The uncapacitated joint location‐inventory problem with shortage costs: a selective extended cutting plane algorithm

Pablo Escalona et al.

International Transactions in Operational Research2026https://doi.org/10.1111/itor.70160article
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This paper addresses the uncapacitated joint location‐inventory problem (JLIP) to design an optimal distribution network for fast‐moving consumer goods where distribution centers (DCs), operating under a continuous review inventory policy, observe uncertain demand from a set of retailers, and shortages are penalized by a cost per unit and time unit. The JLIP is considered intractable, depriving a relevant network design approach when large retail chains charge a fee or issue invoice deductions for backorders. Considering the exact formulation of the inventory on‐hand and backorders at each DC, and using conic programming, we show how to reformulate the JLIP with shortage costs as a convex mixed integer nonlinear problem with second‐order cone constraints. Using convexity‐preserving operations, we develop an extended cutting plane algorithm that takes advantage of a predefined set of linear functions representing the outer linear approximation of the backorders. The computational results highlight the efficiency of the approach in solving small‐, medium‐, and large‐scale instances to optimality compared to the state‐of‐the‐art algorithms available in commercial and open‐source solvers.

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https://doi.org/https://doi.org/10.1111/itor.70160

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@article{pablo2026,
  title        = {{The uncapacitated joint location‐inventory problem with shortage costs: a selective extended cutting plane algorithm}},
  author       = {Pablo Escalona et al.},
  journal      = {International Transactions in Operational Research},
  year         = {2026},
  doi          = {https://doi.org/https://doi.org/10.1111/itor.70160},
}

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0.50

Balanced mode · F 0.40 / M 0.15 / V 0.05 / R 0.40

F · citation impact0.50 × 0.4 = 0.20
M · momentum0.50 × 0.15 = 0.07
V · venue signal0.50 × 0.05 = 0.03
R · text relevance †0.50 × 0.4 = 0.20

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