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I am trying to solve an optimization problem with the Interior Point Method. 4 variables, 8 constraints, of which 2 enforce non-positivity on variables 1 & 2. How would the entirior point method take this non-positivity into account?

For both I assume to need a Barrier function, which affects the Lagrangian (and therefore the KKT conditions). Does anybody know how I can successfully incorporate this in the KKT (Newton-Raphson) matrix form?

Thank you for your time.

$ \begin{bmatrix} ... & ... & ... \\ ... & ... & ...\\ ... & ... & ... \end{bmatrix} \begin{bmatrix} \Delta x\\ \Delta w\\ \Delta y \end{bmatrix} = \begin{bmatrix} ...\\ ...\\ ... \end{bmatrix}$

So far I tried the following solution. Given the dummy problem:

min $f(x1,x2)$

s.t. $h(x) \geq 0$, $x1 \leq 0$

Which for interior point (barrier method) can be rewritten as,

min $f(x1,x2) -\mu*log(-x1) - \mu*log(w)$

s.t. $h(x)-w=0$

$w, -x_1 \geq 0$

Equating the derivative of Lagrangian w.r.t. $x_i$ to 0 yields,

$\nabla f(x) -\nabla h(x)^Ty-\mu*X_n^{-1}e = 0$

where, $X_n = \begin{bmatrix} -x_1 & 0 \\ 0 & 1 \end{bmatrix}$ and $e = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$

Rewriting this results in, $X_n \nabla f(x) -X_n \nabla h(x)^Ty-\mu*e = 0$

To define the Newton step it is required to perturb the system (and neglect higher order terms):

($X_n+\Delta X_n) ( \nabla f(x+\Delta x) $+....

$X_n\nabla f(x) + X_n \nabla^2 f(x)\Delta x +\Delta X_n \nabla f(x) +....$ (Equation I)

Finally, to put this in Matrix form (see the beginning of my post) I have to relate $\Delta X_n$ to $\Delta x$. Noting that,

$\Delta X_n = \begin{bmatrix} \Delta x_1 & 0 \\ 0 & 0 \end{bmatrix}, \Delta x = \begin{bmatrix} \Delta x_1 \\ \Delta x_2 \end{bmatrix} --> \Delta X_n \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \Delta x $

Though this is not a direct relation. As shown in the last equation, I can only find a relation for the transformed column vector of the diagonal matrix. only. Not sure if this is the way to go, or if there is an algebraic trick that I miss.

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    You might find it easier to simply rewrite your problem in terms of nonnegative variables. If you do stay with non-positive variables, keep in mind that the sign restriction on the Lagrange multipliers will be different from what you're used to with non-negative variables. – 2017-02-22
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    Thank you for your time. In the problem I mentioned I did rewrite the system to non-negative values ($-x1 \geq 0$), I will edit my post to make this more clear. Though in the end I am not sure how to implement this for the interior point algorithm (my suggestion seems not to work, though it can be a bug as well). – 2017-02-23

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