Modified control limits
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 This topic has 11 replies, 9 voices, and was last updated 19 years, 2 months ago by marklamfu.

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July 17, 2003 at 11:19 am #32816
David NeagleParticipant@DavidNeagle Include @DavidNeagle in your post and this person will
be notified via email.I have a process that appears to be out of control. I say appears because I belive that the problem to be that the variability of the process is considerably smaller than the specification limits. The CP figure is 2.29, which I believe is six sigma process. I have read somewhere, but cannot recall where, that I can use modified control limits on the chart. However, I do not know how to calculate this. Is there anyone who can help?
David
0July 17, 2003 at 12:50 pm #88044
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.If I understand your point correctly, you are making a big conceptual mistake by mixing stability (equality of behavior over the time) with capability (ability to meet specifications). You don’t mention stability, but you say “out of control” what equals “not stable”.
Stability has NOTHING TO DO with specification limits. So your phrase “I have a process that appears to be out of control. I say appears because I believe that the problem to be that the variability of the process is considerably smaller than the specification limits” just has no meaning.
This subject was discussed several times in this forum before. Check these messages for more clarification.
https://www.isixsigma.com/topic/controlchart3sigmacontrollimits/
https://www.isixsigma.com/topic/whatisastableprocess/
You can also do a forum search.
0July 24, 2003 at 9:17 pm #88310
Omar EsquivelParticipant@OmarEsquivel Include @OmarEsquivel in your post and this person will
be notified via email.If your CP value is as you mentioned higher than 2, than means that your process variability is much less than the variability allowed by the specs, however, since you said that your process is “out of control”, what should be happening is that either your process / machine is not calibrated, needs adjustments or probably is not the right one to meet the specs. This can be seen on your Cpk calculation. This support the previous idea or stability against variability.
You can also be calculating incorrectly the Cp value. What is the value that you have for the Cpk?
0July 24, 2003 at 9:33 pm #88311
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.Omar: How do you relate “Cp>2 and out of control” with “process / machine is not calibrated, needs adjustments or probably is not the right one to meet the specs” Note the David never mentioned that he found a single part to be out of specification.
0July 25, 2003 at 12:37 am #88315What is the purpose of your question?
A control chart is used for a reason. If your process never is going to make defects why do you need to spend resources on the charting process?
Is the variability in your process costing your business money? Eg The length of a tent peg is not very critical but every mm over the nominal length is a waste of material.
If the varibility is costing you money why not use the usual method of control limits to help eliminate waste? Otherwise it may be better not to waste resources on control charting at all.
0July 25, 2003 at 7:21 am #88323
David NeagleParticipant@DavidNeagle Include @DavidNeagle in your post and this person will
be notified via email.I have to admit to being very new to this. We, as a company are moving into the automotive industry. We are going for TS16949 accreditation and part of this is the monitoring and measurement of processes and product. We have not used control charting before but this will now be necessary. As a precursor to this I have been charged with the responsibility of setting up a control chart program. I have taken the most critical features with regard to the function of the product and used those to start with. I have found one or two processes that show very little variation, but have points out of control. As an example, a process has an USL of 10.2mm and a LSL of 10.0mm. The standard deviation is 0.01mm, therefore the distance USL – LSL is 20 sigma, much greater than the 6 sigma natural tolerance limits on the process. As I understand it, this is a six sigma process. As I said in my original post, I have read somewhere about using modified limits which caters for this type of situation, but was unable to locate the text, hence my question. If I have got it wrong, then perhaps you would be kind enough to set me on the right path. As I have said, I am very new to this and wish to learn.
David
0July 25, 2003 at 11:47 am #88332
GabrielParticipant@Gabriel Include @Gabriel in your post and this person will
be notified via email.Welcome to the SPC world. Do you have the AIAG’s SPC handbook? It’s a good start.
I don’t understand why you want to use modified control limits. Control charts are used to detect special causes, i.e. unstability. If you use wider limits you will be just cheating yourself by masking special causes of variation. If you want to do that, just don’t chart the process! This could be the right choice for a process where 20 sigmas fit inside the specification. Remember that the control limits are the voice of the process, and they have nothing to do with specification. Your chart will have the same control limits if your specification is 10 to 10.2mm, 10 to 10.02mm, 10 to 12mm, or even if there is no specification at all.
To better understand your problem, please tell me:
– What kind of SPC chart are you using? Subgroup size? How many points before calculating control limits?
– When you say “have points out of control”, what do you mean exactly? Which kind of outofcontrol signal are you detecting?
– What feature are you measuring? What kind of instrument are you using? Which is the resolution? Have you made an r&R?0July 25, 2003 at 12:12 pm #88333
Rick PastorMember@RickPastor Include @RickPastor in your post and this person will
be notified via email.David:
You maybe confusing three points: Cp and Cpk, control charts, and six sigma.
Cp and Cpk: Let me give an example regarding Cp and Cpk. You have a process whose mean is 5.03, standard deviation, s, is 0.15,USL=10, and LSL=5. From the example below, you can have a great Cp and a poor Cpk. What this mean is that your process has very little variation and therefore has the potential of being a very good process because the Cp is 5.6. On the other hand, the Cpk is 1.11 that means that the process is not centered very well. Therefore you are probably producing lots of bad parts. In this example, all you have to do is shift the mean and that is typically easier than reducing variation.
Example:
Cp is calculated by
Cp=(USL – LSL)/6s where Cp=5.6
Cpk is calculated from
Cpk = min value of CPL and CPU = 1.11
Where CPL=(meanLSL)/3s = 1.11 and CPU=(USLmean)/3s =10
As you can see, you can have a great Cp but a poor Cpk.
Even if The Cp and Cpk do not mean that you have a
Control Charts: Control charts are really different. Even with the above example you could have perfectly goodlooking control charts. Control chart calculate lower and upper control limits (LCL, UCL) based upon your standard deviations. If the process is stable, you could still be producing bad parts because of the statistical fluctuations in the machining operation.
Six Sigma: You probably do not have a six sigma process. Although, the data provided cannot be used to determine that fact. Six sigma requires knowledge of the number of defects generated and the number of defect opportunities. See the six sigma calculator at the top of the web page.0July 25, 2003 at 12:17 pm #88334Dear David,
I too am new to SS, therefore may not be able to help you with the control limits.
However some thought on the sigma calculation – I think there is a bit of an error in the way you have calculated your process to be 20 sigma.
First we do not know where the mean lies – for an ideal situation, let us assume it lies in the centre of the given specs i.e. 10.1 mm.
Next we work out Z(lsl) and Z (usl) which is 10sigma in either case.
Therefore DPMO (usl and also lsl) would be very close to zero (less than one in a billion).
Assuming total DPMO would also be close to zero, we can say Z Bench (short term) would be in the range of just under 11.5 sigma.
If this is true you’re very lucky to be in control of such a process. The whole rationale, of course, depends on your process output following normal distribution.
Bee.
P.S. Anybody out there would like to correct the sigma calculations??0July 25, 2003 at 1:46 pm #88336You are right, I missunderstood David message, what I understood is that he had a Cp>2 process but out of specs.
0July 26, 2003 at 9:09 pm #88381
Bullet BoyParticipant@BulletBoy Include @BulletBoy in your post and this person will
be notified via email.Dave,
It is not possible to equate a Cp value to a Sigma value as Cp is a measure of process potential. It is however, possible to equate a Cpk value to a sigma value, as this takes into account porocess location. A Cpk value of 2 is a 6 sigma process (to convert Cpk to sigma multiply by 3).
Although you could drive a bus through your spec, and your process variability is low relative to it, your process can still most certainly be out of control.
A process is out of control if it violates one or more of the out of control guidlines as defined in AIAG (1 point more than 3sigma from the centre line etc) regardless of its Cpk. Furthermore, a processes Cpk is questionable if the process is not in control, as the variation will consist of both common & special cause variation. Special cause variation is not predictable. Therefore, if a characteristic is deemed to be significant/ critical from an FMEA, and SPC is being used to control, then the process must be brought into control before capability is established and remain in control before characteristic is deleted & SPC ceased.
One reason your process could be out of control, is because it has improved! Consider recalculating your control limits using latest data, and replotting your points you may find that this brings your process under control.
Keep on Truckin’0July 27, 2003 at 1:39 am #88383
marklamfuParticipant@marklamfu Include @marklamfu in your post and this person will
be notified via email.The high CP(2.29) means the process variation is small (the Spec. have cover about 13.8 std.), for 6 sigma process, only CP>2.0 is not enough, at same time, the CPK >1.5( with 1.5 sigma mean shift) and DPPM<=3.4ppm.
Th cause of CP high but process out of control is big mean shift. the fianl index to measure process capability is CPK not CP, CPK =(1K)CP, K is mean shift factor, the more of mean shift , the lower of CPK value .
A few of reminders for your reference:
1. to recheck& confirm whether the process ture average is OK.
2. The setting/location of process machine or tool, the process seem have systemic problem. i suggest you to find the cause of mean shift prior to set control chart.0 
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