Principles Of Operations Management
Principles Of Operations Management
11th Edition
ISBN: 9780135173930
Author: RENDER, Barry, HEIZER, Jay, Munson, Chuck
Publisher: Pearson,
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Chapter 6.S, Problem 7P

a)

Summary Introduction

To determine: The value of x¯¯ .

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 7P

Hence, the value of x¯¯ is 155.16mm.

Explanation of Solution

Given information:

The following information is given:

Day Mean (MM) Range (MM)
1 156.9 4.2
2 153.2 4.6
3 153.6 4.1
4 155.5 5
5 156.6 4.5

Determine the value of x¯¯ :

Day Mean (MM)
1 156.9
2 153.2
3 153.6
4 155.5
5 156.6
Total 775.8

Calculate the x¯¯ :

It is calculated by dividing the sum of mean values and the number of days.

x¯¯=Sum of meanNumber of days=156.9+153.2+153.6+155.5+156.65==775.85=155.16

Hence, the value of x¯¯ is 155.16mm.

b)

Summary Introduction

To determine: The value of R¯ .

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 7P

Hence, the value of R¯ is 4.48mm.

Explanation of Solution

Given information:

The following information is given:

Day Mean (MM) Range (MM)
1 156.9 4.2
2 153.2 4.6
3 153.6 4.1
4 155.5 5
5 156.6 4.5

Determine the value of R¯ :

Day Range (MM)
1 4.2
2 4.6
3 4.1
4 5
5 4.5
Total 22.4

Calculate the R¯ :

It is calculated by dividing the sum of mean values and the number of days.

R¯=Sum of rangeNumber of days=4.2+4.6+4.1+5.0+4.55=22.45=4.48

Hence, the value of R¯ is 4.48mm.

c)

Summary Introduction

To plot: The UCL and LCL for x¯ .

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 7P

Hence, the value of UCL is 156.54mm and LCL is 153.78mm.

Explanation of Solution

Given information:

The following information is given:

Day Mean (MM) Range (MM)
1 156.9 4.2
2 153.2 4.6
3 153.6 4.1
4 155.5 5
5 156.6 4.5

Determine the UCL and LCL of x¯ :

Formulae to determine Upper Control Limit and Lower Control Limit are given as follows:

UCLx¯=x¯¯+A2×R¯

LCLx¯=x¯¯A2×R¯

Here, the overall mean is x¯¯=155.16mm , the average range is R¯=4.48mm and A2 is the Mean factor derived from standard tables showing factors for computing control charts.

Given the sample size of 10, the Mean factor A2=0.308 for σ=3 (refer table S6.1).

Substitute the values in the given formulae:

Upper control limit can be calculated by adding the multiple of average range and mean factor with the overall mean.

UCLx¯=x¯¯+A2×R¯=155.16+0.308×4.48=156.54mm

Lower control limit can be calculated by subtracting the multiple of average range and mean factor from the overall mean.

LCLx¯=x¯¯A2×R¯=155.160.308×4.48=153.78mm

Hence, upper control limit is 156.54mm and 153.78mm.

Plot the values:

Principles Of Operations Management, Chapter 6.S, Problem 7P , additional homework tip  1

d)

Summary Introduction

To plot: The UCL and LCL for R¯ .

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

Answer to Problem 7P

Hence, the value of UCL is 7.96mm and LCL is 0.999mm.

Explanation of Solution

Given information:

The following information is given:

Day Mean (MM) Range (MM)
1 156.9 4.2
2 153.2 4.6
3 153.6 4.1
4 155.5 5
5 156.6 4.5

Determine the UCL and LCL of R¯ :

Formulae to determine Upper Control Limit and Lower Control Limit are given as follows:

UCLR=D4×R¯

LCLR=D3×R¯

Here, the average range is R¯=4.48mm and D3 is the lower range factor and D4 is the upper range factor.

Given the sample size of 10 for σ=3 , the Lower range factor D3=0.223 and the Upper range factor D4=1.777 (refer table S6.1).

Substitute the values in the given formulae:

Upper control limit can be calculated by multiplying upper range factor and average range.

UCLR=D4×R¯=1.777×4.48mm=7.96mm

Upper control limit can be calculated by multiplying lower range factor and average range.

LCLR=D3×R¯=0.223×4.48mm=0.999mm

Hence, upper control limit is 7.96mm and 0.999mm.

Plot the values:

Principles Of Operations Management, Chapter 6.S, Problem 7P , additional homework tip  2

e)

Summary Introduction

To determine: The UCL and LCL for x¯ if the true diameter mean is 155mm.

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

e)

Expert Solution
Check Mark

Answer to Problem 7P

Hence, the value of UCL is 156.38mm and LCL is 153.62mm.

Explanation of Solution

Given information:

The following information is given:

Day Mean (MM) Range (MM)
1 156.9 4.2
2 153.2 4.6
3 153.6 4.1
4 155.5 5
5 156.6 4.5

Given the target value of the diameter μ=155mm , calculate the upper control limits UCLx¯ and LCLx¯ using the formula

UCLx¯=μ+A2×R¯

LCLx¯=μA2×R¯

Here,

The target value of the diameter μ=155mm , the average range R¯=4.48mm and A2 is the Mean factor derived from standard tables showing factors for computing control charts.

Given the sample size of 10, the Mean factor A2=0.308 for σ=3 (refer table S6.1).

Substitute the values in the given formulae:

Upper control limit can be calculated by adding the multiple of average range and mean factor with the overall mean.

UCLx¯=μ+A2×R¯=155+0.308×4.48=156.38mm

Lower control limit can be calculated by subtracting the multiple of average range and mean factor from the overall mean.

LCLx¯=μA2×R¯=1550.308×4.48=153.62mm

Hence, for the x¯ chart, the upper control limit is UCLx¯=156.38mm and the lower control limit is LCLx¯=153.62mm .

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Chapter 6 Solutions

Principles Of Operations Management

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