/* nag_mip_transportation (h03abc) Example Program.
*
* Copyright 2020 Numerical Algorithms Group.
*
* Mark 27.1, 2020.
*
*
*/
#include <nag.h>
#include <stdio.h>
#define COST(I, J) cost[(I)*tdcost + J]
int main(void) {
Integer *dest = 0, exit_status = 0, i, m, maxit, navail, nreq, numit;
Integer *source = 0;
Integer tdcost;
NagError fail;
double *avail = 0, *cost = 0, optcost, *optq = 0, *req = 0, *unitcost = 0;
INIT_FAIL(fail);
printf("nag_mip_transportation (h03abc) Example Program Results\n");
navail = 3;
nreq = 3;
m = navail + nreq;
if (!(cost = NAG_ALLOC(navail * nreq, double)) ||
!(avail = NAG_ALLOC(navail, double)) ||
!(req = NAG_ALLOC(nreq, double)) ||
!(optq = NAG_ALLOC(navail + nreq, double)) ||
!(unitcost = NAG_ALLOC(navail + nreq, double)) ||
!(source = NAG_ALLOC(navail + nreq, Integer)) ||
!(dest = NAG_ALLOC(navail + nreq, Integer))) {
printf("Allocation failure\n");
exit_status = -1;
goto END;
}
tdcost = nreq;
COST(0, 0) = 8.0;
COST(0, 1) = 8.0;
COST(0, 2) = 11.0;
COST(1, 0) = 5.0;
COST(1, 1) = 8.0;
COST(1, 2) = 14.0;
COST(2, 0) = 4.0;
COST(2, 1) = 3.0;
COST(2, 2) = 10.0;
avail[0] = 1.0;
avail[1] = 5.0;
avail[2] = 6.0;
req[0] = 4.0;
req[1] = 4.0;
req[2] = 4.0;
maxit = 200;
/* nag_mip_transportation (h03abc).
* Classical transportation algorithm
*/
nag_mip_transportation(cost, tdcost, avail, navail, req, nreq, maxit, &numit,
optq, source, dest, &optcost, unitcost, &fail);
if (fail.code != NE_NOERROR) {
printf("Error from nag_mip_transportation (h03abc).\n%s\n", fail.message);
exit_status = 1;
goto END;
}
printf("\nGoods From To Number Cost per Unit\n");
for (i = 0; i < m - 1; i++)
printf(" %" NAG_IFMT " %" NAG_IFMT " %8.3f %8.3f\n",
source[i], dest[i], optq[i], unitcost[i]);
printf("\nTotal Cost %8.4f\n", optcost);
END:
NAG_FREE(cost);
NAG_FREE(avail);
NAG_FREE(req);
NAG_FREE(optq);
NAG_FREE(unitcost);
NAG_FREE(source);
NAG_FREE(dest);
return exit_status;
}