5.5.4. Python 的MIQP建模与优化¶
在本节中，我们将使用 MindOpt Python API，以按行输入的形式来建模以及求解 MIQP题示例 中的问题。
首先，引入 Python 包：
27from mindoptpy import *
创建优化模型，并赋予一个名称：
33 model = Model("MIQP_01")
调用 Model.addVar() 来添加四个优化变量，定义其下界、上界、名称和类型（有关函数的详细使用方式，请参考 Python API）:
36 # Step 2. Input model.
37
38 # Add variables.
39 x = []
40 x.append(model.addVar(0.0, 10.0, 0.0, 'I', "x0"))
41 x.append(model.addVar(0.0, float('inf'), 0.0, 'C', "x1"))
接着，我们开始添加线性约束：
43 x.append(model.addVar(0.0, float('inf'), 0.0, 'C', "x3"))
44
45 # Add constraints.
46 # Note that the nonzero elements are inputted in a row-wise order here.
然后，我们来设置目标函数。首先使用类 QuadExpr 创建一个二次表达式，然后有两种方式来构建：
第一种是利用 QuadExpr 中的方法 QuadExpr.addTerms() 分别输入线性部分和二次部分；
第二种是直接输入一个二次表达式。
最后再用 Model.setObjective() 来设置目标函数并将问题设置为 最小化。
47 model.addConstr(1.0 * x[0] + 1.0 * x[1] + 2.0 * x[2] + 3.0 * x[3] >= 1, "c0")
48 model.addConstr(1.0 * x[0] - 1.0 * x[2] + 6.0 * x[3] == 1, "c1")
49
50 # Add objective: 1 x0 + 1 x1 + 1 x2 + 1 x3 + 1/2 [ x0^2 + x1^2 + x2^2 + x3^2 + x0 x1]
51 obj = QuadExpr()
52
53 #option-I
54 obj.addTerms([1.0, 1.0, 1.0, 1.0], [x[0], x[1], x[2], x[3]])
55 obj.addTerms([0.5, 0.5, 0.5, 0.5, 0.5], [x[0], x[1], x[2], x[3], x[0]], [x[0], x[1], x[2], x[3], x[1]])
56
57 #option II
58 # obj = 1*x[0] + 1*x[1] + 1*x[2] + 1*x[3] + 0.5 * x[0]*x[0] + 0.5 * x[1]*x[1] + 0.5 * x[2]*x[2] + 0.5 * x[3]*x[3] + 0.5*x[0]*x[1]
59
问题输入完成后，再调用 Model.optimize() 求解优化问题：
62
然后，通过属性 Status 和属性 ObjVal 来查看优化结果和最优目标值，并通过属性 X 来查看变量的取值。 其他的属性值请查看 属性 章节。
64 model.optimize()
65
66 # Step 4. Retrive model status and objective.
67 # For MIP(MILP,MIQP, MIQCP) problems, if the solving process
68 # terminates early due to reasons such as timeout or interruption,
69 # the model status will indicate termination by timeout (or
70 # interruption, etc.). However, suboptimal solutions may still
最后，我们调用 Model.dispose() 来释放模型：
80 print("Received Mindopt exception.")
示例 mdo_miqp_ex1.py 提供了完整源代码：
1"""
2/**
3 * Description
4 * -----------
5 *
6 * Quadratuc optimization (row-wise input).
7 *
8 * Formulation
9 * -----------
10 *
11 * Minimize
12 * obj: 1 x0 + 1 x1 + 1 x2 + 1 x3
13 * + 1/2 [ x0^2 + x1^2 + x2^2 + x3^2 + x0 x1]
14 * Subject To
15 * c1 : 1 x0 + 1 x1 + 2 x2 + 3 x3 >= 1
16 * c2 : 1 x0 - 1 x2 + 6 x3 = 1
17 * Bounds
18 * 0 <= x0 <= 10
19 * 0 <= x1
20 * 0 <= x2
21 * 0 <= x3
22 * Integers
23 * x0
24 * End
25 */
26"""
27from mindoptpy import *
28
29
30if __name__ == "__main__":
31
32 # Step 1. Create model.
33 model = Model("MIQP_01")
34
35 try:
36 # Step 2. Input model.
37
38 # Add variables.
39 x = []
40 x.append(model.addVar(0.0, 10.0, 0.0, 'I', "x0"))
41 x.append(model.addVar(0.0, float('inf'), 0.0, 'C', "x1"))
42 x.append(model.addVar(0.0, float('inf'), 0.0, 'C', "x2"))
43 x.append(model.addVar(0.0, float('inf'), 0.0, 'C', "x3"))
44
45 # Add constraints.
46 # Note that the nonzero elements are inputted in a row-wise order here.
47 model.addConstr(1.0 * x[0] + 1.0 * x[1] + 2.0 * x[2] + 3.0 * x[3] >= 1, "c0")
48 model.addConstr(1.0 * x[0] - 1.0 * x[2] + 6.0 * x[3] == 1, "c1")
49
50 # Add objective: 1 x0 + 1 x1 + 1 x2 + 1 x3 + 1/2 [ x0^2 + x1^2 + x2^2 + x3^2 + x0 x1]
51 obj = QuadExpr()
52
53 #option-I
54 obj.addTerms([1.0, 1.0, 1.0, 1.0], [x[0], x[1], x[2], x[3]])
55 obj.addTerms([0.5, 0.5, 0.5, 0.5, 0.5], [x[0], x[1], x[2], x[3], x[0]], [x[0], x[1], x[2], x[3], x[1]])
56
57 #option II
58 # obj = 1*x[0] + 1*x[1] + 1*x[2] + 1*x[3] + 0.5 * x[0]*x[0] + 0.5 * x[1]*x[1] + 0.5 * x[2]*x[2] + 0.5 * x[3]*x[3] + 0.5*x[0]*x[1]
59
60 # Set objective and change to minimization problem.
61 model.setObjective(obj, MDO.MINIMIZE)
62
63 # Step 3. Solve the problem and populate optimization result.
64 model.optimize()
65
66 # Step 4. Retrive model status and objective.
67 # For MIP(MILP,MIQP, MIQCP) problems, if the solving process
68 # terminates early due to reasons such as timeout or interruption,
69 # the model status will indicate termination by timeout (or
70 # interruption, etc.). However, suboptimal solutions may still
71 # exist, making it necessary to check the SolCount property.
72 if model.status == MDO.OPTIMAL or model.status == MDO.SUB_OPTIMAL or model.solcount !=0:
73 print(f"Optimal objective value is: {model.objval}")
74 print("Decision variables:")
75 for v in x:
76 print(f"x[{v.VarName}] = {v.X}")
77 else:
78 print("No feasible solution.")
79 except MindoptError as e:
80 print("Received Mindopt exception.")
81 print(" - Code : {}".format(e.errno))
82 print(" - Reason : {}".format(e.message))
83 except Exception as e:
84 print("Received other exception.")
85 print(" - Reason : {}".format(e))
86 finally:
87 # Step 4. Free the model.
88 model.dispose()