The following represents a project that should be scheduled using CPM: IMMEDIATE
ID: 376073 • Letter: T
Question
The following represents a project that should be scheduled using CPM: IMMEDIATE TIMES (DAYS) ACTIVITY PREDECESSORS a m b 2 1 1 3 4 11 1 3 11 2 3 11 F,G b. What is the critical path? O B-E-G-H OA-D-F-H OA-C-F-H O A-D-G-H c. What is the expected project completion time? (Round your answer to 3 decimal places.) Project completion time d. What is the probability of completing this project within 19 days? (Do not round intermediate calculations. Round your answer to 4 decimal places.) Probability daysExplanation / Answer
The expected duration as well as standard deviation of duration of each activity is presented below :
ACTIVITY
a
m
B
Expected duration
Standard deviation of duration
A
1
2
6
2.5
0.83
B
1
2
9
3
1.33
C
2
2
11
3.5
1.50
D
1
8
9
7
1.33
E
1
2
3
2
0.33
F
3
4
11
5
1.33
G
1
3
11
4
1.67
H
2
3
11
4.17
1.50
Following may be noted :
Expected duration of an activity = ( a + 4.m + b)/6
Standard deviation of activity = ( b – a) / 6
The Precedence diagram as follows :
A
B
C
D
E
F
G
H
The possible parallel paths and their corresponding expected durations as follows :
A-C-F-H = 2.5 + 3.5 + 5 + 4.17 = 15.17
A-D-F-H = 2.5 + 7 + 5 + 4.17 = 18.67
A-D-G-H = 2.5 +7 + 4 + 4.17 = 17.67
B-E-G-H = 3+ 2+ 4+ 4.17 = 13.17
Out of above , A-D-F-H has the longest duration and hence forms the critical path. Duration of the critical path is Project Completion time . Hence Project completion time will be 18.67 days
Variance of the duration of critical path
= Sum of Variances of activities A,D,F,H
= 0.83^2 + 1.33^2 + 1.33^2 + 1.5^2
= 0.6889 + 1.7689 + 1.7689 + 2.25
= 6.4767
Hence, standard deviation of the duration of critical path = square root ( 6.4767) = 2.544 days
Let Z value corresponding to probability that project will be complete within 19 days = Z1
Hence,
Expected project completion time ( i.e. expected duration of critical path) + Z1 x Standard deviation of duration of critical path = 19
Or, 18.67 + 2.544.Z1 = 19
Or, 2.544.Z1 = 19 – 18.67 = 0.33
Z1 = 0.33/2.544 = 0.1297 ( 0.13 rounded to 2 decimal places )
Probability corresponding to Z1 = 0.13 will be 0.55172
CRITICAL PATH = A-D-F-H
PROJECT COMPLETION TIME = 18.67 DAYS
PROBABILITY OF COMPLETING THE PROJECT WITHIN 19 DAYS = 0.55172
ACTIVITY
a
m
B
Expected duration
Standard deviation of duration
A
1
2
6
2.5
0.83
B
1
2
9
3
1.33
C
2
2
11
3.5
1.50
D
1
8
9
7
1.33
E
1
2
3
2
0.33
F
3
4
11
5
1.33
G
1
3
11
4
1.67
H
2
3
11
4.17
1.50