Task Name Normal time Normal Crash time Crash cost (slope) Predecessor cost 4. 85 3. 10 5. 100 4. 30 30 4. 25 D AB 4. 90 3. 25 B.C 2 05 21 F. D. 4 100 4 G. D.E 90 20 H. E 80 35 F.G.H 85 Based on the above, the first activity that is selected to reduce the project duration is The project duration at the CRASH POINT (the minimum possible duration for the project) is days. The indirect cost with normal times is $300 and there is a savings of $25 per period reduced. The crash cost (cost at the CRASH POINT) is S The optimum cost-time schedule (the schedule with minimum total cost and corresponding time) is periods and $ Note: Enter the values for each item within square brackets.

Practical Management Science
6th Edition
ISBN:9781337406659
Author:WINSTON, Wayne L.
Publisher:WINSTON, Wayne L.
Chapter2: Introduction To Spreadsheet Modeling
Section: Chapter Questions
Problem 20P: Julie James is opening a lemonade stand. She believes the fixed cost per week of running the stand...
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Task
Name
Crash
cost
(slope)
Normal
Normal
Crash
time
Predecessor
time
cost
4.
35
3.
10
B.
100
4.
30
30
4
25
D
AB
90
3
25
В.С
2.
95
F
D
4
100
4
G
D.E
2
90
20
4
B0
35
F.G.H
85
Based on the above, the first activity that is selected to reduce the project duration is
The
project duration at the CRASH POINT (the minimum possible duration for the project) is
days. The indirect cost with normal times is $300 and there is a savings of $25 per period reduced The crash cost (cost at
the CRASH POINT) is S
The optimum cost-time schedule (the schedule with minumum
total cost and corresponding time) is
periods and $
Note: Enter the values for each item within square brackets.
Transcribed Image Text:Task Name Crash cost (slope) Normal Normal Crash time Predecessor time cost 4. 35 3. 10 B. 100 4. 30 30 4 25 D AB 90 3 25 В.С 2. 95 F D 4 100 4 G D.E 2 90 20 4 B0 35 F.G.H 85 Based on the above, the first activity that is selected to reduce the project duration is The project duration at the CRASH POINT (the minimum possible duration for the project) is days. The indirect cost with normal times is $300 and there is a savings of $25 per period reduced The crash cost (cost at the CRASH POINT) is S The optimum cost-time schedule (the schedule with minumum total cost and corresponding time) is periods and $ Note: Enter the values for each item within square brackets.
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