I am facing difficulty in displaying output of multi objective genetic algorithm.
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This is my fitness function
function t=strength(x)
t(1) = -138.012+0.173.*x(1)+0.113.*x(2)+0.107.*x(3)+0.715.*x(4)-0.301.*x(5)-0.072.*x(6)+0.036.*x(7)+0.083.*x(8)+15.557.*x(9);
t(2) = 0.350-0.002.*x(1)-0.002.*x(2)+0.0.*x(3)+0.0.*x(4)-0.015.*x(5)+0.004.*x(6)-0.001.*x(7)-0.00000827.*x(8)-0.910.*x(9)+0.058.*x(10)+0.008.*x(11)+0.090.*x(12)-0.458.*x(13);
end
and this is my calling function
clear all
clc
opts.PopulationSize = 1000;
rng default;
options = optimoptions('gamultiobj','MutationFcn',@mutationadaptfeasible, 'CrossoverFcn', {@crossoverarithmetic}, 'SelectionFcn', {@selectiontournament,3}, 'PlotFcn', {@gaplotselection, @gaplotscorediversity, @gaplotbestindiv, @gaplotdistance, @gaplotpareto}, 'MaxGenerations', 1e5);
A = []; b = []; Aeq = []; beq = []; lb = [350 0 0 50 2 180 806 983 0.45 5 27.5 23.96 0]; ub = [350 0 0 50 2 180 806 983 0.45 5 27.5 23.96 0]; nonlcon = [];
obj = @(x) (strength(x));
bestx = gamultiobj(obj, 13, A, b, Aeq, beq, lb, ub, nonlcon, options);
format long g
Is their any way i can minimize the t(2), and simultaneous maximize t(1) and also
I want to show the final output of t(1) and t(2) and respective value of their variables ( x(1),x(2),x(3).....etc).
Answers (2)
Walter Roberson
on 17 Apr 2022
obj = @(x) [-1 1].*strength(x);
3 Comments
harsh Brar
on 17 Apr 2022
Walter Roberson
on 17 Apr 2022
Maximizing a function is the same as minimizing the negative of the function.
Walter Roberson
on 17 Apr 2022
[bestx, fval] = gamultiobj(obj, 13, A, b, Aeq, beq, lb, ub, nonlcon, options);
fval = [-1 1].*fval;
The t1 values are now fval(:, 1) and the t2 values are fval(:, 2). For any given row fval(K, :) corresponds to bestx(K, :)
Remember that gamultiobj returns a set of Pareto solutions, places where you cannot improve one of the functions without making the other worse. It is not necessarily possible for there to be a global best solution.
Imagine for example if the equations described a circle and you were trying to maximize x while minimizing y: maximum x is the right of the circle but minimum y is the bottom of the circle, and you cannot have both simultaneously.
You can, of course, min() and max() the fval columns and you might happen to get lucky and find a location that has both at the same time. You just cannot count on it.
harsh Brar
on 18 Apr 2022
1 Comment
Walter Roberson
on 18 Apr 2022
Is there any way i can get single value of the variable
Maybe, but perhaps not.
gamultiobj() would not normally generate duplicates, which suggests to me that the values you are seeing are possibly non-unique in decimal places not shown. For example there might be differences in the 7th decimal place. gamultiobj() would assume those differences are important, since you did not use options to increase the tolerance.
You could try reducing the list by
[~, ia] = uniquetol(bestx, 'byrows', true);
reduced_bestx = bestx(ia,:);
reduced_fval = fval(ia,:);
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on 17 Apr 2022
on 18 Apr 2022
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