Possible bug with coneprog?

조회 수: 6 (최근 30일)
Nick
Nick 2022년 8월 4일
댓글: Nick 2022년 8월 15일
I am wondering if this is a bug in coneprog or expected behavior. The example used to reproduce this is from https://www.mathworks.com/help/optim/ug/discretized-optimal-trajectory-cone-programming.html . When the code below is run as is then the expected result is outputted. However, when a "dummy" constraint is added (by uncommenting the two lines in the setupproblem function) that has nothing to do with the original it changes the answer signicantly.
Thanks in advance for any responses.
clc,clear,close all
p0 = [0 0 0];
pF = [5 10 3];
trajprob = setupproblem(20);
opts = optimoptions('coneprog','Display','iter');
[sol,fval,eflag,output] = solve(trajprob,options=opts)
plottrajandaccel(sol,p0,pF)
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;
%%%%%%%%% NEW DUMMY VARIABLE AND CONSTRAINT %%%%%%%%%
% test_var = optimvar("test_var",N,1);
% trajectoryprob.Constraints.test_cons = test_var >= 0;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
end
function plottrajandaccel(sol,p0,pF)
figure
psol = sol.p;
plot3(psol(:,1),psol(:,2),psol(:,3),'rx')
hold on
plot3(p0(1),p0(2),p0(3),'ks')
plot3(pF(1),pF(2),pF(3),'bo')
hold off
view([18 -10])
xlabel("x")
ylabel("y")
zlabel("z")
legend("Steps","Initial Point","Final Point")
figure
asolm = sol.a;
nasolm = sqrt(sum(asolm.^2,2));
plot(nasolm,"rx")
xlabel("Time step")
ylabel("Norm(acceleration)")
end

채택된 답변

Alan Weiss
Alan Weiss 2022년 8월 5일
I think that you have identified a problem with the default options for trajectory problems: the optimality tolerance should be chosen smaller than default. If you run the problem with a tighter optimality tolerance, ther results look better, and look pretty much the same with or without your dummy variables.
p0 = [0 0 0];
pF = [5 10 3];
trajprob = setupproblem(20);
opts = optimoptions('coneprog','Display','iter','OptimalityTolerance',1e-10);
[sol,fval,eflag,output] = solve(trajprob,options=opts)
Solving problem using coneprog. Iter Fval Primal Infeas Dual Infeas Duality Gap Time 1 1.960000e+01 3.333333e-01 4.000000e-01 2.110223e-02 0.05 2 1.834298e+02 3.226790e-02 3.872148e-02 2.042774e-03 0.06 3 4.671236e+02 3.317278e-03 3.980734e-03 2.100059e-04 0.06 4 1.794310e+02 1.393238e-03 1.671886e-03 8.820130e-05 0.07 5 2.110933e+02 3.292572e-04 3.951086e-04 2.084418e-05 0.07 6 1.970150e+02 9.771943e-05 1.172633e-04 6.186295e-06 0.07 7 1.929653e+02 2.301163e-05 2.761396e-05 1.456790e-06 0.08 8 1.923943e+02 3.491190e-06 4.189428e-06 2.210157e-07 0.08 9 1.923107e+02 8.565044e-07 1.027805e-06 5.422247e-08 0.08 10 1.922943e+02 2.858454e-07 3.430144e-07 1.809593e-08 0.09 11 1.922915e+02 2.171556e-07 2.605867e-07 1.374740e-08 0.09 12 1.922875e+02 9.214488e-08 1.106251e-07 5.833388e-09 0.09 13 1.922872e+02 8.633602e-08 1.036511e-07 5.465649e-09 0.10 14 1.922855e+02 6.106756e-08 7.343547e-08 3.865986e-09 0.10 15 1.922849e+02 5.618335e-08 6.756207e-08 3.556783e-09 0.10 16 1.922831e+02 2.403072e-08 2.936808e-08 1.521302e-09 0.11 17 1.922829e+02 1.899389e-08 2.321210e-08 1.202437e-09 0.11 18 1.922825e+02 1.398995e-08 1.678817e-08 8.856555e-10 0.11 19 1.922819e+02 8.372654e-09 1.004769e-08 5.300445e-10 0.12 20 1.922815e+02 3.353846e-09 4.025421e-09 2.123187e-10 0.12 21 1.922814e+02 2.234111e-09 2.684028e-09 1.414315e-10 0.12 22 1.922813e+02 1.668567e-06 2.709764e-09 9.146051e-11 0.13 23 1.922812e+02 5.259403e-07 5.090527e-07 2.996456e-11 0.13 24 1.922812e+02 3.930634e-07 3.850559e-07 2.266504e-11 0.13 25 1.922812e+02 1.436881e-07 4.543098e-07 9.073223e-12 0.14 Optimal solution found.
sol = struct with fields:
a: [49×3 double] p: [50×3 double] s: [49×1 double] v: [50×3 double] test_var: [50×1 double]
fval = 192.2812
eflag =
OptimalSolution
output = struct with fields:
iterations: 25 primalfeasibility: 1.4369e-07 dualfeasibility: 4.5431e-07 dualitygap: 9.0732e-12 algorithm: 'interior-point' linearsolver: 'prodchol' message: 'Optimal solution found.' solver: 'coneprog'
plottrajandaccel(sol,p0,pF)
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;
%%%%%%%%% NEW DUMMY VARIABLE AND CONSTRAINT %%%%%%%%%
test_var = optimvar("test_var",N,1,"LowerBound",-1,"UpperBound",1);
trajectoryprob.Constraints.test_cons = test_var >= 0;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
end
function plottrajandaccel(sol,p0,pF)
figure
psol = sol.p;
plot3(psol(:,1),psol(:,2),psol(:,3),'rx')
hold on
plot3(p0(1),p0(2),p0(3),'ks')
plot3(pF(1),pF(2),pF(3),'bo')
hold off
view([18 -10])
xlabel("x")
ylabel("y")
zlabel("z")
legend("Steps","Initial Point","Final Point")
figure
asolm = sol.a;
nasolm = sqrt(sum(asolm.^2,2));
plot(nasolm,"rx")
xlabel("Time step")
ylabel("Norm(acceleration)")
end
I will update the example in the documentation.
Alan Weiss
MATLAB mathematical toolbox documentation
  댓글 수: 1
Nick
Nick 2022년 8월 15일
Ok I understand the problem. Thanks Alan!

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