我正在使用Matlab fminsearch来最小化带有两个变量的方程sum((interval-5).^2, 2)*factor间隔是一个包含5个值的向量。 只能从1到30的步长为1的顺序选择它们。因子是0.1到0.9的值。
代码如下。 我认为区间值是正确的,但因子值是错误的。
间隔值:[3 4 5 6 7]因子值:0.6最终输出:6
我认为因子值应为0.1,最终输出应为1作为全局最小值。
%% initialization of problem parameters
minval = 1;
maxval = 30;
step = 1;
count = 5;
minFactor = 0.1;
maxFactor = 0.9;
%% the objective function
fun = @(interval, factor) sum((interval-5).^2, 2)*factor;
%% a function that generates an interval from its initial value
getinterval = @(start) floor(start) + (0:(count-1)) * step;
getfactor =@(start2) floor(start2 * 10)/10;
%% a modified objective function that handles constraints
objective = @(start, start2) f(start, fun, getinterval, minval, maxval, getfactor, minFactor, maxFactor);
%% finding the interval that minimizes the objective function
start = [(minval+maxval)/2 (minFactor+maxFactor)/2];
y = fminsearch(objective, start);
bestvals = getinterval(y(1));
bestfactor = getfactor(y(2));
eval = fun(bestvals,bestfactor);
disp(bestvals)
disp(bestfactor)
disp(eval)
该代码使用以下函数f 。
function y = f(start, fun, getinterval, minval, maxval, getfactor, minFactor, maxFactor)
interval = getinterval(start(1));
factor = getfactor(start(2));
if (min(interval) < minval) || (max(interval) > maxval) || (factormaxFactor)
y = Inf;
else
y = fun(interval, factor);
end
end
我按照亚当的建议尝试了GA函数。 考虑到我的变量来自不同的范围和步骤,我将其更改为两个不同的集合。 这是我的零钱。
step1 = 1;
set1 = 1:step1:30;
step2 = 0.1;
set2 = 0.1:step2:0.9;
% upper bound depends on how many integer used for mapping
ub = zeros(1, nvar);
ub(1) = length(set1);
ub(2) = length(set2);
然后,我改变了目标函数
% objective function
function y = f(x,set1, set2)
% mapping
xmap1 = set1(x(1));
xmap2 = set2(x(2));
y = (40 - xmap1)^xmap2;
end
运行代码后,我想我会得到想要的答案。