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Assuming I have a stochastic MIP that I want to minimize using Benders' Decomposition.

With the objective of the MP being:

$$ Minimize \; cx + \frac{1}{\lvert S \rvert} \sum_{s \in S} \theta_s $$

where $\theta_s$ approximates the second-stage objective.

The objective of the $s$th SP is:

$$ Minimize \; d_s y_s $$

And the dual version of that:

$$ Maximize \; \pi^T (h_s - T_s \bar{x}) $$

which are solved by fixing the fist-stage decisions to $\bar{x}$.

Using Benders' Decomposition, obviously I can keep track of the LB vs. UB by comparing the results of the MP and the SPs, with the MP being a LB on the original MIP and the sum of the SPs being a UB.

If LB = UB, the algorithm terminates:

$$ c\bar{x} + \frac{1}{\lvert S \rvert} \sum_{s \in S} \bar{\theta}_s = c\bar{x} + \frac{1}{\lvert S \rvert} \sum_{s \in S} d_s y_s $$

Now, if I use the Alternative Cut Generation Problem as proposed by Fischetti et al. (2010) , I transform my SPs with the aim to find deeper cuts:

$$ Maximize \; \pi^T (h_s - T_s \bar{x}) - \theta_s \pi_0 $$

But now, the SPs, due to this transformation, will have much smaller values.

At this point, I am unsure how to calculate the UB that the SPs construct.

Considering we basically scaled the $\pi$-variables by using $\pi_0$, due to the L1 normalization of the $\pi$-variables.

Is it a simple as adding back $ + \theta_s \pi_0 $ and then dividing that result by $\pi_0$ to "revert" the normalization of the dual variables?

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The fundamental principle you need is that, for minimization problems, upper bounds come from feasible solutions. Given $\bar{x}$, the subproblems try to find a complete feasible solution with $x=\bar{x}$. Whether you compute a primal or dual form of each subproblem $s$, you want to extract a primal feasible solution $\bar{y}_s$, and then the true objective value of the resulting complete feasible solution is your upper bound: $$c\bar{x}+\frac{1}{|S|}\sum_{s\in S}d_s\bar{y}_s$$

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  • $\begingroup$ Sorry, I forgot to mention that obviously we only get an upper bound if all subproblems are feasible with the current MP solution $\bar{x}. But very good point to use the primal version of the SP to calculate the upper bound! $\endgroup$ Commented Apr 25 at 14:18

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