ASSIGNMENT 7
ASSIGNMENT #7
1.
a. Estimate the potential capability σave = Save / C4 = 1.05 / 0.9400 = σave = 1.118
Cp(ave) = (USL – LSL) / 6*σave = (105 – 85) / 6*1.118 = 20 / 6.708 = Cp(ave) = 2.98
b. Estimate the actual capability
Cpk(ave) = min(Cpl(ave), Cpu(ave))
Cpl(ave) = (µ - LSL) / 3σ = 100-85 / 3*1.118 = 4.47
Cpu(ave) = (USL - µ) / 3σ = 105-100 / 3*1.118 = 1.49
Cpk(ave) = min(Cpl(ave), Cpu(ave)) = min(4.47, 1.49) = Cpk(ave) = 1.49
c. How much could the fallout in the process be reduced be reduced if the process were corrected to operate at the nominal specification?
Zpotential = (LSL - µave) / σave = (85-100) / 1.118 = -13.4
Pave(potential) = φ(Z) = φ(-13.4) = 0
Zactual = (LSL - µave) / 3*σave = (85-100) / 3*1.118 = -4.47
Pave(actual) = φ(Z) = φ(-4.47) = 0.00004
Thus, if the process were corrected to operate at the nominal requirement, the process would be reduced to 0 from 0.00004.
2.
a. Estimate the potential capability of the process. σave = Rave/d2 = 3.5/2.059 = σave = 1.699
Cp(ave) = USL-LSL / 6*σave = 208-192 / 6*1.699 = Cp(ave) = 1.569
b. Estimate the actual process capability.
Cpk(ave) = min(Cpl(ave), Cpu(ave))
Cpl(ave) = (µ - LSL) / 3σ = 199-192 / 3*1.699 = 1.373
Cpu(ave) = (USL - µ) / 3σ = 208-199 / 3*1.699 = 1.766
Cpk(ave) = min(Cpl(ave), Cpu(ave)) = min(1.373, 1.766) = Cpk(ave) = 1.373
c. How much improvement could be made in the process performance if the man could be centered at the nominal value?
Z = LSL-µave / σave = 192-199 / 1.699 = Z = -4.12
Pave(potential) = φ(Z) = φ(-4.12) = 0.00004
Zpotential = (LSL - µave) / σave = (192-199) / 1.699 = -4.12
Pave(potential) = φ(Z) = φ(-4.12) = 0.00004
Zactual = (LSL - µave) / 3*σave = (192-199) /
1.
a. Estimate the potential capability σave = Save / C4 = 1.05 / 0.9400 = σave = 1.118
Cp(ave) = (USL – LSL) / 6*σave = (105 – 85) / 6*1.118 = 20 / 6.708 = Cp(ave) = 2.98
b. Estimate the actual capability
Cpk(ave) = min(Cpl(ave), Cpu(ave))
Cpl(ave) = (µ - LSL) / 3σ = 100-85 / 3*1.118 = 4.47
Cpu(ave) = (USL - µ) / 3σ = 105-100 / 3*1.118 = 1.49
Cpk(ave) = min(Cpl(ave), Cpu(ave)) = min(4.47, 1.49) = Cpk(ave) = 1.49
c. How much could the fallout in the process be reduced be reduced if the process were corrected to operate at the nominal specification?
Zpotential = (LSL - µave) / σave = (85-100) / 1.118 = -13.4
Pave(potential) = φ(Z) = φ(-13.4) = 0
Zactual = (LSL - µave) / 3*σave = (85-100) / 3*1.118 = -4.47
Pave(actual) = φ(Z) = φ(-4.47) = 0.00004
Thus, if the process were corrected to operate at the nominal requirement, the process would be reduced to 0 from 0.00004.
2.
a. Estimate the potential capability of the process. σave = Rave/d2 = 3.5/2.059 = σave = 1.699
Cp(ave) = USL-LSL / 6*σave = 208-192 / 6*1.699 = Cp(ave) = 1.569
b. Estimate the actual process capability.
Cpk(ave) = min(Cpl(ave), Cpu(ave))
Cpl(ave) = (µ - LSL) / 3σ = 199-192 / 3*1.699 = 1.373
Cpu(ave) = (USL - µ) / 3σ = 208-199 / 3*1.699 = 1.766
Cpk(ave) = min(Cpl(ave), Cpu(ave)) = min(1.373, 1.766) = Cpk(ave) = 1.373
c. How much improvement could be made in the process performance if the man could be centered at the nominal value?
Z = LSL-µave / σave = 192-199 / 1.699 = Z = -4.12
Pave(potential) = φ(Z) = φ(-4.12) = 0.00004
Zpotential = (LSL - µave) / σave = (192-199) / 1.699 = -4.12
Pave(potential) = φ(Z) = φ(-4.12) = 0.00004
Zactual = (LSL - µave) / 3*σave = (192-199) /