The Traveling Salesman Problem

1. The Traveling Salesman Problem • Importance: • Variety of scheduling application can be solved as atraveling salesmen problem. • Examples: • Ordering drill position on a drill press. • School bus routing.

2. THE FEDERAL EMERGENCY MANAGEMENT AGENCY • A visit must be made to four local offices of FEMA, going out from and returning to the same main office in Northridge, Southern California.

3. FEMA traveling salesmanNetwork representation

4. 40 2 3 25 35 50 40 50 1 4 65 45 30 80 Home

5. FEMA - Traveling Salesman • Solution approaches • Enumeration of all possible cycles. • This results in (m-1)! cycles to enumerate for a graph with m nodes. • Only small problems can be solved with this approach.

6. FEMA – full enumeration Possible cycles Cycle Total Cost 1. H-O1-O2-O3-O4-H 210 2. H-O1-O2-O4-O3-H 195 3. H-O1-O3-O2-O3-H 240 4. H-O1-O3-O4-O2-H 200 5. H-O1-O4-O2-O3-H 225 6. H-O1-O4-O3-O2-H 200 7. H-O2-O3-O1-O4-H 265 8. H-O2-O1-O3-O4-H 235 9. H-O2-O4-O1-O3-H 250 10. H-O2-O1-O4-O3-H 220 11. H-O3-O1-O2-O4-H 260 12. H-O3-O1-O2-O4-H 260 Minimum For this problem we have (5-1)! / 2 = 12 cycles. Symmetrical problemsneed to enumerate only (m-1)! / 2 cycles.

7. FEMA – optimal solution 40 2 3 25 35 50 40 1 50 4 65 45 30 80 Home

8. The Traveling Salesman Problem • Finding the shortest cycle is different than Dijkstra’s shortest path. It is much harder too, no algorithm exists with polynomial worst-case time complexity! • This means that for large numbers of vertices, solving the traveling salesman problem is impractical. • The problem has theoretical importance because it represents a class of difficult problems known as NP-hard problems. • In these cases, we can use efficient approximation algorithms that determine a path whose length may be slightly larger than the traveling salesman’s path.