# how di i get the UCL LCL

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- This topic has 18 replies, 11 voices, and was last updated 11 years, 11 months ago by Bower Chiel.

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- February 2, 2009 at 4:43 am #51769

colettaParticipant@coletta**Include @coletta in your post and this person will**

be notified via email.Wait time (minutes) in order of occurrence

1

102

173

294

395

556

647

288

69

510

311

3912

4613

3514

3015

616

3217

3318

1119

2020

1321

922

1423

1224

3025

5626

6227

7328

5429

1030

9

I have tried to work it out but cannot. i need to find the UCL FOR THE RANGE AND FOR THE INDIVIDUAL. THE LCL FOR THE INDIVIDUAL. I WANT TO KNOW THE STEPS IN WORKING IT OUT. I GOT AN AVERAGE OF 28.33 AND THE RANGE 700February 2, 2009 at 2:52 pm #180447

MrMHeadParticipant@MrMHead**Include @MrMHead in your post and this person will**

be notified via email.You need to get the USL and LSL from the customer first. Then quantify your process variation, and set your control limits accordingly.

Ex: Customer USL is 1000, your variation is +/- 100, your Control limit should be 900.

This is a very simplistic explanation, but take this idea back to your training material for more detail.0February 2, 2009 at 3:41 pm #180451

KluttzMember@Union-of-Conjoined-Scientists**Include @Union-of-Conjoined-Scientists in your post and this person will**

be notified via email.ummmmm, what?

0February 2, 2009 at 4:03 pm #180454NO mrmhead. NO!(UCL,LCL) and (USL,LSL) are unrelated to eachother: any pairs can be possible and if you know one of them you have ‘no’ info about the second one.

What Production likes is for (UCL,LCL) to be inside (USL, LSL); then the process can perform well compared to what the customer wants.

(UCL,LCL) is calculated from process data to determine what the process considers the area where it will vary inside if there are only common causes happening. The calculation is often automatically done by software and the formula’s used depend on the type of control chart. Sorry Coletta, I don’t know the formula out of my head. Either get them from the help of a software package, from your course material or Google on “Shewhart-rules”.

(USL,LSL) are ‘just a wish’ from the customer: “I want my product like that!” Since the customer generally is considered King you try to delever this to him. And if you can’t you start improvement actions (or decide that losing that customer is okay).

What MrMHead mentions I have seen used before where people: 1>Start with the USL, 2> ‘Determine’ the long-term variation and then 3> calculate what their wished for UCL should be maximally. The same for LCL. And then they start improving their Common Variation until they get UCL and LCL as needed for that customer.

Remi0February 2, 2009 at 4:34 pm #180457Coletta,Wait time are exponentially distributed, although many authors have illustrated wait times on x-bar and R charts.If you’re discerning you might like to read this article:http://www.scipub.org/fulltext/jms2/jms242102-107.pdfI’m not really sure about this as I’ve only used EWMA a few times, and that sometime ago. (I tend to sleep in the afternoons these days :-)

0February 2, 2009 at 5:12 pm #180460Remember, a customer may be a downstream process that MUST have controlled input.

As stated by others, the calculations for these limits are pretty simple and are in many texts. You should see this in 6 Sigma training.

In consideration of real life situations, let me suggest that you look at the inputs that cause the wait time and spend your time controlling them. I.e. If Wait Time is out of control (and I would say that it is from glancing at your data), then what will you do? Look for independent variables/inputs to control.0February 2, 2009 at 5:18 pm #180461Colette,

Since you have sinlge point data I would use the equation for an ImR Chart and plot your data and calcualte your Upper and Lower control limit from your data.

If you do it right you should obtain an UCL of 64.93 and LCL of 8.26

0February 2, 2009 at 5:46 pm #180467UCL – 75.2209

LCL – 1.55501

There are no points out of control – note that you have a right skewed distribution (makes sense since you are naturally bounded on the left).0February 2, 2009 at 6:39 pm #180475

Bower ChielParticipant@Bower-Chiel**Include @Bower-Chiel in your post and this person will**

be notified via email.Hi Coletta, Ron, StanColetta, details of the calculations for the chart limits may be found at http://www.itl.nist.gov/div898/handbook/pmc/section3/pmc322.htm. The handbook is a superb free resource – always an attractive thing for a Scotsman!A quick copy and paste into Minitab yielded 64.93 and -8.26 (negative 8.26) as the UCl and LCL respectively. This throws up an interesting point in that the variable of interest, Wait Time, is one that cannot be negative. Minitab allows you to set a lower bound at zero so that one does not present a silly looking chart.I’d be interested to know how Stan calculated the limits he stated.

Best WishesBower Chiel0February 2, 2009 at 8:42 pm #180480Stan,

Best get out your old test books..you missed the UCL and LCL by a considerable amount.

0February 2, 2009 at 8:48 pm #180482Want to bet on that?

0February 2, 2009 at 8:52 pm #180483Humm, Wait time, I wonder how that is distributed? The reason you got the negative number is you did not consider that

this distribution is skewed right. Skewed right also means the UCL is

going to be further out than you ImR calculations you and everyone

else is using. Use the right distribution and you will get the right

control limits.0February 2, 2009 at 9:43 pm #180485

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

be notified via email.The grasshopper would like to know….

Which transformation did you use?

I used the “optimal” of 0.5 for Box Cox and got …post reverse transformation of 79.5 and 1.01.

0February 3, 2009 at 1:51 am #180493

jane doeParticipant@jane-doe**Include @jane-doe in your post and this person will**

be notified via email.I looked at your data briefly. Since the data is from a rational subgroup it appears is if a lag is present. You can use MINITAB to analyze the lag. With that being done, the correct interval can be used, charted, and the process analyzed. It is typically expected for time interval data to be exponentially distributed. Since your data is not, I would use the Weibull distribution. These data are nearly normal, and the estimates you would in the present form make would be almost useless information at least on the lower end, because it is not possible to get a negative wait time. As the process wait time interval decreases with improvements, data will become more bunched up against the low end (closer to 3 minutes or so) and possibly be exponential. Expect there to be several estimates, over time, as improvements are made.The study of wait times involves queueing theory, and is more complex than what meets the eye.I am not at my computer where I have a stat package for easy, quick analysis. Recall that process averages tend to become normal as the subgroup size increases from 4. With this information, one can use an X bar chart and analyze the standard error to get the UCL & LCL. Analysis of the individuals will yield upper and lower ‘process’ limits.

Hope that helps.0February 3, 2009 at 5:14 pm #180536One of us did the calculations wrong.This must have been a homework problem, it’s not often you will see

a transformation that works out to be a perfect square root.0February 3, 2009 at 5:42 pm #180538Ziggy honey,

Have you figured out how to do this right yet?0February 4, 2009 at 2:12 am #180560Jane,Very impressive sounding but you did not actually say anything.Wow.

0February 4, 2009 at 11:13 am #180574I’m also interested in this topic …

0February 6, 2009 at 7:49 pm #180785

Bower ChielParticipant@Bower-Chiel**Include @Bower-Chiel in your post and this person will**

be notified via email.Hi StanWith the raw data I obtained limits of -8.26 and 64.93, as posted earlier, and a point above the upper limit. The 2 out of 3 and the 4 out of 5 Western Electric Rules also gave signals. Some would argue that therefore you don’t have a stable, predictable scenario and that it is not sensible to think of fitting a distribution.Minitab’s Box-Cox transformation facility returns a rounded value of 0.5. Applying this transformation to the data yields a chart with limits 1.6 and 75.2 (values close to those you quoted)and signals from one of the Western Electric Company Rules – 2 out of 3 consecutive points more than two standard deviations from the centre line, on the same side. The unrounded estimate of 0.39 gave limits of 2.2 and 79.5 and signals. Montgomery reports on work by Nelson advocating the power transformation with exponent 0.2777 for this type of data. This approach yielded limits 2.8 and 85.2 and similar signals.Best Wishes

Bower Chiel0 - AuthorPosts

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