Question

In: Operations Management

A stint for use in coronary surgery requires a special coating. Specifications for this coating call...

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.)

  1. What is the Cp of the process? (2 points)
  2. Does this process meet specifications according to the Cpk index? (2 points)
  3. Assuming that the process mean of 0.09 cannot be changed, what process standard deviation would be required for this process to be a 6-sigma process? (3 points)

Solutions

Expert Solution

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


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