Rc Coleman Managerial Report

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RC Coleman Mnagerial report

Managerial Report
1.
Activity | Expected time| Variance|
A| 6| 0.44|
B| 9| 2.78|
C| 4| 0.44|
D| 12| 7.11|
E| 10| 1.00|
F| 6| 0.44|
G| 8| 7.11|
H| 6| 0.44|
I| 7| 2.78|
J| 4| 0.11|
K| 4| 0.44|
Total Expected time 76 weeks
Activity| ES| EF| LS| LF | Slack| Critical|
A| 0| 6| 3| 9| 3| No|
B| 0| 9| 0| 9| 0| Yes|
C| 9| 13| 9| 13| 0| Yes|
D| 13| 25| 17| 29| 4| No|
E| 13| 23| 13| 23| 0| Yes|
F| 23| 29| 23| 29| 0| Yes|
G| 13| 21| 21| 29| 8| No|
H| 29| 35| 29| 35| 0| Yes|
I| 29| 36| 32| 39| 3| No|
J| 35| 39| 35| 39| 0| Yes|
K| 39| 43| 39| 43| 0| Yes|

Expected time to finish the project is E(t)=B+C+E+F+H+J+K
=9+4+10+6+6+4+4=43 weeks
Standard deviation=the sum of the variances of the critical path =B + C + E + F + H + J + K
=√2.78 + 0.44 + 1.00 + 0.44 + 0.44 + 0.11 + 0.44
= √5.56 = 2.38

At Time =40 z= (40-43)/2.38 = -1.26 Cumulative probability= 0.1038 or ~ 10%

The critical activities are B, C, E, F, H, J and K. The project should be completed (earliest finish) in 43 weeks - therefore R.C. Coleman’s 40 week completion time cannot be achieved. The probability of R.C. Coleman meeting the 40 week deadline is ~10%. This is a low chance, so they should be cautioned if they make 40 weeks their deadline. 2.

80% of a probability of ~0.7995 has a z score of 0.84.
Thus 40-E(t)/√5.67=0.84 E(t)=38 weeks
If the project is shorted to 38 weeks, R.C. Coleman will be able to achieve the goal of 80% completion at 40 weeks. R.C Coleman should crash activities to reduce the expected project completion time to 38 weeks.

3.
Crashing decisions table;
Activity| Normal| Crash| Normal Cost | Crash Cost | Maximum Reduction Time| Crash Cost per Day| A| 6| 4| 1,000| 1,900| 2| 450|
B| 9| 7| 1,000| 1,800| 2|...
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