Chapter 17 – Project Management
4. c. Case 1: Activity-on-Node Diagram
A
D
K
B
E
F
H
I
G
C
M
L
J
P
N
K
R
V
Q
S
T
W
End
End
Start
Start
Chapter 17 – Project Management
5. a.
12
4
28
21
26
18
2
5
8
12
26
28
19
11
4
26
18
28
20
31
24
21
13
7
21
13
26
10
4
7
9
5
2
3
4
8
7
2
11
4
LS
ES
LF
EF
19
12
0
25
25
31
31
10
6
5
6
Chapter 17 – Project Management
5. b.
Summary:
Activity
ES
EF
LS
LF
Slack
1
0
5
0
5
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
55
0
Note: Slack = LS – ES or LF – EF.
18
33
4
3
7
13
11
LS
ES
LF
EF
1
2
8
9
6
0
0
37
23
40
26
23
23
4
5
5
5
5
5
5
23
33
40
33
44
37
44
44
53
53
53
53
55
55
33
33
44
44
20
5
33
18
10
9
2
3
17–14
Education.
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, 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.
17–15
Education.
7. a.
1
6
G
F
D
A
B
b.
Activity
te
2
A
6.00
4/36
B
8.50
9/36
C
8.17
25/36
D
12.00
36/36
E
6.33
16/36
F
6.00
4/36
G
3.50
25/36
H
4.17
1/36
I
6.83
9/36
Path A-C-E:
Expected duration = 6.00 + 8.17 + 6.33 = 20.50
Variance = 36/36 + 4/36 + 25/36 = 65/36
Standard deviation =
Path B-H-I:
Expected duration = 8.50 + 4.17 + 6.83 = 19.50
2
5
7
4
C
E
I
H
Chapter 17 – Project Management
17–16
Specified time = 24 days or less:
Path D-F-G:
Using Appendix B Table B: Probability of completion ≤ 24 days = 0.9686
Path B-H-I:
Specified time = 21 days or less:
Path A-C-E:
Probability of completion ≤ 21 days = 0.6736
Probability of completion ≤ 21 days = 0.9808
Chapter 17 – Project Management
17–17
Education.
c. 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.
In the modified network, the expected times and variances are:
Activity
te
2
C
8.17
25/36
D
6.00
4/36
E
6.33
16/36
F
6.00
4/36
G
3.50
25/36
H
4.17
1/36
I
6.83
9/36
The revised project diagram is:
3
Path C-E:
Expected duration from start of project = 7 + (8.17 + 6.33) = 21.50
Variance = 25/36 + 16/36 = 41/36
Standard deviation =
1
2
4
E (6.33)
F (6.00)
D (6.00)
G (3.50)
I (6.83)
H (4.17)
C (8.17)
End
Chapter 17 – Project Management
17–18
Education.
Path H-I:
Expected duration from start of project = 7 + (4.17 + 6.83) = 18.00
Variance = 1/36 + 9/36 = 10/36
Path C-E:
Using Appendix B Table B: Probability of completion ≤ 24 days = 0.9904
Path D-F-G:
Path H-I:
Probability of completion ≤ 24 days = 1.0000 (z value > +3.00—treat probability of
Using Appendix B Table B: Probability of completion ≤ 21 days = 0.3192
Path D-F-G:
17–19
Education.
Path H-I:
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
First Crash
Second Crash
C
$8
$10
D
$10
$11
E
$9
$10
F
$7
$9
G
$8
$9
H
$7
$8
I
$6
$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
18.00
Critical path is D-F-G.
Rank critical activities according to crash costs:
Activity
Cost per Day
to Crash
F
$7
G
$8
D
$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.
Chapter 17 – Project Management
17–20
Education.
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
Activity
Cost per Day
to Crash
C-E
C
$8
E
$9
D-F-G
G
$8
F
$9
D
$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