local optimisation deterministic optimisation

3 次查看（过去 30 天）
Hazim Hamad
Hazim Hamad 2020-9-22
评论： Matt J 2020-9-23
Hello
I need to get a deterministic optimisation for an equation from text :
f= x(1)* x(2)
bounded by
10 <= x(1)<=100
5<=x(2)<=20
there is a constraints which is
((Th=q=17;
Tc=325;
K=100; are given constant ))
Th = Tc+q*x(2)/K*x(1)
Th<= 345;
The code is :
% ObjectiveFunction = @simple_fitness;
fun = @objfun;
A = [];
b = [];
Aeq = [];
beq = [];
lb = [10 5]; % Lower bound
ub = [100 20]; % Upper bound
% lb=[]
% ub=[]
nvars = 2; % Number of variables
% ConstraintFunction = @t;
% [x,fval] = ga(ObjectiveFunction,nvars,[],[],[],[],LB,UB, ...
% ConstraintFunction);
x0 = [10 20];
[x,fval] = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,@confun);
% [x,fval] =ga(fun,4,A,b,Aeq,beq,lb,ub,@confun);
% display (fval)
%%%%%%%%%%%%%%%%%%%%%%%%
%%% objective function %%%%%%%%%%%
function f = objfun(x)
f = x(1)*x(2);
end
%%%%%%%%%Constraints%%%%%%
function [c,ceq] = confun(x)
% Nonlinear inequality constraints
q=17
Tc=325
K=100
% Th=Tc-q*x(2)/K*x(1);
c=(Tc-q*x(2)/K*x(1))-345;
% Nonlinear equality constraints
ceq = [];
end
%%%%%%%%%%%%%%%
I have got the results that min fvalue = 50 coresponding to this value x1=5 and x2=10
while the answer should be fvalue =212.5 and x1=42.5 ; x2=5.
would you help me to sor out this problem please.
  6 个评论
显示 4更早的评论
Hazim Hamad
Hazim Hamad 2020-9-22
Yes i agree with you ; if i want to use loop and compute the Th based on x1 and x2 obtained from optimsation results for example if can i let the program to compare Th and change the x0 to be x1 and x2 corresponding to minmum f and the loop should compare the Th .
Regards
Hazim
Matt J
Matt J 2020-9-23
Hazim Hamad's comment moved here:
I think the problem with constraints because when I make change on the constraints the solution not affaected.
Regards
Hazim

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Matt J
Matt J 2020-9-22
编辑：Matt J 2020-9-22
fvalue = 50 is clearly a better (i.e. lower) objective value than 212.5, so there is every reason to believe that x1=42.5 x2=5 is not the solution.
  3 个评论
显示 1更早的评论
Hazim Hamad
Hazim Hamad 2020-9-22
it is minmum but the Th should be less the 345 and because the Th obtained from his results is 325.02 and my results Th is 325.85 so his results is bettter .
thanks
Matt J
Matt J 2020-9-22
编辑：Matt J 2020-9-22
If one solution can be better than another based on Th, then it should be made part of the objective function somehow. There is nothing currently in the way the problem is set up to make the optimization prefer a specific value of Th.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Function Creation 的更多信息

标签

产品


版本

R2020a

Community Treasure Hunt

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

Start Hunting!

Translated by