# ASSIGNMENT 7

Topics: Process management, Process capability, Process capability index Pages: 4 (374 words) Published: December 7, 2014
﻿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) /...