CHAPTER 3 PRO J E C T MA N A G E M E N T 21
2
8 5 9
Variance (O) 6 36
æö
–÷
ç
==
÷
ç÷
÷
ç
èø
16
9 2.25 9
Project variance = + + + 1.00
36 36 36 36 »
Project standard deviation = 1.00
Activity
Activity
Time
Early
Start
Early
Finish
Late
Start
Late
Finish
A
2.16
0
2.16
10.13
12.3
B
3.5
0
3.5
11.88
15.38
C
11.83
0
11.83
0
11.83
D
5.16
0
5.16
14.65
19.82
E
3.83
0
3.83
15.98
19.82
F
7
2.17
9.16
12.3
19.3
G
3.92
3.5
7.42
15.38
19.3
H
7.47
11.83
19.3
11.83
19.3
I
10.32
11.83
22.15
14.9
25.22
J
3.83
11.83
15.66
19.98
23.82
K
4
5.16
9.16
19.82
23.82
L
4
3.83
7.83
19.82
23.82
M
5.92
19.3
25.22
19.3
25.22
N
1.23
15.66
16.9
23.82
22.05
O
6.83
25.22
32.05
25.22
32.05
P
7
16.9
23.9
25.05
32.05
(b) As can be seen in the following analysis, the changes do
not have any impact on the critical path or the total project
completion time. A summary of the analysis is below.
A
2.16
B
3.5
C
0
D
5.16
E
3.83
F
7
9.16
G
7.42
H
11.83
0
I
0
11.83
J
0
11.83
K
4
9.16
L
4
7.83
3.24 (a) Probability of completion is 17 months or less:
( )
 
 
17 21
17 2.0
2
1 2.0 1 0.97725 0.0228
P t P z P z
Pz
−

= =  −


= − = − =
(b) Probability of completion in 20 months or less:
( )
 
 
20 21
20 0.5
2
1 0.5 1 0.69146 0.3085
P t P z P z
Pz
−

= =  −


= − = − =
(c) Probability of completion in 23 months or less:
( )
 
23 21
23 1.0 0.84134
2
P t P z P z
−

 = = =


(d) Probability of completion in 25 months or less:
( )
 
25 21
25 2.0 0.97725
2
P t P z P z
−

 = = =


(e)
21 1.645 for a 95% chance of completion
2by the date.
Then 21 2(1.645)
24.29, or 24 months.
x
Pz
x
x
−

=


=+
=
22 CHAPTER 3 PRO J E C T MA N A G E M E N T
3.25 (a)
Project completion time = 14 weeks
Task
Time
ES
EF
LS
LF
Slack
A
3
0
3
0
3
0
B
2
0
2
2
4
2
C
1
0
1
11
12
11
D
7
3
10
3
10
0
E
6
2
8
4
10
2
F
2
1
3
12
14
11
G
4
10
14
10
14
0
(b) To crash to 10 weeks, we follow 2 steps:
1. Crash D by 2 weeks ($150).
2. Crash D and E by 2 weeks each ($100 + $150).
Total crash cost = $400 additional
(c) Using POM for Windows software, minimum project
completion time = 7. Additional crashing cost =
$1,550.
Normal
Time
Crash
Time
Normal
Cost
Crash
Cost
Crash
Cost/Pd
Crash
By
Crashing
Cost
A
3
2
1,000
1,600
600
1
600
B
2
1
2,000
2,700
700
0
0
C
1
1
300
300
0
0
0
D
7
3
1,300
1,600
75
4
300
E
6
3
850
1,000
50
3
150
F
2
1
4,000
5,000
1,000
0
0
G
4
2
1,500
2,000
250
2
500
3.26 (a)
CHAPTER 3 PRO J E C T MA N A G E M E N T 23
3.26 (b)
Task
Time
ES
EF
LS
LF
Slack
A
0.0*
0.0
0.0
0.0
0.0
0.0
B
8.0
0.0
8.0
0.0
8.0
0.0
C
0.1
8.0
8.1
10.4
10.5
2.4
D
1.0
8.0
9.0
12.0
13.0
4.0
E
1.0
8.0
9.0
10.0
11.0
2.0
F
1.0
9.0
10.0
13.0
14.0
4.0
G
2.0
8.0
10.0
9.0
11.0
1.0
H
3.0
8.0
11.0
11.0
14.0
3.0
I
1.0
10.0
11.0
11.0
12.0
1.0
J
4.0
8.0
12.0
8.0
12.0
0.0
K
2.0
12.0
14.0
12.0
14.0
0.0
L
1.0
14.0
15.0
14.0
15.0
0.0
M
0.5
15.0
15.5
15.0
15.5
0.0
N
2.0
11.0
13.0
12.0
14.0
1.0
O
1.0
13.0
14.0
14.5
15.5
1.5
P
1.5
13.0
14.5
14.0
15.5
1.0
Q
5.0
8.1
13.1
10.5
15.5
2.4
R
1.0
15.5
16.5
15.5
16.5
0.0
S
0.5
16.5
17.0
16.5
17.0
0.0
T
1.0
17.0
18.0
17.0
18.0
0.0
U
0.0*
18.0
18.0
18.0
18.0
0.0
*Note: Start (A) and Finish (U) are assigned times of zero.
Critical path is A–B–J–K–L–M–R–S–T–U, for 18 days.
(c) (i) no, transmissions and drivetrains are not on the
critical path.
A
1 day
A
1 day
3.27 (a)
3.28 (a) Expected times for individual activities (using (a + 4m
+ b)/6)). A = 5, B = 6, C = 7, D = 6, E = 3. Expected
project completion time = 15 (Activities A–C–E).
(b) Variance for individual activities (using [(b – a)/6]2).
3.29 (a)
(b)
Activity
Time
ES
EF
LS
LF
Slack
2
A
7
0
7
0
7
0
2
4*
B
3
7
10
13
16
6
1
1*
C
9
7
16
7
16
0
3
9*
D
4
16
20
25
29
9
1
1*
E
5
16
21
16
21
0
1
1*
F
8
21
29
21
29
0
2
4*
G
8
29
37
29
37
0
1
1*
H
6
37
43
37
43
0
2
4*
*Activities on the critical path: A, C, E, F, G, H. Project Comple-
tion time = 43.
(b) 4.8
49 43
= 1 .2 5
4.8
( 4 9 ) .8 9 4 3 5
( 49) (1 .89435)= 0.10565
z
Pt
Pt
=−
=
=
= −
3.30 AON network
B
50
2 days
C
100
1 day
D
150
2 days
E
200
3 days
To crash by 4 days, from 13 days to 9 days,
◼ Crash B by 1 day ($50) to reach 12 days
◼ Crash B by a second day ($50) and C by 1 day ($100) to
24 CHAPTER 3 PRO J E C T MA N A G E M E N T
3.31 (a)
3.32 The overall purpose of Problem 3.32 is to have students use
a network to solve a problem that almost all students face. The
first step is for students to list all courses that they must take,
including possible electives, to get a degree from their particular
college or university. For every course, students should list all the
26 CHAPTER 3 PRO J E C T MA N A G E M E N T
tI = 30
6.67*
44.44
tJ = 10
0.67
0.44
tK = 0
0.10
0.00
44.44
tI = 30
6.67*
44.44
tJ = 10
0.67
0.44
tK = 0
0.10
0.00
44.44
CASE STUDY
SOUTHWESTERN UNIVERSITY: A
1.
Activity Mean
S.D.
Variance
tA = 30
3.33*
11.11
tB = 60
10.00
100.00
tC = 65
8.33*
69.39
tD = 55
11.66*
136.10
tE = 30
1.67
2.78
tF = 0
0.10
0.00
tG = 30
1.67*
2.78
Standard deviation of critical path = 17.87 days
2. P (Completion < 270 days) = P(t 270)
 
−

=  = =

270 260 0.56 0.712 71.2%
P Z P Z
3. Crash to 250 days and to 240 days
Activity
Normal
Time (days)
Crash
Time (days)
Crash
Cost/Day
A
30
20
$1,500
B
60
20
$3,500
C
65
50
$4,000
D
55
30
$1,900
E
30
25
$9,500
F
0*
0
$0
G
30
25
$2,500
H
20
10
$2,000
I
30
20
$2,000
J
10
8
$6,000
K
0*
0
$0
L
30
20
$4,500
*Rounded to zero from 0.1
To crash to 250 days (from the current 260 days), select A at
$1,500/day 10 days = $15,000.
1. Construction project network:
2. The critical path is Activities 1–3–5–6–8–10–11–12–14–16–
17–19–20–21. The project length is 47 months—about four
years from start to finish.
3. Building a hospital is much more complex than an office build-
ing for several reasons. In this case, hundreds of “users” of the
new building had extensive input. Second, the design of the new
layout (circular, pod design) is somewhat radical compared to
4. Since there were 13 months of planning prior to the proposal/
review stage (listed as Activity 1) and the project then took
47 months (for a total of 60 months), 22% of the time was
spent in planning.
28 CHAPTER 3 PRO J E C T MA N A G E M E N T
1.222 Onsite ticketing (M) 1.3 Hire operations manager (T)
1.31 Site plan (U)
1.311 Power, etc. (X)
1.32 Security director (V)
1.321 Set police/fire plan (W)
Answers may vary somewhat at the Level 3 and Level 4.
Level 2 activities should be activities B, C, and T.
1. Determine the expected shutdown time and the probability the
shutdown will be completed 1 week earlier.
2. What are the probabilities that Shale finishes the mainte-
nance project 1 day, 2 days, 3 days, 4 days, 5 days, or 6 days
earlier?
From the precedence data supplied in the problem, we can
develop the following AON network:
22 CHAPTER 3 PRO J E C T MA N A G E M E N T
3.25 (a)
Project completion time = 14 weeks
Task
Time
ES
EF
LS
LF
Slack
A
3
0
3
0
3
0
B
2
0
2
2
4
2
C
1
0
1
11
12
11
D
7
3
10
3
10
0
E
6
2
8
4
10
2
F
2
1
3
12
14
11
G
4
10
14
10
14
0
(b) To crash to 10 weeks, we follow 2 steps:
1. Crash D by 2 weeks ($150).
2. Crash D and E by 2 weeks each ($100 + $150).
Total crash cost = $400 additional
(c) Using POM for Windows software, minimum project
completion time = 7. Additional crashing cost =
$1,550.
Normal
Time
Crash
Time
Normal
Cost
Crash
Cost
Crash
Cost/Pd
Crash
By
Crashing
Cost
A
3
2
1,000
1,600
600
1
600
B
2
1
2,000
2,700
700
0
0
C
1
1
300
300
0
0
0
D
7
3
1,300
1,600
75
4
300
E
6
3
850
1,000
50
3
150
F
2
1
4,000
5,000
1,000
0
0
G
4
2
1,500
2,000
250
2
500
3.26 (a)
CHAPTER 3 PRO J E C T MA N A G E M E N T 23
3.26 (b)
Task
Time
ES
EF
LS
LF
Slack
A
0.0*
0.0
0.0
0.0
0.0
0.0
B
8.0
0.0
8.0
0.0
8.0
0.0
C
0.1
8.0
8.1
10.4
10.5
2.4
D
1.0
8.0
9.0
12.0
13.0
4.0
E
1.0
8.0
9.0
10.0
11.0
2.0
F
1.0
9.0
10.0
13.0
14.0
4.0
G
2.0
8.0
10.0
9.0
11.0
1.0
H
3.0
8.0
11.0
11.0
14.0
3.0
I
1.0
10.0
11.0
11.0
12.0
1.0
J
4.0
8.0
12.0
8.0
12.0
0.0
K
2.0
12.0
14.0
12.0
14.0
0.0
L
1.0
14.0
15.0
14.0
15.0
0.0
M
0.5
15.0
15.5
15.0
15.5
0.0
N
2.0
11.0
13.0
12.0
14.0
1.0
O
1.0
13.0
14.0
14.5
15.5
1.5
P
1.5
13.0
14.5
14.0
15.5
1.0
Q
5.0
8.1
13.1
10.5
15.5
2.4
R
1.0
15.5
16.5
15.5
16.5
0.0
S
0.5
16.5
17.0
16.5
17.0
0.0
T
1.0
17.0
18.0
17.0
18.0
0.0
U
0.0*
18.0
18.0
18.0
18.0
0.0
*Note: Start (A) and Finish (U) are assigned times of zero.
Critical path is A–B–J–K–L–M–R–S–T–U, for 18 days.
(c) (i) no, transmissions and drivetrains are not on the
critical path.
3.27 (a)
3.28 (a) Expected times for individual activities (using (a + 4m
+ b)/6)). A = 5, B = 6, C = 7, D = 6, E = 3. Expected
project completion time = 15 (Activities A–C–E).
(b) Variance for individual activities (using [(b – a)/6]2).
3.29 (a)
(b)
Activity
Time
ES
EF
LS
LF
Slack
2
A
7
0
7
0
7
0
2
4*
B
3
7
10
13
16
6
1
1*
C
9
7
16
7
16
0
3
9*
D
4
16
20
25
29
9
1
1*
E
5
16
21
16
21
0
1
1*
F
8
21
29
21
29
0
2
4*
G
8
29
37
29
37
0
1
1*
H
6
37
43
37
43
0
2
4*
*Activities on the critical path: A, C, E, F, G, H. Project Comple-
tion time = 43.
(b) 4.8
49 43
= 1 .2 5
4.8
( 4 9 ) .8 9 4 3 5
( 49) (1 .89435)= 0.10565
z
Pt
Pt
=−
=
=
= −
3.30 AON network
B
50
2 days
C
100
1 day
D
150
2 days
E
200
3 days
To crash by 4 days, from 13 days to 9 days,
◼ Crash B by 1 day ($50) to reach 12 days
◼ Crash B by a second day ($50) and C by 1 day ($100) to
24 CHAPTER 3 PRO J E C T MA N A G E M E N T
3.31 (a)
3.32 The overall purpose of Problem 3.32 is to have students use
a network to solve a problem that almost all students face. The
first step is for students to list all courses that they must take,
including possible electives, to get a degree from their particular
college or university. For every course, students should list all the
26 CHAPTER 3 PRO J E C T MA N A G E M E N T
CASE STUDY
SOUTHWESTERN UNIVERSITY: A
1.
Activity Mean
S.D.
Variance
tA = 30
3.33*
11.11
tB = 60
10.00
100.00
tC = 65
8.33*
69.39
tD = 55
11.66*
136.10
tE = 30
1.67
2.78
tF = 0
0.10
0.00
tG = 30
1.67*
2.78
Standard deviation of critical path = 17.87 days
2. P (Completion < 270 days) = P(t 270)
 
−

=  = =

270 260 0.56 0.712 71.2%
P Z P Z
3. Crash to 250 days and to 240 days
Activity
Normal
Time (days)
Crash
Time (days)
Crash
Cost/Day
A
30
20
$1,500
B
60
20
$3,500
C
65
50
$4,000
D
55
30
$1,900
E
30
25
$9,500
F
0*
0
$0
G
30
25
$2,500
H
20
10
$2,000
I
30
20
$2,000
J
10
8
$6,000
K
0*
0
$0
L
30
20
$4,500
*Rounded to zero from 0.1
To crash to 250 days (from the current 260 days), select A at
$1,500/day 10 days = $15,000.
1. Construction project network:
2. The critical path is Activities 1–3–5–6–8–10–11–12–14–16–
17–19–20–21. The project length is 47 months—about four
years from start to finish.
3. Building a hospital is much more complex than an office build-
ing for several reasons. In this case, hundreds of “users” of the
new building had extensive input. Second, the design of the new
layout (circular, pod design) is somewhat radical compared to
4. Since there were 13 months of planning prior to the proposal/
review stage (listed as Activity 1) and the project then took
47 months (for a total of 60 months), 22% of the time was
spent in planning.
28 CHAPTER 3 PRO J E C T MA N A G E M E N T
1.222 Onsite ticketing (M) 1.3 Hire operations manager (T)
1.31 Site plan (U)
1.311 Power, etc. (X)
1.32 Security director (V)
1.321 Set police/fire plan (W)
Answers may vary somewhat at the Level 3 and Level 4.
Level 2 activities should be activities B, C, and T.
1. Determine the expected shutdown time and the probability the
shutdown will be completed 1 week earlier.
2. What are the probabilities that Shale finishes the mainte-
nance project 1 day, 2 days, 3 days, 4 days, 5 days, or 6 days
earlier?
From the precedence data supplied in the problem, we can
develop the following AON network: