Why unifrnd(lb,ub,[nPop,D]) gives error?

Why the following code gives error?

lb=[0 0 0 0]; ub=[10 10 pi pi]; nPop=30; D=4;
unifrnd(lb,ub,[nPop,D])

Error using unifrnd

Size information is inconsistent.

[Rik]

Deleted comments can be found in the Google cache (the archived version will require modification of the attributes, as the show/hide comments toggle doesn't execute properly).

Answer 1

[Ameer Hamza]

This cannot be done in single call to unifrnd. You need a for-loop

lb = [0 0 0 0]; ub=[10 10 pi pi]; nPop=30;
D = numel(lb);

M = rand(nPop, D);
for i = 1:D
    M(:,i) = unifrnd(lb(i),ub(i),[nPop,1]);
end

[Sadiq Akbar]

Thank you very much dear Ameer Hamza. But now it gives another error. Can you help me to run the following algorithm for my fitness function given below?

function [xmin,fmin,histout] = GQPSO(fun,D,nPop,lb,ub,maxit,maxeval)

% INPUT:
%   fun     : function handle for optimization
%   D       : problem dimension (number of variables)
%   nPop    : number of particles in the swarm
%   lb      : lower bound constrain
%   ub      : upper bound constrain
%   maxit   : max number of iterations
%   maxeval : max number of function evaluations

% OUTPUT:
%   xmin    : best solution found
%   fmin    : function value at the best solution, f(xmin)
%   histout : record of function evaluations and fitness value by iteration

% EXAMPLE:

% fun = @griewankfcn;
% D = 30;
% nPop = 50;
% lb = -600;
% ub = 600;
% maxit = 1000;
% maxeval = 10000*D;

% [xmin,fmin,histout] = GQPSO(fun,D,nPop,lb,ub,maxit,maxeval);

% OR DIRECTLY:

% [xmin,fmin,histout] = GQPSO(@griewankfcn,30,50,-600,600,1000,10000*30);

% QPSO parameters:
w1 = 0.5;
w2 = 1.0;

c1 = 1.5;
c2 = 1.5;
nPop=30; % By me
% Initializing solution
% x = unifrnd(lb,ub,[nPop,D]);  %This line gave error, so asked question on
 %Mathworks site and see below
% Correction by Ameer Hamza sab from Mathworks site
x = rand(nPop, D);
for i = 1:D
    x(:,i) = unifrnd(lb(i),ub(i),[nPop,1]);
end

% Evaluate initial population
pbest = x;

histout = zeros(maxit,2);

f_x = feval(fun,x);

fval = nPop;

f_pbest = f_x;

[~,g] = min(f_pbest);
gbest = pbest(g,:);
f_gbest = f_pbest(g);

it = 1;

histout(it,1) = fval;
histout(it,2) = f_gbest;

while it < maxit && fval < maxeval

    alpha = (w2 - w1) * (maxit - it)/maxit + w1;
    mbest = sum(pbest)/nPop;

    for i = 1:nPop

        fi = abs(randn(1,D)); % Gaussian QPSO (Coelho, 2010)

        p = (c1*fi.*pbest(i,:) + c2*(1-fi).*gbest)/(c1 + c2);

        u = abs(randn(1,D));  % Gaussian QPSO (Coelho, 2010)
        
        b = alpha*abs(x(i,:) - mbest);
        v = log(1./u);

        if rand < 0.5
            x(i,:) = p + b .* v;
        else
            x(i,:) = p - b .* v;
        end
        
        % Keeping bounds
        x(i,:) = max(x(i,:),lb);
        x(i,:) = min(x(i,:),ub);

        f_x(i) = fun(x(i,:));
        
        fval = fval + 1;

        if f_x(i) < f_pbest(i)
            pbest(i,:) = x(i,:);
            f_pbest(i) = f_x(i);
        end

        if f_pbest(i) < f_gbest
            gbest = pbest(i,:);
            f_gbest = f_pbest(i);
        end

    end
    
    it = it + 1;

    histout(it,1) = fval;
    histout(it,2) = f_gbest;

end

xmin = gbest; 
fmin = f_gbest;

histout(it+1:end,:) = [];

figure
semilogy(histout(:,1),histout(:,2))
title('Gaussian Q-PSO')
xlabel('Function evaluations')
ylabel('Fitness')
grid on

end

And my fitness function is:

function out=myfitness(b)
u=[1 2 3 4];
v=[b(1) b(2) b(3) b(4)];
out1=u-v
out=sum(out1);
end

When I type the followiong in command window, it does not run

>>[xmin,fmin,histout] = GQPSO(@myfitness,4,30,[0 0 0 0],[10 10 pi pi],100,10000*4)

[Walter Roberson]

What error do you encounter?

[Sadiq Akbar]

Thank you very much for your response. When I type the above in command window, I encounter the following error:

>> [xmin,fmin,histout] = GQPSO(@myfitness,4,30,[0 0 0 0],[10 10 pi pi],100,10000*4)
Attempted to access f_pbest(2); index out of bounds because numel(f_pbest)=1.
Error in GQPSO (line 99)
        if f_x(i) < f_pbest(i) 

[Walter Roberson]

Your fitness function is expected to receive a 2D array in which the number of rows corresponds to the population size, and the columns correspond to variables, and it is expected to return a vector which is the same size as the population size. Instead your fitness function expects a vector of four elements as input and returns a scalar.

function out=myfitness(b)
  u = [1 2 3 4];
  out1 = u-b;             %needs R2016b or later
  out = sum(out1.^2,2);
end

[Sadiq Akbar]

Thank you dear Walter Roberson for your 2nd response. But still it gives the following error after your changes:

[xmin,fmin,histout] = GQPSO(@myfitness,4,30,[0 0 0 0],[10 10 pi pi],100,10000*4)
Index exceeds matrix dimensions.
Error in GQPSO (line 101)
        if f_x(i) < f_pbest(i) 

[Walter Roberson]

No error when I tried with the GQPSO you posted in https://www.mathworks.com/matlabcentral/answers/702837-why-unifrnd-lb-ub-npop-d-gives-error#comment_1228747 and the myfitness I posted in https://www.mathworks.com/matlabcentral/answers/702837-why-unifrnd-lb-ub-npop-d-gives-error#comment_1229062

[Sadiq Akbar]

Thanks a lot for your 3rd response dear Walter Roberson. But why does it give the above error to me though I am running it in Matlab R2016b?

[Sadiq Akbar]

Even I ran it for this fitness function also, but it gives me an error:

function e=obj(b)
global u

th=pi/180;
u=[2 7 50*th 85*th]; % angles of all four vectors


%%%%%%%%%%%%%%%
% Swapping vector b. Taken help from Mathworks site
%%%%%%%%%%%%%%%
[~, ix] = sort(u); 
[~, ix1(ix)] = sort(b);  
b = b(ix1);

[~,C]=size(b);
P=C/2;
M=2*C;
%M=10;


xo=zeros(1,M);
for k=1:M 
    for i=1:P 
        xo(1,k)=xo(1,k)+u(i)*exp(-1i*(k-1)*pi*cos(u(P+i))); 
    end 
end


xe=zeros(1,M);
for k=1:M
        for i=1:P
            xe(1,k)=xe(1,k)+b(i)*exp(-1i*(k-1)*pi*cos(b(P+i)));
        end 
end 
abc=0.0;
for m1=1:M
  abc=abc+(abs(xo(1,m1)-xe(1,m1))).^2;  
end 
abc=abc/M; 
e=abc;

And stil it gives me the following error:

Subscripted assignment dimension mismatch.

Error in obj (line 12)

[~, ix1(ix)] = sort(b);

Error in GQPSO (line 56)

f_x = feval(fun,x);

[Walter Roberson]

I see no problem in R2020b with the original code.

Could you confirm you are using R2016b and not R2016a ? The fitness function I posted needs R2016b or later.

[xmin,fmin,histout] = GQPSO(@myfitness,4,30,[0 0 0 0],[10 10 pi pi],100,10000*4)

xmin = 1×4

0.9174 1.6567 3.1416 3.1416

fmin = 0.8816

histout = 100×2

30.0000 9.6990 60.0000 5.5803 90.0000 1.2189 120.0000 1.2189 150.0000 1.2189 180.0000 1.2189 210.0000 1.2189 240.0000 1.2189 270.0000 1.2189 300.0000 1.2189

●

function out=myfitness(b)
  u = [1 2 3 4];
  out1 = u-b;             %needs R2016b or later
  out = sum(out1.^2,2);
end

function [xmin,fmin,histout] = GQPSO(fun,D,nPop,lb,ub,maxit,maxeval)

% INPUT:
%   fun     : function handle for optimization
%   D       : problem dimension (number of variables)
%   nPop    : number of particles in the swarm
%   lb      : lower bound constrain
%   ub      : upper bound constrain
%   maxit   : max number of iterations
%   maxeval : max number of function evaluations

% OUTPUT:
%   xmin    : best solution found
%   fmin    : function value at the best solution, f(xmin)
%   histout : record of function evaluations and fitness value by iteration

% EXAMPLE:

% fun = @griewankfcn;
% D = 30;
% nPop = 50;
% lb = -600;
% ub = 600;
% maxit = 1000;
% maxeval = 10000*D;

% [xmin,fmin,histout] = GQPSO(fun,D,nPop,lb,ub,maxit,maxeval);

% OR DIRECTLY:

% [xmin,fmin,histout] = GQPSO(@griewankfcn,30,50,-600,600,1000,10000*30);

% QPSO parameters:
w1 = 0.5;
w2 = 1.0;

c1 = 1.5;
c2 = 1.5;
nPop=30; % By me
% Initializing solution
% x = unifrnd(lb,ub,[nPop,D]);  %This line gave error, so asked question on
 %Mathworks site and see below
% Correction by Ameer Hamza sab from Mathworks site
x = rand(nPop, D);
for i = 1:D
    x(:,i) = unifrnd(lb(i),ub(i),[nPop,1]);
end

% Evaluate initial population
pbest = x;

histout = zeros(maxit,2);

f_x = feval(fun,x);

fval = nPop;

f_pbest = f_x;

[~,g] = min(f_pbest);
gbest = pbest(g,:);
f_gbest = f_pbest(g);

it = 1;

histout(it,1) = fval;
histout(it,2) = f_gbest;

while it < maxit && fval < maxeval

    alpha = (w2 - w1) * (maxit - it)/maxit + w1;
    mbest = sum(pbest)/nPop;

    for i = 1:nPop

        fi = abs(randn(1,D)); % Gaussian QPSO (Coelho, 2010)

        p = (c1*fi.*pbest(i,:) + c2*(1-fi).*gbest)/(c1 + c2);

        u = abs(randn(1,D));  % Gaussian QPSO (Coelho, 2010)
        
        b = alpha*abs(x(i,:) - mbest);
        v = log(1./u);

        if rand < 0.5
            x(i,:) = p + b .* v;
        else
            x(i,:) = p - b .* v;
        end
        
        % Keeping bounds
        x(i,:) = max(x(i,:),lb);
        x(i,:) = min(x(i,:),ub);

        f_x(i) = fun(x(i,:));
        
        fval = fval + 1;

        if f_x(i) < f_pbest(i)
            pbest(i,:) = x(i,:);
            f_pbest(i) = f_x(i);
        end

        if f_pbest(i) < f_gbest
            gbest = pbest(i,:);
            f_gbest = f_pbest(i);
        end

    end
    
    it = it + 1;

    histout(it,1) = fval;
    histout(it,2) = f_gbest;

end

xmin = gbest; 
fmin = f_gbest;

histout(it+1:end,:) = [];

figure
semilogy(histout(:,1),histout(:,2))
title('Gaussian Q-PSO')
xlabel('Function evaluations')
ylabel('Fitness')
grid on

end

[Walter Roberson]

With regard to your new fitness function obj : As I posted above:

Your fitness function is expected to receive a 2D array in which the number of rows corresponds to the population size, and the columns correspond to variables, and it is expected to return a vector which is the same size as the population size.

Now let us examine your code:

u=[2 7 50*th 85*th]; % angles of all four vectors

Okay, that is a vector of 4 elements.

%%%%%%%%%%%%%%%
% Swapping vector b. Taken help from Mathworks site
%%%%%%%%%%%%%%%
[~, ix] = sort(u); 

That is going to return the vector [3 4 1 2] into ix.

[~, ix1(ix)] = sort(b);  

There you are asking to sort a 2D array with 30 rows and 4 columns. The sort information for that is going to require 30 x 4 output locations, but ix1(ix) would be a vector of 4 locations. Also, sort() by default operates along the first non-singleton dimension, which would be the first dimension for a 30 x 4 array, but it seems more likely that you want to sort along the rows, which would be the second dimension, seeing as you expect an array of length 4.

For this step I would suggest instead

[~, ix1] = sort(b, 2);
ix1 = ix1(:,ix);

which would give you a 2D array, in which each row talks about the sort order of b for its corresponding row

Then

b = b(ix1);

is a problem as it does not do per-row reordering.

Now.. what really is that code trying to do??

Let us experiment:

th=pi/180
u=[2 7 50*th 85*th]
[~, ix] = sort(u)
b = [4 1 3 2]
[~, ix1] = sort(b) 
ix1 = ix1(:,ix)
new_b = b(ix1)

b_2 = [0    17   -11    -6]
[~, ix1_2] = sort(b_2) 
ix1_2 = ix1_2(:,ix)
new_b_2 = b_2(ix1)

ix comes out as [3 4 1 2]

new_b comes out as [3 4 1 2]

new_b_2 comes out as [0 17 -11 -6]

t = sort([b;b_2],2)
t(:,ix)

comes out as

     3     4     1     2
     0    17   -11    -6

so it looks to me as if it wants to sort b per-row and then re-order according to the sorting of u. In that case I would suggest

th=pi/180;
u=[2 7 50*th 85*th];
[~, ix] = sort(u);
sort_b = sort(b,2);
b = sort_b(:,ix);

[Sadiq Akbar]

Thank you very much dear Walter Roberson for your 4th response. Yes indeed I am using Matlalab R2016b. This time I replaced the swapping section by your above code, i.e.

function e=obj(b)
global u
th=pi/180;
u=[2 7 50*th 85*th]; % angles of all four vectors
%%%%%%%%%%%%%%%
% Swapping vector b. Taken help from Mathworks site
%%%%%%%%%%%%%%%
[~, ix] = sort(u); 
sort_b = sort(b,2);
b = sort_b(:,ix);

[~,C]=size(b);
P=C/2;
M=2*C;
%M=10;
xo=zeros(1,M);
for k=1:M 
    for i=1:P 
        xo(1,k)=xo(1,k)+u(i)*exp(-1i*(k-1)*pi*cos(u(P+i))); 
    end 
end
xe=zeros(1,M);
for k=1:M
        for i=1:P
            xe(1,k)=xe(1,k)+b(i)*exp(-1i*(k-1)*pi*cos(b(P+i)));
        end 
end 
abc=0.0;
for m1=1:M
  abc=abc+(abs(xo(1,m1)-xe(1,m1))).^2;  
end 
abc=abc/M; 
e=abc;

Again I ran it but this time it gives the following error:

Index exceeds matrix dimensions.

Error in GQPSO (line 60)

f_pbest = min(f_x(i));

[Sadiq Akbar]

Further, dear Walter Roberson!

I have applied my fitness function on several algorithms. it never gave me any problem. But with this it gives error. So I would request if you can do changes in the algorithm instead of doing changes in my fitness function, it will be very good. Because if my fitness function changes, then it will give problem with other algorithms. I mean to say that there has been no problem in my fitness function for several algorithms till now, but it gives error for this. so if you change it, then again it will give problem for other algorithms. So my request is please do changes in the given algorithm so that it can run my fitness function. I hope you understand. Thank you once again.

[Walter Roberson]

As I posted twice before:

Your fitness function is expected to receive a 2D array in which the number of rows corresponds to the population size, and the columns correspond to variables, and it is expected to return a vector which is the same size as the population size.

Your obj is returning a scalar, not a vector of length 30.

It is not possible to change the algorithm to account for the objective function returning the wrong results.

[Walter Roberson]

            xe(1,k)=xe(1,k)+b(i)*exp(-1i*(k-1)*pi*cos(b(P+i)));

How many rows does b have? How many rows of b does that line access?

Where is your code producing one output for each row in b?

[Sadiq Akbar]

Thanks a lot for your prompt response dear Walter Roberson!

In my fitness function "u" is my desired vector and every algorithm sends one vector of the same size as my u because i give dimension of my u vector to that algorithm. Then that algorithm generates random population of 30 or 50 depends how much i set there. Then out of that population, the algorithm sends one memeber to my algorithm and fitness is determined and saved in a variable there.Then the algorithm sends next memebr of that population to my fitness function and again fitness of that is found and is saved in that old varibale and so on. So up till now my fitness function runs with all the algorithms, but it gives problem here. So can't you make such changes that the algorithm sends one row vector instead of entire population to my fitness funtion as before with other algorithms?

As my u vector is a 4 element row vector, so the algorithm must send "b" the same size as my vector u. Then it will work.

[Walter Roberson]

Read your code!!

x = rand(nPop, D);
for i = 1:D
    x(:,i) = unifrnd(lb(i),ub(i),[nPop,1]);
end

So x is the full matrix, not one row of the matrix.

f_x = feval(fun,x);

And right there you call your function on the entire x matrix.

Your objective function must return one value for each population member.

[Sadiq Akbar]

Thanks for very prompt response. You are absolutely right dear. I understand your reason. But I want to do some changes in the given algorithm so that the algorithm sends one row of the full matrix x in one time, fitness is found, stored, then algorithm sends 2nd row of the full matrix x, finds fitness, stores it and so on.

[Sadiq Akbar]

something like this:

% Initialize the population/solutions
for i=1:n,
    Sol(i,:)=Lb+(Ub-Lb).*rand(1,d);
Fitness(i)=obj(Sol(i,:));  % My fitness function
end

This is some portion I took from some other algorithm to show you.

[Walter Roberson]

% Initialize the population/solutions
for i=1:n,
  Sol(i,:)=Lb+(Ub-Lb).*rand(1,d);
f_x(i)=obj(Sol(i,:));  % My fitness function
end

[Sadiq Akbar]

You mean to say that if I put this code in given algorithm, it will run with my fitness function? If yes, then what portion of the given algorithm should I replace with this above code?

[Walter Roberson]

remove

f_x = feval(fun,x);

insert

for i = n
  f_x(i) = feval(fun,x(i, :)) 
end

[Sadiq Akbar]

Thanks a lot for your continuous help. I did what you told and ran it, but this time it gives the following error:

Undefined function or variable 'n'.

Error in GQPSO (line 59)

for i =n

[Walter Roberson]

for i = nPop
  f_x(i) = feval(fun,x(i, :)) 
end

[Sadiq Akbar]

Thanks for the correction.Now it ran partially as:

f_x =

Columns 1 through 9

0 0 0 0 0 0 0 0 0

Columns 10 through 18

0 0 0 0 0 0 0 0 0

Columns 19 through 27

0 0 0 0 0 0 0 0 0

Columns 28 through 30

0 0 81.9367

Index exceeds matrix dimensions.

Error in GQPSO (line 106)

if f_x(i) < f_pbest(i)

[Walter Roberson]

for i = 1:nPop
  f_x(i) = feval(fun,x(i, :)) 
end

[Sadiq Akbar]

Thanks for your constant support dear. After running it gives the following:

f_x =

65.4690

f_x =

65.4690 144.7119

f_x =

65.4690 144.7119 29.7280

f_x =

65.4690 144.7119 29.7280 45.6022

f_x =

65.4690 144.7119 29.7280 45.6022 177.2501

f_x =

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253

f_x =

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983

f_x =

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193

f_x =

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

f_x =

Columns 1 through 9

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

Column 10

121.0852

f_x =

Columns 1 through 9

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

Columns 10 through 11

121.0852 141.2146

f_x =

Columns 1 through 9

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

Columns 10 through 12

121.0852 141.2146 179.5222

f_x =

Columns 1 through 9

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

Columns 10 through 13

121.0852 141.2146 179.5222 63.1976

f_x =

Columns 1 through 9

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

Columns 10 through 14

121.0852 141.2146 179.5222 63.1976 124.7700

f_x =

Columns 1 through 9

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

Columns 10 through 15

121.0852 141.2146 179.5222 63.1976 124.7700 110.8142

f_x =

Columns 1 through 9

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

Columns 10 through 16

121.0852 141.2146 179.5222 63.1976 124.7700 110.8142 165.7112

f_x =

Columns 1 through 9

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

Columns 10 through 17

121.0852 141.2146 179.5222 63.1976 124.7700 110.8142 165.7112 86.9005

f_x =

Columns 1 through 9

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

Columns 10 through 18

121.0852 141.2146 179.5222 63.1976 124.7700 110.8142 165.7112 86.9005 117.7920

f_x =

Columns 1 through 9

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

Columns 10 through 18

121.0852 141.2146 179.5222 63.1976 124.7700 110.8142 165.7112 86.9005 117.7920

Column 19

158.4781

f_x =

Columns 1 through 9

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

Columns 10 through 18

121.0852 141.2146 179.5222 63.1976 124.7700 110.8142 165.7112 86.9005 117.7920

Columns 19 through 20

158.4781 247.8008

f_x =

Columns 1 through 9

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

Columns 10 through 18

121.0852 141.2146 179.5222 63.1976 124.7700 110.8142 165.7112 86.9005 117.7920

Columns 19 through 21

158.4781 247.8008 52.9352

f_x =

Columns 1 through 9

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

Columns 10 through 18

121.0852 141.2146 179.5222 63.1976 124.7700 110.8142 165.7112 86.9005 117.7920

Columns 19 through 22

158.4781 247.8008 52.9352 143.4848

f_x =

Columns 1 through 9

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

Columns 10 through 18

121.0852 141.2146 179.5222 63.1976 124.7700 110.8142 165.7112 86.9005 117.7920

Columns 19 through 23

158.4781 247.8008 52.9352 143.4848 99.6341

f_x =

Columns 1 through 9

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

Columns 10 through 18

121.0852 141.2146 179.5222 63.1976 124.7700 110.8142 165.7112 86.9005 117.7920

Columns 19 through 24

158.4781 247.8008 52.9352 143.4848 99.6341 11.2423

f_x =

Columns 1 through 9

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

Columns 10 through 18

121.0852 141.2146 179.5222 63.1976 124.7700 110.8142 165.7112 86.9005 117.7920

Columns 19 through 25

158.4781 247.8008 52.9352 143.4848 99.6341 11.2423 67.4893

f_x =

Columns 1 through 9

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

Columns 10 through 18

121.0852 141.2146 179.5222 63.1976 124.7700 110.8142 165.7112 86.9005 117.7920

Columns 19 through 26

158.4781 247.8008 52.9352 143.4848 99.6341 11.2423 67.4893 78.8733

f_x =

Columns 1 through 9

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

Columns 10 through 18

121.0852 141.2146 179.5222 63.1976 124.7700 110.8142 165.7112 86.9005 117.7920

Columns 19 through 27

158.4781 247.8008 52.9352 143.4848 99.6341 11.2423 67.4893 78.8733 90.7319

f_x =

Columns 1 through 9

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

Columns 10 through 18

121.0852 141.2146 179.5222 63.1976 124.7700 110.8142 165.7112 86.9005 117.7920

Columns 19 through 27

158.4781 247.8008 52.9352 143.4848 99.6341 11.2423 67.4893 78.8733 90.7319

Column 28

94.8255

f_x =

Columns 1 through 9

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

Columns 10 through 18

121.0852 141.2146 179.5222 63.1976 124.7700 110.8142 165.7112 86.9005 117.7920

Columns 19 through 27

158.4781 247.8008 52.9352 143.4848 99.6341 11.2423 67.4893 78.8733 90.7319

Columns 28 through 29

94.8255 136.7698

f_x =

Columns 1 through 9

65.4690 144.7119 29.7280 45.6022 177.2501 65.6253 149.4983 184.3193 110.8635

Columns 10 through 18

121.0852 141.2146 179.5222 63.1976 124.7700 110.8142 165.7112 86.9005 117.7920

Columns 19 through 27

158.4781 247.8008 52.9352 143.4848 99.6341 11.2423 67.4893 78.8733 90.7319

Columns 28 through 30

94.8255 136.7698 152.8737

Index exceeds matrix dimensions.

Error in GQPSO (line 106)

if f_x(i) < f_pbest(i)

>>

[Walter Roberson]

I am not sure on the index out of range. I will look at it after I sleep as I am not at my desk.

[Sadiq Akbar]

ok dear. Good night. I will wait for you. Thanks for your help.

[Walter Roberson]

Code seems to be working for me.

Invoke obj_driver or myfit_driver as applicable.

[Sadiq Akbar]

Thanks a lot dear Walter Roberson for your efforts. I ran it. Now its working. I get the graph. But when I removed the semicolons from "xmin" and "fmin" to see their values.In this xmin is the estimated vector which must nearly resemble my u vectro in the "obj.m" file. But I see that its not the same as my u vector. A lot of variations in the values and sometimes their values changes positions as well. As you can see my different runs:

>> myfit_driver
xmin =
    0.9932    2.1081    2.9842    3.1416
fmin =
    0.7489
>> myfit_driver
xmin =
    0.9396    1.9000    3.0854    3.1416
fmin =
    0.7578
>> myfit_driver
xmin =
    0.9186    1.9747    2.9763    3.1416
fmin =
    0.7447
>> myfit_driver
xmin =
    1.0536    2.0371    3.1416    3.1416
fmin =
    0.7612
>> myfit_driver
xmin =
    1.0377    2.0226    2.8995    3.1416
fmin =
    0.7489
>> myfit_driver
xmin =
    0.9598    1.9679    2.8777    3.1416
fmin =
    0.7545
>> myfit_driver
xmin =
    1.0721    1.8975    3.0554    3.1416
fmin =
    0.7556
>> myfit_driver
xmin =
    1.0148    1.8979    2.9469    3.1416
fmin =
    0.7503
>> myfit_driver
xmin =
    0.9659    2.0040    3.1416    3.1416
fmin =
    0.7581
>> myfit_driver
xmin =
    0.9647    2.0892    3.0875    3.1416
fmin =
    0.7537
>> myfit_driver
xmin =
    0.8858    1.9868    3.0025    3.1416
fmin =
    0.7501
>> obj_driver
xmin =
    1.7087    5.7792    1.4902    0.8463
fmin =
    1.9548
>> obj_driver
xmin =
    3.0668    5.6929    0.8875    1.4876
fmin =
    3.0838
>> obj_driver
xmin =
    1.4781    5.9054    1.5567    0.8601
fmin =
    1.7798
>> obj_driver
xmin =
    2.4717    8.4429    0.8501    1.4941
fmin =
    3.8352
>> th=pi/180;
u=[2 7 50*th 85*th]
u =
    2.0000    7.0000    0.8727    1.4835
>> obj_driver
xmin =
    1.4817    7.0154    0.9122    1.5292
fmin =
    0.7388
>> obj_driver
xmin =
    6.6405    2.0153    0.8459    1.4795
fmin =
    0.5694
>> 

Now what to do dear?

[Sadiq Akbar]

Further, when i rant it with my own old fitness function (before your chnages), again it gave error as:

Subscripted assignment dimension mismatch.

Error in fun2sn0 (line 13)

[~, ix1(ix)] = sort(b); % temp stores the randomly generated vector "best" by algorithm

Error in GQPSO (line 54)

f_x(i) = feval(fun,x(i, :));

Error in myfit_driver (line 2)

[xmin,fmin,histout] = GQPSO(@fun2sn0,4,30,[0 0 0 0],[10 10 pi pi],2000,10000*4);