what is this mean: Gradient must be provided for trust-region algorithm?

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Jamal Ahmad
Jamal Ahmad 2013-5-24
Hi,
I have matlab code, when I run it I receive this message:
Warning: Gradient must be provided for trust-region algorithm; using line-search algorithm instead. > In fminunc at 367 In false_alarm_detection_2 at 14
Local minimum found.
Optimization completed because the size of the gradient is less than the selected value of the function tolerance.
criteria details
what is that mean? and how can be solved?
Thanks
  1 个评论
Sean de Wolski
Sean de Wolski 2013-5-24
It means the default algorithm isn't suitable for the way you've called fminunc. You can just ignore the warning, it picked a different algorithm instead.

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采纳的回答

Alan Weiss
Alan Weiss 2013-5-24
编辑：Alan Weiss 2013-5-24
fmincon options describes the restrictions for the trust-region-reflective algorithm. Including Derivatives describes how you include the derivative in the objective function definition.
To avoid the warning without including a derivative, set the Algorithm option to 'interior-point' or some other algorithm:
opts = optimset('Algorithm','interior-point');
x = fmincon(objfn,x0,[],[],[],[],lb,ub,[],opts);
Take a look at the documentation examples for more information on setting gradients and options: <http://www.mathworks.com/help/optim/constrained-optimization.html>
Alan Weiss
MATLAB mathematical toolbox documentation
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Jamal Ahmad
Jamal Ahmad 2013-5-24
编辑：Matt J 2013-5-24
Thank you very much for the answer.
this is the code that I have: can you please tell me how I can solve it? Thanks
clc; clear all; close all;
u = 5;
snr = 20;
pf = [10^-4 5*10^-4 10^-3 5*10^-3 10^-2 5*10^-2 10^-1 5*10^-1 10^0];
for i = 1:length(pf)
% pf(i) = gamma(u,sigma/2) / gamma(u)
% gamma(u,sigma/2) = pf(i) * gamma(u)
vall(i) = pf(i) * gamma(u);
options = optimset('tolfun',10^-10,'tolx',10^-12);
xv = fminunc(@(x)(gammainc(5,x)-(vall(i))).^2,4,options);
sig(i) = xv/2;
g=0;
h=0;
for p=0:(u-2)
gg = (1/factorial(p)) * (sig(i)./2).^p;
g = g + gg;
hh = (1/factorial(p)) * ((sig(i)*snr)/(2*(1+snr)))^p;
h = h + hh;
end
pd(i) = exp(-(sig(i)/2))*g+((1+snr)/snr)^(u-1)*(exp(-(sig(i)/(2+(1+snr))))-exp(-(sig(i)/2))*h);
pm(i) = 1 - pd(i);
end
loglog(pf,pm)

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更多回答（1 个）

Matt J
Matt J 2013-5-24
编辑：Matt J 2013-5-24
FMINUNC seems like overkill for a 1D root finding problem. Why not just use FZERO?
xv = fzero(@(x) gammainc(5,x)- vall(i) ,4);
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Jamal Ahmad
Jamal Ahmad 2013-5-26
编辑：Matt J 2013-5-27
12.830 gammainc(5,12.83) = 0.0024
11.148 gammainc(5,11.148) = 0.012
10.345 gammainc(5,10.345) = 0.024
8.166 gammainc(5,8.166) = 0.12
6.982 gammainc(5,6.982) = 0.240
I believe there must be a solution since;
K>> gammainc(5,5)
ans = 0.5595
K>> gammainc(5,4)
ans = 0.7350
K>> gammainc(5,3)
ans = 0.8753
K>> gammainc(5,2)
ans = 0.9596
K>> gammainc(5,1)
ans = 0.9933
K>> gammainc(5,0)
ans = 1
so, after gammainc(5,6.982) the result will be between 0.24 and 1
Matt J
Matt J 2013-5-27
编辑：Matt J 2013-5-27
so, after gammainc(5,6.982) the result will be between 0.24 and 1
Yes, but your target values of gammainc are not between 0.24 and 1. Here is the stream of vall(i) that your code produces
vall =
0.0024 0.0120 0.0240 0.1200 0.2400 1.2000 2.4000 12.0000 24.0000
As you can see, they exceed 1 for i>=6. As Alan said, it is not possible for gammainc(5,x) to reach a value greater than 1.

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