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
asked
Mar 09 '11 at 15:43
pierre schaus
544●4
Also, I wouldn't be too bothered about the no. of iterations. It's the solve time that really matters.
I didn't know this mailing list. Thank you Gilead for referencing it.