[Ipopt] Why ipopt needs so many iterations when starting from the optimal solution?

[Pierre Schaus]

Hello

I'm making some benchmarking between ipopt and conopt (from ampl) and
strangely when I call solve two consecutive times (the second solve
starting from the optimal solution) ipopt needs 235 iterations to
rediscover this best solution while conopt only 4.

Do you have an idea how to avoid all these iterations when ipopt
starts on the optimal solution or very closely?

Thank you in advance,

Pierre

A fragment of the output when solving again:

iter objective inf_pr inf_du lg(mu) ||d|| lg(rg) alpha_du alpha_pr ls

0 -1.6889364e+03 7.06e+01 1.00e+00 -1.0 0.00e+00 - 0.00e+00 0.00e+00 0

1 -1.8076240e+03 7.06e+01 1.06e+02 -1.0 1.12e+05 - 1.12e-01 1.06e-03f 1

2 -1.7868621e+03 5.70e+01 5.00e+01 -1.0 6.42e+02 - 1.13e-01 1.92e-01h 1

3 -1.7773846e+03 5.15e+01 1.43e+02 -1.0 6.05e+02 - 2.62e-01 9.75e-02h 1

4 -1.7346472e+03 2.86e+01 2.94e+01 -1.0 6.40e+02 - 3.57e-01 4.44e-01h 1

5 -1.7138111e+03 1.80e+01 2.96e+01 -1.0 6.71e+02 - 4.02e-01 3.72e-01h 1

6 -1.6979992e+03 1.06e+01 9.60e+01 -1.0 7.65e+02 - 7.76e-01 4.09e-01h 1

7 -1.6771948e+03 2.75e+00 6.48e+01 -1.0 1.31e+03 - 2.00e-01 7.42e-01h 1

8 -1.6672749e+03 4.10e-01 1.06e+02 -1.0 9.65e+02 - 7.23e-01 9.72e-01h 1

9 -1.6663897e+03 3.55e-01 7.49e+03 -1.0 1.85e+03 - 4.22e-01 1.49e-01f 1

...

iter objective inf_pr inf_du lg(mu) ||d|| lg(rg) alpha_du alpha_pr ls

230 -1.6889365e+03 4.00e-09 1.38e-07 -8.6 6.52e-01 -9.6 1.00e+00 1.00e+00H 1

231 -1.6889365e+03 4.00e-09 5.57e+01 -8.6 3.22e+00 - 1.00e+00 1.56e-02h 7

232 -1.6889365e+03 4.00e-09 5.65e+01 -8.6 4.01e+00 - 1.00e+00 1.19e-07h 24

233 -1.6889365e+03 4.00e-09 2.07e-07 -8.6 3.11e+00 - 1.00e+00 1.00e+00H 1

234 -1.6889365e+03 4.00e-09 5.63e+01 -8.6 5.33e+00 - 1.00e+00 3.91e-03h 9

235 -1.6889365e+03 4.00e-09 5.33e-08 -8.6 9.49e-01 - 1.00e+00 1.00e+00s 22