The following represents a project that should be scheduled using CPM: IMMEDIATE
ID: 364854 • Letter: T
Question
The following represents a project that should be scheduled using CPM: IMMEDIATE PREDECESSORS TIMES (DAYS) ACTIVITY a m b A — 1 2 3 B — 1 4 7 C A 2 8 11 D A 1 5 9 E B 1 2 3 F C,D 1 4 7 G D,E 1 2 3 H F,G 2 3 3 b. What is the critical path? B-E-G-H A-D-F-H A-C-F-H A-D-G-H c. What is the expected project completion time? (Round your answer to 3 decimal places.) Project completion time days d. What is the probability of completing this project within 20 days? (Do not round intermediate calculations. Round your answer to 4 decimal places.) Probability
Explanation / Answer
Following table presents calculated values of Expected duration and standard deviations of durations for all activities :
Activity
TIME ( DAYS)
Expected duration
Standard deviation
a
m
b
e = ( a + 4.m + b)/6
Sd = ( b - a)/6
A
1
2
3
2.00
0.33
B
1
4
7
4.00
1.00
C
2
8
11
7.50
1.50
D
1
5
9
5.00
1.33
E
1
2
3
2.00
0.33
F
1
4
7
4.00
1.00
G
1
2
3
2.00
0.33
H
2
3
3
2.83
0.17
The predecessor diagram as follows :
A
B
C
D
E
F
G
H
The possible paths and their corresponding cumulative expected durations as follows :
A-C-F-H = 2 + 7.5 + 4 + 2.83 = 16.33 days
A-D-F-H = 2 + 5 + 4 + 2.83 = 13.83 days
A-D-G-H = 2 + 5 + 2 + 2.83 = 11.83 days
B-E-G-H = 4 + 2 + 2 + 2.83 = 10.83 days
Out of above, A-C-F-H has the longest duration and hence forms the critical path with expected project completion time = 16.33 days
EXPECTED PROJECT COMPLETION TIME = 16.33 DAYS
Variance of the critical path = Sum of Variances ( t.e Standard deviation squared ) of A , C , F and H
Therefore Variance of critical path
= 0.33^2 + 1.5^2 + 1^2 + 0.17^2
= 0.1089 + 2.25 + 1 + 0.0289
= 3.3878
Hence, Standard deviation of critical path = Square root ( 3.3878 ) = 1.840
Let Z value corresponding to probability of completing the project in 20 days = Z1
Therefore,
Expected project completion time + Z1 x Standard deviation of critical path = 20
Or, 16.33 + 1.840.Z1 = 20
Or, 1.840.Z1 = 20 – 16.33 = 3.67
Or, Z1 = 3.67/1.84 = 1.99 ( rounded to 2 decimal places )
Value of probability for Z = 1.99 as derived from standard normal distribution table =0.9767
PROBABILITY OF COMPLETING THE PROJECT IN 20 DAYS = 0.9767
Activity
TIME ( DAYS)
Expected duration
Standard deviation
a
m
b
e = ( a + 4.m + b)/6
Sd = ( b - a)/6
A
1
2
3
2.00
0.33
B
1
4
7
4.00
1.00
C
2
8
11
7.50
1.50
D
1
5
9
5.00
1.33
E
1
2
3
2.00
0.33
F
1
4
7
4.00
1.00
G
1
2
3
2.00
0.33
H
2
3
3
2.83
0.17
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.