Review Problems of Chapter 17
1.
For the following network diagram, determine both the critical path and the expected
project duration. The numbers on the arrows represent expected activity times.
Solution:
Path
2.
Expected
Path Time
1–2–4–6–8–9
41
1–2–5–6–8–9
48
1–2–5–7–8–9
55*
1–3–7–8–9
40
Chris received new word processing software for her birthday. She also received a
check, with which she intends to purchase a new computer. Chris’s college instructor
assigned a paper due next week. Chris decided that she will prepare the paper on the
new computer. She made a list of the activities she will need to do and their estimated
times.
a.
b.
c.
d.
Arrange the activities into two logical sequences.
Construct an AON diagram.
Determine the critical path and the expected duration time.
What are some possible reasons for the project to take longer than the
expected duration?
Solution:
2.
a.
Choose topic
Library res.
Outline
Shop
Select
Install
Write Paper
Grammar ck
Submit paper
b. AON diagram
Choose
Library
Outline
Start
Write
Shop
c. 7.5
Select
Grammar
Submit
Install
Shop, Select, Install, Write, Grammar, Submit
d. Parallel paths, probabilistic activity durations.
3.
For problem 1, determine the following quantities for each activity: the earliest start
time, latest start time, earliest finish time, latest finish time, and slack time. List the
critical activities, and determine the expected duration of the project.
Solution:
End
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
55
0
LS
ES
37 40
23 26
5 23
5 23
2
0
0
5
5
40 44
33 37
4
3
6
23 33
23 33
18
4
44 53
44 53
5
10
5
1
20 33
5 18
LF
EF
33 44
33 44
89
53 55
53 55
92
13
3
11
7
4.
The project manager of a task force planning the construction of a domed stadium had
hoped to be able to complete construction prior to the start of the next college football
season. After reviewing construction time estimates, it now appears that a certain
amount of crashing will be needed to ensure project completion before the season
opener. Given the following time and cost estimates, determine a minimum-cost
crashing schedule that will shave five weeks off the project length. Note: No activity
can be crashed more than two weeks.
Solution:
B
14
A
12
Start
K
9
End
M
C
10
G
D
17
3
N
15
11
E
18
F
12
H
8
J
12
I
7
Path
A–B–K
5.
N:
P 8
Length after crashing N weeks
0
1
2
35
3
4
5
42
40
C–E–H–P
44
43
42
42
C–D–G–M
45
44
43
43
C–E–H–N
47
46
45
45
45
44
C–F–I–J–P
49
48
47
46
45
44
Activity
Crashed
–
c
c
f
f
e,p
Cost
–
$5,000
5,000
12,000
15,000
36,000
Cum. Cost
–
$5,000
10,000
22,000
37,000
73,000
A construction project has indirect costs totaling $40,000 per week. Major activities in
the project and their expected times are shown in this precedence diagram:
Crashing costs for each activity are
a.
b.
Determine the optimum time-cost crashing plan.
Plot the total-cost curve that describes the least expensive crashing schedule
that will reduce the project length by six weeks.
Solution:
Activity
1–2
1st week
$18 [2]
2nd week 3rd week
$22 [6]
–
2–5
24
25
25
5–7
30
30
35
7–11
15 [1]
20 [3]
–
11–13
30 [4]
33 [5]
36
1–3
12 [6]
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 [5]
15
25
4–6
8 [4]
13
–
6–10
5 [3]
12 [6]
–
15
–
10–12
14
Summary:
Project
Cum. wk.
Length Shortened
35
0
1
2
$18
15
Weeks Crashed
3
4
5
6
$22
20
30
33
12
10
8
5
$15
$18
$25
Path
1–2–5–7–11–13
0
35 wk.
1
34
1–3–8–11–13
32
1–3–9–12–13
20
1–4–6–10–12–13
33
33
Cum. Crash
Costs ($0000)
0
Indirect Costs
($000)
35(40) = 1,400
Total Cost
($000)
1,400
12
$38
$43
$46
Weeks Crashed
2
3
4
33
32
31
5
30
6
29
32
31
30
29
32
31
30
29
34
1
15
34(40) = 1,360
1,375
33
2
33
33(40) = 1,320
1,353
32
3
58
32(40) = 1,280
1,338
31
4
96
31(40) = 1,240
1,336
30
5
139
30(40) = 1,200
1,339
29
6
185
29(40) = 1,160
1,345
1,400
Total
Cost
($000)
1,300
0
30
31
32
33
34
Product Length
(wk)
35
6.
Chuck’s Custom Boats (CCB) builds luxury yachts to customer order. CCB has landed a
contract with a mysterious New York lawyer (Mr. T). Relevant data are shown below.
The complication is that Mr. T wants delivery in 32 weeks or he will impose a penalty
of $375 for each week his yacht is late. Note: No activity can be crashed more than
two weeks.
Develop a crashing schedule.
Solution:
Start
7
L
125, -
9
K
410, 415
5
N
45, 45
4
M
300, 350
6
J
50,-
8
P
-, -
5
Q
200, 225
7
Y
85, 90
6
Z
90, -
Project duration = 39 wk
Project
Shorten
Crash
length
activity
cost
39 wk
–
0
38
Z
90
37
N, L
36
Q
200
35
Q
225
34
M, N
345
170 = (125 + 45)
Stop here; additional crashing will cost more than the $375 weekly penalty.
7.
Given the accompanying network diagram, with times shown in days,
a. Determine the expected duration of the project.
b. Compute the probability that the project will take at least 18 days.
Solution: a. 18.5 (See table in part b.)
b.
Path
Expected Duration
1-2-4-6
5+8.17 +5.33 = 18.5
1-3-5-6
8.33 + 3 + 3.83 =
15.16
Standard
Deviation
1.17
Z17
Probability
–.43
.3336
1.12
2.54
.9945
En
d
1 – .3336 = .67.