Page 114 - Six Sigma for electronics design and manufacturing

20 megaohms (MH) and the UCL x was 23 MH, with a sample size of 9.
A new specification was adopted to keep resistance at a minimum of
16 MH. Assuming that the resistance measurement or process aver-
age = specification nominal (N), describe the Cp and Cpk reject rates
and show the R chart limits.
Example 3.2a solution
Since the process is centered, Cp = Cpk. The distance from the X to
UCL x = 3 s = 3, therefore:
s = 1 Six Sigma and Manufacturing Control Systems – – 83
= s · n = 3
LSL = 16 MH
Process average = 20 MH
Cp = Cpk = (LSL – process average)/3 = (20 – 16)/3 · 3 = 4/9 =
0.444
z = (SL – average)/ = (16 – 20)/3 = 1.33 or z = 3 · Cpk = 1.33
Reject rate = f(–z) = 0.0976 = 91,760 PPM (one-sided rejects only,
below LSL)
R = · d 2 (n = 9) = 3 · 2.97 = 8.91
UCL R = 1.82 · 8.91 = 16.22 MH
LCL R = 0.18 · 8.91 = 1.60 MH
Example 3.2b
A four sigma program was introduced at the company in Example
3.2a. For the surface resistance process, the lower specification limit
(LSL) remained at 16 MH and the process remained the same. De-
scribe the Cp and Cpk reject rates and show the X and R chart limits,
using the same sample size of 9. Repeat for a six sigma program, with
1.5 shift, with the process average and sigma remaining the same.

Example 3.2b solution
The four sigma program implies a specification limit of N ± 4 = N ±
– –
4·3 = N ± 12. The process average (X), which is equal to the nominal
N, is 4 away from the LSL, and is 16 + 12 = 28 MH, given LSL = 16
MH. Cp = Cpk = ± 4 /± 3 = 1.33 and two-sided reject rate from the
z table (Table 2.3) = 64 PPM.
The R chart remains the same as Example 3.4a, since the process
variability did not change. The X chart is centered on X = 28 MH;
LCL x = 28 – 3s = 25 MH; UCL x = 31 MH.
For six sigma, the same methodology applies, except that there is a
±1.5 shift. The specification limits are N ± 6 = N ± 6 · 3 = N ± 18.