Is there a better function to minimize than fminsearch ?
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Hello, I would like to know if it exits a better function to minimize a function than fminsearch ? I have this line :
[X, fval, exitflag, output] = fminsearch(@func, X0, options, params)
I precise I have the optimization toolbox. Thank you for your help !
1 Comment
Walter Roberson
on 18 Oct 2017
Do you need a global optimization or a local optimization? Is the function differentiable? Is its Jacobian known? Is its Hessian known? If you were to pass symbolic variables into the function would it be able to return a symbolic formula in response ?
Answers (2)
Birdman
on 18 Oct 2017
There is a user-written function which contains Hooke-Jeeves algorithm. Maybe this will help you. The inputs and the outputs are clearly defined.
function [X,BestF,Iters] = hookejeeves(N, X, StepSize, MinStepSize, Eps_Fx, MaxIter, myFx)
% Function HOOKEJEEVS performs multivariate optimization using the
% Hooke-Jeeves search method.
%
% Input
%
% N - number of variables
% X - array of initial guesses
% StepSize - array of search step sizes
% MinStepSize - array of minimum step sizes
% Eps_Fx - tolerance for difference in successive function values
% MaxIter - maximum number of iterations
% myFx - name of the optimized function
%
% Output
%
% X - array of optimized variables
% BestF - function value at optimum
% Iters - number of iterations
%
Xnew = X;
BestF = feval(myFx, Xnew, N);
LastBestF = 100 * BestF + 100;
bGoOn = true;
Iters = 0;
while bGoOn
Iters = Iters + 1;
if Iters > MaxIter
break;
end
X = Xnew;
for i=1:N
bMoved(i) = 0;
bGoOn2 = true;
while bGoOn2
xx = Xnew(i);
Xnew(i) = xx + StepSize(i);
F = feval(myFx, Xnew, N);
if F < BestF
BestF = F;
bMoved(i) = 1;
else
Xnew(i) = xx - StepSize(i);
F = feval(myFx, Xnew, N);
if F < BestF
BestF = F;
bMoved(i) = 1;
else
Xnew(i) = xx;
bGoOn2 = false;
end
end
end
end
bMadeAnyMove = sum(bMoved);
if bMadeAnyMove > 0
DeltaX = Xnew - X;
lambda = 0.5;
lambda = linsearch(X, N, lambda, DeltaX, myFx);
Xnew = X + lambda * DeltaX;
end
BestF = feval(myFx, Xnew, N);
% reduce the step size for the dimensions that had no moves
for i=1:N
if bMoved(i) == 0
StepSize(i) = StepSize(i) / 2;
end
end
if abs(BestF - LastBestF) < Eps_Fx
break
end
LastBest = BestF;
bStop = true;
for i=1:N
if StepSize(i) >= MinStepSize(i)
bStop = false;
end
end
bGoOn = ~bStop;
end
function y = myFxEx(N, X, DeltaX, lambda, myFx)
X = X + lambda * DeltaX;
y = feval(myFx, X, N);
% end
function lambda = linsearch(X, N, lambda, D, myFx)
MaxIt = 100;
Toler = 0.000001;
iter = 0;
bGoOn = true;
while bGoOn
iter = iter + 1;
if iter > MaxIt
lambda = 0;
break
end
h = 0.01 * (1 + abs(lambda));
f0 = myFxEx(N, X, D, lambda, myFx);
fp = myFxEx(N, X, D, lambda+h, myFx);
fm = myFxEx(N, X, D, lambda-h, myFx);
deriv1 = (fp - fm) / 2 / h;
deriv2 = (fp - 2 * f0 + fm) / h ^ 2;
diff = deriv1 / deriv2;
lambda = lambda - diff;
if abs(diff) < Toler
bGoOn = false;
end
end
% end
17 Comments
abc abc
on 19 Oct 2017
Walter Roberson
on 19 Oct 2017
Pass in your @func in the myFx argument position.
Birdman
on 19 Oct 2017
Just decide how many variables you have(N) and initial condition(X). You can pass your function from myFx and can easily find the solution. Hope this helps.
abc abc
on 20 Oct 2017
Birdman
on 20 Oct 2017
You define StepSize, MinStepSize, Eps_Fx as
[]
abc abc
on 20 Oct 2017
Birdman
on 20 Oct 2017
Enter it as 100. You can increase or decrease it.
abc abc
on 20 Oct 2017
Birdman
on 20 Oct 2017
How did you define X and N? Write here please.
abc abc
on 20 Oct 2017
Birdman
on 20 Oct 2017
X is the initial condition array and you have to give numerical values instead of a1, a2, etc...
abc abc
on 20 Oct 2017
Birdman
on 20 Oct 2017
Try this:
Enter stepsize=[0.5 ..... 0.5] Minstepsize=[0.01 .... 0.01] Eps_Fx=[1e-6 ..... 1e-6]
abc abc
on 20 Oct 2017
abc abc
on 20 Oct 2017
Birdman
on 20 Oct 2017
I will deal with code and respond to you later.
Firstly, enter the following informations for the *hookejeeves* function.
N=..; X=[a1 .. a11]; StepSize=[0.5 .. 0.5]; MinStepSize=[0.01 .. 0.01]; Eps_Fx=[];%let it be empty
Then, save the following function with the name hookejeeves
function [X,BestF,Iters] = hookejeeves(N, X, StepSize, MinStepSize, MaxIter, myFx)
%Başlangıç atamalarının yapılması. BestF=x(k+1), LastBestF=x(k) gibi
%düşünülebilir.
Xnew = X;
BestF = feval(myFx, Xnew, N);
LastBestF = 100 * BestF + 100;
%bGoOn değişkenine bağlı while döngüsü, maksimum iterasyon sayısına veya
%verilen toleransa ulaşılınca biter.
bGoOn = true;
Iters = 0;
%Civar aramasının gerçekleştiği while döngüsüdür.
while bGoOn
Iters = Iters + 1;
if Iters > MaxIter
break;
end
X = Xnew;
%N=2 değişken için arama yapılmaktadır.
for i=1:N
bMoved(i) = 0;
bGoOn2 = true;
while bGoOn2
xx = Xnew(i);
Xnew(i) = xx + StepSize(i);
F = feval(myFx, Xnew, N);
if F < BestF
BestF = F;
bMoved(i) = 1;
else
Xnew(i) = xx - StepSize(i);
F = feval(myFx, Xnew, N);
if F < BestF
BestF = F;
bMoved(i) = 1;
else
Xnew(i) = xx;
bGoOn2 = false;
end
end
end
end
for i=1:N
bMadeAnyMove(i) = sum(bMoved(i));%Civar araması başarılıysa, bMadeAnyMove(i) değişkeni 0'dan farklı olur.
if bMadeAnyMove(i) > 0
Xnew1(i) = 2*Xnew(i) - X(i);%if bloğunda yeni x(k+1) değeri yukarıda kullanılmak üzere elde edilir.
end
end
BestF = feval(myFx, Xnew1, N);
LastBestF = feval(myFx, Xnew, N);
%Fonksiyonun değeri bir öncekinden daha küçükse, bir önceki değerlerin
%yeni x değerini bulurken kullanılması.
for i=1:N
if BestF < LastBestF
Xnew(i)=Xnew1(i);
X(i)=Xnew(i);
Xnew1(i) = 2*Xnew(i) - X(i);
end
end
%Civar araması başarısızsa, adım sayısı yarıya düşürülür.
for i=1:N
if bMoved(i) == 0
StepSize(i) = StepSize(i) / 2;
end
end
%Adım sayısı, verilen adım sayısından daha küçük olursa iterasyon sonlanır.
bStop = true;
for i=1:N
if StepSize(i) >= MinStepSize(i)
bStop = false;
end
end
bGoOn = ~bStop;
end
Don't worry about the comment lines, they are in turkish. Then enter the following code:
hookejeeves(N,X,StepSize,MinStepSize,Eps_Fx,@intrafunc)
This one should work. Let me know the outcome. If you give 11 variables, it might take a little longer to calculate but do not worry and wait for it to get it done.
Alan Weiss
on 18 Oct 2017
2 votes
You might be interested in the Optimization Decision Table, which exists to help you choose the most appropriate solver.
Alan Weiss
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