linprog optimization of two functions

Hi community,

I am trying to minimize a two stage function in which first value is computed on the bases of X & D and this values is thereafter used in a new minimization function. I think I almost have the correct code, but some errors keep popping up. or should I solve these functions in 2 seperate scripts?Does anyone know how to solve them?

%overall objective function = Min c.'*x+E[Q(x,D)]
%set of parameters
i=1
j=1
c = 3; %s<c
l = 0.25;
q = 6; 
s = 1; 
A1 = 1;
D= 60;
%d = 65; not used in estimation
x= D.'*A1;
%E[Q(x,D)= (l-q).'*z-s.'
%Q(x,D)=(l-q).'*z(*)-s.'*y(*);
%s.t x=y(*)+A1.'*z(*);

%to find value for E[Q(x,D)]
f2 = [-s.',(l-q).']; %[ Y, Z]
E=sum(f2)%
Aeq = [eye(j),A1.'];
beq = x;
lb = [zeros(1,i+j)]; 
ub = [inf(1,j), D(1)];
sol = linprog(f2,[],[],Aeq,beq,lb,ub);
y = sol(1);     
z = sol(2) ;    


%Now objective function is to minimize c.'*x+E
f1= [c.'+E];%or f1=[c.',E] how to define E in the f1 function? outcome of linprog f1 should give a value of x
Aeq = [];%no equality or inequality constraints, only requirement is for x>0, or should constraint of f2 also be added??
beq = [];
lb = [0];%x>0
ub = [Inf]; 
sol = linprog(f1,[],[],Aeq,beq,lb,ub);
x = sol(1) 

I get no solution for x as matlab says the code is unbouded. However, my only constraint for f1 is that x>0, for f2 there is a constraint (x=y+A1.'*z)

hope that anyone knows what I am doing wrong.

Thankyou!

[Torsten]

f1 is negative. Thus x=+Inf minimizes f1.

By the way: min: c'*x and min: c'+x+E have the same vector x as solution.

[bus14]

the value for f2 is indeed negative. However, values of x,c and f1 should be positive. Would increasing the value of c help for x to take a more realistic value?

By the way: min: c'*x and min: c'+x+E have the same vector x as solution.

With this comment, do you mean that setting f1=[ c.'+E] is correct?

[bus14]

problem for me is, if the f1 was set correctly and how I can find a satisfactory value for x which is larger than zero and as upperbound is not much higher than the value of D

[Matt J]

We cannot tell you what the problem should be. Only the solution. There is nothing incorrect about the solution you have obtained, given the problem you've provided.