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CPLEX 호출 가능 라이브러리를 사용하여 lp를 해결했습니다 (VS2010에서). lp는 다음과 같습니다.CPLEX 호출 라이브러리를 사용할 때 lp를 mip으로 변경하는 방법
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
코드는 아래에 있습니다. 이제는 MIP (x에 추가적인 통합 성 제약)을 만들고 싶습니다. 나는 status = CPXlpopt (env, lp);을 status = CPXmipopt (env, lp);으로 바꾸려고 노력했다. 이 작동하지 않고 오류 3003: not a mixed-integer problem 가져옵니다. 아무도 내가 여기서 뭘 놓치고 있는지 알아?
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)
}