Objective problem

[student1]

Dear professor,

I have a problem and I don't know what to do next.

Here I define two variables:

onoff = binvar(1,size(DatenNeu,1),'full');
onoff = sdpvar(1,size(ZielVec,1),1,'full');

than I give constraints

Constraints = [];
for .....
...

than

objective = sum(abs(ZielVec-(Approxy/max(Approxy)))./ZielVec,1);

So this objective doesn't work. I will that objective totalize(sum) the ( ZielVec - (Approxy/the max. value of Approxy)) and this for every Point on Zielvec " ./ZielVec,1 "

But when I use "/max(Approxy)" it doesn't work.

I would be grateful to get your help.

[Johan Löfberg]

First, I guess you mean Approxy./max(Approxy)

If Approxy is a decision variable, you have a very very nasty nonlinear nonconvex objective

[Johan Löfberg]

skip my comment about the elementwise division, obviously not needed

Anyway, horrible objective, and you need to come up with a better objective, or a better formulation

[student1]

I'm sorry, the first two code lines must be=

onoff = binvar(1,size(DatenNeu,1),'full');

Approxy = sdpvar(1,size(ZielVec,1),1,'full');

..

now if I use the objective like this=

objective = sum(abs(ZielVec-Approxy)./ZielVec,1);

it works. But I need Approxy/max(Approxy)

[Johan Löfberg]

that problem is eons easier, at it is a simple LP objective.

If you don't trust me that the objective is horrible, just look at random instances in 2D

for i = 1:10

    y = randn(2,1);

    [X1,X2] = meshgrid(.1:0.011:2,.1:0.011:2);

    for i = 1:length(X1)

     for j = 1:length(X2)

         x = [X1(i,j);X2(i,j)];

         J(i,j) = sum(abs((y - x/max(x))./y));

     end

    end

    mesh(J)

pause(1)

end

[Johan Löfberg]

and 'full' is redundant for vectors (as it cannot be symmetric unless it's 1x1, where is both full and symmetric)

[Johan Löfberg]

Also note that the complexity of this problem depends on how Approxy enters other parts of the model, and what you know about the variable

Effectively, what you have written so far is a model of the following type

x = sdpvar(n,1);
optimize([],sum(abs((y - x/max(x))./y)))

If you know x>=0, this model is equivalent to solving

q = sdpvar(n,1);
optimize([0<=q<=1],sum(abs((y - q)./y)))
x = value(q)*anypositivenumber

etc