Develop a Network Drawing for Hill Construction and Determine the Critical Path. How Long Is the Project Expected to Take?
Autor: byjujoy • April 18, 2016 • Exam • 709 Words (3 Pages) • 3,083 Views
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1. Develop a network drawing for Hill construction and determine the critical path. How long is the project expected to take?
Answer:
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Critical Path[pic 4]
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Critical Path[pic 6]
Critical Path:
A-C-D-G-H-I-L = 30+65+55+30+20+30+30= 260 DAYS
2. What is the probability of finishing in 270 days? What is the probability of finishing in more than 280 days?
Answer:
Project Variance | 319.4444444 |
Project Standard Deviation | √Variance |
√319.4444444 | |
17.87 Days |
Probability to finish in 270 days | |
Z=(Due Date -Expected Date)/Project standard Deviation | (270-260)/17.87 |
Z | 0.56 |
Checking the z=0.56 on the table we can observe that the probability (T<=270) is 71% |
Probability to finish in more than 280 days | |
Z=(Due Date -Expected Date)/Project standard Deviation | (280-260)/17.87 |
Z | 1.12 |
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Checking the z=1.12 on the table we can observe that the probability (T>280) is (1-0.8686) i.e 13.14% |
3. If it is necessary to crash to 240 days, how would Hill do so, and at what costs? As noted in the case, assume that optimistic time estimates can be used as crash time.
Answer:
Critical Path | Crash Costs Per Day on Critical Path | Crash Times (Optimistic) on Critical Path |
| ($) |
|
A | 1500 | 20 |
C | 4000 | 50 |
D | 1900 | 30 |
G | 2500 | 25 |
H | 2000 | 10 |
I | 2000 | 20 |
L | 4500 | 20 |
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Critical Path[pic 8]
Crashing down 20 days will be effective if Hill crashes 10 days in A. Hill cannot crash down more as crash time cannot go below the optimistic value.
Crash down of A by 10days = 10 * 1500 =$15000
For crashing 10 more, we need to crash in the next least Costs Per Day activity i.e D. We will crash 10 days from D.
Crash down of D by 10days = 10 * 1900 =$19000
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