Here is an Activity on Arrow (AOA) network diagram for a project. The lettered A through F represents an activity, as shown in the accompanying precedence diagram.
The probabilistic time for each activity is as follows
Activity |
Time |
Activity |
Time |
A |
4-5-6 |
D |
4-5-6 |
B |
2-5-6 |
E |
2-5-6 |
C |
3-4-5 |
F |
3-4-6 |
What is the expected time and standard deviation of path A-C-D?
What percentage will path A-C-D finish within 13 weeks?
Please type your answer and the steps of computation
To find the expected time, we will use the following formula.
T = (a + 4m + b) / 6
A. (4 + (4 * 5 ) + 6) / 6 = 5
B. (2 + (4 * 5 ) + 6) / 6 = 5
C. (3 + (4 * 4 ) + 5) / 6 = 4
D. (4 + (4 * 5 ) + 6) / 6 = 5
E. (2 + (4 * 5 ) + 6) / 6 = 5
F. (3 + (4 * 4 ) + 6) / 6 = 4
To find the variance, we will use the following formula.
Variance = [(b - a) / 6] ^2
A. [(6 -4) / 6 ] ^ 2 = 0.11
B. [(6 -2) / 6 ] ^ 2 = 0.44
C. [(5 -3) / 6 ] ^ 2 = 0.11
D. [(6 -4) / 6 ] ^ 2 = 0.11
E. [(6 -2) / 6 ] ^ 2 = 0.44
F. [(6 -3) / 6 ] ^ 2 = 0.25
ACTIVITY |
OPTIMISTIC |
MOST LIKELY |
PESSIMISTIC |
EXPECTED TIME |
VARIANCE |
a |
m |
b |
t = (a + 4 m + b) / 6 |
[(b - a) / 6] ^2 |
|
A |
4 |
5 |
6 |
5 |
0.11 |
B |
2 |
5 |
6 |
5 |
0.44 |
C |
3 |
4 |
5 |
4 |
0.11 |
D |
4 |
5 |
6 |
5 |
0.11 |
E |
2 |
5 |
6 |
5 |
0.44 |
F |
3 |
4 |
6 |
4 |
0.25 |
CPM
The ES, EF values are calculated using a forward pass where the ES of the next activity is the maximum EF of all the predecessor activities.
The LS, LF values are calculated using a backward pass where the LS of the next activity is the minimum LF of all the Predecessor activities.
The critical path represents all the activities in a particular chain which have the largest duration and 0 slack for all activities, and due to this fact, any delay in the critical activities can cause a delay for the entire project.
ACTIVITY |
DURATION |
PRE 1 |
PRE 2 |
ES |
EF |
LS |
LF |
SLACK |
A |
5 |
0 |
5 |
0 |
5 |
0 |
||
B |
5 |
0 |
5 |
4 |
9 |
4 |
||
C |
4 |
A |
5 |
9 |
5 |
9 |
0 |
|
D |
5 |
C |
9 |
14 |
9 |
14 |
0 |
|
E |
5 |
B |
5 |
10 |
9 |
14 |
4 |
|
F |
4 |
B |
5 |
9 |
10 |
14 |
5 |
CRITICAL PATH & DURATION:
A - C - D = 14
PROBABILITY
CRITICAL VARIANCE = A - C - D = 0.11 + 0.11 + 0.11 = 0.33
STANDARD DEVIATION = SQRT(CRITICAL VARIANCE) = SQRT(0.33) = 0.57
EXPECTED = 13 DAYS
Z = DUE - EXPECTED / STANDARD DEVIATION
Z = 14 - 13 / 0.57 = 1.754
THE Z VALUE OF 1.754 GIVES 0.9603 OR 96.03% PROBABILITY OF COMPLETION.
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