[#1] First of all, Task C is not on the critical path and does not really need to be crashed. If you rank the crashable tasks by cost in ascending order, then you come up with this sequence: A,B,D. Even if you only look at the incremental crashing cost (the difference between the crashing and normal cost), you would still come up with the same ranking. So then what is the argument for D,B,A?[#2] You are correct on your first statement. C is not in the critical path so it does not need to be taken into account. Answer B (A,D,B) because:
A: 4-2= 2 weeks crash --> 1K$ Increase--> 0.5K per week
B: 3-2= 1 week crash ---> 2K$ Increase--> 2K per week
D: 5-3= 2 week crash ---> 2K$ Increase--> 1K per week So first you choose A, because it provides you with the best crashing time at the lower cost. Then you have to choose between D & B. If we have to choose sequentially and cost is important: chose D that provides you with the most crash time at the lower cost.[#3] In a specific path crashing the activities has nothing to do with the sequence of operations; however the cost/time slope is the factor to consider - i.e. how much it costs you to save one day/week; the least the better. The analysis provided by Vanessa is absolutely correct. Think about it this way: if A, B & D are after each other, and it will cost you $1 to save a week if you crash D and $2 to save one week if you crash B which one to crash first? Obviously D sense it will give you the best value of money

Section: Monitoring and Controlling