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If the quadprog problem is infeasible, how i can solve it with a certain tolerance?

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Mario Bernardo
Mario Bernardo on 24 Jul 2021
Commented: Mario Bernardo on 25 Jul 2021
F1=[0.2700; 0.2300; 0.2500; 0.2400; 0.2400; 0.2400; 0.2500; 0.2500; 0.2400; 0.2500; 0.2400; 0.2600; 0.2600; 0.2300; 0.2600;
0.2700; 0.3100; 0.3200; 0.3500; 0.3700; 0.3700; 0.4000; 0.4100; 0.4600; 0.4800; 0.4400; 0.5000; 0.4800; 0.5000; 0.4800;
0.4800; 0.4700; 0.4500; 0.4700; 0.4500; 0.4700];
F2=[1.8900; 1.8600; 0.3700; 0.7400; 0.1400; 1.8000; 0.6100; 1.6900; 0.4000; 0.3000; -0.9700; 0.4800; 1.3900; -1.7800; -2.3000;
-0.8300; 0; -0.1800; 1.8400; 0.1400; -1.4400; -0.0700; -0.1800; 0.8000; 1.9000; 2.1700; 0.6800; 1.3100; 3.9500; 0.6600; -0.2500;
1.1700; 0.9500; 1.4700; 1.4400; 1.4400];
F3=[0.8100; 1.3600; 2.1100; -2.4200; 2.6500; 0.2700; -0.4200; 3.7900; -0.7900; 2.0800; -1; 1.2200; 3.3800; -2.7100; -4.3900; 1.2900;
1.6300; -2.4700; 3.2800; 4.0800; -2.4500; 2.2500; -3.6500; 1.4600; 2.5800; 3.8800; 2.9500; 2.9400; 3.9500; 2.3100; 3.3000; 0.2500;
4.2300; -0.3600; 4.3800; 1.9300];
F4=[2.0700; -1.4100; 3.3800; -3.3800; 4.0700; 0.7700; 0.4400; 4.3600; 2.0900; 1.2800; -3.0300; 3.3500; 2.9300; -1.4500; -4.5100;
0.8600; -0.4200; -3.0900; 2.6900; 4.8700; -1.0200; 0.5200; -3.9500; 1.2500; 0.9200; 4.0700; 2.2600; 1.8100; 2.2700; 4.6800; 5.8600;
2.5700; 2.7200; -2.7300; 4.1200; 1.1800];
F5=[0; 0.8600; 5.1000; 2.1200; 1.5800; -1.3600; 0.4000; 8.8600; -0.2700; 3.7200; -2.1900; 7.6600; 4.7100; -3.6200; -3.0100; 4.9100;
-4.1100; -1.1100; 5.2900; 2.9600; -4.0800; 4.0800; -3.8400; 0.5200; -0.6000; 1.9700; 3.9400; 4.2000; 2.5600; 1.2100; 5.2200; -3.9700;
3.0300; -0.5700; 0.5800; 3.0800];
F6=[-0.2600; 4.5100; 12.3100; 15.8200; 4.1000; -2.7200; 5.7800; 3.1000; -3.4500; 2.2900; -14.3100; 6.9000; 11.3500; 2.3000; -5.3700;
4.7600; 2.5300; 2.9800; -1.9900; 1.6900; -2.4900; 2.3900; -5.6600; 0.9800; -6.1000; -2.8200; 6.9800; 5.0800; -3.8800; -4.4000;
7.2300; -3.8500; 1; -4.8700; 4.9300; 4.8700];
R=[2.197; 2.707; 3.189; 3.191; 1.62; 0.07845; 1.318; 3.396; -0.9583; 0.6649; -1.762; 2.268; 2.688; -1.746; -4.104; -0.8574; 0.1571;
-1.031; 1.917; 0.9404; -0.7029; -1.181; -1.434; 0.08219; 0; 2.197; 1.682; 1.027; 1.878; 1.091; 1.461; 0.2271; 0.9026;
-0.9091; 2.829; 1.644];
numberOfAssetClass=6;
T=length(F1);
Fmatrix=[F1,F2,F3,F4,F5,F6];
F1mean=mean(F1);
F2mean=mean(F2);
F3mean=mean(F3);
F4mean=mean(F4);
F5mean=mean(F5);
F6mean=mean(F6);
FmeanVector=[F1mean; F2mean; F3mean; F4mean; F5mean; F6mean];
Rmean=mean(R);
H=NaN(numberOfAssetClass,numberOfAssetClass);
for i=1:numberOfAssetClass
for j=1:numberOfAssetClass
H(i,j) = 0;
for t=1:T
H(i,j) = H(i,j) + (Fmatrix(t,i)-FmeanVector(i))*(Fmatrix(t,j)-FmeanVector(j));
end
H(i,j) = H(i,j)/T;
end
end
H=H*2;
f=NaN(numberOfAssetClass,1);
for i=1:numberOfAssetClass
f(i) = 0;
for t=1:T
f(i) = f(i) + 2*(-R(t) + Rmean)*(Fmatrix(t,i)-FmeanVector(i));
end
f(i) = f(i)/T;
end
Aeq=Fmatrix;
beq=R;
lb = zeros(6,1);
ub = ones(size(lb));
b=quadprog(H,f,[],[],Aeq,beq,lb,ub);

Accepted Answer

Matt J
Matt J on 24 Jul 2021
Edited: Matt J on 24 Jul 2021
It never really makes sense to have more equality constraints (here 36) than you do unknowns ( here 6). In order for the problem to be feasible, some of the constraints would have to be redundant.

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