Crew Assignment Problems Dr. AA Trani Associate Professor of Civil
Crew Assignment Problems Dr. AA Trani Associate Professor of Civil
Crew Assignment Problems Dr. AA Trani Associate Professor of Civil
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<strong>Crew</strong> <strong>Assignment</strong> <strong>Problems</strong><strong>Dr</strong>. A.A. <strong>Trani</strong><strong>Associate</strong> <strong>Pr<strong>of</strong>essor</strong> <strong>of</strong> <strong>Civil</strong> EngineeringCEE 5614- Analysis <strong>of</strong> Air Transportation Systems1 <strong>of</strong> 15
General Remarks• Sequentially these problems follow the aircraft assignment problem• Difficult problems because there are substantial constraints associatedwith a crew scheduleHours <strong>of</strong> operation (< 8 flight hours per day, < 14 hours in a dutycycle, ,< 30 hours per week, < 90 hours per month, and < 1,000hours per year)Vacation/rest days (selected by the pilot/crew member)Exogenous parameters• Very important since crew members represent a sizeable amount <strong>of</strong> theDOC cost <strong>of</strong> the airline (senior captains make $240,000 per year at makorU.S. airlines)2 <strong>of</strong> 15
<strong>Crew</strong> Scheduling ProblemA small airline uses LP to allocate crew resources tominimize cost. The following map shows the city pairsto be operated.1Denver2Dallas3New YorkMorningAfternoonNightDEN = Denver,DFW = Dallas4MexicoMEX = Mexico City,JFK = New York3 <strong>of</strong> 15
<strong>Crew</strong> Scheduling Problem.Flight Number O-D Pair Time <strong>of</strong> Day100 DEN-DFW Morning200 DFW-DEN Afternoon300 DFW-MEX Afternoon400 MEX-DFW Night500 DFW-JFK Morning600 JFK-DFW Night700 DEN-JFK Afternoon800 JFK-DEN Afternoon4 <strong>of</strong> 15
<strong>Crew</strong> Scheduling ProblemDefinition <strong>of</strong> terms:a) Rotations consists <strong>of</strong> 1 to 2 flights (to make theproblem simple)b) Rotations cost $2,500 if terminates in the originatingcityc) Rotations cost $3,500 if terminating elsewhereExample <strong>of</strong> a feasible rotations are (100, 200),(500,800),(500), etc.5 <strong>of</strong> 15
<strong>Crew</strong> Scheduling ProblemR iSingleFlightRotationsCost ($) R iTw<strong>of</strong>lightRotationsCost ($)1 100 3,500 9 100,200 2,5002 200 3,500 10 100,300 3,5003 300 3,500 11 500,800 3,5004 400 3,500 12 500,600 2,5005 500 3,500 13 300,400 2,5006 600 3,500 14 200,100 3,5006 <strong>of</strong> 15
R iSingleFlightRotationsCost ($) R iTw<strong>of</strong>lightRotationsCost ($)7 700 3,500 15 600,300 3,5008 800 3,500 16 600,200 3,50017 600,500 3,50018 800,100 3,50019 700,600 3,50020 700,800 3,5007 <strong>of</strong> 15
Decision VariablesDecision variables:⎛R1i = ⎜⎝ 0if i rotation is usedif i rotation is not used8 <strong>of</strong> 15
<strong>Crew</strong> Scheduling ProblemMin Costsubject to: (possible types <strong>of</strong> constraints)a) each flight belongs to a rotationMinz = 3500 R 1 + 3500 R 2 + 3500 R 3 + 3500 R 4 + 3500+R 5R 6 R 7 R 8 R 9 R 103500 + 3500 + 3500 + 2500 + 35009 <strong>of</strong> 15
3500 + 2500 + 2500 + 3500+ 3500 R 15R 11 R 12 R 13 R 143500 R 16 + 3500 R 17 + 3500 R 18 + 3500 R 19 + 3500R 20R 1 R 9 R 10 R 14 R 18s.t. (Flt. 100) + + + + = 1(Flt. 200) R 2 + R 9 + R 14 + R 16= 1(Flt. 300) R 3 + R 10 + R 13 + R 15= 1(Flt. 400) R 4 + R 13= 1(Flt. 500) R 5 + R 11 + R 12 + R 17= 110 <strong>of</strong> 15
(Flt. 600) R 6 + R 12 + R 15 + R 16 + R 17 += 1R 19R 7 R 19 R 20(Flt. 700) + + = 1(Flt. 800) R 8 + R 11 + R 18 + R 20= 111 <strong>of</strong> 15
<strong>Crew</strong> Scheduling ProblemProblem statistics:a) 20 decision variables (rotations)b) 8 functional constraints (one for eachflight)c) All constraints have equality signs12 <strong>of</strong> 15
<strong>Crew</strong> Scheduling Problem (Matlab)Input Fileminmax=1; % Minimization formulationa=[1 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 1 0 00 1 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 00 0 1 0 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 00 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 00 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 1 00 0 0 0 0 1 0 0 0 0 0 1 0 0 1 1 1 0 1 00 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 10 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 1]b=[1 1 1 1 1 1 1 1]'c=[-3500 -3500 -3500 -3500 -3500 -3500 -3500 -3500 -2500 -3500 -3500 -2500-2500 -3500 -3500 -3500 -3500 -3500 -3500 -3500]13 <strong>of</strong> 15
as=[1 2 3 4 5 6 7 8]14 <strong>of</strong> 15
Problem SolutionOptimal Solution (after 8 iterations):bas =[ 9 12 13 4 19 18 20 11]The current basic variable values are :b =1 rotation 9 (100,200) Cost = $2,5001 rotation 12 (500,600) Cost = $2,5001 rotation 13 (300,400) Cost = $2,5000 rotation 4 (400)0 rotation 19 (700,600)0 rotation 18 (800,100)1 rotation 20 (700,800) Cost = $3,5000 rotation 11 (500,800)z = $11,000 dollars to complete all flights (4 crews assigned)15 <strong>of</strong> 15