to 20n of 5. Five cargo ships will be used for shipping goods fro one pot to fiv
ID: 414834 • Letter: T
Question
to 20n of 5. Five cargo ships will be used for shipping goods fro one pot to five ports (labeled 1,2,3,4,5). Any ship can be used for making one port However, because of differences in the ships and transporting, and unloading the goods for the different varies considerably, as shown in the following table. (20 each one) ive trips cargos, the total cost of loading scores, 10 scores or Port $7 $7 $8 $9 $9 $7 $6 $15 S14 S0 $4 $7 $6 $6 $6 $12 $14 $12 S8 S9 $8 $10 $10 Ship in such a way as minimize the total cost for all five shipments. (a) Describe how this problem fits into the general format for the assign problem (b) Give out the optimal shipments for the problem.Explanation / Answer
Ports 1 2 3 4 5 1 4 8 7 15 12 2 7 9 7 14 10 Ship 3 6 9 12 8 9 4 6 7 14 8 10 5 6 6 12 10 9 Assignment matrix Ports 1 2 3 4 5 1 1 0 0 0 0 2 0 0 1 0 0 Ship 3 0 0 0 0 1 4 0 0 0 1 0 5 0 1 0 0 0 Each box in the assignment matrix is a decision variable These can take values 0 or 1, 0 meaning that ship is not matched to that port 1 meaning that ship is matched to that port So this is clear assignment problem Objective function minimize cost 34 Constraints One will be all the decision variables will be binary One ship should be matched with only one port 1 1 = 1 Sum of entire row of assignment matrix for a ship 2 1 = 1 Sum of entire row of assignment matrix for a ship Ship 3 1 = 1 Sum of entire row of assignment matrix for a ship 4 1 = 1 Sum of entire row of assignment matrix for a ship 5 1 = 1 Sum of entire row of assignment matrix for a ship One port can have only one ship 1 1 = 1 Sum of entire column of assignment matrix for a port 2 1 = 1 Sum of entire column of assignment matrix for a port port 3 1 = 1 Sum of entire column of assignment matrix for a port 4 1 = 1 Sum of entire column of assignment matrix for a port 5 1 = 1 Sum of entire column of assignment matrix for a port On solving using solver The optimal assignment is Ship port 1 1 2 3 3 5 4 4 5 2 And the optimal cost is 34
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