggest using two sets of limits on control charts:
• Action limits– Set at 3-sigma– When a point plots outside of this limit, a search for an
assignable cause is made and any necessary corrective action is taken
• Warning limits– Set at 2-sigma– When one or more points fall in between the warning
and action limits or very close to the warning limit, be suspicious
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Typical Out-of-Control Patterns• Point outside control limits
• Hugging the center line
• Hugging the control limits
• Instability
• Sudden shift in process average
• Cycles
• Trends
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Shift in Process Average
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Identifying Potential Shifts
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Cycles
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Trend
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Western Electric Sensitizing Rules:
• One point plots outside the 3-sigma control limits
• Two of three consecutive points plot outside the 2-sigma warning limits
• Four of five consecutive points plot beyond the 1-sigma limits
• A run of eight consecutive points plot on one side of the center line
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Additional sensitizing rules:
• Six points in a row are steadily increasing or decreasing
• Fifteen points in a row with 1-sigma limits (both above and below the center line)
• Fourteen points in a row alternating up and down• Eight points in a row in both sides of the center
line with none within the 1-sigma limits• An unusual or nonrandom pattern in the data• One of more points near a warning or control limit
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Classroom Exercise
• In small groups, choose two of the sensitizing rules. For each of your two rules, make up a (reasonable) situation where that rule would catch a problem and a (reasonable) situation where that rule might falsely identify a problem.
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Final Steps
5. Use as a problem-solving tool– Continue to collect and plot data
– Take corrective action when necessary
6. Compute process capability
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Capability vs. Stability
A process is capable if individual products consistently meet specification
A process is stable only if common variation is present in the process
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Process Capability Calculations
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Process Capability
• The ability of the process to perform at the required level for the given quality characteristic
• Two of the methods for determining:– Find the probability that the process will produce a
part below the LSL plus the probability that the process will produce a part above the USL
– Process Capability RatioPCR = CP = (USL – LSL) / 6σ
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Process Capability Ratio Note
• There are many ways we can estimate the capability of our process
• If σ is unknown, we can replace it with one of the following estimates:– The sample standard deviation S
– R-bar / d2
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PCR and One Sided Specifications
• Upper specification only– CPU = (USL – μ) / 3σ
• Lower specification only– CPL = (μ – LSL) / 3σ
• We can use the same estimate for σ
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PCR and an Off-Center Process
• CPK = min (CPU, CPL)
• Generally, if CP = CPK, then the process is centered at the midpoint of the specifications
• If CP ≠ CPK, then the process is off-center
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Commonly Used Control Charts
• Variables data– x-bar and R-charts
– x-bar and s-charts
– Charts for individuals (x-charts)
• Attribute data– For “defectives” (p-chart, np-chart)
– For “defects” (c-chart, u-chart)
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Control Charts
We assume that the underlying distribution is normal with some mean and some constant but unknown standard deviation .
Letx
x
ni
i
n
1
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Distribution of x
Recall that x is a function of random variables, so it also is a random variable with its own distribution. By the central limit theorem, we know that
where,
x N x ( , )
xnx
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Control Charts
xx
x
x
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Control Charts
x
x
UCL
LCL
UCL & LCL Set atProblem: How do we estimate & ?
3 x
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Control Charts
xx
m
m
ii
1
)(1 fm
RR
m
i
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Control Charts
xx RALCL 2
xx RAUCL 2
RDLCLR 3
RDUCLR 4
xx
m
m
ii
1
)(1 fm
RR
m
i
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Example
• Suppose specialized o-rings are to be manufactured at .5 inches. Too big and they won’t provide the necessary seal. Too little and they won’t fit on the shaft. Twenty samples of 2 rings each are taken. Results follow.
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Part Measurements R Chart X ChartNo. 1 2 x R UCL R UCL LCL Xbar
1 0.502 0.504 0.503 0.002 0.0077 0.002 0.5052 0.4964 0.5032 0.495 0.497 0.496 0.002 0.0077 0.002 0.5052 0.4964 0.4963 0.492 0.496 0.494 0.004 0.0077 0.004 0.5052 0.4964 0.4944 0.501 0.498 0.500 0.003 0.0077 0.003 0.5052 0.4964 0.5005 0.507 0.508 0.508 0.001 0.0077 0.001 0.5052 0.4964 0.5086 0.504 0.504 0.504 0.000 0.0077 0.000 0.5052 0.4964 0.5047 0.497 0.496 0.497 0.001 0.0077 0.001 0.5052 0.4964 0.4978 0.493 0.496 0.495 0.003 0.0077 0.003 0.5052 0.4964 0.4959 0.502 0.501 0.502 0.001 0.0077 0.001 0.5052 0.4964 0.502
10 0.498 0.500 0.499 0.002 0.0077 0.002 0.5052 0.4964 0.49911 0.505 0.507 0.506 0.002 0.0077 0.002 0.5052 0.4964 0.50612 0.502 0.499 0.501 0.003 0.0077 0.003 0.5052 0.4964 0.50113 0.495 0.497 0.496 0.002 0.0077 0.002 0.5052 0.4964 0.49614 0.499 0.496 0.498 0.003 0.0077 0.003 0.5052 0.4964 0.49815 0.503 0.507 0.505 0.004 0.0077 0.004 0.5052 0.4964 0.50516 0.507 0.509 0.508 0.002 0.0077 0.002 0.5052 0.4964 0.50817 0.503 0.501 0.502 0.002 0.0077 0.002 0.5052 0.4964 0.50218 0.497 0.493 0.495 0.004 0.0077 0.004 0.5052 0.4964 0.49519 0.504 0.508 0.506 0.004 0.0077 0.004 0.5052 0.4964 0.50620 0.505 0.503 0.504 0.002 0.0077 0.002 0.5052 0.4964 0.504
Avg = 0.501 0.002x R
Std. = 0.0047
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X-Bar Control Charts
X-Bar Chart
0.490
0.495
0.500
0.505
0.510
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
x
X-bar charts can identify special causes of variation, but they are only useful if the processis stable (common cause variation).
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Control Limits for Range
UCL = D4R = 3.268*.002 = .0065
LCL = D3 R = 0
R Chart
0.000
0.002
0.004
0.006
0.008
0.010
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Observation
Ran
ge
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Special Variables Control Charts
• x-bar and s charts
• x-chart for individuals
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X-bar and S charts
• Allows us to estimate the process standard deviation directly instead of indirectly through the use of the range R
• S chart limits:– UCL = B6σ = B4*S-bar– Center Line = c4σ = S-bar– LCL = B5σ = B3*S-bar
• X-bar chart limits– UCL = X-doublebar +A3S-bar– Center line = X-doublebar– LCL = X-doublebar -A3S-bar
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X-chart for individuals
• UCL = x-bar + 3*(MR-bar/d2)
• Center line = x-bar
• LCL = x-bar - 3*(MR-bar/d2)
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Next Class
• Homework– Ch. 11 (12) Disc. Questions 5, 6, 7– Ch. 11 (12) Problems 5
• Topic– Control Charts, Part II
• Preparation– Chapter 12 (13)