Hello ,
I have a code
a,b -----> variables should be vector(like a=[a1,a2,a3,a4])
x=h1*a^2+h2*b^2+h3;------>objctive fun(h1,h2,h3 are constants let say which has 1*4 value)
---->constraints are also present here.. so I am writting the code like
for i=1:4
x(i)=h1(i)*a^2+h2(i)*b^2+h3(i)
end
obj=sum(x);----->(objective function is basically summation of x )
x0=[1,1];
lb = [];
ub = [];
Aeq = [];
beq =[];
A = [];
b = [];
fun=@(p) obj(p(1),p(2));
nonlcon =@constraintfcn;
opts = optimoptions('fmincon','Display','iter','Algorithm','sqp');
[x_value,fval] = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon,opts);
But I am not getting the ans,I want optimized value as vector .

댓글 수: 2

Matt J
Matt J 2020년 11월 27일
Quite unclear, I'm afraid. What is the for-loop supposed to be doing? Do you have more than one objective to minimize? Why does the right hand side not depend on i?
Soumili Sen
Soumili Sen 2020년 11월 28일
편집: Soumili Sen 2020년 11월 28일
yes, you are right. My objective function is summation of 'x' (I have modified my code), and the unknown optimized veriables are vector. so how can I continue to find the values?

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Matt J
Matt J 2020년 11월 28일
편집: Matt J 2020년 11월 28일

0 개 추천

H1=sum(h1);
H2=sum(h2);
H3=sum(h3);
obj=@(a,b) H1.*a.^2 +H2.*b.^2+H3;
x0=[1,1];
lb = [];
ub = [];
Aeq = [];
beq =[];
A = [];
b = [];
fun=@(p) obj(p(1),p(2));
nonlcon =@constraintfcn;
opts = optimoptions('fmincon','Display','iter','Algorithm','sqp');
[x_value,fval] = fmincon( fun,x0,A,b,Aeq,beq,lb,ub,nonlcon,opts);

댓글 수: 12

Soumili Sen
Soumili Sen 2020년 11월 28일
Thanks for response, but here we are getting a single value of a & b.like
x_value= 1.23,2.3 ---> (values of a and b corresponding)
but I want optimizd value as a=4 values(like a1,a2,a3,a4) and b=4 values.
here I am using unknown veriable declaration as
syms a b;
for my desire output can I write it like
a=sym('a',[1 4]);
If it possible,how can I further proceed?
Matt J
Matt J 2020년 11월 28일
The objective function that you have shown us is only dependent on a single a and b. You have not shown us an 8 variable objective function.
Soumili Sen
Soumili Sen 2020년 11월 28일
ok.sorry for my mistake.how I can resolve this problem according to my desire output?
Matt J
Matt J 2020년 11월 28일
You must show us the objective function that you really want. I still don't know what it is.
As it is, with the objective function shown, the min lies at all zeros for the vector a, since the objective function is just a weighted sum of the squares of the elements in a. That assumes all the elements in h are non-negative. If any element of h is negative, then no minimum exists.
As Matt says, if this does not satisfy the problem (I doubt it will) then more information is needed.
Soumili Sen
Soumili Sen 2020년 11월 28일
편집: Soumili Sen 2020년 11월 28일
ok, below I have modified the representation of my objective function,
h1=rand(1,4);
h2=rand(1,4);
h3=rand(1,4);
a=sym('a',[1 4]);
b=sym('b',[1 4]);
x=h1.*a+h2.*b+h3;
for i=1:4
x1(i)=norm(x(i));
end
obj=sum(x1)
How I can optimized this a values and b values? thanks in advance.
Matt J
Matt J 2020년 11월 28일
편집: Matt J 2020년 11월 28일
Similar to fun() below, nonlcon must accept input in the form of a 4x2 matrix p.
obj=@(a,b) dot(h1,a)+dot(h2,b)+sum(h3);
fun=@(p) obj(p(:,1),p(:,2));
[x_value,fval] = fmincon(fun,ones(4,2),[],[],[],[],nonlcon,opts);
Soumili Sen
Soumili Sen 2020년 11월 29일
yah,the code is running. Thank you so much to clarify the doubts. I have one question i.e,
if I want complex value of 'a' (a,a2,a3,a4 are complex in nature) then how the code will be modified?
Matt J
Matt J 2020년 11월 29일
If a=ar+ac*1i, then you must rewrite the problem with ar and ac (which are real-valued) as the unknowns.
Matt J
Matt J 2020년 11월 29일
You're welcome, but please Accept-click the answer if you consider the matter resolved.
Soumili Sen
Soumili Sen 2020년 11월 29일
yah sure.

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