In: Operations Management
A stint for use in coronary surgery requires a special coating. Specifications for this coating call for it to be at least 0.05 millimeters but no more than 0.15 millimeters. Suppose the process has a normal distribution with a long-run average of 0.09 millimeters and a standard deviation of 0.015 millimeters. (You need to show your work, rather than only a final answer.)
SOLUTION:
Upper Specification Limit (USL) = 0.15mm
Lower Specification Limit (LSL) = 0.05 mm
Process Mean = 0.09mm
Standard Deviation (σ) = 0.015mm
a) Cp = (USL-LSL) / (6σ)
= (0.15-0.05) / (6*0.015)
= 1.11
b) Cpl = (Mean – LSL) / (3 * σ)
= (0.09 - 0.05) / (3*0.015)
= 0.89
Cpu = (USL – Mean) / (3 * σ)
= (0.15 - 0.09) / (3*0.015)
= 1.33
Cpk= Min (Cpl, Cpu) = Min(0.89,1.33) = 0.89
According to the Cpk index for a six sigma process, Cpk > 2 shows that the process is capable and meets specifications.
Since the Cpk, in this case, is 0.89, it does not meet process specifications.
c) Standard deviation for a process to be 6 sigma process.
A Cpk of 2 corresponds to aprroximately a 6 sigma performance. Substituting, Cpl = 2, we get
Cpl = (Mean – LSL) / (3 * σ)
2 = (0.09 - 0.05) / (3*σ)
σ = 0.0067
For σ = 0.0067, Cpu value is
Cpu = (USL – Mean) / (3 * σ)
= (0.15 - 0.09) / (3*0.0067)
= 2.98
Cpk= Min (Cpl, Cpu) = Min(2,2.98) = 2
Process standard deviation of 0.0067 would be required for this process to be a 6-sigma process