Incorrect Sigma to DPMO Conversions

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• #27434

Ken K.
Participant

I have recently noticed that many web sites and other sources publish incorrect or overrounded conversions from Sigma to DPMO.

It appears that many of ignoring the “other” side of the normal distribution when making these calculations. This other side MUST be included in the calculation, unless of course the specification is one-sided.

I don’t have any copies of the famous (infamous) books touting the Six Sigma methodologies, so I don’t know which tables they are using. I tend to not want to make M.H. any richer than he already is.

Consider the following:

S Ls Us LDPMO UDPMO DPMO
6 -7.5 4.5 0.0 3.4 3.4
5 -6.5 3.5 0.0 232.6 233
4 -5.5 2.5 0.0 6209.7 6210
3 -4.5 1.5 3.4 66807.2 66811
2 -3.5 0.5 232.6 308537.5 308770
1 -2.5 -0.5 6209.7 691462.5 697672

Where:
S = Sigma Level (assume shift 1.5 sigma to right)
Ls = LSL position, # of sigmas to left of mean
Us = USL position, # of sigmas to right of mean
LDPMO = Prob of observing defect below LSL, in PPM
UDPMO = Prob of observing defect above USL, in PPM
DPMO = Combined Prob of observing Defect, in PPM

Normal probabilities calculated using MINITAB 13.3’s probability function.

It appears that most just use column UDPMO and ignore column LDPMO. In the higher Sigma levels this doesn’t matter since the LDPMO probabilities are so small, but in the lower Sigma levels it makes a difference.

Any comments? Can anyone see anything wrong with my logic?

I have found many published tables that I feel are incorrect or overrounded. Here are a few websites:

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#67184

Hogg
Participant

Accuracy is relative to the degree of information provided in the calculation and the significance associated (or required) with the result.

My tables are rounded to a very comfortable number of significant figures for the average project. At very low sigma values there is little point in quoting the result to anything more than 1 decimal place or four figures for DPMO. At very high sigma values the issue of sample size and measurement system variation tends to dominate and it becomes almost impossible to distinguish between 5.9 and 6.0 sigma.

Be wary of quoting sigma or DPMO to four decimal places based on a sample of 15 with a measurement variation of 5%! I am always worried by people who divide 2 by 3 on a calculator and quote the result as 0.666666666666

Mulbury consulting, hosted at http://www.longacre.cwc.net

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#67185

“Ken”
Participant

Ken K.,

I tend to support Geoff’s point about the estimation accuracy for dpmo using a standard normal distribution. It’s useful to remember that when estimating these dpmo values we are using the tails of the distribution function. Estimates in these areas are frought with considerable error anyway. As such we will never have enough data to provide an accurate estimate.

To illustrate this point let’s look at two distribution functions that are almost identical, but are still different: the standardize Burr, and the standard normal distribution.

Burr cfd. F(x)= 1-(1+x^c)^-k for x > 0

Normal cfd. F(z)= sigma(c)(k)[1+(mu+sigma(z))^c]^-(k-1)*(mu+sigma(z))^c-1 for z > -mu/sigma

given: c=4.87372, k=6.157568, with mu=0.64472 and sigma=0.16199. Both distributions have a skewness=0, and kurtosis=3.00. Essentially, both distributions approximate the standard normal distribution, but are obviously different. Which one could be used to model your data? Answer, either… But, consider differences in short-term dpmo’s, perfectly centered process:

Sigma Normal Burr
3 2700 >2473
4 64 <96
4.5 6.8 <22
6 0.002 <0.310

Forget about the stuff above the table, but realize either distribution could be used to model the data. If you plotted and overlaid these two distributions, they would line up very close. Yet, selecting one over the other might incur as much as a 15,400% estimation error. What’s the takeaway? We will never know enough about the system, i.e., have enough data to determine which distribution best approximates the data. Therefore, in the final analysis the purpose of these measures is to show improvement, not to get the right answer.

Ref. Advanced Topic in SPC by D. Wheeler, pgs. 195-196

Ken

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#67194

Ken K.
Participant

But even rounding as you do there are errors.

All your values for Sigma 1.8 and less are incorrect. They should be:

0.0 1,000,000
0.1 974,000
0.2 948,000
0.3 921,000
0.4 893,000
0.5 864,000
0.6 834,000
0.7 802,000
0.8 769,000
0.9 734,000
1.0 698,000
1.1 660,000
1.2 621,000
1.3 582,000
1.4 542,000
1.5 501,000
1.6 461,000
1.7 421,000
1.8 383,000
1.9 345,000
2.0 309,000

I don’t buy all this talk of “it just doesn’t matter”. These are published tables based upon the Normal distribution which in most cases are NOT rounded.

I’d accept rounding as you do if they are correctly calculated.

And I don’t accept the “nothing is really normally distributed” excuse. Again, these are published tables based upon a specified, well characterized distribution.

In my business the total number of opportunities per product adds up to hundreds of millions of opportunities per month. With the volumes we run, rounding just 10PPM reflects a difference of thousands of defects. We may miscount a defect here or there, but not by thousands.

In my experience people work so hard to achieve improvements that they want to claim every decimal of Sigma coming to them.

Face it, Mikel got the table wrong over 10 years ago and none of his GE or post-GE disciples have bothered to correct it since then.

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#67197

“Ken”
Participant

Ken K.,

Just curious… Hundreds of millions of opportunities, or defects per opportunity? If you have hundreds of millions of opportunities, then perhaps another question might be how you determine opportunities for your products.

Ken

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#67200

Hogg
Participant

You need to check your figures.

The process sigma is the number of standard deviations that fit between the mean and the nearest process/customer limit.

Typically this is quoted as one-sided error (the nearest side) and is shifted by 1.5 sigma to account for the long-term/short-term difference.

The DPMO figure is derived from the corresponding area under the normal curve that falls outside of the limit(s). For a (long-term) sigma of zero (0) where the limit and mean exactly overlap, then 50% of the curve is outside of the limit – this is the definition of the mean, the half way point. At 0 sigma the 1.5 shift makes this 1.5 sigma (short-term), and equivalent to 500,000 DPMO

1.5 sigma is 500,000 DPMO

If you like, 1.5 sigma is 500,000.0000000000 DPMO
Accurate to 10 dp and without the aid of a caculator.

If you are interested, the table I use is based on Excel formula for the area under the right tail of a standard normal distribution, shifted by 1.5. I think that you will find that this is much what everyone else uses, although I agree that there is some debate as to whether this should be one- or two-sided; and a great deal of debate over the 1.5 shift.

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#67201

Hogg
Participant

Thinking about it and looking at your figures, I suspect that your problem lies in the fact that 0 sigma is not 100,000 DPMO.

Zero sigma (short term) is in fact -1.5 sigma long term, which encompases the area to the right of a line 1.5 standard deviations to the left of the mean. This still leaves the area to the left of -1.5 which is not an error.

Many people do not regard negative sigma as valid, although many processes are so bad that minus sigma figures are no surprise. To deal with this issue some have set 0 sigma as 100% error, and have ‘fiddled’ with the tables to compensate.

Personally I like to use the more accurate mathematical definitions, and just cut off the table at 0 sigma = 933,000 DPMO. Any process operating at less than 0 sigma can safely be regarded as 0!

Note that DPMO is only a convenient approximation to process sigma – the formal definition should be used in preference with an appropriate consideration of sample size, measurement system capablity and accuracy, and the long/short term shift of the process.

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#67202

gt
Participant

Oops – should be 1,000,000 DPMO

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#67203

Jim Parnella
Participant

For what it’s worth I have these values calculated and they are on the web site of Bob Gribus, who wanted to post them there for his readers. These are the Number of Defects (parts per million) for Various Sigma Shifts in the Process Average for Different Sigma Capability Levels, from 3 to 6. They were calculated using Excel, and Bob has included the Excel formulas as well as the calculated values. Here is the web site.

http://www.geocities.com/bobgsma/ParnellaFactor.html

I don’t think that having the two-tail values as opposed to the one-tail values is such a big deal once you get above a sigma level of 2.
Jim
P.S. Does anyone else use anything different than the term “Sigma Level”? Please e-mail me if you call it something different.

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#67225

Hogg
Participant

It is always a good idea to think everything out before replying!

I can see now where your figures come from, however I still believe that my/our table is (more) correct!

You appear to be assuming that an X sigma process is central to two limits 2X standard deviations apart, and then shifted 1.5 sigma making the DPMO equivalent to the area on two tails, one X+1.5 and one X-1.5 sd. The difficulty with your table is that it does not work for one-sided cases or where the distribution is not centred between the limits, which in my experience is almost every case!

The sigma process metric is defined as the distance to the *nearest* limit, and does not relate to the other limit (if there is one). Consider a process centred between two limits, and ignore the 1.5 shift for the moment. If the left and right limits are 2 standard deviations from the mean, this is clearly a 2 sigma process. If however the left limit were 3 sd away, then this would still be a 2 sigma process. Certainly there are less defects experienced with less of the distribution outside of the limits, but you cannot call this a 3 sigma process or something between 2 and 3, it still has to be called a 2 sigma process. Taken to the extreme if the left limit was 10 sd away then this is still a 2 sigma process and effectively becomes a one sided issue, with the defects equal to the area under the normal curve to the right of just the 2 sd limit.

The DMPO equivalent to process sigma is just an approximation for use when the distribution is not particularly normal or the metric is discrete or when it is just easier to count opportunities and defects. It works for one-sided limit cases, and for high values of sigma. For low sigma values and where two sided limits exist it will always be an approximation – one approach is to add both sides of the tails together and calculate the equivalent process sigma from a one-sided limit.

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#67226

Hogg
Participant

I would also like to address the issue of decimal sigma and accuracy down to 10 DPMO.

At 3 sigma the DMPO is (approximately) 60,000. To be accurate to the nearest 10 DPMO you are looking to measure the difference between two proportions

0.06000 and 0.06001

If you have 94 green beads and 6 red ones in a bag, how many beads do you have to take out to accurately or confidently measure the proportion of red? In the case here you would need 94000 green and 6000 red beads to show in the total population a difference of just one red bead – 93999 and 6001 – however sampling from this population would not enable you to confidently detect the difference of just one bead.

If you use Minitab power size calculations for two proportions, one 0.06 and the other 0.06001, and take a power of 0.9 (ie 90% confident that you will find a difference) you will see that the sample size is 1,000,000,000 and that the actual power is only 16%. I would like to suggest that, at 3 sigma, there are more important things to do than count one thousand million opportunities to be 16% confident that your DPMO is accurate to within 10!

On the other hand, if you are using the values for the mean and the standard deviation to (more correctly) work out the process sigma, then you need to realise that to achieve an accuracy of 2 decimal places in your sigma value, the mean and sd must each be calculated to better than +/- 0.5%. If you measure 100 pieces of wood each 1 metre long to the nearest millimetre, then the average will have a potential error of +/- 100 times 0.05% or 5%. Likewise, if you sample 100 items then Minitab will show you that the sample mean and sample standard deviation are only estimates of the true population mean and standard deviation. To achieve the accuracy you are seeking will again require samples close to 1,000,000.

In hard reality, the majority of projects simply show that the process has moved from about 2 sigma to about 3 sigma – a bit like the clock with just an hour hand!

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#67256

Mike B
Participant

Your right, Ken and M.H. has some of the numbers wrong in his book also. There is also a flaw in the overall logic of the numbers. The 3.4 ppm assumes the process is at the “worst case” drift of 1.5s (or -1.5s) but never in-between. In actuality, the DPMO will be somewhere between 3.4 ppm and 2 ppb and probably closer to 2 ppb.

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#67282

Ken K.
Participant

You might need to reconsider your definition of Sigma.

I take mine from the article “The Nature of Six Sigma Quality” by Mikel Harry, published by the Motorola University Press in the early 1990’s (the specific copyright date varies depending on the date of printing.)

By definition, an X Sigma process is one which, when centered halfway between the lower and upper specification limits (LSL & USL), the LSL is located at [mu – X*sigma] (where s is the process standard deviation), and the USL is located at [mu + X*sigma].

A 6 Sigma process is one where, IF the process were centered, the LSL is exactly at [mu – 6*sigma] and the USL is exactly at [mu + 6*sigma).

This is the definition given by Mikel Harry in his article. It’s shown in Figure 8.

When the much debated 1.5*sigma shift occurs (here assuming a shift to the right for the sake of computation), the LSL & USL are no longer as described before, now they are at [mu – (X + 1.5)*sigma] & [mu + (X – 1.5)*sigma], respectively. For a 6 Sigma scenario this is [mu – 7.5*sigma] & [mu + 4.5*sigma], respectivley. This scenario is shown in Figure 9 of the above mentioned article.

In BOTH Figures 8 and 9 Mikel Harry recognizes the propensity for errors in the left tail of the distribution. BOTH tails must be considered.

For a centered distribution the left tail provides a defect probability of 0.001 PPM – this gives rise the slightly famous 0.002 PPM defect probability for a centered 6 Sigma scenario.

For a 1.5*sigma shifted 6 Sigma scenario this left tail defect probability decreases to nearly zero. Mikel actually shows it as “~0” in Figure 9 – the important thing is he does recognize it exists. As the Sigma level decreases the probabilities in the left tail increase:

Sigma Level — Left Tail Probability
6 — ~0
5 — ~0
4 — ~0
3 — 3.4 PPM
2 — 233 PPM
1 — 6210 PPM

The exclusion of the left tail is where the published tables are incorrect.

Ken

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#67283

Ken K.
Participant

Quite simply, we calculate opportunities as:

(# of parts) + (# of electrical connections)

By connection we mean just the soldered connections or wire bonds – not including those internal to parts.

We make fairly complicated products and lots of them. That is exactly why quality is so very critical. One critical defect and someone is likely to be walking home.

Typically we’ll have 1000 to 2500 opportunities per unit, depending on the model.

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#67285

Ken K.
Participant

Like I said, 10 ppm for us relates to roughly a thousand defects per month.

The difference between 0.06000 and 0.06001 you mention is 10 PPM.

Would you like to be one of those walking home after you car stops functioning when we, as a company, decide we “have more important things to do” than worry about those thousand defects?

That’s the whole point. Every defect hurts. Every defect must be measured, understood, and eliminated. It takes a lot of effort to permanently remove a defect and people don’t want those efforts lost in the rounding – regardless of whether their at 2 Sigma or 6 Sigma.

And then again, I always come back to the need to provide accuracy when publishing tables. Leave the rounding to those who are closer to the consequences.

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#67310

Nolan
Participant

This is something that caught my attention also. For a process sigma of z, the yield should be

cdf(z-1.5) – cdf(-z-1.5)

where “cdf” is the cumulative distribution function. However, the text books only calculate the left hand term and ignore the right hand term. This is OK when z is large because the term on the right becomes negligable. But for small process sigma (less than 1) the right hand term becomes more significant. See the table below, which shows the yield for different process sigma values according to the full correct equation and according to the left hand term only (the text book approximation):

Process sigma Yield (whole eqn) Yield (LH term only)
6 0.999996599 0.999996599
5 0.999767327 0.999767327
4 0.993790301 0.99379032
3 0.93318937 0.933192771
2 0.691229794 0.691462467
1.5 0.498650033 0.5
1 0.135905198 0.15865526
0 0 0.066807229

According to the text book approximation (i.e. only taking the left hand side) a zero sigma process has a yield of 6.7%. At negative sigma the curve then tends to minus infinity sigma as the yield tends to zero (you can have negative process sigma by taking just the left hand term).

I believe that it is wrong to approximate the way the text books do. If they had taken the full equation in the first place, then zero sigma would mean zero yield, which makes much more sense. And we wouldn’t have to explain to our inquisitive trainees the reason why negative process sigma exists.

0
#67313

Hogg
Participant

You assume that the process sigma metric is calcuated when the distribution is equally spaced between two limits?

What therefore is the process sigma when
mean = 10
standard distribution = 1

upper customer limit = 8
(optional lower customer limit = 0) ?

The sigma metric is a measure of process performance, not capability. I make that a -2 sigma process, not a 4 sigma process.

0
#67321

Anonymous
Participant

Much ado about nothing

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#67401

Nolan
Participant

I disagree with you Geoff.

In the example you give, mean = 10, standard dev = 1, USL = 8, no LSL. The USL is 2 standard deviations below the mean. This would give a yield of 0.02275 or 2.275%, since cfd(-2) = 0.02275 (where cfd is the cumulative distribution function).

According to the text books, allowing for a 1.5 sigma shift, the process sigma would actually be -0.5, since they calculate cfd(z-1.5), where z is the process sigma. So for this example they give

cfd(-0.5-1.5) = 0.02275

However, the full equation should really be

cfd(z-1.5)-cfd(-z-1.5) = 0.02275

If you calculate the process sigma in this instance (i.e. z) then it comes out to be around 0.088.

The way process sigma is calculated is arbitrary. Those who developed the system decided to use cfd(z-1.5). I’m saying that if they had used cfd(z-1.5)-cfd(-z-1.5), then there would not be negative process sigma. Zero yield would be zero process sigma, which makes more sense to me.

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#67404

Hogg
Participant

The ‘standard’ approach is

where PS is process sigma, Z(u/l) are Z of upper/lower customer limits, and D is the defect equivalent.
and Zu = (UCL-mean)/(standard deviation) etc

PS = Minimum(Zu,Zl) = Zm and
D = cdf(Zm-1.5)

Your approach is

D = cdf(Zu-1.5) – cdf(-Zl-1.5)
D = cdf(PS-1.5) – cdf(-PS-1.5) to reverse calculate PS

The beauty of the first approach to defining the process sigma value is that it is visible and axomatic on the mean, sigma and limit(s).
The problem with the second approach is that it is more complex, not visible (my example – I make it -0.5 sigma, you make it 0.088) and no longer axomatic. The formula assumes a relationship between Zu and Zl through the cdf. I would be interested in seeing your formula for process sigma, based on Zu and Zl and not involving the defect count.

I fully take the point about zero defects and negative sigma, but having thought this through I am happy to take the first approach and leave the DPMO tables as a useful approximation!

0
#67424

Nolan
Participant

This approach gives me problems Geoff!

Just to check, what you are saying is: (firstly I assume you mean USL and LSL for the customer limits, not UCL and LCL which I use for control limits)

Zu = (USL-mean)/Std Dev
Zl = (mean-LSL)/Std Dev
Process sigma PS = min(Zu,Zl) = Zm

Imagine this scenario then. A centred distribution has Zu=Zl=1. The yield from this process would be 68.26%, since the customer spec limits are at 1 standard deviation from the mean. The process sigma, according to your formula would be 1.

Now imagine that the customer decides he doesn’t need a lower specification limit. The yield from the process will go up to 84.13%. But Zm will still be 1, so according to your formula the process sigma will also still be 1.

However, we know that for a yield of 68.26% the process sigma should be just under 2, and for a yield of 84.13% the process sigma will be approx 2.5 (I used my Mulbury pocket look-up table!). But if you apply your simple formula then surely the situation can arise that a particular value of PS can correspond to a range of yields as above (just like Cpk when it looks at only the side with the more defects).

I cannot give a formula for process sigma. I calculate it iteratively or with reference to a look up table.

What do you think to my reasoning above?

0
#67427

Hogg
Participant

Your observation is faultless – I would have been surprised had you not commented (besides, I think I got Yield and Defect confused). Whichever way you look at it there are real problems with defining process sigma!

I was taught to use a one-sided table, and to add the defects from the left spec to those from the right spec, and then to back calculate an equivalent process sigma (using the one-sided table). I can’t say that I like this approach however, as it only relates to an almost virtual one-sided equivalent process and not to any real figures seen on a histogram.

If I use the definition that it is the number of sd to the nearest limit, then this gives the problems you highlight – but then I don’t like the idea of using a table based on two equally spaced limits either!

The issue for me is that (by definition?) a six sigma process should be one where the mean is always at least 6 sd from either limit – centrality is not always possible or desirable – and therefore 4.5 at worst (allowing for the 1.5 hotly debated shift). By extension, a three sigma process should always be at least 3 sd from either limit – 1.5 after shift. The problem is that at low sigma values this does not relate to a well-defined number of defects as you rightly point out.

I would be more than happy to use an alternative definition, providing that the formal definition of process sigma comes first, and that an equivalent DPMO conversion follows. I do have more difficulty with defining a DPMO table based on a theoretical situation with two limits and then defining the process sigma based on a reverse lookup of the table.

Incidentally – the discussion thread began with the issue of rounding. Perhaps in time I will add the extra ‘balanced’ table and let the user select! However, I still go for heavy rounding given that the sampling error for discrete binomial distributions is considerably worse than that for continuous variables.

Which really is my issue – I would like to define process sigma based on continuous mean, limit and sd rather than resorting to discrete proportions! If anyone cares to oblige I am more than happy to convert!

0
#67434

Nolan
Participant

Thanks for the interesting discussion Geoff. It has highlighted some of the dangers when calculating process sigma values.

Further dangers clearly exist when reporting improvements to management. If I improved a process from 1 sigma to 2 sigma my yield would improve from 31% to 69%. If my colleague improved a process from 3 sigma to 6 sigma then his yield would improve from 93% to nearly 100%. The success of the two improvement projects can be given different angles by focusing on different improvement metrics (only 1 sigma improvement but I more than doubled my yield, whereas for my colleague an improvement of 3 sigma but not much of an increase in yield).

I suppose at the end of the day, though, the metrics that really matter are customer satisfaction and bottom line savings.

0
#67435

“Ken”
Participant

Les, Geoff, et. al.,

Sould there be any consideration when using any of these Sigma or DPMO values that we are dealing with estimates?

Not with-standanding the distributional concerns I suggested earlier, do any of you think it would be prudent to consider any of the values cited are at the 50% confidence level?

If one-sided vs. two-sided accuracy seems to be such an important topic, then should we be be reporting a lower 95% confidence bound for sigma value, or an upper 95% confidence bound for dpmo?

If we think this might be prudent, then what reference distribution should we use to compute these confidence bounds?

It seems with all the concerns about one vs two tail estimates, we may have forgotten that these estimates that have error bounds associated with them.

Does anybody have any ideas how to account for the estimation error, and report accordingly?

Ken

0
#67442

Anonymous
Participant

“Not with-standanding” all this math, Webster’s lists
notwithstanding as a single word!

0
#67448

denton
Participant

I’ve read several of the comments on this thread, but not all. If this is something that has already been pointed out, kindly forgive the redundancy.

Converting DPMO to Sigma is worse than a waste of time. It is an impediment to communication.

If I’ve got a process giving me some level of defects, and I express that in Sigma, and tell this number to someone, he then has to convert back to DPMO in order to understand what I mean. Basically, I have encrypted the information I want to transmit. That is good only if I want to keep a lot of people from understanding my message. Usually, you don’t want to do that. Usually, you want a lot of people to understand.

People understand your message better if you just report defects in DPMO.

0
#67556

Nolan
Participant

I fully agree.  People running around here and there quoting sigma values for their processes is one of the most dangerous things for a company.
When someone tells me “My process is 5 sigma” or whatever my first response is “With respect to what?”.  The whole thing revolves around defect opportunities, which are open to abuse.
We can all easily make a process look 1 sigma or 6 sigma depending on what we measure.
Clearly yield is a much better yardstick than process sigma because it is linear.

0
#67791

mcintosh
Participant

Thanks for the interesting topic. I had the same thoughts about the “sigma” value even before I saw your message thread. I completely agree with you, Ken. The definition of “sigma” is flawed if you only account for only one tail of the Normal Distribution and ignore the rest of the defects that could occur in the other tail of the Normal Distribution. Ignoring one tail of the Normal Distribution is just plain wrong.I wish the pioneers at Motorola had implemented a system called 2.00 Cpk since a 2.00 Cpk is equivalent to “six sigma”. Perhaps the best way to calculate your “sigma” value is to multiply your Cpk value by 3. That way we could have avoided all the convoluted thinking that went into the “sigma” value. However that wouldn’t have resolved the debate on calculating the yield from a one tail vs two tails of the Normal distribution. I will state more of my opinion on this later.Also, in calculating a process “sigma,” you need to calculate the process yield. Yield is based on an attribute data distribution, namely the Binomial Distribution. Of course, the Binomial Distribution appears to be Normal when the percent of defects is approximately 50 percent. As you get further away from 50 percent defects, the Binomial Distribution becomes less like the Normal Distributuion. If yield is not Normally Distributed, why are we refering to a Normal Distribution for an estimate of “sigma”? Since most industries don’t have 50 percent defects, I find this blunder in statistical reasoning to be just as disturbing as the one tailed vs. two tailed Normal Distribution debate. I will only comment on the 1.5 “sigma” shift in the calculation of “sigma” as absurd and needlessly confusing.Ultimately, the best metric is yield. You can estimate the confidence limits of the process yield by generating a simple p-chart. I find that you get into trouble any time that you try to calculate the yield from a Cpk or “sigma” value or vice versa. I feel the process yield should be completely divorced from the Cpk or “sigma” values. In my opinion, you should never even try to estimate the yield unless you understand the underlying data distribution. All too often, we are guessing that it is Normally Distributed.Tom

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#67793

Murray
Participant

If you have to worry about the other tail when a process shifts, you are not really in the game anyway. This is picking fly**** out of the pepper. Worry about something important.

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#67799

mcintosh
Participant

Thanks for your viewpoint.
I’ve worked in the aerospace industry.  With the extremely tight tolerances in that field, I was always worried about the other tail.  We could easily get undersized conditions as well as oversized from the same process.
As far as nitpicking, I would have to say that the operational definition of “sigma” is very important.
Tom

0
#67806

Murray
Participant

Let’s see.
Aerospace industry
Tolerances tight
Processes not very capable
Tom wants to debate operational definitions.
Anyone else see the problem with this picture.
Tom, go improve your processes and come debate definitions when you above three sigma. You are not in the game.

0
#67812

mcintosh
Participant

Allen,
I do not appreciate the personal attack.  How can you judge my improvement efforts based only upon a few sentences?  Sometimes, there really is a point in machining where you reach the “state of the art.”
I don’t need to stop my efforts to improve the process to debate operation definitions.  If I see a poorly defined operational definition, I will debate it .

Tom

0
#67816

Ken Myers
Participant

Tom,I believe that Allen is making a reasonable point here, not withstanding his directness. Typically, for well centered processes that are not very capable an estimate from both tails MAY be useful. However, most processes usually deviate greater than about 1/2 sd from the target. A distance of 1/2 sd provides a reasonable bound for practically centered processes. In these cases, the contribution from the low tail, assuming normally distributed data, is a small percentage of the high tail. The take away is we are not trying to provide an accurate estimate of process defect rate, or conversely the yield. To do so would require us to have an understanding of the true distribution of the process data. Instead, we are trying to establish a reasonable measure by which we can track our improvement effort. If you want to provide a reasonable estimate of the defect rate, then use the Poisson distribution on the average number of defects observed over time. I would estimate a 90% upper confidence bound of this point estimate to add some conservatism. I do not believe Six Sigma improvement methods have ever been about making accurate estimates of the defective rate, and/or yield. Instead, they’ve been about using reasonable approximations to base improvement efforts on. Much of the discussion above misses this crucial point. In fact, in his way Allen attempted to make the same point.Ken

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#67843

mcintosh
Participant

Oops, I forgot to mention that in calculating yield, I prefer first pass yield.  By that, I mean that if a part does not meet a tolerance but is reworked, it is still counted as a “defect,” even though it becomes a good part.
I also advocate excluding this part from further manufacturing yield calculations.  In my experience, I have found that a part that has been documented for being out of tolerance early tends to be out of tolerance again later.  That should not come as a surprise, but I felt it should be stated none the less.
As for how a reworked part affect the assembly and/or test yield calculations, I don’t have an opinion to offer.  Any suggestions???
Tom

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#67844

Murray
Participant

Let’s see
Big picture? – Do we “improve” our process by reworking? If this is big picture, I don’t want to see the small ones.
Six Sigma – improve your process by improving your process, not improve your process by reworking an incapable process.
As far as you thinking it was personal — the point is when debating exact DPMO numbers, there is more error in your reporting than there is in tables if people are dumb enoungh to not know how to create them using Excel when you have to worry about both tails. I’ll bet your people don’t know how to center their processes. If they do, tell us how they do it.

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#67847

mcintosh
Participant

Allen,
Your continual chant to improve the process is a gross over-simplification. When you are pushing the limits of the current technology in ID grinding, you will not get a great Cpk or sigma. When you get out of roundness using the best tools available that is one quarter of tolerance, you simply will not get a great Cpk or sigma. We are talking about an ID that is approaching the tolerance of ring gages, which I believe are honed. The only way to “improve the process” will be a paradigm shift in current technology of CNC machining.
Again you oversimplify. Of course, the Machinists center their process by using a SPC chart.
As far as the rounding error using a table vs. an Excel calculation of sigma, your point is well taken. However, using one tail vs. two is still theoretically wrong. This is basic statistics.
Tom

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#67822

melvin
Participant

Tom:
Don’t sweat Allen – he tends to blast everybody…about everything.  I guess he’s a guy with a little bit of smarts that just can’t seem to get people to listen to him, so he resorts to bitterness and “shoving” his opinion down people’s throats.
In any case, I agree that it does tend to get to the point of “beating a dead horse” when some of the more general topics (and their operational definition) are discussed (i.e. sigma level, “stable” process, “opportunity”, et. al.).  But, then again, this is certainly the place to discuss those very topics – in just such detail.
While I agree that the discussion of the operational defintions of “sigma level” and “defects per million opportunities” as performance indices are important to scholars who teach Six Sigma methodology, the most important thing for those of us who use the methodology is the correct use of the tools used to obtain the improved level of performance – whether it is three, four, five, six, seven, ,etc. Sigma.
I think this is basically what Allen was trying to conject – he just has no tact.

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#67828

Ken K.
Participant

I’m the person who started this whole thread.
My point was simple. The DPMO-to-Sigma tables that Mikel Harry originally published many years ago were wrong – plain & simple.
Refering to the tables listed in my original post such as
http://sd.znet.com/~sdsampe/6sigma.htm#conv
If the tables are going to be published with “exact looking” PPM values  (like 308,538 for 2 Sigma, which should really be 308,770) then it should be correct to the dispayed number of significant digits.
For all those who feel 10’s and 100’s of PPM’s don’t really matter at the smaller Sigma levels, I have absolutely no problems with rounding the PPM values, if it is done correctly. In the link mention in my original post
http://www.longacre.cwc.net/sixsigma/sigma_table.htm
The rounded PPM values less than 1.5 Sigma are incorrectly rounded. That darned other tail started to make a difference even in the thousands of PPM.
My point is simple and somewhat academic in nature (as opposed to really making a difference in how, where, or to what extent improvements are made), and certainly I don’t mean to belittle the bizillions of improvement projects that have been done in the name of Six Sigma.
The concern I have is that incorrect tables have multiplied like rabbits. It is just so easy to create correct tables (and correctly rounded tables) that it bugs me to see incorrect tables published by authorities on the subject.

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#67832

howe
Participant

Ken,
Thanks for starting this thread and for your previous post (to which I am replying). I’m not a very theoretical person; much more of a doer. But I do appreciate knowing the limitations of my assumptions, one of which has been the DPMO to sigma conversions that I have done for years. Thanks for clarifying it and raising everyone’s awareness to it.
Cheers,Mike

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#67834

mcintosh
Participant

Thanks for the continued discussion. I completely agree that this has been a good academic discussion.
Like Ken K., I don’t mean to belittle anybody’s six sigma efforts. I fully support everybody’s process improvement efforts. My point is simply that as a Quality Statistician, I am disappointed that the value used to characterize your process, namely “sigma,” is not calculated using a statistically valid methodology. Additionally, there are abuses to calculating “sigma,” such as how opportunities are defined.
Unfortunately, there are many people that abuse the “sigma” value and translate “sigma” to yield. I think that it stems from continually teaching that six sigma translates into 3.4 defects. It has been my experience that people frequently want to know what their “sigma” is and what number of defects (or yield) that translates into being. As we have discussed above, you cannot do that without knowing the underlying distribution.
Lastly, “sigma” can be abused to violate one of Deming’s principles: Eliminate Management by Numbers Alone. Companies or department may compare “sigma” values–an obvious abuse. Earlier in this thread, Allen indicated that anybody with a “sigma” lower than three “needs to get in the game.” This attitude is demeaning, unprofessional, but common. There may be constraints that prevent you from reaching a higher value than three-sigma. For example, do you improve the process by honing all the diameters that don’t have a 1.00 Cpk or three-sigma value? This may not be economically feasible. The cost of this extra processing may outweigh the cost of rework and scrap. You could actually hurt your company by adding cost to the product in this manner. We simply cannot practice quality improvement efforts without looking at “the big picture.”
Thanks for your attention.
Tom

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#67836

Sharyn
Member

Tom
Thanks!! What can we statisticians do to combat this? Any ideas? Sometimes I think it is a loosing battle. Six Sigma is now the latest ‘best practice’. I am fighting but could use some ammunition from ‘outside’. Have any help for me?

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#67839

mcintosh
Participant

Hi Sharyn,
I am not sure exactly what losing battle you are looking to combat. Trying to develop one single value for many different features that have different underlying distributions is tricky at best. I advocate using yield as the best metric to gage process improvements. Perhaps there is a better metric.
I also advocate using the best that six sigma has to offer. The philosophy and tools behind six sigma are proven and excellent. I only have an issue with the way the “sigma” for a process is calculated and the misconceptions associated with “sigma.”
Remember that Six Sigma is a process, and like all processes Six Sigma may be improved. Continue to educate those around you of the positive things. Raise your concerns in a forum like this to question areas that can be improved. Hopefully, you will gain enough support to make those improvements.
Tom

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#67858

Cone
Participant

Interesting reading through this thread.
Using Excel and a simple formula solves the inaccuracy. If people use tables copied from Mikel or any of the other stuff verbatim from his writing, I guess they deserve what anyone deserves who mimics and copies without understanding.
As far as centering the process, I also have my doubts about the ability of the machinists to center their process using SPC. I know a lot of people who claim to use SPC and very, very few who know how to center their process. Most that know how do it in their set up practices, not in their use of SPC. I would be interested in the actual practice of the machinist. I believe many would learn if there is indeed a rational way of centering being used here. Let us know.

Gary

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#67923

mcintosh
Participant

Hi Gary,
You are right about using Excel formulas to help resolve the “sigma” calculation inaccuracies. Unfortunately, I cannot be as indifferent about people who implement six sigma methodology without full understanding. I feel that there are many “Mom and Pop Shops” that will not get the proper training. When it comes to choosing more training globally or making the definition more robust. I personally would choose the latter.
Donald Wheeler is his book “Understanding Statistical Process Control” advocates using an Individual X-Moving Range chart to center the process. In the process described earlier also had trouble with grinding wheel break down. We would lose on average approximately 0.000010 inches per ID. Unfortunately, the grinder won’t recognize an adjustment that small. So we would plot the first (largest) ID on the IX-mR after an adjustment after X number of holes (made as soon as the indexing in the machine would recognize an adjustment). Hope this helps. Maybe somebody else can think of something better.
The point is that six sigma is a worthy goal, but there will be processes where a 1.00 Cpk or three sigma is an achievement in its own right.  To dismiss people that have improved their processes to this level sigma or Cpk as not being “in the game” is an injustice.  I would encourage people to avoid this type of pre-judgment.
Tom

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#114621

JV
Participant

If you have an error rate of 21 defects per 3000 units processed, but the units processed were handled by 3 people before the error was made by person #4.  Can I take 21/3000 and then scale it up to 7000/1000000 or 7000DPMO.  Or is this total misuse of the DPMO concept?

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#114625

Anonymous
Guest

JV,
One of the frustrations of many process engineers working with Six Sigma is that their instructors have not informed them that one of the assumptions of the DPMO ‘model’ is that all defects are ‘point defects.’ In other words, defects are random and independent. If your process does not satisfy this restriction then I should recommend using % defective.
Andy

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#124256

rouble goswami
Member

I have the following observations regarding the veracity of logic posted by the author
Consider the following: S Ls Us LDPMO UDPMO DPMO 6 -7.5 4.5 0.0 3.4 3.4 5 -6.5 3.5 0.0 232.6 233 4 -5.5 2.5 0.0 6209.7 6210 3 -4.5 1.5 3.4 66807.2 66811 2 -3.5 0.5 232.6 308537.5 308770 1 -2.5 -0.5 6209.7 691462.5 697672 Where: S = Sigma Level (assume shift 1.5 sigma to right) Ls = LSL position, # of sigmas to left of mean Us = USL position, # of sigmas to right of mean LDPMO = Prob of observing defect below LSL, in PPM UDPMO = Prob of observing defect above USL, in PPM DPMO = Combined Prob of observing Defect, in PPM Normal probabilities calculated using MINITAB 13.3’s probability function.  In the higher Sigma levels this doesn’t matter since the LDPMO probabilities are so small, but in the lower Sigma levels it makes a difference.

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#124257

rouble goswami
Member

i observe the following fallacies in the author’s claims
Consider the following: S Ls Us LDPMO UDPMO DPMO 6 -7.5 4.5 0.0 3.4 3.4 5 -6.5 3.5 0.0 232.6 233 4 -5.5 2.5 0.0 6209.7 6210 3 -4.5 1.5 3.4 66807.2 66811 2 -3.5 0.5 232.6 308537.5 308770 1 -2.5 -0.5 6209.7 691462.5 697672 Where: S = Sigma Level (assume shift 1.5 sigma to right) Ls = LSL position, # of sigmas to left of mean Us = USL position, # of sigmas to right of mean LDPMO = Prob of observing defect below LSL, in PPM UDPMO = Prob of observing defect above USL, in PPM DPMO = Combined Prob of observing Defect, in PPM Normal probabilities calculated using MINITAB 13.3’s probability function.  In the higher Sigma levels this doesn’t matter since the LDPMO probabilities are so small, but in the lower Sigma levels it makes a difference.

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