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CVJ Systems
AWD Systems
Trans Axle Solutions
eDrive Systems
Capability on Aggregate Processes
2
The Problem
With one machine and a couple of
fixtures, it’s a pretty easy problem. You
do a study on each fixture and show
both are capable. That’s going to be
somewhere between 60 and 200 parts
measured, depending on requirements. Horizontal
Mach
Fixture 1
Fixture 2
But as the copies of the
process grow, so does the
study. If we increase to 6
machines with 2 fixtures, that’s
between 360 and 1,200
measurements IF we agree that
fixtures aren’t switching
around machines. Add in more
operations, and more
opportunities for mixing, and
this problem grows
exponentially.
If we study the AGGREGATE of this output
(assuming there is representation from each
subprocess …) is this good enough?
3
The Problem – Formally Stated
There are two components to a capability study: centeredness (mean location) and
spread (variance/standard deviation). We need to examine the effects of both.
1) Case Study: If I have 3 parallel processes that are capable (high Cp), but they are
not centered (low Cpk), and I randomly draw parts from the AGGREGATE of these
processes, what does the resulting study look like?
• We will create some normal distributions with good Cp values (at least 2.5), center up 1 (Cpk about
2.5) and put the other two at the upper and lower limits. (Cpk close to 1).
• Distributions will all be “in spec” because. If they were all over the place, it’s clear the aggregate
would be bad, we don’t want to test extremes, we want to test borderline cases.
• The question is – given this, will the aggregate show a problem?
2) Case Study: If I have 3 parallel processes that are centered (Cpk ~ Cp) and two
have very capable distributions (low variance, good Cp) with the third having a
large variance, and I randomly draw parts from the AGGREGATE of these
processes, what does the resulting study look like?
• We will generate 3 centered processes (Cpk = Cp). Two distributions will be tight (Cpk = Cp near
2.5) and one of them noisy (Cpk = Cp near 1.33).
• Again, we don’t want to discuss extremes, we want to test a reasonable case and examine the
effects. With the same question – will we see a problem in the aggregate?
4
The Problem – Steps to Making the Models
1) Use Excel to generate random normal data specifying a mean and a standard
deviation. This will simulate the output of the individual processes.
2) Generate thousands of points – simulating a process run at these settings.
3) Grab a subset of these points, as if we were grabbing production parts off the end
of the process for a study. We will grab 40 “parts” at random out of the thousands
with subgrouping.
4) Calculate the capabilities of these experimental draws and confirm they are close to
the desired inputs.
5) Grab a subset from these selected parts at random, as if we then pulled aggregate
parts.
6) Calculate the capabilities of these draws to see what a random study on the
aggregate output shows us.
5
Case Study 1: Three capable distributions, not centered.
Let’s assume we have 3 machines, each trying to make an inside diameter of 30 mm. We do a separate capability
study on each. What happens if we take all the samples and mix them up and do a capability study on the
aggregate? Essentially losing traceability from each of the 3 machines.
5
Set 1 Set 2 Set 3
Name Mach 1 Mach 2 Mach 3
Upper Spec Limit 30.02 30.02 30.02
Lower Spec Limit 29.98 29.98 29.98
n 40 40 40
mean 29.99 30.00 30.01
std dev 0.0023 0.0024 0.0026
Cp 2.86 2.76 2.57
Cpk 1.15 2.71 0.97
Details on the distributions: Notice the standard deviations are all the same (green arrow) – the processes have all
the same noise. The random distributions were all generated with the same standard deviation as an input. And the
calculated sigma of the study confirms the distribution. Close to the input value (.0024), but still random. Notice the
Cps (purple arrow). Because they all have the same noise (generated with the same standard deviation), the Cps are
all roughly the same – they should be. Now examine the Cpks (red arrow). Only the centered process has Cpk~Cp
(it’s centered). The other two have Cpk near 1 because the processes are near the limits. The means in the
generation of the random data were intentionally set here. A few thousand points were generated, and 40 (8
subgroups of 5) were grabbed out of all these random points. Now, if we grab 40 points at random out of THESE
distributions and do a study, what does the aggregate look like?
6
Case Study 1: Three capable distributions, not centered.
The final result looks like this ….
6
The histogram bars from the root distributions (blue, red, green) have been removed for clarity. The purple
histogram shown is from the random pulls of the root distributions. You can sort of tell that there were three
independent distributions that generated this data from the 3 higher bars in the purple histogram.
Notice in the combination, the Cp and Cpk are low (gold box). Notice how high the standard deviation is (green box)
compared to the three source distributions. Notice how non-normal the resulting aggregate distribution is. This
random pull was 12 from Mach 1, 18 from Mach 2, and 10 from Mach 3. (This model was run repeatedly, results were
similar).
In this case, with source distributions different, it had a negative effect on the aggregate distribution. This means
that if an aggregate distribution is capable (regarding centerdness) the underlying distributions are also capable.
Set 1 Set 2 Set 3 Set 4
Name Mach 1 Mach 2 Mach 3 123 Comb
Upper Spec Limit 30.02 30.02 30.02 30.02
Lower Spec Limit 29.98 29.98 29.98 29.98
n 40 40 40 40
mean 29.99 30.00 30.01 30.00
std dev 0.0023 0.0024 0.0026 0.0097
Cp 2.86 2.76 2.57 0.69
Cpk 1.15 2.71 0.97 0.67
7
Case Study 1: The Thought Experiment
7
Let’s recap – we had 3 root distributions, all very capable, but two of them needing
centering. And the aggregate study showed us a very wide, but still centered deviation
we would have interpreted as a fail. Where the sub-processes would have definitely
been a pass (for one of them) and possibly a “pass with centering” on the other two.
Key Points:
1) The result makes sense. Imaging them all aligned in the center, and you drag two
of them left and right from center, you can see in your mind the total distribution
getting wider, but remaining centered.
2) With only 3 distributions, it’s easy to imagine, but what happens if we have more?
What if we had 20 sub-processes, 19 of them a bit right of nominal (perchance well
set up, accounting for tool wear) and the 20th is flubbed. It’s hovering at the lower
limit, what would be the effect?
8
Case Study 1: The Thought Experiment
8
It might look something like this ….
Lower
Limit Upper
Limit
Nominal
19 distributions
well collected
here.
One process is not
set up well, making a
distribution over here.
A resulting aggregate distribution may look like the pink sketch. Not a wide distribution, per se, but one showing two peaks. It may have
an acceptable Cp or Cpk. This effect is because the samples are heavily weighted towards the “good” distribution (19 good vs 1 outlying
process). This effect would be more muted as more good processes were added because of this weighting. Now, 100 parallel processes
aren’t seen too often in manufacturing. But 40 are (multi-cavity injection molding tools come to mind). There are a few takeaways….
1) Were one to attempt an aggregate experiment, one would have to ensure adequate representation from each sub process. (NOT
random). You would want 5 parts from each process at a minimum. And maybe adjust quantities based on where the means of the
initial 5 draws fell on a histogram.
2) It is unlikely the raw Cp/Cpk numbers would be enough to adequately evaluate the results, you would definitely want to plot the results
and convince yourself the numbers made sense. Especially looking for multiple peaks. Something you would NOT expect from sub-
processes that were identical.
Key Point: The more
parallel processes that
contribute to the
distribution over here, the
less you will statistically
notice the stray process…
9
Case Study 2: One process is noisy ….
OK. But what if 2 of the machines are good and one is really noisy? Let’s keep the same parameters. But this time,
Machines 4 and 5 are well centered. Machine 6 is centered, but noisy. A cutter is loose … will this still show up in
an aggregate capability study?
9
The first two machines are running good. Well centered. Tight distributions (blue and maroon). The third machine
is noisy (green). It is ALMOST capable (to a hurdle of 1.33). If we were just considering Mach 6, we would reject it
and tell the supplier that they need to work on the noise.
But let’s assume we have no knowledge of which machine a part came from and we randomly take 40 pieces from
this aggregate population again. What would we get? Will the aggregate be out? Think about the answer before
you continue.
Set 1 Set 2 Set 3
Name Mach 4 Mach 5 Mach 6
Upper Spec Limit 30.02 30.02 30.02
Lower Spec Limit 29.98 29.98 29.98
n 40 40 40
mean 30.00 30.00 30.00
std dev 0.0027 0.0026 0.0051
Cp 2.51 2.52 1.31
Cpk 2.47 2.47 1.24
10
Case Study 2: One process is noisy ….
And here is the result.
10
The aggregate distribution (purple) got worse than the two good machines (4 & 5) because we are including data
from the bad machine (6). But … the two good machines helped the output of the bad one. So the aggregate is a
pass even though one of the subprocesses is NOT.
If we studied each machine individually, we would have caught the “bad apple.” But collectively, it did NOT ruin the
bunch. It made it worse, but not enough to result in a failed study. The likelihood of getting a “tail reading” in the
bad (green) Mach 6 distribution is cut in third by adding the two good machines into the mix. So a similar leveraging
effect works here, too. If we had more and more good processes, the statistical significance of the bad apple would
go down. This does not mean the bad apple is running good, it is just less likely that we would detect it.
Set 1 Set 2 Set 3 Set 4
Name Mach 4 Mach 5 Mach 6 456 Comb
Upper Spec Limit 30.02 30.02 30.02 30.02
Lower Spec Limit 29.98 29.98 29.98 29.98
n 40 40 40 40
mean 30.00 30.00 30.00 30.00
std dev 0.0027 0.0026 0.0051 0.0043
Cp 2.51 2.52 1.31 1.54
Cpk 2.47 2.47 1.24 1.49
11
Resulting Effects
11
When one studies an aggregate sampling from parallel processes,
the ability to detect a stray process in the aggregate is based on
how extreme the errant process is (either process noise or mean
shift) and how many processes are in parallel. In other words:
• The more extreme the response of the stray process is, the more
likely you are to detect it.
• The fewer the processes in the study, the more likely you are to
detect it.
What this means is, you cannot necessarily say:
few many
Number of parallel processes
slight
extreme
Diffe
rence o
f str
ay p
roce
ss to
th
e
rest
of th
e g
rou
p
Ease of detection
“I have a passing capability
study on the aggregate of my
processes, therefore all my
sub processes are OK.”
With two process having
very different means, you
would easily notice a two
peaked distribution. Easy to
detect.
Trivial solution: with two processes
that do not have a detectable
difference … this is what you want,
processes without a detectable
difference.
This corner is also trivial. If I keep
duplicating process that I cannot
detect a difference between, this is
a good thing – multiple processes,
statistically identical.
This is the danger corner,
MAYBE you detect this.
With enough parallel
processes that are good,
they could very much
mask one stray one in an
aggregate study.
12
The Problem (Again)
We are still faced with the main
problem. Given we have
demonstrated that it is possible
for an aggregate study to mask
a stray process, do we then
have to do 30+ piece studies on
every combination? The
answer is no – IF we do a
structured experiment.
If we consider the output of each subprocess as its
own subgroup, we can still detect a stray process
with an aggregate study, but we have to approach this
with a structured, stratified approach.
13
First, here’s a good study.
13
Trial N1 N2 N3 N4 N5
1 20.003 19.986 20.001 19.995 20.004
2 19.998 20.019 19.999 20.017 19.988
3 19.995 19.997 20.003 19.998 19.997
4 20.005 19.996 20.012 20.009 20.009
5 20.008 19.993 19.989 19.994 20.002
6 20.019 19.985 20.003 19.997 20.000
7 19.994 20.005 19.998 19.994 20.012
8 20.000 19.992 19.993 19.998 20.004
9 20.002 19.998 20.002 20.013 20.000
10 20.000 19.988 19.995 20.005 20.000
11 20.007 20.010 19.993 20.010 19.997
12 20.016 19.998 20.009 19.995 19.989
13 19.999 20.001 20.003 20.000 19.998
14 20.009 20.011 20.000 19.990 20.008
15 20.008 20.008 20.001 19.995 20.002
16 19.985 20.001 20.013 20.008 19.992
17 19.995 20.008 20.001 20.004 19.986
18 19.995 19.997 19.999 19.995 20.007
19 20.005 19.998 19.991 20.003 19.993
20 20.003 20.003 19.996 19.999 20.001
Ea
ch
trial is
a d
iffe
ren
t
ma
ch
ine a
nd
fix
ture
We grab 5 parts from each
machine/fixture combination
and keep them controlled
We can generate some data. Let’s assume we want a normal distribution of a feature that’s at 20 ± 0.05 and we want a Cp ~ Cpk = 2.5. We can generate some random, normal data with µ=20 and σ=0.00667 and get such a spread.
Here are the results.
Ppk = 2.212 Cpk = 2.305
Pp = 2.233 Cp = 2.327
Capability Data
This is a little shy of our target Cp of 2.5, but good
enough. What happens if one of the process strays?
This is all centered up. Let’s take trial 5 and move it’s
mean so that it is JUST capable (Cpk = 1.33)
14
Process 5 nudged
14
Ea
ch
trial is
a d
iffe
ren
t
ma
ch
ine a
nd
fix
ture
We grab 5 parts from each
machine/fixture combination
and keep them controlled
We know the lower limit in our case study (19.95) and
we know the σ that generated the data was 0.00667, so
if we set the mean JUST from process 5 to:
19.95 + 0.00667 = 19.97001 this is what we get.
The red arrows are indicating process 5. Now, it is still capable, it is
one stray process riding right at the limit, but still in. You can
almost see the effect in the histogram. But it is obvious in the run
chart. The recommended course of action would be – pass the
processes, BUT machine/fixture 5 needs a separate study.
Trial N1 N2 N3 N4 N5
1 20.003 19.986 20.001 19.995 20.004
2 19.998 20.019 19.999 20.017 19.988
3 19.995 19.997 20.003 19.998 19.997
4 20.005 19.996 20.012 20.009 20.009
5 19.978 19.963 19.959 19.964 19.972
6 20.019 19.985 20.003 19.997 20.000
7 19.994 20.005 19.998 19.994 20.012
8 20.000 19.992 19.993 19.998 20.004
9 20.002 19.998 20.002 20.013 20.000
10 20.000 19.988 19.995 20.005 20.000
11 20.007 20.010 19.993 20.010 19.997
12 20.016 19.998 20.009 19.995 19.989
13 19.999 20.001 20.003 20.000 19.998
14 20.009 20.011 20.000 19.990 20.008
15 20.008 20.008 20.001 19.995 20.002
16 19.985 20.001 20.013 20.008 19.992
17 19.995 20.008 20.001 20.004 19.986
18 19.995 19.997 19.999 19.995 20.007
19 20.005 19.998 19.991 20.003 19.993
20 20.003 20.003 19.996 19.999 20.001
Ppk = 1.567 Cpk = 2.280
Pp = 1.599 Cp = 2.327
Capability Data
Ppk = 2.212 Cpk = 2.305
Pp = 2.233 Cp = 2.327
Capability Data
Before With shifting
15
Process 5 nudged
15
Ea
ch
trial is
a d
iffe
ren
t
ma
ch
ine a
nd
fix
ture
We grab 5 parts from each
machine/fixture combination
and keep them controlled
We know the lower limit in our case study (19.95) and
we know the σ that generated the data was 0.00667, so
if we set the mean JUST from process 5 to:
19.95 + 0.00667 = 19.97001 this is what we get.
Another key point, it is more detectable in Pp/Ppk than Cp/Cpk
because Cp/Cpk calculations are less sensitive to subgroup shifts.
And remember, all these distributions are capable. So if a
subprocess was very errant, this method has an excellent chance of
detecting it.
Trial N1 N2 N3 N4 N5
1 20.003 19.986 20.001 19.995 20.004
2 19.998 20.019 19.999 20.017 19.988
3 19.995 19.997 20.003 19.998 19.997
4 20.005 19.996 20.012 20.009 20.009
5 19.978 19.963 19.959 19.964 19.972
6 20.019 19.985 20.003 19.997 20.000
7 19.994 20.005 19.998 19.994 20.012
8 20.000 19.992 19.993 19.998 20.004
9 20.002 19.998 20.002 20.013 20.000
10 20.000 19.988 19.995 20.005 20.000
11 20.007 20.010 19.993 20.010 19.997
12 20.016 19.998 20.009 19.995 19.989
13 19.999 20.001 20.003 20.000 19.998
14 20.009 20.011 20.000 19.990 20.008
15 20.008 20.008 20.001 19.995 20.002
16 19.985 20.001 20.013 20.008 19.992
17 19.995 20.008 20.001 20.004 19.986
18 19.995 19.997 19.999 19.995 20.007
19 20.005 19.998 19.991 20.003 19.993
20 20.003 20.003 19.996 19.999 20.001
Ppk = 1.567 Cpk = 2.280
Pp = 1.599 Cp = 2.327
Capability Data
Ppk = 2.212 Cpk = 2.305
Pp = 2.233 Cp = 2.327
Capability Data
Before With shifting
16
Why a Stratified Study?
16
Ea
ch
trial is
a d
iffe
ren
t
ma
ch
ine a
nd
fix
ture
Random study, drawn from all
process, no stratification. (What was
process 5 is highlighted)
If we just took all these samples at random, the parts
from machine/fixture 5 would be sprinkled throughout
the data. Similar to what is shown in this table.
This hides the mean shift. The process looks overall less capable
and you lose sight of the fact that there IS a problem with process 5.
If process 5 were actually NOT capable, you may pass this study has
having a few outliers. The conclusion is stratification is key!!
Ppk = 1.567 Cpk = 2.280
Pp = 1.599 Cp = 2.327
Capability Data
Ppk = 2.212 Cpk = 2.305
Pp = 2.233 Cp = 2.327
Capability Data
Before With shifting
Trial N1 N2 N3 N4 N5
1 20.003 19.986 20.001 19.995 20.004
2 19.998 20.019 19.959 20.017 19.988
3 19.995 19.997 20.003 19.998 19.997
4 20.005 19.996 20.012 20.009 20.009
5 20.009 20.001 19.999 19.995 19.972
6 20.019 19.985 20.003 19.997 20.000
7 19.994 20.005 19.998 19.994 20.012
8 20.000 19.992 19.993 19.998 20.004
9 20.002 19.998 20.002 20.013 20.000
10 20.000 19.988 19.995 20.005 20.000
11 20.007 20.010 19.993 20.010 19.997
12 20.016 19.998 20.009 19.964 19.989
13 19.999 20.001 20.003 20.000 19.998
14 19.978 20.011 20.000 19.990 20.008
15 20.008 20.008 20.001 19.995 20.002
16 19.985 19.963 20.013 20.008 19.992
17 19.995 20.008 20.001 20.004 19.986
18 19.995 19.997 19.999 19.995 20.007
19 20.005 19.998 19.991 20.003 19.993
20 20.003 20.003 19.996 19.999 20.001
Ppk = 1.567 Cpk = 1.779
Pp = 1.599 Cp = 1.816
Capability Data
Random pull (NOT stratified) =>
17
Also suspicious – A noisy process
17
As opposed to a mean shift, a stray process due to noise is harder to detect because of only 5 samples from each process. Left
is the original case study (good, Cpk target of 2.5). The middle column has subgroup 5 with a target Cp/Cpk of 1.33 and the right
column has the same data, but with subgroup 5 targeted at Cp/Cpk of 1.00. The issue isn’t detectable in the indices, histograms,
or run charts, but it IS detectable in the R-Chart. Again – the conclusion is an aggregate study is possible IF it is structured and
you are critical of the results. If this were from ONE process, it would be acceptable.
Random pull (NOT stratified) =>
Ppk = 2.212 Cpk = 2.305
Pp = 2.233 Cp = 2.327
Capability Data
Ppk = 2.085 Cpk = 2.178
Pp = 2.115 Cp = 2.209
Capability Data
Ppk = 2.073 Cpk = 2.205
Pp = 2.117 Cp = 2.252
Capability Data
18
Keys to Success
18
An aggregate study could be
successful if:
1) It was controlled and stratified.
Each subgroup representing at least
5 parts from each subprocess, kept
together.
2) You must LOOK AT THE DATA and understand what it is trying to tell you. You cannot just look at
the capability indices. The histogram and especially the run and range charts of the data are crucial to
detect an errant process.
3) You must be willing to investigate errant data points. In this example, we are proving out 20 parallel
process with 100 measurements. Don’t balk at having to do an independent study on process 5. It
warrants it, it looks very suspicious.
4) You need to be reasonably capable overall. This is plain confidence interval logic. If the run chart
above were using more of the tolerance, it would justify doing a full study on a couple of
machine/fixture combinations.
19
Steps to a successful Aggregate Study
1) Create a structured, stratified
experiment.
Guidelines:
• 5 in each subgroup would be the
minimum number.
• If this wasn’t a lot of parts (if we had 2
machines, it would only be 10 parts)
increase the subgroup size until the total
number of parts was at least 40, 100
would be better.
• The minimum of 5 must be maintained. If
there were 100 processes in parallel,
that’s 500 total parts.
19
5 parts from this
machine / fixture
is subgroup 1 n=5, subgroup 2
n=5, subgroup 3 …
And so on …
20
Steps to a successful Aggregate Study
2) Conduct the stratified aggregate study, maintaining subgroup organization
and pay close attention to the Xbar (run) chart and R charts.
20
This would be ideal. Here’s a nice, tight grouping.
Looking at it, you are pretty confident all the
processes are performing the same. From this, you
could safely stop checking. Overall Cp is high. All
sub-processes in the control limits.
This would be ideal. You want to compare the
calculated range control limit of the aggregate (.037)
to the total tolerance (0.1) Here we use a bit more
than a third of the tolerance. Half would be a red
flag. Remember, in the aggregate, we have most
likely grabbed parts within 1 sigma of the mean from
each process.
Exam
ple
1:
21
Steps to a successful Aggregate Study
2) Conduct the stratified aggregate study, maintaining subgroup organization
and pay close attention to the Xbar (run) chart and R charts.
21
This is noisy. The aggregate has a bad Cp/Cpk. This
is too noisy to draw a conclusion on from the
aggregate alone. But it’s hard to detect on the run
chart. All means are within the run chart control
limits.
Exam
ple
2:
Red flag: Range control limit is 0.074 which is more
than half the tolerance. You would want to take the
noisiest process (#20, in this case) and do a full
study. More would be better. Problem noise is
easier to detect in the range chart than the run chart
above. You should also pick the process from the
run chart above that is farthest from nominal and
check that too.
22
Steps to a successful Aggregate Study
2) Conduct the stratified aggregate study, maintaining subgroup organization
and pay close attention to the Xbar (run) chart and R charts.
22
This has one errant process. You would
want to conduct another study focused on
the errant process. Note the errant one is
outside the calculated control limits. (There
may be more than one errant process.)
Exam
ple
3:
The range chart looks good. (It would, the
spread of the subprocess is the same). It
reiterates the fact you need to examine
BOTH.
23
Steps to a successful Aggregate Study
2) Conduct the stratified aggregate study, maintaining subgroup organization
and pay close attention to the Xbar (run) chart and R charts.
23
Here you have a number of points outside the
control limits. Remember – these are all different
processes. This means they are not all centered.
Best solution is to center these up and repeat the
aggregate study (centering fix is the easy fix). Worst
case would be take the 3 FARTHEST subprocesses
and do a full study on each of them. If they are
capable, all of them should be.
Exam
ple
4:
Run chart looks good in this example as well. And it
should, this has the centering problem.
24
Steps to a successful Aggregate Study
3) From the aggregate study, conclude everything is OK OR conduct your sub
process study or studies.
4) Draw your conclusion.
Final remark: There is no substitute for understanding what the graphs and
data are telling you. Take the time to think about how well your model reflects
your processes. One thing is clear – it is investigative which means you need
to conduct an intentional experiment, not a random one.
24
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