VectorAffineFunction-in-SecondOrderCone -> ScalarQuadraticFunction-in-LessThan bridge

[blegat]

The duals won't work for this one though since the dual of ScalarQuadraticFunction-in-LessThan is just a scalar (see jump-dev/MathOptInterface.jl#43), we loose information in comparison with the vector dual of the SOC

[joaquimg]

Once I asked about that and I got the following response "It should be possible to derive the cone dual vector as the gradient of the quadratic cone constraint, scaled by the single dual multiplier". I looked ok, am I missing something?

[blegat]

||Ax + b|| <= c'x + d is rewritten into 2x'A'Ax + (2b'A - c')x + b'b - d for which the gradient is A'Ax + (2b'A - c'). The constraint primal is (c'x + d, Ax + b), by complementary slackness we know that

• if it is zero, the constraint dual could be anything;
• If it is in the interior of the SOC cones, the constraint dual is zero;
• otherwise the constraint dual is at the other side of the cone, i.e. it is a multiple of (c'x + d, -(Ax + b)) (not that the scalar product is with the constraint primal is (c'x + d)^2 - ||Ax + b||^2 which is zero since it is in the boundary if the SOC cone`.

I do not see how to link the gradient with the constraint dual but we might compute (c'x + d, -(Ax + b)) with the variable primal if we are in the third case. However, with the definition of https://github.com/JuliaOpt/MathOptInterface.jl/pull/43/files, in this case the constraint dual of the quadratic constraint is zero so it does not give the necessary scaling information.