COMM 225 Review Problems W2015

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 ? 2 = 0 + 2.25 + 0.69 + 0.44 = 3.38, Standard Deviation = ? = 1.838
For 35 weeks, ?35 =

T−Expected Duration

For 45 weeks, ?45 =

T−Expected Duration

σ σ =

35−40.66

=

45−40.66

=1.838
=1.838

, or ?35 = -3.08  Prob (z ≤ ?35 ) = 0.001
, or ?45 = 2.36  Prob (z ≤ ?45 ) = 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, ?35 =

T−Expected Duration

For 45 weeks, ?45 =

T−Expected Duration

σ σ =

35−39
=1.371

, or ?35 = -2.918  Prob (z ≤ ?35 ) = 0.0018

45−39

= =1.371 , or ?45 = 4.376  Prob (z ≤ ?45 ) = 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