The estimated times and immediate predecessors for the activities in a project a
ID: 410154 • Letter: T
Question
The estimated times and immediate predecessors for the activities in a project at Howard Umrah's retinal scanning company are given in the following table. Assume that the activity times are independent.
Time
(weeks)
Immediate
Time
(weeks)
Immediate
Activity
a
m
b
Predecessor(s)
Activity
a
m
b
Predecessor(s)
A
7
10
11
-----------—
C
9
10
13
A
B
6
10
16
----------—
D
5
8
11
B
This exercise contains only parts a, b, c, d, and e.
a) Based on the activity time estimates, the expected times and variance for each of the activities are (round your response to two decimal places):
Activity
Expected Time
Variance
A
B
C
D
b) The expected completion time of the critical path =
weeks (round your response to two decimal places).
The expected completion time of the path other than the critical path =
weeks (round your response to two decimal places).
c) The variance of the critical path =
weeks
(round your response to two decimal places).
The variance of the path other than the critical path =
weeks
(round your response to two decimal places).
d) If the time to complete the activities on the critical path is normally distributed, then the probability that the critical path will be finished in
23
weeks or less =
(enter as a probability and round your response to two decimal places).
e) If the time to complete the activities on the non critical path is normally distributed, then the probability that the non critical path will be finished in
23
weeks or less =
(enter as a probability and round your response to two decimal places).
Enter your answer in each of the answer boxes.
Time
(weeks)
Immediate
Time
(weeks)
Immediate
Activity
a
m
b
Predecessor(s)
Activity
a
m
b
Predecessor(s)
A
7
10
11
-----------—
C
9
10
13
A
B
6
10
16
----------—
D
5
8
11
B
Explanation / Answer
a) Expected time = (Ta + 4*Tm + Tb)/6
Variance = (Tb-Ta)^2/36
b) Path & Duration:
AC = 9.67+10.33 = 20
BD = 10.33 + 8 = 18.33
So, Critical path is AC
c) Variance:
AC = 0.44+0.44 = 0.88
BD = 2.78+1 = 3.78
d) z value = (T-Expecetd Time)/Mean = (23-20)/0.88 = 3.4
Probability for Z value 3.4 is 0.9997
Times Activity Ta Tm Tb Expected Time Variance A 7 10 11 9.67 0.44 B 6 10 16 10.33 2.78 C 9 10 13 10.33 0.44 D 5 8 11 8.00 1.00Related Questions
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