Index exceeds matrix dimensions.

clc;
clear all;
format short;
tic
% bus no|activepower kW | reactivepower kVar
m=[ 1 0 0 0
 2 100 60 0
 3 90 40 0 
 4 120 80 0 
 5 60 30 0 
 6 -2522.6622 20 0 
 7 200 100 0
 8 200 100 0 
 9 60 20 0 
 10 60 20 0
 11 45 30 0
 12 60 35 0 
 13 60 35 0 
 14 120 80 0 
 15 60 10 0 
 16 60 20 0 
 17 60 20 0 
 18 90 40 0 
 19 90 40 0 
 20 90 40 0 
 21 90 40 0
 22 90 40 0 
 23 90 50 0
 24 420 200 0
 25 420 200 0
 26 60 25 0 
 27 60 25 0 
 28 60 20 0
 29 120 70 0
 30 200 600 0
 31 150 70 0 
 32 210 100 0
 33 60 40 0
 ]; 
 
 %branchno. INbus OUTbus Resistance(ohm) Reactance(ohm)
 l=[ 1 1 2 0.0922 0.0470 
 2 2 3 0.4930 0.2511 
 3 3 4 0.3660 0.1864 
4 4 5 0.3811 0.1941 
 4 5 6 0.8190 0.7070 
 6 6 7 0.1872 0.6188 
 7 7 8 0.7114 0.2351 
 8 8 9 1.0300 0.7400 
 9 9 10 1.0440 0.7400 
 10 10 11 0.1966 0.0650 
 11 11 12 0.3744 0.1238 
 12 12 13 1.4680 1.1550 
 13 13 14 0.5416 0.7129 
 14 14 15 0.5910 0.5260 
 15 15 16 0.7463 0.5450 
 16 16 17 1.2890 1.7210 
 17 17 18 0.7320 0.5740 
 18 2 19 0.1640 0.1565 
 19 19 20 1.5042 1.3554 
 20 20 21 0.4095 0.4784 
 21 21 22 0.7089 0.9373 
 22 3 23 0.4512 0.3083 
 23 23 24 0.8980 0.7091 
 24 24 25 0.8960 0.7011 
 25 6 26 0.2030 0.1034 
 26 26 27 0.2842 0.1447 
 27 27 28 1.0590 0.9337 
 28 28 29 0.8042 0.7006 
 29 29 30 0.5075 0.2585 
 30 30 31 0.9744 0.9630 
 31 31 32 0.3105 0.3619 
 32 32 33 0.3410 0.5302
 33 8  21 2.0000 2.0000
 34 9  15 2.0000 2.0000
 35 12 22 2.0000 2.0000
 36 18 33 0.5000 0.5000
 37 25 29 0.5000 0.5000] ; 
br=length(l); %data branch
no=length(m); %data bus
MVAb=100; %MVA base
KVb=12.66; %KV base
Zb=(KVb^2)/MVAb;
LF=1; %load factor
PF=1; %power factor
% Per unit Values
for i=1:br
 R(i,1)=(l(i,4))/Zb;
 X(i,1)=(l(i,5))/Zb;
end
for i=1:no
 P(i,1)=(LF*(m(i,2))/(1000*MVAb));
 Q(i,1)=(LF*(m(i,3))/(1000*MVAb));
end
% draw bus and line data to a distribution system
R

R = 37×1

0.0575 0.3076 0.2284 0.2378 0.5110 0.1168 0.4439 0.6426 0.6514 0.1227

X

X = 37×1

0.0293 0.1567 0.1163 0.1211 0.4411 0.3861 0.1467 0.4617 0.4617 0.0406

P

P = 33×1

0 0.0010 0.0009 0.0012 0.0006 -0.0252 0.0020 0.0020 0.0006 0.0006

Q

Q = 33×1

1.0e+00 * 0 0.0006 0.0004 0.0008 0.0003 0.0002 0.0010 0.0010 0.0002 0.0002

C=zeros(br,no);
for i=1:br
 a=l(i,2);
 b=l(i,3);
 for j=1:no
 if a==j
 C(i,j)=-1;
 end
 if b==j
 C(i,j)=1;
 end
 end
end
C

C = 37×33

-1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

e=1;
for i=1:no
 d=0;
 for j=1:br
 if C(j,i)==-1
 d=1;
 end
 end
 if d==0
 endnode(e,1)=i;
 e=e+1;
 end
end
endnode

endnode = 2×1

22 33

h=length(endnode);
for j=1:h
 e=2;
 
 f=endnode(j,1);
 % while (f~=1)
 for s=1:no
 if (f~=1)
 k=1; 
 for i=1:br
 if ((C(i,f)==1)&&(k==1))
 f=i;
 k=2;
 end
 end
 k=1;
 for i=1:no
 if ((C(f,i)==-1)&&(k==1));
 f=i;
 g(j,e)=i;
 e=e+1;
 k=3;
 end 
 end
 end
 end
end
for i=1:h
 g(i,1)=endnode(i,1);
end
g;
w=length(g(1,:))

w = 14

for i=1:h
 j=1;
 for k=1:no 
 for t=1:w
 if g(i,t)==k
 g(i,t)=g(i,j);
 g(i,j)=k;
 j=j+1;
 end
 end
 end
end
g;
for k=1:br
 e=1;
 for i=1:h
 for j=1:w-1
 if (g(i,j)==k) 
 if g(i,j+1)~=0
 adjb(k,e)=g(i,j+1); 
e=e+1;
 else
 adjb(k,1)=0;
 end
 end
 end
 end
end
adjb;
for i=1:br-1
 for j=h:-1:1
 for k=j:-1:2
 if adjb(i,j)== adjb(i,k-1)
 adjb(i,j)=0;
 end
 end
 end

how to change the position index can be more than 32 because the data I have to enter is more than 32

end

Index in position 1 exceeds array bounds. Index must not exceed 32.

[KSSV]

Buggy and lengthy code.

[Sri Wahyuni]

I don't understand what you mean. I'm confused, change the program so that it can load more 32 data. can you help me?

Answer 1

[Simon Chan]

Check the following first:

In the following loop, you assign something to variable adjb if g(i,j)==k. However, if g(i,j)~=k, there is nothing assigned to adjb.

So the number of rows on adjb may not reach 'br', which is 37 in your case. If, for example, after k=33, the above condition are all not satisfy, variable adjb would not reach your expected size.

It is better to initialize the variable with a known size, such as adjb = zeros(37,2) or NaN(37,2) depends on your requirement. This is also true for other variables.

for k=1:br
    e=1;
    for i=1:h
        for j=1:w-1
            if (g(i,j)==k)                      % Do something for g(i,j)==k, but do nothing for g(i,j)~=k 
                if g(i,j+1)~=0
                    adjb(k,e)=g(i,j+1); 
                    e=e+1;
                else
                    adjb(k,1)=0;
                end
            end
        end
    end
end

[Sri Wahyuni]

Thank you,sir.

[Sri Wahyuni]

i don't know how to change the g part, so that it reads 37 data, can you help me

[Simon Chan]

I have no idea about any calculation in your code, so you need to decide the expected size of each newly generated variables, such as g and adjb.

Just have a quick look on the code again and you may define g = zeros(2,37) or NaN(2,37) and adjb = zeros(37,2) or NaN(37,2) before assigning any number on these two variables.

[Sri Wahyuni]

thank you, sir