COMM 225 Review Problems W2015

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COMM 225: MIDTERM REVIEW QUESTIONS
TOPIC: PROJECT MANAGEMENT
Q 1.1: Kozar International, Inc. begun marketing a new instant-developing film project. The estimates of R&D activity time (weeks) for Kozar’s project are given in the table below. The project has two paths: AC-E-F and A-B-D-F. Assume the activity times are independent. a) What is the probability that the project will be completed between 35 and 45 days? b) If the time to complete the path A-B-D-F is normally distributed, what is the probability that this path will take at least 38 weeks to be completed?

Activity

Predecessors

A
B
C
D
E
F

A
A
B
C
D, E

Optimistic
time
9
8
9
5
5
10

Time(weeks)
Probable Pessimistic
time
time
9
9
10
12
12
18
8
11
7
10
12
14

Mean

Variance

9
10
12.5
8
7.166
12

0.00
0.44
2.25
1.00
0.69
0.44

Solution:
(a) What is the probability that the project will be completed between 35 and 45 days? The project has two paths:
A-C-E-F:
 Expected Duration = 9 + 12.5 + 7.166 + 12 = 40.66 weeks (Critical Path).  Variance 𝜎 2 = 0 + 2.25 + 0.69 + 0.44 = 3.38, Standard Deviation = 𝜎 = 1.838 For 35 weeks, 𝑧35 =

T−Expected Duration

For 45 weeks, 𝑧45 =

T−Expected Duration

σ
σ

=

35−40.66

=

45−40.66

=1.838
=1.838

, or 𝑧35 = -3.08  Prob (z ≤ 𝑧35 ) = 0.001
, or 𝑧45 = 2.36  Prob (z ≤ 𝑧45 ) = 0.9909

Probability that this path will be completed between 35 and 45 days is 0.9909-0.001 = 0.9899 A-B-D-F:
 Expected Duration = 9 + 10 + 8 + 12 = 39 weeks.
 Variance = 0 + 0.44 + 1.00 + 0.44 = 1.88, Standard Deviation = 1.371 For 35 weeks, 𝑧35 =

T−Expected Duration

For 45 weeks, 𝑧45 =

T−Expected Duration

σ
σ

=

35−39
=1.371

, or 𝑧35 = -2.918  Prob (z ≤ 𝑧35 ) = 0.0018

45−39

= =1.371 , or 𝑧45 = 4.376  Prob (z ≤ 𝑧45 ) = 1

Probability that this path will be completed between 35 and 45 days is 1 – 0.0018 = 0.9982 Hence, the probability of project completion between 35 and 45 days = 0.9899*0.9982= 0.9881 = 98.81%
(b) If the time to complete the path A-B-D-F is normally distributed, what is the probability that this path will take at least 38 weeks to be completed?



The non-critical path A-B-D-F has an expected duration of 39 weeks and standard deviation of 1.371.



This implies 𝑍38 =





This z value corresponds to a probability of 0.2327.
Probability that this path will take less than 38 weeks to be completed is 23.27%. Hence, the probability that this path will take at least 38 weeks to be completed is 76.73%.

T−Expected Duration
σ

=

38−39
1.371

=-0.73.

Q 1.2: Given the following network and time & cost estimates, answer the following questions: (a) What is the project completion time?
(b) What is the total cost required for completing this project on normal time? (c) Crash the network the maximum amount possible and compute the total crash cost.

D

B
A

G

E
C

F
I

Activity

Prede
cessor

H

A
A
B
B,C
C
D,E
G,F
C
H,I

A
B
C
J D
E
F
G
H
I
J
TOTAL =

Activity
duration
(weeks)
Normal Crash
6
4
28
22
29
27
10
5
10
9
10
9
15
14
10
8
2
1
10
8

Activity cost ($)
Normal
10000
5000
20000
4000
2500
1000
1500
600
1000
900
46,500

Crash
16000
9200
20700
6000
3000
7000
7500
10600
2000
8800

Crash
Cost /
Week

#
Days

3000
700
350
400
500
6000
6000
5000
1000
3950

2
6
2
5
1
1
1
2
1
2

Solution:
Parts (a & b)
The project has the following paths and their durations:
 A-B-D-G-H-J, Duration = 6 + 28 + 10 + 15 + 10 + 10 = 79
 A-B-E-G-H-J, Duration = 6 + 28 + 10 + 15 + 10 + 10 = 79
 A-C-E-G-H-J, Duration = 6 + 29 + 10 + 15 + 10 + 10 = 80 (critical path)  A-C-F-H-J, Duration = 6 + 29 + 10 + 10 + 10 = 65
 A-C-I-J, Duration = 6 + 29 + 2 + 10 = 47
The project completion time = length of critical path = 80 weeks Normal total cost = sum of the normal cost for all the...
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