CD16-1
CD Chapter 16 Pert/CPM Models for Project Management
Review Questions
16.1-2 He has decided to focus on meeting the deadline of 47 weeks.
16.1-4 (1) the activities of the project; (2) the immediate predecessors of the activities; and (3) the
estimated duration of the activities
16.2-1 (1) activity information; (2) precedence relationships; and (3) time information (duration)
16.3-1 (a) A path through a project network is one of the routes following the arrows (arcs) from
the start node to the finish node; (b) the length of a path is the sum of the estimated
durations of the activities on the path; (c) the longest path is called the critical path.
16.3-4 A forward pass is the process of starting with the initial activities and working forward in
time toward the final activities.
16.3-7 A backward pass starts with the final activities and works backward in time toward the
initial activities instead of starting with the initial activities.
CD16-2
CD16-3
16.4-3 It is assumed that the mean critical path will turn out to be the longest path through the
project network.
16.4-6 p2=sum of the variances of the durations for the activities on the mean critical path.
16.4-8 It is usually higher than the true probability.
16.5-1 Using overtime, hiring additional labor, and using special materials or equipment are all
ways of crashing an activity.
16.5-3 No, only crashing activities on the critical path will reduce the duration of the project.
16.5-4 Crash costs per week saved are being examined.
16.5-6 An activity cannot start until its immediate predecessor starts and then completes its
duration.
16.6-1 PERT/Cost is a systematic procedure to help the manager plan, schedule, and control
project costs.
16.6-3 A common assumption is that the costs of performing an activity are incurred at a constant
rate throughout its duration.
16.6-4 A work package is a group of related activities.
CD16-4
16.7-2 Computer implementation has allowed for application to larger projects, faster revisions in
project plans and effortless updates and changes in schedules.
16.7-4 The technique of computer simulation to approximate the probability that the project will
meet its deadline is an alternative for improving on PERT/CPM.
16.7-5 The Precedence Diagramming Method has been developed as an extension of PERT/CPM
to deal with overlapping activities.
16.7-7 It encourages effective interaction between the project manager and subordinates that leads
to setting mutual goals for the project.
Problems
16.1 a)
St ar t
AB
C D
E
0
23
22
3
CD16-6
b) Start → A → C → Finish Length = 4 weeks
c)
Activity
ES
EF
LF
Slack
Critical Path
Start
0
0
0
0
Yes
A
0
2
3
1
No
B
0
3
3
0
Yes
C
3
5
8
3
No
D
3
5
5
0
Yes
E
5
8
8
0
Yes
Finish
8
8
8
0
Yes
d) No, this will not shorten the length of the project because the activity is not on the
critical path.
CD Chapter 16 – Pert/CPM Models for Project Management
CD16-7
16.2 a)
St art
AB
GC
I
Fi ni sh
0
DEF H K
J
21
23 2
11 2 4
1
3
0
b) Start → A → D → Finish Length = 4 weeks
Start → A → E → Finish Length = 5 weeks
CD16-8
c)
Activity
ES
EF
LF
Slack
Critical Path
Start
0
0
0
0
Yes
A
0
2
2
0
Yes
B
0
1
1
0
Yes
C
1
3
3
0
Yes
D
2
4
8
4
No
E
3
6
8
2
No
F
2
4
4
0
Yes
G
2
3
3
0
Yes
H
3
4
4
0
Yes
I
4
5
5
0
Yes
J
5
8
8
0
Yes
K
4
8
8
0
Yes
Finish
8
8
8
0
Yes
Critical Paths: Start → A → F → K → Finish
d) No, this will not shorten the length of the project because A is not on all of the critical
paths.
CD16-9
16.3 a)
St art
A
B
G
C
I
Fi ni sh
D E
F
H
K
J
M
L
N
0
1
2
432
35
14
35
23
4
0
CD16-10
b, c, & d)
Activity
ES
EF
LF
Slack
Critical Path
Start
0
0
0
0
Yes
A
0
1
1
0
Yes
B
1
3
3
0
Yes
C
3
7
7
0
Yes
D
3
6
6
0
Yes
E
3
5
6
1
No
F
7
10
10
0
Yes
G
6
11
11
0
Yes
H
10
11
11
0
Yes
I
11
15
15
0
Yes
J
15
17
17
0
Yes
K
15
18
19
1
No
L
17
20
20
0
Yes
M
18
23
24
1
No
N
20
24
24
0
Yes
Finish
24
24
24
0
Yes
CD16-11
16.4 a)
St art
A
B
G
I
Fi ni sh
D
EK
J
L
0
30
53
715
2
30
10
15
0
C2
F25 H10
b) Start → A → B → J → L → Finish Length = 75 minutes *critical path
CD16-12
c, d & e)
Activity
ES
EF
LF
Slack
Critical Path
Start
0
0
0
0
Yes
A
0
30
30
0
Yes
B
30
35
35
0
Yes
C
0
2
32
30
No
D
2
5
35
30
No
E
0
7
10
3
No
F
7
32
35
3
No
G
0
15
23
8
No
H
15
25
33
8
No
I
25
27
35
8
No
J
35
45
45
0
Yes
K
0
15
45
30
No
L
45
75
75
0
Yes
Finish
75
75
75
0
Yes
16.5 a) Start → A → D → H → M → Finish Length = 19 weeks
Start → B → E → J → M → Finish Length = 20 weeks *critical path
CD16-13
b)
Activity
ES
EF
LF
Slack
Critical Path
Start
0
0
0
0
Yes
A
0
6
7
1
No
B
0
3
3
0
Yes
C
0
4
4
0
Yes
D
6
10
11
1
No
E
3
10
10
0
Yes
F
4
8
12
4
No
G
4
10
10
0
Yes
H
10
13
14
1
No
I
6
11
14
3
No
J
10
14
14
0
Yes
K
8
11
15
4
No
L
10
15
15
0
Yes
M
14
20
20
0
Yes
N
15
20
20
0
Yes
Finish
20
20
20
0
Yes
Ken will be able to meet his deadline if no delays occur.
Focus attention on activities with 0 slack (those in the critical paths).
d) If activity I takes 2 extra weeks there will be no delay because its slack is 3. If activity
16.6 a)
St art
AB
E C
F
Fi ni sh
0
51
6 2
3
0
D
4
b)
Activity
ES
EF
LS
LF
Slack
Critical Path
Start
0
0
0
0
0
Yes
A
0
5
0
5
0
Yes
B
0
1
11
12
11
No
C
1
3
12
14
11
No
D
5
9
7
11
2
No
E
5
11
5
11
0
Yes
F
11
14
11
14
0
Yes
Finish
14
14
14
14
0
Yes
c) 6 months
CD16-15
16.7
Activity
ES
EF
LS
LF
Slack
Critical Path
Start
0
0
0
0
0
Yes
A
0
3
0
3
0
Yes
B
3
11
3
11
0
Yes
C
11
29
11
29
0
Yes
D
29
39
29
39
0
Yes
E
29
34
30
35
1
No
F
34
44
35
45
1
No
G
39
50
39
50
0
Yes
H
50
67
50
67
0
Yes
I
29
38
36
45
7
No
J
44
53
45
54
1
No
K
53
57
57
61
4
No
L
53
60
54
61
1
No
M
67
70
67
70
0
Yes
N
60
69
61
70
1
No
Finish
70
70
70
70
0
Yes
16.8
Activity
ES
EF
LS
LF
Slack
Critical Path
Start
0
0
0
0
0
Yes
A
0
1
0
1
0
Yes
B
1
3
1
3
0
Yes
C
3
9
3
9
0
Yes
D
9
13
11
15
2
No
E
9
10
9
10
0
Yes
F
10
14
10
14
0
Yes
G
13
18
15
20
2
No
H
18
23
20
25
2
No
I
9
12
11
14
2
No
J
14
17
14
17
0
Yes
K
17
21
17
21
0
Yes
L
17
18
20
21
3
No
M
23
24
25
26
2
No
N
21
26
21
26
0
Yes
Finish
26
26
26
26
0
Yes
16.9
Activity
ES
EF
LS
LF
Slack
Critical Path
Start
0
0
0
0
0
Yes
A
0
1
0
1
0
Yes
B
1
3
1
3
0
Yes
C
3
10
3
10
0
Yes
D
10
14
13
17
3
No
E
10
13
10
13
0
Yes
F
13
16
13
16
0
Yes
G
14
18
17
21
3
No
H
18
24
21
27
3
No
I
10
15
11
16
1
No
J
16
22
16
22
0
Yes
K
22
25
22
25
0
Yes
L
22
25
22
25
0
Yes
M
24
25
27
28
3
No
N
25
28
25
28
0
Yes
Finish
28
28
28
28
0
Yes
16.11 a) Start → A → E → I → Finish Length = 17 months
Start → A → C → F → I → Finish Length = 17 months
CD16-17
c) Start → A → E → I → Finish
฀
d−
p
p
2=22−17
25 =1
By Table 16.7, P(T ≤ 22 months) = 0.84.
Start → A → C → F → I → Finish
฀
d−
p
p
2=22−17
27 =0.96
By Table 16.7, P(T ≤ 22 months) is just less than 0.84.
Start → B → D → G → J → Finish
฀
d−
p
p
2=22−17
28 =0.95
By Table 16.7, P(T ≤ 22 months) is just less than 0.84.
d) There is somewhat less than a 77% chance that the drug will be ready in 22 weeks.
3
4
5
6
7
8
9
10
11
12
13
14
B C D E F G H I J K
Time Estimates On Mean
Activity o m p Critical Path
A 1.5 2 15 * 4.0833 5.0625 Mean Critical
B 2 3.5 21 6.1667 10.027778 Path
C 1 1.5 18 4.1667 8.0277778 = 17.08333
D 0.5 1 15 3.25 5.8402778
= 25.34028
E 3 5 24 * 7.8333 12.2500
F 1 2 16 4.1667 6.25 P(T< = d) = 0.8356
G 0.5 1 14 3.0833 5.0625 where
H 2.5 3.5 25 6.9167 14.0625 d = 22
I 1 3 18 * 5.1667 8.0277778
J 2 3 18 5.3333 7.1111111
3
4
5
6
7
8
9
10
11
12
13
14
B C D E F G H I J K
Time Estimates On Mean
Activity o m p Critical Path
A 1.5 2 15 * 4.0833 5.0625 Mean Critical
B 2 3.5 21 6.1667 10.027778 Path
C 1 1.5 18 * 4.1667 8.0277778 = 17.58333
D 0.5 1 15 3.25 5.8402778
= 27.36806
E 3 5 24 7.8333 12.2500
F 1 2 16 * 4.1667 6.25 P(T<=d) = 0.8007
G 0.5 1 14 3.0833 5.0625 where
H 2.5 3.5 25 6.9167 14.0625 d = 22
I 1 3 18 * 5.1667 8.0277778
J 2 3 18 5.3333 7.1111111
Start → B → D → G → J → Finish
3
4
5
6
7
8
9
10
11
12
13
14
B C D E F G H I J K
Time Estimates On Mean
Activity o m p Critical Path
A 1.5 2 15 4.0833 5.063 Mean Critical
B 2 3.5 21 * 6.1667 10.028 Path
C 1 1.5 18 4.1667 8.028 = 17.83333
D 0.5 1 15 * 3.25 5.840
= 28.04167
E 3 5 24 7.8333 12.250
F 1 2 16 4.1667 6.250 P(T< = d) = 0.7843
G 0.5 1 14 * 3.0833 5.063 where
H 2.5 3.5 25 6.9167 14.063 d = 22
I 1 3 18 5.1667 8.028
J 2 3 18 * 5.3333 7.111
Start → B → H → J → Finish
3
4
5
6
7
8
9
10
11
12
13
14
B C D E F G H I J K
Time Estimates On Mean
Activity o m p Critical Path
A 1.5 2 15 4.0833 5.063 Mean Critical
B 2 3.5 21 * 6.1667 10.028 Path
C 1 1.5 18 4.1667 8.028 = 18.41667
D 0.5 1 15 3.25 5.840
= 31.20139
E 3 5 24 7.8333 12.250
F 1 2 16 4.1667 6.250 P(T< = d) = 0.7394
G 0.5 1 14 3.0833 5.063 where
H 2.5 3.5 25 * 6.9167 14.063 d = 22
I 1 3 18 5.1667 8.028
J 2 3 18 * 5.3333 7.111
Then, based on these spreadsheets, the answers to (a), (b), (c), and (d) would be
a) Start → A → E → I → Finish Length = 17.08 months
Start → A → C → F → I → Finish Length = 17.58 months
d) There is approximately a 73% chance that the drug will be ready in 22 weeks.
CD Chapter 16 – Pert/CPM Models for Project Management
CD16-19
16.13 a)
St art
A
B
E
C
Fi ni sh
D
b)
Activity
2
A
4
0.111
B
2
0
C
4.83
0.25
D
3
0.444
E
3.17
0.25
c) Start → A → B → C → Finish Length = 10.83 weeks *critical path
CD16-20
16.14 a)
Activity
2
A
12
0
B
23
16
C
15
1
D
27
9
E
18
4
F
6
4
฀
p
2=57−50
25 =1.4
By Appendix A, P(T ≤ 57 days) = 0.9192.
e) (0.9772)(0.9192)=0.8982.
16.15 a)
Activity
2
A
32
1.78
B
27.7
2.78
C
36
11.1
D
16
0.444
E
32
0
F
53.7
32.1
G
16.7
4
H
20.3
2.78
I
34
7.11
J
17.7
9