Also solve for the minimum total cost
associated with finishing the project in 13 weeks and the total
project's duration (and of each activity according to the minimum
cost schedule)
Homework Answers
slope of the activites
slope = crash cost - normal cost / normal time - crash time
| activity | slope | Max crash days |
| A | 3500 | 2 |
| B | N | 0 |
| C | 14000 | 2 |
| D | 10000 | 1 |
| E | N | 0 |
| F | 4000 | 2 |
| G | 7000 | 1 |
The critical path is ACEF with durAtion 13.
cost of this schedule = 56000+10000x13 + 1x8000 = 194000
crash schedule
(i) crashing the activity A by one day would bring the duration to 12. cost is 56000+3500+120000 =179500. Now there are 2 critical paths which are ACEF and BDF
(ii) Crashing F for one day will cost 4000 and bring the duration to 11. cost is 56000+7500+110000 =173500
(iii) Crashing F by one more day at 4000, will bring the duration to 10. Cost is 56000+11500+100000 = 167500
(iv) Crashing of A and D by one day each ( cost 13500) would reduce the duration to 9 days. Cost if 56000+25000+90000 =1714000
No more crashing is possible. Least cost is obtained with duration 10 with total cost of 167500
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