Indexed exceeds matrix dimensions

clc;
close all;  % Close all figures (except those of imtool.)
clear;  % Erase all existing variables. Or clearvars if you want.
workspace;  % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 25;
figure;
imshow('pc3.png');
hold on
hold on
%legend('Relay', 'Sink', 'Sensor');
% placed 13 node according to the co-ordinates 
x1 = [105,107,58,58,75,146,118,18,48,55,108,80,80];
y1 = [190,280,190,280,145,153,103,153,103,65,65,50,10];
hold on
A=100;
B=140;
%plot the output node 
plot(A,B, 'mo-', 'MarkerSize', 10,'LineWidth', 2);
hold on
% placed 50 point randomly
Xmin = 60;
Xmax = 99;
x2 = Xmin+rand(1,50)*(Xmax-Xmin);
Ymin = 10;
Ymax= 260;
y2 = Ymin+rand(1,50)*(Ymax-Ymin);
plot(x2, y2, 'b*');
%numPoints2 = 50;
%x2 = 167*rand(numPoints2, 1);
%y2 = 302*rand(numPoints2, 1);

% Plot set 1 in red
plot(x1, y1, 'r.', 'MarkerSize', 13);
% Plot set 2 in blue
hold on;
plot(x2, y2, 'b*', 'MarkerSize', 5);
grid on;
for i=1:13
    for j=1:50
    
% Find distances between every point in set 1 to every point in set #2.
distances(i,j) = pdist2([x1(i),y1(i)], [x2(j), y2(j)], 'euclidean');
    end 
    minDistance(i) = min(distances(i,:));
end
disp (distances);
dist1 = sqrt((x2-A).^2+(y2-B).^2)
%disp (dist1);
dist2 =sqrt((x1-A).^2+(y1-B).^2)
%disp (dist2);
% Find min distance
disp (minDistance);
a=1:50;
%choose 10 node randomly from 50 node
out=a(randperm(numel1(a),10))
s=250;
Et=16.7;
Eamp=1.97;
%caculate the energy consumption of 13 sensor to that choosen 10 random node which store in out
%?_? (?, ?_?? )=?(?t+ ???? ???^2)
for i=1:10
for k=1:13
Ets(i,k)=s*(Et+Eamp*(distances(k,out(i)).^2));
end 
end
disp(Ets);
N=10;
data_agg=1;
Er=36.1;
%calculate the energy consumption of 10 relay node while transmiitting to 1 output device
%?_??=???? (?, ???/d??? ) )+????t
%?_??=(?−1) ?r(?)
%?_? (?)= ?r(?)
for i=1:10
for k= 1:1
Etf(i,k)= N*data_agg*Et*(s*dist1(k,out(i)))+N*s*data_agg*Et; %transmission energy of relay node
Erf=(N-1)*Er*s;%reception energy of relay node
Ef_total(i,k)=Etf(i,k)+Erf %total energy consumption of relay
end 
end
d0=40;
c=10;
%find out fitness function of that 10 relay node eith respect to 13 sensor node
%Fitness =c*∑j∈R*cost(j)*xj +c*(∑i∈S,j∈R Ei * xij+∑j∈R xj*Er)+ c* √(∑i∈S,j∈R(dij-d0)^2 ))/(∑i∈S,j∈R xij)+c*(∑i∈S,j∈R xij )/(⋃i∈S,j∈R xij) 
% xj=1,if relay exist otherwise 0
%S=13 sensor
%R=10 relay stored in out
for i=1:10
for k=1:13
fitness(i,k)= (c*out(i)*1)+(c*(Ets(k,out(i))*1+Ef_total(k,out(i))*1))+(c*sqrt((distances(k,out(i))-d0).^2))+c;
end 
end
disp(fitness)

[Luna]

Please define your question. You have too many syntax errors, end end together in the same line. Also, what is distances variable?

What do you want to get from this code?

[Sushree Patra]

distance is defined and this one is the part of code my problem is when i calculate the fitness function then it shows indexed exceeds matrxmatrix dimensions

[Sushree Patra]

due to shortage of space i put end in same line. please help me out.

if the code is not clear then i will send the total code

Answer 1

[Walter Roberson]

Ets is 10 by 13.

Your fitness calculation includes Ets(k,out(i)) . But out(i) can be up to 50 and k can be up to 13 . You probably want Ets(i,k)

[Sushree Patra]

Thank you for you answer. out(i) is that select 10 node from 50 node randomly

my objective is to find out energy consumption of 13 predefined node with respect to selected 10 node which are stored in out.

the i have to find out Etf which is the energy consumption of that choosen 10 node to 1 out put node having distance and stored in dist2.

then calculate the fitness function

[Walter Roberson]

the value out(i) can be up to 50 because it is a random node number . You try to use it to index an array in a position that has 13 columns. You also try to use row number up to 13 even through the array has only 10 rows.

[Sushree Patra]

means out where i choose 10 number i have to choose 50 random number ?

[Sushree Patra]

which part i have to change can you please help me in my code

[RAJENDRA KUMAR SHARMA]

how to make objective function of coordination relay please anybody explain

[Walter Roberson]

objective function of relay coordination should be asked as a new topic; http://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer

Answer 2

[Luna]

Your out and Ets variable is below:

out = [21 20 24 26 39 36 1 28 33 15]  % 1x10 array
Ets = [.. ..] % 10x13 matrix
Ef_total = [.. ..] % 10x1 array

When you get in for loop for k = 1:10 and i =1:13, out(i) return to you the first, second, third element of the out array.

For example i = 1. out(i) will be 21. Your Ets only have 10 rows and 13 columns. You can't get the 21st column element you are asking for in bolded below. Because there is only 13 columns. Same for Ef_total.

(Ets(k,out(i)) will be Ets(k,21) ... etc.. % you get the idea)

The below formula in for loop definetely cause error.

fitness(i,k)= (c*out(i)*1)+(c*(Ets(k,out(i))*1+Ef_total(k,out(i))*1))+(c*sqrt((distances(k,out(i))-d0).^2))+c;

You should change them as follows but check if it does the correct calculation:

fitness(i,k)= (c*out(i)*1)+(c*(Ets(i,k)*1+Ef_total(i,1)))+(c*sqrt((distances(k,out(i))-d0).^2))+c;

[Sushree Patra]

Thank you for the answer. But 21st is out(i) first element

Ets and Ef_total value will calculated

and if i increase the column value then its also showing same index exceeds matrix dimensions.

can you please help me solve it in my code where to do changes I am unable

[Sushree Patra]

yes i got the point.

Thank you very much

[Sushree Patra]

one more think as i placed the blue star in body randomly but they are not covered all over the body only in between the define range. also that 50 blue star distributed uniformly randomly

[Luna]

Did it fix your code?

[Sushree Patra]

yes it fixed in my code