Pp and Ppk indices - Is it statistically possible to have a Ppk greater than Pp?

The Taz!

Quite Involved in Discussions
#11
Darius said:
I don't really understand

How can you have Pp if you have a unilateral tolerance?
The equation is: pp = (USL - LSL) / (6 Sigma)

If a software make calculations of pp without one of the specs, IMO is wrong.
:bonk:
My point to start with. . . I'm out of here
 
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D

Darius

#12
Taz, I readed your first poll in this and found that it wasn't necessary to say anything else.
:agree1:

But this topic goes on and on, it's a nonesense, Bev D, sorry Donald Wheeler said that Cp is greater than pp but one exception (bad sampling), nothing about pp being bigger than ppk.

I agree with Taz,
To answer your question, "Not on this planet" . . not even with New Math. . .
As for the ASQ guy, look for another Guru.

:bonk:
 
D

Darius

#13
FYI

I readed an article where it commented that in their example cpk being bigger than the cp, I still don't tink it to be possible but..., I checked on the formulas (because only variances, and indicators appear, no limit, no target, and no original data), and I found that the example was unilateral spec LSL and cp was calculated with te next formula:

cp = (Target - LSL)/(3 Sigma)

I tink that cp MUST NOT BE CALCULATED FOR UNILATERAL SPECS, but the formula in escence looked OK.

and cpk in the traditional sort for unilateral LSL spec

cpk= (Mean - LSL)/(3 Sigma)

I started to tink how to obtain cpk > cp (with that calculations, THAT I AM NOT AGREE), and is clear that is possible if the target is nearer than the mean from the spec limit. So some guys can tink it to be posible.

:bonk:
 

The Taz!

Quite Involved in Discussions
#14
Thanx

Darius,

My point was for Unilateral tolerances only and that there is no Cp or Pp for unilateral tolerances. If it does not exist, how can another index be compared to it.

Thanx for the follow-up.
 
D

Darius

#15
:thanx: Tanks Taz, I tried to show why some guys can tink it posible, as I said in the lasts posts, I agree with you in every point of this topic.

Cp or pp CAN NOT BE CALCULATED FOR UNILATERAL, because the Cp, Cpk, pp, ppk take in account that the specs are centered on the target, so, if you calculate it with

cp = (Target - LSL)/(3 Sigma) or cp = (USL - Target)/(3 Sigma)

you implicate that the other spec limit exists, and there is two possible ways:

1- that the tolerance is just from the target to the limit, desto, the other spec limit is on the target (the target not centered on specs, and so violate the consideration that the target is centered on specs).
2- That the other limit is at the same distance from the "official spec", so it's wrong to calculate the cpk/ppk with only the "official spec".
:lmao:
 
D

Dave Dunn

#16
I think a lot of confusion comes into play when describing unilateral/bilateral tolerances, and describing a dimension that due to its nature is "unilateral/bilateral". Tolerance describes what is "permitted", the physical actuality of the dimension describes what is "possible".

For example:
Widget A is specified at 1.000±.005" This is a dimension that I would consider to be "bilateral", and the tolerance also is bilateral.

Widget B is specified at 1.002+.003/-.007" this is a "bilateral" dimension with an unequally distributed bilateral tolerance.

Widget C is specified at 1.005+0/-.010" This is a bilateral dimension, but a unilateral tolerance. Note however that the resulting tolerance permitted is the same as the first example. Tolerance might be shifted to the lower side due to design intent.

Widget D is specified at 1.000" max. This is a unilateral dimension. It is not "permitted" to be larger than 1" in length, however it is not physically possible for something to be of a size smaller than zero. This type of dimension does not have Cp or Pp calculated, but can have Cpk and Ppk calculated from the maximum tolerance.

What it boils down to is, if it's possible for your measurement to exceed your upper and lower limit, you can calculate Cp/Pp. If one or the other is not possible, then your dimension is one-sided and only Cpk/Ppk apply.
 

Paul F. Jackson

Quite Involved in Discussions
#17
Pp and Cp make little or no sense for examining the potential capability of a unilateral tolerance specification where one limit of the tolerance can only be approached but never exceeded (zero-perfection or infinity). The target for the distribution of a unilateral geometric tolerance (flatness, position, profile, parallelism, perpendicularity) that has a maximum value specified is zero.

There can however be a difference between the potential capability and the current capability of distributions of unilateral geometric tolerances. The measurement data for many of these specifications is a reflection of the size of the tolerance zone required to contain the deviation. Therefore if a tolerance zone for surface profile was given as 1.0 and a point above the basic (or theoretically exact) profile measured +0.25 then it would take a tolerance zone of 0.5 equally displaced from the basic profile + & - 0.25 to contain it.

The data that we analyze for Cpu reflects the value 0.5 as it relates to the specification 1.0. If 30 of the measured points ranged from +.15 to +.35 the capability to the specification would be figured from a distribution with a range of 0.3 to 0.7. The potential capability for that distribution however lies in examining the distribution as if it was centered on the basic profile. A mean shift for the distribution of these points might yield a range of -.1 to +.1 then none of the resultant zones would exceed 0.2 as compared to the specification 1.0.

This "Potential Cpu" an examination of the potential capability can be accomplished with position specifications by monitoring the individual cartesian coordinates and re-figuring the position deviations with candidate mean shifts for X,Y,& Z.

So I say that Cp and Pp make no sense for these types of distributions but you can predict a potential capability for them. Naturally as the distribution approaches the boundary its shape becomes less-normal and one must use the appropriate statistical analysis or transformation to make the prediction.
 
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