Variable f coeffficient depends on x for linprog

조회 수: 20(최근 30일)
Ahmet Dogan
Ahmet Dogan 2022년 5월 4일
댓글: Torsten 2022년 5월 27일 19:24
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?
  댓글 수: 5
Torsten
Torsten 2022년 5월 27일 19:24
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.

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답변(2개)

Bruno Luong
Bruno Luong 2022년 5월 4일
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
  댓글 수: 2
Alan Weiss
Alan Weiss 2022년 5월 4일
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

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Torsten
Torsten 2022년 5월 4일
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)
  댓글 수: 4
Bruno Luong
Bruno Luong 2022년 5월 7일
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.

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