projectCrash

projectCrash - Project scheduling and crash problem This...

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Unformatted text preview: Project scheduling and crash problem This workbook contains several worksheets dealing with the scheduling and crash problems on page 97 of course pack. Here are descriptions of each worksheet: worksheet name 1 node-arc min finish time 2 node-arc min start time 3 node-arc path 4 node-arc finish time crash 5 node-arc start time crash description node-arc representation of minimum completion time of project using finish times node-arc representation of minimum completion time of project using start times node-arc representation of critical path problem (mandatory reading) node-arc representation of crash problem using finish times (mandatory reading) node-arc representation of crash problem using start times (mandatory reading) project using finish times (mandatory reading) project using start times (mandatory reading) tory reading) es (mandatory reading) es (mandatory reading) Find the earliest finish time of each activity s A B C D E sa -1 1 sb -1 1 AC -1 1 AD -1 1 BC -1 1 BD -1 1 CF -1 DE -1 EF EG Gf Ff Duration finish time start time 0 0 0 6 6 0 9 9 0 8 26 18 7 16 9 F Gf time 6 <= 9 <= 8 <= 7 <= 8 <= 7 <= 12 <= 10 <= 12 <= 6 <= 0 <= 0 <= 1 1 -1 -1 1 1 -1 -1 10 12 6 1 1 0 38 earliest finish time 38 16 26 38 26 38 38 finish time diff 6 9 20 10 17 7 12 10 12 12 0 0 Farid Alizadeh: The formula here attempts to add the amount coming into the node from this arc. This will ensure that finish times are taken into account. The If part does this job. Please note that this If expression must return an array of values not just a single value. For this reason when you enter this formula you should enter it by typing Ctrl-Shift-Enter. Just pushing Enter is incorrect. Farid Alizadeh: This column contains the difference between start time of the head and the tail of the arc. This quantity should not be less than time limit Find the earliest start time of each activity s A B C D E sa -1 1 sb -1 1 AC -1 1 AD -1 1 BC -1 1 BD -1 1 CF -1 DE -1 EF EG Gf Ff F G f time 0 <= 0 <= 6 <= 6 <= 9 <= 9 <= 8 <= 7 <= 10 <= 10 <= 6 <= 12 <= 1 1 -1 -1 1 1 -1 -1 1 1 start time diff 0 0 18 9 18 9 8 7 10 16 6 12 Farid Alizadeh Note the differ expression here only the amoun will ensure tha included only. Otherwise the m previous formu Duration start time 0 0 6 0 9 0 8 18 7 9 10 12 16 26 6 32 0 38 earliest finish time 38 start time diff Farid Alizadeh: Note the difference in the If expression here. Now we are adding only the amount leaving the tail. This will ensure that start times are included only. Otherwise the model is identical to the previous formulation Find the critical path s A sa -1 sb -1 AC AD BC BD CF DE EF EG Gf Ff balance balance LHS -1 -1 B 1 1 -1 -1 -1 -1 C 1 D E F G f in path 0 1 0 0 0 1 0 1 0 1 1 0 time 0 0 6 6 9 9 8 7 6 10 12 10 1 1 1 -1 -1 1 -1 -1 1 1 1 -1 -1 0 0 path length 0 0 1 1 1 1 38 0 0 0 0 0 0 0 0 0 0 Farid Alizadeh: To find the critical path we need to find the longest path from start to finish node in the network. This is quite similar to the transshipment problem with the start point treated as a supply point with supply of 1, and the end node as a demand node with demand of 1. All other nodes act is transshipment nodes. This along with the condition that variables on arcs are 0 or 1 ensures that the set of arcs chosen is in fact a path. However, it is not necessary to explicitly add the 0-1 constraint into the network, the solution of the linear program in this case is automatically 0-1. Also note that unlike usual transshipment problems we are maximizing length rather than minimizing it. lizadeh: the critical path we need to find the path from start to finish node in the k. This is quite similar to the ipment problem with the start point as a supply point with supply of 1, and node as a demand node with demand of ther nodes act is transshipment nodes. ong with the condition that variables on 0 or 1 ensures that the set of arcs chosen ct a path. However, it is not necessary to tly add the 0-1 constraint into the k, the solution of the linear program in e is automatically 0-1. te that unlike usual transshipment s we are maximizing length rather than zing it. Find the least cost schedule that finishes within 25 days s A B C D E F G sa -1 1 sb -1 1 AC -1 1 AD -1 1 BC -1 1 BD -1 1 CF -1 1 DE -1 1 EF -1 1 EG -1 Gf Ff -1 Duration finish time cost of improve improve 0 0 6 4 9 4 8 13 7 6 10 13 12 25 f 1 -1 required time 4 <= 4 <= 8 <= 2 <= 8 <= 2 <= 12 <= 7 <= 12 <= 6 <= 1 0 <= 1 0 <= 0 25 F T ex im du im in fi 25 6 25 <= $15 25 $10 $20 0 <= 5 improved duration 0 4 2 <= 5 4 5 <= 5 8 $3 $30 0 <= 5 2 5 <= 5 7 $40 $50 3 <= 5 12 0 <= 5 6 0 cost: 0 390 time used by arc 4 4 9 2 9 2 12 7 12 12 0 0 Farid Alizadeh: This model is similar to worksheet 1) except that required time is calculated from improved duration rather that the original duration. The objective is now the cost of improvement and a new constraint indicating the project finish time (which is finish time of node f) should not exceed 25. Find the least cost schedule that finishes within 25 days s A B C D E F G sa -1 1 sb -1 1 AC -1 1 AD -1 1 BC -1 1 BD -1 1 CF -1 1 DE -1 1 EF -1 1 EG -1 Gf Ff -1 Duration finish time cost of improve improve max improv allowed improved duration 0 0 0 6 0 9 0 8 5 7 4 10 6 12 13 f 1 -1 required time 0 <= 0 <= 4 <= 4 <= 4 <= 4 <= 8 <= 2 <= 7 <= 7 <= 1 6 <= 1 12 <= 0 25 6 19 <= $15 25 $10 $20 0 <= 5 4 2 <= 5 4 5 <= 5 8 $3 $30 0 <= 5 2 5 <= 5 7 $40 $50 3 <= 5 12 0 <= 5 6 0 cost: 0 390 time used by arc 0 0 5 4 5 4 8 2 7 13 6 12 Farid Alizadeh: This model is similar to worksheet 1) except that required time is calculated from improved duration rather than the original duration. The objective is now the cost of improvement and a new constraint indicating the project finish time (which is finish time of node f) should not exceed 25. ...
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