[Ipopt] Which particular set of equations does Ipopt solve in practise for a given problem instance?
Martin Neuenhofen
martinneuenhofen at googlemail.com
Tue Apr 25 15:39:40 EDT 2017
Dear all,
I would be very thankful if someone knew the answer. I am afraid it is a
very stupid question. I apologize in advance. I literally ask which root
Ipopt actually attempts to compute.
Background:
I want to implement a sensitivity computation for inaccurate Hessians ( I
just approximate DF*d-w=r=0 by r(d)=[F(z+eps*d)-F(z)]/eps-w with Dr~ the
hesse approximated DF at the solution, where F is the
eps-KKT-conditions, and solve for d with a few Quasi-Newton steps) but I
need to know the accurate set of equations F=0 that Ipopt solves.
Unfortunately it does not help to let Ipopt solve the conditions very
accurately and then assume the obtained Ipopt solution would also
accurately satisfy another set of eps-KKT conditions because both sets of
equations are so badly conditioned that the solution of either will still
lead to large residuals when plugged into the other. So I need to know
Ipopt's set of eps-KKT-equations.
Issue:
In the papers there are reformulated KKT conditions for a reformulated
problem, but when output.{x,lambda,zL,zU} are given back, I need to know
exactly which size([x;lambda;zL;zU],1) equations they satisfy, expressed in
terms of the NLP callback functions/constants f,Df,c,Dc,cL,cU,xL,xU. Can
someone name them?
I know there is an sIpopt2 that gives sensitivities, too, but for me doing
some math is simpler than trying to get compiled and interfaced a new
solver and get it perform well again.
Sought information:
An answer of the following format would be superb, however I know the
following example is wrong since c=0 does not hold (since only cL<=c(x)<=cU
must hold).
F(x,lambda,zL,zU;mu) = [ Df' + Dc' * lambda + (zU-zL) ; c ; diag(x-xL)*zL -
mu ; diag(xU-x)*zU - mu ]
... I just wonder where all the slacks in the output can be found. There
should be a vector s of size mg<=m, where mg is nnz(cU-cL) and two
additional Lagrange multipliers for them.
Kind regards
Martin
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