Qurious Cube
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Maybe this can work? Numbering is same as sorting. Disregard the layers and treat the whole thing as a single graph. Use Cuthill-Mckee (if the bandwidth is low) or other heuristics for the graph ...

According to the equality constraint (the equal to 4 one), at least two of the $x_{?}$ are 1. Therefore, the slack for the 1st constraint is at most 3. According to the equality constraint, the worst ...

Treat nodes in 1st and 3nd layer as extra edges in the 2st layer to make a merged layer. Minimize bandwidth of 2nd layer. Repeat for the other layers. layer 2: o o o \ / \ / layer 1: @ ...

Here is a small variation of RobPratt's answer. I will use two sets, as an example. Two sets of constants are given: $S_1 = \lbrace x_{1,1}, x_{1,2}, x_{1,3} \rbrace$ and $S_2 = \lbrace x_{2,1}, x_{2,... View answer 1 votes For timing within the gurobi solver, maybe you can call <optimizer>.solve() method with report_timing=True, as described in pyomo's documentation at https://pyomo.readthedocs.io/en/stable/... View answer 1 votes Minimize$x^2$where$1 \le x \le 2. \begin{aligned} \min_{x} \quad & f(x)\\ \textrm{s.t.} \quad & h_{1}(x) \le 0\\ &h_{2}(x) \le 0 \\ \end{aligned} where \begin{align} f(x) &= x^... View answer 0 votes Transform the optimizing variablesx$and$y$in everything (the whole model) to$u = \frac{ax}{\ln(b + cy)}$and$v = x$Transformed constraint$u + dv \le A\$ is linear. Therefore, the transformed ...