CHAPTER 3 PRO J E CT MA N A G E M EN T 23
3.25* (a)
E
7.0
3.2
10.2
23.0
30.0
19.8
No
F
10.0
30.0
40.0
30.0
40.0
0.0
Yes
Activity
Time
ES
EF
LS
LF
Slack
A
7
0
7
0
7
0
2
B
3
7
10
13
16
6
1
C
9
7
16
7
16
0
3
D
4
16
20
25
29
9
1
E
5
16
21
16
21
0
1
F
8
21
29
21
29
0
2
G
8
29
37
29
37
0
1
1**
tion time = 43.
(b) 4.8
( 49) (1 .89435)= 0.10565
Pt
=−
= −
3.26* (a)
(b) Expected times and variances:
Activity
a
m
b
Expected Time
Variance
A
8
10
12
10.0
0.44
B
6
7
0.25
C
3
3
0.03
D
20
30
20.0
11.11
E
6
7
0.11
F
9
10
11
10.0
0.11
G
6
7
10
0.44
H
15
16
15.0
0.11
I
11
13
11.2
0.25
J
6
7
0.11
K
4
7
0.44
(c) ES, EF, LS, LF, and slack times:
Activity
Time
ES
EF
LS
LF
Slack
Critical
A
10.0
0.0
10.0
0.0
10.0
0.0
Yes
B
7.2
0.0
7.2
22.8
30.0
22.8
No
C
3.2
0.0
3.2
19.8
23.0
19.8
No
D
20.0
10.0
30.0
10.0
30.0
0.0
Yes
G
7.3
30.0
37.3
47.7
55.0
17.7
No
H
15.0
40.0
55.0
40.0
55.0
0.0
Yes
I
11.2
40.0
51.2
50.8
62.0
10.8
No
J
7.0
55.0
62.0
55.0
62.0
0.0
Yes
K
6.7
62.0
68.7
62.0
68.7
0.0
Yes
L
2.2
55.0
57.2
66.5
68.7
11.5
No
(d) The critical path is given by the activities
Project variance = Sum of variances of activities
( )
70 68.7
70 0.37 0.644
3.51
P t P z P z
−

= = =


(f) Probability of completion in 80 days or less:
( )
−

= = =


80 68.7
80 3.22 0.99934
3.51
P t P z P z
3.51


3.27* The construction company problem involves 23 separate
activities. These activities, their immediate predecessors, and time
estimates were given in the problem. One of the most difficult
24 CHAPTER 3 PRO J E C T MA N A G E M E N T
Table for Problem 3.27
Times
Activity
Opt
Most
Pess
E(t)
ES
EF
LS
LF
Slack
1
1.0
4.00
5.00
3.67
0.67
0.00
3.67
9.00
12.67
9.00
2
2.0
3.00
4.00
3.00
0.33
0.00
3.00
16.50
19.50
16.50
3
3.0
4.00
5.00
4.00
0.33
0.00
4.00
14.50
18.50
14.50
4
7.0
8.00
9.00
8.00
0.33
0.00
8.00
3.50
11.50
3.50
5
4.0
4.00
5.00
4.17
0.17
3.67
7.83
12.67
16.83
9.00
6
1.0
2.00
4.00
2.17
0.50
4.00
6.17
18.50
20.67
14.50
7
4.0
5.00
6.00
5.00
0.33
8.00
13.00
11.50
16.50
3.50
8
1.0
2.00
4.00
2.17
0.50
13.00
15.17
16.50
18.67
3.50
9
3.0
4.00
4.00
3.83
0.17
7.83
11.67
16.83
20.67
9.00
10
1.0
1.00
2.00
1.17
0.17
3.00
4.17
19.50
20.67
16.5
11
18.0
20.00
26.00
20.67
1.33
0.00
20.67
0.00
20.67
0.00
12
1.0
2.00
3.00
2.00
0.33
15.17
17.17
18.67
20.67
3.50
13
1.0
1.00
2.00
1.17
0.17
20.67
21.83
20.67
21.83
0.00
14
0.1
0.14
0.16
0.14
1.00
21.83
21.97
21.83
21.97
0.00
15
0.2
0.30
0.40
0.30
0.03
21.97
22.27
24.84
25.14
2.87
16
1.0
1.00
2.00
1.17
0.17
21.97
23.14
21.97
23.14
0.00
17
1.0
2.00
3.00
2.00
0.33
23.14
25.14
23.14
25.14
0.00
18
3.0
5.00
7.00
5.00
0.67
25.14
30.14
25.14
30.14
0.00
19
0.1
0.10
0.20
0.12
0.02
30.14
30.25
30.14
30.25
0.00
20
0.1
0.14
0.16
0.14
0.00
30.25
30.39
33.33
33.47
3.08
21
2.0
3.00
6.00
3.33
0.67
30.25
33.59
30.25
33.59
0.00
22
0.1
0.10
0.20
0.12
0.02
30.39
30.50
33.47
33.58
3.08
23
0.0
0.20
0.20
0.17
0.03
33.59
33.75
33.59
33.75
0.00
As you can see, the expected project length is about 34 weeks. The activities along the critical path are activities
11, 13, 14, 16, 17, 18, 19, 21, and 23.
3.28 Helps to modify the AON with the lowest costs to crash.
2. Now both paths through are critical. We would need to
shorten A or F, or shorten C and either B/E. This is not
worth it, so we would not bother to crash any further.
3.29
3.30 Crash costs per unit time are $600 for A, $900 for B, and
3.31 (a) Project completion time = 16 (Activities A–D–G)
Figure for Problem 3.31
26 CHAPTER 3 PRO J E C T MA N A G E M E N T
1
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**
69.44
tD = 55
11.66**
136.11
tE = 30
1.67
2.78
tF = 0.1
0.00
0.00
tG = 30
1.67**
2.78
tH = 20
3.33**
11.11
tI = 30
6.67**
44.44
tJ = 10
0.67
0.44
tK = 0.1
0.00
0.00
tL = 30
6.67**
44.44
**Critical path
Standard deviation of critical path = 17.87 days
LO 3.2: Draw AOA and AON networks
LO 3.4: Determine a critical path
LO 3.5: Calculate the variance of activity times
AACSB: Analytical thinking
2. P (Completion < 270 days) = P(t 270)
−

= = =


270 260 0.56 0.712 71.2%
17.87
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.
To crash to 240 days now (from the current 250 days), select D at
$1,900/day 10 days = $19,000.
Total cost to crash to 240 days = $34,000.
LO 3.6: Crash a project
AACSB: Analytical thinking
VIDEO CASE STUDIES
PROJECT MANAGEMENT AT ARNOLD
PALMER HOSPITAL
The Arnold Palmer Hospital video for this case (8 minutes) is
available from Pearson. Also note that the Global Company
Profile in Chapter 6 highlights this hospital.
1. Construction project network:
− − − − − −
+
Critical path = A C D G H I L (260 days)
Variance of critical path = 11.11 + 69.44 + 136.11 + 2.78
+ 11.11 44.44 + 44.44 = 319.43
CHAPTER 3 PRO J E CT MA N A G E M EN T 27
2
impacted on design/construction. Fifth, hospitals, by their very
nature, are more complex from a safety, health hazard, security,
quiet, serenity perspective than an office building.
LO 3.4: Determine a critical path
AACSB: Application of knowledge
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.
G
2
25
27
25
27
0
H
4
10
14
16
20
6
I
4
14
18
20
24
6
J
18
28
24
34
6
K
2
10
12
21
23
11
M
8
18
26
26
34
N
6
10
16
28
34
O
7
27
34
27
34
0
P
10
30
14
34
4
Q
4
10
14
19
23
R
4
14
18
23
27
MANAGING HARD ROCK’S ROCKFEST
There is a short (9-minute) video available from Pearson and filmed
specifically for this text that supplements this case.
Hard Rock’s ES, EF, LS, LF, and slack:
Activity
time
Early
Start
Early
Finish
Late Start
Late
Finish
Slack
A
7
0
7
0
7
0
B
3
7
10
7
10
0
C
3
7
10
18
21
11
D
5
10
15
10
15
0
E
6
15
21
15
21
0
F
4
21
25
21
25
0
S
3
25
28
31
34
6
T
4
7
11
13
17
6
U
6
11
17
17
23
6
V
7
11
18
23
30
12
W
4
18
22
30
34
12
X
8
17
25
23
31
6
Y
6
10
16
22
28
12
Z
6
16
22
28
34
12
Network diagram
28 CHAPTER 3 PRO J E C T MA N A G E M E N T
3. Major challenges a project manager faces in a project like
one of the key managers or promoters, and many others.
4. Work breakdown structure, with example of Level 1, 2, 3, 4
tasks:
1.0 Rockfest event, with site selected (A) [Level 1]
1.1 Select local promoter (B) [Level 2]
1.1.5 Staff travel (P)
1.1.6 Merchandise deals (Y)
1.1.6.1 Online sales of merchandise (Z)
1.1.7 Hire sponsor coordinator (Q)
1.1.7.1 Finalize sponsors (R)
1.1.7.2 Signage for sponsors (S)
1.2 Hire production manager (C)
ADDITIONAL CASE STUDY
1. Determine the expected shutdown time and the probability the
LO 3.4: Determine a critical path
AACSB: Analytical thinking
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
CHAPTER 3 PRO J E CT MA N A G E M EN T 29
Activity
2
1
0.25
0.0625
2
0.17
0.0289
3
0.33
0.1089
8
0.33
0.1089
16
2.33
5.4289
21
0.67
0.4489
27
0.17
0.0289
28
0.67
0.4489
29
1.17
1.3689
Variance for critical path:
8.0337
Therefore,
== 8.0337 2.834.
As an approximation, we can
use the customary equation for the normal distribution:
−
=
Due date ( )Et
z
(Note: This might be a good time to discuss the difference
between a continuous and a discrete probability distribution and
the appropriate procedure for using a continuous distribution as an
approximation to a discrete, if you have not already done so.)
Finish Time
z
Probability
1 day early
–0.353
36.3
2 days early
–0.706
24.0
3 days early
–1.058
14.5
4 days early
–1.411
7.9
5 days early
–1.764
3.9**
6 days early
–2.117
1.7
7 days early
–2.470
0.7
**The appropriate procedure for using the normal distribution gives 3.0%—
roughly a 30% difference.
There is, by the approximate procedure used, a 3.9% probability
of finishing 5 days, or 1 week, early.
LO 3.5: Calculate the variance of activity times
AACSB: Analytical thinking
The following table indicates the expected times, variances, and slacks needed to complete the rest of the case:
Activity
Opt
Most
Likely
Pess
E(t)
ES
EF
LS
LF
Slack
1
1.0
2.0
2.5
1.92
0.25
0.00
1.92
0.00
1.92
0.000
2
1.5
2.0
2.5
2.00
0.17
1.92
3.92
1.92
3.92
0.000
3
2.0
3.0
4.0
3.00
0.33
3.92
6.92
3.92
6.92
0.000
4
1.0
2.0
3.0
2.00
0.33
3.92
5.92
22.50
24.50
18.580
5
1.0
2.0
4.0
2.17
0.50
3.92
6.08
10.25
12.42
6.333
6
2.0
2.5
3.0
2.50
0.17
3.92
6.42
13.42
15.92
9.500
7
2.0
4.0
5.0
3.83
0.50
3.92
7.75
29.58
33.42
25.670
8
1.0
2.0
3.0
2.00
0.33
6.92
8.92
6.92
8.92
0.000
9
1.0
1.5
2.0
1.50
0.17
5.92
7.42
26.67
28.17
20.750
10
1.0
1.5
2.0
1.50
0.17
5.92
7.42
24.50
26.00
18.580
11
2.0
2.5
3.0
2.50
0.17
6.08
8.58
19.92
22.42
13.830
12
15.0
20.0
30.0
20.83
2.50
6.08
26.92
12.42
33.25
6.330
13
1.0
1.5
2.0
1.50
0.17
6.42
7.92
15.92
17.42
9.500
14
3.0
5.0
8.0
5.17
0.83
6.42
11.58
28.08
33.25
21.670
15
3.0
8.0
15.0
8.33
2.00
7.75
16.08
33.42
41.75
25.670
16
14.0
21.0
28.0
21.00
2.33
8.92
29.92
8.92
29.92
0.000
17
1.0
5.0
10.0
5.17
1.50
7.42
12.58
28.17
33.33
20.750
18
2.0
5.0
10.0
5.33
1.33
7.42
12.75
26.00
31.33
18.580
19
5.0
10.0
20.0
10.83
2.50
8.58
19.42
22.42
33.25
13.830
20
10.0
15.0
25.0
15.83
2.50
7.92
23.75
17.42
33.25
9.500
21
4.0
5.0
8.0
5.33
0.67
29.92
35.25
29.92
35.25
0.000
22
1.0
2.0
3.0
2.00
0.33
12.75
14.75
31.33
33.33
18.580
23
1.0
2.0
2.5
1.92
0.25
14.75
16.67
33.33
35.25
18.580
24
1.0
2.0
3.0
2.00
0.33
26.92
28.92
33.25
35.25
6.330
25
1.0
2.0
3.0
2.00
0.33
23.75
25.75
33.25
35.25
9.500
26
2.0
4.0
6.0
4.00
0.67
16.08
20.08
41.75
45.75
25.670
27
1.5
2.0
2.5
2.00
0.17
35.25
37.25
35.25
37.25
0.000
28
1.0
3.0
5.0
3.00
0.67
37.25
40.25
37.25
40.25
0.000
29
3.0
5.0
10.0
5.50
1.17
40.25
45.75
40.25
45.75
0.000
From the table, we can see that the expected shutdown time is 45.75, or 46 days. There are nine activities on the critical path.
30 CHAPTER 3 PRO J E C T MA N A G E M E N T
3. Shale Oil is considering increasing the budget to shorten the
shutdown. How do you suggest the company proceed?
In order to shorten the shutdown, Shale Oil would have to
determine the costs of decreasing the activities on the critical