Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side is 0-by-0

clc;
cp=4186;
Tground=15;
node_i=xlsread('Energy Networks.xlsx','Branches','B6:B35');
node_o=xlsread('Energy Networks.xlsx','Branches','C6:C35');
P=xlsread('Energy Networks.xlsx','Foglio1','O36:O64');
G=xlsread('Energy Networks.xlsx','Foglio1','N36:N65');
L=xlsread('Energy Networks.xlsx','Branches','E6:E35');
nNodes=size(P,1);
nBranches=size(G,1)
Tsupply=95;
TInletCondition=zeros(nNodes,1);
TInletCondition(1)=Tsupply;
TOutletCondition=zeros(nNodes,1);
nBoundaryNodes=0; %initialization
U=xlsread('Energy Networks.xlsx','Branches','M6:M35');
Perimeter=xlsread('Energy Networks.xlsx','Branches','N6:N35');
K=zeros(nNodes,nNodes);
f=zeros(nNodes,1);
A=zeros(nNodes,nBranches);
for i=1:nBranches
    in=node_i(i,1);
    out=node_o(i,1);
    A(in,i)=1;
    A(out,i)=-1;
end

for j=1:nBranches
    if G(j)<0
        A(:,j)=-A(:,j);
    end
end



% assembling of K and f


for j=1:nBranches
    iInlet=find(A(1:nNodes,j)==1);
    iOutlet=find(A(1:nNodes,j)==-1); 
    K(iOutlet,iInlet)=-G(j)*cp;
end
for i=1:nNodes
    jInlet=find(A(i,1:nBranches)==1);
    jOutlet=find(A(i,1:nBranches)==-1);
    if isempty(jOutlet)==0 && isempty(jInlet)==0
        jTotal=[jInlet jOutlet];
        K(i,i)=sum(G(jOutlet)*cp+0.5*sum(L(jTotal).*Perimeter(jTotal).*U(jTotal)));
    end
    for j=1:nBranches
        f(i)=f(i)+0.5*abs(A(i,j)).*L(j).*Perimeter(j).*U(j).*Tground;
    end
end

%Boundary conditions
for i=1:nNodes
    counter=0;
    summation=0;
    for j=1:nBranches
        if A(i,j)~=0
            counter=counter+1;
        end
    end
    summation=sum(A(i,:));
    if (counter==1) && (summation==-1)
     TOutletCondition(i)=1;
    end
end    
for i=1:nNodes
  if TInletCondition(i)>0
     K(i,:)=0;
     K(i,i)=1;
     f(i)=TInletCondition(i);
  end
   if TOutletCondition(i)>0
       joutletBranch=find(A(1,1:nBranches)==-1);
       iPreviousnode=find(A(1:nNodes,joutletBranch)==1);
       K(i,iPreviousnode)=1-(L(joutletBranch).*U(joutletBranch).*Perimeter(joutletBranch))/(2.*cp.*G(joutletBranch));
       K(i,i)=-(1+(L(joutletBranch).*U(joutletBranch).*Perimeter(joutletBranch))/(2.*cp.*G(joutletBranch)));
   end
end

Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side is 0-by-0.

Error in ThermalProblem (line 80)

K(i,i)=-(1+(L(joutletBranch).*U(joutletBranch).*Perimeter(joutletBranch))/(2.*cp.*G(joutletBranch)));

Help me please

Answer 1

[Walter Roberson]

       joutletBranch=find(A(1,1:nBranches)==-1);

What happens if there are no matches? Then joutletBranch would be empty, and you would be computing emptiness, and trying to assign it to the definite scalar position K(i,i)

What happens if there are more than one match? Then you have an / operator between two vectors, and the result is going to depend on the orientation of the vector. row vector / row vector of the same length returns a scalar. column vector / column vector of the same length returns a square matrix. So we need to know the orientation of L(joutletBranch)

MATLAB had an oddity that when you index a vector by a vector, then no matter what the orientation is of the index, the result will have the same orientation of the vector being indexed. So regardless of whether the find() is returning a row or column vector, we need to know the orientation of L to know what shape L(joutletBranch) would have.

L=xlsread('Energy Networks.xlsx','Branches','E6:E35');

and that is a column vector. So L(joutletBranch) is going to return a column vector whenever jOutputBranch is a vector with more than one element, and as mentioned above, column vector / column vector is going to result in a square matrix, and that will not fit into the scalar destination.

[Salvatore Calabrese]

thankyou so much!!

joutletBranch=find(A(1,1:nBranches)==-1);

in this line I had to write i instead of 1 like this

joutletBranch=find(A(i,1:nBranches)==-1);

now my code is running whitout errors!