Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.



Re: fsolve trustregion radius question
Posted:
Apr 9, 2013 10:21 AM


On 4/9/2013 8:11 AM, Anna wrote: > Hello, > > > I have a system of nonlinear equations and i am trying to solve if > with fsolve. > > I get a solution as function value is approaching 0, which is correct. > But trustregion radius value increases towards the end. Does it mean > that solution is not valid? > > > My output: > > > Iteration Funccount f(x) step optimality radius > 0 4 1.81987e+17 1.79e+17 1 > 1 8 1.20317e+16 1 2.34e+16 1 > 2 12 7.51981e+14 2.5 2.92e+15 2.5 > 3 16 4.69988e+13 6.25 3.65e+14 6.25 > 4 20 2.93743e+12 15.625 4.57e+13 15.6 > 5 24 1.83589e+11 39.0625 5.71e+12 39.1 > 6 28 1.14744e+10 96.1258 7.14e+11 97.7 > 7 32 7.17156e+08 79.7071 8.92e+10 240 > 8 36 4.48229e+07 39.8535 1.12e+10 240 > 9 40 2.80148e+06 19.9267 1.39e+09 240 > 10 44 175094 9.96322 1.74e+08 240 > 11 48 10942.3 4.98133 2.18e+07 240 > 12 52 683.465 2.49012 2.72e+06 240 > 13 56 42.5875 1.24396 3.41e+05 240 > 14 60 2.62715 0.619798 4.26e+04 240 > 15 64 0.155563 0.305586 5.32e+03 240 > 16 68 0.00783099 0.144585 660 240 > 17 72 0.000214572 0.0586426 73.3 240 > 18 76 8.39263e07 0.0144368 4.03 240 > 19 80 5.05294e11 0.00102595 0.031 240 > 20 84 4.49694e16 8.03849e06 9.24e05 240 > 21 88 3.90081e21 2.39824e08 2.72e07 240 > > Equation solved. > > fsolve completed because the vector of function values is near zero > as measured by the default value of the function tolerance, and > the problem appears regular as measured by the gradient. > > > y = > > 2.3911 0.3713 0.0006 > > > fval = > > 1.0e10 * > > 0.0000 > 0.0000 > 0.6246 > > > exitflag = > > 1 > > > Thanks in advance!
A large value of the trustregion radius means the solver believes that its internal model of the function is valid for a large region. Worry about a small trustregion radius, and be happy with a large one.
Alan Weiss MATLAB mathematical toolbox documentation



