to the Traveling Salesman Problem 1 Susanne Timsj Applied Optimization and Modeling Group (TOM) Department of Mathematics and Physics
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1 An Application of Lagrangian Relaxation to the Traveling Salesman Problem 1 Susanne Timsj Applied Optimization and Modeling Group (TOM) Department of Mathematics and Physics M lardalen University SE , V ster s, Sweden Technical Report IMa-TOM June 15, 1999 Abstract The paper deals with the traveling salesman problem, i.e. the problem to nd the length of the shortest closed tour visiting a number of cities. The problem is formulated as an optimization problem. A solution strategy based on Lagrangian relaxation is presented. Applying Lagrangian relaxation will generate a convex, but non-smooth dual problem. For the solution of the dual problem a subgradient method is used. Heuristics are applied to modify the relaxed solution into a primal feasible one. An algorithm is implemented and presented together with some computational results. 1 Introduction The Traveling Salesman Problem (TSP) is a well-known mathematical programming problem, see e.g. [1], [2]. The present paper consider a variant know as the Symmetric Traveling Salesman Problem (STSP). The problem might be outlined as follows. A Salesman will visit a number of cities, starting from one city, visiting every city exactly once. After visiting all cities he will return to the initial city. A trip which fullls this requirement is called a tour. The salesman wants to travel on the shortest (cheapest) tour. He knows the distance between every city, and he can thereby decide which way is the shortest, but as the number of cities increases this becomes dicult. The problem may then be formulated as an optimization problem, solvable with mathematical programming techniques. To nd the solution of the symmetric traveling salesman problem a number of optimization techniques have been suggested and implemented, including heuristic methods, branch-and-bound, Lagrangian relaxation, simulated annealing and genetic algorithms. A survey is given in [2]. The present paper presents a solution strategy based on Lagrangian relaxation. The technique was rst introduced by Held and Karp [3]. The methodology is to relax 'hard' constraints, which will generate a relaxed problem which is easy to solve. Applying Lagrangian relaxation will also generate a convex, but non-smooth dual problem. For the solution of the dual problem methods for non-smooth optimization, e.g. subgradient methods, must be used. Since the original problem (primal problem) is an integer one a zero duality gap is not guarantied and therefore a heuristic method is used to generate primal feasible solutions. In Section 2 the problem considered is formulated and in Section 3, an algorithm solving the dual problem is presented. Heuristics generating primal feasible solutions are developed in Section 4. Section 5 gives some computational results, and nally, in Section 6, some conclusions are given. 1 Written as a ve weeks undergraduate course in mathematics supervised by Erik Dotzauer. 1
2 2 The Symmetric Traveling Salesman Problem In this section the problem to be considered is formulated. Let N be the number of cities. Dene c ij as the distance between city i and city j. It is assumed that c ij = c ji, i = 1; :::; N, and j = 1; :::; N. Moreover, let x ij be a binary variable indicating if the path between city i and city j is in the tour. If the path is in the tour, x ij is equal to one, otherwise, x ij is zero. In mathematical terms a city is referred to as a node and a path between to cities is referred to as an arc. In order to simplify the exposition, dene the following. First, dene a tree on the nodes f2; :::; N g. A tree is a graph, i.e. a number of nodes connected with arcs, without cycles, i.e. in a tree there exists one and only one path between two nodes. This gives a tree with N? 2 arcs. Further, add two distinct arcs from node one to two distinct nodes in the tree. This agrees with the denition of a 1-tree, i.e. a 1-tree dened on N nodes is a connected graph with N arcs and exactly one cycle. Dene degree of node i as the number of arcs connected to the node. Thus, a 1-tree where all nodes have degree two is a tour. Dene the symmetric traveling salesman problem as the following mathematical programming problem, f = min x s:t: 2 X N 4 f(x) = j=1 x ij + x is a 1-tree, where x i = (x i;1 ; :::; x i;n ) and x = (x 1 ; :::; x N ). The constraints x ij + c ij x ij 3 5 (1) x ji = 2, j = 1; :::; N; x ji = 2, j = 1; :::; N; (2) ensure that all nodes have degree two, i.e. that the 1-tree is a tour. The objective is to minimize the total distance on such tour. In (1), f(x) is the objective function, and f its optimal value. 3 A Solution Algorithm Based on Lagrangian Relaxation This section presents a solution strategy for problem (1) based on Lagrangian relaxation [4]. The technique was rst introduced by Held and Karp [3]. The methodology is to relax 'hard' constraints, which will generate a relaxed problem which is easy to solve. Applying Lagrangian relaxation will also generate a convex, but non-smooth dual problem. Problem (1) is called the primal problem. The objective function value of the dual problem forms a lower bound on the objective function value of every feasible solution of the primal problem. The technique is to maximize the dual, and since the dual might be non-smooth, methods for non-smooth optimization must be used. Since problem (1) is an integer one, a zero duality gap is not guarantied, and therefore heuristics are used to generate primal feasible solutions. Lagrangian relaxation is performed by introducing multipliers = ( 1 ; :::; N ). Combining with (2) and adding to the objective in (1) gives the relaxed problem () = min s:t: x " j=1 x is a 1-tree, c ij x ij + j ( 2? x ij? j=1 where P () is dened as the dual objective function. The objective in (3) may equivalently be expressed N as problem is easy to solve [3]. P N j=1 bc ij x ij + 2 P N j=1 j, where bc ij = c ij? i? j is dened as the reduced cost. The relaxed The dual problem is dened as x ji ) # (3) = max [()] ; (4) 2
3 where is the optimal dual objective function value. By duality theory, the optimal value of the dual objective is a lower bound on the objective function value of every feasible solution of the primal problem, i.e. relation f(x) is valid, and from (1), (3) and (4) then f(x) f () follows. The duality gap is dened as the dierence between the optimal primal objective f and the optimal dual objective. However, to simplify the presentation we will refer to the dierence f(x)? () as the duality gap. When the primal problem (1) is a integer one, as in our case, the dual problem is in general non-smooth. Due to this fact, a zero duality gap is not guaranteed. Since the dual problem is non-smooth, a solution method for non-smooth optimization [5] must be used. An algorithm frequently used for the solution of such problems is the subgradient method. In each iteration k of the algorithm, a new set of dual variables [] k+1 is computed from [] k, such that we get an improving direction of the dual objective (). The direction chosen is dened from the subgradient, g, with elements g j, j = 1; :::; N, given by The new update [] k+1 is computed as g j = 2? x ij? x ji. (5) where the step length is dened by [] k+1 = [] k + k [g] k, (6) k = k? ([] k ) ; 0 < k jj[g] k 2. (7) jj Expression (7) is knows as the Polyak rule II. Normally, the subgradient is normed to stabilize the algorithm. Since the optimal dual objective is unknown, the best primal feasible solution obtained so far is used instead. A common choice is to start the sequence k with k = 2 and reduce k with a factor of two whenever ([] k ) has failed to increase in a specied number of iterations. The complete solution algorithm for problem (1) is stated as: Step 0. Let k = 0. Choose 0. Let [] k = 0. Step 1. Given [] k, solve the relaxed problem. Step 2. Generate a primal feasible solution. Step 3. If the duality gap is small, then stop. Step 4. Compute [] k+1, let k = k + 1 and go to Step 1. In Step 0 of the algorithm, initial values for the dual variables are chosen. To generate a primal feasible solution in Step 2 the method presented in Section 4 is used. The update of in Step 4 is a step in the solution procedure of the dual problem (4). 4 Computing a Primal Feasible Solution Since the primal problem (1) is an integer problem, a zero duality gap is not guaranteed. As a consequence, given the optimal multipliers, the solution of the relaxed problem (3) may not be a feasible solution of the primal problem. Therefore so called Lagrangian heuristics are applied, i.e. heuristics that modies the relaxed solution to a primal feasible one. In this section, an algorithm implementing such heuristics is developed. In brief the heuristic algorithm is performed as follows. From the solution of the relaxed problem (3), rst a node with degree one is chosen. Such node exists since the relaxed solution is a 1-tree but not a tour (if it was a tour it had already been primal feasible). From this node a unique path to the other 3
4 nodes is constructed by in each iteration of the algorithm adding a new node with corresponding arc. The construction is performed by, as far as possible, using arcs present in the relaxed solution. In the path, all nodes will have degree two. Finally, when all nodes are in the path, a last arc from the last node to the rst node is added. This will generate a tour, i.e. a primal feasible solution. The heuristic algorithm is stated as: Step 0. k = 2. Compute deg(i). Choose a node i with deg(i) = 1, i.e. a node i with one arc (i; j). Let n first = i and n current = j. Dene the set of nodes M k = fi : i = 1; :::; N gnfn first ; n current g and the set of arcs T k = f(i; j)g: Step 1. If k = N, then let T k+1 = T k [ f(n current ; n first )g and stop. Step 2. Let i = n current. If deg(i) = 1, then go to Step 3, else if deg(i) = 2, then go to Step 4, else (deg(i) > 2) go to Step 5. Step 3. (deg(i) = 1) Choose a node j 2 M k with deg(j) 1. Let M k+1 = M k nfjg and T k+1 = T k [ f(i; j)g. Let deg(i) = 2 and deg(j) = deg(j) + 1. Let n current = j, k = k + 1 and go to Step 1. Step 4. (deg(i) = 2) There exists exactly one node j 2 M k which in the relaxed solution is connected to node i. Let M k+1 = M k nfjg and T k+1 = T k [ f(i; j)g. Let n current = j, k = k + 1 and go to Step 1. Step 5. (deg(i) > 2) Choose a node j 2 M k which in the relaxed solution is connected to node i. Let M k+1 = M k nfjg and T k+1 = T k [ f(i; j)g. For all nodes j 0 2 M k, j 0 6= j, which in the relaxed solution are connected to node i, let deg(j 0 ) = deg(j 0 )?1. Let deg(i) = 2. Let n current = j, k = k+1 and go to Step 1. In Step 0, deg(i) is dened as the degree of node i in the relaxed solution, which is given as input to the algorithm. In Step 3 and Step 5, j 2 M k is chosen as the node with the corresponding smallest reduced cost bc ij. When the algorithm terminates, T k+1 in Step 1 is a feasible solution of problem (1). 5 Computational Results The algorithm for the solution of problem (1) presented in Section 3 is applied to 20 test problems from the test problem data base TSPLIB [6]. Initial in Step 0 is chosen equal to zero. During the rst ten iterations no primal feasible solution is computed, i.e. Step 2 is not performed when k < 10. The choice for the sequence k in (7) is to start with k = 0:1 and reduce k with a factor of two whenever ([] k ) has failed to increase in three iterations. The calculations are interrupted after 100 iterations. The result is shown in Table 1. In the table, fknown best is the best known primal feasible solution (PFS), f best best is the best primal feasible solution and kbest is the iteration index in which f was rst computed. Also the duality gap in iteration 25, 50 and 100 are shown. The duality gap is given in percent of the best dual objective so far. The last row shows the mean value of each column. From Table 1 we se that duality gaps less than one percent were for 17 of the 20 data sets. In 10 cases duality gaps less than 0:5 percent were in 25 iterations. In average, the duality gap is 0:69 percent after 25 iterations and 0:42 percent after 100 iterations. In one case (ulysses16) the best known solution was. 6 Conclusions The paper has discussed a solution strategy based on Lagrangian relaxation for the symmetric traveling salesman problem. A heuristic method to nd primal feasible solutions has been developed. An algorithm was implemented and presented together with some computational results. The results show that duality gaps less one percent are in most cases in less than twenty-ve iterations. 4
5 Data Set N f best known Best PFS Found f best Duality Gap in Iteration k k best k = 25 k = 50 k = 100 burma :0574 0:0574 0:0574 ulysses :0781 0:0369 0:0065 ulysses :1152 0:0373 0:0340 swiss :1430 0:0973 0:0245 berlin :1716 0:1124 0:0281 st :2297 0:0528 0:0433 eil :1309 0:0575 0:0299 gr :5164 0:2389 0:1030 rat :2924 0:2017 0:0926 lin :5185 0:2128 0:0777 pr :6696 0:2632 0:1733 pr :7591 0:3091 0:1073 bier :0860 1:0860 1:0860 ch :4165 1:4165 1:4165 pr :2903 0:1824 0:1456 gr :5112 0:1847 0:1314 kroa :7955 0:3374 0:1824 u :0076 4:0076 4:0076 rat :6740 1:1184 0:5875 krob :3667 0:2014 0:1462 Average :6915 0:5106 0:4240 Table 1: Computational results for the algorithm applied to test problems from TSPLIB. References [1] G. L. Nemhauser and L. A. Wolsey, Integer and Combinatorial Optimization, Wiley, [2] Laurence A. Wolsey, Integer Programming, Wiley, [3] M. Held and R. M. Karp, The Traveling Salesman Problem and Minimum Spanning Trees: Part II, Mathematical Programming 1, pp. 6-25, [4] Marshall L. Fisher, An Applications Oriented Guide to Lagrangian Relaxation, Interfaces 15:2, pp , [5] N. Z. Shor, Minimization Methods for Non-Dierentiable Functions, Springer-Verlag, [6] Gerhard Reinelt, TSPLIB 95, Technical Report, Universit t Heidelberg,
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