1. TSP.1 9.4 Travelling Salesperson Problem (TSP) Very famous problem Many practical applications Very easy to describe Very difficult to solve (Curse of Dimensionality) We shall consider the dynamic programming (DP) approach Other approaches: see 620-362

2. TSP.2 Problem Formulation There are many ways to describe this problem. We shall consider the following: –English version –Linear Programming oriented version –Linear Programming Free version –Dynamic programming version

3. TSP.3 English Version You are given a set of n cities You are given the distances between the cities You start and terminate your tour at your home city You must visit each other city exactly once. Your mission is to determine the shortest tour.

4. TSP.4 Maths versions We shall consider two Maths Version The first is LP-based The second is LP-free The first version dominates the OR literature

5. TSP.5 TSP Version 1 (LP) Decision variable: A boolean matrix x interpreted as follows: x(i,j):= 1, iff we go from city i to city j. x(i,j) := 0, otherwise

6. TSP.6 Example This matrix represents the tour (1,2,3,4,1)

7. TSP.7 Objective function d(i,j) = (direct) distance between city i and city j.

8. TSP.8 Constraints Each city must be “exited” exactly once Each city must be “entered” exactly once

9. TSP.9 Is this enough ?

10. TSP.10 No! The first two constraints allow sub- tours Thus, we have to add a constraint that will prevent sub-tours

11. TSP.11 Explanation: sub-tours Two subtour: (1,2,1) and (3,4,3) This solution is not feasible for the TSP

12. TSP.12 If we start at the home city n=1, we will not visit city 3 and 4. We must go from city 2 to either city 3 or city 4. 1 2 3 4

13. TSP.13 Subtour elimination constraint S = subset of cities |S| = cardinality of S (# of elements in S) There are 2 n such sets !!!!!!!

14. TSP.14 Example Consider S={1,2}, |S|=2 Hence the sub-tour elimination constraint is not satisfied. Indeed, thee are two subtours in this solution

15. TSP.15 Thus, LP Version

16. TSP.16 LP-Free Version Decision variables: x j := j-th city on the tour, j=1,2,…,n Example: x=(1,3,2,4,1) We start at city 1, then go to city 3, then go to city 2 then go to city 4 then return to city 1.

17. TSP.17 ASSUMPTION Assume that 0 is the home city, and that there are n other cities

18. TSP.18 Objective function

19. TSP.19 Constraints The constraint basically says that x is a permutation of the cities (1,2,3,…,n) Make sure that you appreciate the role of { } in this formulation.

20. TSP.20 LP-Free Formulation There are n! feasible solutions

21. TSP.21 Which one do you prefer?

22. TSP.22 LP Version

23. TSP.23 LP Free Version

24. TSP.24 DP Solution Let, f(i,s) := shortest sub-tour given that we are at city i and still have to visit the cities in s (and return to home city) Then clearly,

25. TSP.25 Explanation Then clearly, ….. (i,s) We are at city i and still have to visit the cities in s Suppose we decide that from here we go to city j Then we shall travel the Distance d(i,j) (j,s\{j}) We are now at city j and still have to visit the cities in s\{j}

26. TSP.26 Example (Winston, p. 751) Distance (miles) Cities: New York, Miami, Dallas, Chicago

27. TSP.27 Initialization (s=  ) f(1,  ) = d(1,0) = 1334 f(2,  ) = d(2,0) = 1559 f(3,  ) = d(3,0) = 809

28. TSP.28 Iteration (on, i and s) We shall generate s systematically by its “size”. Size = 1: Possible values for s: {1}, {2}, {3}. s = {1} : f(2,{1})= ? ; f(3,{1})= ? s = {2} : f(1,{2})= ? ; f(3,{2})= ? s = {3} : f(1,{3})= ? ; f(2,{3})= ?

29. TSP.29 |s|=1 f(i,{j}) = d(i,j)+f(j,  ) f(2,{1})= d(2,1) + f(1,  ) = 1343 + 1334 = 2677 f(3,{1})= d(3,1) + f(1,  ) = 1397 + 1334 = 2731 f(1,{2})= d(1,2) + f(2,  ) = 1343 + 1559 = 2902 f(3,{2})= d(3,2) + f(2,  ) = 921 + 1559 = 2480 f(1,{3})= d(1,3) + f(3,  ) = 1397 + 809 = 2206 f(2,{3})= d(2,3) + f(3,  ) = 921 + 809 = 1730

30. TSP.30 |s| = 2 f(i,s)= min{d(i,j)+f(j,s\{j}): j in s} Size = 2: Possible values for s: {1,2}, {1,3}, {2,3} Thus, we have to determine the values of f(3,{1,2}) = ? ; f(2,{1,3}) = ? ; f(1,{2,3}) = ? Eg: f(3,{1,2}) = min {d(3,j) + f(j,s\{j}): j in {1,2} } = min {d(3,1) + f(1,{2}), d(3,2) + f(2,{1})} = min {1397+2902,921+2677} = min {4299,3598} = 3598, N(3,{1,2})={2}

31. TSP.31 |s| = 3 In this case there is only one feasible s, namely s={1,2,3}. Thus, there is only one equation to solve, namely for i=0, s={1,2,3}. The value of f(0,{1,2,3}) is the shortest tour. Note that in this case f(0,{1,2,3})=min {d(0,j) + f(j,{1,2,3}\{j}: j in {1,2,3} = min {d(0,1)+f(1,{2,3}), d(0,2)+ f(2,{1,3}), d(0,3)+f(3,{1,2})} =min {1334+3073, 1559+3549, 809 + 3598} = min {4407,5108,4407} = 4407, N(0,{1,2,3})={1,3}

32. TSP.32 Recovery S=(0,{1,2,3}), N(s)={1,3}, c=1 S={1,{2,3}}, N(s)={2}, c=(1,2) S={2,{3}}, N(s)={3}, c=(1,2,3). Hence: x*=(0,1,2,3,0), z*=4407