What happens when I give a partial x0?

[othman moumni]

Hi,
I have a High level model in YALMIP  and I want to know if using the option usex0 works when I assign partially my variables because it is difficult to give a complete x0 that respects all the constraints.

Thanks!

ps: I use IPOPT as a solver

[Johan Löfberg]

>> sdpvar x y

>> assign(x,3);

>> solvesdp([],x^2+y^2,sdpsettings('usex0',1,'solver','ipopt'))

means x will start at 3 and y will be initialized to 0 when ipopt is told to start from a current solution

[othman moumni]

OK, but what happens when I have a constraint like:

>> sdpvar x y

>> assign(x,3);

>> solvesdp([x+y>=4],x^2+y^2,sdpsettings('usex0',1,'solver','ipopt'))

the initial point with x=3 and y=0 is infeasible. Does the solver take into account the partial assignment?

[Johan Löfberg]

Depends on the solver. Some solvers will complain that the supplied initial point is infeasible, some will try to start from the infeasible point anyway

[othman moumni]

Ok,
Maybe this is not the case for Ipopt as it uses interior points methods. I will look into Ipopt documentation.

Thank you for your help

[Johan Löfberg]

Being an IP method does not mean it never works with infeasible points. Problems are shifted and homogenized etc in these algorithms. How would ipopt be able to start from scratch otherwise?

Haven't you even run the code? It runs perfectly fine from infeasible initial point

%Let ipopt start from where it wants 

sdpvar x y

solvesdp([x^4+y^4<=4],(x-1)^2+(y-1)^2,sdpsettings('solver','ipopt'))

% Start from infeasible

assign(x,3);

assign(y,0);

solvesdp([x^4+y^4<=4],(x-1)^2+(y-1)^2,sdpsettings('usex0',1,'solver','ipopt'))

% Start from optimal

solvesdp([x^4+y^4<=4],(x-1)^2+(y-1)^2,sdpsettings('usex0',1,'solver','ipopt'))

[othman moumni]

I have run the code but I thought that maybe IPOPT simply ignore the starting point given.