In an NLP, I have a constraint that I would like to formulate in a convex manner preferably without introducing binary variables and/or big M formulations if possible. The actual problem is non-convex and I am happy with a local solution but I want to avoid any integer variables as there is not a good choice of open-source solvers for nonlinear problems. But of course, if it is not possible to avoid binary variables then it should be fine. The issue is to reformulate IF-THEN-ELSE condition of the type below to determine the appropriate parameter to use: $ IF \hspace{0.5mm}\ \sum_{i=1}^{n}\left(x_i{\ c}_i^t\ \ -{\ L}_t\right)\ <0$ Then $\hspace{0.5mm}\ r_t\ =\ s_t \hspace{0.5mm}\ Else \hspace{0.5mm}\ r_t\ =\ g_t$
where
- r, s , g and L are constant parameters (time dependent)
- $x_i$ is a vector of optimisation variables (continuous)
- ${\ c}_i^t$ are coefficients of $x_i$
The actual constraint is given by: $\sum_{t=1}^{T}{EP}_t=0\ $
where ${EP}_0\ =\ 0$
${EP}_1\ =\ \sum_{i=1}^{n}{x_ic_i^1} - L_1$
${EP}_2\ =\ \sum_{i=1}^{n}{x_ic_i^2} - L_2\ +\ {EP}_1\ r_1\ $
${EP}_3\ =\ \sum_{i=1}^{n}{x_ic_i^3} - L_3\ +\ {EP}_2\ r_2\ $
${EP}_t\ =\ \sum_{i=1}^{n}{x_ic_i^t} - L_t\ +\ {EP}_{t-1}\ r_t\ $
$If\ {EP}_{t-1}\ <0\ then \hspace{0.5mm}\ r_t\ =\ s_t \hspace{0.5mm}\ Else \hspace{0.5mm}\ r_t\ =\ g_t$
where
- $r_t, \hspace{0.5mm}\ s_t , \hspace{0.5mm}\ g_t and \hspace{0.5mm}\ L_t$ are constants
- $0\leq x_i \leq 1$ $\hspace{5mm}\ i \in {1,2...n}$
- ${\ c}_i^t$ are coefficients of $x_i$
- Only $x_i$ are the optimisation variables
How can this constraint be reformulated either using Big M /binary variables or using some other formulation? I don't want to use IF in the constraint as it will make the problem hard to solve. Please note that there are a number of other nonlinear constraints in the model. I was thinking that we may be able to write it using min and max, e.g., ${EP}_t\ =\ \sum_{i=1}^{n}{x_ic_i^t}\ +\ min\left\{0,{EP}_{t-1}\ \right\}\ s_t\ \ +\ max\left\{0,{EP}_{t-1}\ \right\}\ g_t$ and then introduce some auxiliary variables but I am not sure where to go next.
I have given a numerical example in the attached image which explains how this constraint is constructed and where we have determined value of x.
$c = \begin{pmatrix} 1 & 20 & 30 & 40\\ 1 & 2 & 60 & 2\\ 5 & 3 & 2 & 1\\ 10 & 8 & 5 & 40\\ 3 & 6 & 2 & 3 \end{pmatrix} , \hspace{0.5mm}\ L = \begin{pmatrix} 30\\ 60\\ 30\\ 20\\ 1 \end{pmatrix} , \hspace{0.5mm}\ x = \begin{pmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{pmatrix}, s = \begin{pmatrix} 0.001\\ 0.002\\ 0.003\\ 0.004\\ 0.005 \end{pmatrix} , g = \begin{pmatrix} 0.006\\ 0.050\\ 0.009\\ 0.010\\ 0.020 \end{pmatrix}$
The above constraint will be satisfied for the following values of $x = \begin{pmatrix} 0.363034965\\ 0.514820246\\ 0.539183467\\ 0.856864726\\ \end{pmatrix}$
