Download PRODUCTIONS/OPERATIONS MANAGEMENT

Chapter 10
Quality Control
1
Phases of Quality Assurance
Inspection
before/after
production
Acceptance
sampling
The least
progressive
Corrective
action during
production
Process
control
Quality built
into the
process
Continuous
improvement
The most
progressive
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Inspection: Appraisal of good/service quality
Cost
• How Much (sample size) /How Often (hourly, daily)
Total Cost
Cost of
inspection
(appraisal and
Prevention cost)
Optimal
Amount of Inspection
Cost of
passing
defectives
(failure cost)
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Inspection
• Where/When
• Raw materials
• Finished products
Inputs
Acceptance
sampling
Transformation
Process
control
Outputs
Acceptance
sampling
• Before a costly operation, PhD comp. exam before candidacy
• Before an irreversible process, firing pottery
• Before a covering process, painting, assembly
• Centralized vs. On-Site, my friend checks quality at cruise lines
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Examples of Inspection Points
Type of
business
Fast Food
Inspection
points
Cashier
Counter area
Eating area
Building
Kitchen
Hotel/motel Parking lot
Accounting
Building
Main desk
Supermarket Cashiers
Deliveries
Characteristics
Accuracy
Appearance, productivity
Cleanliness
Appearance
Health regulations
Safe, well lighted
Accuracy, timeliness
Appearance, safety
Waiting times
Accuracy, courtesy
Quality, quantity
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Statistical Process Control (SPC)
• SPC: Statistical evaluation
of the output of a process during production
• The Control Process
–
–
–
–
–
–
Define
Measure
Compare to a standard
Evaluate
Take corrective action
Evaluate corrective action
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Statistical Process Control
• Shewhart’s classification of variability:
common cause vs. assignable cause
• Variations and Control
– Random variation: Natural variations in the
output of process, created by countless minor
factors, e.g. temperature, humidity variations.
– Assignable variation: A variation whose source
can be identified. This source is generally a
major factor, e.g. tool failure.
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Mean and Variance
• Given a population of numbers, how to
compute the mean and the variance?
Population  {x1 , x2 ,..., x N }
N
Mean   
x
i 1
i
N
N
Variance   2 
2
(
x


)
 i
i 1
N
Standard deviation  
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Statistical Process Control
• From a large population of goods or
services (random if possible) a sample is
drawn.
– Example sample: Midterm grades of BA3352
students whose last name starts with letter R
{60, 64, 72, 86}, with letter S {54, 60}
•
•
•
•
Sample size= n
Sample average or sample mean= x
Sample range= R
Standard deviation of sample means=
x 

n
where  : Standard deviation of the population
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Sampling Distribution
Sampling distribution is the distribution of sample means.
Sampling distribution
Variability of the average scores of
people with last name R and S
Process distribution
Variability of the scores
for the entire class
Mean
Grouping reduces the variability.
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Normal Distribution
normdist(x,.,.,1)
normdist(x,.,.,0)
Probab


Mean
x


95.44%
99.74%
Excel statistica l functions : normdist ( x, mean, st _ dev,0) normal pdf at x.
Excel statistica l functions : normdist ( x, mean, st _ dev,1) normal cdf at x.
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Cumulative Normal Density
1
prob
normdist(x,mean,st_dev,1)
0
x
norminv(prob,mean,st_dev)
Excel statistica l functions :
Cumulative function (cdf) at x : normdist ( x, mean, st _ dev,1)
Inverse function of cdf at " prob": norminv ( prob, mean, st _ dev)
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Normal Probabilities: Example
• If temperature inside a firing oven has a
normal distribution with mean 200 oC and
standard deviation of 40 oC, what is the
probability that
– The temperature is lower than 220 oC
=normdist(220,200,40,1)
– The temperature is between 190 oC and 220oC
=normdist(220,200,40,1)-normdist(190,200,40,1)
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Control Limits
Process is in control if sample mean is between control limits.
These limits have nothing to do with product specifications!
Sampling
distribution
Process
distribution
Mean
LCL
Lower
control
limit
UCL
Upper
control
limit
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Setting Control Limits:
Hypothesis Testing Framework
• Null hypothesis: Process is in control
• Alternative hypothesis: Process is out of control
• Alpha=P(Type I error)=P(reject the null when it is true)=
P(out of control when in control)
• Beta=P(Type II error)=P(accept the null when it is false)
P(in control when out of control)
• If LCL decreases and UCL increases what happens to
– Alpha ?
– Beta?
• Not possible to target alpha and beta simultaneously,
control charts target a desired level of Alpha.
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Type I Error=Alpha
/2
/2
Mean
Probability
of Type I error
LCL
UCL
LCL  norminv( /2, mean, st_dev)
UCL  norminv(1 - /2, mean, st_dev)
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Control Chart
Abnormal variation
due to assignable sources
Out of
control
UCL
Mean
Normal variation
due to chance
LCL
Abnormal variation
due to assignable sources
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15
Sample number
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Observations from Sample Distribution
UCL
LCL
1
2
3
4
Sample number
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Control Charts
• Control charts for variables (measurable
quantities), e.g. length, temperature
– Mean control charts
• To check mean
– Range control charts
• To check variability
• Control charts for attributes, e.g. fit, defective
– p-charts
• To check proportion of defectives (occurrences)
– c-charts
• To check the number of defectives (occurrences)
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Mean control chart
Grand mean x  average of x
UCL  x  z x  grand mean plus a multiple of standard deviation
LCL  x  z x  grand mean minus a multiple of standard deviation
UCL  x norminv(1 - /2, x,  x )  x
z

Most often z is set to 2 or 3.
x
x
If the standard deviation of the sample means is not known,
use the average of sample ranges to get the limits:
R  average of sample ranges R
UCL  x  A2 R  grand mean plus a multiple of the average of sample ranges
LCL  x  A2 R  grand mean minus a multiple of the average of sample ranges
Multiplier A_2 depends on n and is available in Table 10-2.
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Range Control Chart
UCL  D4 R  A multiple of the average of sample ranges
LCL  D3 R  A multiple of the average of sample ranges
Multipliers D_4 and D_3 depend on n and are available in Table 10-2.
EX: In the last five years, the range of GMAT scores of incoming PhD class is
88, 64, 102, 70, 74. If each class has 6 students, what are UCL and LCL for
GMAT ranges?
R  (88  64  102  70  74) / 5  79.6. For n  6, D 4  2, D3  0.
UCL  D4 R  2 * 79.6  159.2 LCL  D3 R  0 * 79.6  0
Are the GMAT ranges in control?
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Mean and Range Charts: Which?
(process mean is
shifting upward)
Sampling
Distribution
UCL
Detects shift
x-Chart
LCL
UCL
R-chart
Does not
detect shift
LCL
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Mean and Range Charts: Which?
Sampling
Distribution
(process variability is increasing)
UCL
Does not
reveal increase
x-Chart
LC
L
UCL
R-chart
Reveals increase
LC
L
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Use of p-Charts
• p=proportion defective, assumed to be known
• When observations can be placed into two categories.
– Good or bad
– Pass or fail
– Operate or don’t operate
– Go or no-go gauge
UCL  p  z p
where  p 
LCL  p  z p
p(1  p)
, z as before
n
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Use of c-Charts
• c=number of occurrences per unit
• Use only when the number of occurrences per
unit can be counted.
•
•
•
•
•
Scratches, chips, dents, or errors per item
Cracks or faults per unit of distance
Breaks or Tears per unit of area
Bacteria or pollutants per unit of volume
Calls, complaints, failures per unit of time
UCL  c  z c
LCL  c  z c
if c is not known, use the average c
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C-chart Example
• While the nuclear submarine Kursk was being raised in the
Barents sea (between Svalbard, No and Novaya Zemlya, Ru),
which took 15 hours, engineers took a reading of number of
Geiger counts per hour to detect any increase in radiation
levels. Should they have stopped before 5th or 10th hour given
3-sigma control and the readings data: 42, 48, 50, 45, 52, 66,
64, 84, 92, 76.
At the 5th hour, average number of counts=47.4, stdev of counts=6.88,
UCL=47.4+3*6.88=68.05, LCL=47.4-3*6.88=26.75. Do not stop.
At the 10th hour, average number of counts=61.9, stdev of counts=7.87,
UCL=61.9+3*7.87=85.51, LCL=61.9-3*7.87=38.29. Stop, 9th reading is
out of control.
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Up and Down Run Charts
• If all readings are in control, is the process
really in control?
• There could be trends in readings even
when they are in control.
Counting Up/Down Runs
U
U
D
U
(r=8 runs)
D
U
D U
U D
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Up and Down Run Charts
UCL  E (r )  z r  Expected runs plus a multiple of stdev of runs
LCL  E (r )  z r  Expected runs minus a multiple of stdev of runs
2K - 1
16 K  29
and  r 
3
90
K  Number of samples
E(r) 
EX: What are 3-sigma UCL and LCL for the number of runs in 50 samples?
2K - 1
16 K  29
K  50, E(r) 
 33 and  r 
 2.92
3
90
UCL  E (r )  z r  33  3 * 2.92
LCL  E (r )  z r  33 - 3 * 2.92
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Process Capability
• Tolerances/Specifications
– Requirements of the design or customers
• Process variability
– Natural variability in a process
– Variance of the measurements coming from the process
• Process capability
– Process variability relative to specification
– Capability=Process specifications / Process variability
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Process Capability:
Specification limits are not control chart limits
Lower
Specification
Upper
Specification
Process variability matches
specifications
Lower
Specification
Sampling
Distribution
is used
Upper
Specification
Process variability well within
Lower
Upper
specifications
Specification Specification
Process variability exceeds
specifications
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Process Capability Ratio
When the process is centered, process capability ratio
Cp =
Upper specification – lower specification
6
A capable process has large Cp.
Example: The standard deviation, of sample averages of the
midterm 1scores obtained by students whose last names start
with R, has been 7. The SOM management requires the
scores not to differ by more than 50% in an exam. That is the
highest score can be at most 50 points above the lowest
score. Suppose that the scores are centered, what is the
process capability ratio?
Answer: 50/42
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Process Capability Ratio
When the process is not centered, process capability ratio
Cpk= Min{Process mean - lower spec , Upper spec - Process mean}
3
When the process is not centered, the closest spec to mean determines
the capability of the process because that spec is likely to be
more of a limiting factor than the other.
Example: Suppose that the process is not centered in the previous example
and the SOM wants all the scores to fall within 50% and 100%. What is the
Capability ratio if the average score was 70?
Answer: From the lower limit, we have (70-50)/21
From the upper limit, we have (100-70)/21
Then the ratio is 20/21
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3 Sigma and 6 Sigma Quality
Upper
specification
Lower
specification
Process
mean
+/- 3 Sigma
+/- 6 Sigma
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Chapter 10 Supplement
Acceptance
Sampling
34
Acceptance Sampling
• Acceptance sampling: Is a lot of N products good
if a random sample of n (n<N) products contain
only c defects?
– For example take a sample of 10(=n) milk bottles out
of every 100(=N). If 1(=c) or more bottles do not fit
specifications, reject the entire lot of 100 bottles.
• c is determined to balance type I and type II
errors.
• This is a smart compromise between 100%
inspection and no inspection.
• Generally used for input/output inspection.
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Why not to emphasize
Acceptance Sampling (AS)
• AS plans have no clearly stated economic objective.
They target some levels of type I and II errors.
• AS incorporate an attitude of punishment by
rejecting entire lots after examining small samples.
This feeds the mistrust between supplier and the
customer.
• AS does not attempt to find the root cause of
defectives. It merely detects defectives. Real
problem is actually finding the root cause. Some
people say that:
– “AS provides elegant solutions to balance type I and II
errors by making a type III error: solving the wrong
problem”.
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