Operations Management (McGraw-Hill Series in Operations and Decision Sciences)
Operations Management (McGraw-Hill Series in Operations and Decision Sciences)
12th Edition
ISBN: 9780078024108
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:

Operations Management (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:

Operations Management (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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Ergonomics problem A work measurement analyst in the Electrical Company took 10 observations (for each element) of an operation job. The operation consists of four elements of work. Result of observation was shown in a Table attached. a. Determine the standard time and standard output if the total allowance time is 12% b. Check whether number of observation is sufficient or not for a 95 percent confidence level and 5 percent significant level.
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Q2 (a) A time study analyst timed an assembly operation for 30 cycles, and then computed the average time per cycle, which was 18.75 minutes. The analyst assigned a performance rating of 0.96 and decided that an appropriate allowance was 15 percent. Assume the allowance factor is based on the workday. Determine the following: (i) Observed time (OT) (ii) The Normal times (NT) (iii) The standard time (ST)                                                             (b) If the analyst wants to estimate the number of observations that will be needed to achieve a specified maximum error, with a confidence of 95.5 %. A preliminary study yielded a mean of 5.2 minutes and a standard deviation of 1.1 minutes. Determine the total number of observation needed for these two cases: (a) A maximum error of + or - 6 % (b) A maximum error of 40 minutes

Chapter 7 Solutions

Operations Management (McGraw-Hill Series in Operations and Decision Sciences)

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