Cpk relation to Reliability

[d.conroy]

Hi, Im a little confused can anyone help.

If I need to maintain a 95% reliability on parts dose that mean I need to maintain a true 0.66Cpk.

I'm thinking on a normal distribution curve 2 sigma away from the mean will contain 95% of values.

So then as long as my Cpk is greater than 0.66 (2 sigma away from the limit) the process will meet the 95% reliability requirement.

I know there are confidence intervals on the Cpk also, so the indicated Cpk from a sample will need to be greater than 0.66 to have any confidence of meeting the reliability.

Am I correct?

 

[Bev D]

the term reliability has several different meanings...which meaning are you using?

my assumption at this point is that you must prove a 95% conformance (1-defect rate) with XX% confidence. If your usage is different then our answer will be different....

 

[Darius]

Hi, Im a little confused can anyone help.

If I need to maintain a 95% reliability on parts dose that mean I need to maintain a true 0.66Cpk.

You are really confused, I think what you want is the lower limit of the Cpk estimation to be 0.66 at 95% of confidence.
note added: when you obtain the value of Cpk from a data set, the value is just an estimate for such sample, and there are some formulas to determine the range in wich the real value (for such sample) must lay, so it's a minimum and a maximum value for such interval where the real value must be, such interval is calculated in function with the %confidence specified for such calculus.

To answer this question, sample size is needed, and you didn't provide such information, so I will give you the way to do it, but you will need to add VBA code in Excel. It will give the value of Cpk needed to obtain a minimum value at 95% of 0.66

Function CpkNotLessThan(Cpk_Min As Single, Conf100 As Byte, Sample_Size) As Single
Dim N As Long, BIG as boolean, Cpk_Obj as Single, Delta as Single
  Cpk_Obj = Cpk_Min: Delta = 0.5: BIG = False
  Do
    DIF = Cpk_Min - (Cpk_Obj - Application.WorksheetFunction.NormSInv((1 - (1 - Conf100 / 100) / 2)) * (1 / (9 * Sample_Size) + (Cpk_Obj) ^ 2 / (2 * Sample_Size - 2)) ^ 0.5)
    If Abs(DIF) < 0.0001 Then GoTo 1
    If DIF > 0 Then
      If BIG Then Delta = Delta / 2
      BIG = False
      Cpk_Obj = Cpk_Obj + Delta
    Else
      If Not (BIG) Then Delta = Delta / 2
      BIG = True
      Cpk_Obj = Cpk_Obj - Delta
    End If
  Loop Until False
1: CpkNotLessThan = Cpk_Obj
End Function

after writing the code and enabling VBA Macros, in an Excel Sheet, you will need to write =CpkNotLessThan(0.66,95,50), for sample size of 50.

 

[Darius]

Well, I think maybe my point was not clear enought, but you can read it in Size Matters by Mark L. Crossley, an excellent article.

I must add that in my previous post, I wrote the formula:

DIF = Cpk_Min - (Cpk_Obj - Application.WorksheetFunction.NormSInv((1 - (1 - Conf100 / 100) / 2)) * (1 / (9 * Sample_Size) + (Cpk_Obj) ^ 2 / (2 * Sample_Size - 2)) ^ 0.5)

And must be (because we only care about the lower bound)

DIF = Cpk_Min - (Cpk_Obj - Application.WorksheetFunction.NormSInv(Conf100 / 100) * (1 / (9 * Sample_Size) + (Cpk_Obj) ^ 2 / (2 * Sample_Size - 2)) ^ 0.5)

That macro obtain the puntual Cpk if you specify the "not less than ..." value, my appologies if I was not clear.

 

[d.conroy]

Thank you very much Darius.

Your answer was the same as what I was thinking (A little easier to understand though)
The Excel sheet works well too, this gave results to that agreed with the calculator I created.

Regards