What is the meaning of this code line? th1(2:NB)=​inv(BM(2:N​B,2:NB))*P​1.';

clear all
tic
%                  branch no.|from |  to |              reactance               
linedata = [               1     1      2                0.20          70     1;
    2     1      4                0.20          100    1;
    3     1      5                0.30          100    1;
    4     2      3                0.25          30     1;
    5     2      4                0.10          90     1;
    6     2      5                0.30          40     1;
    7     2      6                0.20          40     1; 
    8     3      5                0.26          45     1;
    9     3      6                0.10          90     1;
    10    4      5                0.40          15     1;
    11    5      6                0.30          50     1];


P1=[0.5 0.6 -0.7 -0.7 -0.7];

SB=2; %Seller bus or source bus
BB=5; %Buyer bus or sink bus

P2=zeros(1,length(P1));

for i=1:length(P1)
    if i==SB-1
        P2(i)=P1(i)+0.1;
    elseif i==BB-1
        P2(i)=P1(i)-0.1;
    else
        P2(i)=P1(i);
    end
end

X=linedata(:,4);
L=linedata(:,5);
ls=linedata(:,6); % line status

for k=1:size(linedata,1)
    h(k)=linedata(k,2);       % h is a column vecror of 'FROM BUS'
    x(k)=linedata(k,3);       % x is a column vecror of 'TO BUS'
    LL(h(k),x(k))=L(k);
    
end
ss=[h x];
NB=max(unique(ss));

b=zeros(NB);
for k=1:size(linedata,1)
    if (ls(k)==1)
        
        b(h(k),x(k))=-inv(X(k));
        b(x(k),h(k))=-inv(X(k));
    else
        continue;
    end
end

for i=1:NB
    for j=1:NB
        if(i~=j)
            B(i,j)=-b(i,j); % CALCULATION OF  OFF-DIAGONAL ELEMENTS
        end
    end
end

BM=zeros(NB);   

for i=1:NB
    for j=1:NB
        if i~=j
            BM(i,i)=BM(i,i)+B(i,j); %CALCULATION OF DIAGONAL ELEMENTS
        else
            continue;
        end
    end
end

for i=1:NB
    for j=1:NB
        if(i~=j)  
            BM(i,j)=-B(i,j);
        end
    end
end
% calculation of the B-matrix completed

th1(2:NB)=inv(BM(2:NB,2:NB))*P1.'; 
th2(2:NB)=inv(BM(2:NB,2:NB))*P2.'; 

for k=1:size(linedata,1)
    PF1(h(k),x(k))=(th1(h(k))-th1(x(k)))*B(h(k),x(k))*100;
    PF2(h(k),x(k))=(th2(h(k))-th2(x(k)))*B(h(k),x(k))*100;
end

PTDF=(PF2-PF1)/10;
   
    tt=zeros(1,size(linedata,1));
    t1=zeros(1,size(linedata,1));
    t2=zeros(1,size(linedata,1));
    t3=zeros(1,size(linedata,1));
    
     for k=1:size(linedata,1)
       if (ls(k)==1)
           LL(h(k),x(k))=L(k);
       else
            LL(h(k),x(k))=0;
       end
     end
    
    for k=1:size(linedata,1)
        t1(k)=PTDF(h(k),x(k));
        t2(k)=PF1(h(k),x(k));
        t3(k)=LL(h(k),x(k));
    end
    
    for k=1:size(linedata,1)
    if t1(k)< 0
    tt(k)=(-t3(k)-t2(k))./t1(k);
    elseif t1(k)> 0
        tt(k)=(t3(k)-t2(k))./t1(k);
    end
    end
    
    
    %trial code to find overloaded line (not necessary for only ATC calculation)
    TL=zeros(5,6);
    for k=1:size(linedata,1)
        if PTDF(h(k),x(k))<0
            TL(h(k),x(k))=(-LL(h(k),x(k))-PF1(h(k),x(k)))./PTDF(h(k),x(k));
        elseif PTDF(h(k),x(k))>0
            TL(h(k),x(k))=(LL(h(k),x(k))-PF1(h(k),x(k)))./PTDF(h(k),x(k));
        end
    end
   %trial code completed (not necessary for only ATC calculation)      
        
   
   ATC=min(nonzeros(tt));
   fprintf('\nATC BETWEEN BUS %d(SELLER) AND BUS %d(BUYER) IS= %f MW (WITH PTDF-based method)\n',SB,BB,ATC);
   [sb,bb]=find(TL==ATC);
   fprintf('\nOVERLOADED LINE=%d-%d\n', sb,bb);
   toc        

Answer 1

[Sam Chak]

Hi @shravya boggulapally

I think the code means , which solves a system of linear equations, .

For more info, please check mldivide:

https://www.mathworks.com/help/matlab/ref/mldivide.html