Finding close-to-linear solution

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Tintin Milou
Tintin Milou 2022년 5월 16일
편집: Tintin Milou 2022년 5월 17일
I am using fsolve within fsolve. The most time spent in my code is on the inner fsolve function which matlab calls 1.5 million times.
The inner function that needs to be solved is as follows:
function F = solveE(E,c)
F = E - log(c'*exp(0.99*E));
Here, E is an Nx1 vector and c is an NxN matrix. N is typically around 400. Values for c are all positive and real. Values for E are negative and real. Is there a faster way of solving this?
Within the outer fsolve, I call this function using
E0 = fsolve(@(E) solveE(E,c),E0,options);
One thing I am already doing is to use my solution E0 as an initial guess for the next iteration (when matlab adjusts its guess for the variable that solves the outer fsolve, which then will change the value for c).
Thanks for any suggestions.

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Matt J
Matt J 2022년 5월 16일
편집: Matt J 2022년 5월 16일
Pre-transpose c before the optimization to avoid repeatng the tranpose every iteration.
ct=c';
function outer(p,ct)
E0 = fsolve(@(E) solveE(E,ct),E0,options);
end
Also, for the inner problem, supply the Jacobian of F,
options.SpecifyObjectiveGradient=true;
E0 = fsolve(@(E) solveE(E,ct),E0,options);
function [F,J] = solveE(E,ct)
N=length(ct);
expE=exp(0.99*E)';
tmp=ct*expE(:);
F = E - log(tmp);
if nargout>1
J=eye(N)-0.99*(ct.*expE)./tmp;
end
end
  댓글 수: 7
Tintin Milou
Tintin Milou 2022년 5월 17일
편집: Tintin Milou 2022년 5월 17일
That might help. The Jacobian works now and it cuts the running time in half. So that's pretty good!

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추가 답변(1개)

Catalytic
Catalytic 2022년 5월 16일
편집: Catalytic 2022년 5월 16일
Using fsolve inside of fsolve is a doubtful-sounding thing to do. Why not just combine the equations from solveE with the equations in your outer problem and solve a single system?
As a simpler example, instead of having something like this as your equation function...
function F=Equations(x,z0)
z=fsolve(@(z) z^3-x, z0);
F(1)=x+1-z;
end
...it is probably better to have this
function F=Equations(xz)
x=xz(1); z=xz(2);
F(1)=z^3-x;
F(2)=x+1-z;
end
  댓글 수: 5
Matt J
Matt J 2022년 5월 17일
편집: Matt J 2022년 5월 17일
I don't know what "well-behaved" refers to here. It seems very hard to know in advance whether the result of the inner fsolve will be differentiable with respect to the outer unknowns. It certainly isn't guaranteed by the smoothness of solveE. You can see in Catalytic's example that the inner equation function @(z) z^3-x is a highly smooth, polynomial function of both z and x. Yet, solving for z leads to a non-differentiable function of x.

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