6. process capability analysis (variable data)

QUALITY TOOLS &
TECHNIQUES
By: -
Hakeem–Ur–Rehman
IQTM–PU 1
TQ T
PROCESS CAPABILITY ANALYSIS
(VARIABLE DATA)
USING MINITAB

PROCESS CAPABILITY
 The inherent ability of a process to meet the expectations of
the customer without any additional efforts. (or)
 The ability of a process to meet product design/technical
specifications
– Design specifications for products (Tolerances)
 upper and lower specification limits (USL, LSL)
– Process variability in production process
 natural variation in process (3 from the mean)
 Provides insight as to whether the process has a :
 Centering Issue (relative to specification limits)
 Variation Issue
 A combination of Centering and Variation
 Allows for a baseline metric for improvement.
2

3
Y1
Y2
Y3
Op i Op i + 1
Analysis ScrapOff-Line
Correction
Correctable
?
Verified
?
The X’s
(Inputs)
X1
X2
X3
X4
X5
Data for
Y1…Yn
10.16
10.11
10.05
10.33
10.44
9.86
10.07
10.29
10.36
10.16
10.11
10.05
10.33
10.44
9.86
10.07
10.29
10.36
9.87
9.99
10.12
10.43
10.21
10.01
10.15
10.44
10.03
10.33
10.15
9.87
9.99
10.12
10.43
10.21
10.01
10.15
10.44
10.03
10.33
10.15
10.16
10.11
10.05
10.33
10.44
9.86
10.07
10.29
10.36
The Y’s
(Outputs)
NoYes
Y = f(X) (Process Function)
Frequency
Variation – “Voice of
the Process”
10.410.310.210.110.09.90 10.59.80
Critical X(s):
Any variable(s)
which exerts an
undue influence on
the important
outputs (CTQ’s) of a
process
Y1
Y2
Y3
Op i Op i + 1
Correction
Correctable
?
Verified
?
The X’s
(Inputs)
X1
X2
X3
X4
X5
Data for
Y1…Yn
10.16
10.11
10.05
10.33
10.44
9.86
10.07
10.29
10.36
10.16
10.11
10.05
10.33
10.44
9.86
10.07
10.29
10.36
9.87
9.99
10.12
10.43
10.21
10.01
10.15
10.44
10.03
10.33
10.15
9.87
9.99
10.12
10.43
10.21
10.01
10.15
10.44
10.03
10.33
10.15
10.16
10.11
10.05
10.33
10.44
9.86
10.07
10.29
10.36
The Y’s
(Outputs)
NoYes
Y1
Y2
Y3
Op i Op i + 1
Correction
Correctable
?
Verified
?
The X’s
(Inputs)
X1
X2
X3
X4
X5
X1
X2
X3
X4
X5
Data for
Y1…Yn
10.16
10.11
10.05
10.33
10.44
9.86
10.07
10.29
10.36
10.16
10.11
10.05
10.33
10.44
9.86
10.07
10.29
10.36
9.87
9.99
10.12
10.43
10.21
10.01
10.15
10.44
10.03
10.33
10.15
9.87
9.99
10.12
10.43
10.21
10.01
10.15
10.44
10.03
10.33
10.15
10.16
10.11
10.05
10.33
10.44
9.86
10.07
10.29
10.36
The Y’s
(Outputs)
NoYes
Y = f(X) (Process Function)Y = f(X) (Process Function)
Frequency
Variation – “Voice of
the Process”
10.410.310.210.110.09.90 10.59.80 10.410.310.210.110.09.90 10.59.80
Critical X(s):
Any variable(s)
which exerts an
undue influence on
the important
process
Critical X(s):
Any variable(s)
which exerts an
undue influence on
the important
process
LSL = 9.96 USL = 10.44
DefectsDefects
10.410.310.210.110.09.90 10.5 10.69.809.70
-1-3-4-5-6 -2 +6+4+3+2+1 +5
Requirements – “Voice
of the Customer”
10.16
10.11
10.05
10.33
10.44
9.86
10.07
10.29
10.36
9.87
9.99
10.12
10.43
10.21
10.01
10.15
10.44
10.03
10.33
10.15
10.16
10.11
10.05
10.33
10.44
9.86
10.07
10.29
10.36
Data - VOP
Percent Composition
LSL = 9.96LSL = 9.96 USL = 10.44USL = 10.44
DefectsDefects
10.410.310.210.110.09.90 10.5 10.69.809.70 10.410.310.210.110.09.90 10.5 10.69.809.70
-1-3-4-5-6 -2 +6+4+3+2+1 +5
Requirements – “Voice
of the Customer”
10.16
10.11
10.05
10.33
10.44
9.86
10.07
10.29
10.36
9.87
9.99
10.12
10.43
10.21
10.01
10.15
10.44
10.03
10.33
10.15
10.16
10.11
10.05
10.33
10.44
9.86
10.07
10.29
10.36
Data - VOP
10.16
10.11
10.05
10.33
10.44
9.86
10.07
10.29
10.36
9.87
9.99
10.12
10.43
10.21
10.01
10.15
10.44
10.03
10.33
10.15
10.16
10.11
10.05
10.33
10.44
9.86
10.07
10.29
10.36
10.16
10.11
10.05
10.33
10.44
9.86
10.07
10.29
10.36
9.87
9.99
10.12
10.43
10.21
10.01
10.15
10.44
10.03
10.33
10.15
10.16
10.11
10.05
10.33
10.44
9.86
10.07
10.29
10.36
Data - VOP
Percent Composition
Capability Analysis Numerically
Compares the VOP to the VOC

Stability
A Stable Process is consistent over time. Time Series Plots and Control
Charts are the typical graphs used to determine stability.
At this point in the Measure Phase there is no reason to assume the
process is stable.
Index
PCData
48043238433628824019214496481
70
60
50
40
30
Time Series Plot of PC Data

TWO KINDS OF VARIABILITY
 Inherent variability:-
 Inherent in machine/process (design, construction and nature of
operation).
 Assignable variability.
 Variability where causes can be identified.
 Assignable variability eliminated / minimized by Process Capability
Study.
FOR A CAPABLE PROCESS:
INHERENT + ASSIGNABLE < TOLERANCE 5

Check for Normality.
By looking at the “P-values”
the data look to be Normal
since P is greater than .05

PROCESS
CAPABILITY
ANALYSIS FOR
VARIABLE DATA
7

Process Capability Index:
Cp -- Measure of Potential Capability
6variationprocess
variationprocess LSLUSL
actual
allowable
Cp


Cp = 1
Cp < 1
Cp > 1
LSL USL
8

Process Capability Index:
Cpk -- Measure of Actual Capability





 

 3
,
3
min
XUSLLSLX
Cpk
“σ” is the standard deviation of the production process
Cpk considers both process variation () and process
location (X)
9

 WHY Cpk IS NEEDED?
Cpk TELLS U ABOUT THE POSITIONING / LOCATION OF THE CURVE
 IS Cp NOT ENOUGH?
Cp TELLS U ONLY ABOUT THE SMARTNESS OF CURVE
10CHANGE IN MEAN OVER THE TIME

PROCESS CAPABILITY INDEX
EXAMPLE
A manufacturing process produces a certain part
with a mean diameter of 2 inches and a standard
deviation of 0.03 inches. The lower and upper
engineering specification limits are 1.90 inches and
2.05 inches.
56.0]56.0,11.1min[
)03.0(3
205.2
,
)03.0(3
90.12
min
3
,
3
min






 





 


XUSLLSLX
Cpk
83.0
)03.0(6
90.105.2
6






LSLUSL
Cp
Therefore, the process is not capable (the variation
is too large and the process mean is not on target)11

PROCESS CAPABILITY INDEX
EXERCISE
Consider the two processes shown here:
Process A Process B
µ = 105
σ = 3
µ = 110
σ = 1
Specifications area at 100 ± 4: Calculate Cp, Cpk and
interpret these ratios. Which Process would prefer to
use?
12

PROCESS CAPABILITY ANALYSIS:
EXAMPLE#1 (Minitab)
13
The length of a camshaft for an automobile engine is specified at 600 + 2 mm. To
avoid scrap / rework, the control of the length of the camshaft is critical.
The camshaft is provided by an external supplier. Access the process capability for
this supplier.
Filename: “camshaft.mtw”
Stat > quality tools > capability analysis (normal)

EXAMPLE#1 (Minitab) (Cont…) – LENGTH
14

EXAMPLE#1 (Minitab) (Cont…) – LENGTH
15

EXAMPLE#1 (Minitab) (Cont…)
16
The length of a camshaft for an automobile engine is specified at 600 + 2 mm. To
avoid scrap / rework, the control of the length of the camshaft is critical.
The camshaft is provided by an external supplier. Access the process capability for this
supplier.
Filename: “camshaft.mtw”
Stat > quality tools > capability analysis (normal)
Process Capability Indices & Sigma Quality Level

EXAMPLE#1 (Minitab) (Cont…)
17
Process Capability Indices & Sigma Quality Level
Sigma Quality Level = Z.Bench (Potential (Within) Capability) + 1.5
= 1.26 + 1.5
= 2.76

EXERCISE#1 (Minitab)–BOTH SUPPLIERS
18
Histogram of camshaft length suggests mixed
populations.
Further investigation revealed that there are two
suppliers for the camshaft. Data was collected over
camshafts from both sources.
Are the two suppliers similar in performance?
If not, What are your recommendations?
FILENAME: “camshafts.mtw”

19
 PROCESS CAPABILITY STUDY ASSUMPTIONS:
1. The performance measure data reflects statistical
control when plotted over a control chart (i.e.: X–Bar &
Range Chart)
2. The performance measure data distributed normally.
 NORMALLY TEST:
o Generate a normal probability plot and performs a
hypothesis test to examine whether or not the
observations follow a normal distribution. For the
normality test, the hypothesis are,
o Ho: Data follow a normal distribution Vs H1: Data do
not follow a normal distribution
o If ‘P’ value is > alpha; Accept Null Hypothesis (Ho)

20
NORMALITY TEST:
 In an operating engine, parts of the
crankshaft move up and down.
AtoBDist is the distance (in mm) from
the actual (A) position of a point on
the crankshaft to a baseline (B)
position. To ensure production quality,
a manager took five measurements
each working day in a car assembly
plant, from September 28 through
October 15, and then ten per day
from the 18th through the 25th.
 You wish to see if these data follow a
normal distribution,
 so you use Normality test.
 Open the worksheet CRANKSH.MTW

21
INTERPRETING THE RESULTS:
The graphical output is a plot of normal probabilities versus the data. The data
depart from the fitted line most evidently in the extremes, or distribution tails.
 The Anderson–Darling test’s ‘p–value’ indicates that, at a levels greater than
0.022, there is evidence that the data do not follow a normal distribution.
 There is a slight tendency for these data to be lighter in the tails than a normal
distribution because the smallest points are below the line and the largest point is
just above the line.
 A distribution with heavy tails would show the opposite pattern at the extremes.

PROCESS CAPABILITY FOR
NON–NORMAL DATA
22
TO ADDRESS NON–NORMAL DATA IS TO
NORMALIZE THE DATA USING A BOX–COX
TRANSFORMATION.
In Box – Cox Transformation, values (Y)
are transformed to the power of ‘λ’ i.e.:
Yλ
If λ = 0Y transformed into ln (Y)

NON–NORMAL DATA
23
For Example:
Open the File: Boxcox.mtw and prepare a Histogram
IS this
Normal
Data?

NON–NORMAL DATA
24
Box – Cox Transformation can be used to normalized the data.

NON–NORMAL DATA
25
Box-Cox Transformation can be used to normalized the data.

NON–NORMAL DATA
26
Histogram comparison of Non–normal data after normalization

NON–NORMAL DATA
27
EXAMPLE:
Suppose you work for a company that manufactures floor tiles and are
concerned about warping in the tiles. To ensure production quality, you
measure warping in ten tiles each working day for ten days.
A Histogram shows that your data do not follow a normal distribution, so you
decide to use the Box–Cox power transformation to try to make the data
“more normal”.
Open Worksheet: Tiles.mtw
First you need to find the
optimal lambda value for the
transformation. Then you will
do the capability analysis,
performing the Box – Cox
transformation with that
value.

NON–NORMAL DATA
28
 Choose Stat  Control Charts  Box-Cox Transformation
 Choose All observations for a chart are in one column, then enter
“Warping”
 In Subgroup Sizes, enter 10, Click ok
The best estimate of lambda is
0.43, but practically speaking,
you may want a lambda value
that corresponds to an intuitive
transformation, such as the
square root (a lambda of 0.5).
So you will run the Capability
Analysis with a Box–Cox
transformation, Using
lambda = 0.5.
EXAMPLE (Cont…)

NON–NORMAL DATA
29
EXAMPLE (Cont…)

NON–NORMAL DATA
30
EXAMPLE (Cont…)

NON–NORMAL DATA
31
Second approach to address non – normal data is to
identify exact type of distribution other than normal
distribution
 INDIVIDUAL IDENTIFICATION OF DISTRIBUTION
 Use to evaluate the optimal distribution for your data
based on the probability plots and goodness-of-fit
tests prior to conducting a capability analysis study.
Choose from 14 distributions.
 You can also use distribution identification to
transform your data to follow a normal distribution
using a Box–Cox transformation or a Johnson
transformation.

NON–NORMAL DATA
32
EXAMPLE:
Suppose you work for a company that manufactures floor tiles and are
concerned about warping in the tiles. To ensure production quality, you
measure warping in 10 tiles each working day for 10 days.
The distribution of the data is unknown. Individual Distribution Identification
allows you to fit these data with 14 parametric distributions and 2
transformations.
Open Worksheet: Tiles.mtw

NON–NORMAL DATA
33
Distribution ID Plot for Warping
Descriptive Statistics
N N* Mean StDev Median Minimum Maximum Skewness Kurtosis
100 0 2.92307 1.78597 2.60726 0.28186 8.09064 0.707725 0.135236
Box-Cox transformation: Lambda = 0.5
Johnson transformation function:
0.882908 + 0.987049 * Ln( ( X + 0.132606 ) / ( 9.31101 - X ) )
Goodness of Fit Test:
Distribution AD P LRT P
Normal 1.028 0.010
Box-Cox Transformation 0.301 0.574
Lognormal 1.477 <0.005
3-Parameter Lognormal 0.523 * 0.007
Exponential 5.982 <0.003
2-Parameter Exponential 3.892 <0.010 0.000
Weibull 0.248 >0.250
3-Parameter Weibull 0.359 0.467 0.225
Smallest Extreme Value 3.410 <0.010
Largest Extreme Value 0.504 0.213
Gamma 0.489 0.238
3-Parameter Gamma 0.547 * 0.763
Logistic 0.879 0.013
Loglogistic 1.239 <0.005
3-Parameter Loglogistic 0.692 * 0.085
Johnson Transformation 0.231 0.799
Best fit
distribution will
be having p–
value greater
than 0.05. But
The Best fit is
Johnson
Transformation.

NON–NORMAL DATA
34

NON–NORMAL DATA
35

6. process capability analysis (variable data)

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