Transcript Document

TM 620: Quality Management
Session Seven – 9 November 2010
• Control Charts, Part I
– Variables
Recall: What is Quality?
• Juran – Quality is fitness for use
=> we should be able to determine a set of
measurable characteristics which define
quality
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
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
Recall: Cost of Quality
Traditional View
Costs
Total Cost
Cost of
Control
Cost of Failure
Quality Level
Traditional Loss Function
LSL
USL
x
T
LSL
T
USL
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)
Loss per unit (San Diego)
= $0.44
= $1.33
How can this be?
Sullivan, “Reducing Variability: A New Approach to Quality,” Quality
Progress, 17, no.7, 15-21, 1984.
Example
LSL
USL
x
T
U.S. Plant (2 = 8.33)
Japanese Plant (2 = 2.78)
Taguchi Loss Function
x
T
x
T
Taguchi Loss Function
L(x)
k(x - T)2
T
L(x) = k(x - T)2
x
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.
Estimating Loss Function
L(x)
400
10
400 = k(10.05 - 10.00)2
= k(.0025)
10.05
Estimating Loss Function
L(x)
400
10
400 = k(10.05 - 10.00)2
= k(.0025)
k = 160,000
10.05
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.
Example 2
L(x)
f(x)
25
1
1.5
25 = k(1 - 1.5)2
k = 100
Example 2
L(x)
f(x)
25
1
1.5
25 = k(1 - 1.5)2
k = 100
Single Sided Loss Functions
Smaller is better
L(x) = kx2
Larger is better
L(x) = k(1/x2)
Example 2
L(x)
f(x)
25
1
Example 2
L(x)
f(x)
25
1
25 = k(1)2
k = 25
Expected Loss
Expected Loss: Piston Diameter
Diameter
9.925
9.950
9.975
10.000
10.025
10.050
10.075
Probability
Process A
0.000
0.200
0.200
0.200
0.200
0.200
0.000
Probability
Process B
0.025
0.075
0.200
0.400
0.200
0.075
0.025
Expected Loss
Expected Loss: Piston Diameter
Diameter
9.925
9.950
9.975
10.000
10.025
10.050
10.075
Loss
900
400
100
0
100
400
900
Probability
Process A
0.000
0.200
0.200
0.200
0.200
0.200
0.000
Probability
Process B
0.025
0.075
0.200
0.400
0.200
0.075
0.025
Expected Loss
Expected Loss: Piston Diameter
Diameter
9.925
9.950
9.975
10.000
10.025
10.050
10.075
Loss
900
400
100
0
100
400
900
Probability Weighted Probability
Process A
Loss
Process B
0.000
0.0
0.025
0.200
80.0
0.075
0.200
20.0
0.200
0.200
0.0
0.400
0.200
20.0
0.200
0.200
80.0
0.075
0.000
0.0
0.025
Expected Loss
Expected Loss: Piston Diameter
Diameter
9.925
9.950
9.975
10.000
10.025
10.050
10.075
Loss
900
400
100
0
100
400
900
Probability Weighted Probability Weighted
Process A
Loss
Process B
Loss
0.000
0.0
0.025
22.5
0.200
80.0
0.075
30.0
0.200
20.0
0.200
20.0
0.200
0.0
0.400
0.0
0.200
20.0
0.200
20.0
0.200
80.0
0.075
30.0
0.000
0.0
0.025
22.5
Expected Loss
Expected Loss: Piston Diameter
Diameter
9.925
9.950
9.975
10.000
10.025
10.050
10.075
Loss
900
400
100
0
100
400
900
Probability Weighted Probability Weighted
Process A
Loss
Process B
Loss
0.000
0.0
0.025
22.5
0.200
80.0
0.075
30.0
0.200
20.0
0.200
20.0
0.200
0.0
0.400
0.0
0.200
20.0
0.200
20.0
0.200
80.0
0.075
30.0
0.000
0.0
0.025
22.5
Exp. Loss =
200.0
145.0
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 ] }
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 ] }
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 + 22 - 2T + T2 }
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 + 22 - 2T + T2 }
=k {2 + ( - T)2 }
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 + 22 - 2T + T2 }
=k {2 + ( - T)2 }
= k { 2 + ( x - T)2 } = k (2 +D2 )
Example
Since for our piston example, x = T,
D2 = (x - T)2 = 0
L(x) = k2
Example (Piston Diam.)
Expected Loss: Piston Diameter
Diameter
9.925
9.950
9.975
10.000
10.025
10.050
10.075
(x-)
0.0056
0.0025
0.0006
0.0000
0.0006
0.0025
0.0056
2
Probability Weighted Probability Weighted
2
2
Process A
(x-)
Process B
(x-)
0.000
0.0000
0.025
0.0001
0.200
0.0005
0.075
0.0002
0.200
0.0001
0.200
0.0001
0.200
0.0000
0.400
0.0000
0.200
0.0001
0.200
0.0001
0.200
0.0005
0.075
0.0002
0.000
0.0000
0.025
0.0001
Var =
0.0013
0.0009
E[LA(x)] = .0013*k E[LB(x)] = .0009*k
=
200
=
145
Example (Sony)
LSL
USL
x
T
U.S. Plant (2 = 8.33)
Japanese Plant (2 = 2.78)
E[LUS(x)] = 0.16 * 8.33 = $1.33
E[LJ(x)] = 0.16 * 2.78 = $0.44
Tolerance (Pistons)
Recall,
L(x)
400
10
400 = k(10.05 - 10.00)2
= k(.0025)
k = 160,000
10.05
Tolerance
L(x)
Suppose repair for
an engine which will
400
fail during warranty
200
can be made for only
$200
LSL
10.05
10 USL
Tolerance
L(x)
Suppose repair for an
engine which will fail
during warranty can be
made for only $200
200
=160,000(tolerance)2
400
200
LSL
10.05
10 USL
Tolerance
L(x)
Suppose repair for an
engine which will fail
during warranty can be
made for only $200
400
200
200 = 160,000(tolerance)2
tolerance
= (200/160,000)1/2
= .0354
LSL
10.05
10 USL
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
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
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
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
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”
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
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
Terms
•
•
•
•
•
•
•
Specifications
Target (or Nominal) Value
Upper Specification Limit
Lower Specification Limit
Random Variation
Non-random Variation
Process stability
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
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
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
Types of Inspection
• Receiving
• In Process
• Final
• None
• One Hundred Percent
• Acceptance Sampling
Quality Design & Process Variation
Lower Spec
Limit
Upper Spec
Limit
Acceptance
Sampling
60
80
100
120
140
Statistical
Process
Control
60
140
Experimental
Design
60
140
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.
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
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
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)
Statistical Process Control
Tools
• The magnificent seven
• The tool most often associated with
Statistical Process Control is Control
Charts
Common
Causes
Special
Causes
Histograms do
not take into
account
changes over
time.
Control charts
can tell us when a
process changes
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.
Capability Versus Control
Control
Capability
Capable
Not Capable
In Control
IDEAL
Out of Control
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
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
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
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
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
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
Shift in Process Average
Identifying Potential Shifts
Cycles
Trend
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
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
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.
Final Steps
5. Use as a problem-solving tool
– Continue to collect and plot data
– Take corrective action when
necessary
6. Compute process capability
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
Process Capability
Calculations
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 Ratio
PCR = CP = (USL – LSL) / 6σ
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
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
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
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)
Control Charts
  




We assume that the underlying distribution is normal
with some mean  and some constant but unknown
standard deviation .
Let
n
xi
x
i 1 n
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
x  N (  , x )
where,
x 
x
n
Control Charts
x
  




x
x

x
Control Charts
x
UCL

x
LCL
UCL & LCL Set at 3 x
Problem: How do we estimate  &  ?
Control Charts
x
m
x
i 1
i
m

m
R
R
i 1
m
 f ( )  
Control Charts
x
m
x
i 1
i
m

m
R
R
i 1
m
 f ( )  
UCLx  x + A2 R
UCLR  D4 R
LCLx  x  A2 R
LCLR  D3 R
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.
Part
No.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Measurements
1
2
0.502 0.504
0.495 0.497
0.492 0.496
0.501 0.498
0.507 0.508
0.504 0.504
0.497 0.496
0.493 0.496
0.502 0.501
0.498 0.500
0.505 0.507
0.502 0.499
0.495 0.497
0.499 0.496
0.503 0.507
0.507 0.509
0.503 0.501
0.497 0.493
0.504 0.508
0.505 0.503
Avg =
Std. =
x
0.503
0.496
0.494
0.500
0.508
0.504
0.497
0.495
0.502
0.499
0.506
0.501
0.496
0.498
0.505
0.508
0.502
0.495
0.506
0.504
0.501
x
0.0047
R
0.002
0.002
0.004
0.003
0.001
0.000
0.001
0.003
0.001
0.002
0.002
0.003
0.002
0.003
0.004
0.002
0.002
0.004
0.004
0.002
0.002
R
R Chart
UCL
R
0.0077 0.002
0.0077 0.002
0.0077 0.004
0.0077 0.003
0.0077 0.001
0.0077 0.000
0.0077 0.001
0.0077 0.003
0.0077 0.001
0.0077 0.002
0.0077 0.002
0.0077 0.003
0.0077 0.002
0.0077 0.003
0.0077 0.004
0.0077 0.002
0.0077 0.002
0.0077 0.004
0.0077 0.004
0.0077 0.002
X Chart
UCL
0.5052
0.5052
0.5052
0.5052
0.5052
0.5052
0.5052
0.5052
0.5052
0.5052
0.5052
0.5052
0.5052
0.5052
0.5052
0.5052
0.5052
0.5052
0.5052
0.5052
LCL
0.4964
0.4964
0.4964
0.4964
0.4964
0.4964
0.4964
0.4964
0.4964
0.4964
0.4964
0.4964
0.4964
0.4964
0.4964
0.4964
0.4964
0.4964
0.4964
0.4964
Xbar
0.503
0.496
0.494
0.500
0.508
0.504
0.497
0.495
0.502
0.499
0.506
0.501
0.496
0.498
0.505
0.508
0.502
0.495
0.506
0.504
X-Bar Control Charts
X-Bar Chart
0.510
x
0.505
0.500
0.495
0.490
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19 20
X-bar charts can identify special causes of
variation, but they are only useful if the process
is stable (common cause variation).
Control Limits for Range
UCL = D4R = 3.268*.002 = .0065
LCL = D3 R = 0
R Chart
0.010
Range
0.008
0.006
0.004
0.002
0.000
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19 20
Observation
Special Variables Control
Charts
• x-bar and s charts
• x-chart for individuals
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
X-chart for individuals
• UCL = x-bar + 3*(MR-bar/d2)
• Center line = x-bar
• LCL = x-bar - 3*(MR-bar/d2)
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)