coneprog: Infeasibilities are satisfied in last step from iterative display but exitflag is still -2 (NoFeasiblePointFound)

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Nick 2022-7-8

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https://ww2.mathworks.cn/matlabcentral/answers/1756245-coneprog-infeasibilities-are-satisfied-in-last-step-from-iterative-display-but-exitflag-is-still-2

评论： Torsten 2022-8-4

I am trying to solve a trajectory optimization problem with coneprog under the problem-based approach and am having convergence issues. The last iteration of the iterative display appears to show that the problem is feasibly solved:

Iter           Fval  Primal Infeas    Dual Infeas    Duality Gap    Time
 0.000000e+00   2.989297e-01   2.661135e-01   2.396722e-03    0.01
 1.671582e+00   1.371375e-01   1.220827e-01   1.099524e-03    0.02
 3.247714e+00   4.683911e-02   4.169716e-02   3.755409e-04    0.02
 5.466823e+00   2.962828e-02   2.637572e-02   2.375500e-04    0.02
 6.748515e+00   1.679716e-02   1.495319e-02   1.346743e-04    0.02
 3.367365e+00   1.788542e-03   1.592198e-03   1.433995e-05    0.03
 5.484853e+00   4.005449e-06   3.565735e-06   3.211443e-08    0.03
 5.484859e+00   6.005173e-09   5.346126e-09   4.817329e-11    0.04
 5.484864e+00   1.200136e-11   1.091778e-11   1.141043e-13    0.04

However, I am still getting a -2 exit flag and the message:

Problem is infeasible.

Is there anyway to extract anymore information about why this solution is infeasible? The functions where the solver operations occur are .p files so I can't find what's going wrong on the inside. Also, when a solution is returned infeasible all output fields are empty except the error message. I am using optimality and constraint tolerances of 1e-8 for reference, but even when I drop them down to 1e-1 just for the sake of trying to get a feasible solution returned it still does not work. Does this mean maybe that the solution converges but upper/lower bounds are violated? Any advice would be much appreciated.

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Alan Weiss 2022-8-4

The solver is designed to return empty outputs for infeasible problems. Otherwise, people are tempted to use the outputs as solutions, which they are not.

If you want to investigate why your problem is infeasible, you can try relaxing constraints until you get a feasible solution. There are several ways of relaxing constraints. You can give a linear constraint more room, such as writing A*x <= b + 10, or some other positive added constant other than 10. For cone constraints you can include them one at a time until your problem becomes infeasible. Or you can relax them similarly to the linear constraint: \norm(A*x - b) <= d*x - gamma + 10, or some other positive added constant instead of 10. Sometimes it is an interaction between bounds and other constraints that makes a problem infeasible.

You mentioned that you are doing a trajectory optimization problem. If it is similar to the one in the documentation, then I suggest that you model your problem on the problem in the documentation, which works, and change things one at a time until you reach an issue.

Good luck,

Alan Weiss

MATLAB mathematical toolbox documentation

Nick 2022-8-4

Ok I see. Thank you for such quick and helpful responses!

Torsten 2022-8-4

@Bruno Luong

Can you show how to "make a wrapper around the p function and capture the last call input" ? I honestly don't know how this can be done in the case of "coneprog".

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Answer 1

Alan Weiss 2022-7-10

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https://ww2.mathworks.cn/matlabcentral/answers/1756245-coneprog-infeasibilities-are-satisfied-in-last-step-from-iterative-display-but-exitflag-is-still-2#answer_1004065

I think that you have a misunderstanding about what the iterative display shows. For example, consider the documentation example about trajectory optimization Discretized Optimal Trajectory, Problem-Based. If you insert iterative display in an infeasible problem there, you get the following display:

p0 = [0 0 0];
pF = [5 10 3];
myprob = setupproblem(1);
opts = optimoptions("coneprog",Display="iter");
[solm,fvalm,eflagm,outputm] = solve(myprob,Options=opts);
Solving problem using coneprog.
Iter           Fval  Primal Infeas    Dual Infeas    Duality Gap    Time
   1   9.800000e-01   3.681614e-01   3.311258e-01   3.148330e-03    0.10
   2   2.051010e+00   1.708611e-01   1.536731e-01   1.461117e-03    0.12
   3   1.886027e+00   8.505728e-02   7.650085e-02   7.273667e-04    0.12
   4   2.587887e+00   1.176815e-02   1.058432e-02   1.006352e-04    0.12
   5   2.287761e+00   2.885400e-05   2.595136e-05   2.467444e-07    0.13
   6   2.286884e+00   2.348651e-07   3.892967e-08   3.701098e-10    0.13
Problem is infeasible.
function trajectoryprob = setupproblem(T,air)
    if nargin == 1
        air = false;
    end
    N = 50;
    g = [0 0 -9.81];
    p0 = [0 0 0];
    pF = [5 10 3];
    Amax = 25;
    t = T/N;
    p = optimvar("p",N,3);
    v = optimvar("v",N,3);
    a = optimvar("a",N-1,3,"LowerBound",-Amax,"UpperBound",Amax);
    trajectoryprob = optimproblem;
    s = optimvar("s",N-1,"LowerBound",0,"UpperBound",3*Amax);
    trajectoryprob.Objective = sum(s)*t;
    scons = optimconstr(N-1);
    for i = 1:(N-1)
        scons(i) = norm(a(i,:)) <= s(i);
    end
    acons = optimconstr(N-1);
    for i = 1:(N-1)
        acons(i) = norm(a(i,:)) <= Amax;
    end
    vcons = optimconstr(N+1,3);
    vcons(1,:) = v(1,:) == [0 0 0];
    if air
        vcons(2:N,:) = v(2:N,:) == v(1:(N-1),:)*exp(-t) + t*(a(1:(N-1),:) + repmat(g,N-1,1));
    else
        vcons(2:N,:) = v(2:N,:) == v(1:(N-1),:) + t*(a(1:(N-1),:) + repmat(g,N-1,1));
    end
    vcons(N+1,:) = v(N,:) == [0 0 0];
    pcons = optimconstr(N+1,3);
    pcons(1,:) = p(1,:) == p0;
    if air
        pcons(2:N,:) = p(2:N,:) == p(1:(N-1),:) + t*(1+exp(-t))/2*v(1:(N-1),:) + t^2/2*(a(1:(N-1),:) + repmat(g,N-1,1));
    else
        pcons(2:N,:) = p(2:N,:) == p(1:(N-1),:) + t*v(1:(N-1),:) + t^2/2*(a(1:(N-1),:) + repmat(g,N-1,1));
    end
    pcons((N+1),:) = p(N,:) == pF;
    trajectoryprob.Constraints.acons = acons;
    trajectoryprob.Constraints.scons = scons;
    trajectoryprob.Constraints.vcons = vcons;
    trajectoryprob.Constraints.pcons = pcons;
end

The iterative display looks a lot like yours. This problem is definitely infeasible, it is not a question of twiddling some options to make it feasible. I know that this problem is infeasible because setupproblem(1) specifies too small a time for the problem to be solved with the given limits on applied acceleration (see the example for details). Perhaps something like that is going on with your example.

Alan Weiss

MATLAB mathematical toolbox documentation

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Torsten 2022-7-10

No chance to get "solm" after the 6th iteration ?

Nick 2022-8-4

Ok yes I had a false understanding of what the iterative display is showing. Thank you for this example. As @Torsten mentioned, do you know if we should be getting any output at iterations that are not converged? Like if we were to set the maximum iterations to something smaller?

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