Chapter 17, Problem 45CE

In painting an automobile, the thickness of the color coat has a lower specification limit of 0.80 mil and an upper specification limit of 1.20 mils. Find the C_p and C_pk capability indexes if (a) the process mean is 1.00 mil and the process standard deviation is 0.05 mil; (b) the process mean is .90 mil and the process standard deviation is 0.05 mil.

To determine

Find the $C_p$ and $C_{pk}$ indexes for a process with lower specification limit of 0.80 mil and an upper specification limit of 1.20 mils, process mean is 1.00 mil and the process standard deviation is 0.05.

Answer to Problem 45CE

The value of $C_p$ is 1.33.

The value of $C_{pk}$ is 1.33.

Explanation of Solution

Calculation:

The given information is that, lower specification limit is 0.80 mil and an upper specification limit is 1.20 mils, process mean is 1.00 mil and the process standard deviation is 0.05.

The formula for $C_p$ is,

$C_p=\frac{USL-LSL}{6\sigma }$

In the formula, USL denotes the upper specification limit, LSL denotes the lower specification limit, $\sigma$ denotes the magnitude of the process variation.

The formula for $C_{pk}$ is,

$C_{pk}=\frac{\min \left(\mu -LSL,USL-\mu \right)}{3\sigma }$

In the formula, USL denotes the upper specification limit, LSL denotes the lower specification limit, $\sigma$ denotes the magnitude of the process variation, $\mu$ is the center line.

$C_p$:

Substitute, $\text{LSL}=0.80,\text{USL}=1.20,\sigma =0.05,\mu =1.00$ is the formula of $C_p$.

$\begin{array}{c}{C}_p=\frac{1.20-0.80}{6\left(0.05\right)}\\ =\frac{0.4}{0.3}\\ =1.33\end{array}$

Hence, the value of $C_p$ is 1.33.

$C_{pk}$:

Substitute, $\text{LSL}=0.80,\text{USL}=1.20,\sigma =0.05,\mu =1.00$ is the formula of $C_{pk}$.

$\begin{array}{c}{C}_{pk}=\frac{\min \left(1.00-0.80,1.20-1.00\right)}{3\left(0.05\right)}\\ =\frac{\min \left(0.20,0.20\right)}{0.15}\\ =\frac{0.20}{0.15}\\ =1.33\end{array}$

Hence, the value of $C_{pk}$ is 1.33.

To determine

Find the $C_p$ and $C_{pk}$ indexes for a process with lower specification limit of 0.80 mil and an upper specification limit of 1.20 mils, process mean is 0.90 mil and the process standard deviation is 0.05.

Answer to Problem 45CE

The value of $C_p$ is 1.33.

The value of $C_{pk}$ is 0.67.

Explanation of Solution

Calculation:

The given information is that, lower specification limit is 0.80 mil and an upper specification limit is 1.20 mils, process mean is 0.90 mil and the process standard deviation is 0.05.

The formula for $C_p$ is,

$C_p=\frac{USL-LSL}{6\sigma }$

In the formula, USL denotes the upper specification limit, LSL denotes the lower specification limit, $\sigma$ denotes the magnitude of the process variation.

The formula for $C_{pk}$ is,

$C_{pk}=\frac{\min \left(\mu -LSL,USL-\mu \right)}{3\sigma }$

In the formula, USL denotes the upper specification limit, LSL denotes the lower specification limit, $\sigma$ denotes the magnitude of the process variation, $\mu$ is the center line.

$C_p$:

Substitute, $\text{LSL}=0.80,\text{USL}=1.20,\sigma =0.05,\mu =0.90$ is the formula of $C_p$.

$\begin{array}{c}{C}_p=\frac{1.20-0.80}{6\left(0.05\right)}\\ =\frac{0.4}{0.3}\\ =1.33\end{array}$

Hence, the value of $C_p$ is 1.33.

$C_{pk}$:

Substitute, $\text{LSL}=0.80,\text{USL}=1.20,\sigma =0.05,\mu =0.90$ is the formula of $C_{pk}$.

$\begin{array}{c}{C}_{pk}=\frac{\min \left(0.90-0.80,1.20-0.90\right)}{3\left(0.05\right)}\\ =\frac{\min \left(0.10,0.30\right)}{0.15}\\ =\frac{0.10}{0.15}\\ =0.67\end{array}$

Hence, the value of $C_{pk}$ is 0.67.