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6
Project Management
Answers to Questions
6-1. CPM/PERT is popular because it provides a picture
of the steps of a project, it is easy for managers and
participants to understand, and it is easy to apply.
6-2. The goal is to show the precedence relationship of
activities in a project.
6-3. A dummy activity is used to show a precedence relationship without the passage of time. It is used most
frequently to complete a precedence relationship so
that two activities will not have the same start and
end nodes.
6-4. The critical path is the longest path in the network. It
can be computed by summing the activity times along
each path and then seeing which path is the longest. It
also is the path with no slack available.
6-5. Slack is the amount of time an activity can be delayed
without affecting the overall project duration. It is
computed by subtracting the earliest start time for an
activity from the latest start time or the earliest finish
time from the latest finish time for an activity.
6-6. The mean activity time is computed as t = (a + 4m + b)/6,
where a is the optimistic activity time, m is the most
likely time, and b is the pessimistic time. The variance
is computed as v2 = [(b − a)/6]2.
6-7. Total project variance is computed by summing the
variances of the critical path activities.
6-8. The purpose of project crashing is to shorten the project duration at the least possible cost.
6-9. See which activity on the critical path has the minimum crash cost, and reduce this activity duration by
the maximum amount or until another path becomes
critical. If more than one path is critical, both paths
must be reduced by the same amount simultaneously.
Repeat this process until the crashing objective is
reached.
6-10. The preference should depend on the project, including the perceived degrees of variability in project activities, the ability to determine probabilistic time estimates, and the degree to which probabilistic analysis is
required.
6-11. The Gantt chart is a graphical technique with a bar or
time line displayed for each activity in the project.
However, while it will indicate the precedence relationships between activities, these relationships are
usually not as easy to discern (i.e., visualize) as with a
network.
6-12. Indirect costs include the cost for facilities, equipment, and machinery, interest on investment, utilities,
6-13.
6-14.
6-15.
6-16.
labor, personnel costs, etc. Direct costs are financial
penalties for not completing a project on time. In general, project crashing costs and indirect costs have an
inverse relationship; crashing costs are highest when
the project is shortened, whereas indirect costs increase
as the project duration increases.
A heavy reliance by the project manager on the network can mask errors in the precedence relationships
or missing activities can be overlooked. Attention to
critical path activities can be excessive to the point of
neglecting other crucial project activities. Obtaining
both deterministic and probabilistic time estimates
can be difficult. The time estimates can be overly
optimistic or pessimistic.
For an activity-on-node network, nodes represent
project activities and branches show precedence relationships, whereas on an activity-on-arrow network,
branches represent activities and nodes are events
specifying the end of one activity and the beginning
of another.
The project team consists of a group of individuals
selected because of their special skills, expertise, and
experience related to the project. Project planning
includes the identification of all project activities and
their precedence relationships the determination of
activity times, the determination of project duration,
comparison of the project time with objectives, and
the determination of resource requirements to meet
objectives. Project control includes making sure all
activities have been identified, making sure the activities are completed in their proper sequence, identifying resources as they are required, and adjusting the
schedule to reflect changes.
The WBS depends on the “project” the student uses.
Figure 6.1 should be used as a guideline.
Solutions to Problems
6-1.
Weeks
Activity 0 1 2 3 4 5 6 7 8 9 10
1–2
1–3
2–4
3–4
39
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Project Management
6-2.
Weeks
Activity 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
1–2
1–3
2–4
3–4
3–5
4–6
5–6
Project completion time = 17 weeks
Activity
Slack (weeks)
1–2
1–3
2–4
3–4
3–5
4–6
5–6
Paths: 1 → 2 → 4 → 6
4 + 8 + 2 = 14
1→2→4→5→6
4 + 8 + 5 + 6 = 23 *
1→2→5→6
4 + 3 + 6 = 13
1→3→5→6
7 + 9 + 6 = 22
1→3→6
7 + 5 = 12
6-4. Paths: 1 → 3 → 5 → 6 ⇒ 10 + 4 + 2 = 16
1 → 3 → 4 → 5 → 6 ⇒ 10 + 5 + 3 + 2 = 20 *
1 → 2 → 4 → 5 → 6 ⇒ 7 + 6 + 3 + 2 = 18
6-5.
Activity Time
11
0
11
4
0
4
0
1–2
1–3
2–4
3–4
3–5
4–5
5–6
7
10
6
5
4
3
2
ES
EF
LS
LF
Slack
0
0
7
10
10
15
18
7
10
13
15
14
18
20
2
0
9
10
14
15
18
9
10
15
15
18
18
20
2
0
2
0
4
0
0
Critical path = 1-3-4-5-6
The critical path activities have no slack.
6-3. a.
Weeks
Activity 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30
1–2
1–3
2–4
2–5
3–5
4–5
4–6
5–6
3–6
Project completion time = 23 weeks
Activity
Slack (weeks)
1–2
1–3
2–4
2–5
3–5
4–5
4–6
5–6
3–6
0
1
0
10
1
0
9
0
11
8
6
5
1
7
9
3
5
8
10
9
6
4
5
0
7
12
15
18
4
7
Activity Time
2
5
1–2
1–3
1–4
2–5
2–6
3–5
4–5
4–7
5–7
5–8
6–9
7–8
8–9
ES
EF
LS
LF
Slack
0
0
0
8
8
10
9
9
15
15
12
27
31
8
10
9
14
12
15
9
16
27
30
30
31
38
1
0
6
9
16
10
15
20
15
16
20
27
31
9
10
15
15
20
15
15
27
27
31
38
31
38
1
0
6
1
8
0
6
11
0
1
8
0
0
6-7.
4
3
4
Activity Time
Critical path = 1-3-5-7-8-9
Project duration = 38 mo.
b.
2
6-6.
6
1–2
1–3
2–4
2–6
3–4
3–5
4–6
5–6
6–7
10
7
4
15
6
12
7
0
9
ES
EF
LS
LF
Slack
0
0
10
10
7
7
14
19
25
10
7
14
25
13
19
21
19
34
0
5
14
10
12
13
18
25
25
l0
12
18
25
18
25
25
25
34
0
5
4
0
5
6
4
6
0
Critical path = 1-2-6-7
Project completion time = 34 wk.
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CHAPTER 6
6-8.
6-10.
Select and hire Train personnel
personnel
4
Make personnel
Determine
d
b
assignments
survey objectives
4
3
f
a
1
6
5
2
2
3
5
3
Design
c
e
questionnaire
Select target
3
audience
a.
Paths
Events
Duration (days)
1→2→4→5→6
1→2→3→4→5→6
1→2→3→5→6
A
B
C
12
14
13
Path B is the critical path.
b. ESij = maximum(EFi)
EFij = ESij + tij
LSij = LFij − tij
LFij = minimum(LSj)
ES = 3, EF = 6
LS = 5, LF = 8
1
ES = 8, EF = 12
LS = 8, LF = 12
4
ES = 0, EF = 3
LS = 0, LF = 3
4
3
ES = 12, EF = 14
LS = 12, LF = 14
5
2
3
ES = 3, EF = 8
LS = 3, LF = 8
5
0
6
2
Activity
4
3
2 S=0
5
S=0
S=0
4
3
S=0
5
6
2
3
S=1
6-9.
Activity Time
1–2
2–3
2–4
2–5
3–4
3–5
3–6
4–7
5–6
6–8
6–9
7–9
8–9
9–10
2
5
2
4
0
1
3
4
3
3
1
2
0
1
8
12
3
0
9
0
2
3
0
12
7
30
21
20
0
5
16
5
17
6
0
3
ES
EF
LS
LF
Slack
0
0
0
8
12
3
3
3
5
21
21
21
21
33
28
51
42
58
58
53
63
75
8
12
3
8
21
3
5
6
5
33
28
51
42
53
28
56
58
63
75
59
63
78
4
0
9
12
12
12
19
18
21
40
46
23
21
52
53
53
42
70
58
72
75
75
12
12
12
12
21
12
21
21
21
52
53
53
42
72
53
58
58
75
75
78
75
78
4
0
9
4
0
9
16
15
16
19
25
2
0
19
25
2
0
12
0
19
12
0
6-11.
c. Sij = LSij − ESij
S=0
3
1–2
1–3
1–4
2–3
3–6
4–3
4–5
4–6
5–6
6–7
6–8
6–9
6–10
7–12
8–9
9–11
10–11
11–13
11–14
12–15
13–14
14–15
Critical path = 1-3-6-10-11-14-15
Project completion time = 78 wk.
ES = 8, EF = 11
LS = 9, LF = 12
ES = 8, EF = 8
LS = 8, LF = 8
S=2
Activity Time
3
3
1
41
Project Management
ES
EF
LS
LF
Slack
0
2
2
2
7
7
7
7
8
11
11
11
14
14
2
7
4
6
7
8
10
11
11
14
12
13
14
15
0
2
6
4
8
7
8
8
8
11
13
12
14
14
2
7
8
8
8
8
11
12
11
14
14
14
14
15
0
0
4
2
1
0
1
1
0
0
2
1
0
0
Critical path: 1-2-3-5-6-8-9-10.
Time until general is ready to battle = 15 days.
a
b
c
d
e
f
dummy
g
h
dummy
1–2
2–3
2–5
2–4
3–6
5–6
3–5
3–7
6–7
4–5
Time
ES
EF
LS
LF
Slack
2
6
4
2
4
5
0
3
2
0
0
2
2
2
8
8
8
8
13
4
2
8
6
4
12
13
8
11
15
4
0
2
4
6
9
8
8
12
13
8
2
8
8
8
13
13
8
15
15
8
0
0
2
4
1
0
0
4
0
4
Critical path = a-b-dummy-f-h or 1-2-3-5-6-7
Project completion time = 15 wk
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Project Management
6-12.
Activity
a
m
b
t
ES
EF
LS
LF
Slack
S2
1–2
1–3
1–4
2–3
2–5
2–6
3–6
4–6
5–7
4–7
6–7
6
2
4
3
7
4
3
5
3
4
2
10
7
8
10
9
12
6
9
20
12
9
15
16
11
15
20
15
9
16
35
16
14
10.16
7.66
7.83
9.66
10.50
11.16
6.00
9.50
19.66
11.33
8.66
0
0
0
10.16
10.16
10.16
19.83
7.83
20.66
7.83
25.83
10.16
7.66
7.83
19.83
20.66
21.33
25.83
17.33
40.33
19.16
34.50
0
18.00
14.33
16.00
10.16
20.50
25.66
22.16
20.66
29.00
31.66
10.16
25.66
22.16
25.66
20.66
31.66
31.66
31.66
40.33
40.33
40.33
0
18.00
14.33
5.83
0
10.33
5.83
14.33
0
21.16
5.83
2.25
5.43
1.35
4.00
4.67
3.35
1.00
3.35
28.41
4.00
4.00
ES
EF
LS
LF
Slack
S2
0
0
9
9
16
16
19
19
25
25
22
25
25
9
11
16
19
17
21
25
22
25
25
23
27
33
0
7
2
0
2
10
0
2
6
0
2
6
0
4
4
1
9
0
4
1
1
0
0
0
0
1
Expected project completion time = 40.33 wk
s = 5.95
Critical path = 1-2-5-7
6-13.
f
3
j
6
9
b
1
e
c
d
2
a
b
c
d
e
f
g
h
dummy
dummy
i
j
k
1–2
1–3
2–3
2–4
3–4
3–6
4–5
4–7
5–6
5–8
7–8
6–9
8–9
8
g
a
Activity
k
5
4
i
7
h
a
m
b
t
5
3
4
5
1
1
3
1
0
0
1
2
5
8
12
7
8
1
4
6
2.5
0
0
1
2
8
17
15
10
23
1
13
9
7
0
0
1
2
11
9
11
7
10
1
5
6
3
0
0
1
2
8
0
7
11
9
18
26
19
21
31
25
24
31
25
9
18
18
19
19
31
25
24
31
25
25
33
33
Critical path = a-d-g-dummy-k or 1-2-4-5-8-9
Expected project completion time = 33 wk
s = 3.87
r(x ≤ 40)
x − µ 40 − 33
=
= 1.81
σ
3.87
P = ( x ≤ 40) = 0.9649
Z=
m= 33 x = 40
Time (days)
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43
Project Management
6-14. a, b, c, d.
Activity
a
m
b
t
ES
EF
LS
LF
Slack
S2
1–2
1–3
1–4
2–5
2–6
3–5
4–5
4–7
5–8
5–7
7–8
6–9
8–9
4
6
2
3
1
3
0
2
9
5
5
7
3
8
10
10
6
4
6
0
8
15
12
6
20
8
12
15
14
9
13
18
0
12
22
21
12
25
20
8.00
10.16
9.33
6.00
5.00
7.50
0
7.66
15.16
12.33
6.83
18.66
9.16
0
0
0
8.00
8.00
10.16
9.33
9.33
17.66
17.66
30.00
13.00
36.83
8.00
10.16
9.33
14.00
13.00
17.66
9.33
17.00
32.83
30.00
36.83
31.66
46.00
3.66
0
8.33
11.66
22.33
10.16
17.66
22.33
21.66
17.66
30.00
27.33
36.83
11.66
10.16
17.66
17.66
27.33
17.66
17.66
30.00
36.83
30.00
36.83
46.00
46.00
3.66
0
8.33
3.66
14.33
0
8.33
13.00
4.00
0
0
14.33
0
1.77
2.25
4.00
1.00
4.00
6.25
0
2.76
4.67
7.08
1.35
9.00
8.01
ES
EF
LS
LF
Slack
S2
0
2.50
2.50
2.50
8.00
8.00
8.00
8.00
9.50
12.50
12.50
12.50
17.00
17.00
2.50
5.50
8.00
9.33
8.00
9.50
11.50
12.50
12.50
14.50
14.16
17.00
17.00
18.00
0
7.50
2.50
2.66
10.50
8.00
9.00
10.50
9.50
15.00
15.33
12.50
17.00
17.00
2.50
10.50
8.00
9.50
10.50
9.50
12.50
15.00
12.50
17.00
17.00
17.00
17.00
18.00
e. Critical path = 1-3-5-7-8-9
f. Expected project completion time = 46 mo
s = 5 mo
6-15. a, b, c, d.
Activity
a
m
b
1–2
2–4
2–3
2–5
3–4
3–5
3–6
4–7
5–6
7–9
6–9
6–8
8–9
9–10
1
1
3
3
0
1
2
2
1
1
1
2
0
1
2
3
5
6
0
1.5
3
4
3
2
1
4
0
1
6
5
10
14
0
2
7
9
5
3
5
9
0
1
t
2.50
3.00
5.50
6.83
0
1.50
3.50
4.50
3.00
2.00
1.66
4.50
0
1.00
e. Critical path = 1-2-3-5-6-8-9-10
f. Expected project completion time = 18 months
s = 1.97
6-16.
Z=
0
5.00
0
0.16
2.50
0
1.00
2.50
0
2.50
2.83
0
0
0
0.694
0.436
1.35
3.35
0
0.026
0.689
1.35
0.436
0.109
0.436
1.35
0
0
x − µ 50 − 46 4
=
= = 0.80
σ
5
5
From normal table, p = 0.2881
0.5000 − 0.2881 = 0.2119 probability that the project
will exceed 50 mo.
6-17. a, b.
d
2
1
m = 46
s = 40
Time (months)
50
e
a
p=?
b
6
5
3
h
0
7
f
9
i
c
j
4
g
8
k
10
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CHAPTER 6
Activity
a
b
c
d
e
f
g
Dummy
Dummy
h
Dummy
i
j
k
1–2
1–3
1–4
2–6
2–5
3–7
4–8
5–6
6–7
6–9
7–8
7–9
8–9
9–10
Page 44
Project Management
a
m
b
t
ES
EF
LS
LF
Slack
1
2
1
4
3
10
5
0
0
2
0
1
2
2
2
5
3
10
7
15
9
0
0
3
0
4
5
2
3
8
5
25
12
25
14
0
0
7
0
6
10
2
2.00
5.00
3.00
11.50
7.16
15.83
9.16
0
0
3.50
0
3.83
5.33
2
0
0
0
2.00
2.00
5.00
3.00
9.16
13.50
13.50
20.83
20.83
20.83
26.16
2.00
5.00
3.00
13.50
9.16
20.83
12.16
9.16
13.50
17.00
20.83
24.66
26.16
28.16
7.33
0
8.66
9.33
13.66
5.00
11.66
20.83
20.83
22.66
20.83
22.33
20.83
26.16
9.33
5.00
11.66
20.83
20.83
20.83
20.83
20.83
20.83
26.16
20.83
26.16
26.16
28.16
7.33
0
8.66
7.33
11.66
0
8.66
11.66
7.33
9.16
0
1.50
0
0
c. Critical path = b-f-dummy j-k or 1-3-7-8-9-10
d. Expected project completion time = 28.17 weeks;
s = 3.00
e.
S2
1.09
1.00
0.436
12.25
2.25
6.25
2.25
0
0
0.689
0
0.689
1.77
0
6-18.
p =?
15
m = 18
s = 1.97
Time (days)
p=?
Z=
m = 28.16
s=3
Time (weeks)
35
Z=
x − µ 35 − 28.16 6.84
=
=
= 2.28
σ
3
3.00
From normal table, p = 0.4887
0.5000 − 0.4887 = 0.0113 probability that the company will be fined.
x − µ 15 − 18
=
= −1.52
σ
1.97
From normal table, p = 0.4357
0.5000 − 0.4357 = 0.0643 probability that preparations
would be in time.
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CHAPTER 6
45
Project Management
6-19.
Activity
a
m
b
t
ES
EF
LS
LF
Slack
S2
1–2
1–4
1–6
2–3
3–4
4–7
4–8
4–5
3–9
5–8
6–8
6–13
7–8
8–10
8–11
9–13
10–12
11–12
12–13
1
4
20
4
2
8
10
5
6
0
1
5
0
5
4
5
0
5
1
3
6
35
7
3
12
16
9
8
0
2
8
0
10
7
7
0
9
3
5
10
50
12
5
25
21
15
14
0
2
12
0
15
10
12
0
20
7
3.00
6.33
35.00
7.33
3.16
13.50
15.83
9.33
8.66
0
1.83
8.16
0
10.00
7.00
7.50
0
10.16
3.33
0
0
0
3.00
10.33
13.50
13.50
13.50
10.33
22.83
35.00
35.00
27.00
36.83
36.83
19.00
46.83
43.83
54.00
3.00
6.33
35.00
10.33
13.50
27.00
29.33
22.83
19.00
22.83
36.83
43.16
27.00
46.83
43.83
26.50
46.83
54.00
57.33
7.50
14.66
0
10.50
17.83
23.33
21.00
27.50
41.16
36.83
35.00
49.16
36.83
44.00
36.83
49.83
54.00
43.83
54.00
10.50
21.00
35.00
17.83
21.00
36.83
36.83
36.83
49.83
36.83
36.83
57.33
36.83
54.00
43.83
57.33
54.00
54.00
57.33
7.50
14.66
0
7.50
7.50
9.83
7.50
14.00
30.83
14.00
0
14.16
9.83
7.16
0
30.83
7.16
0
0
0.436
1.00
25.00
1.77
0.25
8.01
3.35
2.76
1.77
0
0.029
1.16
0
1.66
1.00
1.16
0
2.50
1.00
To be 90 percent certain of delivering the part on
time, RusTech should probably specify at least 50.3 or
51 weeks in the contract bid.
Expected completion time = 57.33 days
σ = 5.77
x − µ 67 − 57.33
Z=
=
= 1.68
σ
5.77
6-21.
P(x ≤ 67) = 0.9535
8
l
11
6-20.
e
3
1
b
1
a
2
6
g
c
4
11
h
l
7
d
5
j
9
n
Critical path = a-d-j-n-o
Expected project completion time = 45 wk
σ = 4.10
Since probability is 0.90, Z = 1.29.
x − 45
4.10
x − 45 = 5.29
x = 50.3
1.29 =
2 c
d
3
b
10
i
6
9
m
12
q
13
j
e
4
7
f
5
o
i
a
m
k
o
h
8
f
g
n
k
p
10
The activity schedule, critical path, expected project
time and variance determined using POM for Windows
as follows.
Project completion time = 118.6667
Project standard deviation = 5.86
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CHAPTER 6
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
Page 46
Project Management
Activity
Time
Early
Start
Early
Finish
Late
Start
Late
Finish
Slack
15
8.8333
24.1667
19.5
8.1667
13.6667
20.1667
25
14.6667
23
8.6667
7.1667
5
4.3333
7
5.5
20.83333
0
0
15
39.1667
39.1667
47.3333
61
58.6667
58.6667
58.6667
81.6667
83.6667
73.3333
90.8333
90.8333
90.3333
97.8333
15
8.8333
39.1667
58.6667
47.3333
61
81.1667
83.6667
73.3333
81.6667
90.3333
90.8333
78.3333
94.6667
97.8333
95.8333
118.6667
0.0
64.5
15
39.1667
51.5
59.6667
73.3333
58.6667
78.1667
61.8333
84.8333
83.6667
92.8333
93.5
90.8333
113.1667
97.8333
15
73.3333
39.1667
58.6667
59.6667
73.3333
93.5
83.6667
92.8333
84.8333
93.5
90.8333
97.8333
97.8333
97.8333
118.6667
118.6667
0
64.5
0
0
12.3333
12.3333
12.3333
0
19.5
3.1667
3.1667
0
19.5
3.1667
0
22.8333
0
Activity std dev
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
1.6667
1.1667
2.5
2.1667
1.1667
2.3333
1.5
3.3333
2
2.3333
1.3333
1.5
0.6667
1
1
0.8333
2.5
Critical path = a-c-d-h-l-o-q
µ = 118.67
σ = 5.85
x − µ 120 − 118.67
Z=
=
0.227
σ
5.85
P( x ≤ 120) = 0.59
b. Present (normal) critical path = 1-2-4
Normal critical path time = 30 wk
Crash critical path (all crash time) = 1-4
Maximum possible project crash time = 20 wk
c. Normal cost = 3950
Crashed project cost = 4700
Start
node
End
node
Normal
time
Crash
time
Normal
cost
Crash
cost
1
1
1
2
3
2
4
3
4
4
20
24
14
10
11
8
20
7
6
5
1,000
1,200
700
500
550
3950
1,480
1,400
1,190
820
730
5
a
b
c
d
e
Crash cost/pd
a
b
c
d
e
Crash by
Crashing cost
10
4
0
0
5
400
200
0
0
150
750
40
50
70
80
30
6-23.
c
2
6-22. a.
20
2
a
14
3
d
10
24
1
4
4
g
5
1
h
11
b
e
3
6
f
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CHAPTER 6
Normal critical path = 1 → 2 → 5 → 6
Normal critical path time = 36 wk
Project completion time = 36
Normal cost = 14400
Minimum project completion time = 22
Crash cost = 23,250
a
b
c
d
e
f
g
h
Crash
time
Normal
cost
Crash
cost
Crash
cost/pd
Crash
by
16
14
8
5
4
6
10
15
8
9
6
4
2
4
7
10
2,000
1,000
500
600
1,500
800
3,000
5,000
4,400
1,800
700
1,300
3,000
1,600
4,500
8,000
300
160
100
700
750
400
500
600
8
5
2
1
1
0
2
5
Normal
time
End
node
1
1
2
2
3
3
4
2
3
3
4
4
5
5
a
b
c
d
e
f
g
2,400
800
200
700
750
0
1,000
3,000
Normal
time
Crash
time
8
10
5
3
6
3
4
5
7
3
1
4
3
3
Crashing
cost/pd
Crash
by
100
50
200
100
75
0
200
3
2
2
0
2
0
1
a
b
c
d
e
f
g
Crashing cost
a
b
c
d
e
f
g
h
i
j
k
47
6-24. Project completion time = 23
Normal cost = 1500
Minimum project completion time = 15
Crash cost = 2650
Normal
time
a
b
c
d
e
f
g
h
Project Management
Normal
cost
100
250
400
200
150
100
300
Crash
cost
400
400
800
400
300
100
500
Crashing
Cost
300
100
400
0
150
0
200
6-25. Project completion time = 33
Normal cost = 28800
Minimum project completion time = 26
Crash cost = 33,900
Start
node
End
node
Normal
time
Crash
time
9
11
7
10
1
5
6
3
1
2
8
7
9
5
8
1
3
5
3
1
2
6
4,800
9,100
3,000
3,600
0
1,500
1,800
0
0
0
5,000
6,300
15,500
4,000
5,000
0
2,000
2,000
0
0
0
7,000
Normal
cost
750
3,200
500
700
0
250
200
0
0
0
1,000
Crash
by
Crashing
cost
2
0
0
2
0
0
1
0
0
0
2
1,500
0
0
1,400
0
0
200
0
0
0
2,000
Critical path = 1-2-4-5-8-9
Crashing cost = $5,100
Total network cost = $33,900
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CHAPTER 6
Page 48
Project Management
6-26.
b
a
1
d
2
e
3
4
f
6
h
j
8
c
9
i
5
Project
a
b
c
d
e
f
g
h
i
j
g
7
Activity
Time
ES
EF
LS
23
3
3.167
4.167
2.833
5
1.833
5.833
3.833
4.167
2.167
0
0
0
3
5.833
10.833
10.833
12.667
16.667
20.833
3
3.167
4.167
5.833
10.833
12.667
16.667
16.5
20.833
23
0
7.667
6.667
3
5.833
15.167
10.833
17
16.667
20.833
LF
Slack
3
10.833
10.833
5.833
10.833
17
16.667
20.833
20.833
23
0
7.667
6.667
0
0
4.333
0
4.333
0
0
Probability the project will be completed in 21 days?
x − µ 21 − 23
=
= −1.18
σ
1.70
P( x ≤ 21) = .119
Z=
6-27.
7
h
k
i
2
6
b
a
1
4
e
5
12
l
f
n
8
g
c
r
11
j
d
3
p
o
15
q
13
10
m
u
s
9
t
14
Standard
Deviation
2.028
0.667
0.5
0.833
0.5
1
0.167
0.833
0.5
0.5
0.5
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CHAPTER 6
a-e-f-g-j-o-p
or
1-4-5-8-10-11-12-15
Project duration
= 91.667
σ = 3.3082
From January 20 to April 29 is 101 days.
x−µ
Z
101 − 91.667
=
3.3082
= 2.82
P( x ≤ 101) = .9976
P( x ≤ 101) =
CASE: Bloodless Coup Concert
t
v
1–2
1–3
2–3
2–4
2–5
5–6
5–7
4–9
6–9
7–9
3–9
3–8
8–9
4.17
5.33
2.17
5.50
3.50
4.17
3.00
3.00
3.33
2.00
6.33
5.50
0.00
0.69
0.44
0.25
1.37
1.37
0.69
0.69
0.11
0.44
0.11
2.79
3.35
0.00
ET = 9.67
LT = 12.17
4
Hotel
arrangements
Press
conference
s = 2.50
5.50
s = 2.50
3.00
ET = 4.17
LT = 4.17
ET = 15.17
LT = 15.17
ET = 11.84
LT = 11.84
3.33
2 Union
9
6
Hire
Secure
Set up
negotiations stagehands
s=0
stage
auditorium
3.50
Assign
s=0
s=0
s
=
0
ushers
4.17
5
s = 2.50
s = 2.50
Hire
2.00
1
ushers
2.17
ET = 7.67
3.00
LT = 7.67
s = 3.31
ET = 0
s
=
2.50
0 Dummy
LT = 0
7
Print
Advertise
tickets
promotion
ET = 10.67
Hire
LT = 13.17
preliminary
concert act
6.33
5.33
3
ET = 6.34
LT = 8.84
49
Activity “n,” send out acceptance letters, has ES =
45.83 (March 6) and LF = 66.17 (March 20), so it
appears the club would meet the deadline of March
30 to send out acceptance letters.
Activity “q,” send out schedules, has ES = 66.16
(March 26) and LS = 83.50 (April 14) and LF = 91.67,
so it seems likely the club would meet the deadline of
April 15 for sending out game schedules.
The critical path is:
Activity
Project Management
Sell
tickets
5.50
s = 3.33
8
ET = 11.84
LT = 15.17
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CHAPTER 6
Page 50
Project Management
Activity
a
m
b
1–2
1–3
2–3
2–4
2–5
5–6
5–7
4–9
6–9
7–9
3–9
3–8
8–9
2
4
1
3
1
2
1
2
2
1
2
1
0
4
5
2
5
3
4
3
3
3
2
6
5
0
7
8
4
10
8
7
5
4
6
3
12
12
0
t
4.16
5.33
2.16
5.50
3.50
4.16
3.00
3.00
3.33
2.00
6.33
5.5
0
ES
EF
LS
LF
Slack
sd(S)
0
0
4.16
4.16
4.16
7.66
7.66
9.66
11.83
10.66
6.33
6.33
11.83
4.16
5.33
6.33
9.66
7.66
11.83
10.66
12.66
15.16
12.66
12.66
11.83
11.83
0
3.50
6.66
6.66
4.16
7.66
10.16
12.16
11.83
13.16
8.83
9.66
15.16
4.16
8.83
8.83
12.16
7.66
11.83
13.16
15.16
15.16
15.16
15.16
15.16
15.16
0
3.50
2.50
2.50
0
0
2.50
2.50
0
2.50
2.50
3.33
3.33
0.83
0.66
0.50
1.16
1.16
0.83
0.66
0.33
0.66
0.33
1.66
1.83
0
Expected project completion time = 15.17 days
σ = 1.79
x − µ 18 − 15.17
Z=
=
= 1.58
σ
1.79
P( x ≤ 18) = 0.9429
CASE SOLUTION: Moore Housing
Contractors
Following is the CPM/PERT network for building a house.
13
Grading
Gutters
Drain
Tiles
o
d
p
12
15
Landscaping
Roof
n
1
a
2
b
c
3
Excavate Foundation
f
4
Framing
6
Insulation
e
Furnace
5
g
7
k
i
9
l
r
11
Windows
Brickwork
j Electrical
Sheetrock Sub-floors
m
q
8
10
Finish
Carpentry
14
Project completion time = 45.8333
Project standard deviation = 2.409472
Bathrooms
u
16
20
Lights
x
Paint 18
t
Kitchen
h
Floors
w
s
Basement
floor
Plumbing
19
17
v
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CHAPTER 6
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
s
t
u
v
w
x
Project Management
Activity
time
Early
Start
Early
Finish
Late
Start
Late
Finish
Slack
Activity
std dev
4.1667
3.1667
3.8333
2.1667
2
3.8333
3.1667
4.1667
2.8333
2.1667
5.1667
6.5
8.3333
3.3333
2.3333
3.5
4.1667
6.3333
5.8333
4.3333
3.3333
6.3333
5
2.8333
0
4.1667
7.3333
7.3333
7.3333
11.1667
9.3333
9.3333
15
15
12.5
17.8333
17.6667
24.3333
27.6667
30
26
33.5
30.1667
30.1667
30.1667
34.5
40.8333
40.8333
4.1667
7.3333
11.1667
9.5
9.3333
15
12.5
13.5
17.8333
17.1667
17.6667
24.3333
26
27.6667
30
33.5
30.1667
39.8333
36
34.5
33.5
40.8333
45.8333
43.6667
0.0
4.1667
7.8333
33.83333
7.3333
11.6667
9.3333
13.5
21
15.5
12.5
23.8333
17.6667
30.3333
33.6667
36
26
39.5
35
30.1667
31.17
34.5
40.8333
43
4.1667
7.3333
11.6667
36
9.3333
15.5
12.5
17.6667
23.8333
17.6667
17.6667
30.3333
26
33.6667
36
39.5
30.1667
45.8333
40.8333
34.5
34.5
40.8333
45.8333
45.8333
0.0
0.0
05
26.5
0.0
0.5
0.0
4.1667
6
0.5
0.0
6
0.0
6
6
6
0
6
4.8333
0
12.3333
0
0
2.1667
05
0.5
0.5
0.5
0.3333
0.5
0.5
0.8333
0.5
0.5
0.8333
0.8333
1
0.6667
0.6667
0.8333
0.5
1
1.5
1
0.6667
1
1
0.5
51
Critical Path for Moore Contractors
Critical path: a-b-e-g-k-m-q-t-v-dumvw-w.
Notice that the expected completion time is 45.83 days which is very close to the realtor’s due date for completion.
The probability of finishing in 45 days is 0.3647
36.47% is not a very high probability that the contractor will complete a house within 45 days thus the Moores should probably inflate their bid.