Variable f coeffficient depends on x for linprog

I have a minimization problem and I wonder that if i can solve the problem with linprog or how can i solve within matlab framework?

E.g. f= [f(1) f(2)]; x = linprog(f,A,b,Aeq,beq);

In that problem i have variable f coefficient depend on x value. f(1)=0, if x(1)<0 / f(1)=50, if x(1)>0 .

How can i solve this minimization problem?

[John D'Errico]

Moved an answer to a comment, from @Ahmet Dogan

"Thank you for your responses but i cannot solve the problems with your suggestions. I am still searching about that. Maybe, I have to change the problem definition. "

I would say this just means you probably do not understand how to use the solutions posed, not that they do not solve your problem.

So if your problem is not what you stated, then explain why it is not, HERE. As a comment.

[Ahmet Dogan]

Yes,Maybe i did not understand. I have an idea for solution but i need your help.

x(1),x(2),...,x(36)>=0 fpoz=[fpoz1,fpoz2,...,fpoz36] lb=0, ub=250

y(1),y(2),...,y(36)=<0 fneg=[0,0,...,0] lb=-250, ub=0

Fcost=x(1)*fpoz1+...+x(36)*fpoz36

In this solution, at least one of x or y should equal to zero; x(1)*y(1)=0...x(36)*y(36)=0 I dont know how to do this equality constraints. Can you please help for that?

Inequality (add also y)

x(1)+x(2)+ y(1)+y(2)>250

x(1)+x(2)+x(3)+ y(1)+y(2)+y(3)>250

....

x(1)+x(2)+x(3)+ y(1)+y(2)+y(3)+...x(35)+y(36)>250

x(1)+x(2)+ y(1)+y(2)<1000

x(1)+x(2)+x(3)+ y(1)+y(2)+y(3)<1000

.....

x(1)+x(2)+x(3)+...+ y(1)+y(2)+y(3)+...x(35)+y(35)<1000

Equality constraints x(1)+x(2)+x(3)+y(1)+y(2)+y(3)+...x(36)+y(36)=1000

[Torsten]

Set

x(37) = y(1), x(38) = y(2) ,...,x(72) = y(36)

Then your "new" problem should read:

f_con = [0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.27 0.27 0.27 0.29 0.29 0.29 0.29 0.30 0.30 0.30 0.30 0.30 0.30 0.30 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.27 0.27];
f = [zeros(1,36),f_con];

x(1)+x(2) >= 250
x(1)+x(2)+x(3) >= 250
....
x(1)+x(2)+x(3)+...+x(35)>=250
    
x(1)+x(2) <= 1000
x(1)+x(2)+x(3) <= 1000
.....
x(1)+x(2)+x(3)+...+x(35) <= 1000
    
x(1)-x(37) <= 0
x(2)-x(38) <= 0
...
x(36)-x(72) <= 0

x(1)+x(2)+x(3)+...+x(36) = 1000
    
lb = [-250*ones(1,36),zeros(1,36)];
ub = [250*ones(1,36),Inf(1,36)];

[Ahmet Dogan]

Thank you for your helps in advance. I just solved the problem using nonlinear inequality constraints as follows;

https://www.mathworks.com/help/optim/ug/nonlinear-inequality-constraints.html

x(1)+x(2) >= 250

x(1)+x(2)+x(3) >= 250

....

x(1)+x(2)+x(3)+...+x(71)>=250

x(1)+x(2) <= 1000

x(1)+x(2)+x(3) <= 1000

.....

x(1)+x(2)+x(3)+...+x(71) <= 1000

x(1)+x(2)+x(3)+y(1)+y(2)+y(3)+...x(71)+y(72)=1000

Following constarints provide that only positive values are added to cost function while positive and negative values can be used for constraints.

x(1)*x(37)=0; at less one of the variables should be zero.

....

x(36)*x(72)=0;

lb=[0] ub=[250] for x(1)....x(36)

lb=[-250] ub=[0] for x(37)....x(72)

c = [x(1)*x(37);

(x(2)*x(38);

. . .

x(36)*x(72);

];

f_con = [0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.27 0.27 0.27 0.29 0.29 0.29 0.29 0.30 0.30 0.30 0.30 0.30 0.30 0.30 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.27 0.27];

fobj= x(1)*f_con(1)+ ...+x(36)*f_con(36)

[Torsten]

Sorry, but I don't see the equivalence of your original problem and the one you write above.

Even if they were: I don't understand why you choose such a complicated non-linear reformulation. The linear approach I suggested is simple and elegant if your f-vector only has non-negative entries.

Answer 1

[Bruno Luong]

Solve 2 problems

f= [0 f2]; x = linprog(f,A,b,Aeq,beq,lb1,up1); with lb1 = -inf(size(x)), up1 = zeros(size(x));

g= [50 f2]; x = linprog(g,A,b,Aeq,beq,lb2,up2); with lb2 = zeros(size(x)), up2 = inf(size(x));

Then pick the best solution

[Ahmet Dogan]

Thank you for the answer but I had simplified the question. In my main problem, I have 36 x variables with equality and inequality constraints.

Some of my inequality constraints

x(1)+x(2)>250

x(1)+x(2)+x(3)>250

....

x(1)+x(2)+x(3)+...x(35)>250

x(1)+x(2)<1000

x(1)+x(2)+x(3)<1000

.....

x(1)+x(2)+x(3)+...x(35)<1000

Equality constraints

x(1)+x(2)+x(3)+...x(36)=1000

lb=-250

ub=250

f_con=[0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.27 0.27 0.27 0.29 0.29 0.29 0.29 0.30 0.30 0.30 0.30 0.30 0.30 0.30 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.27 0.27];

f(i)=f_con(i) if x(i)>0

f(i)=0 if x(i)<0

According to that some of x variables should be less than zero and some of them should be greter than zero. So, I think I need a combined solution.

[Alan Weiss]

I think that you can do what you want using binary indicator variables for each condition as described in Integer and Logical Modeling. This involves creating new binary variables and new constraints based on the Big-M formulation, and then using intlinprog to solve the resulting problem.

Alan Weiss

MATLAB mathematical toolbox documentation

Answer 2

[Torsten]

According to that some of x variables should be less than zero and some of them should be greter than zero. So, I think I need a combined solution.

Add new variables

y(i) (i=1,...,36)

and constraints

y(i) >= x(i)
y(i) >= 0

and define the objective as

f_con*y

For the interpretation:

y(i) = max(x(i),0)

[Bruno Luong]

+1 this is probably the simplest reformulation. Simply add lower bound of 0 to x.

[Bruno Luong]

Actually thinking more. Torsten reformulation is not equivalent in general. Assuming that

f = (1,-1)
Aeq = (1,1)
beq = 1
A = []
b = []

Torsten solution is x=(0,1), but the original problem is unbounded: x=(x1,-x1+1) with x1 arbitrary small (x1 = -inf).

[Torsten]

My reformulation would be:

min: y1 - y2
x1 + x2 = 1
y1 >= x1
y2 >= x2
y1 >= 0 
y2 >= 0

This problem is also unbounded :

y1 = 0, y2 = +Inf, x1 = 0, x2 = 1

But I didn't think in depth if this reformulation is always correct if f has negative components.

Maybe one can find a counterexample.

[Bruno Luong]

I guess you get an idea, if I change A and b to

A = (-1,0)
b = 10;

(x >= -10 if you will) and keep the rest identical,

now your problem is unbounded and the original problem is not. Fact is the halfspace { x1 <= 0 } won't be capture by you refomulation by working on other half-space halfspace { x1 >= 0 } alone.