CPLEX呼び出し可能ライブラリ（VS2010）を使用してlpを解決しました。lp は次のとおりです。
Maximize
obj: x1 + 2 x2 + 3 x3
Subject To
c1: - x1 + x2 + x3 <= 20
c2: x1 - 3 x2 + x3 <= 30
Bounds
0 <= x1 <= 40
End
コードを以下に示します。今、私はそれを MIP (x の追加の完全性制約) にしたいと思います。に変えてみstatus = CPXlpopt (env, lp);ましstatus = CPXmipopt (env, lp);た。これは機能せず、エラーが発生します3003: not a mixed-integer problem。ここで何が欠けているか知っている人はいますか？
int main ()
{
/* Declare and allocate space for the variables and arrays where we
will store the optimization results including the status, objective
value, variable values, dual values, row slacks and variable
reduced costs. */
int solstat;
double objval;
double *x = NULL;
double *pi = NULL;
double *slack = NULL;
double *dj = NULL;
CPXENVptr env = NULL;
CPXLPptr lp = NULL;
int status = 0;
int i, j;
int cur_numrows, cur_numcols;
/* Initialize the CPLEX environment */
env = CPXopenCPLEX (&status);
/* Turn on output to the screen */
status = CPXsetintparam (env, CPX_PARAM_SCRIND, CPX_ON);
/* Turn on data checking */
status = CPXsetintparam (env, CPX_PARAM_DATACHECK, CPX_ON);
/* Create the problem. */
lp = CPXcreateprob (env, &status, "lpex1");
/* Now populate the problem with the data. */
#define NUMROWS 2
#define NUMCOLS 3
#define NUMNZ 6
/* To populate by column, we first create the rows, and then add the columns. */
int status = 0;
double obj[NUMCOLS];
double lb[NUMCOLS];
double ub[NUMCOLS];
char *colname[NUMCOLS];
int matbeg[NUMCOLS];
int matind[NUMNZ];
double matval[NUMNZ];
double rhs[NUMROWS];
char sense[NUMROWS];
char *rowname[NUMROWS];
CPXchgobjsen (env, lp, CPX_MAX); /* Problem is maximization */
/* Now create the new rows. First, populate the arrays. */
rowname[0] = "c1";
sense[0] = 'L';
rhs[0] = 20.0;
rowname[1] = "c2";
sense[1] = 'L';
rhs[1] = 30.0;
status = CPXnewrows (env, lp, NUMROWS, rhs, sense, NULL, rowname);
if ( status ) goto TERMINATE;
/* Now add the new columns. First, populate the arrays. */
obj[0] = 1.0; obj[1] = 2.0; obj[2] = 3.0;
matbeg[0] = 0; matbeg[1] = 2; matbeg[2] = 4;
matind[0] = 0; matind[2] = 0; matind[4] = 0;
matval[0] = -1.0; matval[2] = 1.0; matval[4] = 1.0;
matind[1] = 1; matind[3] = 1; matind[5] = 1;
matval[1] = 1.0; matval[3] = -3.0; matval[5] = 1.0;
lb[0] = 0.0; lb[1] = 0.0; lb[2] = 0.0;
ub[0] = 40.0; ub[1] = CPX_INFBOUND; ub[2] = CPX_INFBOUND;
colname[0] = "x1"; colname[1] = "x2"; colname[2] = "x3";
status = CPXaddcols (env, lp, NUMCOLS, NUMNZ, obj, matbeg, matind, matval, lb, ub, colname);
/* Optimize the problem and obtain solution. */
status = CPXlpopt (env, lp);
cur_numrows = CPXgetnumrows (env, lp);
cur_numcols = CPXgetnumcols (env, lp);
x = (double *) malloc (cur_numcols * sizeof(double));
slack = (double *) malloc (cur_numrows * sizeof(double));
dj = (double *) malloc (cur_numcols * sizeof(double));
pi = (double *) malloc (cur_numrows * sizeof(double));
status = CPXsolution (env, lp, &solstat, &objval, x, pi, slack, dj);
/* Write the output to the screen. */
printf ("\nSolution status = %d\n", solstat);
printf ("Solution value = %f\n\n", objval);
for (i = 0; i < cur_numrows; i++) {
printf ("Row %d: Slack = %10f Pi = %10f\n", i, slack[i], pi[i]);
}
for (j = 0; j < cur_numcols; j++) {
printf ("Column %d: Value = %10f Reduced cost = %10f\n",
j, x[j], dj[j]);
}
/* Finally, write a copy of the problem to a file. */
status = CPXwriteprob (env, lp, "lpex1.lp", NULL);
/* Free up the solution */
... (additional code to free up the solution)...
return(status)
}