Aircraft stand allocation

Task

Approach

1. 01Costs and windows computed up front

   For every pair of flight and stand: whether the bridge applies, the handling time for the aircraft class and mode, the number of buses (one per 80 passengers), and the cost. The stand is busy from runway time plus taxiing through handling, so each pair gets its own occupancy window.

   $$c_{fs}=r^{\text{taxi}}{x}_s+b_{fs}{r}^{\text{bridge}}{h}_{k(f)}^{\text{bridge}}+(1-b_{fs})({\rho }_s{h}_{k(f)}^{\text{remote}}+n_f{r}^{\text{bus}}{\beta }_{s,\tau (f)})$$

2. 02Only the cheapest stands per flight

   Each flight keeps its 90 cheapest stands. That leaves about 99 thousand binary variables instead of 301 thousand, and the mode (bridge or bus) is fixed by the pair, so it needs no variable of its own.

3. 03One assignment model, conflicts only where they can occur

   Every flight gets exactly one stand; a stand holds at most one aircraft per minute; neighboring stands never hold two wide-bodies at once. Rows are generated only for stand-minute pairs where at least two candidates could clash.

   $$\min \sum _{(f,s)\in P}{c}_{fs}{z}_{fs}\text{s.t.}\sum _s{z}_{fs}=1\forall f,\sum _{f\in F_{s\theta }}{z}_{fs}\le 1\forall (s,\theta ),\sum _{f\in W_{s\theta }}{z}_{fs}+\sum _{f\in W_{s+1,\theta }}{z}_{f,s+1}\le 1$$

4. 04From stand groups to single stands

   The first version grouped identical stands into 22 groups on five-minute slots, with at most as many aircraft in a group as it has stands. That is exact for capacity, because overlapping time windows can always be spread over identical stands. The private data gave every stand its own taxi and bus times, so the final model works per stand, at one-minute resolution.

5. 05Open-source solver, bounded time

   Data preparation in SQL with window functions, the model in Pyomo, solved by CBC with a 90-minute limit and a 5% gap target.

Result

Stack