How can I get binary output instead of fraction from gamultiobj optimization?

1 次查看(过去 30 天)
I am using the below code to get a binary decision output. Instead I am getting fractions. How can I fix this?
D = [1,2,3,1,2,3,4,1,2,2,1,2,3,4,4,1,2,3,4,3];
F = [4,3,2,4,4,3,2,4,2,2,4,4,4,2,2,5,4,2,2,2];
T = 21;
%H = round(rand(20,20));
H = zeros(20,20);
H(1,2) = 1;
H(1,3) = 1;
H(2,3) = 1;
H(4,5) = 1;
H(4,6) = 1;
H(4,7) = 1;
H(5,6) = 1;
H(5,7) = 1;
H(6,7) = 1;
H(8,9) = 1;
H(8,10) = 1;
H(11,12) = 1;
H(11,13) = 1;
H(12,13) = 1;
H(13,14) = 1;
H(13,15) = 1;
H(16,17) = 1;
H(17,18) = 1;
H(17,19) = 1;
H(17,20) = 1;
fitnessfcn = @(x)[sum(x), -sum(D.*x)];
nvars = 20;
A = -F;
b = -T;
Aeq = [];
Beq = [];
lb = zeros(20,1);
ub = ones(20,1);
[x, fval] = gamultiobj(fitnessfcn, nvars, A, b, Aeq, Beq, lb, ub, @(x) mycon(x, H));
function xsum= constraint2(x,H)
xsum = 0;
for i = 1:length(H)
for j = i+1:length(H)
xsum = xsum + H(i,j) * x(i) * x(j);
end
end
end
function [c,ceq] = mycon(x,H)
c = [];
ceq = constraint2(x,H);
end

采纳的回答

Walter Roberson
Walter Roberson 2018-3-3
gamultiobj() does not do integer constraints explicitly. You have constrained that the values must be at least 0 and no greater than 1, but you have not constrained that they must be exactly 0 or exactly 1.
In order to implement integer constraints in gamultiobj, you need to use a custom initial population and custom mutation function and custom cross-over function that just happen to only return binary values.
  5 个评论
Md Enamul Haque
Md Enamul Haque 2018-3-4
I changed the name and still the same error. However sometimes it runs and gives the following message.
Infeasible individuals are present in the final population.
Optimization terminated: average change in the spread of Pareto solutions less than options.FunctionTolerance.
Constraints are not satisfied within constraint tolerance.
Walter Roberson
Walter Roberson 2018-3-4
The initial population that you provide through the int_pop function must satisfy the linear constraints (I would need to dig deeper to see if it had to satisfy the nonlinear constraints.)
Your int_pop function, given that lb and ub, has a 50% chance of returning a matrix of all 0, and a 50% chance of returning a matrix of all 1, with all of the population members being that same 0 or 1.
Inside the int_pop it will find that the span between the 0 lowerbound and 1 upperbound is 1. It will multiply the span by rand() and add that to the lowerbound of 0, getting you a matrix of values that turn out to be the values of rand() unchanged. Then the code generates a single random number, and 50% chance takes the floor() of the entire matrix, and 50% chance takes the ceil() of the entire matrix. Well, rand() is not able to generate either 0 or 1 exactly, so the range of data that has been generated is (0, 1) open-open interval -- none of the values as large as 1, none of the values as small as 0. floor() of such a matrix would be all 0, and ceil() of such a matrix would be all 1.
This is not a situation in which you need to generalize. You can construct a function:
function Population = int_pop(GenomeLength, ~, options)
A = options.LinearConstr.Aineq;
b = options.LinearConstr.bineq;
curpop = zeros(0, GenomeLength);
popsize = options.PopulationSize;
while size(curpop,1) < popsize
more_needed = popsize - size(curpop,1);
thisbatch = round(rand(more_needed, GenomeLength));
passfail = all(A * thisbatch.' <= b, 1);
goodones = thisbatch(passfail,:);
curpop = [curpop; goodones];
end
Population = curpop;

请先登录,再进行评论。

更多回答(0 个)

类别

Help CenterFile Exchange 中查找有关 Solver Outputs and Iterative Display 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by