Chapter 17 - Project Management
CHAPTER 17
PROJECT MANAGEMENT
Solutions
1.
Use the network diagrams given in the text to determine both the critical path and the expected
project duration. The critical path and the expected project duration are indicated below with an
“*”. Expected duration for each path equals the sum of the activity times on the path.
a.
Path
b.
c.
Expected Duration
1–2–4–7–10–12
4 + 9 + 5 + 2 + 3 = 23
1–2–5–8–10–12
4 + 8 + 7 + 2 + 3 = 24
1–3–6–9–11–12
10 + 6 + 4 + 5 + 6 = 31*
Path
Expected Duration
1–2–4–6–8–9
5 + 18 + 3 + 4 + 9 + 2 = 41
1–2–5–6–8–9
5 + 18 + 10 + 4 + 9 + 2 = 48
1–2–5–7–8–9
5 + 18 + 10 + 11 + 9 + 2 = 55*
1–3–7–8–9
5 + 13 + 11 + 9 + 2 = 40
Path
Expected Duration
1–2–5–12–16
10 + 14 + 13 + 7 = 44*
1–3–6–13–16
14 + 15 + 6 + 4 = 39
1–3–7–14–16
14 + 11 + 13 + 3 = 41
1–4–8–9–10–11–15–16
3 + 8 + 1 + 0 + 7 + 6 + 10 = 35
1–4–8–10–11–15–16
3 + 8 + 4 + 7 + 6 + 10 = 38
Path
Expected Duration
A-D
7 + 8 = 15
B-E
3+4=7
C-F-E
5 + 7 + 4 = 16
C-F-G
5 + 7 + 6 = 18*
d.
17-1
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Education.
Chapter 17 - Project Management
2.
a.
Ch
Lib
Out
Sh
Sel
Inst
Write
Ck
Sub
Note: The answers for this problem assume that Chris will prepare the paper on the new
computer, but will prepare the outline by hand.
b. 1. AOA diagram (All times are in hours)
Lib
Ch
Start
Out
2.0
0.6
0.4
Sh
Inst
2.0
Sel
Write
Ck
Sub
3.0
0.5
0.2
Write 3.0
Ck 0.5
Sub 0.2
End
0.8
1.0
2. AON diagram (All times are in hours)
Ch 0.6
Lib 2.0
Out 0.4
Start
Sh 2.0
Sel 1.0
End
Inst 0.8
c. The paths and their expected duration times are given below. Expected duration for each path
equals the sum of the activity times on the path.
Path
Expected
Duration
Ch-Lib-Out-Write-Ck-Sub
6.7 hours
Sh-Sel-Inst-Write-Ck-Sub
7.5 hours*
Conclusion: Critical path is Sh-Sel-Inst-Write-Ck-Sub and the expected duration is 7.5
hours.
d. The project could take longer than the expected duration if any activity on the critical path is
delayed or if the other activities on the parallel path (Ch-Lib-Out) are delayed by more than
0.8 hours (7.5 – 6.7).
17-2
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Education.
Chapter 17 - Project Management
Blank forms to distribute to students if desired:
3a. Bank location.
Activity
1-2
1-3
2-4
2-5
3-5
4-5
5-6
2
4
6
8
Weeks after start
10
12
14
4
6
8
10
Weeks after start
12
14
16
18
4
6
8
10
Weeks after start
12
14
16
18
4
6
8
10
Weeks after start
12
14
16
18
16
18
20
b. Solved Problem #2.
Activity
1-2
2-5
2-4
1-3
3-4
4-5
2
3. a. Bank location.
Activity
1-2
1-3
2-4
2-5
3-5
4-5
5-6
2
20
b. Solved Problem #2.
Activity
1-2
2-5
2-4
1-3
3-4
4-5
2
17-3
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Education.
Chapter 17 - Project Management
4.
a.
C
A
E
Dummy
B
F
Immediate
Predecessor
–
–
Activity
A
B
C
E
F
End
A
C
A,B
E,F
17-4
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Education.
Chapter 17 - Project Management
4.
b. Case 1: Activity on-Arrow Diagram
D
2
A
B
1
5
3
E
K
F
H
6
C
End
8
I
G
4
7
Case 2: Activity-on-Arrow Diagram
L
5
M
2
N
6
J
1
K
V
R
3
Q
9
End
7
W
S
4
P
T
8
17-5
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Education.
Chapter 17 - Project Management
4.
c. Case 1: Activity-on-Node Diagram
A
D
K
Start
B
E
End
F
C
H
G
I
Case 2: Activity-on-Node Diagram
L
M
N
P
J
K
R
V
Start
S
End
Q
T
W
17-6
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Education.
Chapter 17 - Project Management
5.
a.
21
13
12
4
11
4
7
0
2
5
7
4
LS
ES
26
18
2
2
7
5
1
3
31
24
26
19
26
19
19
12
28
21
28
21
8
12
6
10
11
10
10
3
5
6
4
10
10
9
6
16
16
Summary:
Activity ES
1–2
0
20
20
16
16
EF
4
LS
7
LF
11
Slack
7
1–3
0
10
0
10
0
2–4
4
13
12
21
8
2–5
4
12
11
19
7
3–6
10
16
10
16
0
4–7
13
18
21
26
8
5–8
12
19
19
26
7
6–9
16
20
16
20
0
7–10
18
20
26
28
8
8–10
19
21
26
28
7
9–11
20
25
20
25
0
10–12
21
24
28
31
7
11–12
25
31
25
31
Note: Slack = LS – ES or LF – EF.
0
LF
EF
28
20
10
8
11
4
4
26
18
9
19
12
0
0
21
13
31
31
25
25
25
25
20
20
Conclusion: Critical path is 1-3-6-9-11-12. Expected project duration is 31.
17-7
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Education.
Chapter 17 - Project Management
5.
b.
Summary:
Activity
1
ES
0
EF
5
LS
0
LF
5
Slack
0
2
5
23
5
23
0
3
5
18
20
33
15
4
23
26
37
40
14
5
23
33
23
33
0
6
33
37
40
44
7
7
33
44
33
44
0
8
44
53
44
53
0
9
53
55
53
Note: Slack = LS – ES or LF – EF.
55
0
Conclusion: Critical path is 1-2-5-7-8-9. Expected project duration is 55.
LS
ES
37 40
23 26
5
5
23
23
LF
EF
40 44
33 37
4
3
2
6
23 33
23 33
18
0
0
4
5
5
44 53
44 53
5
53 55
53 55
10
1
5
20
5
33
18
33
33
44
44
8
9
9
2
3
7
13
11
17-8
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Education.
Chapter 17 - Project Management
6.
The network diagram is given in Problem 1, Part a. After 12 weeks, activities 1-2, 1-3, and 2-4
have been finished; activity 2-5 is 75% finished; and activity 3-6 is 50% finished. How many
weeks after the original start date time should the project be finished. The original path expected
durations are given below:
Path
Expected Duration
1–2–4–7–10–12
4 + 9 + 5 + 2 + 3 = 23
1–2–5–8–10–12
4 + 8 + 7 + 2 + 3 = 24
1–3–6–9–11–12
10 + 6 + 4 + 5 + 6 = 31*
One way of solving this problem is to look at the effects on each path:
Path 1-2-4-7-10-12:
Activities 1-2 and 2-4 were estimated to require take 4 + 9 = 13 weeks. They finished in 12
weeks. In other words, Activity 2-4 finished 1 week early. The net effect is to reduce the expected
time of this path by 1 week to 22 weeks.
Path 1-2-5-8-10-12:
Activities 1-2 and 2-5 were estimated to require 4 + 8 = 12 weeks. Activity 2-5 is 75%
completed; therefore, 25% of the activity time (.25 * 8 = 2 weeks) remains. Activity 2-5 should
have been completed at the end of 12 weeks but is running 2 weeks late. The net effect is to
increase the expected time of this path by 2 weeks to 26 weeks.
Path 1-3-6-9-11-12:
Activities 1-3 and 3-6 were estimated to require 10 + 6 = 16 weeks. Activity 3-6 is 50%
completed; therefore, 50% of the activity time (.50 * 6 = 3 weeks) remains. This means that this
activity should be completed 1 week early because it should be completed by the end of week 15.
The net effect is to reduce the expected time of this path by 1 week to 30 weeks.
Conclusion: The longest path (1-3-6-9-11-12) is estimated to take 30 weeks. Therefore, the
expected project duration is 30 weeks. The original estimated project duration was 31 weeks. We
estimate that the project should be completed 1 week early.
17-9
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Education.
Chapter 17 - Project Management
7.
a.
C
5
2
E
A
Start
F
3
D
G
6
End
8
1
B
I
H
4
b.
Activity
A
B
C
D
E
F
G
H
I
te
6.00
8.50
8.17
12.00
6.33
6.00
3.50
4.17
6.83
7
2
4/36
9/36
25/36
36/36
16/36
4/36
25/36
1/36
9/36
Path A-C-E:
Expected duration = 6.00 + 8.17 + 6.33 = 20.50
Variance = 4/36 + 25/36 + 16/36 = 45/36
45
Standard deviation = √36 = 1.118
Path D-F-G:
Expected duration = 12.00 + 6.00 + 3.50 = 21.50
Variance = 36/36 + 4/36 + 25/36 = 65/36
65
Standard deviation = √36 = 1.344
Path B-H-I:
Expected duration = 8.50 + 4.17 + 6.83 = 19.50
Variance = 9/36 + 1/36 + 9/36 = 19/36
19
Standard deviation = √36 = 0.726
𝑧=
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑒𝑑 𝑡𝑖𝑚𝑒 − 𝑃𝑎𝑡ℎ 𝑚𝑒𝑎𝑛
𝑃𝑎𝑡ℎ 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
17-10
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Education.
Chapter 17 - Project Management
Specified time = 24 days or less:
Path A-C-E:
𝑧=
24 − 20.50
= 3.13
1.118
Probability of completion ≤ 24 days = 1.00 (z value > +3.00—treat probability of completion
as being = 1.00)
Path D-F-G:
𝑧=
24 − 21.50
= 1.86
1.344
Using Appendix B Table B: Probability of completion ≤ 24 days = 0.9686
Path B-H-I:
𝑧=
24 − 19.50
= 6.20
0.726
Probability of completion ≤ 24 days = 1.00 (z value > +3.00—treat probability of completion
as being = 1.00)
Probability of completion ≤ 24 days = 1.00 x 0.9686 x 1.00 = 0.9686
Specified time = 21 days or less:
Path A-C-E:
𝑧=
21 − 20.50
= 0.45
1.118
Probability of completion ≤ 21 days = 0.6736
Path D-F-G:
𝑧=
21 − 21.50
= −0.37
1.344
Using Appendix B Table B: Probability of completion ≤ 21 days = 0.3557
Path B-H-I:
𝑧=
21 − 19.50
= 2.07
0.726
Probability of completion ≤ 21 days = 0.9808
Probability of completion ≤ 21 days = 0.6736 x 0.3557 x 0.9808 = 0.2350
17-11
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Education.
Chapter 17 - Project Management
Assume it is the end of the 7th day. Activities A and B have been completed while
Activity D is 50% completed. Time estimates for the completion time of Activity D are
5, 6, and 7. Activities C and H are ready to begin. We can re-draw the network as shown
below.
c.
Revised D: te = [5 + (4*6) + 7]/6 = 36/6 = 6.00
Revised Variance of D = (7 – 5)2 / 36 = 4/36
In the modified network, the expected times and variances are:
Activity
C
D
E
F
G
H
I
te
8.17
6.00
6.33
6.00
3.50
4.17
6.83
2
25/36
4/36
16/36
4/36
25/36
1/36
9/36
The revised project diagram is:
2
C (8.17)
End
7th
y
1
E (6.33)
D (6.00)
F (6.00)
3
G (3.50)
5
H (4.17)
6
I (6.83)
4
Path C-E:
Expected duration from start of project = 7 + (8.17 + 6.33) = 21.50
Variance = 25/36 + 16/36 = 41/36
41
Standard deviation = √36 = 1.067
Path D-F-G:
Expected duration from start of project = 7 + (6.00 + 6.00 + 3.50) = 22.50
Variance = 4/36 + 4/36 + 25/36 = 33/36
33
36
Standard deviation = √
= 0.957
17-12
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Education.
Chapter 17 - Project Management
Path H-I:
Expected duration from start of project = 7 + (4.17 + 6.83) = 18.00
Variance = 1/36 + 9/36 = 10/36
10
Standard deviation = √36 = 0.527
𝑧=
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑒𝑑 𝑡𝑖𝑚𝑒 − 𝑃𝑎𝑡ℎ 𝑚𝑒𝑎𝑛
𝑃𝑎𝑡ℎ 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
Specified time = 24 days or less:
Path C-E:
𝑧=
24 − 21.50
= 2.34
1.067
Using Appendix B Table B: Probability of completion ≤ 24 days = 0.9904
Path D-F-G:
𝑧=
24 − 22.50
= 1.57
0.957
Using Appendix B Table B: Probability of completion ≤ 24 days = 0.9418
Path H-I:
𝑧=
24 − 18.00
= 11.39
0.527
Probability of completion ≤ 24 days = 1.0000 (z value > +3.00—treat probability of
completion as being = 1.0000)
Probability of completion ≤ 24 days = 0.9904 x 0.9418 x 1.0000 = 0.9328
Specified time = 21 days or less:
Path C-E:
𝑧=
21 − 21.50
= −0.47
1.067
Using Appendix B Table B: Probability of completion ≤ 21 days = 0.3192
Path D-F-G:
𝑧=
21 − 22.50
= −1.57
0.957
Using Appendix B Table B: Probability of completion ≤ 21 days = 0.0582
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Education.
Chapter 17 - Project Management
Path H-I:
𝑧=
21 − 18.00
= 5.69
0.527
Probability of completion ≤ 21 days = 1.0000 (z value > +3.00—treat probability of
completion as being = 1.0000)
Probability of completion ≤ 21 days = 0.3192 x 0.0582 x 1.0000 = 0.0186
d.
The partners want to shorten the project by 2 days as long as it does not cost more than
$20,000. The cost per day (in thousands) to crash activities is shown below:
Activity
C
D
E
F
G
H
I
First Crash
$8
$10
$9
$7
$8
$7
$6
Second Crash
$10
$11
$10
$9
$9
$8
$8
Step 1:
The paths and their expected duration are shown below:
Path
Expected Duration
C-E
21.50
D-F-G
22.50
H-I
Critical path is D-F-G.
18.00
Rank critical activities according to crash costs:
Activity
F
G
D
Cost per Day
to Crash
$7
$8
$10
Activity F should be shortened 1 day because it has the lowest crashing cost ($7,000).
Path D-F-G now will be 21.50 days as shown below.
17-14
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Education.
Chapter 17 - Project Management
Step 2:
The paths and their expected duration are shown below:
Path
Expected Duration
C-E
21.50
D-F-G
21.50
H-I
18.00
Critical paths are C-E and D-F-G.
Rank critical activities according to crash costs:
Path
C-E
D-F-G
Activity
C
E
G
F
D
Cost per Day
to Crash
$8
$9
$8
$9
$10
Choose one activity on each path to crash:
Activity C should be shortened 1 day because it has the lowest crashing cost ($8,000) on
Path C-E.
Activity G should be shortened 1 day because it has the lowest crashing cost ($8,000) on
Path D-F-G.
Path C-E now will be 20.50 days.
Path D-F-G now will be 20.50 days.
Conclusion: Total crashing cost is $7,000 + $8,000 + $8,000 = $23,000, which is $3,000
over budget. The partners have to determine whether the benefits of crashing the project
by 1 day or by 2 days are worth the extra cost.
17-15
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Education.
Chapter 17 - Project Management
8.
Path
A
B
C
D
E
Expected
Duration
10
8
12
15
14
Variance
1.21
2.00
1.00
2.89
1.44
Std. Dev.
1.100
1.414
1.000
1.700
1.200
a. Probability (Time ≤ 16 weeks):
𝒛=
𝑺𝒑𝒆𝒄𝒊𝒇𝒊𝒆𝒅 𝒕𝒊𝒎𝒆 − 𝑷𝒂𝒕𝒉 𝒎𝒆𝒂𝒏
𝑷𝒂𝒕𝒉 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏
Path A:
16 − 10
𝑧=
= 5.45
1.100
Probability of completion ≤ 16 weeks = 1.0000 (z value > +3.00—treat probability of completion
as being = 1.0000)
Path B:
16 − 8
𝑧=
= 5.66
1.414
Probability of completion ≤ 16 weeks = 1.0000 (z value > +3.00—treat probability of completion
as being = 1.0000)
Path C:
16 − 12
𝑧=
= 4.00
1.000
Probability of completion ≤ 16 weeks = 1.0000 (z value > +3.00—treat probability of completion
as being = 1.0000)
Path D:
16 − 15
𝑧=
= 0.59
1.700
Using Appendix B Table B: Probability of completion ≤ 16 weeks = 0.7224
Path E:
16 − 14
𝑧=
= 1.67
1.200
Using Appendix B Table B: Probability of completion ≤ 16 weeks = 0.9525
Conclusion: Probability (T  16) = 1.0000 x 1.0000 x 1.0000 x 0.7224 x 0.9525 = 0.6881
17-16
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Education.
Chapter 17 - Project Management
b. Probability (Time ≤ 15 weeks):
𝒛=
𝑺𝒑𝒆𝒄𝒊𝒇𝒊𝒆𝒅 𝒕𝒊𝒎𝒆 − 𝑷𝒂𝒕𝒉 𝒎𝒆𝒂𝒏
𝑷𝒂𝒕𝒉 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏
Path A:
15 − 10
𝑧=
= 4.55
1.100
Probability of completion ≤ 15 weeks = 1.0000 (z value > +3.00—treat probability of completion
as being = 1.0000)
Path B:
15 − 8
𝑧=
= 4.95
1.414
Probability of completion ≤ 15 weeks = 1.0000 (z value > +3.00—treat probability of completion
as being = 1.0000)
Path C:
15 − 12
𝑧=
= 3.00
1.000
Using Appendix B Table B: Probability of completion ≤ 15 weeks = 0.9987
Path D:
15 − 15
𝑧=
= 0.00
1.700
Using Appendix B Table B: Probability of completion ≤ 15 weeks = 0.5000
Path E:
15 − 14
𝑧=
= 0.83
1.200
Using Appendix B Table B: Probability of completion ≤ 15 weeks = 0.7967
Conclusion: Probability (T  15) = 1.0000 x 1.0000 x 0.9987 x 0.5000 x 0.7967 = 0.3978
17-17
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Education.
Chapter 17 - Project Management
c. Probability (Time ≤ 13 weeks):
𝒛=
𝑺𝒑𝒆𝒄𝒊𝒇𝒊𝒆𝒅 𝒕𝒊𝒎𝒆 − 𝑷𝒂𝒕𝒉 𝒎𝒆𝒂𝒏
𝑷𝒂𝒕𝒉 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏
Path A:
13 − 10
𝑧=
= 2.73
1.100
Using Appendix B Table B: Probability of completion ≤ 13 weeks = 0.9968
Path B:
13 − 8
𝑧=
= 3.54
1.414
Probability of completion ≤ 13 weeks = 1.0000 (z value > +3.00—treat probability of completion
as being = 1.0000)
Path C:
13 − 12
𝑧=
= 1.00
1.000
Using Appendix B Table B: Probability of completion ≤ 13 weeks = 0.8413
Path D:
13 − 15
𝑧=
= −1.18
1.700
Using Appendix B Table B: Probability of completion ≤ 13 weeks = 0.1190
Path E:
13 − 14
𝑧=
= −0.83
1.200
Using Appendix B Table B: Probability of completion ≤ 13 weeks = 0.2033
Conclusion: Probability (T  13) = 0.9968 x 1.0000 x 0.8413 x 0.1190 x 0.2033 = 0.0203
17-18
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Education.
Chapter 17 - Project Management
9.
Given:
Path
Activity
Mean
Standard
Deviation
A
C
5
1.3
D
4
1.0
E
8
1.6
B
Probability (T > 10 weeks):
𝒛=
𝑺𝒑𝒆𝒄𝒊𝒇𝒊𝒆𝒅 𝒕𝒊𝒎𝒆 − 𝑷𝒂𝒕𝒉 𝒎𝒆𝒂𝒏
𝑷𝒂𝒕𝒉 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏
Path A:
Expected duration = 5 + 4 = 9
Standard deviation:
We must find the variance of the path first by squaring the individual standard deviations to find
the variance for each activity, add the variances, and take the square root of the sum.
Variance = 1.32 + 1.02 = 1.690 + 1.000 = 2.690
Standard deviation = √2.690 = 1.640
𝑧=
10 − 9
= 0.61
1.640
Using Appendix B Table B: Probability of completion ≤ 10 weeks = 0.7291
Path B:
Expected duration = 8
Standard deviation = 1.600
10 − 8
= 1.25
1.600
Using Appendix B Table B: Probability of completion ≤ 10 weeks = 0.8944
𝑧=
Conclusion: Probability (T ≤ 10 weeks) = 0.7291 x 0.8944 = 0.6521
Probability (T > 10 weeks) = 1.0000 – 0.6521 = 0.3479
17-19
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Education.
Chapter 17 - Project Management
10.
Given:
The project described below is scheduled to be completed in 11 weeks.
Path
Activity
Mean
Standard
Deviation
A
C
4
0.70
D
6
0.90
E
3
0.62
F
9
1.90
B
a. If you were the manager of this project, would you be concerned? Explain.
Probability (T ≤ 11 weeks):
𝒛=
𝑺𝒑𝒆𝒄𝒊𝒇𝒊𝒆𝒅 𝒕𝒊𝒎𝒆 − 𝑷𝒂𝒕𝒉 𝒎𝒆𝒂𝒏
𝑷𝒂𝒕𝒉 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏
Path A:
Expected duration = 4 + 6 = 10
Standard deviation:
We must find the variance of the path first by squaring the individual standard deviations to
find the variance for each activity, add the variances, and take the square root of the sum.
Variance = 0.702 + 0.902 = 0.490 + 0.810 = 1.300
Standard deviation = √1.300 = 1.140
11 − 10
= 0.88
1.140
Using Appendix B Table B: Probability of completion ≤ 11 weeks = 0.8106
𝑧=
Path B:
Expected duration = 3 + 9 = 12
Standard deviation:
We must find the variance of the path first by squaring the individual standard deviations to
find the variance for each activity, add the variances, and take the square root of the sum.
Variance = 0.622 + 1.902 = 0.384 + 3.610 = 3.994
Standard deviation = √3.994 = 1.998
11 − 12
= −0.50
1.998
Using Appendix B Table B: Probability of completion ≤ 11 weeks = 0.3085
𝑧=
Conclusion: Probability (T ≤ 11 weeks) = 0.8106 x 0.3085 = 0.2501. The manager should be
concerned because there is only a 0.2501 probability of finishing on time.
17-20
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Education.
Chapter 17 - Project Management
b. If there is a penalty of $5,000 a week for each week the project is late, what is the probability
of incurring a penalty of at least $5,000?
This question is asking what the probability will be that the project will take at least 12 weeks
(which would be 1 week late).
Probability (T ≥ 12 weeks):
𝒛=
𝑺𝒑𝒆𝒄𝒊𝒇𝒊𝒆𝒅 𝒕𝒊𝒎𝒆 − 𝑷𝒂𝒕𝒉 𝒎𝒆𝒂𝒏
𝑷𝒂𝒕𝒉 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏
Path A:
12 − 10
= 1.75
1.140
Using Appendix B Table B: Probability of completion < 12 weeks = 0.9599
𝑧=
Path B:
12 − 12
= 0.00
1.998
Using Appendix B Table B: Probability of completion < 12 weeks = 0.5000
𝑧=
Conclusion: Probability (T < 12 weeks) = 0.9599 x 0.5000 = 0.4800
Probability (T ≥ 12 weeks) = 1.0000 – 0.4800 = 0.5200
17-21
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Education.
Chapter 17 - Project Management
11.
Given:
We are given the following times (optimistic, most likely, and pessimistic) for each activity in the
table below.
a. Determine the expected completion time for each path and its variance.
Activity
Optimistic
Most
Likely
1-2
8
8
8
8.00
0/36
2-3
11
12
13
12.00
4/36
2-4
5
6
7
6.00
4/36
2-5
10
12
14
12.00
16/36
3-8
9
10
12
10.17
9/36
4-6
14
18
26
18.67
144/36
4-7
13
13
13
13.00
0/36
5-9
7
10
12
9.83
25/36
6-11
8
10
14
10.33
36/36
7-11
10.5
13
15.5
13.00
25/36
8-11
5
7
10
7.17
25/36
9-10
10
11
12
11.00
4/36
10-11
6
6
6
6.00
0/36
Pessimistic
Mean
Variance
Path 1-2-3-8-11:
Expected completion time = 8.00 + 12.00 + 10.17 + 7.17 = 37.34
Variance = 0/36 + 4/36 + 9/36 + 25/36 = 38/36
38
Standard deviation =√36 = 1.027
Path 1-2-4-6-11:
Expected completion time = 8.00 + 6.00 + 18.67 + 10.33 = 43.00
Variance = 0/36 + 4/36 + 144/36 + 36/36 = 184/36
184
Standard deviation =√ 36 = 2.261
Path 1-2-4-7-11:
Expected completion time = 8.00 + 6.00 + 13.00 + 13.00 = 40.00
Variance = 0/36 + 4/36 + 0/36 + 25/36 = 29/36
29
Standard deviation =√36 = 0.898
17-22
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Education.
Chapter 17 - Project Management
Path 1-2-5-9-10-11:
Expected completion time = 8.00 + 12.00 + 9.83 + 11.00 + 6.00 = 46.83
Variance = 0/36 + 16/36 + 25/36 + 4/36 + 0/36 = 45/36
45
Standard deviation =√36 = 1.118
b. Determine the probability that the project will require more than 49 weeks.
Probability (T > 49 weeks):
Path 1-2-3-8-11:
49 − 37.34
𝑧=
= 11.35
1.027
Probability of completion ≤ 49 weeks = 1.0000 (z value > +3.00—treat probability of
completion as being = 1.0000)
Path 1-2-4-6-11:
49 − 43.00
𝑧=
= 2.65
2.261
Using Appendix B Table B: Probability of completion ≤ 49 weeks = 0.9960
Path 1-2-4-7-11:
49 − 40.00
𝑧=
= 10.02
0.898
Probability of completion ≤ 49 weeks = 1.0000 (z value > +3.00—treat probability of
completion as being = 1.0000)
Path 1-2-5-9-10-11:
49 − 46.83
𝑧=
= 1.94
1.118
Using Appendix B Table B: Probability of completion ≤ 49 weeks = 0.9738
Conclusion: Probability (T ≤ 49 weeks) = 1.0000 x 0.9960 x 1.0000 x 0.9738 = 0.9699
Probability (T > 49 weeks) = 1.0000 – 0.9699 = 0.0301
17-23
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Education.
Chapter 17 - Project Management
c. Determine the probability that the project will be completed in 46 weeks or less.
Probability (T ≤ 46 weeks):
Path 1-2-3-8-11:
46 − 37.34
𝑧=
= 8.43
1.027
Probability of completion ≤ 46 weeks = 1.0000 (z value > +3.00—treat probability of
completion as being = 1.0000)
Path 1-2-4-6-11:
46 − 43.00
𝑧=
= 1.33
2.261
Using Appendix B Table B: Probability of completion ≤ 46 weeks = 0.9082
Path 1-2-4-7-11:
46 − 40.00
𝑧=
= 6.68
0.898
Probability of completion ≤ 46 weeks = 1.0000 (z value > +3.00—treat probability of
completion as being = 1.0000)
Path 1-2-5-9-10-11:
46 − 46.83
𝑧=
= −0.74
1.118
Using Appendix B Table B: Probability of completion ≤ 46 weeks = 0.2296
Conclusion: Probability (T ≤ 46 weeks) = 1.0000 x 0.9082 x 1.0000 x 0.2296 = 0.2085
17-24
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Education.
Chapter 17 - Project Management
12.
Given:
We are given the following times (optimistic, most likely, and pessimistic) for each activity in the
table below.
Activity
Optimistic
Most
Likely
Pessimistic
Mean
Variance
A
2
4
6
4.00
16/36
D
6
8
10
8.00
16/36
E
7
9
12
9.17
25/36
H
2
3
5
3.17
9/36
F
3
4
8
4.50
25/36
G
5
7
9
7.00
16/36
B
2
2
3
2.17
1/36
I
2
3
6
3.33
16/36
J
3
4
5
4.00
4/36
K
4
5
8
5.33
16/36
C
5
8
12
8.17
49/36
M
1
1
1
1.00
0/36
N
6
7
11
7.50
25/36
O
8
9
13
9.50
25/36
End
a. Construct a network diagram.
D
E
5
2
9
F
A
G
6
Start
1
H
12
B
End
K
C
3
I
J
7
M
4
10
N
8
O
11
17-25
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Education.
Chapter 17 - Project Management
b. The project manager will receive a bonus of $1,000 if the project is finished within 26 weeks.
P (T ≤ 26):
Path A-D-E-H:
Expected completion time = 4.00 + 8.00 + 9.17 + 3.17 = 24.34
Variance = 16/36 + 16/36 + 25/36 + 9/36 = 66/36
66
Standard deviation =√36 = 1.354
26 − 24.34
= 1.23
1.354
Using Appendix B Table B: Probability of completion ≤ 26 weeks = 0.8907
𝑧=
Path A-F-G:
Expected completion time = 4.00 + 4.50 + 7.00 = 15.50
Variance = 16/36 + 25/36 + 16/36 = 57/36
57
Standard deviation =√36 = 1.258
26 − 15.50
= 8.35
1.258
Probability of completion ≤ 26 weeks = 1.0000 (z value > +3.00—treat probability of
completion as being = 1.0000)
𝑧=
Path B-I-J-K:
Expected completion time = 2.17 + 3.33 + 4.00 + 5.33 = 14.83
Variance = 1/36 + 16/36 + 4/36 + 16/36 = 37/36
37
Standard deviation =√36 = 1.014
26 − 14.83
= 11.02
1.014
Probability of completion ≤ 26 weeks = 1.0000 (z value > +3.00—treat probability of
completion as being = 1.0000)
𝑧=
Path C-M-N-O:
Expected completion time = 8.17 + 1.00 + 7.50 + 9.50 = 26.17
Variance = 49/36 + 0/36 + 25/36 + 25/36 = 99/36
99
Standard deviation =√36 = 1.658
26 − 26.17
= −0.10
1.658
Using Appendix B Table B: Probability of completion ≤ 26 weeks = 0.4602
Conclusion: The probability of receiving a bonus of $1,000 = Probability (T ≤ 26 weeks) =
0.8907 x 1.0000 x 1.0000 x 0.4602 = 0.4099
𝑧=
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Education.
Chapter 17 - Project Management
The project manager will receive a bonus of $500 if the project is finished within 27 weeks.
P (T ≤ 27):
Path A-D-E-H:
27 − 24.34
𝑧=
= 1.96
1.354
Using Appendix B Table B: Probability of completion ≤ 27 weeks = 0.9750
Path A-F-G:
27 − 15.50
𝑧=
= 9.14
1.258
Probability of completion ≤ 27 weeks = 1.0000 (z value > +3.00—treat probability of
completion as being = 1.0000)
Path B-I-J-K:
27 − 14.83
𝑧=
= 12.00
1.014
Probability of completion ≤ 27 weeks = 1.0000 (z value > +3.00—treat probability of
completion as being = 1.0000)
Path C-M-N-O:
27 − 26.17
𝑧=
= 0.50
1.658
Using Appendix B Table B: Probability of completion ≤ 27 weeks = 0.6915
Conclusion: The probability of receiving a bonus of $500 = Probability (T ≤ 27 weeks) =
0.9750 x 1.0000 x 1.0000 x 0.6915 = 0.6742
17-27
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Education.
Chapter 17 - Project Management
13.
Given:
We have the following list of activities for a project and crashing costs:
Path
Top
Activity
A
B
C
Duration (weeks)
5
6
3
First Crash
$8
$7
$14
Second Crash
$10
$9
$15
Middle
D
E
C
3
7
3
$9
$8
$14
$11
$9
$15
Bottom
F
G
H
5
5
5
$10
$11
$12
$15
$13
$14
Determine which activities should be crashed to shorten the project by 3 weeks as cheaply as
possible. Draw the precedence diagram first.
B
2
4
C
A
E
7
D
3
1
H
6
F
G
5
17-28
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Education.
Chapter 17 - Project Management
Step 1:
The paths and their expected duration are shown below:
Path
Expected Duration
A-B-C
14
D-E-C
13
F-G-H
15
Critical path is F-G-H.
Rank critical activities according to crash costs:
Activity
F
G
H
Cost per Week
to Crash
$10
$11
$12
Activity F should be shortened 1 week because it has the lowest crashing cost ($10). Path
F-G-H now will be 14 weeks as shown below.
Activities crashed to this point: Step 1: F (first crash).
Step 2:
The paths and their expected duration are shown below:
Path
Expected Duration
A-B-C
14
D-E-C
13
F-G-H
Critical paths are A-B-C and F-G-H.
14
Rank critical activities according to crash costs:
Path
A-B-C
F-G-H
Activity
B
A
C
G
H
F
Cost per Week
to Crash
$7
$8
$14
$11
$12
$15
17-29
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Education.
Chapter 17 - Project Management
Choose one activity on each path to crash:
Activity B should be shortened 1 week because it has the lowest crashing cost ($7) on
Path A-B-C.
Activity G should be shortened 1 week because it has the lowest crashing cost ($11) on
Path F-G-H.
Path A-B-C now will be 13 weeks as shown below.
Path D-F-G now will be 13 weeks as shown below.
Activities crashed to this point: Step 1: F (first crash). Step 2: B (first crash) & G (first
crash).
Step 3:
The paths and their expected duration are shown below:
Path
Expected Duration
A-B-C
13
D-E-C
13
F-G-H
Critical paths are A-B-C, D-E-C, and F-G-H.
13
Rank critical activities according to crash costs:
Path
A-B-C
D-E-C
F-G-H
Activity
A
B
C
E
D
C
H
G
F
Cost per Week
to Crash
$8
$9
$14
$8
$9
$14
$12
$13
$15
Choose one activity on each path to crash:
At first glance, it appears that we should crash Activity A on Path A-B-C and Activity E
on Path D-E-C. The combined crashing cost would be $8 + $8 = $16. However, Activity
C is on both paths and could be crashed for $14.
Activity C should be shortened 1 week at a crashing cost of $14.
Activity H should be shortened 1 week because it has the lowest crashing cost ($12) on
Path F-G-H.
Paths A-B-C, D-E-C, and F-G-H will decrease to 12 weeks.
Activities crashed to this point: Step 1: F (first crash). Step 2: B (first crash) & G (first
crash). Step 3: C (first crash) & H (first crash).
17-30
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Education.
Chapter 17 - Project Management
A summary of the crashing steps is shown below:
Path
A-B-C
D-E-C
F-G-H
Initial time
14 wk.
13 wk.
15 wk.
Step 1 Crashing
Activity
Cost
F
$10
Length after
crashing 1 week
14 wk.
13 wk.
14 wk.
Length after
crashing 2 weeks
13 wk.
13 wk.
13 wk.
Step 2 Crashing
Activity
Cost
B
$7
G
11
$18
Length after
crashing 3 weeks
12 wk.
12 wk.
12 wk.
Step 3 Crashing
Activity
Cost
C
$14
H
12
$26
Conclusion: Expected project duration = 12 weeks. Total crashing cost = $10 + $18 + $26 = $54.
17-31
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Education.
Chapter 17 - Project Management
14.
Given:
We have the activities and crashing costs shown below:
Immediate
Predecessor
--A
--C
C
C
D
E
F
I
B
G
H
H, J
K, M, N, P
Activity
A
B
C
D
E
F
G
H
I
J
K
M
N
P
End
Normal Time
(weeks)
12
14
10
17
18
12
15
8
7
12
9
3
11
8
Crashing Cost ($000)
First Week
$15
$10
$5
$20
$16
$12
$24
--$30
$25
$10
--$40
$20
Crashing Cost ($000)
Second Week
$20
$10
$5
$21
$18
$15
$24
----$25
$10
----$20
Construct a network diagram:
B
A
K
Start
End
M
C
G
D
E
N
P
H
F
J
I
17-32
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Education.
Chapter 17 - Project Management
Determine a minimum-cost crashing schedule that will shave five weeks off the project length:
Step 1:
The paths and their expected duration are shown below:
Path
Expected Duration
A-B-K
35
C-D-G-M
45
C-E-H-N
47
C-E-H-P
44
C-F-I-J-P
49
Critical path is C-F-I-J-P.
Rank critical activities according to crash costs:
Cost per Week
to Crash
Activity
C
$5
F
$12
P
$20
J
$25
I
$30
Activity C should be shortened 1 week because it has the lowest crashing cost ($5).
Paths C-D-G-M, C-E-H-N, C-E-H-P, and C-F-I-J-P all will decrease 1 week as shown below.
Activities crashed to this point: Step 1: C (first week).
Step 2:
The paths and their expected duration are shown below:
Path
Expected Duration
A-B-K
35
C-D-G-M
44
C-E-H-N
46
C-E-H-P
43
C-F-I-J-P
48
Critical path is C-F-I-J-P.
Rank critical activities according to crash costs:
Cost per Week
to Crash
Activity
C
$5
F
$12
P
$20
J
$25
I
$30
Activity C should be shortened 1 week because it has the lowest crashing cost ($5).
Paths C-D-G-M, C-E-H-N, C-E-H-P, and C-F-I-J-P all will decrease 1 week.
Activities crashed to this point: Step 1: C (first week). Step 2: C (second week).
17-33
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Education.
Chapter 17 - Project Management
Step 3:
The paths and their expected duration are shown below:
Path
Expected Duration
A-B-K
35
C-D-G-M
43
C-E-H-N
45
C-E-H-P
42
C-F-I-J-P
47
Critical path is C-F-I-J-P.
Rank critical activities according to crash costs:
Cost per Week
to Crash
Activity
F
$12
P
$20
J
$25
I
$30
Activity F should be shortened 1 week because it has the lowest crashing cost ($12).
Path C-F-I-J-P will decrease 1 week as shown below.
Activities crashed to this point: Step 1: C (first week). Step 2: C (second week). Step 3: F (first
week).
Step 4:
The paths and their expected duration are shown below:
Path
Expected Duration
A-B-K
35
C-D-G-M
43
C-E-H-N
45
C-E-H-P
42
C-F-I-J-P
46
Critical path is C-F-I-J-P.
Rank critical activities according to crash costs:
Cost per Week
to Crash
Activity
F
$15
P
$20
J
$25
I
$30
Activity F should be shortened 1 week because it has the lowest crashing cost ($15).
Path C-F-I-J-P will decrease 1 week as shown below.
Activities crashed to this point: Step 1: C (first week). Step 2: C (second week). Step 3: F (first
week). Step 4: F (second week).
17-34
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Education.
Chapter 17 - Project Management
Step 5:
The paths and their expected duration are shown below:
Path
Expected Duration
A-B-K
35
C-D-G-M
43
C-E-H-N
45
C-E-H-P
42
C-F-I-J-P
45
Critical paths are C-E-H-N and C-F-I-J-P.
Rank critical activities according to crash costs:
Cost per Week
to Crash
Activity
E
$16
N
$40
C-F-I-J-P
P
$20
J
$25
I
$30
Activity E should be shortened 1 week because it has the lowest crashing cost ($16) on Path C-EH-N. Paths C-E-H-N and C-E-H-P each will decrease 1 week.
Activity P should be shortened 1 week because it has the lowest crashing cost ($20) on Path C-FI-J-P. Paths C-E-H-P and C-F-I-J-P each will decrease 1 week as shown below.
Activities crashed to this point: Step 1: C (first week). Step 2: C (second week). Step 3: F (first
week). Step 4: F (second week). Step 5: E (first week) & P (first week).
Path
C-E-H-N
The final duration for each path is shown below:
Path
Expected Duration
A-B-K
35
C-D-G-M
43
C-E-H-N
44
C-E-H-P
40
C-F-I-J-P
44
17-35
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Education.
Chapter 17 - Project Management
A summary of the crashing steps is shown below:
Path
A-B-K
C-D-G-M
C-E-H-N
C-E-H-P
C-F-I-J-P
Initial
time
35 wk.
45 wk.
47 wk.
44 wk.
49 wk.
Step 1 Crashing
Activity
Cost
C
$5
Length after
crashing 1
week
35 wk.
44 wk.
46 wk.
43 wk.
48 wk.
Step 2 Crashing
Activity
Cost
C
$5
Length after
crashing 2
weeks
35 wk.
43 wk.
45 wk.
42 wk.
47 wk.
Length after
crashing 3
weeks
35 wk.
43 wk.
45 wk.
42 wk.
46 wk.
Step 3 Crashing
Activity
Cost
F
$12
Length after
crashing 4
weeks
35 wk.
43 wk.
45 wk.
42 wk.
45 wk.
Length after
crashing 5
weeks
35 wk.
43 wk.
44 wk.
40 wk.
44 wk.
Step 4 Crashing Step 5 Crashing
Activity
Cost Activity
Cost
F
$15 E
$16
P
20
$36
Conclusion: Expected project duration = 44 weeks. Total crashing cost = $5 + $5 + $12 + $15 + $36 =
$73(000) = $73,000.
17-36
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Education.
Chapter 17 - Project Management
15.
Given:
We have the following activities and crashing costs ($000):
Crashing Costs ($000)
Activity
First
Week
Second
Week
Third
Week
1-2
$18
$22
---
2-5
$24
$25
$25
5-7
$30
$30
$35
7-11
$15
$20
---
11-13
$30
$33
$36
1-3
$12
$24
$26
3-8
---
---
---
8-11
$40
$40
$40
3-9
$3
$10
$12
9-12
$2
$7
$10
12-13
$26
---
---
1-4
$10
$15
$25
4-6
$8
$13
---
6-10
$5
$12
---
10-12
$14
$15
---
Indirect costs total $40(000) per week.
17-37
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Education.
Chapter 17 - Project Management
a. Determine the optimum time-cost crashing plan.
We will continue crashing as long as the crashing cost at a given step ≤ $40.
Step 1:
The paths and their expected duration are shown below:
Path
Expected Duration
1-2-5-7-11-13
35
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.
Rank critical activities according to crash costs:
Cost per Week
to Crash
Activity
7-11
$15
1-2
$18
2-5
$24
5-7
$30
11-13
$30
Activity 7-11 should be shortened 1 week because it has the lowest crashing cost ($15) and
this cost is ≤ $40.
Path 1-2-5-7-11-13 will decrease 1 week as shown below.
Activities crashed to this point: Step 1: 7-11 (first week).
Step 2:
The paths and their expected duration are shown below:
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.
Rank critical activities according to crash costs:
Cost per Week
to Crash
Activity
1-2
$18
7-11
$20
2-5
$24
5-7
$30
11-13
$30
Activity 1-2 should be shortened 1 week because it has the lowest crashing cost ($18) and
this cost is ≤ $40. Path 1-2-5-7-11-13 will decrease 1 week as shown below.
Activities crashed to this point: Step 1: 7-11 (first week). Step 2: 1-2 (first week).
17-38
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Education.
Chapter 17 - Project Management
Step 3:
The paths and their expected duration are shown below:
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 paths are 1-2-5-7-11-13 & 1-4-6-10-12-13.
Rank critical activities according to crash costs:
Cost per Week
to Crash
Path
Activity
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 should be shortened 1 week because it has the lowest crashing cost ($20) on
Path 1-2-5-7-11-13, which will decrease 1 week as shown below.
Activity 6-10 should be shortened 1 week because it has the lowest crashing cost ($5) on
Path 1-4-6-10-12-13, which will decrease 1 week as shown below.
The combined crashing cost = $25 and is ≤ $40.
Activities crashed to this point: Step 1: 7-11 (first week). Step 2: 1-2 (first week). Step 3: 711 (second week) & 6-10 (first week).
17-39
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Education.
Chapter 17 - Project Management
Step 4:
The paths and their expected duration are shown below:
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 paths are 1-2-5-7-11-13, 1-3-8-11-13, & 1-4-6-10-12-13.
Rank critical activities according to crash costs:
Cost per Week
to Crash
Path
Activity
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 should be shortened 1 week on Paths 1-2-5-7-11-13 & 1-3-8-11-13 because it
has the lowest crashing cost ($30) to shorten both paths. Both paths will decrease 1 week as
shown below.
Activity 4-6 should be shortened 1 week because it has the lowest crashing cost ($8) on Path
1-4-6-10-12-13, which will decrease 1 week as shown below.
The combined crashing cost = $38 and is ≤ $40.
Activities crashed to this point: Step 1: 7-11 (first week). Step 2: 1-2 (first week). Step 3: 711 (second week) & 6-10 (first week). Step 4: Activity 11-13 (first week) & Activity 4-6
(first week).
17-40
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Education.
Chapter 17 - Project Management
Step 5:
The paths and their expected duration are shown below:
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 paths are 1-2-5-7-11-13, 1-3-8-11-13, & 1-4-6-10-12-13.
Rank critical activities according to crash costs:
Cost per Week
to Crash
Path
Activity
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 shortened 1 week on Paths 1-2-5-7-11-13 & 1-3-8-11-13 because it
has the lowest crashing cost ($33) to shorten both paths. Both paths would decrease 1 week.
Activity 1-4 could be shortened 1 week because it has the lowest crashing cost ($10) on Path
1-4-6-10-12-13, which would decrease 1 week.
The combined crashing cost = $43 and is > $40. Stop.
The marginal cost of crashing > marginal benefit of crashing. We stop crashing after Step 4.
17-41
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Education.
Chapter 17 - Project Management
A summary of the crashing steps is shown below:
Path
1-2-5-7-11-13
1-3-8-11-13
1-3-9-12-13
1-4-6-10-12-13
Initial
time
35
32
20
33
Step 1 Crashing
Activity
Cost
7-11
$15
Length after
crashing 1
week
34
32
20
33
Length after
crashing 2
weeks
33
32
20
33
Step 2 Crashing
Activity
Cost
1-2
$18
Length after
crashing 3
weeks
32
32
20
32
Step 3 Crashing
Activity
Cost
7-11
$20
6-10
$ 5
$25
Length after
crashing 4
weeks
31
31
20
31
Step 4 Crashing
Activity
Cost
11-13
$30
4-6
$ 8
$38
Conclusion: Expected project duration = 31 weeks (shortened by 4 weeks).
Total crashing cost = $15 + $18 + $25 + $38 = $96.
Total indirect costs = 31 x $40 = $1,240.
Total cost over 31 weeks = $96 + $1,240 = $1,336(000) = $1,336,000
b. Plot the total-cost curve that describes the least expensive crashing schedule that will reduce
the project length by 6 weeks:
In Part a of this problem, we reduced the project length by 4 weeks. We must decrease the
project length by 2 more weeks. We will start with Step 5 from Part a above.
17-42
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Education.
Chapter 17 - Project Management
Step 5:
The paths and their expected duration are shown below:
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 paths are 1-2-5-7-11-13, 1-3-8-11-13, & 1-4-6-10-12-13.
Rank critical activities according to crash costs:
Cost per Week
to Crash
Activity
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 should be shortened 1 week on Paths 1-2-5-7-11-13 & 1-3-8-11-13 because it
has the lowest crashing cost ($33) to shorten both paths. Both paths will decrease 1 week as
shown below.
Activity 1-4 should be shortened 1 week because it has the lowest crashing cost ($10) on
Path 1-4-6-10-12-13, which will decrease 1 week as shown below.
The combined crashing cost = $43.
Activities crashed to this point: Step 1: 7-11 (first week). Step 2: 1-2 (first week). Step 3: 711 (second week) & 6-10 (first week). Step 4: Activity 11-13 (first week) & Activity 4-6
(first week). Step 5: Activity 11-13 (second week) & Activity 1-4 (first week).
Path
1-2-5-7-11-13
17-43
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Education.
Chapter 17 - Project Management
Step 6:
The paths and their expected duration are shown below:
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 paths are 1-2-5-7-11-13, 1-3-8-11-13, & 1-4-6-10-12-13.
Rank critical activities according to crash costs:
Cost per Week
to Crash
Activity
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 should be shortened 1 week because it has the lowest crashing cost ($22) on
Path 1-2-5-7-11-13, which will decrease 1 week as shown below.
Activity 1-3 should be shortened 1 week because it has the lowest crashing cost ($12) on
Path 1-3-8-11-13, which will decrease 1 week as shown below.
Activity 6-10 should be shortened 1 week because it has the lowest crashing cost ($12) on
Path 1-4-6-10-12-13, which will decrease 1 week as shown below.
The combined crashing cost = $46.
Activities crashed to this point: Step 1: 7-11 (first week). Step 2: 1-2 (first week). Step 3: 711 (second week) & 6-10 (first week). Step 4: Activity 11-13 (first week) & Activity 4-6
(first week). Step 5: Activity 11-13 (second week) & Activity 1-4 (first week). Step 6:
Activity 1-2 (second week), Activity 1-3 (first week), & Activity 6-10 (second week).
Path
1-2-5-7-11-13
Path
Expected Duration
1-2-5-7-11-13
29
1-3-8-11-13
29
1-3-9-12-13
20
1-4-6-10-12-13
29
17-44
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Education.
Chapter 17 - Project Management
A summary of the crashing steps is shown below:
Path
Initial
time
1-2-5-7-11-13
1-3-8-11-13
1-3-9-12-13
1-4-6-10-12-13
35
32
20
33
Step 1
Crashing
Activity Cost
7-11
$15
Length
after
crashing 1
week
34
32
20
33
Step 2
Crashing
Activity Cost
1-2
$18
Length
after
crashing 2
weeks
33
32
20
33
Step 3
Crashing
Activity Cost
7-11
$20
6-10
$ 5
$25
Length
after
crashing 3
weeks
32
32
20
32
Length
after
crashing 4
weeks
31
31
20
31
Step 4
Crashing
Activity Cost
11-13
$30
4-6
$ 8
$38
Length
after
crashing 5
weeks
30
30
20
30
Step 5
Crashing
Activity Cost
11-13
$33
1-4
$10
$43
Length
after
crashing 6
weeks
29
29
20
30
Step 6
Crashing
Activity Cost
1-2
$22
1-3
$12
6-10
$12
$46
Conclusion: Expected project duration = 29 weeks (shortened by 6 weeks).
Total crashing cost = $15 + $18 + $25 + $38 + $43 + $46 = $185.
Total indirect costs = 29 x $40 = $1,160.
Total cost over 29 weeks = $185 + $1,160 = $1,345(000) = $1,345,000
The table below summarizes the cumulative crashing cost, indirect cost, and total cost for
project lengths of 35, 34, 33, 32, 31, 30, and 29 weeks:
Cumulative
Project
Weeks
Length Shortened
35
0
Cumulative
Crash Cost
($000)
0
Indirect Cost Total Cost
($000)
($000)
35(40) = 1,400
1,400
34
1
15
34(40) = 1,360
1,375
33
2
15 + 18 = 33
33(40) = 1,320
1,353
32
3
33 + 25 = 58
32(40) = 1,280
1,338
31
4
58 + 38 = 96
31(40) = 1,240
1,336
30
5
96 + 43 = 139
30(40) = 1,200
1,339
29
6
139 + 46 = 185 29(40) = 1,160
1,345
17-45
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Education.
Chapter 17 - Project Management
The total cost-curve is plotted below:
1,400
Total
Cost
($000)
1,300
0
29
30
31
32
33
34
35
Project Length
(weeks)
17-46
Copyright © 2015 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill
Education.
Chapter 17 - Project Management
16.
Given:
We have the following crashing costs for activities:
Activity
Crash Cost
1st Week
Crash Cost
2nd Week
K
$410
$415
L
$125
---
N
$45
$45
M
$300
$350
J
$50
---
Q
$200
$225
P
---
---
Y
$85
$90
Z
$90
---
The network diagram is shown below:
M
L
Start
P
Q
K
N
Z
Y
End
J
Develop a crashing schedule given that Mr. T wants delivery in 32 weeks or he will impose a
penalty of $375 for each week his yacht is late:
We will crash as long as the crash cost at a step ≤ $375.
17-47
Copyright © 2015 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill
Education.
Chapter 17 - Project Management
Step 1:
The paths and their expected duration are shown below:
Path
Expected Duration
K-L-M-Q-P-Z
39
K-L-M-Q-Y
32
K-N-J-Q-P-Z
39
K-N-J-Q-Y
32
Critical paths are K-L-M-Q-P-Z & K-N-J-Q-P-Z.
Rank critical activities according to crash costs:
Path
K-L-M-Q-P-Z
K-N-J-Q-P-Z
Activity
Z
L
Q
M
K
N
J
Z
Q
K
Cost per Week
to Crash
$90
$125
$200
$300
$410
$45
$50
$90
$200
$410
Activity Z should be shortened 1 week because it has the lowest crashing cost ($90) to crash
both critical paths, which both will decrease by 1 week as shown below. This cost is ≤ $375.
Activities crashed to this point: Step 1: Activity Z (1st week).
17-48
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Education.
Chapter 17 - Project Management
Step 2:
The paths and their expected duration are shown below:
Path
Expected Duration
K-L-M-Q-P-Z
38
K-L-M-Q-Y
32
K-N-J-Q-P-Z
38
K-N-J-Q-Y
32
Critical paths are K-L-M-Q-P-Z & K-N-J-Q-P-Z.
Rank critical activities according to crash costs:
Path
K-L-M-Q-P-Z
K-N-J-Q-P-Z
Activity
L
Q
M
K
N
J
Q
K
Cost per Week
to Crash
$125
$200
$300
$410
$45
$50
$200
$410
Activity L should be shortened 1 week because it has the lowest crashing cost ($125) on Path
K-L-M-Q-P-Z, which will decrease Paths K-L-M-Q-P-Z & K-L-M-Q-Y 1 week as shown
below.
Activity N should be shortened 1 week because it has the lowest crashing cost ($45) on Path
K-N-J-Q-P-Z, which will decrease Paths K-N-J-Q-P-Z & K-N-J-Q-Y 1 week as shown
below.
Combined crashing cost = $170. This cost is ≤ $375.
Activities crashed to this point: Step 1: Activity Z (1st week). Step 2: Activity L (1st week) &
Activity N (1st week).
17-49
Copyright © 2015 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill
Education.
Chapter 17 - Project Management
Step 3:
The paths and their expected duration are shown below:
Path
Expected Duration
K-L-M-Q-P-Z
37
K-L-M-Q-Y
31
K-N-J-Q-P-Z
37
K-N-J-Q-Y
31
Critical paths are K-L-M-Q-P-Z & K-N-J-Q-P-Z.
Rank critical activities according to crash costs:
Path
K-L-M-Q-P-Z
K-N-J-Q-P-Z
Activity
Q
M
K
N
J
Q
K
Cost per Week
to Crash
$200
$300
$410
$45
$50
$200
$410
Activity Q should be shortened 1 week because it has the lowest crashing cost ($200) to
crash both critical paths. This cost is ≤ $375. All four paths will decrease by 1 week as shown
below.
Activities crashed to this point: Step 1: Activity Z (1st week). Step 2: Activity L (1st week) &
Activity N (1st week). Step 3: Activity Q (1st week).
17-50
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Education.
Chapter 17 - Project Management
Step 4:
The paths and their expected duration are shown below:
Path
Expected Duration
K-L-M-Q-P-Z
36
K-L-M-Q-Y
30
K-N-J-Q-P-Z
36
K-N-J-Q-Y
30
Critical paths are K-L-M-Q-P-Z & K-N-J-Q-P-Z.
Rank critical activities according to crash costs:
Path
K-L-M-Q-P-Z
K-N-J-Q-P-Z
Activity
Q
M
K
N
J
Q
K
Cost per Week
to Crash
$225
$300
$410
$45
$50
$225
$410
Activity Q should be shortened 1 week because it has the lowest crashing cost ($225) to
crash both critical paths. This cost is ≤ $375. All four paths will decrease by 1 week as shown
below.
Activities crashed to this point: Step 1: Activity Z (1st week). Step 2: Activity L (1st week) &
Activity N (1st week). Step 3: Activity Q (1st week). Step 4: Activity Q (2nd week).
17-51
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Education.
Chapter 17 - Project Management
Step 5:
The paths and their expected duration are shown below:
Path
Expected Duration
K-L-M-Q-P-Z
35
K-L-M-Q-Y
29
K-N-J-Q-P-Z
35
K-N-J-Q-Y
29
Critical paths are K-L-M-Q-P-Z & K-N-J-Q-P-Z.
Rank critical activities according to crash costs:
Path
K-L-M-Q-P-Z
K-N-J-Q-P-Z
Activity
M
K
N
J
K
Cost per Week
to Crash
$300
$410
$45
$50
$410
Activity M should be shortened 1 week because it has the lowest crashing cost ($300) on
Path K-L-M-Q-P-Z, which will decrease Paths K-L-M-Q-P-Z & K-L-M-Q-Y 1 week as
shown below.
Activity N should be shortened 1 week because it has the lowest crashing cost ($45) on Path
K-N-J-Q-P-Z, which will decrease Paths K-N-J-Q-P-Z & K-N-J-Q-Y 1 week as shown
below.
Combined crashing cost = $345. This cost is ≤ $375.
Activities crashed to this point: Step 1: Activity Z (1st week). Step 2: Activity L (1st week) &
Activity N (1st week). Step 3: Activity Q (1st week). Step 4: Activity Q (2nd week). Step 5:
Activity M (1st week) & Activity N (2nd week).
17-52
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Education.
Chapter 17 - Project Management
Step 6:
The paths and their expected duration are shown below:
Path
Expected Duration
K-L-M-Q-P-Z
34
K-L-M-Q-Y
28
K-N-J-Q-P-Z
34
K-N-J-Q-Y
28
Critical paths are K-L-M-Q-P-Z & K-N-J-Q-P-Z.
Rank critical activities according to crash costs:
Path
K-L-M-Q-P-Z
K-N-J-Q-P-Z
Activity
M
K
J
K
Cost per Week
to Crash
$350
$410
$50
$410
Activity M could be shortened 1 week because it has the lowest crashing cost ($350) on Path
K-L-M-Q-P-Z, which would decrease Paths K-L-M-Q-P-Z & K-L-M-Q-Y 1 week.
Activity J could be shortened 1 week because it has the lowest crashing cost ($50) on Path
K-N-J-Q-P-Z, which would decrease Paths K-N-J-Q-P-Z & K-N-J-Q-Y 1 week.
However, combined crashing cost = $400. This cost is > $375. Stop.
Do not perform this crashing. We are finished crashing after Step 5.
17-53
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Chapter 17 - Project Management
A summary of the crashing steps is shown below:
Path
Initial
time
K-L-M-Q-P-Z
K-L-M-Q-Y
K-N-J-Q-P-Z
K-N-J-Q-Y
39
32
39
32
Step 1
Crashing
Activity Cost
z
$90
Length
after
crashing 1
week
38
32
38
32
Step 2
Crashing
Activity Cost
L
$125
N
$ 45
$170
Length
after
crashing 2
weeks
37
31
37
31
Step 3
Crashing
Activity Cost
Q
$200
Length
after
crashing 3
weeks
36
30
36
30
Length
after
crashing 4
weeks
35
29
35
29
Step 4
Crashing
Activity Cost
Q
$225
Length
after
crashing 5
weeks
34
28
34
28
Step 5
Crashing
Activity Cost
M
$300
N
$ 45
$345
Conclusion: Expected project duration = 34 weeks (shortened by 5 weeks).
Total crashing cost = $90 + $170 + $200 + $225 + $345 = $1,030.
Mr. T’s yacht will be 2 weeks late.
Total penalty cost = 2 x $375 = $750.
Total Cost = Crashing Cost + Penalty Cost = $1,030 + $750 = $1,780.
17-54
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Education.
Chapter 17 - Project Management
17.
Given:
We have the following activities and time estimates:
Path
Top
Activity
A
B
C
Optimistic
4
7
3
Most Likely
5
8
5
Pessimistic
6
10
9
Bottom
D
E
F
7
2
1
8
3
4
11
4
6
Draw the network:
B
2
4
C
A
Start
6
1
End
F
D
E
3
5
Path
Top
Activity
A
B
C
Optimistic
4
7
3
Most Likely
5
8
5
Pessimistic
6
10
9
Mean
5.00
8.17
5.33
Standard
Deviation
4/36
9/36
36/36
Bottom
D
E
F
7
2
1
8
3
4
11
4
6
8.33
3.00
3.83
16/36
4/36
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
Variance = 4/36 + 9/36 + 36/36 = 49/36
Path D-E-F:
Expected completion time = 8.33 + 3.00 + 3.83 = 15.16
Variance = 16/36 + 4/36 + 25/36 = 45/36
Conclusion: Expected project completion = 18.50 days.
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Chapter 17 - Project Management
b. Determine the probability that the project will take at least 18 days.
Probability (T ≥ 18):
Path A-B-C:
49
Standard deviation =√36 = 1.167
18 − 18.50
= −0.43
1.167
Using Appendix B Table B: Probability of completion < 18 days = 0.3336
𝑧=
Path D-E-F:
45
Standard deviation =√36 = 1.118
18 − 15.16
= 2.54
1.118
Using Appendix B Table B: Probability of completion < 18 days = 0.9945
𝑧=
Conclusion: Probability (T < 18) = 0.3336 x 0.9945 = 0.3318
Probability (T ≥ 18) = 1.0000 – 0.3318 = 0.6682.
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Education.
Chapter 17 - Project Management
18.
Given:
We are given the events, probabilities, and costs in the table below:
Event
1
2
3
4
5
6
7
Probability
.25
.35
.20
.80
.10
.40
.60
Cost ($000)
15
25
55
10
77
55
50
Create a risk matrix for this project. Use a vertical scale of $0 to $80.
$80
$60
5
3
6
7
Cost
$40
$20
0
2
1
4
.25
.50
.75
Probability of Occurring
1.00
Conclusion: The manager should be most concerned about Event #7, which has a greater than
50% probability of occurring and a relatively high cost.
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Education.
Chapter 17 - Project Management
19.
Given:
We are given the events, probabilities, and costs in the table below:
Event
1. Equipment breakdown
2. Vendor is late with key segment.
3. Subcontractor has labor issues.
4. Weather problems.
5. Funding delays.
6. Testing delays.
Probability
.20
.60
.30
Unknown
.40 to .60
.40
Cost ($000)
40
200
140
15
50
20
Create a matrix for this project:
2
$200
$150
Cost
3
$100
5
$50
1
6
0
4
.25
.50
.75
1.00
Probability of Occurring
The weather problems (4) and funding delays (5) are placed using the highest probabilities
because we are conservative (i.e., we will assume the worst case).
Conclusion: The manager should be most concerned about Event 2 (vendor is late with key
segment) due to its greater than 50% probability and its high cost.
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Education.
Chapter 17 - Project Management
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
subsidiary started because this was a one-time, unique effort requiring the planning and
coordination of diverse activities. In addition, much of the work cut across functional boundaries.
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
undoubtedly had different priorities based on how they were evaluated, not on the success or
failure of this particular project. Rather, their evaluations would concern ongoing activities
directly related to their divisions.
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
are fully content to operate in a static environment. Moreover, experienced workers might have
witnessed similar projects that involved project personnel leaving the organization or moving into
less desirable jobs once the project had been completed. Thus, Bert Mill might have wanted to
avoid that possibility.
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
certain amount of resentment, which would bode ill for any future dealings he might have with
these people. In fact, that is just what happened when he took the disagreement that engineer Bob
Cates was having with the Mexican engineer on layout to Bob’s former boss, Sam Sargis.
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
would translate into these workers keeping their full-time positions. On that basis, more people
like Bert Mill might be willing to work on the project. In addition, more people would have an
opportunity to expand their job horizons.
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Education.
Chapter 17 - Project Management
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):
Path A: 10 + (1.645)(4) = 16.58
Path B: 14 + (1.645)(2) = 17.29
Path C: 13 + (1.645)(2) = 16.29
We know that we need all three probabilities to be > 0.9500; therefore we can start by focusing on the
longest path (Path B) and specify the project completion time as 17.29 days as a starting point.
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Chapter 17 - Project Management
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.
Row
1
2
3
4
5
6
A
Path
A
B
C
B
Mean
10
14
13
Column
C
D
Sigma
z value
4
1.8225
2
1.645
2
2.145
Completion =
Prob. of Completion =
E
Prob.
0.9658
0.9500
0.9840
17.29
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)
Cell E3: =ROUND(NORMSDIST(D3),4)
Cell E4: =ROUND(NORMSDIST(D4),4)
Cell E6: =ROUND(E2*E3*E4,4)
Note: We will use Cell E5 to enter the desired completion time.
As shown above, the probability of completing the project within 17.29 days is 0.9028, which is less than
0.9500.
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.
After experimentation, a project completion time of 18.02 days gives us a probability of project
completion of 0.9501 as shown in the table below.
Row
1
2
3
4
5
6
A
Path
A
B
C
B
Mean
10
14
13
Column
C
D
Sigma
z value
4
2.005
2
2.01
2
2.51
Completion =
Prob. of Completion =
E
Prob.
0.9775
0.9778
0.9940
18.02
0.9501
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Chapter 17 - Project Management
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.
We know that when we started with a project completion of 17.29 days above, the probability of
completion was 0.9028 (just above 0.9000).
We can lower the project completion time from 17.29 until the value in Cell E6 is close to 0.9000. As
shown in the table below, a project completion time of 17.26 days provides a probability of completion of
0.9002.
Row
1
2
3
4
5
6
A
Path
A
B
C
B
Mean
10
14
13
Column
C
D
Sigma
z value
4
1.815
2
1.63
2
2.13
Completion =
Prob. of Completion =
E
Prob.
0.9652
0.9484
0.9834
17.26
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
shown in the table below, a project completion time of 16.78 days provides a probability of completion of
0.8506.
Row
1
2
3
4
5
6
A
Path
A
B
C
B
Mean
10
14
13
Column
C
D
Sigma
z value
4
1.695
2
1.39
2
1.89
Completion =
Prob. of Completion =
E
Prob.
0.9550
0.9177
0.9706
16.78
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.
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