5.1.2. LP Modeling and Optimization in CΒΆ

In this section, we will utilize MindOpt C API to model and solve the linear optimization problem in Example of Linear Programming.

First of all, include the header files:

24#include "Mindopt.h"

Create an optimization model m:

54    CHECK_RESULT(MDOemptyenv(&env));
55    CHECK_RESULT(MDOstartenv(env));
56    CHECK_RESULT(MDOnewmodel(env, &m, MODEL_NAME, 0, NULL, NULL, NULL, NULL, NULL));

Next, we set the optimization sense to minimization via MDOsetintattr() and add four decision variables using MDOaddvar() (please refer to C API for the detailed usages of MDOsetintattr() and MDOaddvar()):

61    /* Change to minimization problem. */
62    CHECK_RESULT(MDOsetintattr(m, MODEL_SENSE, MDO_MINIMIZE));
63
64    /* Add variables. */
65    CHECK_RESULT(MDOaddvar(m, 0, NULL, NULL, 1.0, 0,         10.0, MDO_CONTINUOUS, "x0"));
66    CHECK_RESULT(MDOaddvar(m, 0, NULL, NULL, 2.0, 0, MDO_INFINITY, MDO_CONTINUOUS, "x1"));
67    CHECK_RESULT(MDOaddvar(m, 0, NULL, NULL, 1.0, 0, MDO_INFINITY, MDO_CONTINUOUS, "x2"));
68    CHECK_RESULT(MDOaddvar(m, 0, NULL, NULL, 1.0, 0, MDO_INFINITY, MDO_CONTINUOUS, "x3"));

Then we set the constraint matrix \(A\). There are two rows in \(A\) and we use the following four arrays to represent this matrix. In this context, row1_idx and row2_idx represent the positions of the non-zero elements in the first and second row of \(A\), repsectively. Meanwhile, row1_val and row2_val hold the values of corresponding non-zero elements.

46    int    row1_idx[] = { 0,   1,   2,   3   };
47    double row1_val[] = { 1.0, 1.0, 2.0, 3.0 };
48    int    row2_idx[] = { 0,    2,   3   };
49    double row2_val[] = { 1.0, -1.0, 6.0 };

We call MDOaddconstr() to input the linear constraints into the model m:

70    /* Add constraints. */
71    CHECK_RESULT(MDOaddconstr(m, 4, row1_idx, row1_val, MDO_GREATER_EQUAL, 1.0, "c0"));
72    CHECK_RESULT(MDOaddconstr(m, 3, row2_idx, row2_val, MDO_EQUAL,         1.0, "c1"));

Once the model is constructed, we call MDOoptimize() to solve the problem:

77    CHECK_RESULT(MDOoptimize(m));

Then, we can retrieive the optimal objective value and solutions as follows:

78    CHECK_RESULT(MDOgetintattr(m, STATUS, &status));
79    if (status == MDO_OPTIMAL) 
80    {
81        CHECK_RESULT(MDOgetdblattr(m, OBJ_VAL, &obj));
82        printf("The optimal objective value is: %f\n", obj);
83        for (i = 0; i < 4; ++i) 
84        {
85            CHECK_RESULT(MDOgetdblattrelement(m, X, i, &x));
86            printf("x[%d] = %f\n", i, x);
87        }
88    } 
89    else 
90    {
91        printf("No feasible solution.\n");
92    }

Lastly, we call MDOfreemodel() and MDOfreeenv() to free the model:

27#define RELEASE_MEMORY  \
28    MDOfreemodel(m);    \
29    MDOfreeenv(env);
97    RELEASE_MEMORY;

Complete example codes are provided in MdoLoEx1.c.

  1/**
  2 *  Description
  3 *  -----------
  4 *
  5 *  Linear optimization (row-wise input).
  6 *
  7 *  Formulation
  8 *  -----------
  9 *
 10 *  Minimize
 11 *    obj: 1 x0 + 2 x1 + 1 x2 + 1 x3
 12 *  Subject To
 13 *   c0 : 1 x0 + 1 x1 + 2 x2 + 3 x3 >= 1
 14 *   c1 : 1 x0        - 1 x2 + 6 x3 = 1
 15 *  Bounds
 16 *    0 <= x0 <= 10
 17 *    0 <= x1
 18 *    0 <= x2
 19 *    0 <= x3
 20 *  End
 21 */
 22#include <stdio.h>
 23#include <stdlib.h>
 24#include "Mindopt.h"
 25
 26/* Macro to check the return code */
 27#define RELEASE_MEMORY  \
 28    MDOfreemodel(m);    \
 29    MDOfreeenv(env);
 30#define CHECK_RESULT(code) { int res = code; if (res != 0) { fprintf(stderr, "Bad code: %d\n", res);  RELEASE_MEMORY; return (res); } }
 31#define MODEL_NAME  "LP_01"
 32#define MODEL_SENSE "ModelSense"
 33#define STATUS      "Status"
 34#define OBJ_VAL     "ObjVal"
 35#define X           "X"
 36
 37int main(void)
 38{
 39    /* Variables. */
 40    MDOenv *env;
 41    MDOmodel *m;
 42    double obj, x;
 43    int status, i;
 44
 45    /* Model data. */
 46    int    row1_idx[] = { 0,   1,   2,   3   };
 47    double row1_val[] = { 1.0, 1.0, 2.0, 3.0 };
 48    int    row2_idx[] = { 0,    2,   3   };
 49    double row2_val[] = { 1.0, -1.0, 6.0 };
 50
 51    /*------------------------------------------------------------------*/
 52    /* Step 1. Create environment and model.                            */
 53    /*------------------------------------------------------------------*/
 54    CHECK_RESULT(MDOemptyenv(&env));
 55    CHECK_RESULT(MDOstartenv(env));
 56    CHECK_RESULT(MDOnewmodel(env, &m, MODEL_NAME, 0, NULL, NULL, NULL, NULL, NULL));
 57
 58    /*------------------------------------------------------------------*/
 59    /* Step 2. Input model.                                             */
 60    /*------------------------------------------------------------------*/
 61    /* Change to minimization problem. */
 62    CHECK_RESULT(MDOsetintattr(m, MODEL_SENSE, MDO_MINIMIZE));
 63
 64    /* Add variables. */
 65    CHECK_RESULT(MDOaddvar(m, 0, NULL, NULL, 1.0, 0,         10.0, MDO_CONTINUOUS, "x0"));
 66    CHECK_RESULT(MDOaddvar(m, 0, NULL, NULL, 2.0, 0, MDO_INFINITY, MDO_CONTINUOUS, "x1"));
 67    CHECK_RESULT(MDOaddvar(m, 0, NULL, NULL, 1.0, 0, MDO_INFINITY, MDO_CONTINUOUS, "x2"));
 68    CHECK_RESULT(MDOaddvar(m, 0, NULL, NULL, 1.0, 0, MDO_INFINITY, MDO_CONTINUOUS, "x3"));
 69
 70    /* Add constraints. */
 71    CHECK_RESULT(MDOaddconstr(m, 4, row1_idx, row1_val, MDO_GREATER_EQUAL, 1.0, "c0"));
 72    CHECK_RESULT(MDOaddconstr(m, 3, row2_idx, row2_val, MDO_EQUAL,         1.0, "c1"));
 73
 74    /*------------------------------------------------------------------*/
 75    /* Step 3. Solve the problem and populate optimization result.      */
 76    /*------------------------------------------------------------------*/
 77    CHECK_RESULT(MDOoptimize(m));
 78    CHECK_RESULT(MDOgetintattr(m, STATUS, &status));
 79    if (status == MDO_OPTIMAL) 
 80    {
 81        CHECK_RESULT(MDOgetdblattr(m, OBJ_VAL, &obj));
 82        printf("The optimal objective value is: %f\n", obj);
 83        for (i = 0; i < 4; ++i) 
 84        {
 85            CHECK_RESULT(MDOgetdblattrelement(m, X, i, &x));
 86            printf("x[%d] = %f\n", i, x);
 87        }
 88    } 
 89    else 
 90    {
 91        printf("No feasible solution.\n");
 92    }
 93
 94    /*------------------------------------------------------------------*/
 95    /* Step 4. Free the model.                                          */
 96    /*------------------------------------------------------------------*/
 97    RELEASE_MEMORY;
 98
 99    return 0;
100}