Homework#6: Project Management
1. (PERT/CPM Problem)
Tom Schriber, a director of personal of Management Resources, Inc. is in the
process of designing a program that is customers can use in the job-finding
process. Some of the activities include preparing resumes, writing letters,
making appointments to see prospective employers, researching companies
and industries, and so on. Some of the information on the activities is shown
in the following table:
Activity Days
A
B
C
D
E
F
G
H
I
J
K
L
I
a m b
8 10 12
6 7 9
3 3 4
10 20 30
6 7 8
9 10 11
6 7 10
14 15 16
10 11 13
6 7 8
4 7 8
1 2 4
b)
Immediate
Predecessors
tlatcimtb ,
A
C
B, D , E
B, D, E
F
F
G, H
I, J
G, H
10
6
BY
rag /
41g
7. 2
114
3.2
1136
20
1001g
7
11g
10
119
7. 3
419
IS
119
II. 2
114
7
b. 7
2.2
?
119
419
114
a) Construct a network for this problem.
b) Determine the expected time and variance for each activity.
c) Determine EST, EFT, LST, LFT and slack for each activity.
.
.
d) Determine the critical path and project completion time.
e) Determine the probability that the project will be finished in 70 days or
less.
f) Determine the probability that the project will be finished in 80 days or
less.
g) Determine the probability that the project will be finished in 90 days or
less.
0
A)
,
C)
O
0
10
10
A
D
30
7
2030
10
10
10
0
>
B
O
O
7.2
O
22.87.230
O
✓
✓
,
19.8
✓
40
IS
55
0
F
40
40
3.2
3.2
C-
10.2
23
7
30
I
19.8
critical
path
project completion
3.2
23
A
:
time
-
:
D
-
F
-
-
J
-
K
✓
30
10
µ ,
30
G
7
37.3
47.77.355
L
57.2
66.52.2
68.7
55
J
-
g
É
,
J
62
55
7
62
.
4-gtlgttgttgttg.tt
todays
3.51
:
probability
or
0.51-0.1443
70-68.7
:
0.37
From
2=7
:
3. SI
=
O
v68.7 ET 68.7
>
:
:
6.768.7
68.70
68.7
101-201-10 1-151-71-6.7=68.7
:
top
62
✓
11.5
✓
55
i.
z :X
68.7
0
40
days
68.7
E--68.7
e)
H
?
51,2
→
1%8
C
K
10.8
,
d)
O
50.811.262
,
,
O
55
62
30
22.8
ST
H
\
v0
0
40
,
0
n
o
normal distribution table
2 :O -3 >
→
probability
-
-
0.1443
0.6443 #
of finish
less
project
in
E- 68.7
-
f
°
2
:
"
"
+
GP
=
table
days
80
3.23
?
distribution
3.23
:
probability
i.
→
probability
of finish
=
0.994
project
in
less
or
0.51-0.4994
:
0.9994
:
g
z
80-68.7
3.51
=
normal
from
#
5--68.7
°
2
:
?
✗
From
normal
GP
table
2=6.07
"
"
=
→
90-68.7
3.51
=
distribution
i.
days
:O St
.
:
of finish
probability
90
6.07
or
less
0.499g
0.9999
#
probability
=
0.4999
project
in
2. (Project Crashing Problem)
Assume the project information that follows. Compute the total direct cost
for each project duration. If the indirect costs for each project duration are
$450 (21 time units), $400 (20), $350 (19), 300(18), $250 (17), and $200
(16), compute the total project cost for each project duration. What duration
represents the lowest total project cost? What is the cost?
Activity
Immediate
Predecessors
A
B
C
D
E
F
G
A
A
A
C
D
B, E, F
Time (Day) Direct Costs ($)
4
/
g
0
A
0
S
5
B
14
9
9
18
°
I
1
13
C
13
C-
17
→
→
8
6
S
-
14
14
0
,
4
18
12
12
F
18
12
6
18
→
5
critical path
total direct cost
project completion
total
project
A
-
day
:
:
80
18
G
21
18
3
21
D
-
F
G -51-7+61-3=21
-
-
901-1001-801-60 1-701-501-200=650
=
cost
12
7
=
%
,
0
D
S
✓
20
0
7
\
÷
40
650
S
S
crash cost /
Normal Crash Normal Crash
5
4
90
120
9
5
100
140
8
7
80
120
7
6
60
80
4
2
70
190
6
4
50
190
3
2
200
280
21
days
6501-410
:
1,100
indirect cost
day
( 1st
crash
3
D)
S
B
14
8
9
17
/
O
0
A
0
S
13
C
S
5
o
o
o
C-
13
17
→
→
8
5
S
4
13
17
G
20
17
3
20
,
-
13
o
17
7
o
\
o
D
g
11
-
critical path
A C
:
project
duration
-
0
4
6
17
option
A
G
-
c-
-
p
I:-c
D
-
E
60
=
130
=
110
80
=
G
✓
30
=
c- D=
6701-400=1,070
80
:
9
/
A
11
G
-
A)
o
17
days
:
'
0
F
F
6501-20=670
:
project cost
( 2nd crash
D
-
20
=
total direct cost
total
E
-
-
A
11
6
5
11
)
4
B
13
7
9
16
4
4
o
o
C
C-
16
→
→
8
4
4
12
12
o
-
12
12
4
o
16 G
19
3
19
,
16
16
7
O
\
4
o
D
10
F
10
6
10
lb
→
6
4
critical path
:
A C
-
A
project
duration
=
total direct cost
total
project cost
-
D
19
-
-
E
F
-
-
10
G
G
days
6701-30=700
:
:
7001-350=1,050
16
option
A
=
c- D=
c-
-
p
i:-c
D
G
-
E
30
60
=
130
=
110
=
:
80
80
✓
lrionoisniirioboiooloizoid
cgrd
crash G)
3
/
g
0
A
B
13
7
9
16
12
12
C
4
8
C-
16
→
→
4
o
o
o
4
4
4
0
4
12
-
4
12
,
o
lb G
16
16
18
18
2
7
0
\
0
D
4
10
10
F
16
10
6
16
→
6
4
critical path
A c
:
-
A
project
duration
14th
D
-
F
-
-
option
G
A)
G
c-
-
☒
=
F
I:-c
*
7801-300=1,080
:
=
c-
7001-80=780
:
project cost
E
days
18
=
total direct cost
total
-
-
10
-
E
30
60
=
130
=
110
=
✓
80
80
:
crash 1=807
>
/
0
A
0
4
4
B
13
6
9
15
u
4
4
C
C-
11
11
Is
→
→
4
4
7
-
4
11
11
15
IS
G
17
15
2
17
,
7
\
4
D
10
10
F
IS
10
5
IS
→
4
critical path
:
A c
-
A
project
duration
=
total direct cost
total
project cost
-
D
17
-
-
E
F
-
-
=
10
option
G
A)
G
c-
days
7801-110
:
6
c=
890
8901-250=1,140
-
=
☒
F
I:-c
☒
-
E
=
30
60
=
130
=
110
=
,
80
go
✓
ugh crash
£
& "
)
y
0
A
0
4
4
B
13
g
g
14
u
4
4
C
4
4
11
11
C-
14
→
→
7
11
-
11
3
14
14
G
16
14
2
16
,
7
\
4
D
10
10
F
14
10
4
14
→
4
project
duration
total direct cost
project cost
total
i.
19
days
The cost
:
10
days
8901-130=1,020
:
:
1.0201-200=1,220
lowest total
duration represents the
cost
project
16
=
6
#
$1,050 #