Aircraft Routing and Scheduling
DETAILS EXAMPLE 11.3.1 IN TEXT
Initial master Problem:
T=2
M1 = 2
M2 = 2
S1 =
1-1
0
0
0
0
0
0
0
0
0
0
0
0
2-1
0
0
0
0
1
0
1
0
0
1
0
1
3-1
1
0
0
1
1
0
1
0
0
0
0
0
4-1
1
0
1
0
1
0
0
1
0
0
0
0
5-1
0
0
0
0
0
1
1
0
0
1
1
0
6-1
0
0
0
0
0
0
0
0
1
1
1
1
7-1
0
0
0
0
0
0
1
0
0
1
1
1
8-1
0
1
0
1
0
1
0
0
1
0
0
0
9-1
0
0
0
0
0
0
0
0
0
1
1
1
10-1
0
1
0
0
0
1
0
1
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
0
0
1
0
0
1
0
0
0
0
1
0
0
1
0
0
1
0
0
0
0
1
0
1
0
0
0
1
0
0
0
1
0
0
0
0
0
0
1
0
0
1
0
0
0
1
0
0
0
0
1
0
0
1
0
0
1
0
0
Oih
0
0
0
0
Dih
0
0
0
0
S2 =
1-2
0
0
0
0
0
0
0
0
0
0
0
0
2-2
1
0
0
1
1
0
1
0
0
0
0
0
3-2
0
0
0
0
0
1
1
0
0
1
1
0
4-2
1
0
1
0
0
1
0
1
0
0
0
0
5-2
0
0
0
0
0
0
0
0
1
1
1
1
6-2
0
0
0
0
0
0
1
0
0
1
1
1
7-2
0
1
0
1
0
1
0
0
1
0
0
0
8-2
0
0
1
0
0
0
0
0
0
1
1
1
9-2
0
1
0
0
0
1
0
1
1
0
0
0
Oih
0
0
0
0
1
0
0
0
0
0
0
1
1
0
0
0
0
0
1
0
0
0
0
1
0
0
1
0
0
1
0
0
0
0
1
0
0
0
0
1
1
0
0
0
0
0
1
0
0
0
1
0
0
0
0
1
0
0
1
0
0
1
0
0
Dih
0
0
0
0
1
0
0
0
ij
1
1
2
3
4
5
6
7
8
9
10
11
12
450
300
500
400
900
900
900
900
1500
1500
1500
1500
2
1
2
3
4
5
6
7
8
9
10
11
12
450
450
500
500
1000
1000
1000
1000
1350
1350
1350
1350
li
S1
1-1
-375
2-1
4800
3-1
2650
4-1
2750
5-1
4800
6-1
6000
7-1
5400
8-1
3100
9-1
4500
2-2
2950
3-2
4700
4-2
2950
5-2
5400
6-2
5050
7-2
3300
8-2
4550
9-2
3800
10-1
3600
S2
1-2
0
Solving the master problem through linear programming will result in the optimal solution of schedules 4-1, 7-1
and 8-1 from S1 with values of 2/3, and schedules 1-2 = 1 , 2-2, 8-2 and 9-2 from S2 with values of 1, 1/3,
1/3,and 1/3 with a resulting profit of 11,267. An optimal integer solution exists for this problem with schedules
1-1 and 4-1 from S1, and 6-2 and 7-2 from S2, with profit 10725.
x14  x17  x18 
2
3
x12  1
x22  x28  x29  13
UB  11267
1  41x
x14  0
0  41x
x14  1
x17  x18  0.5
1  31x  11x
x12  1
1  92 x  82 x
x26  x27  0.5
52601  BU
UB  11175
x17  1
x17  0
x`41  1
x14  1
x11  1
x17  1
x 26  x
INFEASIBLE
UB  1
INTEG
Figure 11.3
Branch-and-Bound Tree for Aircraft Scheduling Problem
Sub-problem
Dual Variables:
In this problem, the dual variables associated with the flight legs j are as follows:
1
2
3
4
5
0
-616.66
216.66
2650 2550
6
7
8
9
10
2733.33
-2250
0
1500 4333.33
11
12
0
183.33
The dual variable  for type 1 is –16.66. The dual variable for type 2 aircraft is 0.
The dual variable for the origin and destination airports 0,0,0,3150 for large planes and 0,0,183.33,2966.66 for
small planes for sfo, lax, nyc and sea.
Candidate schedules
Now that dual variables are known candidate schedules to be included in the master problem can be considered.
Potential schedules: (all turn times standard and included in flight leg duration)
Duration of flight legs:
1
2
3
4
5
7
12
1.5
1.5
1.5
1.5
3
3
6
7
12
Windows of flight legs (earliest departure, latest arrival)
1
2
3
4
5
0500,1200
1:
1200,2359
0500,1200
1200,2359
0500,1200
0500,1200
1200,0500
The attached directed graph will show the following results for the following potential schedules:
Schedule A2: (450-0) + (500-216.67) + (900-2733.33) + (900-0) - -16.66 – 0 = -183.34
Schedule A3: (500-216.67) + (1500-4333.33) + (1500-0) + (1500-183.33) –16.66 – 0 = 283.33
Other schedules are checked for the same, and ones with positive results are included.
The following schedules are generated from the directed graph, ones with positive results included in the master
problem.
Profit
Dual
variable
Potential
profit
Oih
0
1
0
0
0
0
0
1
0
0
1
0
0
0
0
1
0
0
0
1
Dih
0 0 0 0 0
11-1
0
0
1
0
0
0
0
0
0
1
1
1
5000
4733.33
12-1
1
0
1
0
0
1
0
1
0
0
0
0
2750
2933.34
13-1
0
0
0
0
0
1
1
0
0
1
1
0
4800
4800
14-1
1
0
1
0
0
1
1
0
0
0
0
0
2750
683.34
15-1
1
0
1
0
0
1
1
0
0
1
0
1
5750
5200.00333
283.33
-183.33
0
2066.66
549.996667
0 0 0 0 0
1 0 1 0 0
0 1 0 1 1
2.
Generating small plane schedules is as follows, calculated in the same manner as above:
10-2
11-2
0
1
1
1
0
1
1
1
1
1
0
0
1
0
0
1
0
0
0
0
0
0
0
0
Profit
2950
3900
Dual variable 2233.33 4700
Potential profit 716.667 -800
12-2
0
0
0
0
0
1
1
0
0
1
1
0
4700
4816.67
-116.67
13-2
1
0
1
0
0
1
1
0
0
0
0
0
2950
700
2250
14-2
1
0
1
0
0
1
1
0
0
1
0
1
5650
5216.67
433.333
15-2
1
1
1
1
0
0
0
0
0
0
0
0
1900
2250
-350
Oih
1
0
0
0
0
0
1
0
0
0
1
0
0
0
0
1
0
0
0
1
1
0
0
0
0
0
1
0
0
0
0
1
0
0
0
1
1
0
0
0
Dih
1
0
0
0
0
0
1
0
The schedules with positive potential profit are included in the master problem.
Master Problem
The new master problem when solves through linear programming has the following schedules. Some of the schedule numbers have
changed from the notation given in the potential schedules, ie 14-1 became 12-1 as potential schedules 12 and 13 were either negative
or zero for the potential profit.
S1:
0
0
1
1
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0
0
1
0
1
0
0
0
0
0
0
0
0
0
0
1
0
1
0
0
0
1
0
0
0
1
0
0
1
0
0
1
0
0
0
0
0
0
0
0
0
1
0
1
0
0
1
0
1
1
0
1
0
0
0
0
0
1
0
0
1
0
0
0
0
0
1
1
0
0
1
1
0
0
0
0
0
1
1
1
1
0
0
1
0
0
1
1
1
0
1
0
0
1
0
0
0
0
0
0
0
0
1
1
1
0
1
0
1
1
0
0
0
0
0
0
0
0
1
1
1
0
1
1
0
0
0
0
0
0
1
1
0
0
1
0
1
S2:
0
0
0
0
0
0
0
0
0
0
0
0
ij
1
0
0
1
1
0
1
0
0
0
0
0
0
0
0
0
0
1
1
0
0
1
1
0
1
0
1
0
0
1
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
1
0
0
1
1
1
0
1
0
1
0
1
0
0
1
0
0
0
0
0
1
0
0
0
0
0
0
1
1
1
0
1
0
0
0
1
0
1
1
0
0
0
0
1
0
1
1
0
1
0
0
0
0
0
1
0
1
0
0
1
1
0
0
0
0
0
1
0
1
0
0
1
1
0
0
1
0
1
1
1
2
3
4
5
6
7
8
9
10
11
12
450
300
500
400
900
900
900
900
1500
1500
1500
1500
1
2
3
4
5
6
7
8
9
10
11
12
450
450
500
500
1000
1000
1000
1000
1350
1350
1350
1350
2
li
1
1-1
2-1
3-1
4-1
5-1
6-1
7-1
8-1
9-1
10-1
11-1
12-1
13-1
-375
4800
2650
2750
4800
6000
5400
3100
4500
3600
5000
2750
5750
1-2
2-2
3-2
4-2
5-2
6-2
7-2
8-2
9-2
10-2
13-2
14-2
0
2950
4700
2950
5400
5050
3300
4550
3800
2950
2950
5650
2
When the new master problem is solved through linear programming the result is:
X4-1,6-1,7-1,8-1,4-2,11-2 = 0.5 and X2-1 = 1
The branch and bound tree for the master problem is as follows:
x14  x16  x17  x18  0.5
x 12  1
x 24  x 11
2  0.5
TARGET  11575
1  41x
x14  0
x14  1
x14  0
x13  x16  x110  x111  0.5
x17  x18  0.5
x 24  x 11
2  0.5
x 26  x 27  0.5
x 12  1
x 12  1
UB  11575
UB  11175
x16  0
x14  0
x16  1
x17  0
x17  1
x16  0
x110  x111  1
x14  0
x16  1
x14  1
x 12  x 22  1
x11  x 24  x 11
2 1
x17  1
UB  11550
INTEGER OPTIMAL
UB  11525
INFEASIBLE
SOLUTION
x14  1
x17  0
x11  x 26  x 27  1
UB  10725
Figure 11.5
Branch-and-Bound Tree for Aircraft Scheduling Problem