1. Vehicle Routing
John H. Vande Vate
Fall, 2002 1

2. Shared Transportation Capacity
Large shipments reduce transportation costs but increase inventory costs
EOQ trades off these two costs
Reduce shipment size without increasing transportation costs?
Combine shipments on one vehicle 2

3. TL vs LTL     
Transport
Inventory
Inventory
Transport    3

4. Shared Loads ?         Inventory Transport? Inventory4

5. Issues Design Routes that Minimize the transportation cost
Respect the capacity of the vehicle
This may require several routes
Consider inventory holding costs
This may require more frequent visits 5

6. Our Approach Minimize Transportation Cost (Distance)
Traveling Salesman Problem
Respect the capacity of the Vehicle
Multiple Traveling Salesmen
Consider Inventory Costs
Estimate the Transportation Cost
Estimate the Inventory Cost
Trade off these two costs. 6

7. The Traveling Salesman Problem
Minimize Distance
s.t.
start at the depot,
visit each customer exactly once,
return to the depot
A single vehicle no capacity
Only issue is distance. 7

8. IP Formulation Set Cities; param d{Cities, Cities};
var x{Cities, Cities} binary;
minimize Distance:
sum{f in Cities, t in Cities}d[f,t]x[f,t];
s.t. DepartEachCity{f in Cities}:
sum{t in Cities}x[f,t] = 1;
s.t. ArriveEachCity{t in Cities}:
sum{f in Cities}x[f,t] = 1; 8

9. SubTours
3 cities
3 edges        9

10. Eliminating Subtours S.t. SubTourElimination {S subset Cities:
card(S) > 0 and
card(S) < card(Cities)}:
sum{f in S, t in S} x[f,t] <= card(S) - 1; 10

11. An Equivalent Statement
3 cities
No edges out        11

12. An Equivalent Formulation
S.t. SubTourElimination
{S subset Cities:
card(S) > 0 and
card(S) < card(Cities)}:
sum{f in S, t not in S} x[f,t] >= 1; 12

13. How Many Constraints? How many subsets of N items? 2N Omit 2 subsets:
All N items
The empty set
2N - 2 13

14. OK, Half of that...
If we have an edge out of S, we must have an edge out of N\S.
Why? The edge out of S is an edge into N\S. But there are exactly |N\S| edges into cities in N\S. Since one of them comes from S, not all the edges from cities in N\S can lead to cities in N\S. At least one must go to S. 14

15. Still Lots! How many subsets of N items? 2N Omit 2 subsets: 2N - 2
All N items
The empty set
2N - 2
Half of that: 2N-1 - 1 15

16. How Many?
N N-1 - 1
Too Many! 16

17. Optimization is Possible But...
It is difficult
Few 100 cities is the limit
Is it appropriate?
Other approaches…. 17

18. Heuristics The Strip Heuristic Partition the region into narrow strips
Routing in each strip is easy ~ 1-Dimensional
Paste the routes together 18

19. The Strip Heuristic x x x x x x x x x x x x x x x x 19

20. Nearest Neighbor x x x x x x x x x x x x x x x x 20

21. Clark-Wright
Shortcut a “tour” by finding the greatest “savings” x x x x x x 21

22. Clark-Wright
Shortcut a “tour” by finding the greatest “savings” x x x x x x 22

23. Nearest Insertion x x x x x x x x x x x x x x x x 23

24. Nearest Insertion x x x x x x x x x x x x x x x x 24

25. Nearest Insertion x x x x x x x x x x x x x x x x 25

26. Nearest Insertion x x x x x x x x x x x x x x x x 26

27. Nearest Insertion x x x x x x x x x x x x x x x x 27

28. Nearest Insertion x x x x x x x x x x x x x x x x 28

29. Nearest Insertion x x x x x x x x x x x x x x x x 29

30. Nearest Insertion x x x x x x x x x x x x x x x x 30

31. Nearest Insertion x x x x x x x x x x x x x x x x 31

32. Nearest Insertion x x x x x x x x x x x x x x x x 32

33. Nearest Insertion x x x x x x x x x x x x x x x x 33

34. Nearest Insertion x x x x x x x x x x x x x x x x 34

35. Nearest Insertion x x x x x x x x x x x x x x x x 35

36. Nearest Insertion x x x x x x x x x x x x x x x x 36

37. Nearest Insertion x x x x x x x x x x x x x x x x 37

38. Improvement Heuristics
2-Opt x x x x x x x x x x x x x x x x 38

39. Local Minima
No improvement found, but…
Tour still isn’t good 39

40. Probabilistic Methods
Simulated Annealing
With probability that reduces over time, accept an exchange that makes things worse (gets you out of local minima). 40

41. Optimization-Based Heuristics
Minimum Spanning Tree Heuristic
Build a minimum spanning tree on the edges between customers
Double the tree to get an Eulerian Tour (visits everyone perhaps several times and returns to the start)
Short cut the Eulerian Tour to get a Hamilton Tour (Traveling Salesman Tour) 41

42. The Spanning Tree Is Easy to construct Is a Lower bound on the TSP
Use the Greedy Algorithm
Add edges in increasing order of length
Discard any that create a cycle
Is a Lower bound on the TSP
Drop one edge from the TSP and you have a spanning tree
It must be at least as long as the minimum spanning tree 42

43. The Spanning Tree x x x x x x x x x x x x x x x x 43

44. Double the Spanning Tree
Duplicate each edge in the Spanning Tree
The resulting graph is connected
The degree at every node must be even
That’s an Eulerian Graph (you can start at a city, walk on each edge exactly once and return to where you started)
It’s no more than twice the length of the shortest TSP 44

45. The Spanning Tree x x x x x x x x x x x x x x x x 14 15 13 12 11 19 1820 10 9 x x 17 x x 16 5 6 21 x 4 8 22 x x 7 23 3 27 x 26 28 x 25 24 2 29 x x 1 x 45 30

46. Short Cut the Eulerian Tour14 15 x 13 x x 12 11 x 19 18 20 10 9 x x 17 x x 16 5 6 21 x 4 8 22 x x 7 23 3 27 x 26 28 x 25 24 2 29 x x 1 x 46 30

47. Short Cut the Eulerian Tour Short Cut the Eulerian Tour14 15 x 13 x x 12 11 x 19 18 20 10 9 x x 17 x x 16 5 6 21 x 4 8 22 x x 7 23 3 27 x 26 28 x 25 24 2 29 x x 1 x 47 30

48. Spacefilling Curves
There are no more points in the unit square than in the interval from 0 to 1!? 48

49. Proof Each point (X,Y) on the map Express X = string of 0’s and 1’s
1*24+0*23+0*22+0*21+0*20+1*2-1 +0*2-2
Express Y = string of 0’s and 1’s
Y = 9.75 =
0*24+1*23+0*22+0*21+1*20+1*2-1 +1*2-2
Space Filling Number - interleave bits
(X,Y) = 49

50. So,... Each pair of points X = 16.5 = 10000.10 Y = 9.75 = 01001.11
maps to a unique point
(X,Y) = 50

51. How to Use this? A mapping of (X,Y) into the unit interval
Think of this as the inverse mapping of the unit interval onto the square (our super tour)
For a given customer (X,Y) is the fraction of the way along the super tour where it lies
Visit the customers in the order of (X,Y) (short cut the super tour to visit our customers) 51

52. The TSP For More on SpaceFilling Curves visit
There are several books on the TSP …. 52

53. Our Approach Minimize Transportation Cost (Distance)
Traveling Salesman Problem
Respect the capacity of the Vehicle
Multiple Traveling Salesmen
Consider Inventory Costs
Estimate the Transportation Cost
Estimate the Inventory Cost
Trade off these two costs. 53

54. Different Approaches Route First - Cluster Second
Build a TSP tour
Partition it to meet capacity
Cluster First - Route Second
Decide who gets served by each route
Then build the routes 54

55. Route First Vehicle Cap: 15 x x x x x x x x x x x x x x x x 4 6 6 2 23 x x 3 x x 3 5 x 5 x 4 x 6 x x x 5 5 55

56. Cluster First Sweep Heuristic Vehicle Cap: 15 x x x x x x x x x x x x4 6 x x x 6 x 2 2 5 x x 3 x x 3 x 3 x 5 x 5 x 4 x 6 x x x 5 5 56

57. Our Approach Minimize Transportation Cost (Distance)
Traveling Salesman Problem
Respect the capacity of the Vehicle
Multiple Traveling Salesmen
Consider Inventory Costs
Estimate the Transportation Cost
Estimate the Inventory Cost
Trade off these two costs. 57