Loose Leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences)
Loose Leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences)
13th Edition
ISBN: 9781260152203
Author: William J Stevenson
Publisher: McGraw-Hill Education
Question
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Chapter 7, Problem 10P

a)

Summary Introduction

To determine: The standard time for the operation.

Introduction: The amount of the dependency on human effort by an organization in terms of achieving its goals is given by the work design. It is directly linked to the productivity of an organization where good work design helps in achieving high productivity.

a)

Expert Solution
Check Mark

Answer to Problem 10P

The standard time for the operation is 5.80 minutes.

Explanation of Solution

Given information:

ElementPerformance ratingObservations (minutes per cycles)Allowance
1 2 3 4 5 6
1 1.1 1.2 1.17 1.16 1.22 1.24 1.15 0.15
2 1.15 0.83 0.87 0.78 0.82 0.85 0 0.15
3 1.05 0.58 0.53 0.52 0.59 0.6 0.54 0.15

Formula:

Standardtime=Normaltime×Allowancefactor=NT×AFNT=Observedtime×PerformanceRatingAFjob=1+A

Calculation of standard time of operation:

ElementPerformance ratingObservations (minutes per cycles)AllowanceObserved timeNormal timeAfjobStandard time
1 2 3 4 5 6
1 1.1 1.2 1.17 1.16 1.22 1.24 1.15 0.15 1.198 1.317 1.15 1.52
2 1.15 0.83 0.87 0.78 0.82 0.85 0 0.15 0.83 0.954 1.15 1.10
3 1.05 0.58 0.53 0.52 0.59 0.6 0.54 0.15 0.564 0.592 1.15 0.68
Standard time for operation 3.29

Excel Worksheet:

Loose Leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences), Chapter 7, Problem 10P , additional homework tip  1

Element 1:

Observed time is calculated by taking mean for the 1.2, 1.17, 1.16, 1.22, 1.24 and 1.15 which gives 1.198.

Normal time is calculated by multiplying observed timing, 1.198 and performance rating, 1.1 which yields 1.317 minutes.

Allowance factor is calculated by adding 1 with the allowance factor 0.15 to give 1.15.

ST=NT×AFjob=1.317minutes×1.15=1.52minutes

Standard time for element 1 is calculated by multiplying normal time of 1.317 minutes with allowance factor of 1.15 which gives 1.52 minutes.

Same process applies for element 2 and 3 which yields the standard times as 1.10 and 0.68. The standard time for operation is obtained by adding 1.52, 1.10 and 1.68 minutes which gives 3.29 minutes.

Hence, the standard time for the operation is 3.29 minutes.

b)

Summary Introduction

To determine: The number of observations for element 2.

Introduction: The amount of the dependency on human effort by an organization in terms of achieving its goals is given by the work design. It is directly linked to the productivity of an organization where good work design helps in achieving high productivity.

b)

Expert Solution
Check Mark

Answer to Problem 10P

The number of observations for element 2 is 67.

Explanation of Solution

Given information:

ElementPerformance ratingObservations (minutes per cycles)Allowance
1 2 3 4 5 6
1 1.1 1.2 1.17 1.16 1.22 1.24 1.15 0.15
2 1.15 0.83 0.87 0.78 0.82 0.85 0 0.15
3 1.05 0.58 0.53 0.52 0.59 0.6 0.54 0.15

Confidence= 95.5%

1% of true value

Formula:

n=(zsax¯)2

z=Number ofnormalstandarddeviationsfordesiredconfidences=Samplestandarddeviationx¯=Sample meana=Desired accuracy

Calculation of number of observations for element A:

In the above formula the sample standard deviation is calculated by,

Calculation of standard deviation:

Element 2 Differences Square of differences Standard deviation
0.83 0 00.0339
0.87 0.04 0.0016
0.78 -0.05 0.0025
0.82 -0.01 0.0001
0.85 0.02 0.0004
Mean=0.83 SSQ=0.0046

Excel worksheet:

Loose Leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences), Chapter 7, Problem 10P , additional homework tip  2

Z value for confidence interval of 95.5% is 2.00.

The confidence interval 0.952 gives 0.4775, which is in the midway of 0.4772 (z = 2.00) and 0.4778 (z = 2.01).

  • Using z = 2.00:

n=(2.00×0.03390.01×0.83)2=66.7272

The number of observations from the standard z table for confidence level of 95.5% is calculated by dividing the product of 2.00 and 0.0339 with product of 0.01 and 0.83 and squaring the resultant which gives 66.7272.

  • Using z = 2.01:

n=(2.01×0.03390.01×0.83)2=67.3962

The number of observations from the standard z table for confidence level of 95.5% is calculated by dividing the product of 2.01 and 0.0339 with product of 0.01 and 0.83 and squaring the resultant which gives 67.3962.

Hence, the number of observation is 68.

c)

Summary Introduction

To determine: The number of observations needed to estimate the mean time for element 2 within 0.01 minute of its true value.

Introduction: The amount of the dependency on human effort by an organization in terms of achieving its goals is given by the work design. It is directly linked to the productivity of an organization where good work design helps in achieving high productivity.

c)

Expert Solution
Check Mark

Answer to Problem 10P

The number of observations needed to estimate the mean time for element 2 within 0.01 minute of its true value is 47.

Explanation of Solution

Given information:

ElementPerformance ratingObservations (minutes per cycles)Allowance
1 2 3 4 5 6
1 1.1 1.2 1.17 1.16 1.22 1.24 1.15 0.15
2 1.15 0.83 0.87 0.78 0.82 0.85 0 0.15
3 1.05 0.58 0.53 0.52 0.59 0.6 0.54 0.15

Confidence= 95.5%

0.10 minute of actual value.

Formula:

n=(zse)2

z=Number ofnormalstandarddeviationsfordesiredconfidences=Samplestandarddeviatione=Maximumacceptableamountoftimeerror

Calculation of number of observations for element C:

In the above formula the sample standard deviation is calculated by,

Z value for confidence interval of 95.5% is 2.00.

The confidence interval 0.952 gives 0.4775, which is in the midway of 0.4772 (z = 2.00) and 0.4778 (z = 2.01).

  • Using z = 2.00:

n=(2.00×0.03390.01)2=45.9684

The number of observations from the standard z table for confidence level of 95.5 is calculated by dividing the product of 2.00 and 0.0339 with 0.01 and squaring the resultant which gives 45.9684.

  • Using z = 2.00:

n=(2.01×0.03390.01)2=46.4292

The number of observations from the standard z table for confidence level of 95.5 is calculated by dividing the product of 2.01 and 0.0339 with 0.01 and squaring the resultant which gives 46.4292.

Hence, the number of observations needed to estimate the mean time for element 2 within 0.01 minute of its true value is 46.

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Chapter 7 Solutions

Loose Leaf for Operations Management (The Mcgraw-hill Series in Operations and Decision Sciences)

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