# Scheduling Flight Landings

## Contents

## Problem description

The plane movements in large airports are subject to numerous security constraints. The problem presented in this section consist of calculating a schedule for flight landings on a single runway. More general problems have been studied but they are fairly complicated (dynamic cases, for instance through planes arriving late, instances with several runways, etc.).

Ten planes are due to arrive. Every plane has an earliest arrival time (time when the plane arrives above the zone if traveling at maximum speed) and a latest arrival time (influenced among other things by its fuel supplies). Within this time window the airlines choose a target time, communicated to the public as the flight arrival time. The early or late arrival of an aircraft with respect to its target time leads to disruption of the airport and causes costs. To take into account these cost and to compare them more easily, a penalty per minute of early arrival and a second penalty per minute of late arrival are associated with every plane. The time windows (in minutes from the start of the day) and the penalties per plane are given in the following table.

Characteristics of flight time windows

+-----------------+---+---+---+---+---+---+---+---+---+---+ |Plane | 1| 2| 3| 4| 5| 6| 7| 8| 9| 10| +-----------------+---+---+---+---+---+---+---+---+---+---+ |Earliest arrival |129|195| 89| 96|110|120|124|126|135|160| |Target time |155|258| 98|106|123|135|138|140|150|180| |Latest Arrival |559|744|510|521|555|576|577|573|591|657| |Earliness penalty| 10| 10| 30| 30| 30| 30| 30| 30| 30| 30| |Lateness penalty | 10| 10| 30| 30| 30| 30| 30| 30| 30| 30| +-----------------+---+---+---+---+---+---+---+---+---+---+

Due to turbulence and the duration of the time during which a plane is on the runway, a security interval has to separate any two landings. An entry in line p of column q in the following table denotes the minimum time interval (in minutes) that has to lie between the landings of planes p and q, even if they are not consecutive. Which landing schedule minimizes the total penalty subject to arrivals within the given time windows and the required intervals separating any two landings?

Matrix of minimum intervals separating landings

+--+--+--+--+--+--+--+--+--+--+--+ | | 1| 2| 3| 4| 5| 6| 7| 8| 9|10| +--+--+--+--+--+--+--+--+--+--+--+ | 1| –| 3|15|15|15|15|15|15|15|15| | 2| 3| –|15|15|15|15|15|15|15|15| | 3|15|15| –| 8| 8| 8| 8| 8| 8| 8| | 4|15|15| 8| –| 8| 8| 8| 8| 8| 8| | 5|15|15| 8| 8| –| 8| 8| 8| 8| 8| | 6|15|15| 8| 8| 8| –| 8| 8| 8| 8| | 7|15|15| 8| 8| 8| 8| –| 8| 8| 8| | 8|15|15| 8| 8| 8| 8| 8| –| 8| 8| | 9|15|15| 8| 8| 8| 8| 8| 8| –| 8| |10|15|15| 8| 8| 8| 8| 8| 8| 8| –| +--+--+--+--+--+--+--+--+--+--+--+

## Variables

costs Cost or being late mintimes Minimal intervals between landings var1 30 plus var2 30 minus

## Reference

Applications of optimization... Gueret, Prins, Seveaux

% Marcus Edvall, Tomlab Optimization Inc, E-mail: tomlab@tomopt.com % Copyright (c) 2005-2009 by Tomlab Optimization Inc., $Release: 7.2.0$ % Written Oct 7, 2005. Last modified Apr 8, 2009.

## Problem setup

costs = [ 129 195 89 96 110 120 124 126 135 160;... 155 258 98 106 123 135 138 140 150 180;... 559 744 510 521 555 576 577 573 591 657;... 10 10 30*ones(1,8);... 10 10 30*ones(1,8)]; mintimes = [ 0 3 15 15 15 15 15 15 15 15;... 3 0 15 15 15 15 15 15 15 15;... 15 15 0 8 8 8 8 8 8 8;... 15 15 8 0 8 8 8 8 8 8;... 15 15 8 8 0 8 8 8 8 8;... 15 15 8 8 8 0 8 8 8 8;... 15 15 8 8 8 8 0 8 8 8;... 15 15 8 8 8 8 8 0 8 8;... 15 15 8 8 8 8 8 8 0 8;... 15 15 8 8 8 8 8 8 8 0]; p = size(costs,2); q = p; prec = tom('prec',p,q,'int'); land = tom('land',p,1,'int'); early = tom('early',p,1,'int'); late = tom('late',p,1,'int'); BIG = 1e4; % All variables are integers. bnds = {costs(1,:)' <= land <= costs(3,:)'; 0 <= prec <= 1; 0 <= early <= costs(2,:)'-costs(1,:)'; 0 <= late <= costs(3,:)'-costs(2,:)' prec(1:p+1:p^2) == 0}; % q is less than p con1 = []; for i=2:p con1{i-1} = {prec(i,i+1:end) == 0}; end % Disjunctive constr 1 con2 = []; counter = 1; for i=1:p-1 % q is less than p for j=i+1:q con2{counter} = {land(i) + mintimes(i,j) <= ... land(j) + BIG*prec(j,i)}; counter = counter + 1; end end % Disjunctive constr 2 con3 = []; counter = 1; for j=1:p-1 % p is less than q for i=j+1:p con3{counter} = {land(i) + mintimes(i,j) <= ... land(j) + BIG*(1-prec(i,j))}; counter = counter + 1; end end % Landing constraint con4 = {land == costs(2,:)' - early + late}; % Objective objective = costs(4,:)*early + costs(5,:)*late; constraints = {bnds, con1, con2, con3, con4}; options = struct; options.solver = 'cplex'; options.name = 'Scheduling Flight Landings'; sol = ezsolve(objective,constraints,[],options); PriLev = 1; if PriLev > 0 planes = 10; % number of planes [times,idx] = sort(sol.land); late = sol.late; early = sol.early; for p = 1:planes, time = times(p); plane = idx(p); if (late(plane) > 0.5), disp(['plane ' num2str(plane) ' will arrive minute ' ... num2str(time) ' (' num2str(late(plane)) ' minute(s) late)']) elseif (early(plane) > 0.5), disp(['plane ' num2str(plane) ' will arrive minute ' ... num2str(time) ' (' num2str(early(plane)) ' minute(s) early)']) else disp(['plane ' num2str(plane) ' will arrive minute ' ... num2str(time) ' (on time)' ]) end end end % MODIFICATION LOG % % 051021 med Created. % 060113 per Added documentation % 060125 per Moved disp to end % 060203 med Print level updated % 090308 med Converted to tomSym

Problem type appears to be: mip ===== * * * =================================================================== * * * TOMLAB - Tomlab Optimization Inc. Development license 999001. Valid to 2010-02-05 ===================================================================================== Problem: --- 1: Scheduling Flight Landings f_k 699.999999999999890000 f(x_0) 0.000000000000000000 Solver: CPLEX. EXIT=0. INFORM=101. CPLEX Branch-and-Cut MIP solver Optimal integer solution found FuncEv 105 Iter 30 CPU time: 0.015625 sec. Elapsed time: 0.016000 sec. plane 3 will arrive minute 98 (on time) plane 4 will arrive minute 106 (on time) plane 5 will arrive minute 118 (5 minute(s) early) plane 7 will arrive minute 126 (12 minute(s) early) plane 6 will arrive minute 134 (1 minute(s) early) plane 8 will arrive minute 142 (2 minute(s) late) plane 9 will arrive minute 150 (on time) plane 1 will arrive minute 165 (10 minute(s) late) plane 10 will arrive minute 180 (on time) plane 2 will arrive minute 258 (on time)