Bil Fredlund, president of Lincoln Log Construction, is considering placing a bi
ID: 340426 • Letter: B
Question
Bil Fredlund, president of Lincoln Log Construction, is considering placing a bid on a building project. Bill has determined that five tasks would need to be performed to carry out the proj ect. Using the PERT three-estimate approach, Bill has obtained the estimates in the next table for how long these tasks will take. Also shown are the precedence relationships for these tasks. management Time Required Optimistic Most Likely PessimisticImmediate Task Estimate Estimate Estimate Predecessors 3 weeks 2 weeks 3 weeks 1 week 2 weeks 4 weeks 2 weeks 5 weeks 3 weeks 3 weeks 5 weeks 2 weeks 6 weeks 5 weeks 5 weeks B, D There is a penalty of $500,000 if the project is not completed in 11 weeks. Therefore, Bill is very interested in how likely it is that his company could finish the project in time. a) Construct the project network for this project. (b) Find the estimate of the mean and variance of the duration of each activity (c) Find the mean critical path. (d) Find the approximate probability of completing the project within 11 weeks.Explanation / Answer
Answer to question a :
The Precedence diagram as follows :
A
B
D
C
E
Answer to question b:
The estimates of mean and variance of duration of each activity as follows :
Task
Optimistic (O)
Most likely ( m)
Pessimistic(P)
Mean
Variance
A
3
4
5
4.00
0.11
B
2
2
2
2.00
0.00
C
3
5
6
4.83
0.25
D
1
3
5
3.00
0.44
E
2
3
5
3.17
0.25
It is to be noted :
Mean value of a task = ( Optimistic duration + 4 x Most likely duration + Pessimistic Duration)/6
Variance of a task = ( Pessimistic time – Optimistic time ) ^2/36
Answer to question C:
The possible paths and their mean durations as follows :
A-B-C = 4 + 2 + 4.83 = 10.83
A-B-E = 4 + 2 + 3.17 = 9.17
A-D-E = 4 + 3 + 3.17 = 10.17
Since A-B-C has the longest duration, it forms the CRITICAL PATH
MEAN CRITICAL PATH = A-B-C
Answer to question d:
Variance of the critical path duration
= sum of variances of A, B and C
= 0.11 + 0 + 0.25
= 0.36
Standard deviation of critical path = Square root ( Variance of critical path ) = Square root ( 0.36) = 0.6 weeks
Let z value corresponding to probability that the project will complete within 11 weeks = Z1
Therefore ,
Mean duration of critical path + Z1 x standard deviation of duration of critical path = 11
Or. 10.83 + 0.6.Z1 = 11
Or, 0.6.Z1 = 0.17
Or, Z1 = 0.17/0.6
Or, Z1 = 0.28
Corresponding probability for Z = 0.28 as derived from standard normal distribution table 0.61026
PROBABILITY FOR COMPLETING THE PROJECT WITHIN 11 WEEKS = 0.61026
A
B
D
C
E
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