J0444
OPERATION MANAGEMENT
SPC
Universitas Bina Nusantara
Process Capability and Statistical Quality
Control



Process Variation
Process Capability
Process Control Procedures
–
–

Variable data
Attribute data
Acceptance Sampling
–
Operating Characteristic Curve
Basic Forms of Variation


Assignable variation is caused by
factors that can be clearly identified
and possibly managed.
Common variation is inherent in the
production process.
Taguchi’s View of Variation
High
High
Incremental
Cost of
Variability
Incremental
Cost of
Variability
Zero
Zero
Lower Target
Spec
Spec
Upper
Spec
Traditional View
Lower
Spec
Target
Spec
Upper
Spec
Taguchi’s View
Process Capability

Process limits

Tolerance limits

How do the limits relate to one
another?
Process Capability Index, Cpk
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.
C pk
 X  LTL
UTL - X 


= min 
or

3

3



Shifts in Process Mean
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
Statistical Process Control
(SPC) Charts
UCL
Normal Behavior
LCL
1
2
3
4
5
6
Samples
over time
UCL
Possible problem, investigate
LCL
1
2
3
4
5
6
Samples
over time
UCL
Possible problem, investigate
LCL
1
2
3
4
5
6
Samples
over time
Control Limits are based on the Normal
Curve
x
m
-3
-2
-1
Standard
deviation
units or “z”
units.
0
1
2
3
z
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.
99.7%
x
LCL
UCL
Example of Constructing a p-Chart:
Required Data
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
100
100
100
100
100
100
100
100
100
100
100
100
100
100
100
4
2
5
3
6
4
3
7
1
2
3
2
2
8
3
Statistical Process Control Formulas:
Attribute Measurements (p-Chart)
Given:
T o ta l N u m b e r o f D e fe c tiv e s
p =
T o ta l N u m b e r o f O b s e rv a tio n s
sp =
p (1 - p)
n
Compute control limits:
UCL = p + z sp
LCL = p - z sp
Example of Constructing a p-chart: Step 1
1. Calculate the
sample proportions, p
(these are what can
be plotted on the pchart) for each
sample.
Sample
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
n Defectives
100
4
100
2
100
5
100
3
100
6
100
4
100
3
100
7
100
1
100
2
100
3
100
2
100
2
100
8
100
3
p
0.04
0.02
0.05
0.03
0.06
0.04
0.03
0.07
0.01
0.02
0.03
0.02
0.02
0.08
0.03
Example of Constructing a p-chart:
Steps 2&3
2. Calculate the average of the sample proportions.
55
p =
1500
= 0.036
3. Calculate the standard deviation of the sample
proportion
sp =
p (1 - p)
=
n
.036(1- .036)
= .0188
100
Example of Constructing a p-chart: Step 4
4. Calculate the control limits.
UCL = p + z sp
LCL = p - z sp
.036  3(.0188)
UCL = 0.0924
LCL = -0.0204 (or 0)
Example of Constructing a p-Chart: Step 5
5. Plot the individual sample proportions, the average
of the proportions, and the control limits
0.16
0.14
0.12
UCL
0.1
p 0.08
0.06
0.04
0.02
LCL
0
1
2
3
4
5
6
7
8
9
O b servation
10
11
12
13
14
15
Example of x-Bar and R Charts:
Required Data
Sample
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Obs 1
10.68
10.79
10.78
10.59
10.69
10.75
10.79
10.74
10.77
10.72
10.79
10.62
10.66
10.81
10.66
Obs 2
10.689
10.86
10.667
10.727
10.708
10.714
10.713
10.779
10.773
10.671
10.821
10.802
10.822
10.749
10.681
Obs 3
10.776
10.601
10.838
10.812
10.79
10.738
10.689
10.11
10.641
10.708
10.764
10.818
10.893
10.859
10.644
Obs 4
10.798
10.746
10.785
10.775
10.758
10.719
10.877
10.737
10.644
10.85
10.658
10.872
10.544
10.801
10.747
Obs 5
10.714
10.779
10.723
10.73
10.671
10.606
10.603
10.75
10.725
10.712
10.708
10.727
10.75
10.701
10.728
Example of x-bar and R charts: Step 1. Calculate
sample means, sample ranges, mean of means, and
mean of ranges.
Sample
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Obs 1
10.68
10.79
10.78
10.59
10.69
10.75
10.79
10.74
10.77
10.72
10.79
10.62
10.66
10.81
10.66
Obs 2
10.689
10.86
10.667
10.727
10.708
10.714
10.713
10.779
10.773
10.671
10.821
10.802
10.822
10.749
10.681
Obs 3
10.776
10.601
10.838
10.812
10.79
10.738
10.689
10.11
10.641
10.708
10.764
10.818
10.893
10.859
10.644
Obs 4
10.798
10.746
10.785
10.775
10.758
10.719
10.877
10.737
10.644
10.85
10.658
10.872
10.544
10.801
10.747
Obs 5
10.714
10.779
10.723
10.73
10.671
10.606
10.603
10.75
10.725
10.712
10.708
10.727
10.75
10.701
10.728
Averages
Avg
10.732
10.755
10.759
10.727
10.724
10.705
10.735
10.624
10.710
10.732
10.748
10.768
10.733
10.783
10.692
Range
0.116
0.259
0.171
0.221
0.119
0.143
0.274
0.669
0.132
0.179
0.163
0.250
0.349
0.158
0.103
10.728 0.220400
Example of x-bar and R charts: Step 2.
Determine Control Limit Formulas and
Necessary Tabled Values
x Chart Control Limits
UCL = x + A 2 R
LCL = x - A 2 R
R Chart Control Limits
UCL = D 4 R
LCL = D 3 R
n
2
3
4
5
6
7
8
9
10
11
A2
1.88
1.02
0.73
0.58
0.48
0.42
0.37
0.34
0.31
0.29
D3
0
0
0
0
0
0.08
0.14
0.18
0.22
0.26
D4
3.27
2.57
2.28
2.11
2.00
1.92
1.86
1.82
1.78
1.74
Example of x-bar and R charts: Steps 3&4. Calculate x-bar Chart
and Plot Values
UCL = x + A 2 R  10.728 - .58(0.2204 ) = 10.856
LCL = x - A 2 R  10.728 - .58(0.2204 ) = 10.601
10.900
UCL
10.850
M ean s
10.800
10.750
10.700
10.650
10.600
LCL
10.550
1
2
3
4
5
6
7
8
Sam ple
9
10
11
12
13
14
15
Example of x-bar and R charts: Steps 5&6.
Calculate R-chart and Plot Values
UCL = D 4 R  ( 2.11)(0.2204)  0.46504
LCL = D3 R  (0)(0.2204)  0
0 .8 0 0
0 .7 0 0
0 .6 0 0
0 .5 0 0
R
UCL
0 .4 0 0
0 .3 0 0
0 .2 0 0
0 .1 0 0
LCL
0 .0 0 0
1
2
3
4
5
6
7
8
S a m p le
9
10
11
12
13
14
15
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)
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)
Acceptance Sampling

Disadvantages
–
–
–
Risks of accepting “bad” lots and
rejecting “good” lots
Added planning and documentation
Sample provides less information than
100-percent inspection
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.
Risk




Acceptable Quality Level (AQL)
– Max. acceptable percentage of defectives
defined by producer.
a (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.
Probability of acceptance
Operating Characteristic Curve
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
a = .05 (producer’s risk)
n = 99
c=4
 =.10
(consumer’s risk)
1
2
AQL
3
4
5
6
7
8
LTPD
Percent defective
9
10 11 12
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.
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.
What you need to determine
your sampling plan is “c” and
“n.”
Example: Step 2. Determine “c”
First divide LTPD by AQL.
LTPD
.03
=
= 3
AQL
.01
Then find the value for “c” by selecting the value in the TN7.10
“n(AQL)”column that is equal to or just greater than the ratio above.
Exhibit TN 7.10
c
0
1
2
3
4
LTPD/AQL
44.890
10.946
6.509
4.890
4.057
So, c = 6.
n AQL
0.052
0.355
0.818
1.366
1.970
c
5
6
7
8
9
LTPD/AQL
3.549
3.206
2.957
2.768
2.618
n AQL
2.613
3.286
3.981
4.695
5.426
Example: Step 3.
Determine Sample Size
Now given the information below, compute the sample size
in units to generate your sampling plan.
c = 6, from Table
n (AQL) = 3.286, from Table
AQL = .01, given in problem
n(AQL/AQL) = 3.286/.01 = 328.6, or 329 (always round up)
Sampling Plan:
Take a random sample of 329 units from a lot.
Reject the lot if more than 6 units are defective.