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
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Chapter 17, Problem 15P

PRINTED BY: 92248ddb24ccbc6@placeholder.10274.edu. Printing is for personal, private use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted.

A construction project has indirect costs totaling $40,000 per week. Major activities in the project and their expected times in weeks are shown in this precedence diagram.

Chapter 17, Problem 15P, PRINTED BY: 92248ddb24ccbc6@placeholder.10274.edu. Printing is for personal, private use only. No , example  1

Crashing costs for each activity are:

Chapter 17, Problem 15P, PRINTED BY: 92248ddb24ccbc6@placeholder.10274.edu. Printing is for personal, private use only. No , example  2

a. Determine the optimum time–cost crashing plan.

b. Plot the total-cost curve that describes the least expensive crashing schedule that will reduce the project length by six weeks.

a)

Expert Solution
Check Mark
Summary Introduction

To determine: The optimum cost-saving plan.

Introduction:

Project crashing:

It is method to shorten the total time taken for a project by reducing the time taken for one or more activities on the critical path. The reduction in the normal time taken is known as crashing.

Answer to Problem 15P

The activities to be crashed are 7-11, 1-2, 6-10, 11-13, and 4-6. The total crashing cost is $1,336,000.

Explanation of Solution

Given information:

  • Indirect cost is $40,000 per week.

Operations Management (McGraw-Hill Series in Operations and Decision Sciences), Chapter 17, Problem 15P , additional homework tip  1

Activity Crash cost first week ($000) Crash cost second week ($000) Crash cost third week ($000)
1 to 2 18 22
2 to 5 24 25 25
5 to 7 30 30 35
7 to 11 15 20
11 to 13 30 33 36
1 to 3 12 24 26
3 to 8
8 to 11 40 40 40
3 to 9 3 10 12
9 to 12 2 7 10
12 to 13 26
1 to 4 10 15 25
4 to 6 8 13
6 to 10 5 12
10 to 12 14 15

Project crashing:

Calculation of expected duration of each path:

Path 1-2-5-7-11-13:

Expected duration=5+8+7+4+11=35

Path 1-3-8-11-13:

Expected duration=4+12+5+11=32

Path 1-3-9-12-13:

Expected duration=4+6+9+1=20

Path 1-4-6-10-12-13:

Expected duration=3+12+9+8+1=33

Step 1:

Critical path is 1-2-5-7-11-13.

The activities are ranked according to the cost per week to crash.

Activity Cost ($)
7-11 15
1-2 18
2-5 24
5-7 30
11-13 30

Activity 7-11 will be crashed first by 1 week since it has the lowest crashing cost ($15) and this cost is ≤ 40. Path 1-2-5-7-11-13 will decrease by 1 week.

Step 2:

The paths and new expected duration are:

Path Expected Duration
1-2-5-7-11-13 34
1-3-8-11-13 32
1-3-9-12-13 20
1-4-6-10-12-13 33

Critical path is 1-2-5-7-11-13.

The activities are ranked according to the cost per week to crash.

Activity Cost ($)
1-2 $18
7-11 $20
2-5 $24
5-7 $30
11-13 $30

Activity 1-2 will be crashed first by 1 week since it has the lowest crashing cost ($18) and this cost is ≤ 40. Path 1-2-5-7-11-13 will decrease by 1 week.

Step 3:

The paths and new expected duration are:

Path Expected Duration
1-2-5-7-11-13 33
1-3-8-11-13 32
1-3-9-12-13 20
1-4-6-10-12-13 33

Critical path is 1-2-5-7-11-13 and 1-4-6-10-12-13.

The activities are ranked according to the cost per week to crash.

Path Activity Cost ($)
1-2-5-7-11-13 7-11 $20
1-2 $22
2-5 $24
5-7 $30
11-13 $30
1-4-6-10-12-13 6-10 $5
4-6 $8
1-4 $10
10-12 $14
12-13 $26

Activity 7-11 will be crashed first by 1 week since it has the lowest crashing cost ($20). Path 1-2-5-7-11-13 will decrease by 1 week.

Activity 6-10 will be crashed first by 1 week since it has the lowest crashing cost ($5). Path 1-4-6-10-12-13 will decrease by 1 week.

The combined crash cost ($25) is ≤ $40.

Step 4:

The paths and new expected duration are:

Path Expected Duration
1-2-5-7-11-13 32
1-3-8-11-13 32
1-3-9-12-13 20
1-4-6-10-12-13 32

Critical path is 1-2-5-7-11-13, 1-3-8-11-13, and 1-4-6-10-12-13.

The activities are ranked according to the cost per week to crash.

Path Activity Cost ($)
1-2-5-7-11-13 1-2 $22
2-5 $24
5-7 $30
11-13 $30
1-3-8-11-13 1-3 $12
11-13 $30
8-11 $40
1-4-6-10-12-13 4-6 $8
1-4 $10
6-10 $12
10-12 $14
12-13 $26

Activity 11-13 will be crashed first by 1 week since it has the lowest crashing cost ($30). Paths 1-2-5-7-11-13 and 1-3-8-11-13 will decrease by 1 week.

Activity 4-6 will be crashed first by 1 week since it has the lowest crashing cost ($8). Path 1-4-6-10-12-13 will decrease by 1 week.

The combined crash cost ($38) is ≤ $40.

Step 5:

The paths and new expected duration are:

Path Expected Duration
1-2-5-7-11-13 31
1-3-8-11-13 31
1-3-9-12-13 20
1-4-6-10-12-13 31

Critical path is 1-2-5-7-11-13, 1-3-8-11-13, and 1-4-6-10-12-13.

The activities are ranked according to the cost per week to crash.

Path Activity Cost ($)
1-2-5-7-11-13 1-2 $22
2-5 $24
5-7 $30
11-13 $33
1-3-8-11-13 1-3 $12
11-13 $33
8-11 $40
1-4-6-10-12-13 1-4 $10
6-10 $12
4-6 $13
10-12 $14
12-13 $26

Activity 11-13 could be crashed first by 1 week since it has the lowest crashing cost ($33). Paths 1-2-5-7-11-13 and 1-3-8-11-13 will decrease by 1 week.

Activity 1-4 could be crashed first by 1 week since it has the lowest crashing cost ($10). Path 1-4-6-10-12-13 will decrease by 1 week.

The combined crash cost ($43) is ≥ $40.

Since the marginal cost of crashing is greater than the marginal benefit of crashing, crashing will be stopped at step 4.

The final project duration time is 31 weeks. The activities that are crashed are:

Activity 7-11 (First week)

Activity 7-11 (Second week)

Activity 1-2

Activity 6-10

Activity 11-13

Activity 4-6

Calculation of total crashing cost:

The total crashing cost is calculated by summing the crashing cost involved all the steps and the indirect costs every week.

Total crashing cost=15+18+25+38+(31 weeks×$40)=96+1,240=$1,336

The activities to be crashed are: 7-11, 1-2, 6-10, 11-13, and 4-6. The total crashing cost is $1,336,000.

b)

Expert Solution
Check Mark
Summary Introduction

To Plot: The cost curve with the least expensive crashing which will reduce the project by 6 weeks.

Introduction:

Project crashing:

It is method to shorten the total time taken for a project by reducing the time taken for one or more activities on the critical path. The reduction in the normal time taken is known as crashing.

Answer to Problem 15P

Cost curve:

Operations Management (McGraw-Hill Series in Operations and Decision Sciences), Chapter 17, Problem 15P , additional homework tip  2

Explanation of Solution

Given information:

  • Indirect cost is $40,000 per week.

Operations Management (McGraw-Hill Series in Operations and Decision Sciences), Chapter 17, Problem 15P , additional homework tip  3

Activity Crash cost first week ($000) Crash cost second week ($000) Crash cost third week ($000)
1 to 2 18 22
2 to 5 24 25 25
5 to 7 30 30 35
7 to 11 15 20
11 to 13 30 33 36
1 to 3 12 24 26
3 to 8
8 to 11 40 40 40
3 to 9 3 10 12
9 to 12 2 7 10
12 to 13 26
1 to 4 10 15 25
4 to 6 8 13
6 to 10 5 12
10 to 12 14 15

Project crashing:

Calculation of expected duration of each path:

Path 1-2-5-7-11-13:

Expected duration=5+8+7+4+11=35

Path 1-3-8-11-13:

Expected duration=4+12+5+11=32

Path 1-3-9-12-13:

Expected duration=4+6+9+1=20

Path 1-4-6-10-12-13:

Expected duration=3+12+9+8+1=33

Step 1:

Critical path is 1-2-5-7-11-13.

The activities are ranked according to the cost per week to crash.

Activity Cost ($)
7-11 15
1-2 18
2-5 24
5-7 30
11-13 30

Activity 7-11 will be crashed first by 1 week since it has the lowest crashing cost ($15) and this cost is ≤ 40. Path 1-2-5-7-11-13 will decrease by 1 week.

Step 2:

The paths and new expected duration are:

Path Expected Duration
1-2-5-7-11-13 34
1-3-8-11-13 32
1-3-9-12-13 20
1-4-6-10-12-13 33

Critical path is 1-2-5-7-11-13.

The activities are ranked according to the cost per week to crash.

Activity Cost ($)
1-2 $18
7-11 $20
2-5 $24
5-7 $30
11-13 $30

Activity 1-2 will be crashed first by 1 week since it has the lowest crashing cost ($18) and this cost is ≤ 40. Path 1-2-5-7-11-13 will decrease by 1 week.

Step 3:

The paths and new expected duration are:

Path Expected Duration
1-2-5-7-11-13 33
1-3-8-11-13 32
1-3-9-12-13 20
1-4-6-10-12-13 33

Critical path is 1-2-5-7-11-13 and 1-4-6-10-12-13.

The activities are ranked according to the cost per week to crash.

Path Activity Cost ($)
1-2-5-7-11-13 7-11 $20
1-2 $22
2-5 $24
5-7 $30
11-13 $30
1-4-6-10-12-13 6-10 $5
4-6 $8
1-4 $10
10-12 $14
12-13 $26

Activity 7-11 will be crashed first by 1 week since it has the lowest crashing cost ($20). Path 1-2-5-7-11-13 will decrease by 1 week.

Activity 6-10 will be crashed first by 1 week since it has the lowest crashing cost ($5). Path 1-4-6-10-12-13 will decrease by 1 week.

The combined crash cost ($25) is ≤ $40.

Step 4:

The paths and new expected duration are:

Path Expected Duration
1-2-5-7-11-13 32
1-3-8-11-13 32
1-3-9-12-13 20
1-4-6-10-12-13 32

Critical path is 1-2-5-7-11-13, 1-3-8-11-13, and 1-4-6-10-12-13.

The activities are ranked according to the cost per week to crash.

Path Activity Cost ($)
1-2-5-7-11-13 1-2 $22
2-5 $24
5-7 $30
11-13 $30
1-3-8-11-13 1-3 $12
11-13 $30
8-11 $40
1-4-6-10-12-13 4-6 $8
1-4 $10
6-10 $12
10-12 $14
12-13 $26

Activity 11-13 will be crashed first by 1 week since it has the lowest crashing cost ($30). Paths 1-2-5-7-11-13 and 1-3-8-11-13 will decrease by 1 week.

Activity 4-6 will be crashed first by 1 week since it has the lowest crashing cost ($8). Path 1-4-6-10-12-13 will decrease by 1 week.

The combined crash cost ($38) is ≤ $40.

Step 5:

The paths and new expected duration are:

Path Expected Duration
1-2-5-7-11-13 31
1-3-8-11-13 31
1-3-9-12-13 20
1-4-6-10-12-13 31

Critical path is 1-2-5-7-11-13, 1-3-8-11-13, and 1-4-6-10-12-13.

The activities are ranked according to the cost per week to crash.

Path Activity Cost ($)
1-2-5-7-11-13 1-2 $22
2-5 $24
5-7 $30
11-13 $33
1-3-8-11-13 1-3 $12
11-13 $33
8-11 $40
1-4-6-10-12-13 1-4 $10
6-10 $12
4-6 $13
10-12 $14
12-13 $26

Activity 11-13 will be crashed first by 1 week since it has the lowest crashing cost ($33). Paths 1-2-5-7-11-13 and 1-3-8-11-13 will decrease by 1 week.

Activity 1-4 will be crashed first by 1 week since it has the lowest crashing cost ($10). Path 1-4-6-10-12-13 will decrease by 1 week. The combined crash cost is ($43).

Step 6:

The paths and new expected duration are:

Path Expected Duration
1-2-5-7-11-13 30
1-3-8-11-13 30
1-3-9-12-13 20
1-4-6-10-12-13 30

Critical path is 1-2-5-7-11-13, 1-3-8-11-13, and 1-4-6-10-12-13.

The activities are ranked according to the cost per week to crash.

Path Activity Cost ($)
1-2-5-7-11-13 1-2 $22
2-5 $24
5-7 $30
11-13 $36
1-3-8-11-13 1-3 $12
11-13 $36
8-11 $40
1-4-6-10-12-13 6-10 $12
4-6 $13
10-12 $14
1-4 $15
12-13 $26

Activity 1-2 will be crashed first by 1 week since it has the lowest crashing cost ($22). Path 1-2-5-7-11-13 will decrease by 1 week.

Activity 1-3 will be crashed first by 1 week since it has the lowest crashing cost ($12). Path 1-3-8-11-13 will decrease by 1 week.

Activity 6-10 will be crashed first by 1 week since it has the lowest crashing cost ($12). Path 1-4-6-10-12-13 will decrease by 1 week. The combined crash cost is ($436.

The final project duration time is 29 weeks. The activities that are crashed are:

Activity 7-11 (First week)

Activity 7-11 (Second week)

Activity 1-2 (First week)

Activity 1-2 (Second week)

Activity 11-13 (First week)

Activity 11-13 (Second week)

Activity 4-6

Activity 6-10

Activity 1-4

Activity 1-3

Calculation of total crashing cost:

The total crashing cost is calculated by summing the crashing cost involved all the steps and the indirect costs every week.

Total crashing cost=15+18+25+38+(29 weeks×$40)=185+1,160=$1,345

Summarization of total costs for different project lengths:

Project Length Cumulative Weeks shortened Cumulative crash cost ($000) Indirect cost ($000) Total cost ($000)
A B C (D=A×40) E = C+D
35 0  $                              -    $   1,400.00  $ 1,400.00
34 1  $                         15.00  $   1,360.00  $ 1,375.00
33 2  $                         33.00  $   1,320.00  $ 1,353.00
32 3  $                         58.00  $   1,280.00  $ 1,338.00
31 4  $                         96.00  $   1,240.00  $ 1,336.00
30 5  $                       139.00  $   1,200.00  $ 1,339.00
29 6  $                       185.00  $   1,160.00  $ 1,345.00

Cost curve:

The cost curve is plotted by taking the project length on the X-axis and the total cost on the Y-axis.

Operations Management (McGraw-Hill Series in Operations and Decision Sciences), Chapter 17, Problem 15P , additional homework tip  4

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Chapter 17 Solutions

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

Ch. 17 - Why might a person wish to be involved with a...Ch. 17 - Prob. 12DRQCh. 17 - What are some aspects of the project managers job...Ch. 17 - What is the main benefit of a project organization...Ch. 17 - What trade-offs are associated with time and cost...Ch. 17 - Who needs to be involved in assessing the cost of...Ch. 17 - Prob. 3TSCh. 17 - Project management techniques have been used...Ch. 17 - Give three examples of unethical conduct involving...Ch. 17 - For each of the following network diagrams,...Ch. 17 - Chris received new word processing software for...Ch. 17 - Prepare a Gantt chart for each of the following in...Ch. 17 - a. Develop a list of activities and their...Ch. 17 - For each of the problems listed, determine the...Ch. 17 - PRINTED BY: 92248ddb24ccbc6@placeholder.10274.edu....Ch. 17 - Three recent college graduates have formed a...Ch. 17 - The new director of special events at a large...Ch. 17 - PRINTED BY: 92248ddb24ccbc6@placeholder.10274.edu....Ch. 17 - The project described in the following table is...Ch. 17 - The following precedence diagram reflects three...Ch. 17 - A project manager has compiled a list of major...Ch. 17 - Here is a list of activity times for a project as...Ch. 17 - The project manager of a task force planning the...Ch. 17 - PRINTED BY: 92248ddb24ccbc6@placeholder.10274.edu....Ch. 17 - Chucks Custom Boats (CCB) builds luxury yachts to...Ch. 17 - Prob. 17PCh. 17 - Create a risk matrix in the style of Figure 17.13...Ch. 17 - Create a risk matrix for this project: Explain...Ch. 17 - CASE: The Case of the Mexican crazy quilt 1. The...Ch. 17 - CASE: The Case of the Mexican crazy quilt 2. The...Ch. 17 - CASE: The Case of the Mexican crazy quilt 3. The...Ch. 17 - CASE: The Case of the Mexican crazy quilt 4. The...Ch. 17 - CASE: The Case of the Mexican crazy quilt 5. The...Ch. 17 - B. Smitty Smith is a project manager for a large...
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