Chapter 17 – Project Management
17–60
Education.
1
6
A
D
B
C
E
F
17. Given:
We have the following activities and time estimates:
Path
Activity
Optimistic
Most Likely
Pessimistic
Top
A
4
5
6
B
7
8
10
C
3
5
9
Bottom
D
7
8
11
E
2
3
4
F
1
4
6
Draw the network:
Path
Activity
Optimistic
Most Likely
Pessimistic
Mean
Standard
Deviation
Top
A
4
5
6
5.00
4/36
B
7
8
10
8.17
9/36
C
3
5
9
5.33
36/36
Bottom
D
7
8
11
8.33
16/36
E
2
3
4
3.00
4/36
F
1
4
6
3.83
25/36
a. Determine the expected duration of the project.
Path A-B-C:
Expected completion time = 5.00 + 8.17 + 5.33 = 18.50
2
3
4
5
Chapter 17 – Project Management
17–61
Education.
b. Determine the probability that the project will take at least 18 days.
Probability (T ≥ 18):
Path A-B-C:
Path D-E-F:
Standard deviation =√
=.
Chapter 17 – Project Management
17–62
Education.
$80
$40
$20
$60
18. Given:
We are given the events, probabilities, and costs in the table below:
Event
Probability
Cost ($000)
1
.25
15
2
.35
25
3
.20
55
4
.80
10
5
.10
77
6
.40
55
7
.60
50
5
3 6
7
2
1
4
Cost
Chapter 17 – Project Management
17–63
$50
$100
$150
$200
19. Given:
We are given the events, probabilities, and costs in the table below:
Event
Probability
Cost ($000)
1. Equipment breakdown
.20
40
2. Vendor is late with key segment.
.60
200
3. Subcontractor has labor issues.
.30
140
4. Weather problems.
Unknown
15
5. Funding delays.
.40 to .60
50
6. Testing delays.
.40
20
Create a matrix for this project:
3
2
Cost
Chapter 17 – Project Management
17–64
Education.
Case: The Mexican Crazy Quilt
The case combines behavioral considerations in project management with development of a foreign
subsidiary. Although some students will have minor problems getting all the names straight, I think you
will find that the case produces good discussion as well as a realistic view of the sorts of difficulties often
encountered on projects.
1. Linderman Industries was correct to adopt a project organization for getting the Mexican
2. Naturally, the division managers would be opposed to releasing their best people to work on the
project. In fact, while the project might be important to top management, the division managers
3. Many people are quite content to work in a stable environment, where there is little chance of
unexpected events that might upset the established routine. These people are risk-adverse; they
4. Conway realized that he might “go to the well” once too often: not every argument would be
decided in his favor. Moreover, he undoubtedly recognized that these arguments stirred up a
5. To begin with, firms must recognize the potential problems, and then set up a mechanism to deal
with them before they occur. One possible approach to the problem might be to rotate people in
and out of the project. This would involve shorter worker absences from their regular jobs, which
17–65
Education.
Case: Time, Please
Given:
Path
Expected
Duration (weeks)
Standard
Deviation
A
10
4
B
14
2
C
13
2
Project duration for Smitty to include in the proposal for a 5% risk of not delivering the project on
time:
This means we would like a 0.9500 probability of completion within the specified time. When we take the
probability of completion within the specified time for each path and multiply these probabilities, we need
to obtain a joint probability of at least 0.9500.
We start by looking for a Z value in Appendix B Table B corresponding to 0.9500:
z = 1.645.
Next, we can analyze the length of each path if we use this corresponding z value (1.645):
Chapter 17 – Project Management
17–66
We can set up a spreadsheet in Excel to automate our calculations. To find a probability for a z value
using Excel, we enter the following formula: =NORMSDIST(z), where z = the cell where we have
calculated z.
Column
Row
A
B
C
D
E
1
Path
Mean
Sigma
z value
Prob.
2
A
10
4
1.8225
0.9658
3
B
14
2
1.645
0.9500
4
C
13
2
2.145
0.9840
5
Completion =
17.29
6
Prob. of Completion =
0.9028
We enter the following formulas in the spreadsheet above:
Cell D2: =($E$5-B2)/C2
Cell D3: =($E$5-B3)/C3
Cell D4: =($E$5-B4)/C4
Cell E2: =ROUND(NORMSDIST(D2),4)
From here, we can experiment by entering project completion times greater than 17.29 days until the
value in Cell E6 is close to 0.9500. An alternate approach would be to create a simple Excel Solver model
to solve this. Note: We will carry the project completion time to two decimals.
Column
Row
A
B
C
D
E
1
Path
Mean
Sigma
z value
Prob.
2
A
10
4
2.005
0.9775
3
B
14
2
2.01
0.9778
4
C
13
2
2.51
0.9940
5
Completion =
18.02
6
Prob. of Completion =
0.9501
Chapter 17 – Project Management
17–67
Education.
Project duration for Smitty to include in the proposal for a 10% risk of not delivering the project
on time:
This means we would like a 0.9000 probability of completion within the specified time.
Column
Row
A
B
C
D
E
1
Path
Mean
Sigma
z value
Prob.
2
A
10
4
1.815
0.9652
3
B
14
2
1.63
0.9484
4
C
13
2
2.13
0.9834
5
Completion =
17.26
6
Prob. of Completion =
0.9002
Project duration for Smitty to include in the proposal for a 15% risk of not delivering the project
on time:
This means we would like a 0.8500 probability of completion within the specified time.
We know that when we used a project completion of 17.26 days above, the probability of completion was
0.9002.
We can lower the project completion time from 17.26 until the value in Cell E6 is close to 0.8500. As
Column
Row
A
B
C
D
E
1
Path
Mean
Sigma
z value
Prob.
2
A
10
4
1.695
0.9550
3
B
14
2
1.39
0.9177
4
C
13
2
1.89
0.9706
5
Completion =
16.78
6
Prob. of Completion =
0.8506
The pros of quoting project times aggressively are the ability to secure new business or to win project
approval.
The cons of quoting project times aggressively are the increased probability of missing the deadline,
which could lead to severe consequences for the project manager.
The pros of quoting project times conservatively are the increased probability of meeting the deadline.
The cons of quoting project times conservatively are the reduced likelihood of being able to secure new
business or to gain project approval.