2017-08-17 15:18:19
viya
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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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