Is there some memory in the solver?

[mss]

Hi @Muhammad

@Muhammad

I expect the initial orientation of the resulted trajectory is 0.75 rad = 45 deg since it has been setted in init_x.
How to set the initial position of the solution? Is it using constr_x0 or using init_x.

This is constr_x0. init_x is only the starting point for the solver from where he tries to find a solution.

@Muhammad

My goal to understand what is the reason of the behavior of the resulted trajectory is the next plot. Why does the trajectory choose to turn to right and not to left.

Probably because your cost is lower on this way. Some questions to come closer:

• Do you have enough iterations for your solver that it can converge?
• Is the accuracy high/low enough that the solution is feasible?
• Do you get the same behavior if you have everything rotated by 45° (your ref is on the x-axis)?
  • Might the influence of u in your cost function try to reduce the used energy?
  • Other approaches: almost no weight on u or penalizing delta u ==> your actual task needs no control weighting, but not sure if weight == 0 works properly.

@Muhammad

Is it because the initial states or something another?

That you will find out, if you solve the problem once by init_x = zeros( ), init_u = zeros( ) [all of the below mentioned]

@FreyJo

In Matlab, you can use

ocp.set('init_x', *);
ocp.set('init_u', *);
ocp.set('init_lam', *);
ocp.set('init_pi', *);

and then by solving it with e.g. your reference values for init_x. Create the solver each time new, for the case of other internal states.