How to invoke and use GA tool in Matlab R2022b?

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Sadiq Akbar 2022-12-14

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1877662-how-to-invoke-and-use-ga-tool-in-matlab-r2022b

评论： Star Strider 2022-12-20

I typed optimtool in Matlab R2022b to invoke the GA Tool but its' not there. Can anybody guide me how to invoke and use GA tool in Matlab R 2022b because I usually tune the setting for the best results but since its not here, so what to do and how to do?

Regards,

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Sadiq Akbar 2022-12-14

在 MATLAB Online 中打开

Thanks a lot for your kind response dear Star Strider. But I am not so much expert to code it by my own. Can you demonstrate it by getting my desired solution. My functions are as below:

clear all
clc
global u
global Noise
u=[-35, 35, -5, 5, -25, 25, 45, -45];% My desired vector
nvars=length(u);% No. of varibles
lb=-90*ones(1,nvars);%Lower bound
ub=90*ones(1,nvars);%Upper bound
Population=250;
iterations=500;
Noise=20;
nn=0;
for n=1:10 
    nn=nn+1;
    % Invoke GA to get executed and returns us best solution x, fitness value fval and execution time i.e.,
    [x,fval,time] = ga(???????);
    one(nn)=fval;   
    temp(nn,:)=x; 
    time1(nn)=time;
    
    [~, ix] = sort(u); 
    [~, ix1(ix)] = sort(temp(nn,:));  
    two(nn,:) = temp(nn,ix1); 
end
%%%%%%%%%%%%%%%%%%%%%
% GA code here
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
% My function called by GA for fitness
%%%%%%%%%%%%%%%%%%%%%
function e=myfun(b)
global u
global Noise
M=6; %Constant1
N=6; %Constant2
K=length(u);%Constant3
d = 0.5; %Constant4
vec = @(MAT) MAT(:); 
vecH = @(MAT) MAT(:).'; 
steerVecT = @(ang) exp(1j*2*pi*d*(0:M-1).'*sin(vecH(ang))); 
steerVecR = @(ang) exp(1j*2*pi*d*(0:N-1).'*sin(vecH(ang))); 
%%%%%%%%%%%%%%%%%%%%
% Swapping vector b
%%%%%%%%%%%%%%%%%%%%
[~, ix] = sort(u); 
[~, ix1(ix)] = sort(b);  
b = b(ix1);
A = ones(K, 1);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Calculation of yo and ye
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
yo = yMatTR(deg2rad(u), steerVecT, steerVecR); 
yo=awgn(yo,Noise);
ye = yMatTR(deg2rad(b), steerVecT, steerVecR); 
%%%%%%%%%%%%%%%%%%
% MSE
%%%%%%%%%%%%%%%%%%
e=norm(yo-ye).^2/(M*N);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% User defined function called within myfun
%%%%%%%%%%%%%%%%%%%%%%%%5%%%
function y = yMatTR(targetAngle, steerVecT, steerVecR)
   steerA = steerVecT(targetAngle); 
   steerB = steerVecR(targetAngle); 
   y=sum( steerA.*permute(steerB,[3,2,1]) ,2);
   y=y(:);
 %%%%%%%%%%%%% End of my code %%%%%%%%%%%%%%%%%%%%%
 
 

I want to run it and the GA returns me the three values i.e., the best solution, fitness and time. But the best solution should be equal to or nealry equal to my desired vector u.

Star Strider 2022-12-14

First, there is no reason to put the ga call in a loop unless you are varying specific parameters between calls to it.

Second, I have no idea what the fitness function is or the problem you are optimising.

Third, it is never advisable to use global variables. See the documentation on Passing Extra Parameters to understand how to do that correctly.

Sadiq Akbar 2022-12-14

Reason to put ga in loop: Because I want to get 10 independent run values for the same parameters.

The fitness function is a user defined function defined for the problem. My problem is to find the desired solution x whose value matches the value of my desired vector u or nearly matches it.

Use of global variables: If I don't use global variables, then how will it be visible to my fitness function?

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Answer 1

Star Strider 2022-12-14

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1877662-how-to-invoke-and-use-ga-tool-in-matlab-r2022b#answer_1127642

在 MATLAB Online 中打开

If you are doing parameter estimation, this approach will likely work —

x = 1:0.1:10;

y = 2.5*exp(-(x-5).^2/2)+randn(size(x))*0.25;

objfcn = @(b,x) b(1).*exp(-(x-b(2)).^2*b(3));

fitnessfcn = @(b) norm(y-objfcn(b,x));

Parms = 3;

opts = optimoptions('ga', 'MaxGenerations',1000);

[B,fval,exitflag,output] = ga(fitnessfcn, Parms,[],[],[],[],[],[],[],opts);

Optimization terminated: average change in the fitness value less than options.FunctionTolerance.

Parameters = B

Parameters = 1×3

2.5311 4.8704 0.5030

BestFitness = fval

BestFitness = 2.2760

Generations = output.generations

Generations = 129

figure

plot(x, y, '.', 'DisplayName','Data')

hold on

plot(x, objfcn(B,x), '-r', 'DisplayName','Function Fit')

hold off

grid

legend('Location','best')

This uses ga to estimate the parameters of the function fitting the data.

With respect to eliminating global variables, see the ‘Passing Extra Parameters’ documentation I already linked to.

.

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Sadiq Akbar 2022-12-15

编辑：Sadiq Akbar 2022-12-15

在 MATLAB Online 中打开

Thank you very much dear Star Strider for your kind response. Yes you may be write. You can say its a type of parameter estimation. But what is this reproduced below?

x = 1:0.1:10;
y = 2.5*exp(-(x-5).^2/2)+randn(size(x))*0.25;
objfcn = @(b,x) b(1).*exp(-(x-b(2)).^2*b(3));
fitnessfcn = @(b) norm(y-objfcn(b,x));

As far as I understand your code, you have a desired vector [2.5 5 0.25] like mine u but yours has 3 values and my u has got 4 values. Your function is y=2.5*exp(-(x-5).^2/2) and then you are adding noise to it. But mine has two or three functions as can be seen from "myfun.m". b is the random vector coming one by one from the random population generated by ga The same is the case with mine. But I don't understand the 3rd and 4th lines in your code. Can you explain them please?

Now coming to my problem. I want that when ga runs and stops after completion, it should return me an estimated vector "two" which must be equal to my desired vector u or if not equal, then nearly equal to u. So how will we do that then? Can you do that for me by removing global variables because I am not too much expert in Matlab by using my above function "myfun.m"?

Star Strider 2022-12-18

在 MATLAB Online 中打开

My pleasure!

I am not certain what parameter you are optimising. In that event, it needs to be the input to the fitness function:

[B,fval,exitflag,output] = ga(@(b)myfun(b,u,Noise), Parms,[],[],[],[],[],[],[],opts)

This selects ‘b’ as the argument to be optimised that ga passes to ‘myfun’.

With that change, this runs without error —

clear all
clc
u=[-35 35 50];
dim=length(u);
lb=-90*ones(1,dim);
ub=90*ones(1,dim);
Noise=35;
Parms = 3;
opts = optimoptions('ga', 'MaxGenerations',1000);
[B,fval,exitflag,output] = ga(@(b)myfun(b,u,Noise), Parms,[],[],[],[],[],[],[],opts)
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
B = 1×3
  -35.0195   50.0848   35.0961
fval = 8.4405e-04
exitflag = 1
output = struct with fields:
      problemtype: 'unconstrained'
         rngstate: [1×1 struct]
      generations: 200
        funccount: 10050
          message: 'Optimization terminated: average change in the fitness value less than options.FunctionTolerance.'
    maxconstraint: []
       hybridflag: []
function e=myfun(b,u,Noise)
M=6; %Constant1
N=6; %Constant2
K=length(u);%Constant3
d = 0.5; %Constant4
vec = @(MAT) MAT(:); 
vecH = @(MAT) MAT(:).'; 
steerVecT = @(ang) exp(1j*2*pi*d*(0:M-1).'*sin(vecH(ang))); 
steerVecR = @(ang) exp(1j*2*pi*d*(0:N-1).'*sin(vecH(ang))); 
%%%%%%%%%%%%%%%%%%%%
% Swapping vector b
%%%%%%%%%%%%%%%%%%%%
[~, ix] = sort(u); 
[~, ix1(ix)] = sort(b);  
b = b(ix1);
A = ones(K, 1);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Calculation of yo and ye
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
yo = yMatTR(deg2rad(u), steerVecT, steerVecR); 
yo=awgn(yo,Noise);
ye = yMatTR(deg2rad(b), steerVecT, steerVecR); 
%%%%%%%%%%%%%%%%%%
% MSE
%%%%%%%%%%%%%%%%%%
e=norm(yo-ye).^2/(M*N);
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% User defined function called within myfun
%%%%%%%%%%%%%%%%%%%%%%%%5%%%
function y = yMatTR(targetAngle, steerVecT, steerVecR)
   steerA = steerVecT(targetAngle); 
   steerB = steerVecR(targetAngle); 
   y=sum( steerA.*permute(steerB,[3,2,1]) ,2);
   y=y(:);   
end

If you want ‘lb’ aand ‘ub’ to be parameter constraints (although it does not appear to make much difference here, since the results are well within those ranges), put them in the appropriate places in the argument list.

I still do not understand what you are doing, since the estimated parameters in a curve-fitting optimisation make the objective function values approximate the supplied data when the original fitting function is evaluated with those parameters, similar to the code I originally posted as an example. Here, the results appear to be the actual function fit, rather than the parameters (whatever they are) that provide that fit.

If it does what you want, it is not necessary that I understand it or how it actually works.

.

Star Strider 2022-12-19

My pleasure.

See the options section of the ga documentation for suppressing the output messages and providing the population size as an input. All that is possible using an optimoptions options structure.

With regard to the parameter bounds, see lb and ub in the documentation. They are input arguments to ga and so are part of the argument list.

The ga function itself will not return the execution time. With respect to the time, I have this as part of a standard ga call I use:

t0 = clock;

fprintf('\nStart Time: %4d-%02d-%02d %02d:%02d:%07.4f\n', t0)

[theta,fval,exitflag,output,population,scores] = ga(ftns, Parms, [],[],[],[],lb,ub,[],[],opts);

t1 = clock;

fprintf('\nStop Time: %4d-%02d-%02d %02d:%02d:%07.4f\n', t1)

GA_Time = etime(t1,t0)

DT_GA_Time = datetime([0 0 0 0 0 GA_Time], 'Format','HH:mm:ss.SSS');

fprintf('\nElapsed Time: %23.15E\t\t%s\n\n', GA_Time, DT_GA_Time)

Using tic and toc instead of the clock and etime calls are another option. Using datetime as I use it here makes it easier to parse the elapsed time in seconds and fractional seconds into hours, minutes, and seconds, since some of my ga optimisations run for several hours.

The ‘exitflag’ and ‘output’ and other outputs are for my benefit, since they contain information about the optimisation (such as the number of generations) that I believe are important. If you do not need the information they provide, then eliminate them from the output list. Returning their information is not required. The only output information that is required is the first output, since it contains the parameters that have been optimised. The second output is the optimal fitness value, usually considered to be important.

.

Star Strider 2022-12-19

在 MATLAB Online 中打开

The loop addressing appeasrs to be the problem.

Try this —

% clear all
% clc
u=[-35 35 50];
dim=length(u);
lb=-90*ones(1,dim);
ub=90*ones(1,dim);
% Noise=35;
% PopulationSize_Data=200;
PopulationSize_Data = 20;
MaxGenerations_Data=1000;
FunctionTolerance_Data=1e-6;
ConstraintTolerance_Data=1e-6;
options = optimoptions('ga','PopulationSize', PopulationSize_Data);
options = optimoptions('ga','MaxGenerations', MaxGenerations_Data);
options = optimoptions('ga','FunctionTolerance', FunctionTolerance_Data);
options = optimoptions('ga','ConstraintTolerance', ConstraintTolerance_Data);
options = optimoptions('ga','CreationFcn', @gacreationuniform);
options = optimoptions('ga','CrossoverFcn', {  @crossoverheuristic [] });
options = optimoptions('ga','MutationFcn', @mutationadaptfeasible);
options = optimoptions('ga','Display', 'off');
Noise=10:12;
% for mm=1:numel(Noise)
%     [B,fval] = ga(@(b)myfun(b,u,Noise(mm)), dim,[],[],[],[],lb,ub,[],options)
%     one(mm)=fval;
%     two(mm,:)=B;
% end
NrCallsToGA = 5;                            % Number Of Times To Run The Inner Loop
for mm=1:numel(Noise)
    for nn = 1:NrCallsToGA
        [B,fval] = ga(@(b)myfun(b,u,Noise(mm)), dim,[],[],[],[],lb,ub,[],options);
        onev(nn,:) = fval;
        twov(nn,:) = B;
    end
    [~,idx] = min(onev);                    % Choose Lowest 'fval' For Each Run
    one(mm,:)=onev(idx);                    % Save That 'fval' Value
    two(mm,:)=twov(idx,:);                  % Save That 'B' Value
end
one
one = 3×1
    0.0620
    0.0513
    0.0436
two
two = 3×3
   49.8298   34.9975  -34.8970
  -34.8288   50.1081   34.9002
  -34.8867   50.1524   35.0214
function e=myfun(b,u,Noise)
M=6; %Constant1
N=6; %Constant2
K=length(u);%Constant3
d = 0.5; %Constant4
vec = @(MAT) MAT(:); 
vecH = @(MAT) MAT(:).'; 
steerVecT = @(ang) exp(1j*2*pi*d*(0:M-1).'*sin(vecH(ang))); 
steerVecR = @(ang) exp(1j*2*pi*d*(0:N-1).'*sin(vecH(ang))); 
%%%%%%%%%%%%%%%%%%%%
% Swapping vector b
%%%%%%%%%%%%%%%%%%%%
[~, ix] = sort(u); 
[~, ix1(ix)] = sort(b);  
b = b(ix1);
A = ones(K, 1);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Calculation of yo and ye
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
yo = yMatTR(deg2rad(u), steerVecT, steerVecR); 
yo=awgn(yo,Noise);
ye = yMatTR(deg2rad(b), steerVecT, steerVecR); 
%%%%%%%%%%%%%%%%%%
% MSE
%%%%%%%%%%%%%%%%%%
e=norm(yo-ye).^2/(M*N);
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% User defined function called within myfun
%%%%%%%%%%%%%%%%%%%%%%%%5%%%
function y = yMatTR(targetAngle, steerVecT, steerVecR)
   steerA = steerVecT(targetAngle); 
   steerB = steerVecR(targetAngle); 
   y=sum( steerA.*permute(steerB,[3,2,1]) ,2);
   y=y(:);   
end

This corrected version runs without error.

I do not understand what you were doing originally. If the nested loops were supposed to do something specific, that was not obvious, so I initially replaced them with a single loop and one loop counter, that being ‘mm’.

I assume that you want to run ga several times with a specific value of ‘Noise’. If so, something like this may be more appropriate:

Noise=10:12;

NrCallsToGA = 5; % Number Of Times To Run The Inner Loop

for mm=1:numel(Noise)

for nn = 1:NrCallsToGA

[B,fval] = ga(@(b)myfun(b,u,Noise(mm)), dim,[],[],[],[],lb,ub,[],options);

onev(nn,:) = fval;

twov(nn,:) = B;

end

[~,idx] = min(onev); % Choose Lowest 'fval' For Each Run

one(mm,:)=onev(idx); % Save That 'fval' Value

two(mm,:)=twov(idx,:); % Save That 'B' Value

end

I ran this to test it, and it works as I intend it to work. (I checked the interim results to be sure, however I did not include that additional code here.) Change ‘NrCallsToGA’ value to be whatever works best (and that you have time for).

I reduced the population size so it would run in the 55 seconds allowed here.

.