Download Production and Operations Management: Manufacturing and Services

PROCESS CAPABILITY AND
SPC
Chapter 9A
9A-2
OBJECTIVES
1.
2.
3.
Explain what statistical quality control is.
Calculate the capability of a process.
Understand how processes are monitored with
control charts.
Types of Situations where SPC can be
Applied




How many paint defects are there in the finish of a
car?
How long does it take to execute market orders?
How well are we able to maintain the dimensional
tolerance on our ball bearing assembly?
How long do customers wait to be served from our
drive-through window?
LO 1
9A-4
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
Example: A poorly trained
employee that creates
variation in finished
product output.
Example: A molding
process that always leaves
“burrs” or flaws on a
molded item.
Testing examples
Taguchi’s View of Variation


LO 1
Traditional view is that quality within the range is
good and that the cost of quality outside this range
is constant
Taguchi views costs as increasing as variability
increases, so seek to achieve zero defects and that
will truly minimize quality costs
Process Capability


Taguchi argues that tolerance is not a yes/no
decision, but a continuous function
Other experts argue that the process should be so
good the probability of generating a defect should
be very low
LO 2
9A-8
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 Normal Behavior
Control
(SPC) Charts
9A-9
UCL
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
9A-10
Control Limits are based on the Normal Curve
x
m
-3
-2
-1
Standard
deviation
units or “z”
units.
0
1
2
3
z
9A-11
Control Limits
We establish the Upper Control Limits (UCL)
and the Lower Control Limits (LCL) with plus
or minus 3 standard deviations from some xbar or mean value. Based on this we can
expect 99.7% of our sample observations to
fall within these limits.
99.7%
LCL
UCL
x
Process Control with Attribute
Measurement: Using ρ Charts


Created for good/bad attributes
Use simple statistics to create the control
limits
Total number of defects from all samples
p
Number of samples  Sample size
sp 

p 1 p
n
UCL  p  zs p
LCL  p  zs p
LO 3

9A-13
Example of Constructing a p-Chart:
Required Data
Sample
No. of
No.
Samples
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
Number of
defects found
in each sample
4
2
5
3
6
4
3
7
1
2
3
2
2
8
3
9A-14
Example of Constructing a p-chart: Step 1
1. Calculate the
sample proportions,
p (these are what
can be plotted on the
p-chart) 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
9A-15
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
9A-16
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)
9A-17
Example of Constructing a p-Chart: Step 5
5. Plot the individual sample proportions, the average
of the proportions, and the control limits
How to Construct x and R Charts
X Chart
UCL X  X  A2 R
LCL X  X  A2 R
R Chart
UCL R  D4 R
LCL R  D3 R
LO 3
9A-19
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
9A-20
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
9A-21
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
9A-22
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
UCL = D 4 R  ( 2.11)(0.2204)  0.46504
LCL = D3 R  (0)(0.2204)  0
Then plot both graphs: Means to
the Mean chart and Ranges to the
Range chart.
9A-23
ANY QUESTIONS?