用matlab求解一个优化问题（目标函数非线性，约束函数线性）

来源：百度知道 编辑：UC知道 时间：2024/05/10 11:08:35
min f(x)=12*x(1)+21*x(2)+21*x(3)+13*x(4)+20*x(5)+23*x(6)+15*x(7)+17*x(8)+27*x(9)+17*x(10)+19*x(11)+31*x(12)+200*[(x(1)+x(2)+x(3))^0.6+(x(4)+x(5)+x(6))^0.6+(x(7)+x(8)+x(9))^0.6+(x(10)+x(11)+x(12))^0.6]
s.t. x(1)+x(2)+x(3)<=183.55;
x(4)+x(5)+x(6)<=283.55;
x(7)+x(8)+x(9)<=383.55;
x(10)+x(11)+x(12)<=183.55;
x(1)+x(4)+x(7)+x(10)>=116.45;
x(2)+x(5)+x(8)+x(11)>=316.45;
x(3)+x(6)+x(9)+x(12)>=416.45;
x(1),…x(12)>=0;
我是这样求的：
目标函数（运行显示下标不对）
function f=myobj(x)
c = [12 21 21; 13 20 23; 15 17 27; 17 19 31];
size(x) = [4,3];
y = c.*x;
y = sum(y(:));
s = 0;
for i = 1:4
sg = sum(x(i,:));
s = s+sg^0.6;
end
f = y+200*s;
command window
A= [1,1,1,0,0,0,0,0,0,0,0,0;
0,0,0,1,1,1,0,0,0,0,0,0;
0,0,0,0,0,0,1,1,1,0,0,0;
0,0,0,0,0,0,0,0,0,1,1,1;
-1,0,0,-1,0,0,-1,0,0,-1,0,0;
0,-1,0,0,-1,0,0,-1,0,0,-1

程序编的非常好，只有一句：size(x) = [4,3]改为x =zeros(4,3) 即可。
%=================================
function fff
clear;clc;
A= [1,1,1,0,0,0,0,0,0,0,0,0;
0,0,0,1,1,1,0,0,0,0,0,0;
0,0,0,0,0,0,1,1,1,0,0,0;
0,0,0,0,0,0,0,0,0,1,1,1;
-1,0,0,-1,0,0,-1,0,0,-1,0,0;
0,-1,0,0,-1,0,0,-1,0,0,-1,0;
0,0,-1,0,0,-1,0,0,-1,0,0,-1]
b=[200-1.645*10;
300-1.645*10;
400-1.645*10;
200-1.645*10;
-100-1.645*10;
-300-1.645*10;
-400-1.645*10]
x0=[0,0,0,0,0,0,0,0,0,0,0,0];
lb=[0,0,0,0,0,0,0,0,0,0,0,0];
ub=[];
[X,FVAL,EXITFLAG]=fmincon(@myobj,x0,A,b,[],[],lb,ub)

function f=myobj(x)
c = [12 21 21; 13 20 23; 15 17 27; 17 19 31];
x =zeros(4,3)
y = c.*x;
y = sum(y(:));
s = 0;
for i = 1:4
sg = sum(x(i,:));
s = s+sg.^0.6;
end
f = y+200*s;
%=============================

运行结果：
X =

19.5167 69.5167 94.5167