Chapter 17, Problem 44CE

(a)

To determine

Find the $C_p$ and $C_{pk}$ indexes for a process with $\text{LSL}=540,\text{USL}=550,\sigma =1.25,\mu =545$.

Answer to Problem 44CE

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, $\text{LSL}=540,\text{USL}=550,\sigma =1.25,\mu =545$.

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, $\text{LSL}=540,\text{USL}=550,\sigma =1.25,\mu =545$ is the formula of $C_p$.

$\begin{array}{c}{C}_p=\frac{550-540}{6\left(1.25\right)}\\ =\frac{10}{7.5}\\ =1.33\end{array}$

Hence, the value of $C_p$ is 1.33.

$C_{pk}$:

Substitute, $\text{LSL}=540,\text{USL}=550,\sigma =1.25,\mu =545$ is the formula of $C_{pk}$.

$\begin{array}{c}{C}_{pk}=\frac{\min \left(545-540,550-545\right)}{3\left(1.25\right)}\\ =\frac{\min \left(5,5\right)}{3.75}\\ =\frac{5}{3.75}\\ =1.33\end{array}$

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

(b)

To determine

Find the $C_p$ and $C_{pk}$ indexes for a process with $\text{LSL}=540,\text{USL}=550,\sigma =1.25,\mu =543$.

Answer to Problem 44CE

The value of $C_p$ is 1.33.

The value of $C_{pk}$ is 0.80.

Explanation of Solution

Calculation:

The given information is that $\text{LSL}=540,\text{USL}=550,\sigma =1.25,\mu =543$.

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 $\text{LSL}=540,\text{USL}=550,\sigma =1.25,\mu =543$ is the formula of $C_p$.

$\begin{array}{c}{C}_p=\frac{550-540}{6\left(1.25\right)}\\ =\frac{10}{7.5}\\ =1.33\end{array}$

Hence, the value of $C_p$ is 1.33.

$C_{pk}$:

Substitute, $\text{LSL}=540,\text{USL}=550,\sigma =1.25,\mu =543$ is the formula of $C_{pk}$.

$\begin{array}{c}{C}_{pk}=\frac{\min \left(543-540,550-543\right)}{3\left(1.25\right)}\\ =\frac{\min \left(3,7\right)}{3}\\ =\frac{3}{3.75}\\ =0.80\end{array}$

Hence, the value of $C_{pk}$ is 0.80.