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

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

[Sean de Wolski]

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.

Answer 1

[Alan Weiss]

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

[Jamal Ahmad]

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)

Answer 2

[Matt J]

FMINUNC seems like overkill for a 1D root finding problem. Why not just use FZERO?

   xv = fzero(@(x) gammainc(5,x)- vall(i) ,4);

[Sean de Wolski]

@Matt, I suggested FMINUNC in Jamal's other thread because fzero can overstep and pass a negative value into gammainc. This throws an error.

[Matt J]

OK, but then, why not use FMINBND? Or, if you don't know an upper bound, you could do something like this

xv = fzero(@(z) gammainc(5,z.^2)- vall(i) , 2).^2;

[Alan Weiss]

gammainc(5,x) is a monotone decreasing function with value 1 at x=0 and value 0 at x=Inf. Jamal wants to find a value of x so that gammainc(5,x) = gamma(5)*10^0 = 24. This is impossible. It is also not possible for gamma(5)*5*10^-1 or gamma(5)*10^-1 or gamma(5)*5*10^-2.

For values y such that 0 < y < 1, it is not hard to find a value x so that gammainc(5,x) = y:

 x = fzero(@(x)gammainc(5,x)-y,[0,100]);

Why? Because

 gammainc(5,100) = 5.9919e-91

This is close enough to 0 for all practical purposes, and fzero will not try to step outside the initial interval [0,100].

Alan Weiss

MATLAB mathematical toolbox documentation

[Jamal Ahmad]

Many thanks for all reputable responses.

so, is there any way to find the values of sig(i) when the value of pf is between (5*10^-2 to 10^0)? according to below matlab code I get NaN, and the follwoing message appears:

Exiting fzero: aborting search for an interval containing a sign change because no sign change is detected during search. Function may not have a root.

for the values of pf between (10^-4 to 10^-2), there are solutions which are (25.661 22.297 20.691 16.333 13.965).

clc; clear all; close all;
u = 5;
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)
      % gamma(5) = 24

      vall(i) = pf(i) * gamma(u);

      options = optimset('tolfun',10^-10,'tolx',10^-12);
      xv = fzero(@(z) gammainc(5,z.^2)- vall(i) , 2).^2;
      sig(i) = xv*2;
  end

Thanks again

Jamal

[Jamal Ahmad]

    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]

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.