# From DPMO to Sigma Level

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- This topic has 12 replies, 5 voices, and was last updated 3 months ago by Strayer.

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- February 7, 2018 at 11:47 am #55936

Erik 2018Participant@Erik2018**Include @Erik2018 in your post and this person will**

be notified via email.I have a question regarding DPMO. If you calculate DPMO, why would you want to look up the corresponding sigma level? And does it have any meeting as the DPMO can consist of multiple items with (often) non-normal distribution of some of these items.

0February 7, 2018 at 7:59 pm #202254

StrayerParticipant@Straydog**Include @Straydog in your post and this person will**

be notified via email.It’s easier to compare small numbers. More importantly, the sigma level is statistically more meaningful – Look at a bell curve. The differences get finer as you approach the tails and sigma level accounts for that while DPMO doesn’t. For instance the difference between 2 and 3 sigma is 42,851 DPMO. The difference between 3 and 4 is just 2,636. For your second question, if you aren’t looking at a single specification it’s mixing apples and oranges. But it doesn’t really matter as long as you’re accurately counting defects and opportunities and you’re clear about what’s included. The distribution is irrelevant.

0February 8, 2018 at 6:13 am #202256

Chris SeiderParticipant@cseider**Include @cseider in your post and this person will**

be notified via email.If the opportunities are truly the value added ones, then DPMO is a fantastic tool to measure across services, product lines, etc. And since sigma level is well known, it’s great.

ONLY problem is the question has to be asked–did they shift the number or do straight read from Z table. And….some tables are shifted already “being helpful”.

0February 9, 2018 at 11:11 am #202260

Gabriel GómezParticipant@GomezMGab**Include @GomezMGab in your post and this person will**

be notified via email.The Sigma level with DPMO is a approximation to normal distribution from binomial distribution; according with central limit theory, its due to the DPMO measure the number of trials and success (%) in each sample or success (%) in the Critical characteristic of the part. Remember that a part could have many opportunities to evaluate (CTQ’s).

Example : Suppose that we have a process with this characteristic:

Example:

Units/shift = 30,000

Defective parts = 300

Defects observed = 350

Opportunities = 15

Dpu = 350/30,000 = .011

DPMO = (.011/15) x 1,000,000 = 777

Yield : 1-(0.011/15) = 99.9922%

ZBench : 3.164 *From normal distribution tables

Sigma Level: 4.664

CPk Process: 1.550February 14, 2018 at 5:02 am #202272

Erik 2018Participant@Erik2018**Include @Erik2018 in your post and this person will**

be notified via email.@straydog Thanks. Never thought about that. This implies that it is more and more difficult to achieve higher sigma scores, the higher your current score is. Am i right?

@GomezMGab Thanks. But this implies that the transformation from binominal to a normal distribution can only be done when p is not that high, isn’t?

0February 14, 2018 at 6:06 pm #202273

StrayerParticipant@Straydog**Include @Straydog in your post and this person will**

be notified via email.@Erik2018 That’s sort of right. Keep in mind that the closer you get to zero defects the smaller the change necessary to raise sigma level. Improving a few DPMO won’t make much difference if your sigma level is low. But it will the closer you get to 6 sigma or even better where those small DPMO improvements get more and more difficult.

0March 15, 2018 at 8:11 pm #202368

Gabriel GómezParticipant@GomezMGab**Include @GomezMGab in your post and this person will**

be notified via email.you do not need P to perform the calculation, the approximation assumes that if the size of the sample is large enough, it will approach normal …

0March 26, 2018 at 12:38 am #202389

Erik 2018Participant@Erik2018**Include @Erik2018 in your post and this person will**

be notified via email.Thanks! By the way, why is it based on a binominal distribution? Isn’t it about defectives (a part could have many ctq’s/defectives) which corresponds with a Poisson distribution?

0April 1, 2018 at 12:10 pm #202426

Mike CarnellParticipant@Mike-Carnell**Include @Mike-Carnell in your post and this person will**

be notified via email.@Erik2018 You need to remember that there are a lot of metrics that get used and they can all represent the same number. The opportunity count is what is important (as much as I hate to say that because it opens the door for people to play with jacking up opportunity counts rather than fixing things). The Op count reflects complexity. If I was Jack Welch in 1996 and I had GE lighting and GE Aircraft Engines and I am measuring first pass yield then I have that whole apples to oranges issue. If I use DPMO now it is normalized and I can make the comparison. converting it to a sigma value doesn’t change anything it represents but it confuses a lot of people. I only use it with management and only when I have to. When you are dealing with operations people I use DPU. It is a number that makes sense in the context of their jobs.

I wouldn’t waste a lot of time on this. It doesn’t move you any closer to fixing anything.

Just my opinion.

0April 1, 2018 at 4:02 pm #202428

StrayerParticipant@Straydog**Include @Straydog in your post and this person will**

be notified via email.Well said, @Mike-Carnell Pretty much the only thing DPMO or Sigma Level give us is a metric that can be theoretically used for objective comparison between unlike products/processes. And that’s likely to be misleading since we can always miscount or “game” the counts.

0April 3, 2018 at 8:01 am #202433

Mike CarnellParticipant@Mike-Carnell**Include @Mike-Carnell in your post and this person will**

be notified via email.@straydog That has always been the issue with DPMO. If you have defects/opportunities there are 2 ways to play the game 1. reduce defects which is what we are supposed to do 2. Increase opportunities which is the game a lot play particularly management. The whole opportunity counting game goes all the way back into Motorola. Unless a person is completely stupid and you see rework everywhere and they line is supposed to be at 4 sigma or greater then they are playing the denominator game.

0November 13, 2019 at 11:03 am #243478

Erik 2018Participant@Erik2018**Include @Erik2018 in your post and this person will**

be notified via email.One question regarding your example:

Example : Suppose that we have a process with this characteristic:

Example:

Units/shift = 30,000

Defective parts = 300

Defects observed = 350

Opportunities = 15

Dpu = 350/30,000 = .011

DPMO = (.011/15) x 1,000,000 = 777

Yield : 1-(0.011/15) = 99.9922%

ZBench : 3.164 *From normal distribution tables

Sigma Level: 4.664

CPk Process: 1.55If you calculate z-score based on current yield, why do you add 1.5 tot get short term sigma? Isn’t the z-score (based on current defects) the short term sigma and the long term sigma 3.164-1.5 = 1.6164?

0November 15, 2019 at 12:24 am #243510

StrayerParticipant@Straydog**Include @Straydog in your post and this person will**

be notified via email.@Erik2018 The 1.5 shift is controversial. You’ll find much discussion here and elsewhere about it, including whether or not 1.5 is the right value to account for long term drift. Most of us just accept it.

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