数学建模 二次规划(普通求解方式 && 罚函数法)

matlab一般求解方式

示例

可以看出
$\frac12x^THx=2x_1^2-4x_1x_2+4x_2^2$
即
$\left[ \begin{matrix} x_1 & x_2 \\ \end{matrix} \right] \left[ \begin{matrix} a&b \\ c&d \\ \end{matrix} \right] \left[ \begin{matrix} x_1 \\ x_2 \\ \end{matrix} \right]=ax_1^2+(b+c)x_1x_2+dx_2^2=4x_1^2-8x_1x_2+8x_2^2$
即

$\left\{ \begin{aligned} a=4\\ d=8\\ b+c=-8 \end{aligned} \right.$

但是H必须是对称矩阵（二次型），所以b=c=-4

所以matlab代码

% 二次规划
clc
h=[4,-4;-4,8];
f=[-6;-3];
a=[1 1;4 1];
b=[3;9];
[x,value]=quadprog(h,f,a,b,[],[],zeros(2,1))

结果

Minimum found that satisfies the constraints.

Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the default value of the optimality tolerance,
and constraints are satisfied to within the default value of the constraint tolerance.

<stopping criteria details>


x =

    1.9500
    1.0500


value =

  -11.0250

>> 

外罚函数法

罚函数类似于拉格朗日对偶问题里面的松弛变量（KKT乘子），把非线性的约束放到目标函数里去，变成一个无约束的极值问题

$P(x,M)$被称为增广目标函数，原优化问题被转化为以$P(x,M)$为目标函数的无约束极值问题
$\min P(x,M)$
两个问题的最优解一样

非常聪明的方式

结果

>> main
警告: Gradient must be provided for trust-region algorithm; using quasi-newton
algorithm instead. 
> In fminunc (line 395)
  In main (line 1) 

Local minimum possible.

fminunc stopped because the size of the current step is less than
the default value of the step size tolerance.

<stopping criteria details>


Local minimum possible.

fminunc stopped because it cannot decrease the objective function
along the current search direction.

<stopping criteria details>


x =

    1.2205
    0.8829


y =

   10.2692

% 主函数
[x,y]=fminunc('fa',rand(2,1))

% 外罚函数法求解二次规划
function f=fa(x);
M=50000;
g=x(1)^2+x(2)^2+8;
f=g-M*min(x(1),0)-M*min(x(2),0)-M*min(x(1).^2-x(2),0)+M*abs(-x(1)-x(2).^2+2);

来源：https://blog.csdn.net/qq_36607894/article/details/100107267