Constraints definded by epigraph of univariate convex functions

DDS accepts constraints of the form \begin{eqnarray} \sum_{i=1}^\ell \alpha_i f_i(a_i^\top x + \beta_i) + g^\top x + \gamma \leq 0, \ \ \ a_i, g \in \mathbb R^{n}, \ \ \beta_i, \gamma \in \mathbb R, \ \ i \in \{1,\ldots,\ell\}, \end{eqnarray} where \(\alpha_i \geq 0\) and \(f_i(x)\), \(i \in \{1,\ldots,\ell\}\), can be any function from the following Table

type	f(z)
1	\(-\ln(z)\)
2	\(e^z\)
3	\(z \ln(z)\)
4	\(\|z\|^p, \ p \geq 1\)
5	\(-z^p, \ z>0, \ 0 \leq p \leq 1\)
6	\(\frac 1z, \ z >0 \)

By using this simple structure, we can model many interesting optimization problems. Geometric programming (GP) and entropy programming (EP) with many applications in engineering are constructed with constraints of the above form when \(f_i(z)=e^z\) for \(i\in\{1,\ldots,\ell \}\) and \(f_i(z)=z\ln(z)\) for \(i\in\{1,\ldots,\ell \}\), respectively. The other functions with \(p\) powers let us solve optimization problems related to \(p\)-norm minimization.

Let us assume that we want to add the following \(s\) constraints to our code \begin{eqnarray*} \sum _{type} \sum_{i=1}^{\ell_{type}^j} - \alpha_i^{j,type} f_{type}((a_i^{j,type})^\top x + \beta_i^{j,type}) + g_{j}^ \top x + \gamma_{j} \leq 0, \ \ \ \ j=1,\ldots,s. \end{eqnarray*} The \(type\) of each function is given in the above table. The abbreviation we use for these constraints is TD. Hence, if the \(k\)th input block are the constraints of the bove form, then we have cons{k,1}='TD'. cons{k,2} is a cell array of MATLAB with \(s\) rows, each row represents one constraint. For the \(j\)th constraint we have:

cons{k,2}{j,1} is a matrix with two columns: the first column shows the type of a function from the above table and the second column shows the number of that function in the constraint. Let us say that in the \(j\)th constraint, we have \(l_{2}^j\) functions of type 2 and \(l_{3}^j\) functions of type 3, then we have \begin{eqnarray} \text{cons\{k,2\}\{j,1\}} =\left [\begin{array} {ccc} 2 & l_{2}^j \\ 3 & l_3^j \end{array} \right ]. \end{eqnarray} The types can be in any order, but the functions of the same type are consecutive and the order must match with the rows of \(A\) and \(b\).
cons{k,2}{j,2} is a vector with the coefficients of the functions in a constraint. Note that the coefficients must be in the same order as their corresponding rows in \(A\) and \(b\). If in the \(j\)th constraint we have 2 functions of type 2 and 1 function of type 3, it starts as \begin{eqnarray} \text{cons\{k,2\}\{j,2\}}=[\alpha_1^{j,2}, \alpha_2^{j,2}, \alpha_1^{j,3}, \cdots]. \end{eqnarray}

To add the rows to \(A\), for each constraint \(j\), we first add \(g_{j}\), then \(a_i^{j,type}\)'s in the order that matches cons{k,2}. We do the same thing for vector \(b\) (first \(\gamma_j\), then \(\beta_i^{j,type}\)'s). The part of \(A\) and \(b\) corresponding to the \(j\)th constraint is as follows if we have for example five types \begin{eqnarray} \text{A}=\left [ \begin{array} {c} g_j^\top \\[.3em] a_1^{j,1} \\ \vdots \\ a_{l_1^j}^{j,1}\\ \vdots \\ a_1^{j,5} \\ \vdots \\ a_{l_5^j}^{j,5} \end{array}\right], \ \ \ \text{b}=\left [ \begin{array} {c} \gamma_j \\[.3em] \beta_1^{j,1} \\ \vdots \\ \beta_{l_1^j}^{j,1} \\ \vdots \\ \beta_1^{j,5} \\ \vdots \\ \beta_{l_5^j}^{j5} \end{array}\right] \end{eqnarray}

Example

Assume that we want to solve \begin{eqnarray} &\min& c^\top x \nonumber \\ &\text{s.t.}& -\ln(x_2+2x_3+55) + 2e^{x_1+x_2+1} + x_1 -2 \leq 0, \nonumber \\ && -3\ln(x_1+2x_2+3x_3-30) + e^{-x_3-3} -x_3 +1 \leq 0, \nonumber \\ && x \geq 0. \end{eqnarray} For this problem, we define: \begin{eqnarray*} &&\text{cons\{1,1\}='LP'}, \ \ \text{cons\{1,2\}}=[3], \\ && \text{cons\{2,1\}='TD'}, \ \ \text{cons\{2,2\}}=\left\{ \left[\begin{array} {ccc} 1 & 1 \\ 2 & 1\end{array} \right ], [1 \ \ 2] \ ; \ \ \left[\begin{array} {ccc} 1 &1 \\ 2 & 1\end{array} \right ], [3 \ \ 1] \right\}, \\ &&\text{A\{1,1\}}=\left [ \begin{array} {ccc}-1& 0& 0 \\ 0 & -1 & 0 \\0 & 0 & -1 \end{array}\right], \ \ \ \text{b\{1,1\}}=\left [ \begin{array} {c} 0 \\ 0 \\ 0 \end{array}\right], \\ &&\text{A\{2,1\}}=\left [ \begin{array} {ccc} 1 & 0 & 0\\ 0 & 1 & 2 \\ 1 & 1 & 0 \\ 0 & 0 & -1\\ 1 & 2 & 3\\ 0 & 0 & -1 \end{array}\right], \ \ \ \text{b\{2,1\}}=\left [ \begin{array} {c} -2 \\ 55 \\ 1 \\ 1 \\ -30 \\ -3 \end{array}\right]. \end{eqnarray*}