Solving the gate assignmentat Amsterdam Airport Schiphol

Han Hoogeveen

May 19, 2015

Gate assignment at Schiphol

Gate assignment consists of three stages:

I Seasonal planning

I One-day-ahead planning

I Tactical planning

We look at the One-day-ahead planning problem

Problem description

For each flight we know:

I Its arrival and departure time;

I The type of plane that flies it;

I Origin and destination;

I The preferences of the airline;

I The handling company.

Problem description (2)

Possibility of splitting flights

I Stays longer than 3 hours may be split

I Stays with different origin and destination region may besplit

Problem description (3)

Goal: find a robust solution subject to

I a number of single-gate constraints;

I a number of multiple-gate constraints;

The problem instance cannot be partitioned without loss ofcapacity!

Model - Cost function

Several possibilities: all based on idle time

I Square idle time (Based on variance)

I Any well-chosen function: we use the arc tangent.

Cost function refinement:

I Punctuality-factor(i, j)

You need to know the immediate predecessor of each plane atthe gate.

Arc tangent function

Approach: ILP

I Standard formulation: assign a flight to a position on agate; this requires many variables.

I Multiple-gate constraints yield many constraints

I Problems with 30 flights are solvable

We need a better formulation.

Two-phase ILP approach

I Phase 1: find for each set of gates of the same type theright number of ‘gate plans’, while taking into account thesingle-gate constraints.

I Phase 2: assign the gate plans to the actual gates, whiletaking into account the multiple-gate constraints.

We need to distinguish between several types of gates becauseof size, security, etc.

Phase 1 - Model

A gate plan of type h is feasible if

I the minimum idle time is at least 20 minutes

I each flight can be assigned to a gate of type h

Phase 1 - Model (2)

Suppose all feasible gate plans of all types are known.

Gate plan j is characterized by

gij =

{1 if gate plan j contains flight i;0 otherwise.

thj =

{1 if gate plan j is of type h;0 otherwise.

Model - Constraints

Basically two constraints are needed:

I Each flight must be assigned exactly once

I the number of gateplans of type h is equal to the numberof gates of type h

Model - Constraints (2)

Minimisen∑

j=1

cjxj s.t.

n∑j=1

gijxj = 1 for i = 1, . . . ,m

n∑j=1

thjxj = Th for h = 1, . . . ,H

xj ∈ {0, 1} (j = 1, . . . , n)

Model - Constraints (3)

Possibility of preferences:

I El Al has designated gates

I Some airlines have own gates

n∑j=1

m∑i=1

H∑h=1

pihrthjgijxj ≥ Pr for r = 1, . . . , R

Model - Constraints (4)

Possibility of not assigning flights:

n∑j=1

gijxj + Ui = 1 for i = 1, . . . ,m

This is only used for finding a feasible solution quickly.

Solving - First phase

Overview solving first phase:

I Relax original GA problem to LP-relaxation

I Solve LP-relaxation with column generation

I Solve ILP problem for the set of gate plans discovered insolving the LP-relaxation

ZLP ≤ ZGA ≤ ZILP

Pricing problem

Use graph Gh = (V,E) for representing gate plans for gate oftype h

II Vertex for every possible flight on gate of type h

I Start and End vertex

I Arc for all possible successive flights

I Path through graph is feasible gate plan

Pricing problem (2)

Set the cost of the arcs such that the length of the path isequal to the reduced cost of the corresponding gate plan.

I Assumption: Flights are sorted on arrival times

I Result: flight indices form topological ordening

I Finding shortest path: O(|V |+ |E|)

Enhancement

Creating additional columns

1. Remove one flight in the currently optimal gate plan fromthe graph (once for each flight)

2. Recalculate shortest path

3. Put newly found gate plan in additional pool

I May lead to a better solution

I More importantly: this speeds up the ILP tremendously.

Solving - Second phase

Assigning gate plans to physical gates

I Impossible constraints in first phase:I Overlapping wingspansI Combining 2 smaller gatesI Flights with same departure time on neighboring gatesI Flights with same departure time acrossI ‘Opposing groups of passengers’ close to each other

I Usage of additional gate details

Testing

Implementation of algorithm tested with:

I Random data

I Seasonal data provided by Schiphol

I Detailed data for one day, provided by Schiphol

Results

I Maximum needed time for solving LP: 1000 seconds.

I Maximum generated columns: 2600

I Size extra columnpool: 190.000

I Maximum needed time for solving ILP: 2300 seconds.

I Typical integrality-gap: 0.00%− 0.04%

Acceptable running times (speed-up with a factor of 3 by usingnew version of CPLEX)

Conclusion

I New ILP formulation

I Realistic constraints

I Integrality gap relatively small

I Acceptable running time

I Promising results with real life input data