1.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
     1.2 +++ b/examples/csv/transp_csv.mod	Mon Dec 06 13:09:21 2010 +0100
     1.3 @@ -0,0 +1,70 @@
     1.4 +# A TRANSPORTATION PROBLEM
     1.5 +#
     1.6 +# This problem finds a least cost shipping schedule that meets
     1.7 +# requirements at markets and supplies at factories.
     1.8 +#
     1.9 +#  References:
    1.10 +#              Dantzig G B, "Linear Programming and Extensions."
    1.11 +#              Princeton University Press, Princeton, New Jersey, 1963,
    1.12 +#              Chapter 3-3.
    1.13 +
    1.14 +set I;
    1.15 +/* canning plants */
    1.16 +
    1.17 +set J;
    1.18 +/* markets */
    1.19 +
    1.20 +set K dimen 2;
    1.21 +/* transportation lane */
    1.22 +
    1.23 +set L;
    1.24 +/* parameters */
    1.25 +
    1.26 +param a{i in I};
    1.27 +/* capacity of plant i in cases */
    1.28 +
    1.29 +param b{j in J};
    1.30 +/* demand at market j in cases */
    1.31 +
    1.32 +param d{i in I, j in J};
    1.33 +/* distance in thousands of miles */
    1.34 +
    1.35 +param e{l in L};
    1.36 +/* parameters */
    1.37 +
    1.38 +param f;
    1.39 +/* freight in dollars per case per thousand miles */
    1.40 +
    1.41 +table tab_plant IN "CSV" "plants.csv" :
    1.42 +  I <- [plant], a ~ capacity;
    1.43 +
    1.44 +table tab_market IN "CSV" "markets.csv" :
    1.45 +  J <- [market], b ~ demand;
    1.46 +
    1.47 +table tab_distance IN "CSV" "distances.csv" :
    1.48 +  K <- [plant, market], d ~ distance;
    1.49 +
    1.50 +table tab_parameter IN "CSV" "parameters.csv" :
    1.51 +  L <- [parameter], e ~ value ;
    1.52 +
    1.53 +param c{i in I, j in J} := e['transport cost'] * d[i,j] / 1000;
    1.54 +/* transport cost in thousands of dollars per case */
    1.55 +
    1.56 +var x{(i,j) in K} >= 0;
    1.57 +/* shipment quantities in cases */
    1.58 +
    1.59 +minimize cost: sum{(i,j) in K} c[i,j] * x[i,j];
    1.60 +/* total transportation costs in thousands of dollars */
    1.61 +
    1.62 +s.t. supply{i in I}: sum{(i,j) in K} x[i,j] <= a[i];
    1.63 +/* observe supply limit at plant i */
    1.64 +
    1.65 +s.t. demand{j in J}: sum{(i,j) in K} x[i,j] >= b[j];
    1.66 +/* satisfy demand at market j */
    1.67 +
    1.68 +solve;
    1.69 +
    1.70 +table tab_result{(i,j) in K} OUT "CSV" "result.csv" :
    1.71 +  i ~ plant, j ~ market, x[i,j] ~ shipment;
    1.72 +
    1.73 +end;