Examples of Quality Control and the Capability Index

Question 1: The Cookie Cuttery

The Cookie Cuttery produces cookies to an outer diameter (OD) specification of 4 ± 0.1 inches. The baker takes multiple sample measurements over time and finds the sample mean for the outer diameter to be 4.02 inches with a standard deviation of 0.03 inch.

a) Calculate the process capability index.

b) What does this figure tell you about the process?

Solution:

(a) Since the process is not centered, the process capability index is computed as

\[{{C}_{pk}}=\min \left\{ \frac{4.1-4.02}{3\times 0.03},\,\,\frac{4.02-3.9}{3\times 0.03} \right\}=\min \left\{ 0.88889,\,\,1.33333 \right\}=0.88889\]

(b) The capability of the process is less than 1, which means that the inherent variability of the process doesn’t fall within the acceptable range of variability allowed by the design specifications.

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Question 2: Med-Tech.

Med-Tech engineers are trying to improve the design of a pump shaft that has an outer diameter specification of 13 inches with a tolerance of 0.003 inch. Available inspection data from the past year indicate that the mean value of the diameter has been 13.001 with a standard deviation of 0.0025 inch. The gear sells for $125. The estimated loss to society is $20 for any gear that has a diameter at the upper or lower tolerance limit. Annual sales of the gear amount to 40,000 units.

a) Calculate the average loss per unit of production.

b) What is the expected loss per year?

c) What happens to the average loss per unit and the expected loss per year if the mean is shifted to the target value of 13 inches?

Solution:

(a) Let X be the random diameter. Then, from the data provided we know that

\[X\tilde{\ }N\left( 13.001,\,\,{{0.0025}^{2}} \right) \]

The loss function (loss to society) per unit is

\[L\left( X \right)=\left\{ \begin{array}{ll} & 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,12.997\le X\le 13.003 \\ & -20\,\,\,\,\,\,\,\,\,\,\,\,\,\text{else} \\ \end{array} \right. \]

This means that the expected value is

\[E\left( L\left( X \right) \right)=-20\left[ 1-\Pr \left( 12.997\le X\le 13.003 \right) \right] \] \[ = -20\left[ 1-\Pr \left( \frac{12.997-13.001}{0.0025}\le \frac{X-13.001}{0.0025}\le \frac{13.003-13.001}{0.0025} \right) \right]=-20\left( 1-\Pr \left( -1.6\le Z\le 0.8 \right) \right) \] \[=-20\left( 1-0.733345 \right)=-5.33309\]

So, the expected loss per unit to society is $5.33309.

(b) The yearly demand is 40,000, which means that the expected loss per year to society is

\[40,000\times 5.33309=\$213,324\]

(c) In this case we have

\[E\left( L\left( X \right) \right)=-20\left[ 1-\Pr \left( 12.997\le X\le 13.003 \right) \right] \] \[ = -20\left[ 1-\Pr \left( \frac{12.997-13}{0.0025}\le \frac{X-13}{0.0025}\le \frac{13.003-13}{0.0025} \right) \right]=-20\left( 1-\Pr \left( -1.2\le Z\le 1.2 \right) \right) \] \[x=-20\left( 1-0.769861 \right)=-4.60279\]

So, the expected loss per unit to society is $4.60279. The expected loss per year to society is

\[40,000\times 4.60279=\$184,112\]

So, a good portion of loss would be saved if the mean is shifted to the target.

Question 3: McDougals.

McDougals is considering the purchase of a new machine to cut french fries to a width

specification of 0.4 ± 0.03 inches (McDougals advertises it has the best fries, and has found that french fries must be precisely cut and then cooked for an "exact" amount of time if they are to attain the ultimate crispness.) After a trial run on this machine, Mcnougats has determined that the machine has a sample mean of 0.401 inches with a standard deviation of 0.002 inches.

a) Calculate the Cpk for this machine.

b) Should McDougals purchase this machine? Why or why not?

Solution: (a) The capability index is

\[{{C}_{pk}}=\min \left\{ \frac{0.43-0.401}{3\times 0.002},\,\,\frac{0.401-0.37}{3\times 0.002} \right\}=\min \left\{ 4.83333,\,\,5.16667 \right\}=4.833\]

(b) The capability index is well above 1, which means that the machine is within the acceptable limits of variability, and hence, McDouglas should buy it.

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