Solving SDP problem with cxv - difference between MATLAB and R solution

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Breden Anwar
Breden Anwar 2021-11-15
I solved the following Linear Matrix Inequality (LMI) problem using cvx in Matlab:
Lhs = [19.467593196, 1.82394007, 0.1625838, 0.01685267, 0.002495194;
1.823940068, 1.78664305, 0.9845668, 0.32951706, 0.010431878;
0.162583843, 0.98456679, 1.2333818, 0.92276329, 0.132643463;
0.016852668, 0.32951706, 0.9227633, 1.55698000, 0.848190932;
0.002495194, 0.01043188, 0.1326435, 0.84819093, 0.638889503];
S = [0.001, -0.001, 0, 0, 0;
-0.001, 0.001, 0, 0, 0;
0, 0, 0, 0, 0;
0, 0, 0, 0.001 -0.001;
0, 0, 0, -0.001, 0.001];
cvx_begin sdp
variable t
minimize t
Lhs+t*S/2 >= 0;
cvx_end
where Lhs and S are appropriate matrices. The result makes sense.
I need to solve the same problem in R. As far as I understood, it can't be expressed as a LMI. Thus, I exploited the dual formulation to write the problem as
Lhs = matrix(c(19.467593196, 1.82394007, 0.1625838, 0.01685267, 0.002495194,
1.823940068, 1.78664305, 0.9845668, 0.32951706, 0.010431878,
0.162583843, 0.98456679, 1.2333818, 0.92276329, 0.132643463,
0.016852668, 0.32951706, 0.9227633, 1.55698000, 0.848190932,
0.002495194, 0.01043188, 0.1326435, 0.84819093, 0.638889503), ncol = 5, byrow = T)
S = matrix(c(0.001, -0.001, 0, 0, 0,
-0.001, 0.001, 0, 0, 0,
0, 0, 0, 0, 0,
0, 0, 0, 0.001, -0.001,
0, 0, 0, -0.001, 0.001), ncol = 5, byrow = T)
X = Variable(k, k, PSD = T)
constr = list(matrix_trace(S%*%X) == 1,
X >= 0)
prob = Problem(Maximize(-matrix_trace(Lhs%*%X)), constr)
Unfortunately, the result is totally wrong. Where is the mistake?

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