There are two machines for sale that you are considering purchasing for your sawmill to produce hardwood flooring. You want to find the one that has a higher process capability index, or Cpk. The goal is to produce flooring that is between 46 and 50 millimeters thick. The first machine is more accurate on average, producing to a mean of 48 millimeters but unfortunately it has more variation with a standard deviation of 0.7 millimeters. The second machine is not as accurate, with a mean of 47mm, but does deliver a more consistent output, with standard deviation of .3mm.

Required:
a. What is the Cpk of machine 1?
b. What is the Cpk of machine 2?
c. If your goal is to be capable', what would you do?
d. If (somehow) you could combine the best of both machines (the centering or average of machine 1 coupled with the constancy or standard deviation of machine 2, what would the Cpk be?

That is the Upper specification limit (USL) = 50 mm and the Lower specification limit (LSL) = 46 mm. Mean (μ) = 48 mm and standard deviation (σ) = 0.7

[tex]cpk=min(\frac{USL-\mu}{3\sigma},\frac{\mu -LSL}{3\sigma} )\\cpk=min(\frac{50-48}{3*0.7} ,\frac{48-46}{3*0.7} )=min(0.952,0.952)\\cpk=0.952[/tex]

b) Given that for machine 2:

The goal is between 46 and 50 millimeters. That is the Upper specification limit (USL) = 50 mm and the Lower specification limit (LSL) = 46 mm. Mean (μ) = 47 mm and standard deviation (σ) = 0.3

[tex]cpk=min(\frac{USL-\mu}{3\sigma},\frac{\mu -LSL}{3\sigma} )\\cpk=min(\frac{50-47}{3*0.3} ,\frac{47-46}{3*0.3} )=min(3.33,1.11)\\cpk=1.11[/tex]

c)

To improve the Cpk value we either move the average or reduce the variation in the results.

d) If we combine the best of the two machines; That is the Upper specification limit (USL) = 50 mm and the Lower specification limit (LSL) = 46 mm. Mean (μ) = 48 mm and standard deviation (σ) = 0.3

[tex]cpk=min(\frac{USL-\mu}{3\sigma},\frac{\mu -LSL}{3\sigma} )\\cpk=min(\frac{50-48}{3*0.3} ,\frac{48-46}{3*0.3} )=min(2.22,2.22)\\cpk=2.22[/tex]