Question

用matlab求解一个优化问题（目标函数非线性，约束函数线性）

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,-1,0;
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]=fmincon(@myobj,x0,A,b,[],[],lb,ub)
运行显示：
Warning: Large-scale (trust region) method does not currently solve this type of problem,
switching to medium-scale (line search).
> In fmincon at 260
??? Error using ==> fmincon
FMINCON cannot continue because user supplied objective function failed with the following error:
Subscript indices must either be real positive integers or logicals.
请热心人士帮忙看下怎么改？谢谢！
fval不可能是0.我觉得求出来的x是1*12矩阵，需转换为4*3矩阵，代到目标函数中算才有结果，但我不知道怎么转换。

Answer 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 38.7083 88.7083 113.7083 38.7083 88.7083 113.7083 19.5167 69.5167 94.5167

FVAL =

0

EXITFLAG =

1

Answer 2

Global optimal solution found.
Objective value: 112550.5
Objective bound: 112550.4
Infeasibilities: 0.000000
Extended solver steps: 83
Total solver iterations: 4753
Variable Value Reduced Cost
X1 50.65000 0.000000
X2 0.000000 7.000000
X3 132.9000 0.000000
X4 0.000000 3.438348
X5 0.000000 8.438348
X6 283.5500 0.000000
X7 65.80000 0.000000
X8 316.4500 0.000000
X9 0.000000 0.1050611E+09
X10 0.000000 2.000000
X11 0.000000 2.000000
X12 0.000000 0.1050611E+09
Row Slack or Surplus Dual Price
1 112550.5 -1.000000
2 0.000000 3.000000
3 0.000000 5.438348
4 1.300000 0.000000
5 183.5500 0.000000
6 0.000000 -215.0000
7 0.000000 -217.0000
8 0.000000 -40.97349