Optimization: Unable to perform assignment because value of type 'optim.pro​blemdef.Op​timization​Expression​' is not convertible to 'double'.

Hello,

I would like to perform a multiplication with a decision variable Yi. When doing so, I get the following error:

Unable to perform assignment because value of type 'optim.problemdef.OptimizationExpression' is not convertible to 'double'.

What's the Problem and how could I convert Yi to an double?

The code is:

Yi = optimvar('Yi', ns, 1,'type','integer','LowerBound',0,'UpperBound',1);
Dij = csvread("test_distanzmatrix_20.csv",1,1);
ns = size(Dij,1);
[...]
    Yi_transponiert = transpose(Yi);
    YD_1 = zeros(ns);
    YD_2 = zeros(ns);
    for i = 1:ns
    YD_1(:,i) = Yi(:) .* Dij(:,i);  
    end 
    for j = 1:ns
        YD_2(j,:) = Yi_transponiert(:) .* Dij(j,:); 
    end
    LP_Distanz = YD_1 .* YD_2; 
    for i = 1:ns
        LP_Distanz (i,i) = 0; 
    end 
    logistikpunktsuche.Constraints.cons9 = LP_Distanz >= D^2; 

Answer 1

[Walter Roberson]

https://www.mathworks.com/help/optim/ug/optimexpr.html

Do not preallocate YD_1 as zeros, use optimexpr()

[Torsten]

Yi_transponiert(:) is always a column vector, Dij(j,:) is a row vector.

The elementwise product of a column and a row vector gives a matrix by implicit expansion.

You can't assign a matrix to YD_2(j,:) which is a single row.

[Walter Roberson]

What is the error message?

[Walter Roberson]

Your array sizes are confused. And your code only works if Dij is a square matrix (well, or if the number of columns exceeds the number of rows maybe it would not fail.)

I cannot tell what sizes you are intending... and I am left wondering whether perhaps you should just be calculating something like YD_1 = Yi .* Dij

    %Dij = csvread("test_distanzmatrix_20.csv",1,1);
    Dij = rand(5, 5);
    
    ns = size(Dij,1);
    
    Yi = optimvar('Yi', ns, 'type', 'integer', 'LowerBound', 0, 'UpperBound', 1);

    Yi_transponiert = transpose(Yi);
    YD_1 = optimexpr([ns,1]);
    YD_2 = optimexpr([1,ns]);
    for i = 1:ns
        YD_1(:,i) = Yi(:) .* Dij(:,i);  
    end 
    for j = 1:ns
        YD_2(j,:) = Yi_transponiert .* Dij(j,:); 
    end
    LP_Distanz = YD_1 .* YD_2; 
    for i = 1:ns
        LP_Distanz (i,i) = 0; 
    end 
    

[Walter Roberson]

    Yi = optimvar('Yi', ns, 'type', 'integer', 'LowerBound', 0, 'UpperBound', 1);

That generates a column vector

    Yi_transponiert = transpose(Yi);

That transposes it to be a row vector.

    YD_2(j,:) = Yi_transponiert(:) .* Dij(j,:); 

That (:) reshapes the row vector to be a column vector. When you use (:) then no matter what size the input was, the output will be a column vector. Which is a problem if you are multiplying by a row vector.

[Torsten]

A(:) will never be a row vector, independent of what A is.

A = [1 2 3]

A = 1×3

1 2 3

A(:)

ans = 3×1

1 2 3

A = [1;2;3]

A = 3×1

1 2 3

A(:)

ans = 3×1

1 2 3

[Walter Roberson]

The code I posted at https://www.mathworks.com/matlabcentral/answers/1890467-optimization-unable-to-perform-assignment-because-value-of-type-optim-problemdef-optimizationexpre#comment_2552437 already has the correction for both of those problems.

[Laura Grönniger]

I stand corrected... Any factor other than 0 should be greater than D. Can this be represented? So: If not 0, then greater than D.

[Walter Roberson]

I do not understand what you are requesting about facters other than 0 should be greater than D ??

Are you looking at

logistikpunktsuche.Constraints.cons9 = LP_Distanz >= D^2;

and saying that the real constraint is that the value is permitted to be 0 if some corresponding entry is 0, otherwise has to be >= D^2 ?

If so then I am not clear as to which value to refer to for the "factor" that is permitted to be 0 ?

If you had an array of factors the same size as LD_Distanz then

logistikpunktsuche.Constraints.cons9 = (FactorArray == 0) | (LP_Distanz >= D^2);

Answer 2

[Matt J]

Here's a way to express the constraints linearly:

    Dij=rand(5); D=1;  %hypothetical input data

    ns = size(Dij,1);
    
    Yi = optimvar('Yi', ns, 'type', 'integer', 'LowerBound', 0, 'UpperBound', 1);
    M  = optimvar('M', [ns,ns], 'type', 'integer', 'LowerBound', 0, 'UpperBound', 1);

    YY=Yi*ones(1,ns);
    
    logistikpunktsuche.Constraints.Mupper= M<=(YY+YY.')/2;%Mupper and Mlower together
    logistikpunktsuche.Constraints.Mlower= M>=(YY+YY.')-1;%equivalent to M==Yi&Yi.'
    
    logistikpunktsuche.Constraints.LP_Distanz =  Dij.*(1-eye(ns)).*M >= D^2.*M

logistikpunktsuche = struct with fields:
    Constraints: [1×1 struct]