Rc Coleman Managerial Report
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
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