Chapter 6.S, Problem 6P

Summary Introduction

To develop: Appropriate control chart and determine whether there is any cause for concern in the cutting process.

Introduction: Control charts used to determine whether the process is under control or not. Attributes and variables are the factors under the control charts.

Explanation of Solution

Given information:

The following information is given:

Develop appropriate control chart:

To check whether the process is in control, create the appropriate control charts for the mean $\overline{x}$ and the range $\overline{R}$.

Determine mean of the averages:

It can be calculated by taking the average of all given values for $\overline{x}$.

$\begin{array}{c}\text{Mean of the average}=\frac{\text{Sum of }\overline{x}\text{ values}}{\text{Total hours}}\\ =\frac{\left(\begin{array}{l}3.25+3.10+3.22+3.39+3.07+2.86+3.05+2.65+\\ 3.02+2.85+2.83+2.97+3.11+2.83+3.12+2.84+\\ 2.86+2.74+3.41+2.89+2.65+3.28+2.94+2.64\end{array}\right)}{24}\\ =\frac{71.57}{24}\\ =2.982\end{array}$

Hence, the mean of averages $\overline{\overline{x}}=2.982$

Determine mean of the ranges:

It can be calculated by taking the average of all given values for R.

$\begin{array}{c}\text{Mean of the average}=\frac{\text{Sum of }\overline{x}\text{ values}}{\text{Total hours}}\\ =\frac{\left(\begin{array}{l}0.71+1.18+1.43+1.26+1.17+0.32+0.53+1.13+\\ 0.71+1.33+1.17+0.40+0.85+1.31+1.06+0.50+\\ 1.43+1.29+1.61+1.09+1.08+0.46+1.58+0.97\end{array}\right)}{24}\\ =\frac{24.57}{24}\\ =1.024\end{array}$

Hence, the mean of ranges is $\overline{R}=1.024$.

Determine the upper control limit and lower control limit of $\overline{x}$-chart:

Determine the upper control limit (UCL) and the lower control limit (LCL) of the $\overline{x}$ chart using the following equations:

$\text{UCL}=\overline{\overline{x}}+{\text{A}}_2\times \overline{R}$ (1)

$\text{LCL}=\overline{\overline{x}}-{\text{A}}_2\times \overline{R}$ (2)

Here, the overall mean $\overline{\overline{x}}=2.982$, the average range $\overline{R}=1.024$ and A₂ is the Mean factor derived from standard tables showing factors for computing control charts.

Given the sample size of 4, the Mean factor is ${\text{A}}_2=0.729$(Refer the S6.1 table (Factors for computing control chart limits (3 sigma))).

Substitute into the equation (1),

$\begin{array}{c}\text{UCL}=\overline{\overline{x}}+{\text{A}}_2\times \overline{R}\\ =2.982\text{in}\text{.}+0.729\times 1.024\text{in}\text{.}\\ \text{=3}\text{.7285}\text{in}\text{.}\end{array}$

Substitute into the equation (2),

$\begin{array}{c}\text{LCL}=\overline{\overline{x}}-{\text{A}}_2\times \overline{R}\\ =2.982\text{in}\text{.}-0.729\times 1.024\text{in}\text{.}\\ \text{=2}\text{.2355}\text{in}\text{.}\end{array}$

Hence, the upper control limit is $\text{UCL=3}\text{.7285}\text{in}\text{.}$ and the lower control limit is $\text{LCL=2}\text{.2355}\text{in}\text{.}$for mean.

Determine the upper control limit and lower control limit of $\overline{R}$-chart:

Determine the upper control limit (UCL_R) and lower control limit (LCL_R) of the $\overline{R}$ chart using equations:

${\text{UCL}}_{\text{R}}={\text{D}}_{\text{4}}\times \overline{R}$ (3)

${\text{LCL}}_{\text{R}}={\text{D}}_3\times \overline{R}$ (4)

Here,

${\text{D}}_{\text{3}}$ refers to the lower range factor

${\text{D}}_{\text{4}}$ refers to the upper range factor

$\overline{R}$ refers to the average range

Given the sample size of 4 for $\sigma =3$, Lower range factor ${\text{D}}_{\text{3}}=0$ and Upper range factor ${\text{D}}_{\text{4}}=2.282$(Refer the S6.1 table (Factors for computing control chart limits (3 sigma))).

Substitute into the equation (3),

$\begin{array}{c}{\text{UCL}}_{\text{R}}={\text{D}}_{\text{4}}\times \overline{R}\\ =2.282\times 1.024\text{in}\text{.}\\ =2.3368\text{in}\text{.}\end{array}$

Similarly, substitute in equation (4),

$\begin{array}{c}{\text{LCL}}_{\text{R}}={\text{D}}_3\times \overline{R}\\ =0\times 1.024\text{in}\text{.}\\ =0\text{in}\text{.}\end{array}$

Hence, the upper control limit is ${\text{UCL}}_{\text{R}}=\text{2}\text{.3368}\text{in}\text{.}$ and the lower control limit is ${\text{LCL}}_{\text{R}}=\text{0}\text{in}\text{.}$.

Draw the control chart for the sample means:

The process mean is $\overline{\overline{x}}=2.982$, $\text{UCL=3}\text{.7285}\text{in}\text{.}$, and $\text{LCL=2}\text{.2355}\text{in}\text{.}$.

Observe that all the sample means are well within the control limits. The first five observations are above the mean of the averages.

Draw the control chart for the sample range:

The average range is $\overline{R}=1.024$, ${\text{UCL}}_{\text{R}}\text{=2}\text{.3368}\text{in}\text{.}$, and ${\text{LCL}}_{\text{R}}\text{=0}\text{in}\text{.}$.

Observe that all the sample ranges are well within the control limits

Hence, the process seems to be in “statistical control”. (However, a process for making precision wires obviously should not have so much of variation in the mean or in the range.)