EBK PRINCIPLES OF OPERATIONS MANAGEMENT
EBK PRINCIPLES OF OPERATIONS MANAGEMENT
11th Edition
ISBN: 9780135175859
Author: Munson
Publisher: VST
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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.

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Explanation of Solution

Given information:

The following information is given:

EBK PRINCIPLES OF OPERATIONS MANAGEMENT, Chapter 6.S, Problem 6P , additional homework tip  1

Develop appropriate control chart:

To check whether the process is in control, create the appropriate control charts for the mean x¯ and the range R¯ .

Determine mean of the averages:

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

Mean of the average=Sum of x¯ valuesTotal hours=(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)24=71.5724=2.982

Hence, the mean of averages x¯¯=2.982

Determine mean of the ranges:

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

Mean of the average=Sum of x¯ valuesTotal hours=(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)24=24.5724=1.024

Hence, the mean of ranges is R¯=1.024 .

Determine the upper control limit and lower control limit of x¯ -chart:

Determine the upper control limit (UCL) and the lower control limit (LCL) of the x¯ chart using the following equations:

UCL=x¯¯+A2×R¯ (1)

LCL=x¯¯A2×R¯ (2)

Here, the overall mean x¯¯=2.982 , the average range 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 A2=0.729 (Refer the S6.1 table (Factors for computing control chart limits (3 sigma))).

Substitute into the equation (1),

UCL=x¯¯+A2×R¯=2.982in.+0.729×1.024in.=3.7285in.

Substitute into the equation (2),

LCL=x¯¯A2×R¯=2.982in.0.729×1.024in.=2.2355in.

Hence, the upper control limit is UCL=3.7285in. and the lower control limit is LCL=2.2355in. for mean.

Determine the upper control limit and lower control limit of R¯ -chart:

Determine the upper control limit (UCLR) and lower control limit (LCLR) of the R¯ chart using equations:

UCLR=D4×R¯ (3)

LCLR=D3×R¯ (4)

Here,

D3 refers to the lower range factor

D4 refers to the upper range factor

R¯ refers to the average range

Given the sample size of 4 for σ=3 , Lower range factor D3=0 and Upper range factor D4=2.282 (Refer the S6.1 table (Factors for computing control chart limits (3 sigma))).

Substitute into the equation (3),

UCLR=D4×R¯=2.282×1.024in.=2.3368in.

Similarly, substitute in equation (4),

LCLR=D3×R¯=0×1.024in.=0in.

Hence, the upper control limit is UCLR=2.3368in. and the lower control limit is LCLR=0in. .

Draw the control chart for the sample means:

The process mean is x¯¯=2.982 , UCL=3.7285in. , and LCL=2.2355in. .

EBK PRINCIPLES OF OPERATIONS MANAGEMENT, Chapter 6.S, Problem 6P , additional homework tip  2

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 R¯=1.024 , UCLR=2.3368in. , and LCLR=0in. .

EBK PRINCIPLES OF OPERATIONS MANAGEMENT, Chapter 6.S, Problem 6P , additional homework tip  3

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.)

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Chapter 6 Solutions

EBK PRINCIPLES OF OPERATIONS MANAGEMENT

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