## Erik 2018

@erik2018Member since February 7, 2018

was active active 2 months, 2 weeks ago## Forum Replies Created

## Forum Replies Created

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- March 24, 2020 at 3:20 am #246796

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

be notified via email.@Mike-Carnell: sorry forgot your name

0March 24, 2020 at 3:20 am #246795

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

be notified via email.@Straydog: sorry posted a wrong name.

0March 23, 2020 at 11:25 am #246786

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

be notified via email.@Strayer. Okay but is DPMO always long term? Does not it depend on your sample size? Otherwise both DPMO and sigma level are short term and no 1.5 conversion have to be carried out

0March 23, 2020 at 11:23 am #246785

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

be notified via email.Looking back at these answers I am still interested why it is not poisson instead of binominal?

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?

0April 10, 2019 at 2:52 am #238350

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

be notified via email.Thanks for all your replies. At the moment we use a control chart just for analysis of a few weeks of production. Based on an histogram and a control chart we could determine out of control situations, variety within the control limits and track both back to certain X-s. You’re right that using output just for control without analysis is useless, even more when the desired output changes each hour. However, in this case it is the output on the bottleneck, which needs to produce a certain amount of products each hour. An hour lost on the bottleneck is an hour lost for the company (that’s what Goldratt learned me). In that case, I was interested in which control chart to use.

Speaking of tracking X-s, what do you mean exactly if I may ask. E.g. we can regard cycle time as the dependent variable (amount of minutes per hour/desired output per hour), which is maybe a more useful performance indicator than output. What I was wondering, what do you regard as tracking X-s? Or do you mean tracking a subdivision of the Y. E.g. I can imagine that we track downtime as it could be an important part of the cycle time of a product. However, this still isn’t an X, but part of the Y (the same as waiting time is not an X but part of throughput time (throughput time = waiting time + production time + …). What would you track in those cases e.g. in a run/control chart?

0March 27, 2019 at 6:38 am #237590

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

be notified via email.Thanks for all your responses. Indeed, it depends maybe on your aim of making a SIPOC (and luckily no jail if you do it otherwise:)). In our case we have students carrying out a LSS green belt assignment in practice. As they are unfamiliar with the process to improve at the start of their internship, we think it is very useful to have a SIPOC like described before. If later on in the analysis phase they focus on part of the processSIPOC, they can easily find the inputs of that specific phase + the direct suppliersof these inputs. If you have the inputs summarized for alle the process steps, you need to filter them afterwards.

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?

0February 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?0 - AuthorPosts