formulating the objective function for MILP

I have an optimization problem as follows:

Min ((C_1*x_1)+(C_2*x_2)+(C_3*x_3)+(C_4*min(X))).

s.t "some constraints which I don't think they are relevant to my question at this point, All linear of course"

x_1, x_2, and x_3, as well as all "C"s, are scalars

"X" in that last term is a vector. I want to get the minimum value of that vector into the objective function but I couldn't work it out

Any suggestion?

Abdullah

Answer 1

[Alan Weiss]

Perhaps you can try solving three different problems, one with C_4 multiplying x_1, one multiplying x_2, and one multiplying x_3. Then take the appropriate solution, the solution that has the multiplier of C_4 as the minimum.

Of course, this might not work, but if it does then I think you are done.

Alan Weiss

MATLAB mathematical toolbox documentation

[Abdullah Almehizia]

I think my question was not introduced right

"x"s are the variables and "C"s are just coefficients

So I don't see the point of multiplying C_4 with the other variables?! By the way, the vector X is defined as follows

X=[x_4;x_5;...;x_8760] I know this is a very large matrix, but I couldn't formulate my problem any other way

[Alan Weiss]

I am not sure that I understand you even now, because you say that this is an MILP, but I see no integer constraints.

Even so, I suppose that you could try using fmincon, but it might not work because the objective is not smooth. I suppose that the constraints that you do not mention keep the problem in a bounded region.

Good luck,

Alan Weiss

MATLAB mathematical toolbox documentation

[Abdullah Almehizia]

My god I'm so bad at writing questions

x_1, x_2 and x_3 are integers and x_4,...,xn are continuous variables

Thanks Alan for your constant response

Answer 2

[Matt J]

If C_4<0, the problem can be reformulated linearly as follows

Min ((C_1*x_1)+(C_2*x_2)+(C_3*x_3)+(C_4*z)).

s.t. all your original constraints and

z<=x_4
z<=x_5
   ...
z<=x_8760

As a result of the above constraints, z<=min(X). If C_4<0, this must be satisfied with equality at the optimum, because the optimizer must try to push z as close as it can get to +inf.

If C_4>0, then the solution lies at min(X)=-inf unless you have constraints in place to prevent that. We would have to see those constraints to decide how to modify the above solution.

[Abdullah Almehizia]

%
tf=5e3;
L="an 8760 by 1 vector" 
C_tank=300;
vec_tank=repmat(C_tank,[N*24,1]); 
slack_coe_zeros=zeros((N*24)-1,3);
slack_coe=zeros((N*24)-1,N*24);                                                      % Grid energy coeficients
m=2;
for i=1:1:(N*24)-1
  for j=1:1:m
  slack_coe(i,j)=1;
  end
  m=m+1;
  end
slack=[slack_coe_zeros,slack_coe];
coe=eye(N*24);
f_5=[C_pv; C_wt; C_csp; vec_tank]; 
a_9=slack;          b_9=zeros((N*24)-1,1);
a_10=[P_pv_year/tf, P_wt_year/tf, P_csp_year/tf, coe];              b_10=L; 
f=f_5; 
intcon=[1,2,3];                                                            % Variable type
A=a_9; b=b_9;                                                              % Inequality constraint
Aeq=a_10;  beq=b_10;                                                       % Equality constraint
lb=[zeros(3,1);repmat(-inf,[N*24,1])];                                     % Lower bounds
ub=inf((N*24)+3,1);                                                        % Upper bounds
[x, fval]=intlinprog(f,intcon,A,b,Aeq,beq,lb,ub);

This is the code I'm trying to execute. I'm not quite sure if it is easy to read or not. Of course, this is not what I want to do exactly. As previously mentioned I want the fourth term of the objective function "f_5" to take the minimum value from the whole vector

[Matt J]

I can't tell from the code how C_4 would be defined in the actual problem.

[Abdullah Almehizia]

C_4=C_tank

Answer 3

[Matt J]

You could use ga() to solve the problem instead of trying to pose it as an MILP, but you would have to eliminate the equality constraints, since ga cannot simultaneously process equality and integer constraints.

You can eliminate the equality constraints from the problem by using them to solve for some of the variables in terms of the others. If you have the patience to read through a really long thread on this, see here

http://www.mathworks.com/matlabcentral/answers/104739#answer_114013

But it's basically very simple. The equalities

     Aeq*x=beq

should have Aeq full row rank, otherwise some of your equality constraints are redundant. Since Aeq is full rank, the equalties can be partitioned, possibility after re-ordering the variables, as

     [P,Q]*[xp;xq]=beq

where P is a square nonsingular sub-matrix and x=[xp;xq] is a corresponding partition of the unknowns x. Then the sub-partition xp can be written in terms of xq as

    xp=P\(beq-Q*xq)

Thus, you can eliminate the variables xp by using the above substitution formula to replace them with an expression in xq everywhere in your problem. This leads to a reformulation of the entire problem in terms of xq only and no equality constraints (just inequalities).