Setting up properly the fminunc function

Dear all,

I am trying to maximize this function

T=1000;
z=randn(T,1);
u=randn(T,1);
k1=0.01;
k2=0.01;
options=optimset('LargeScale','off','display','off','TolFun',0.0001,'TolX',0.0001,...
    'GradObj','off', 'Hessian','off','DerivativeCheck','off');

for t=1:T
 [x,fval,exitflag,output,G_sum,H]=fminunc('funct',u(t),options,...
 z,k1, k2,t, u,T);
end

where

 function  LL= funct(x,z,k1, k2,t, u,T)
if t==1
     u=[x; u(2:T) ];
 elseif t==T
  u=[u(1:T-1);x ];   
 else
  u=[u(1:t-1);x;u(t+1:T) ]; 
end

  e2=(z-u).^2;
  
kk= - 0.5*x^2/k2;
LL = -( -.5/k1*sum(e2(t:T))+kk) ;

end

However, I am not sure if the function is properly set up in terms of 'x'. As you can see 'x' changes position within 'u'.

I suspect that an alternative approach would be

 function  LL= funct(x,z,k1, k2,t, u,T)
u(t)=x;

  e2=(z-u).^2;
  
kk= - 0.5*x^2/k2;
LL = -( -.5/k1*sum(e2(t:T))+kk );

end

Which of the two is more efficient? Is there any alternative solution?

Thanks in advance.

[Walter Roberson]

http://www.mathworks.com/help/matlab/math/parameterizing-functions.html

[ektor]

Dear Walter,

Thank you for this link.

I am not sure how to do that.

It should be something like:

function y = funn(z,k1, k2,t, u,T, valueofx)

function  LL= funct(x)
u(t)=x;
  e2=(z-u).^2;
  
kk= - 0.5*x^2/k2;
LL = -( -.5/k1*sum(e2(t:T))+kk );
end
end

[ektor]

Any ideas, please?

Answer 1

[Walter Roberson]

T=1000;
z=randn(T,1);
u=randn(T,1);
k1=0.01;
k2=0.01;
options = optimset('LargeScale','off','display','off','TolFun',0.0001,'TolX',0.0001,...
    'GradObj','off', 'Hessian','off','DerivativeCheck','off');

x = zeros(1, T); fval = zeros(1,T); exitflag = zeros(1,T);
output = cell(1,T); G_sum = cell(1,T); H = cell(1,T);

for t = 1:T
  [x(t), fval(t), exitflag(t) ,output{t}, G_sum{t}, H{t}] = fminunc(@(x) funct(x, z, k1, k2, t, u, T), u(t), options);
end

function  LL = funct(x, z, k1, k2, t, u, T)
  u(t) = x;
  e2 = (z-u).^2;
  kk = - 0.5*x^2/k2;
  LL = -( -.5/k1*sum(e2(t:T))+kk );
end

If you only include e2(t:T) in your sum, then it is not clear to me why you are calculating e2 over the whole vector? Why not, for example,

for t = 1:T
  [x(t), fval(t), exitflag(t) ,output{t}, G_sum{t}, H{t}] = fminunc(@(x) funct(x, z, k1, k2, u(t:end)), u(t), options);
end

function  LL = funct(x, z, k1, k2, urest)
  urest(1) = x;
  e2 = (z-urest).^2;
  kk = - 0.5*x^2/k2;
  LL = -( -.5/k1*sum(e2)+kk );
end

[ektor]

THank you so much!!

May I ask why cant' I have

for t = 1:T
  [x, fval, exitflag ,output, G_sum, H] = fminunc(@(x) funct(x, z, k1, k2, t, u, T), u(t), options);
end

instead of

for t = 1:T
  [x(t), fval(t), exitflag(t) ,output{t}, G_sum{t}, H{t}] = fminunc(@(x) funct(x, z, k1, k2, t, u, T), u(t), options);
end

Is the latter faster than the former?

[Walter Roberson]

The first of those calculates for all t values, throwing away all of the intermediate results and keeping only the last result. It would not leave you with an x for t = 1, an x for t = 2, and so on, only with an x for t = T.

[ektor]

Thank you very much Walter!