Question: Assume the network and data that follow: ACTIVITY NORMAL TIME (WEEKS) NORMAL COST CRASH TIME (WEEKS) CRASH COST IMMEDIATE PREDECESSORS A 5 $ 70 4

Assume the network and data that follow:

ACTIVITY NORMAL TIME (WEEKS) NORMAL COST CRASH TIME (WEEKS) CRASH COST IMMEDIATE PREDECESSORS
A 5 $ 70 4 $ 80
B 4 50 2 90 A
C 4 90 1 120 A
D 5 40 3 70 A
E 6 120 4 180 B
F 8 60 7 120 D
G 5 140 4 160 C, E, F

b. Indicate the critical path when normal activity times are used.
A-D-F-G
A-D-E-G
A-C-G
A-B-E-G

c.

Demonstrate the crashing steps needed to minimize the cost of crashing down to various project lengths. Consider project lengths of 23, 22, 21, 20, 19, and 18 weeks. For each possible step down in project length, specify which activity would be crashed to achieve that length, compute the resultant length of each path in the project, and compute the crashing costs as shown in the table below.(Leave no cells blank - be certain to enter "0" wherever required.)

Project Duration: 23 Weeks 22 Weeks 21 Weeks 20 Weeks 19 Weeks 18 Weeks
Length of path ABEG
Length of path ACG
Length of path ADFG
Activity crashed (Click to select)AGB & FFNoneD (Click to select)GFDNoneAB & F (Click to select)FGAB & FNoneD (Click to select)FB & FADNoneD (Click to select)DNoneGAGB & F (Click to select)FGDNoneAF
Crash cost for this step
Cumulative crash cost
Total direct cost

d-1.

If the indirect costs for each project duration are $400 (23 weeks), $350 (22 weeks), $300 (21 weeks), $250 (20 weeks), $200 (19 weeks), and $150 (18 weeks), what is the total project cost for each duration?

Duration Total Costs
23 $
22
21
20
19
18

d-2. Indicate the minimum total project cost duration.
19 weeks
13 weeks
14 or 15 weeks
17 weeks

 Assume the network and data that follow: ACTIVITY NORMAL TIME (WEEKS)

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