Optimization terminated: first-order optimality less than OPTIONS.TolFun, and no negative/zero curvature detected in trust region model.

When TolFun is 1e-6, I got the above message in 0th iteration. So no further iteration is not going on. How i will solve that problem.

Answer 1

[Matt J]

It might be good news. The optimization might have succeeded. Call whatever function it is you're using with more output arguments, so you can check the value of the gradient and other diagnostic information, like the EXITFLAG, output.firstorderopt, and output.stepsize

[smitirupa]

Actually i can not find out where is the problem. I know problem is on firstorderopt. But i can not sort out this.Can u help in this regard. I cant guess what is the problem

[Matt J]

Show us the code and its output.

[smitirupa]

    *M-function file*
    function er_norm=obj_fun(constants,xdata,ydata)
    % x=xdata;
    % y=ydata;
    C0=constants(1);C1=constants(2);C2=constants(3);C3=constants(4);C4=constants(5)
    % C4=constants(5);
    % c5=constants(4);
    ynew=C0+C1*xdata.^1+C2*xdata.^2+C3*xdata.^3+C4*xdata.^4
    % er_norm=norm(ynew-ydata)

    Run file
    clear;
    clc;
    format long
    data=xlsread('uic60.xlsx');
    options = optimset('Display','iter','MaxIter',1e6,'TolFun',1e-32,'MaxFunEvals',1e4,'TolX',1e-12);
    xdata=data(:,1);
    ydata=data(:,2);
    C0=0.0013e-5; C1=0.05e-5; C2=0.050e-8; C3=1e-8; C4=1e-8;
    constants=[C0 C1 C2 C3 C4 ];
     x0=[C0 C1 C2 C3 C4];
    f=@(x,xdata)C0+C1*xdata.^1+C2*xdata.^2+C3*xdata.^3+C4*xdata.^4

    C0=constants(1);    C1=constants(2);   C2=constants(3); C3=constants(4);   C4=constants(5); 

    [x,resnorm,residual,exitflag,output]=lsqcurvefit(f,x0,xdata,ydata,[],[],options)
    hold on
    plot(xdata,ydata,xdata,f(x,xdata))
    hold off

    Output

     Iteration  Func-count     f(x)          step          optimality   CG-iterations
         0          6       0.0213194                             0
    Optimization terminated: first-order optimality less than OPTIONS.TolFun,
     and no negative/zero curvature detected in trust region model.

    x =

      1.0e-006 *

       0.013000000000000   0.500000000000000   0.000500000000000   0.010000000000000   0.010000000000000

    resnorm =

       0.021319448979382

    residual =

       0.022935647666543
       0.023656562500035
       0.024356871699866
       0.025019867046445
       0.025633290174737
       0.026185655588680
       0.026669444268175
       0.027079640891809
       0.027413127048670
       0.027667330252872
       0.027840305551303
       0.027930623282206
       0.027937480382255
       0.027860616651143
       0.027700403307839
       0.027458049835533
       0.027135791605277
       0.026741063395763
       0.026286298265767
       0.025780306693696
       0.025228612201459
       0.024635093285392
       0.024001373315168
       0.023328973281281
       0.022619515997451
       0.021874698169094
       0.021096298908960
       0.020286164636195
       0.019446186558797
       0.018578369810777
       0.017684771000595
       0.016768921428978
       0.015839440091998
       0.014902604350151
       0.013962464085410
       0.013024009034124
       0.012094644594510
       0.011183955871576
       0.010300513176566
       0.009450913772099
       0.008640223741933
       0.007873802280339
       0.007156439816903
       0.006492522579311
       0.005885429179882
       0.005337132152756
       0.004847230419930
       0.004412992162858
       0.004028749502575
       0.003687592949644
       0.003381492624863
       0.003103786830481
       0.002849529871807
       0.002615211321742
       0.002398800414927
       0.002198575461979
       0.002013065540527
       0.001840938989762
       0.001681007933138
       0.001532206494037
       0.001393597821352
       0.001264345981194
       0.001143720532711
       0.001031070393152
       0.000925826840403
       0.000827481436292
       0.000735587976429
       0.000649744912417
       0.000569596612284
       0.000494819910255
       0.000425124919974
       0.000360244998115
       0.000299937263116
       0.000243975168336
       0.000192148814180
       0.000144259366930
       0.000100119206205
       0.000059547532453
       0.000022370360660
      -0.000011583201362
      -0.000042481120883
      -0.000070491855813
      -0.000095784700387
      -0.000118533101906
      -0.000138915199198
      -0.000157117092201
      -0.000173333578265
      -0.000187771275313
      -0.000200649542713
      -0.000212203214292
      -0.000222683663747
      -0.000232360766126
      -0.000241524032413
      -0.000250483286458
      -0.000259569749041
      -0.000269134790273
      -0.000279550789851
      -0.000291207202182
      -0.000304510957187
      -0.000319878960482
      -0.000337737729934
      -0.000358511358458
      -0.000382620009492
      -0.000410462318839
      -0.000442412301963
      -0.000478795192309
      -0.000519882244883
      -0.000565859108261
      -0.000616776225174
      -0.000671469488819

thanx

[Matt J]

LSQCURVEFIT is trying to minimize your objective function

 f=@(x,xdata)C0+C1*xdata.^1+C2*xdata.^2+C3*xdata.^3+C4*xdata.^4

as a function of 'x'. However, the function you've provided has no actual dependence on x, only on xdata. So, since all points have zero gradient, lsqcurvefit stops with any initial point x0 that you choose.

If your goal is to fit xdata,ydata to a one-dimensional 4th order polynomial, you should just use POLYFIT,

 constants=polyfit(xdata,ydata,4);