[Ipopt] ipopt with l-bfgs: slow progress on unbounded linear objective
Stefan Vigerske
stefan at math.hu-berlin.de
Sat Jun 20 06:53:32 EDT 2015
Hi,
I cannot reproduce this behavior (and I assume you mean that x is
increasing only slowly, not decreasing).
Even when putting x>=0 as a constraint instead of just a variable bound,
Ipopt reports unboundedness after 35 iterations.
The values for x in all iterations are
curr_x[ 1]{x}= 0.0000000000000000e+00
curr_x[ 1]{x}= 1.0000000000000000e+04
curr_x[ 1]{x}= 4.0000000000000000e+04
curr_x[ 1]{x}= 1.3000000000000000e+05
curr_x[ 1]{x}= 4.0000000000000000e+05
curr_x[ 1]{x}= 1.2100000000000000e+06
curr_x[ 1]{x}= 3.6400000000000005e+06
curr_x[ 1]{x}= 1.0930000000000002e+07
curr_x[ 1]{x}= 3.2800000000000007e+07
curr_x[ 1]{x}= 9.8410000000000030e+07
curr_x[ 1]{x}= 2.9524000000000012e+08
curr_x[ 1]{x}= 8.8573000000000036e+08
curr_x[ 1]{x}= 2.6572000000000010e+09
curr_x[ 1]{x}= 7.9716100000000029e+09
curr_x[ 1]{x}= 2.3914840000000011e+10
curr_x[ 1]{x}= 7.1744530000000046e+10
curr_x[ 1]{x}= 2.1523360000000012e+11
curr_x[ 1]{x}= 6.4570081000000037e+11
curr_x[ 1]{x}= 1.9371024400000010e+12
curr_x[ 1]{x}= 5.8113073300000039e+12
curr_x[ 1]{x}= 1.7433922000000012e+13
curr_x[ 1]{x}= 5.2301766010000031e+13
curr_x[ 1]{x}= 1.5690529804000012e+14
curr_x[ 1]{x}= 4.7071589413000044e+14
curr_x[ 1]{x}= 1.4121476824000012e+15
curr_x[ 1]{x}= 4.2364430472100040e+15
curr_x[ 1]{x}= 1.2709329141640014e+16
curr_x[ 1]{x}= 3.8127987424930048e+16
curr_x[ 1]{x}= 1.1438396227480014e+17
curr_x[ 1]{x}= 3.4315188682441050e+17
curr_x[ 1]{x}= 1.0294556604732416e+18
curr_x[ 1]{x}= 3.0883669814197350e+18
curr_x[ 1]{x}= 9.2651009442592154e+18
curr_x[ 1]{x}= 2.7795302832777658e+19
curr_x[ 1]{x}= 8.3385908498332975e+19
curr_x[ 1]{x}= 1.8338590849833298e+20
I attach the full log.
If I define x>=0 as variable bound, then Ipopt needs only 6 iterations.
Stefan
On 06/18/2015 05:49 PM, Ipopt User wrote:
> When I try to solve the simple scalar problem min{-x : x >= 0} with IPOPT
> and use hessian_approximation = limited-memory, the progress is very slow.
> Essentially, x is reduced by 1 in each iteration.
>
> From what I understand, this is due to the Hessian approximation, which in
> this case is just the identity matrix (plus 1/x due to the log barrier
> term, which is significant only in the first few iterations). Solving Hs =
> -g, with H the Hessian, s the step and g the gradient, it finds s = 1.
> Indeed when I set limited_memory_init_val = 0.1, x is reduced by 10 in each
> iteration.
>
> This seems undesirable to me. Other L-BFGS based solvers like SNOPT and
> L-BFGS-B figure out that the problem is unbounded in a few iterations.
>
>
>
> _______________________________________________
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>
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