problem in my matlab code 'Index in position 1 exceeds array bounds (must not exceed 1)'

Hello everyone,

please how can i solve this problem:Index in position 1 exceeds array bounds (must not exceed 1).

Error in Sphere (line 51)

SinrC(i,j)=Pcellular(i,j)*calculate_gain(PositionsBS(1),PositionsBS(2), PositionsC(i,1),PositionsC(i,2)

)/(Icv+Noise);

Error in GeneticAlgorithm (line 11)

population.Chromosomes(i).fitness = obj( population.Chromosomes(i).Gene(:) );

Error in Main (line 50)

[BestChrom] = GeneticAlgorithm (M , N, MaxGen , Pc, Pm , Er , Problem.obj , visualization).

This is the code where i have SinrC(i,j):

throughputC=zeros(C);
  SinrC=zeros(C,RB);
for i=1:C
    Icv=0;
     for j=1:RB
                if binc(i,j)==1  
           
           for v=1:V   
           
           if binv(v,j)==1
           Icv=Icv+Pv(v,j)*calculate_gain(PositionsBS(1),PositionsBS(2), PositionsV(v,1),PositionsV(v,2) );
           end
           end
           SinrC(i,j)=Pcellular(i,j)*calculate_gain(PositionsBS(1),PositionsBS(2), PositionsC(i,1),PositionsC(i,2) )/(Icv+Noise);
            throughputC(i)=  throughputC(i)+W*log2(1+ SinrC(i,j));
            
                end
     end

[Walter Roberson]

That code has a number of variables whose size we do not know, so it is difficult for us to say.

I note by the way tyat ou refer to throughputC(i) with a single subscript, after having defined throughputC=zeros(C ) . zeros(C ) defines a CxC matrix, equivalent to zeros(C,C) . This would not cause any problem in the part of the code you posted, but could potentially cause a problem in other parts of your code.

[brahmi ibtissem]

The problem appear whene i put a value of C greator than 1.

here:

PositionsC=rand(1,2);%*XX;
PositionsV=rand(1,2);%*XX;
PositionsBS=rand(1,2);%*XX;
% Problem preparation 
C=2;
V=3;

RB=C;

whene C+, no problem it gives to me the value of fitness function but whene i put like here C=2 or plus, i have this problem:

Index in position 1 exceeds array bounds (must not exceed 1).

Error in Sphere (line 51)

SinrC(i,j)=Pcellular(i,j)*calculate_gain(PositionsBS(1),PositionsBS(2), PositionsC(i,1),PositionsC(i,2)

)/(Icv+Noise);

Error in GeneticAlgorithm (line 11)

population.Chromosomes(i).fitness = obj( population.Chromosomes(i).Gene(:) );

Error in Main (line 50)

[BestChrom] = GeneticAlgorithm (M , N, MaxGen , Pc, Pm , Er , Problem.obj , visualization)

[KALYAN ACHARJYA]

>>whos PositionsC?? Please share

[brahmi ibtissem]

whos PositionsC

Name Size Bytes Class Attributes

PositionsC 1x2 16 double global

Answer 1

[Walter Roberson]

You define PositionsC as a row vector but you access it as a column vector.

[brahmi ibtissem]

how can i access to PositionsC in order to solve the problem please.

[Walter Roberson]

% Problem preparation 
C=2;

PositionsC=rand(C,2);%*XX;
PositionsV=rand(C,2);%*XX;
PositionsBS=rand(C,2);%*XX;

[brahmi ibtissem]

I still have this problem

Matrix dimensions must agree.

Error in Sphere (line 89)

c=a+b;

Error in GeneticAlgorithm (line 11)

population.Chromosomes(i).fitness = obj( population.Chromosomes(i).Gene(:) );

Error in Main (line 51)

[BestChrom] = GeneticAlgorithm (M , N, MaxGen , Pc, Pm , Er , Problem.obj , visualization)

[Walter Roberson]

c=a+b does not appear in the code you posted, so we can only make guesses about it.

One of my guesses is that you are using R2018a or earlier.

Answer 2

[brahmi ibtissem]

THis is the code of c=a

=b

a=sum(throughputV);

b=sum(throughputC);

c=a+b;

I calculate throughputC and throughputV by using there instructions:

SinrC=zeros(C,RB);
for i=1:C
    Icv=0;
     for j=1:RB
                if binc(i,j)==1  
           
           for v=1:V   
           
           if binv(v,j)==1
           Icv=Icv+Pv(v,j)*calculate_gain(PositionsBS(1),PositionsBS(2), PositionsV(v,1),PositionsV(v,2) );
           end
           end
           SinrC(i,j)=Pcellular(i,j)*calculate_gain(PositionsBS(1),PositionsBS(2), PositionsC(i,1),PositionsC(i,2) )/(Icv+Noise);
            throughputC(i)=  throughputC(i)+W*log2(1+ SinrC(i,j));
            
                end
    end
if throughputC(i)<SINRcth
    constraintSatified=false;
end
end
throughputV=zeros(V);
SinrV=zeros(V,RB);    
for v=1:V
    Ivc=0;  
    for j=1:RB
            if binv(i,j)==1  
            for c=1:C
           if binc(c,j)==1
           Ivc=Ivc+Pcellular(i,j)*calculate_gain(PositionsC(i,1),PositionsC(i,2), PositionsV(v,1),PositionsV(v,2) );
            end
           end
           SinrV(v,j)=Pv(v,j)*calculate_gain(PositionsV(v,1),PositionsV(v,2),PositionsC(c,1),PositionsC(c,2) )/(Ivc+Noise);
         throughputV(v)=  throughputV(v)+W*log2(1+ SinrV(v,j));
         disp( throughputV(v));
            end
    end
   if throughputV(v)< SINRvth
       
constraintSatified=false;
   end
end

[Walter Roberson]

Fishing through your postings as you do not show all of your code in one place, we see the lines

C=2;
V=3;
throughputC=zeros(C);
throughputV=zeros(V);

So throughputC is 2 x 2 and sum(throughputC) would be 1 x 2. Meanwhile throughputV is 3 x 3, and sum(throughputV) would be 1 x 3. You cannot add a 1 x 2 vector and a 1 x 3 vector.

I refer back to my first response,

I note by the way that you refer to throughputC(i) with a single subscript, after having defined throughputC=zeros(C ) . zeros(C ) defines a CxC matrix, equivalent to zeros(C,C) . This would not cause any problem in the part of the code you posted, but could potentially cause a problem in other parts of your code.

If you want a 1 x C vector, then you should be using zeros(1,C) not zeros(C )

[brahmi ibtissem]

Thank you for your help