Interpretation of Cp and Cpk

A

alpina850

#1
Dear all,

Is there anyone could help me whether the below statement is correct or not:

Average shrinkage = 0.26%
Standard Deviation (σ)of shrinkage = 0.024%
Dimension of molded part = 21.03mm
Considering of only changes of molded shrinkage caused by material,
and if we think it’s within the control (it means Cpk1.33(±4σ)
the changes of dimension should be within the figure as follows:

21.03X8X0.024 = 0.04mm

It means that even if the dimension changes according to the changes ±4σ of molded shrinkage caused by material itself, the changes of dimension should be within the tolerance of ±0.05mm

Could you please share with me of why it is correct or incorrect.

Thank you very much
 
Elsmar Forum Sponsor
#2
I am not sure I am getting this right: Assuming that we start at 21.03mm, the average shrinkage 0.26% would take us down to 20.98, which would put the average part on the lower tolerance limit... ...or is it 21.03mm after shrinking?

/Claes
 

Celtic Warrior

Involved In Discussions
#3
I am not sure I get it either...
If the target dimension is 21.03, taking account of the average shrinkage would take it to 20.98, however in worst case the 4x SD should be applied also (4x0.024) to allow for the maximum possible shrinkage, result could be 20.884 after shrinking.:confused:
 

Miner

Forum Moderator
Staff member
Admin
#5
It is not only a shift in the average. Your final variatiation will also increase. You started with a certain variation in product dimensions. Since not all product shrinks by the same amount this variation builds upon your starting variation.

You could normally add variances (StDev^2), but in your situation, I would use a Monte Carlo analysis to determine the increase.

You can perform simple MC analyses in Excel without an add-in. Read this website: An Excel Tutorial: An Introduction to Excel's Normal Distribution Functions. Towards the end, it shows how to combine the Rand() function with the normal distribution functions to generate random data from a normal distribution.

You can then generate random data columns for the part and the shrinkage. Then multiply the part column by (1-shrinkage) column to generate the after shrinkage dimensions.
 
A

alpina850

#6
Sorry for all, I think I need to re-phrase again the statement above, please ignore my previous post:-

1) Average of material shrinkage is 0.26%
2) The standard deviation of material shrinkage is 0.024%

*Remark: Please ignore the group of data which to generate the said average and standard deviation of material shrinkage.

If we assume that the Cpk for the material shrinkage is 1.33. Thus, could we say that the material shrinkage should be shrunk within ±0.05mm from the dimension of 21.03mm.

Besides of this, could we use the below formula? or how do we describe below formula?

Formula : 21.03X8X0.024 = 0.04mm

Sorry again to all of you for the misguiding before.

Thank you
 
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