5.4.3. C++ 的QCP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 二次约束规划问题示例 中的问题。
首先，引入头文件：
25#include "MindoptCpp.h"
并创建优化模型：
32 /*------------------------------------------------------------------*/
33 /* Step 1. Create environment and model. */
34 /*------------------------------------------------------------------*/
35 MDOEnv env = MDOEnv();
36 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，以及其在目标函数中线性项的系数（有关 MDOModel::set() 和 MDOModel::addVar() 的详细使用方式，请参考 C++ API）：
43 /* Change to minimization problem. */
44 model.set(MDO_IntAttr_ModelSense, MDO_MINIMIZE);
45
46 /* Add variables. */
47 std::vector<MDOVar> x;
48 x.push_back(model.addVar(0.0, 10.0, 0.0, MDO_CONTINUOUS, "x0"));
49 x.push_back(model.addVar(0.0, MDO_INFINITY, 0.0, MDO_CONTINUOUS, "x1"));
50 x.push_back(model.addVar(0.0, MDO_INFINITY, 0.0, MDO_CONTINUOUS, "x2"));
51 x.push_back(model.addVar(0.0, MDO_INFINITY, 0.0, MDO_CONTINUOUS, "x3"));
接着我们开始设置二次目标函数。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，obj_nnz 代表线性部分的非零元的个数。
54 /* Linear part in the objective: 1 x0 + 1 x1 + 1 x2 + 1 x3 */
55 int obj_nnz = 4;
56 MDOVar obj_idx[] = { x[0], x[1], x[2], x[3] };
57 double obj_val[] = { 1.0, 1.0, 1.0, 1.0 };
85 /* Create objective. */
86 MDOQuadExpr obj = MDOQuadExpr(0.0);
87 /* Add linear terms */
88 obj.addTerms(obj_val, obj_idx, obj_nnz);
然后，调用 MDOQuadExpr::addTerms 来设置目标的二次项系数 \(Q\)。
其中，qo_values 表示要添加的二次项的系数，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，qo_nnz 表示二次项中的非零元个数。
58 /* Quadratic part in the objective: 1/2 [ x0^2 + x1^2 + x2^2 + x3^2 + x0 x1] */
59 int qo_nnz = 5;
60 MDOVar qo_col1[] = { x[0], x[1], x[2], x[3], x[0] };
61 MDOVar qo_col2[] = { x[0], x[1], x[2], x[3], x[1] };
62 double qo_values[] = { 0.5, 0.5, 0.5, 0.5, 0.5 };
89 /* Add quadratic terms */
90 obj.addTerms(qo_values, qo_col1, qo_col2, qo_nnz);
最后，我们调用 MDOModel::setObjective 设定优化目标与方向。
104 /* Set optimization sense. */
105 model.setObjective(obj, MDO_MINIMIZE);
现在我们开始添加二次约束。构建二次约束中的二次表达式的方式与二次目标函数类似。
64 /* Linear part in the first constraint: 1 x0 + 1 x1 + 2 x2 + 3 x3 */
65 int c1_nnz = 4;
66 MDOVar c1_idx[] = { x[0], x[1], x[2], x[3] };
67 double c1_val[] = { 1.0, 1.0, 2.0, 3.0 };
68 /* Quadratic part in the first constraint: - 1/2 [ x0^2 + x1^2 + x2^2 + x3^2 + x0 x1] */
69 int qc1_nnz = 5;
70 MDOVar qc1_col1[] = { x[0], x[1], x[2], x[3], x[0] };
71 MDOVar qc1_col2[] = { x[0], x[1], x[2], x[3], x[1] };
72 double qc1_values[] = { -0.5, -0.5, -0.5, -0.5, -0.5 };
93 MDOQuadExpr c1 = MDOQuadExpr(0.0);
94 c1.addTerms(c1_val, c1_idx, c1_nnz);
95 c1.addTerms(qc1_values, qc1_col1, qc1_col2, qc1_nnz);
74 /* Linear part in the second constraint: 1 x0 - 1 x2 + 6 x3 */
75 int c2_nnz = 3;
76 MDOVar c2_idx[] = { x[0], x[2], x[3] };
77 double c2_val[] = { 1.0, -1.0, 6.0 };
78 /* Quadratic part in the second constraint: 1/2 [x1^2] */
79 int qc2_nnz = 1;
80 MDOVar qc2_col1[] = { x[1] };
81 MDOVar qc2_col2[] = { x[1] };
82 double qc2_values[] = { 0.5 };
99 MDOQuadExpr c2 = MDOQuadExpr(0.0);
100 c2.addTerms(c2_val, c2_idx, c2_nnz);
101 c2.addTerms(qc2_values, qc2_col1, qc2_col2, qc2_nnz);
然后，我们调用 MDOModel::addQConstr 将二次约束添加至模型中。
96 model.addQConstr(c1, MDO_GREATER_EQUAL, 1.0, "c0");
102 model.addQConstr(c2, MDO_LESS_EQUAL, 1.0, "c1");
问题输入完成后，调用 MDOModel::optimize() 求解优化问题：
110 /* Solve the problem. */
111 model.optimize();
示例 MdoQcoEx1.cpp 提供了完整源代码：
1/**
2 * Description
3 * -----------
4 *
5 * Quadratically constrained quadratic 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 *
14 * Subject To
15 * c0 : 1 x0 + 1 x1 + 2 x2 + 3 x3 - 1/2 [ x0^2 + x1^2 + x2^2 + x3^2 + x0 x1] >= 1
16 * c1 : 1 x0 - 1 x2 + 6 x3 + 1/2 [x1^2] <= 1
17 * Bounds
18 * 0 <= x0 <= 10
19 * 0 <= x1
20 * 0 <= x2
21 * 0 <= x3
22 * End
23 */
24#include <iostream>
25#include "MindoptCpp.h"
26#include <vector>
27
28using namespace std;
29
30int main(void)
31{
32 /*------------------------------------------------------------------*/
33 /* Step 1. Create environment and model. */
34 /*------------------------------------------------------------------*/
35 MDOEnv env = MDOEnv();
36 MDOModel model = MDOModel(env);
37
38 try
39 {
40 /*------------------------------------------------------------------*/
41 /* Step 2. Input model. */
42 /*------------------------------------------------------------------*/
43 /* Change to minimization problem. */
44 model.set(MDO_IntAttr_ModelSense, MDO_MINIMIZE);
45
46 /* Add variables. */
47 std::vector<MDOVar> x;
48 x.push_back(model.addVar(0.0, 10.0, 0.0, MDO_CONTINUOUS, "x0"));
49 x.push_back(model.addVar(0.0, MDO_INFINITY, 0.0, MDO_CONTINUOUS, "x1"));
50 x.push_back(model.addVar(0.0, MDO_INFINITY, 0.0, MDO_CONTINUOUS, "x2"));
51 x.push_back(model.addVar(0.0, MDO_INFINITY, 0.0, MDO_CONTINUOUS, "x3"));
52
53 /* Prepare model data. */
54 /* Linear part in the objective: 1 x0 + 1 x1 + 1 x2 + 1 x3 */
55 int obj_nnz = 4;
56 MDOVar obj_idx[] = { x[0], x[1], x[2], x[3] };
57 double obj_val[] = { 1.0, 1.0, 1.0, 1.0 };
58 /* Quadratic part in the objective: 1/2 [ x0^2 + x1^2 + x2^2 + x3^2 + x0 x1] */
59 int qo_nnz = 5;
60 MDOVar qo_col1[] = { x[0], x[1], x[2], x[3], x[0] };
61 MDOVar qo_col2[] = { x[0], x[1], x[2], x[3], x[1] };
62 double qo_values[] = { 0.5, 0.5, 0.5, 0.5, 0.5 };
63
64 /* Linear part in the first constraint: 1 x0 + 1 x1 + 2 x2 + 3 x3 */
65 int c1_nnz = 4;
66 MDOVar c1_idx[] = { x[0], x[1], x[2], x[3] };
67 double c1_val[] = { 1.0, 1.0, 2.0, 3.0 };
68 /* Quadratic part in the first constraint: - 1/2 [ x0^2 + x1^2 + x2^2 + x3^2 + x0 x1] */
69 int qc1_nnz = 5;
70 MDOVar qc1_col1[] = { x[0], x[1], x[2], x[3], x[0] };
71 MDOVar qc1_col2[] = { x[0], x[1], x[2], x[3], x[1] };
72 double qc1_values[] = { -0.5, -0.5, -0.5, -0.5, -0.5 };
73
74 /* Linear part in the second constraint: 1 x0 - 1 x2 + 6 x3 */
75 int c2_nnz = 3;
76 MDOVar c2_idx[] = { x[0], x[2], x[3] };
77 double c2_val[] = { 1.0, -1.0, 6.0 };
78 /* Quadratic part in the second constraint: 1/2 [x1^2] */
79 int qc2_nnz = 1;
80 MDOVar qc2_col1[] = { x[1] };
81 MDOVar qc2_col2[] = { x[1] };
82 double qc2_values[] = { 0.5 };
83
84 /* Construct model. */
85 /* Create objective. */
86 MDOQuadExpr obj = MDOQuadExpr(0.0);
87 /* Add linear terms */
88 obj.addTerms(obj_val, obj_idx, obj_nnz);
89 /* Add quadratic terms */
90 obj.addTerms(qo_values, qo_col1, qo_col2, qo_nnz);
91
92 /* Add 1st quadratic constraint. */
93 MDOQuadExpr c1 = MDOQuadExpr(0.0);
94 c1.addTerms(c1_val, c1_idx, c1_nnz);
95 c1.addTerms(qc1_values, qc1_col1, qc1_col2, qc1_nnz);
96 model.addQConstr(c1, MDO_GREATER_EQUAL, 1.0, "c0");
97
98 /* Add 2nd quadratic constraint. */
99 MDOQuadExpr c2 = MDOQuadExpr(0.0);
100 c2.addTerms(c2_val, c2_idx, c2_nnz);
101 c2.addTerms(qc2_values, qc2_col1, qc2_col2, qc2_nnz);
102 model.addQConstr(c2, MDO_LESS_EQUAL, 1.0, "c1");
103
104 /* Set optimization sense. */
105 model.setObjective(obj, MDO_MINIMIZE);
106
107 /*------------------------------------------------------------------*/
108 /* Step 3. Solve the problem and populate optimization result. */
109 /*------------------------------------------------------------------*/
110 /* Solve the problem. */
111 model.optimize();
112
113 if(model.get(MDO_IntAttr_Status) == MDO_OPTIMAL)
114 {
115 cout << "Optimal objective value is: " << model.get(MDO_DoubleAttr_ObjVal) << endl;
116 cout << "Decision variables:" << endl;
117 int i = 0;
118 for (auto v : x)
119 {
120 cout << "x[" << i++ << "] = " << v.get(MDO_DoubleAttr_X) << endl;
121 }
122 }
123 else
124 {
125 cout<< "No feasible solution." << endl;
126 }
127
128 }
129 catch (MDOException& e)
130 {
131 std::cout << "Error code = " << e.getErrorCode() << std::endl;
132 std::cout << e.getMessage() << std::endl;
133 }
134 catch (...)
135 {
136 std::cout << "Error during optimization." << std::endl;
137 }
138
139 return static_cast<int>(MDO_OKAY);
140}