Crew Assignment Problems
Dr. A.A. Trani
Associate Professor of Civil Engineering
CEE 5614- Analysis of Air Transportation Systems
1 of 15
General Remarks
• Sequentially these problems follow the aircraft assignment problem
• Difficult problems because there are substantial constraints associated
with a crew schedule
Hours of operation (< 8 flight hours per day, < 14 hours in a duty
cycle, ,< 30 hours per week, < 90 hours per month, and < 1,000
hours per year)
Vacation/rest days (selected by the pilot/crew member)
Exogenous parameters
• Very important since crew members represent a sizeable amount of the
DOC cost of the airline (senior captains make $240,000 per year at makor
U.S. airlines)
2 of 15
Crew Scheduling Problem
A small airline uses LP to allocate crew resources to
minimize cost. The following map shows the city pairs
to be operated.
1
New York
Denver
3
2
DEN = Denver,
DFW = Dallas
Dallas
4
Mexico
Morning
Afternoon
Night
MEX = Mexico City,
JFK = New York
3 of 15
Crew Scheduling Problem.
Flight Number
O-D Pair
Time of Day
100
DEN-DFW
Morning
200
DFW-DEN
Afternoon
300
DFW-MEX
Afternoon
400
MEX-DFW
Night
500
DFW-JFK
Morning
600
JFK-DFW
Night
700
DEN-JFK
Afternoon
800
JFK-DEN
Afternoon
4 of 15
Crew Scheduling Problem
Definition of terms:
a) Rotations consists of 1 to 2 flights (to make the
problem simple)
b) Rotations cost $2,500 if terminates in the originating
city
c) Rotations cost $3,500 if terminating elsewhere
Example of a feasible rotations are (100, 200),
(500,800),(500), etc.
5 of 15
Crew Scheduling Problem
Ri
Single
Flight
Rotations
Ri
Twoflight
Rotations
Cost ($)
Cost ($)
1
100
3,500
9
100,200
2,500
2
200
3,500
10
100,300
3,500
3
300
3,500
11
500,800
3,500
4
400
3,500
12
500,600
2,500
5
500
3,500
13
300,400
2,500
6
600
3,500
14
200,100
3,500
6 of 15
Ri
Single
Flight
Rotations
Ri
Twoflight
Rotations
Cost ($)
Cost ($)
7
700
3,500
15
600,300
3,500
8
800
3,500
16
600,200
3,500
17
600,500
3,500
18
800,100
3,500
19
700,600
3,500
20
700,800
3,500
7 of 15
Decision Variables
Decision variables:
 1
Ri = 
 0
if i rotation is used
if i rotation is not used
8 of 15
Crew Scheduling Problem
Min Cost
subject to: (possible types of constraints)
a) each flight belongs to a rotation
Min
z = 3500
R +
R1
+ 3500
R2
+ 3500
R3
+ 3500
R4
+ 3500
5
3500
R6
+ 3500
R7
+ 3500
R8
+ 2500
R9
+ 3500
R 10
9 of 15
3500 R
+ 3500
3500
11
+ 2500
R 12
+ 2500
R 13
+ 3500
R 14
R 17
+ 3500
R 18
+ 3500
R 19
R 15
R 16
+ 3500
+ 3500
R 20
s.t. (Flt. 100)
R1
+
R9
+
R 10
+
R 14
+
R 18
=1
(Flt. 200)
R2
+
R9
+
R 14
+
R 16
=1
(Flt. 300)
R3
+
R 10
+
R 13
+
R 15
=1
(Flt. 400)
R4
+
R 13
(Flt. 500)
R5
+
R 11
=1
+
R 12
+
R 17
=1
10 of 15
(Flt. 600)
R =1
R6
+
R 12
+
R 15
(Flt. 700)
R7
+
R 19
+
R 20
(Flt. 800)
R8
+
R 11
+
R 18
+
R 16
+
R 17
+
19
=1
+
R 20
=1
11 of 15
Crew Scheduling Problem
Problem statistics:
a) 20 decision variables (rotations)
b) 8 functional constraints (one for each
flight)
c) All constraints have equality signs
12 of 15
Crew Scheduling Problem (Matlab)
Input File
minmax=1; % Minimization formulation
a=[1 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 1 0 0
01000000100001010000
00100000010010100000
00010000000010000000
00001000001100000010
00000100000100111010
00000010000000000011
0 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 of 15
bas=[1 2 3 4 5 6 7 8]
14 of 15
Problem Solution
Optimal 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)
1
rotation 12 (500,600)
1
rotation 13 (300,400)
0
rotation 4 (400)
0
rotation 19 (700,600)
0
rotation 18 (800,100)
1
rotation 20 (700,800)
0
rotation 11 (500,800)
Cost = $2,500
Cost = $2,500
Cost = $2,500
Cost = $3,500
z = $11,000 dollars to complete all flights (4 crews assigned)
15 of 15