9
You can add an extra binary that equals $1$ if and only if the first constraint is satisfied:
\begin{align}
x_1+x_2+x_3 &\ge \delta\\
x_1+x_2+x_3 &\le 3\delta\\
x_4+x_5+x_6 &\ge 1 - \delta\\
x_4+x_5+x_6 &\le 3(1 - \delta)\\
\delta &\in \{0,1\}
\end{align}
If $\delta=1$, the first two constraints become:
$$
1 \le x_1+x_2+x_3 \le 3
$$
And ...
5
This is the transportation problem, a network flow problem (in a directed bipartite network) and hence solvable exactly as an LP. If you prefer a combinatorial algorithm, this might be a good exercise to learn the primal-dual method, and your heuristic approach can provide the initial solution.
4
It seems to me that this problem can be reduced to a variant of the generalized assignment problem (GAP). If you set $U_{\max} = 1, \forall u$, then your problem is indeed GAP with equal weights for items. I know GAP is NP-hard, but not sure about this variant of GAP. I guess you can find some good stuff if you look for variants of GAP.
4
Let $w_o$ denote the weight of object $o$, and let $c_b$ denote the capacity of bin $b$.
You can interpret this as a job shop scheduling problem. The correspondence is that each object is a job, with duration $w_o$, each bin is a machine that is available for only $c_b$ time units, and $z$ is the makespan.
It is also a special case of the bottleneck ...
3
I did a few computational trials on a small example, and it appears that you can solve the Lagrangean relaxation via a gradient based method. If we reverse the constraint requiring all users to be assigned at least one provider, to $$-\sum_{c=1}^C d_{u,c}\ge -1\,\forall u$$(so that all multipliers are nonnegative), the Lagrangean problem is$$\min_{\lambda,\...
2
I assume one way to gain good level of interpretation of solver's result in optimization (especially in a high dimensional setting) is by doing something called Scenario Discovery. You can read more about this from linked papers here.
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