The formulation of the transportation problem is AMPL is a straightforward translation of the mathematical programme for the transportation problem. We will build the file `transportation.mod`.

The sets and are declared as `SUPPLY_NODES`

and `DEMAND_NODES`

respectively:

set SUPPLY_NODES; set DEMAND_NODES;

The supply and demand are declared as **integer** parameters:

param Supply {SUPPLY_NODES} >= 0, integer; param Demand {DEMAND_NODES} >= 0, integer;

The cost is declared over the `SUPPLY_NODES`

and `DEMAND_NODES`

:

param Cost {SUPPLY_NODES, DEMAND_NODES};

Now, the mathematical programme follows directly:

var Flow {SUPPLY_NODES, DEMAND_NODES} >= 0, integer; minimize TotalCost: sum {i in SUPPLY_NODES, j in DEMAND_NODES} Cost[i, j] * Flow[i, j]; subject to UseSupply {i in SUPPLY_NODES}: sum {j in DEMAND_NODES} Flow[i, j] = Supply[i]; subject to MeetDemand {j in DEMAND_NODES}: sum {i in SUPPLY_NODES} Flow[i, j] = Demand[j];Note that we assume the transportation is balanced.

In the main discussion of transportation problems, we saw that adding bounds to the flow variables allowed us to easily either bound the transportation of good from a supply node to a demand node or remove an arc from the problem altogether.

We can add bounds to our AMPL formulation by declaring 2 new parameters with defaults:

param Lower {SUPPLY_NODES, DEMAND_NODES} integer default 0; param Upper {SUPPLY_NODES, DEMAND_NODES} integer default Infinity;and adding them to the

`Flow`

variable declaration:
var Flow {i in SUPPLY_NODES, j in DEMAND_NODES} >= Lower[i, j], <= Upper[i, j], integer;

Also, since some arcs no longer exist you should set a default of 0 for `Cost`

(thus you don't have to define costs for non-existent arcs).

param Cost {SUPPLY_NODES, DEMAND_NODES} default 0;

Balanced transportation models are preferred as there is no confusion about the relational operators for the supply and demand constraints. We can use a script file (`transportation.run`) to balance any transportation problem automatically:

reset; model transportation.mod; param costFromDummy {DEMAND_NODES} default 0; param costToDummy {SUPPLY_NODES} default 0; param difference; # Add the problem date file here # e.g., data brewery.dat; let difference := (sum {s in SUPPLY_NODES} Supply[s]) - (sum {d in DEMAND_NODES} Demand[d]); if difference > 0 then { let DEMAND_NODES := DEMAND_NODES union {'Dummy'}; let Demand['Dummy'] := difference; let {s in SUPPLY_NODES} Cost[s, 'Dummy'] := costToDummy[s]; } else if difference < 0 then { let SUPPLY_NODES := SUPPLY_NODES union {'Dummy'}; let Supply['Dummy'] := - difference; let {d in DEMAND_NODES} Cost['Dummy', d] := costFromDummy[d]; }; # else the problem is balanced # Make sure the problem is balanced check : sum {s in SUPPLY_NODES} Supply[s] = sum {d in DEMAND_NODES} Demand[d]; option solver cplex; solve; display Flow;Note the

`costToDummy`

and `costFromDummy`

allow for the definition of costs on any flow from/to a dummy node in the data file.
-- MichaelOSullivan - 02 Apr 2008

I | Attachment | History | Action | Size | Date | Who | Comment |
---|---|---|---|---|---|---|---|

mod | transportation.mod | r2 r1 | manage | 1.2 K | 2008-04-02 - 12:58 | MichaelOSullivan | |

run | transportation.run | r1 | manage | 0.9 K | 2008-04-02 - 10:42 | MichaelOSullivan |

Topic revision: r6 - 2008-04-02 - MichaelOSullivan

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