How Many Expected Defective Assemblies?
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 This topic has 14 replies, 5 voices, and was last updated 16 years, 5 months ago by AB.

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March 3, 2006 at 10:40 am #42594
M GreenawayParticipant@MGreenaway Include @MGreenaway in your post and this person will
be notified via email.Hi AllI am having to kinda justify why things go wrong, and why sadly on occasion a product is shipped that does not work. I would like to approach this discussion based on hard statistical methodology, and what is generally considered ‘best in class’ performance based on the SIx Sigma approach.So lets assume my finished product has 200 opportunities in its manufacturing process for something to go wrong. The product is not very complex, having about 10 different components, but each is manufactured under its own process, which will have its own process capability at various stages of production, and these will then be assembled, which will also have its own potential flaws and resultant capability. Any of these 200 potential problems would result in a final product being considered ‘defective’ (although not safety critical).Hence if each of these processes operates at a six sigma level of performance for each of the 200 opportunities for error, how many defective final assemblies should I expect ?
0March 3, 2006 at 11:19 am #134582
Adrian P. SmithParticipant@AdrianP.Smith Include @AdrianP.Smith in your post and this person will
be notified via email.To make the math slightly easier, let’s make the following assumptions:
You have 10 processes to produce the components, which all operate at 99% yield. You also have a final assembly process which also operates at 99% yield (not Six Sigma of course, more like 4).
It seems to me that the 200 opportunities are not relevant for this. We just assume that the 10 processes are under control and consistently produce good components 99% of the time.
So, when we are about to assemble a product, we select a component from the first process and have a 99% chance of selecting one which is defect free. We then select a component from the second process and again have a 99% chance of selecting one which is defect free.
At this point, the probability of having 2 defect free components is 0,99 x 0,99 = 98,01%.
You can see from this that as we continue to select components from the remaining 8 processes, we always have a 99% chance of picking a good one, and that our overall probability of having no defective components is the probability of having zero defects at the previous step multiplied by 99%.
So, once we have picked our 10 components we have a 0,99 to the power 10 = 90,4% chance of having all good components.
Now as we go through assembly, assuming we have all good components, the assembly only has a 99% chance of not introducing any new defects.
So the overall probability of having a defect free product is again 0,904 x 0,99 = 89,5%
So despite that fact that our individual processes are running at 99% yield, we will nevertheless see 10,5% defective product coming out of the final assembly process.
(We could have got the same answer of course by just saying we have 11 steps and doing 0,99 to the power 11.)
Clearly if all the processes are operating at Six Sigma, the yield for each would be 99,99966% and therefore the overall yield of good product would be (99,99966 to the power 11) 99,99626%.0March 3, 2006 at 11:28 am #134583
M GreenawayParticipant@MGreenaway Include @MGreenaway in your post and this person will
be notified via email.Thanks Adrian
That basic statistics I can follow very easily, however there is a slight complication I think. OK so are 10 processes have a 99% yield, as such an assembly containing a defect from process 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9 or 10 is .01*10, yes
But also our assembly could contain a defect from process 1 AND 2, or 2 AND 3, or 3 AND 4, etc, etc
Also it could contain defects from processes 1 AND 2 AND 3, or 2 AND 3 AND 4, etc, etc, etc
How do I also include these probabilities ?0March 3, 2006 at 11:41 am #134584
Mark HarlandParticipant@MarkHarland Include @MarkHarland in your post and this person will
be notified via email.If I’m understanding your question correctly, you have 10, 6 Sigma processes running concurrently. (The opporyunities would already have been taken into account in reaching the Sigma rating). These combine into a final assembly process.
At 6 Sigma you have a yield score of 99.99966% efficient, and so if you calculate the rolled yeild (99.9966% to the power of 11) you have a rolled sigma yield of 99.99626%. This reads across to a DPMO of 5.4 or 48 Defects per Million Oppprtunities.
So if you know the number of units you are likely to produce you can calculate how many will be defects you can expect. The calculation would be Number of Units divided by 1,000,000 and then multiplied by 48.
Hope this helps
Regards
Mark Harland
0March 3, 2006 at 11:54 am #134585
Adrian P. SmithParticipant@AdrianP.Smith Include @AdrianP.Smith in your post and this person will
be notified via email.Those probabilities are already included.
Take the situation when we have selected 2 components. As we said the probability of having no defectives is 98,01%.
The other possibilities are:
Component from process 1 is defective (1% chance), component from process 2 is good. Probability of this outcome is 0,01 x 0,99 = 0,99%
Component from process 2 is good, component from process 2 is defective. Probability = 0,99% (same as first case).
Both components are defective. Probability = 0,01 x 0,01 = 0,01%
0,99% + 0,99% + 0,01% = 1,99% chance of having AT LEAST one defect.
100% – 1,99% = 98,01%, or the chance of having zero defects (same as we calculated before).
This continues for the whole 11 steps. By calculating the probability of getting a perfect product, the probability which is left is the probability of having 1 OR MORE defects in any possible combination.
Hope this makes sense.
0March 3, 2006 at 12:05 pm #134586
Mark HarlandParticipant@MarkHarland Include @MarkHarland in your post and this person will
be notified via email.Further to your reply to another respondant, and asking the possibility of considering multiple errors appearing on the same unit from different processes.
At 6 sigma the process produces 3.4 defects per million opportunities and so the probability of an output unit having errors from two different processes is 0.0000034 multiplyed by 0.0000034 or 0.00000000001186. The probability of getting an output unit with a defect from 3 diffrerent processes is 0.000000000000000000003558, in other words so small as to not be worth considering. By the time you get to the combinations on an output unit containing a defect from all 10 processes plus your final assembly process, your computer would explode.
Any way the upshot is that the whole sum total of these possibilities would add up to way less than 1 chance in a million. In theory it could happen but quite frankly you have more chance of winning the lottery.
0March 3, 2006 at 12:23 pm #134587
M GreenawayParticipant@MGreenaway Include @MGreenaway in your post and this person will
be notified via email.True
But my processes are probably more like 3 or 4 sigma, which could be as much as approx 68000 defects in a million, hence the AND calculation becomes more significant does it not ?0March 3, 2006 at 1:11 pm #134590
Mark HarlandParticipant@MarkHarland Include @MarkHarland in your post and this person will
be notified via email.If all your processes are 4 Sigma your rolled yield would be 93.38% or 66807 DPMO, if all your processes are running at 3 Sigma the rolled yield would be 46.63% or 539,828 DPMO.
As stated by Adrian earlier, the chance of an output unit being a result of defects in two processes is taken into account in the rolled yield. I was showing you the calculations behind this.
Do you know the Sigma rating for each of your 11 processes or is this hypothetical at the moment?0March 3, 2006 at 1:41 pm #134593
M GreenawayParticipant@MGreenaway Include @MGreenaway in your post and this person will
be notified via email.This is purely hypothetical at the moment. I guess I am trying to demonstrate that the more opportunity for error that exists in the component processes, the more chance you have of a defective final product. This is kinda obvious but I want to show this with some sound statistical argument. I want to show ‘best in class’ i.e. 6 sigma processes, and ‘normal class’ if you like, i.e. 3 and 4 sigma processes.
0March 3, 2006 at 1:51 pm #134595
Adrian P. SmithParticipant@AdrianP.Smith Include @AdrianP.Smith in your post and this person will
be notified via email.It might help if you let us know the purpose of your demonstration.
Are you trying to convince someone to launch Six Sigma? To buy into it? Or that every process on earth should be at Six Sigma?
From my point of view, this is not a “statistical” argument. It is basic permutation and probability theory.
Or are you asking for the statistical demonstration that 6 sigma is better than 3 or 4?0March 3, 2006 at 3:11 pm #134608
M GreenawayParticipant@MGreenaway Include @MGreenaway in your post and this person will
be notified via email.Well I can only repeat my previous posts, I am having to argue if you like that things go wrong – or if you like that a certain level of problems will occur.
I also want to show that even for world class excellent processes problems still occur. That is the basic argument, and to illustrate this I want to use statistics, and where necessary quote authoratative sources / methodology, such as Six Sigma, this website and various puiblications.
I also really want to demonstrate that near excellent subprocesses will combine to produce a lower yield than each has in isolation when they combine into an assembly.0March 3, 2006 at 4:04 pm #134614It’s sad that these folks don’t know how to answer your basic question.
200 opportunities all running at 6 sigma would give you 680 PPM defective assemblies. So 680 out of 1,000,000 assemblies would have defects.0March 3, 2006 at 4:04 pm #134615
Adrian P. SmithParticipant@AdrianP.Smith Include @AdrianP.Smith in your post and this person will
be notified via email.OK
Your purpose might be better served by analogy than by restating a proof of probability theory.
“Even for world class excellent processes problems still occur.”
Analogy – planes crash. The process is way above 6 sigma at about 1 DPMO – they are most definitely world class. However, we all know that planes crash from time to time.
“Near excellent processes will combine to produce a lower yield than each has in isolation.”
Again, this is basic probability, and it doesn’t just apply to near excellent processes. Any sequence of processes will inevitably produce lower yield than each step in isolation.
Take the analogy of rolling a sixsided dice.
Assume that rolling a six is “good” and anything else is a defect.
When you roll one dice, you have a 1 in 6 chance of getting a six.
Roll 2 and you only have a 1 in 36 chance of getting 2 sixes.
This should help understand the idea of cumulative probability, and then it shouldn’t be too much of a leap to state that anything less than perfection when run in series with other less than perfect things will inevitably produce a lower overall yield.
Who are your audience that they have such high demands of proof??0March 3, 2006 at 4:12 pm #134616
Adrian P. SmithParticipant@AdrianP.Smith Include @AdrianP.Smith in your post and this person will
be notified via email.Thanks Stan
Could you explain your logic a bit more?
Seems to me that even if what I think your logic is were correct, you would have 680 defects and not neccessarily 680 defective units.
You could have 3 units with 200 defects each or 680 units with 1 defect each.
??0March 3, 2006 at 4:52 pm #134617I think you may benefit from articulating the question as “What is the probability that your final product has at least one defect given:
a) Individual probabilities of each component to be defective
b) The probability of introducing an error during the assembly process”
Asy ou add one high quality component to another high quality component with a high quality process, the overall quality is not better than the individuals but rather as good as “the weakest link” at best and as bad as the “combined weakness” at worst.
The concept of cumulative probability of a set of events is probably is the best explanation for this.
Hope this helps
AB0 
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