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# Quantitative Math for Cost/ Day Activities

3. Determine the cost/day for each activity and complete the time-Cost trade-off function for the various possible plans.

Immediate TIME COST
Activity Predecessor Normal Crash Normal Crash Cost/Day

A --- 7 6 \$400 \$460

B A 5 3 500 600

C A 4 2 600 780

D B 3 1 200 320

E C 6 3 500 710

F D,E 6 5 500 630

Plan Time Cost

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#### Solution Preview

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3. Determine the cost/day for each activity and complete the time-Cost trade-off function for the various possible plans.

Immediate TIME COST
Activity Predecessor Normal Crash Normal Crash Crash Cost/Day

A --- 7 6 \$400 \$460 \$60

B A 5 3 500 600 \$50

C A 4 2 600 780 \$90

D B 3 1 200 320 \$60

E C 6 3 500 710 \$70

F D,E 6 5 500 630 \$130

First draw the network diagram:

The critical path is ACEF and the normal project completion time is 23 days.
The other path is ABDF with a time length of 21 days.
Using normal time, the slack analysis for the project is as below

Activity On Critical Activity Earliest Earliest Latest Latest Slack
Name Path Time Start Finish Start Finish (LS-ES)
1 A Yes 7 0 7 0 7 0
2 B no 5 7 12 9 14 2
3 C Yes 4 7 11 7 11 0
4 D no 3 12 15 14 17 2
5 E Yes 6 11 17 11 17 0
6 F Yes 6 17 23 17 23 0

Project Completion Time = 23 Days
Total Cost of Project = \$2,700

Since the next longest path (after the Critical path) is 21 days long, we can crash upto 2 days without changing the critical path. On the critical path, the minimum cost of crashing per day is for activity at at\$60 per day. So we should crash the activity ...

#### Solution Summary

The solution determines the cost/day for each activity. The Time-Cost trade-off function for the various possibilities is determined.

\$2.19