当目标函数受约束时，使用fmincon MATLAB求解优化

Question

我想解决：

我使用以下MATLAB代码，但它不起作用。

有人可以指导我吗？

function f=objfun

f=-f;

function [c1,c2,c3]=constraint(x)
a1=1.1; a2=1.1; a3=1.1;
c1=f-log(a1)-log(x(1)/(x(1)+1)); 
c2=f-log(a2)-log(x(2)/(x(2)+1))-log(1-x(1)); 
c3=f-log(a3)-log(1-x(1))-log(1-x(2));


x0=[0.01;0.01]; 
[x,fval]=fmincon('objfun',x0,[],[],[],[],[0;0],[1;1],'constraint')

Answer 1

你需要稍微解决问题。您正在尝试找到使{3} LHS函数的最小值最小的点x（(l_1,l_2)）。所以，你可以用伪代码重写你的问题

maximise, by varying x in [0,1] X [0,1]
       min([log(a1)+log(x(1)/(x(1)+1)) ...
            log(a2)+log(x(2)/(x(2)+1))+log(1-x(1)) ...
            log(a3)+log(1-x(1))+log(1-x(2))])

由于Matlab有fmincon，所以将其重写为最小化问题，

minimise, by varying x in [0,1] X [0,1]
       max(-[log(a1)+log(x(1)/(x(1)+1)) ...
             log(a2)+log(x(2)/(x(2)+1))+log(1-x(1)) ...
             log(a3)+log(1-x(1))+log(1-x(2))])

所以实际的代码是

F=@(x) max(-[log(a1)+log(x(1)/(x(1)+1)) ...
             log(a2)+log(x(2)/(x(2)+1))+log(1-x(1)) ...
             log(a3)+log(1-x(1))+log(1-x(2))])
[L,fval]=fmincon(F,[0.5 0.5])

返回

L =
    0.3383    0.6180
fval =
    1.2800

Answer 2

还可以使用以下MATLAB代码在convex optimization package CVX中解决此问题：

cvx_begin
 variables T(1);
 variables x1(1);
 variables x2(1);

 maximize(T)
 subject to:
  log(a1) + x1 - log_sum_exp([0, x1]) >= T;
  log(a2) + x2 - log_sum_exp([0, x2]) + log(1 - exp(x1)) >= T;
  log(a3) +  log(1 - exp(x1)) +  log(1 - exp(x2)) >= T;
  x1 <= 0;
  x2 <= 0;
cvx_end
l1 = exp(x1); l2 = exp(x2);

要使用CVX，每个约束和目标函数必须使用CVX的规则集以可证明凸的方式编写。替换x1 = log(l1)和x2 = log(l2)允许人们这样做。请注意：log_sum_exp([0,x1]) = log(exp(0) + exp(x1)) = log(1 + l1)

这也会返回答案：l1 = .3383，l2 = .6180，T = -1.2800