The Vehicle Routing Problem with
Time Windows and Scheduled
Departure.
Ho-Chi-Minh City, March-2010
Cristian Oliva San Martín.
Universidad Católica de la Santísima Concepción.
Chile.
INTRODUCTION
Algorithms
Remarks
VRPTW
VRP
VRPTWSL
Create and Implement an algorithm
for solving the VRPTWSL.
Methods of Solution
Exact Methods
Approximation Methods
Heuristics
•Construction.
•Local search.
Metaheuristics
•Simulated Annealing
•Genetic algorithms
•Ant Colony Optimization
•Tabu-Search

The vehicle is loaded in the depot, This time
is calculated as:
load time = load factor * service time.

Arrival Time:
Load time + Travel time + departure time.

Departure time:
Arrival time + Service time.
Distances
[0,20]
[20,24]
[24,25]
[25,0]
[0,5]
[5,7]
[7,0]
[0,3]
[3,18]
[18,19]
[19,15]
[15,16]
[16,0]
Customer
0
20
24
25
5
7
3
18
19
15
16
X coord.
40
30
25
25
42
40
42
15
15
20
20
Y coord.
50
50
50
52
65
66
66
75
80
80
85
Demand
0
10
10
40
10
20
10
20
10
40
40
10
5
2
17
17
3
16
18
36
5
5
5
55
Ready Time
0
10
65
169
15
170
65
179
278
384
475
Arrival time Wait time
37
0
132
0
224
0
331
0
62
0
155
15
276
0
108
0
234
0
329
0
424
0
519
0
664
0
Due Date
1236
73
144
224
67
225
146
254
345
429
528
Service Time
0
90
90
90
90
90
90
90
90
90
90
24
2
5
25
20
10
3
18
36
17
5
55
16
18
16
5
3
5
7
15
5
19
24
2
5
25
20
10
[0,27]
3
18
36
17
5
55
16
18
16
5
3
5
7
15
5
19
24
2
5
25
[37,127]
20
10
[0,27]
3
18
36
17
5
[27,45]
16
55
18
16
5
3
5
7
15
5
19
[132,222]
24
2
5
25
[37,127]
20
10
[0,27]
3
18
36
[45,90]
17
[27,45]
55
18
[62,152]
5
16
16
5
3
5
7
15
5
19
[132,222]
24
2
[224,314]
5
25
[37,127]
[108,198]
20
10
[0,27]
3
18
36
[45,90]
17
[27,45]
55
18
[62,152]
5
16
16
5
3
5
7
[170,260]
15
5
19
[132,222]
Route 2: [276,1236]
24
Route 1: [331,1236]
2
[224,314]
5
25
[37,127]
[108,198]
20
10
[0,27]
3
18
36
[45,90]
17
[27,45]
[234,324]
55
18
[62,152]
5
16
16
5
3
5
7
15
5
19
[132,222]
Route 2: [276,1236]
24
Route 1: [331,1236]
2
[224,314]
5
25
[37,127]
[108,198]
20
10
[0,27]
3
18
36
[45,90]
17
[234,324]
[27,45]
55
18
[62,152]
5
16
16
5
3
5
7
[170,260]
15
5
19
[329,419]
[132,222]
Route 2: [276,1236]
24
Route 1: [331,1236]
2
[224,314]
5
25
[37,127]
[108,198]
20
10
[0,27]
3
18
36
[45,90]
17
[27,45]
[234,324]
55
18
[62,152]
5
16
16
5
3
5
7
[170,260]
15
[424,514]
5
19
[329,419]
[132,222]
Route 2: [276,1236]
24
Route 1: [331,1236]
2
[224,314]
5
25
[37,127]
[108,198]
20
10
[0,27]
3
18
36
[45,90]
17
[234,324]
[27,45]
55
18
[62,152]
5
16
16
[519,609]
5
3
5
7
[170,260]
15
[424,514]
5
19
[329,419]
[132,222]
Route 2: [276,1236]
24
Route 1: [331,1236]
2
[224,314]
5
Route 3: [664,1236]
25
[37,127]
[108,198]
20
10
[0,27]
3
18
36
[45,90]
17
[234,324]
[27,45]
55
18
[62,152]
5
16
16
[519,609]
5
3
5
7
[170,260]
15
[424,514]
5
19
[329,419]
VRPTW
SL
Simple Heuristics for VRPTWSL
1. Nearest Neighbor (NN) [for TSP]
(By Rosenkrantz, Sterns, Lewis, SIAM
J. Computing 6(1977) 563-581)
Step 1. Start with node 0 as the beginning of the TSP tour
Step 2. - Find the node closest to the last node added to
the tour satisfying all the constraints.
- Add this node to the tour
Step 3. - Repeat Step 2 until all nodes are added to the tour
- Then join the first and the last nodes
READY DUE SERVICE
CLIENT XCOORD YCOORD DEMAND TIME DATE TIME
0
40
50
0
0
1236
0
1
42
66
10
65
146
90
2
18
75
20
99
148
90
3
45
68
10
912
967
90
4
38
68
20
255
324
90
0
LOAD
TIME
0
9
9
9
9
DISTANCES
0
18
47
20
23
READY DUE SERVICE
CLIENTE XCOORD YCOORD DEMANDA TIME DATE TIME
0
40
50
0
0
1236
0
1
42
66
10
65
146
90
2
18
75
20
99
148
90
3
38
68
20
255
324
90
4
45
68
10
912
967
90
departure= 65+90=155
1
Arrival= 9+18=27
pi= 9
0
Arrival= 155+18=173
LOAD
TIME
0
9
9
9
9
DISTANCES
0
18
47
20
23
READY DUE SERVICE
CLIENT XCOORD YCOORD DEMAND TIME DATE TIME
0
40
50
0
0
1236
0
1
42
66
10
65
146
90
2
18
75
20
99
148
90
3
38
68
20
255
324
90
4
45
68
10
912
967
90
departure= 65+90=155
1
Arrival = 18+18=36
arrival= 155+6=161
3
departure=
255+90=345
pi= 18
0
arrival =345+20= 371
LOAD
TIME
0
9
9
9
9
DISTANCES
0
18
47
20
23
READY DUE SERVICE
CLIENT XCOORD YCOORD DEMAND TIME DATE TIME
0
40
50
0
0
1236
0
1
42
66
10
65
146
90
2
18
75
20
99
148
90
3
38
68
20
255
324
90
4
45
68
10
912
967
90
departure= 65+90=155
1
arrival = 27+18=45
arrival= 155+6=161
pi= 27
3
departure= 255+90=345
arrival=1002+23=1025
arrival= 345+7=352
4
departure= 912+90=1002
0
LOAD
TIME
0
9
9
9
9
DISTANCES
0
18
47
20
23
READY DUE SERVICE
CLIENT XCOORD YCOORD DEMAND TIME DATE TIME
0
40
50
0
0
1236
0
1
42
66
10
65
146
90
2
18
75
20
99
148
90
3
38
68
20
255
324
90
4
45
68
10
912
967
90
departure= 65+90=155
1
arrival = 27+18=45
2
pi= 27
departure= 255+90=345
arrival=1002+23=1025
arrival= 345+2=347
4
departure= 912+90=1002
DISTANCES
0
18
47
20
23
arrival=36+47=83
arrival= 155+2=157
3
LOAD
TIME
0
9
9
9
9
0
departure=99+90=189
pi= 36
arrival=189+57=246
Instances v/s Total Costs
Heurística basada en Distancias
Heurística basada en Ventanas de Tiempo
50000
45000
35000
30000
25000
20000
15000
10000
5000
C208
C207
C206
C205
C204
C203
C202
C201
C208
C207
C206
C205
C204
C203
C202
C201
C109
C108
C107
C106
C105
C104
C103
C208
C207
C206
C205
C204
C203
C202
C201
C109
C108
C107
C106
C105
C104
C103
C102
0
C101
Total Costs
40000
25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50100100100100100100100100
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Instance
Instances v/s Number of vehicles
Nº Veh.Heurística basada en Distancias
Nº Veh.Heurística basada Ventanas de Tiempo
30
20
15
10
5
C208
C207
C206
C205
C204
C203
C202
C201
C208
C207
C206
C205
C204
C203
C202
C201
C109
C108
C107
C106
C105
C104
C103
C208
C207
C206
C205
C204
C203
C202
C201
C109
C108
C107
C106
C105
C104
C103
C102
0
C101
Nº of Vehicles
25
25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50100100100100100100100100
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Instance
Results
Instances v/s Nº of loaders
Nº Carg.Heurística basada en Distancias
Nº Carg. Heurística basada en Ventanas de Tiempo
3.5
2.5
2
1.5
1
0.5
C208
C207
C206
C205
C204
C203
C202
C201
C208
C207
C206
C205
C204
C203
C202
C201
C109
C108
C107
C106
C105
C104
C103
C208
C207
C206
C205
C204
C203
C202
C201
C109
C108
C107
C106
C105
C104
C103
C102
0
C101
Nº of loaders
3
25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50100100100100100100100100
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Instances
Local Search Heuristics (LSH)
• Simple heuristics:
- generates one solution only
• Local search heuristics:
- generates many solutions
- choose the best one
Definition: Given a solution S, its neighborhood
N(S) is defined as the set of all solutions than
can be obtained through a well-defined
modification of S
Basic idea of Local Search Heuristics
• Start with a solution
(initial solution obtained
by a simple heuristic)
S1
S2
N(S2)
• Search its
neighborhood
• Select a solution from
the neighborhood
• Either reject or
accept the new
solution
N(S1)
S3
S4
Three components of a LSH
1. Neighborhood design
2. Search process in the neighborhood
3. Acceptance-rejection criterion
N(S3)
K-Opt heuristic
• Lin, Bell Systems Technical J. 44(1965) 2245-2269
• Lin & Kernigham, Operations Research 21(1973) 498-516
Definition: K-exchange is a procedure that
replaces K arcs in a given TSP tour by K
new arcs so that the resulting tour is still
a TSP tour.
K-Opt heuristic
B
C
A
B
A
C
E
D
B
A
“2-exchange”
D
E
Replace
(A, B) & (C, D)
by
(A, C) & (B, D)
Replace
(A, B), (C, D) & (D, E)
by
(B, D), (A, D) & (C, E)
“3-exchange”
C
E
D
B
C
A
E
D
8
8
4
4
1
1
7
7
6
6
5
2
3
5
2
Initial Solution X
Solution X’
Total cost: 42
Total cost: 36
Waiting time: 25
Waiting time: 24
3
(1 , 2)  42  36  0.005(24  25)
(1 , 2)  6.005
2. Simulated Annealing (SA)
Step 1: Find an initial solution using any heuristic, S.
Let the incumbent solution S*=S.
Select the cooling parameter T > 0.
Step 2: Select a candidate solution Sc from N(S).
If F(Sc)<F(S*), update the incumbent, S*=Sc.
If F(Sc)<F(S), accept Sc, i.e. let S = Sc.
If F(Sc)>F(S), accept Sc with probability
exp((F(S)-F(S
c))/T),
if Sc is accepted, then let S = Sc,
o/w, S is not updated.
Step 3: - Update b such that it becomes smaller.
- Repeat Step 2 until a certain stopping
rule is satisfied (e.g. stop after 1000
iterations).
Note: Three things may affect effectiveness of SA
- neighborhood design
- the way a candidate solution is selected
- cooling parameter
3. Tabu Search (TS) (1 of 3)
•
Similar to SA with a different acceptance-rejection
criterion
• Tabu list of mutations (prohibited moves)
Step 1: - Find an initial solution using any
heuristic, S.
- Let the incumbent solution S*=S.
- Set tabu list length L. Initially the tabu
list is empty.
3. Tabu Search (TS) (2 of 3)
.
Step 2: (1) Select a candidate solution Sc from N(S) such
that the move from S to Sc is not a mutation on
the tabu list.
(2) Enter the reverse mutation of the move at the
top of the tabu list, push all other entries in the
list one position down, & delete the entry at the
bottom of list if the list length exceeds L.
(3) Accept Sc, i.e. let S = Sc.
(4) If F(Sc)<F(S*), update the incumbent, S*=Sc.
3. Tabu Search (TS) (3 of 3)
Step 3: Repeat Step 2 until some stopping rule is
satisfied (e.g. stop after 1000 iterations).
Three things may affect effectiveness of TS
- neighborhood design
- the way a candidate solution is selected
- Length parameter