2017-08-17 15:18:19
viya
101

These examples were developed to show the relationship of variable control charts and six sigma. They can be used as guidelines for communications between an enterprise and its suppliers.

A variable control chart for PCB surface resistance was created. There is only one minimum specification for resistance. The X bar was 20 megaohms (MH) and the UCL^ 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 average =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 UCLx = 35 = 3, therefore:

s=1

σ = s * Vn = 3

LSL=16 MH

Process average = 20 MH

Cp = Cpk = (LSL - process average)/3a = (20 -16)/3 * 3 = 4/9 = 0.444

z = (SL - average)/a = (16 - 20)/3 = 1.33 OTZ = 3 • Cpk = 1.33 Reject rate =f[-z) = 0.0976 = 91,760 PPM (one-sided rejects only, below LSL)

R= σ•d2 (n = 9) = 3 • 2.97 = 8.91

UCLR = 1.82 • 8.91 = 16.22 MH

LCLR = 0.X8-8.91 = 1.60 MH

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 a remained the same. Describe the Cp and Cpk reject rates and show the X and i? chart limits, using the same sample size of 9. Repeat for a six sigma program, with 1.5 σ shift, with the process average and sigma remaining the same.

Example 3.2b solution

The four sigma program implies a specification limit ofN±4(r = N±4•3 = N ± 12. The process average which is equal to the nominal N，is 4 tr away from the LSL，and is 16 + 12 = 28 MH，given LSL = 16 MH. Cp = Cpk = ± 4 CT/± 3 〇• = 2.33 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 σ did not change. The X chart is centered on X = 28 MH; ICL, = 28 - 3s = 25 MH; UCLX = 31 MH.

For six sigma, the same methodology applies, except that there is a *1.5 σ shift. The specification limits are N ± 6σ~ N ± 6 • 3 = N ± 18.

Given the LSL = 16 MH, the specification nominal N is 16 + 18 = 34 MH. Therefore, Cp = 2; Cpk « 1.5; reject rate from previous tables (±1.5 cr shift) = 3.4 PPM.

Assuming that the shift is toward the lower specification, then the process average could be +4.5 <T from the LSL or —1.5 (T from the nominal: 34 ~ 1.5 ■ 3 = 29.5 MH; or 16 + 4.5 3 = 29.5 MH. '

The R chart remains the same as_Example 3.4a, since the process variability <r did not change. If the X chart is centered on 5 = 29.5 then LCLV = 29.5 - 3 s = 26.5 MH and UCL^ = 32.5 MH. ’

TAG: six sigma control chart

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