[未答复] 非线性优化中约束含有整数选择什么求解器

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jingyi.yang 发表于 2021-11-24 16:54:54
本帖最后由 jingyi.yang 于 2021-11-24 16:58 编辑
  1. <blockquote>%目标函数用m文件
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jingyi.yang 发表于 2021-11-24 16:59:07
%目标函数用m文件
fun = @eta;

%第14到17变量为整数
intcon = 14:17;
x0 = [1e5,0.002,0.002,0.002,0.002,1.05,1.05,1.05,1.05,0.05,0.1,0.1,0.05,2,10,10,2,0,0.01,0];    %初始量

%线性不等式约束
A = [
    -1,zeros(1,19);
    1,zeros(1,19);
    0,-1,zeros(1,18);
    0,1,zeros(1,18);
    zeros(1,2),-1,zeros(1,17);
    zeros(1,2),1,zeros(1,17);
    zeros(1,3),-1,zeros(1,16);
    zeros(1,3),1,zeros(1,16);
    zeros(1,4),-1,zeros(1,15);
    zeros(1,4),1,zeros(1,15);
    zeros(1,5),-1,zeros(1,14);
    zeros(1,5),1,zeros(1,14);
    zeros(1,6),-1,zeros(1,13);
    zeros(1,6),1,zeros(1,13);
    zeros(1,7),-1,zeros(1,12);
    zeros(1,7),1,zeros(1,12);
    zeros(1,8),-1,zeros(1,11);
    zeros(1,8),1,zeros(1,11);
    zeros(1,9),-1,zeros(1,10);
    zeros(1,9),1,zeros(1,10);
    zeros(1,10),-1,zeros(1,9);
    zeros(1,10),1,zeros(1,9);
    zeros(1,11),-1,zeros(1,8);
    zeros(1,11),1,zeros(1,8);
    zeros(1,12),-1,zeros(1,7);
    zeros(1,12),1,zeros(1,7);
    zeros(1,13),-1,zeros(1,6);
    zeros(1,13),1,zeros(1,6);
    zeros(1,14),-1,zeros(1,5);
    zeros(1,14),1,zeros(1,5);
    zeros(1,15),-1,zeros(1,4);
    zeros(1,15),1,zeros(1,4);
    zeros(1,16),-1,zeros(1,3);
    zeros(1,16),1,zeros(1,3);
    zeros(1,18),-1,zeros(1,1);
    zeros(1,18),1,zeros(1,1);
    ];         
b = [
    -1e5;
    1e6;
    -0.002;
    0.004;
    -0.002;
    0.004;
    -0.002;
    0.004;
    -0.002;
    0.004;
    -1.05;
    4;
     -1.05;
    4;
     -1.05;
    4;
     -1.05;
    4;
    -0.05;
    0.08;
    -0.1;
    0.15;
    -0.1;
    0.15;
    -0.05;
    0.08;
    -1;
    3;
    -10;
    20;
    -10;
    20;
    -1;
    3;
    -0.01;
    0.15;
   
   
    ];

%等式约束
Aeq = [
        zeros(1,17),1,0,0;
        zeros(1,19),1;
        ];
beq = [
        0;
        0;
        ];

%用MILP的话目标函数必须是double输入
%[x,fval] = intlinprog(fun,intcon,A,b,Aeq,beq,[],[],x0);

jingyi.yang 发表于 2021-11-24 17:00:16
详细代码见一楼,不知道为什么帖子没发出来
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