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Fmincon inside a for loop, values not changing!

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motleyfool
motleyfool el 21 de Sept. de 2018
Comentada: motleyfool el 21 de Sept. de 2018
I have a task to run an optimization for every row with 100 columns, This is how I have formulated. values of all other parameters are calculated(Code and matrix attached if needed). When i run the loop, I do not get any response (it gives me my initial conditions as final answer)
Here is my script
Aeq = ones(1,n);
Beq = 1;
a0 = rand(1,n);
options = optimoptions(@fmincon,'Display','off');
w0 = 1/n*ones(1,n);
tic
for i = 3:t % trial
[a_opt] = fmincon(@(w) longsharpe(a0, avrcc, avrco, avroc, avroo, avrrvp,...
avrtvl, rcc, rco, roc, roo, tvl, rvp, i, n),w0,[],[],Aeq,Beq,[],[],[],options);
w_strat2(i,:) = a_opt;
end
and Here is the function,
function [y] = longsharpe(a, avrcc, avrco, avroc, avroo, avrrvp,...
avrtvl, rcc, rco, roc, roo, tvl, rvp, i, n)
% find weight vector for each day
w = zeros(1,n);
ret = rcc(i-1:i,:);
mu = mean(ret);
Q = cov(ret);
for j = 1:n
w(1,j) = a(1)*(rcc(i-1,j)-avrcc(i-1))./n...
+ a(2)*(roo(i,j)-avroo(i))./n...
+ a(3)*(roc(i-1,j)-avroc(i-1))./n ...
+ a(4)*(rco(i,j)-avrco(i))./n...
+ a(5)*(tvl(i-1,j)/avrtvl(i-1,j))*(rcc(i-1,j)-avrcc(i-1))./n...
+ a(6)*(tvl(i-1,j)/avrtvl(i-1,j))*(roo(i,j)-avroo(i))./n...
+ a(7)*(tvl(i-1,j)/avrtvl(i-1,j))*(roc(i,j)-avroc(i-1))./n...
+ a(8)*(tvl(i-1,j)/avrtvl(i-1,j))*(rco(i,j)-avrco(i))./n...
+ a(9)*(rvp(i-1,j)/avrrvp(i-1,j))*(rcc(i-1,j)-avrcc(i-1))./n...
+ a(10)*(rvp(i-1,j)/avrrvp(i-1,j))*(roo(i,j)-avroo(i))./n...
+ a(11)*(rvp(i-1,j)/avrrvp(i-1,j))*(roc(i-1,j)-avroc(i-1))./n...
+ a(12)*(rvp(i-1,j)/avrrvp(i-1,j))*(rco(i,j)-avrco(i))./n;
end
%setting up sharpe ratio with risk free rate 4%
y = -(mu*w'-0.04/252)/sqrt(w*(Q*w'));
end

Respuesta aceptada

Matt J
Matt J el 21 de Sept. de 2018
Editada: Matt J el 21 de Sept. de 2018
Perhaps you meant to have,
a_opt = fmincon(@(a) longsharpe(a,avrcc....
In any case, you are currently giving fmincon an objective function f(w) that is constant as a function of w, so it's no surprise that it is always optimal at the initial point.

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