PRIN.OF OPERATIONS MANAGEMENT-MYOMLAB
PRIN.OF OPERATIONS MANAGEMENT-MYOMLAB
11th Edition
ISBN: 9780135226742
Author: HEIZER
Publisher: PEARSON
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Chapter 6.S, Problem 11P

a)

Summary Introduction

To determine: The control limits for the mean and range chart and the overall means

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

a)

Expert Solution
Check Mark

Answer to Problem 11P

Hence, the UCL for x¯ -chart is 10.00714in and LCL is 9.993865in. The UCL for R chart is 0.024323in and LCL is 0in. The overall means for x¯ -chart is 10.0005in and R chart is 0.0115in.

Explanation of Solution

Given information:

The following information is given:

Sample Sample mean (in.) Range (in.)
1 10.002 0.011
2 10.002 0.014
3 9.991 0.007
4 10.006 0.022
5 9.997 0.013
6 9.999 0.012
7 10.001 0.008
8 10.005 0.013
9 9.995 0.004
10 10.001 0.011
11 10.001 0.014
12 10.006 0.009

Twelve samples that contain five parts each were taken.

Calculate the average for sample and range:

Sample Sample mean (in.) Range (in.)
1 10.002 0.011
2 10.002 0.014
3 9.991 0.007
4 10.006 0.022
5 9.997 0.013
6 9.999 0.012
7 10.001 0.008
8 10.005 0.013
9 9.995 0.004
10 10.001 0.011
11 10.001 0.014
12 10.006 0.009
Total 120.006 0.138
Average 10.0005 0.0115

Working note:

Average for sample:

It is calculated by dividing the total of sample and number of samples. Hence, the value of x¯¯ is 10.0005.

Average=Total of SampleNumber of samples=120.00612=10.0005

Average for range:

It is calculated by dividing the total of range and number of samples. Hence, the value of R¯ is 0.0115.

Average=Total of SampleNumber of samples=0.13812=0.0115

Determine the UCL and LCL for mean:

Formulae to calculate control limits:

UCL=x¯¯+A×R¯ (1)

LCL=x¯¯A×R¯ (2)

Here, the overall mean x¯¯=10.0005in. , the average range R¯=0.0115in. and A2 is the Mean factor derived from standard tables showing factors for computing control charts. Given the sample size of 5, the Mean factor A2=0.577 for σ=3 .

Substitute the values in equation (1) to determine the value of UCL as follows:

UCL=x¯¯+A×R¯=10.0005in+0.577×0.0115in.=10.00714

Hence, the UCL value is 10.00714.

Substitute the values in equation (2) to determine the value of LCL as follows:

LCL=x¯¯A×R¯=10.0005in.0.577×0.0115in.=9.993865in.

Hence, the LCL value is 9.993865.

Therefore for the x¯ chart, the upper control limit is UCLx¯=10.00714 and the lower control limit is LCLx¯=9.993865in .

Determine the UCL and LCL for range:

Formulae to calculate control limits:

UCLR=D4×R¯ (3)

LCLR=D3×R¯ (4)

Here,

The average range is R¯=0.0115in.

D3 refers to the lower range factor

D4 refers to the upper range factor

R¯ refers to the average range

Substitute the values in equation (3) to determine the value of UCL as follows:

UCLR=D4×R¯=2.115×0.0115in.=0.024323in.

Hence, the UCL value is 0.024323.

Substitute the values in equation (4) to determine the value of LCL as follows:

LCLR=D3×R¯=0×0.0115in.=0

Hence, the LCL value is 0.

Therefore, for the R-chart, the upper control limit is UCLR=0.024323 and the lower control limit is LCLR=0 .

b)

Summary Introduction

To plot: The values of sample means and ranges in the chart.

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

b)

Expert Solution
Check Mark

Answer to Problem 11P

Control chart has been plotted for sample means and ranges.

Explanation of Solution

Given information:

The following information is given:

Sample Sample mean (in.) Range (in.)
1 10.002 0.011
2 10.002 0.014
3 9.991 0.007
4 10.006 0.022
5 9.997 0.013
6 9.999 0.012
7 10.001 0.008
8 10.005 0.013
9 9.995 0.004
10 10.001 0.011
11 10.001 0.014
12 10.006 0.009

Twelve samples that contain five parts each were taken.

Plot the sample mean values in the x¯ control chart where UCLx¯=10.0005 and LCLx¯=9.993865

PRIN.OF OPERATIONS MANAGEMENT-MYOMLAB, Chapter 6.S, Problem 11P , additional homework tip  1

Note: Observe that sample 3 mean value is lower than the LCLx¯=9.993865in.

Plot the sample mean values in theR-control chart where UCLR=0.024323in and LCLR=0

PRIN.OF OPERATIONS MANAGEMENT-MYOMLAB, Chapter 6.S, Problem 11P , additional homework tip  2

c)

Summary Introduction

To determine: Whether the process is in control

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

c)

Expert Solution
Check Mark

Answer to Problem 11P

Process is out of control during sample #3 in x¯ -chart.

Explanation of Solution

Given information:

The following information is given:

Sample Sample mean (in.) Range (in.)
1 10.002 0.011
2 10.002 0.014
3 9.991 0.007
4 10.006 0.022
5 9.997 0.013
6 9.999 0.012
7 10.001 0.008
8 10.005 0.013
9 9.995 0.004
10 10.001 0.011
11 10.001 0.014
12 10.006 0.009

Twelve samples that contain five parts each were taken.

Determine whether the process is in control:

Since, the mean for sample #3 is outside the control limits of the x¯ control chart, it cannot be concluded that the process is in statistical control.

d)

Summary Introduction

To determine: Why the process is not in control

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

d)

Expert Solution
Check Mark

Explanation of Solution

Given information:

The following information is given:

Sample Sample mean (in.) Range (in.)
1 10.002 0.011
2 10.002 0.014
3 9.991 0.007
4 10.006 0.022
5 9.997 0.013
6 9.999 0.012
7 10.001 0.008
8 10.005 0.013
9 9.995 0.004
10 10.001 0.011
11 10.001 0.014
12 10.006 0.009

Twelve samples that contain five parts each were taken.

Determine why the process is not in control:

Further investigations are necessary to check whether the mean value of sample #3 is a freak incident outside the three sigma limits (which has a 0.27% probability). Perhaps 12 more samples can be drawn and examined whether such an incident occurs again. In case it happens again, the process needs to be examined in detail.

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