How to solve "Conversion to logical from optim.prob​lemdef.Opt​imizationI​nequality is not possible." for logical statement for variable.

I have tried to find the optimal location (D) to install the renewables on the contact wire of the railway in which the lowest line losses are produced and lowest power from the grid is required (Pgrid). The objective function is to minimize sum(Pgrid). The error occured while calculated the contact wire loss (function: CW_calculation_1) due to logical operations of variable D.

Contact wire losses are calculated according to train location (left or right of D, train travels from left to right in this example).

xa(i) is train location at time i (Nx1vector), N is the number of time step. Ptr and Pres are parameters (Nx1 vector).

I have read all questions and answers in this community but still could not solve this problem. Is there any way to solve this issue? Please kindly advise. Thanks.

% Main optimization
% Decision variables
Pgrid = optimvar('Pgrid',N);
D = optimvar('D','LowerBound',0,'UpperBound',Dts);

% Minimize Contact wire loss
prob.ObjectiveSense = 'minimize';
prob.Objective = sum(Pgrid*DeltaT)

% Satisfy power load with power from PV, grid and battery
prob.Constraints.loadBalance = Pres + Pgrid == Ptr+CW(Ptr,Pres,Itr,Ires,xa,Dts,D,N,r);

Function CW below:

for i=1:1:N
  if  A>0  %case2.1
        if D-xa(i)>=0   % this creates error.
            %calculate losses while train is on the left-hand side of D
                B=optimexpr(N);
                B(i)=10*xa(i);
        elseif xa(i)-D>=0 
            %%calculate losses while train is on the right-hand side of D              
                C=optimexpr(N);
               C(i)=10*(xa(i)-D);            
        end
  elseif  A<0
       do something
  end
  CW=B+C;
end

Below is the error I encountered:

Conversion to logical from optim.problemdef.OptimizationInequality is not possible.

Error in CW_calculation_1 (line 30)

if D-xa(i)>=0

Answer 1

[Alan Weiss]

You need to take care of logical constraints using a roundabout sort of modeling construct. The documentation is here: Integer and Logical Modeling

An example here: Optimal Dispatch of Power Generators: Problem-Based

Good luck,

Alan Weiss

MATLAB mathematical toolbox documentation

[Nakaret Kano]

Hi Alan, Thank you for your answer. I have tried to model based on the Big-M formulation but I still get the same error. I used the following Express logical constraints:

My main code is:

prob = optimproblem;
M=27;
% Decision variables
Pgrid = optimvar('Pgrid',N);
Dres = optimvar('Dres','LowerBound',0,'UpperBound',Dts);

%Binary variables 
z = optimvar('z',N,'Type','integer','LowerBound',0,'UpperBound',1);
z1= optimvar('z1',N,'Type','integer','LowerBound',0,'UpperBound',1);

% Minimize Contact wire loss
prob.ObjectiveSense = 'minimize';
prob.Objective = sum(Pgrid*DeltaT)+sum(z+z1);

% Constraints for logical statements big-M
prob.Constraints.cons1 = Dres-xa <= M*z;
prob.Constraints.cons2 = Dres-xa >= -M*z1;
prob.Constraints.cons3 = z+z1 ==1 ;

% Satisfy power load + losses with power from RESs and grid
prob.Constraints.loadBalance = Pres + Pgrid == Ptr+CW_calculation_2_short(Ptr,Pres,Itr,Ires,xa,Dts,Dres,N,r,z,z1);

% Solve the linear program
options = optimoptions(prob.optimoptions,'Display','none');
[value,sol,fval] = solve(prob,'Options',options);

And code for function CW_calculation_2_short is:

function [Pcw_total]=CW_calculation_2_short(Ptr,Pres,Itr,Ires,xa,Dts,Dres,N,r,z,z1)

    %% Create Zeros vector for Contact wire for each scenario
    Pcw1=zeros(N,1); Pcw5=zeros(N,1);Pcw9=zeros(N,1);Pcw13=zeros(N,1);
    Pcw2=zeros(N,1);Pcw6=zeros(N,1);Pcw10=zeros(N,1);Pcw14=zeros(N,1);
    Pcw3=zeros(N,1);Pcw7=zeros(N,1);Pcw11=zeros(N,1);Pcw15=zeros(N,1);
    Pcw4=zeros(N,1);Pcw8=zeros(N,1);Pcw12=zeros(N,1);
    Pcw_total=zeros(N,1);

    %% Calculate Contact wire losses
for i=1:1:N
    
  if  Pres(i)>0 && Ptr(i)>0 && Pres(i)-Ptr(i)>0  %case2.1

            %calculate losses while train is on the left-hand side of Dres 
            if Dres-xa(i)>=0
     
                Pcw3=optimexpr(N);
                Pcw3(i)=Itr(i)^2*r*(Dres-xa(i));
                Pcw4=optimexpr(N);
                Pcw4(i)=(Ires(i)-Itr(i))^2*r*(Dts-Dres);

            %%calculate losses while train is on the right-hand side of Dres
           elseif Dres-xa(i)<=0
    
                Pcw5=optimexpr(N);
                Pcw5(i)=Ires(i)^2*r*(xa(i)-Dres);
                Pcw6=optimexpr(N);
                Pcw6(i)=(Ires(i)-Itr(i))^2*r*(Dts-xa(i));

            end
  end  

    Pcw_total=optimexpr(N);
    Pcw_total=Pcw3+Pcw4+Pcw5+Pcw6;
end

Please kindly advise on this issue. Thanks.

[Alan Weiss]

I have not gone through all your code in detail, but one line struck me as being inadmissible:

if  Pres(i)>0 && Ptr(i)>0 && Pres(i)-Ptr(i)>0  %case2.1

As I tried to explain briefly before, you cannot have this kind of line in your code. The strightforward test: if you are using an if statement with optimization variables, then you did it wrong.

It is difficult to give straightforward recipes for converting code from "if" statement form to "logical model" form. I tried to give the building blocks in Integer and Logical Modeling. But it is easy to get confused and not be sure how to proceed.

You need to create binary variables tied to your real optimization variables, and use the binary variables to encapsulate the "if" constraints.

You have a few other lines of code that are not sensible. For example, in a loop on i you have

Pcw4=optimexpr(N);
Pcw4(i)=(Ires(i)-Itr(i))^2*r*(Dts-Dres);

The line Pcw4=optimexpr(N) zeros out the Pcw4 variable, and so every time through the loop this command erases whatever came before. This leads to another rule of thumb: if you have an optimexpr line in a loop, you did it wrong. Use optimexpr the same way you would use zeros: to initialize a variable to zeros of the right size before running a loop, not to zero out variables in the middle of a loop.

Good luck,

Alan Weiss

MATLAB mathematical toolbox documentation