How to use optimization variable in if condition and how to multiple an optimization variable with an optim. expression?

Hello everybody, I'm trying to solve a unit commitment problem using intlinprog with problem-based optimization method. I have two problems. 1) I need to multiply an optimization variable that shows the commitment status of any unit with an optimization expression that shows the time off and I'm getting an error: "Error using optim.internal.problemdef.Times.getTimesOperator At least one argument must be numeric."

ans = state(i,j).*t_off(i,j)

Is there any way to multiply an optimization variable with an optimization expression?

2) The most serious problem I have encountered when I'm trying to create a logical operator in an if, using the optimization expression toff, and I'm getting an error: "Conversion to logical from optim.problemdef.OptimizationConstraint is not possible" . What I'm trying to do for example :

if toff<10
   statement1
end
if toff>=10
   statement2
end

Does anyone have any ideas that could help??? Thanks in advance

Answer 1

[Mary Fenelon]

Multiplying two optimization variables results in a quadratic expression and these aren't allowed in a mixed-integer linear program.

There is an example of a unit commitment problem here that might give you some ideas on how to formulate the constraint you need without multiplying variables. There is a more complex version of a unit commitment problem here .

Look here for a way to model the logical constraint.

[Alex Alex]

Thanks a lot for your answer it was really helpful. To be a little more specific after changing my model in order to avoid multiplying an optimization variable with an optimization expression I came up with those 2 following functions. With an optimization variable state(5,48)

I had in mind to use a code like (apart from the first column which would be filled manually)

for i=1:5
  for j=1:48
     ...
  end
end

What i tried to do before posting was:

for i=1:5
    if state(i,1)==1
        toff(i,1)=0
    end
    if state(i,1)==0
        toff(i,1)=tstart(i)+1
    end
end

for i=1:5
    for j=2:48
        if state(i,j)==1
        toff(i,j)=0
        end
    end
    if state(i,j)==0
        toff(i,j)=toff(i,j-1)+1
    end
end

And the second one:

The difference from your example is the use of the recursive method in the i,j loop and not finding (0,1) values for the optimization expression. Could this be modeled in the same way?? Thanks a lot for your time and support.

[Mary Fenelon]

Yes, you can specify linear constraints to model those logical constraints using the same pattern x - My <= 0. For the first:

• t_i_j - M1*(1-state_i_j) <= 0 where 0 <= t_i_j <= M1
• t_i_j - toff_i_j-1 - 1 - M2*state_i_j <= 0 where 0 <= t_i_j - t_i_j-1 - 1 <= M2

For the second, you need to add a binary variable y that is 1 when t > 10:

• (t - 10) - M3*y <= 0 where 0 <= t - 10 <= M3

Use y to compute the value of C.

[Sanyam Maheshwari]

I am also facing the issue with my optimization problem.

global SL V p r k m n C Z s 
SL = [0.75 0.75 0.75 0.75];
V = 94100;
p = [0.07,0.18,0.2,0.3];
r = [55 55 55 55;
     47 47 47 47;
     45 45 45 45;
     49 49 49 49];
k = [33 33 33 33;
     28 28 28 28;
     29 29 29 29;
     30 30 30 30];
m = 4;
n= 4;
C = [78,69,70,73;
    64,68,56,59;
    34,39,42,41;
    52,47,48,45];
Z =[250	250	250	250;
    320	320	320	320;
    440	440	440	440;
    350	350	350	350];
s = [110,95,99,100];
global SL V r k m n C Z s 
% Generating Variables
Q = optimvar('Q',m,n,'LowerBound',0,'UpperBound',Z);
b = optimvar('b',m,n,'Type','integer','LowerBound',0,'UpperBound',1);
rng
y = rand(4);
x = sym('x',[4,1]);
% limit of integration
l = y*Q; 
q = sum(l(:));
% Constraints
% Budget Constraint
B = C*b*y*Q;
budget = sum(B(:)) <= V;
% normal constraint
normal = int(normpdf(x, 400, 100),q,Inf) <=SL;
% each product connects to exactly one supplier
con4 = sum(b,1) == ones(m,1)';

% Objective Function:-
% optimization problem
demandprob = optimproblem;
% Revenue 
revenue = sum(s*b*x,1);
% Cost
cost = sum(sum(C*b*y*Q,1),2);
%Salvage
salvage = sum(b*r*(sum(sum(y*Q - x,1),2)),1);
% Revenue when demand is more
revenue2 = sum(s*b*y*Q,2);
% Salvage when demand is more
salvage2 = sum(b*k*(x - sum(sum(y*Q,1),2)),1);
% The objective function to maximize the below Expected Profit
demandprob.Objective = int(((revenue - cost + salvage).*normpdf(x, 400, 100)),0,q); + int(((revenue2 - cost - salvage2).*normpdf(x, 400, 100)),q,Inf);
% Include the constraints in the problem.
demandprob.Constraints.budget = budget;
demandprob.Constraints.normal = normal;
demandprob.Constraints.con4 = con4;

opts = optimoptions('intlinprog','Display','off','PlotFcn',@optimplotmilp);
% Call the solver to find the solution.
[sol,fval,exitflag,output] = solve(demandprob,'options',opts);

Also I want to maximize this optimization problem

What should I do further.

please help