5.3.3. C++ 的QP建模和优化¶
在本节中，我们将使用 MindOpt C++ API，以按行输入的形式来建模以及求解 二次规划问题示例 中的问题。
首先，引入头文件：
24#include "MindoptCpp.h"
并创建优化模型：
31 /*------------------------------------------------------------------*/
32 /* Step 1. Create environment and model. */
33 /*------------------------------------------------------------------*/
34 MDOEnv env = MDOEnv();
35 MDOModel model = MDOModel(env);
接下来，我们通过 MDOModel::set() 将目标函数设置为 最小化，并调用 MDOModel::addVar() 来添加四个优化变量，定义其下界、上界、名称和类型（有关 MDOModel::set() 和 MDOModel::addVar() 的详细使用方式，请参考 C++ API）：
42 /* Change to minimization problem. */
43 model.set(MDO_IntAttr_ModelSense, MDO_MINIMIZE);
44
45 /* Add variables. */
46 std::vector<MDOVar> x;
47 x.push_back(model.addVar(0.0, 10.0, 0.0, MDO_CONTINUOUS, "x0"));
48 x.push_back(model.addVar(0.0, MDO_INFINITY, 0.0, MDO_CONTINUOUS, "x1"));
49 x.push_back(model.addVar(0.0, MDO_INFINITY, 0.0, MDO_CONTINUOUS, "x2"));
50 x.push_back(model.addVar(0.0, MDO_INFINITY, 0.0, MDO_CONTINUOUS, "x3"));
接着，我们开始添加线性约束：
52 /* Add constraints. */
53 model.addConstr(1.0 * x[0] + 1.0 * x[1] + 2.0 * x[2] + 3.0 * x[3], MDO_GREATER_EQUAL, 1.0, "c0");
54 model.addConstr(1.0 * x[0] - 1.0 * x[2] + 6.0 * x[3], MDO_EQUAL, 1.0, "c1");
然后，我们创建一个二次表达式 MDOQuadExpr, 再调用 MDOQuadExpr::addTerms 来设置目标函数线性部分。
obj_idx 表示线性部分的索引，obj_val 表示与 obj_idx 中的索引相对应的非零系数值，整数 4 代表线性部分的非零元的个数。
56 /*Create a QuadExpr. */
57 MDOQuadExpr obj = MDOQuadExpr(0.0);
58
59 /* Add objective linear term.*/
60 const MDOVar obj_idx[] = { x[0], x[1], x[2], x[3]};
61 const double obj_val[] = { 1.0, 1.0, 1.0, 1.0};
62 obj.addTerms(obj_val, obj_idx, 4);
然后，调用 MDOQuadExpr::addTerms 来设置目标的二次项系数 qo_col1 和 qo_col2 表示与qo_values相对应的二次项的第一个变量和第二个变量，第四个参数表示
Note
为了确保
64 /* Add quadratic objective matrix Q.
65 *
66 * Note.
67 * 1. The objective function is defined as c^Tx + 1/2 x^TQx, where Q is stored with coordinate format.
68 * 2. Q will be scaled by 1/2 internally.
69 * 3. To ensure the symmetricity of Q, user needs to input only the lower triangular part.
70 *
71 * Q = [ 1.0 0.5 0 0 ]
72 * [ 0.5 1.0 0 0 ]
73 * [ 0.0 0.0 1.0 0 ]
74 * [ 0 0 0 1.0 ]
75 */
76
77 const double qo_values[] =
78 {
79 1.0,
80 0.5, 1.0,
81 1.0,
82 1.0
83 };
84 const MDOVar qo_col1[] =
85 {
86 x[0],
87 x[1], x[1],
88 x[2],
89 x[3]
90 };
91 const MDOVar qo_col2[] =
92 {
93 x[0],
94 x[0], x[1],
95 x[2],
96 x[3]
97 };
98
99 obj.addTerms(qo_values, qo_col1, qo_col2, 5);
100
101 model.setObjective(obj, MDO_MINIMIZE);
问题输入完成后，再调用 MDOModel::optimize() 求解优化问题：
107 model.optimize();
示例 MdoQoEx1.cpp 提供了完整源代码：
1/**
2 * Description
3 * -----------
4 *
5 * Linear optimization (row-wise input).
6 *
7 * Formulation
8 * -----------
9 *
10 * Minimize
11 * obj: 1 x0 + 1 x1 + 1 x2 + 1 x3
12 * + 1/2 [ x0^2 + x1^2 + x2^2 + x3^2 + x0 x1]
13 * Subject To
14 * c0 : 1 x0 + 1 x1 + 2 x2 + 3 x3 >= 1
15 * c1 : 1 x0 - 1 x2 + 6 x3 = 1
16 * Bounds
17 * 0 <= x0 <= 10
18 * 0 <= x1
19 * 0 <= x2
20 * 0 <= x3
21 * End
22 */
23#include <iostream>
24#include "MindoptCpp.h"
25#include <vector>
26
27using namespace std;
28
29int main(void)
30{
31 /*------------------------------------------------------------------*/
32 /* Step 1. Create environment and model. */
33 /*------------------------------------------------------------------*/
34 MDOEnv env = MDOEnv();
35 MDOModel model = MDOModel(env);
36
37 try
38 {
39 /*------------------------------------------------------------------*/
40 /* Step 2. Input model. */
41 /*------------------------------------------------------------------*/
42 /* Change to minimization problem. */
43 model.set(MDO_IntAttr_ModelSense, MDO_MINIMIZE);
44
45 /* Add variables. */
46 std::vector<MDOVar> x;
47 x.push_back(model.addVar(0.0, 10.0, 0.0, MDO_CONTINUOUS, "x0"));
48 x.push_back(model.addVar(0.0, MDO_INFINITY, 0.0, MDO_CONTINUOUS, "x1"));
49 x.push_back(model.addVar(0.0, MDO_INFINITY, 0.0, MDO_CONTINUOUS, "x2"));
50 x.push_back(model.addVar(0.0, MDO_INFINITY, 0.0, MDO_CONTINUOUS, "x3"));
51
52 /* Add constraints. */
53 model.addConstr(1.0 * x[0] + 1.0 * x[1] + 2.0 * x[2] + 3.0 * x[3], MDO_GREATER_EQUAL, 1.0, "c0");
54 model.addConstr(1.0 * x[0] - 1.0 * x[2] + 6.0 * x[3], MDO_EQUAL, 1.0, "c1");
55
56 /*Create a QuadExpr. */
57 MDOQuadExpr obj = MDOQuadExpr(0.0);
58
59 /* Add objective linear term.*/
60 const MDOVar obj_idx[] = { x[0], x[1], x[2], x[3]};
61 const double obj_val[] = { 1.0, 1.0, 1.0, 1.0};
62 obj.addTerms(obj_val, obj_idx, 4);
63
64 /* Add quadratic objective matrix Q.
65 *
66 * Note.
67 * 1. The objective function is defined as c^Tx + 1/2 x^TQx, where Q is stored with coordinate format.
68 * 2. Q will be scaled by 1/2 internally.
69 * 3. To ensure the symmetricity of Q, user needs to input only the lower triangular part.
70 *
71 * Q = [ 1.0 0.5 0 0 ]
72 * [ 0.5 1.0 0 0 ]
73 * [ 0.0 0.0 1.0 0 ]
74 * [ 0 0 0 1.0 ]
75 */
76
77 const double qo_values[] =
78 {
79 1.0,
80 0.5, 1.0,
81 1.0,
82 1.0
83 };
84 const MDOVar qo_col1[] =
85 {
86 x[0],
87 x[1], x[1],
88 x[2],
89 x[3]
90 };
91 const MDOVar qo_col2[] =
92 {
93 x[0],
94 x[0], x[1],
95 x[2],
96 x[3]
97 };
98
99 obj.addTerms(qo_values, qo_col1, qo_col2, 5);
100
101 model.setObjective(obj, MDO_MINIMIZE);
102
103 /*------------------------------------------------------------------*/
104 /* Step 3. Solve the problem and populate optimization result. */
105 /*------------------------------------------------------------------*/
106 /* Solve the problem. */
107 model.optimize();
108
109 if(model.get(MDO_IntAttr_Status) == MDO_OPTIMAL)
110 {
111 cout << "Optimal objective value is: " << model.get(MDO_DoubleAttr_ObjVal) << endl;
112 cout << "Decision variables:" << endl;
113 int i = 0;
114 for (auto v : x)
115 {
116 cout << "x[" << i++ << "] = " << v.get(MDO_DoubleAttr_X) << endl;
117 }
118 }
119 else
120 {
121 cout<< "No feasible solution." << endl;
122 }
123
124 }
125 catch (MDOException& e)
126 {
127 std::cout << "Error code = " << e.getErrorCode() << std::endl;
128 std::cout << e.getMessage() << std::endl;
129 }
130 catch (...)
131 {
132 std::cout << "Error during optimization." << std::endl;
133 }
134
135 return static_cast<int>(MDO_OKAY);
136}