Restricted Delivery Problems on a Network. December 17, Abstract

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1 Restricted Delivery Problems on a Network Esther M. Arkin y, Refael Hassin z and Limor Klein x December 17, 1996 Abstract We consider a delivery problem on a network one is given a network in which nodes have supplies or demands for certain products, and arcs have lengths satisfying the Triangle Inequality. A vehicle of innite capacity, travels through the network, carrying products to their destinations, and is limited in that it can carry only a single type of product at a time. The general problem asks for a shortest delivery route of all products from their origin to their destination. Here we consider certain restrictions on the delivery paths allowed, and compare the quality of the solution of the unrestricted problem to that of the restricted one. Both the general and restricted problems are NP-hard, and we discuss approximation algorithms. We also give a constant factor approximation algorithm for the Clustered Traveling Salesman Problem. Keywords: Traveling salesman problem, approximation algorithm. 1 Introduction In this paper we are concerned with a delivery problem on a network. We are given a network (a directed or undirected graph) G = (N; E) in which arcs have lengths (travel distances). We assume throughout that the lengths satisfy the triangle inequality. Every node in the graph is characterized by the type and quantity of product that it demands, and the type and quantity of product it supplies. Nodes in which there is a supply of some product are called warehouses, and nodes in which there is a demand of some product are called demand nodes. A node may be a warehouse for more than one product and also a demand node for more than one product. We assume that for each product type, its total supply is equal to its total demand. The deliveries are carried out by a vehicle of innite capacity, starting at a given node, traveling through the network, and terminating at a given node, carrying items to their destinations. When more than one product type is involved, the total distance traveled may be shortened when a product is placed for temporary storage at a node that is not its demand node. Such a placement is called a drop. In many practical applications there is an important limitation on the deliveries: at most one product type can be carried at a time. Such instances occur when the dierent product types may mix and this is undesirable, such as dierent grain types, soil, or liquids. The y Department of Applied Mathematics and Statistics, SUNY Stony Brook, Stony Brook, NY USA. estie@ams.sunysb.edu. Partially supported by NSF Grants CCR and CCR z Department of Statistics and Operations Research, School of Mathematical Sciences, Tel-Aviv University, Tel-Aviv 69978, Israel. hassin@math.tau.ac.il x Department of Statistics and Operations Research, School of Mathematical Sciences, Tel-Aviv University, Tel-Aviv 69978, Israel. 1

2 general problem asks for a shortest route to deliver products from their origins to their destinations. It is easy to see that this problem is NP-hard, since it generalizes the Traveling Salesman Problem (TSP). Often, in practice, additional restrictions are placed on the allowable delivery routes. For instance, the delivery of products of each type must be done consecutively, if the cost of preparing the vehicle to switch the type of product carried is very high. In other cases, drops may be too expensive or technically dicult and consequently may be excluded. Another reason for such restrictions might be administrative convenience. It is not hard to see that the restricted versions of the problem we consider are also NP-hard, however, in practice they may be \easier" than the general problem, or more amenable to approximation algorithms. We are therefore motivated to compare the quality of an optimal solution to the unrestricted problem to an optimal solution to the restricted problems. We use r to denote the ratio of the optimal solution to a restricted problem to the optimal solution to the unrestricted problem. By denition, r 1. We note that the requirement of the delivery route to be a cycle or path holds for both the restricted and the unrestricted problems. Observation 1.1 Note that for the restriction of consecutive deliveries, if the delivery route is required to be a cycle, then a trivial upper bound on r is p, the number of product types: By repeating an unrestricted optimal solution p times, each time carrying a dierent product, we obtain a feasible restricted solution. In particular, this procedure applies even to the general case of directed graphs, in which the unrestricted version of the problem permits drops. If the cycle delivery route has a given start node, the same bound holds: If this start node is a warehouse, the proof is as before. If it is not a warehouse, we simply walk along the unrestricted solution until visiting the rst warehouse, noting that this unrestricted solution cannot be delivering any products before it visited the rst warehouse. In fact, by the triangle inequality, the unrestricted solution visits a warehouse at the very next node it visits. We then repeat the unrestricted solution p times, except that the last time we omit the arc between the start node and the rst warehouse, thus stopping at the start node (which is the same as the designated end node). In the following example we show that the above ratio is tight (and hence not bounded by a constant), even for undirected graphs, and such that the unrestricted problem does not permit drops: Example 1.2 Consider a graph containing a Hamiltonian cycle with kp nodes, v 1 ; :::; v kp, and edges of unit lengths. The other edges in the graph have length given by their distance along the cycle. There are p product types, and each product type has k demand nodes and k warehouses, with supply and demand of 1 unit. Specically, for i = 1; :::; k, node v ip has a demand for product p and supply of product 1. For each i = 1; :::; k and j = 1; :::; p? 1 node v ip+j has a demand for product j and supply of product j + 1. The unrestricted optimal solution consists of traversing the cycle once while the restricted solution traverses it p times. Hence the ratio r is equal to r = kp2 = p. kp Observation 1.3 An observation similar to Observation 1.1 can be made if the required delivery route is a path, and the graph is undirected. In this case r 2p? 1, since by 2

3 traversing the unrestricted optimal path from its beginning to its end, and then back to its beginning p? 1 times, and then one last time from the beginning to the end, we obtain a feasible solution to the restricted problem. The question of whether this bound is tight remains open. For a directed graph, the path problem may have an unbounded ratio even for 2 products, in the simpler cases we consider, and we give such an example in Section 3. In this paper we will discuss two variants of the problem for which we will show a bounded ratio. There is just one product type, and we impose a restriction that all pick-ups must be made before any deliveries. In this case we show in Section 2 that the ratio of the restricted optimal solution to the unrestricted optimal solution is 2 (resp. 3) if the delivery route is required to be a cycle (resp. path), and the graph is undirected. For a directed path, the same bound (2) for the case of a cycle holds, however, we give a simple example showing that the ratio for the case of a path may be unbounded. In Section 3 each product type is available at a single warehouse. When the graph is undirected and deliveries of each product type are restricted to be consecutive, we show that the ratio of the restricted optimal solution to the unrestricted optimal solution is 3. The same bound is obtained when drops are not permitted (but the consecutiveness restriction is not imposed). We then investigate the eect of restricting the deliveries of each product to be consecutive when drops are already forbidden. We prove a bound of 2 using a theorem about strongly connected graphs which is proven in Section 4. The theorem is interesting in its own right. Finally, for a directed graph the ratios in all cases are bounded by p, the number of products in the graph, as is true for the more general problem, and we give examples showing that this bound is asymptotically tight. We can approximate the optimal solution to some of the unrestricted problems above, by approximating (whenever possible) the solution to the corresponding restricted problems. Recall that r is dened to be the ratio of the optimal solution to the restricted problem to the optimal solution to the unrestricted problem. Let R TSP be the factor by which a TSP can be approximated. The unrestricted single product problem can be approximated, in polynomial time, by a factor of r R TSP. See Section 2, in which we also give approximation algorithms with better bounds to the unrestricted problem. In Section 5, we give (the rst) approximation algorithms for the Clustered Traveling Salesman Problem, in which the nodes of a graph are partitioned into \clusters", and a salesman must visit nodes of each cluster consecutively. Let R CTSP be the factor by which a Clustered TSP with xed start points in each cluster, can be approximated. Then an unrestricted single warehouse per product delivery problem can be approximated by a factor of rr CTSP in polynomial time, see Section 3. The Clustered TSP has many other applications, one of which is the following: Holes are to be drilled in locations v 1 ; v 2 ; : : : ; v n, each hole must have a specied diameter from a set of known dierent diameters. Since it is costly to change the diameter of the drill, the problem is to nd a short route for the drill, such that each of the dierent diameters is drilled consecutively. 3

4 We conclude this introduction with a brief survey of related work. The problem of arranging or rearranging items in a graph was studied by [1, 4, 2]. The rst two papers assume that the delivery vehicle has a unit capacity, while the third assumes innite capacity. They give approximation algorithms for certain versions of the problem. [7, 20] study the rearrangement problems in a warehouse with a vehicle of unit capacity, and a warehouse in which items can be stored temporarily. Rearrangement problems are also studied by [16] who consider capacitated vehicles and assume that the transportation cost through an edge depends on the number of vehicles traversing it. Other related work concerns the Stacker Crane Problem in which one seeks a shortest tour in a mixed graph, such that a set of directed arcs must be traversed. The problem can be thought of as an instance of the delivery problem with drops forbidden in which for each product there is a single warehouse and a single demand node. Connecting the warehouse to the demand node by a directed arc, results in the Stacker Crane Problem, to which [11] give a 9/5 approximation algorithm. The Clustered TSP had been studied by [5, 19, 13], who describe (exponential time) algorithms for its solution. [8] study the Chinese Postman Problem, in which there are additional precedence restrictions, and give a polynomial time algorithm for its solution. Finally, there are many other optimization problems in which it is of interest to examine the eect of restrictions on the solution's value. One such example is the relationship between the optimal makespan with preemption and the optimal makespan without preemption in a scheduling problem. This issue was investigated in [18] in the context of scheduling unrelated parallel machines, and a bound of 4 on the ratio of the two solutions was proved. 2 The Single Product Case In this section we discuss the delivery problem of a single product type from a set of warehouse nodes W to a set of demand nodes D. Let jw j = n and jdj = m. Without loss of generality we may assume that W \ D = ;. Furthermore, since the vehicle has innite capacity, and the triangle inequality holds on the distances, drops cannot shorten a delivery route, and we may assume that W [ D = N. In the restricted version of this problem we require the vehicle rst to visit each warehouse, thus collecting all the supply, and then to visit each of the demand nodes, thereby distributing the product. The problem has 4 variants, dened by the following: 1. The delivery route is required to be a cycle or a path. 2. The graph is directed or undirected. Remark 2.1 We will restrict our attention to the case in which the start and end nodes of the route are given. The case in which they may be chosen has the same bounds on r and very similar proofs. We begin by showing that both the restricted and the unrestricted problems are NP-hard for the case of the delivery route being a cycle. Similar reductions can be used to show that all other versions of both restricted and unrestricted problem are NP-hard. 4

5 Theorem 2.2 The restricted (and unrestricted) single product problem, for a delivery route which is a cycle, with a given start node that is the same as the end node, is NP-hard. Proof: We use a reduction from TSP: Given an instance of TSP, select an arbitrary node to be a warehouse with supply n? 1 (n is the number of nodes in the graph) and all other nodes are demand points for 1 unit each. Let the designated start node of the delivery route be the single warehouse. Clearly, an optimal solution to the unrestricted problem is also an optimal solution to the restricted problem and is also an optimal tour on the original TSP instance. Remark 2.3 It is known that TSP has a lower bound of a xed constant performance ratio achievable in polynomial time (assuming P 6= N P ) [3]. The reduction above shows the bound applies to the delivery problem as well. We now show the bound of r 2 for a directed graph in which the delivery route is a cycle. The same bound also holds for undirected graphs since they can be thought of as a special case of a directed graph in which the length matrix is symmetric. We later show the bound is tight, even for the special case of an undirected graph. Theorem 2.4 For the delivery problem of a single product in a directed graph with a cycle delivery route, the ratio of the optimal solution to a restricted problem to the optimal solution of an unrestricted problem, is at most 2, i.e., r 2. Proof: Let OPT be the optimal solution to the unrestricted problem. We construct a solution feasible to the restricted problem by traversing OPT twice. During the rst traversal, all the supply is collected, and in the second traversal all the demand is delivered. It is perhaps surprising that this bound is tight, as we see in the following simple example: Example 2.5 We construct an undirected graph that is a complete graph on 2n nodes. On one of its Hamiltonian cycles, C, we give all edges length 1. All other edges of the graph have lengths given by the length of a shortest path between the endpoints of the edge, in the graph described above. We refer to such lengths as induced by the triangle inequality. Along C the nodes alternate being supply and demand nodes, each with unit supply or demand. Let one of the supply nodes be the designated start node of the delivery route. The unrestricted optimal solution traverses the cycle C once, and has length 2n, whereas a restricted solution has length 2 2n since it must traverse the length of the cycle twice. Hence r = 2 in this case. Next, we show that for the problem of a path with a given start and end node in a directed graph, the ratio of the optimal solution to the restricted problem to the optimal solution of the unrestricted problem can be unbounded. Example 2.6 Construct a directed graph on 4 nodes. The start node 1 is a warehouse, as is node 3. Nodes 2 and 4 are demand nodes, and node 4 is the nal node on the delivery route. All supplies and demands are 1 unit. The lengths of the arcs are dened as follows: For i < j the directed arc (i; j) has length 1, and the directed arc (j; i) has length M. The unrestricted optimal solution has length 3, and the restricted optimal solution has length M + 4. Thus the ratio r tends to innity as M does. 5

6 In contrast to the above example, we now show the bound of r 3 for an undirected graph in which the delivery route is a path. We later show the bound is tight. Theorem 2.7 For the delivery problem of a single product in an undirected graph with a delivery route which is a path with start node s and end node t, the ratio of the optimal solution to a restricted problem to the optimal solution of an unrestricted problem, is at most 3, i.e., r 3. Proof: Let OPT be the optimal solution to the unrestricted problem. As in the proof of Theorem 2.4, we construct a solution feasible to the restricted problem using OPT. From s traverse OPT to t, collecting all the demand. Return from t to s (along OPT), and traverse OPT from s to t again, this time delivering all demand. Clearly, the length of this feasible solution to the restricted problem is at most 3 times the length of OPT. To see that this bound is asymptotically tight we have the following simple example: Example 2.8 Construct an undirected graph on 4 nodes, The start node 1 is a warehouse, as is node 3. Nodes 2 and 4 are demand nodes, and node 4 is the nal node on the delivery route. All supplies and demands are 1 unit. The lengths of edges (1; 2) and (3; 4) are 0, and all other edge lengths are M. The unrestricted optimal solution has length M, and the restricted optimal solution has length 3M. Thus r = 3. We summarize our bounds in the following table. The numbers in the table are r, the ratio of an optimal solution to the restricted problem to an optimal solution to the unrestricted problem. In all instances, there is an example that matches the upper bound. cycle path directed graph 2 1 undirected graph 2 3 Remark 2.9 The unrestricted single product delivery problem can be approximated by a factor of r R TSP in polynomial time. Currently for directed graphs, the best known TSP approximation factor is log jv j, and for undirected graphs 5/3 for a path with 2 given endpoints and 3/2 otherwise [14, 12]. A better approximation algorithm for undirected graphs can be obtained, by considering the unrestricted problem directly. Here R TSP is the factor by which a shortest tour in an undirected graph can be approximated. Theorem 2.10 The unrestricted delivery problem for a single product, with a cycle (resp. path) delivery route and general start and end nodes in an undirected graph, can be approximated by a factor of R TSP (resp. 2R TSP ). If start and end nodes are specied, the approximation factor increases by 1, (resp. 2) to R TSP + 1 (resp. 2R TSP + 2). Proof: The algorithm for a cycle delivery route is simple: Consider all the warehouses and demand nodes as input to any algorithm that obtains an approximation algorithm for the traveling salesman problem. For any tour on all warehouses and demand nodes, there is a warehouse such that if the tour is traversed, starting (and ending) at that node, the total 6

7 supply collected is always greater than or equal to the total demand delivered. See, for example, an observation by Lin and Kernighan [LK pp ]. Hence, the tour delivery problem can be approximated by a factor of R TSP (currently 3=2). For a path delivery route, we construct the tour as for the cycle delivery route. Since the length of an optimal tour is at most twice the length of an optimal path, we obtain a 2R TSP factor approximation (currently 2 3=2 = 3). For the case of specied start and end nodes, we construct an approximation as before, but have to add to it an edge from the designated start node to the arbitrary start node of the approximation. Similarly, at the end, an edge from the arbitrary end point of the approximation to the designated end node. In the case of a cycle delivery route, the length of each such additional edge is no more than half the optimal solution, by the triangle inequality, since an optimal solution must visit both endpoints of such an edge. For a path delivery route, we can bound the length of each edge by at most the length of the unrestricted optimal solution. 3 The Single Warehouse per Product Case Let G = (N; E) be a graph with a set W of p warehouses such that each warehouse contains a dierent product. Thus, the number of products is equal to the number of warehouses. In this section we study the restricted version of the problem, in which deliveries of each product type must be consecutive. Recall that a drop is the placement of a product for temporary storage at a node that is not its demand node. Since the delivery vehicle has innite capacity, all feasible solutions to the restricted problem do not benet (get shorter) if we are allowed to use drops. This, however, is not the case for the unrestricted problem, in which drops may shorten the length of a delivery route. We begin by showing that the problems of nding an optimal restricted, and an optimal unrestricted delivery route are NP-hard, whether drops are allowed or not. Theorem 3.1 The restricted (and unrestricted) single warehouse per product problem for a cycle delivery route is NP-hard. Proof: Use a reduction from TSP: Given an instance of TSP, double each node i into a warehouse and a demand node for product i = 1; : : : ; p, with distance zero between them. Our next two theorems assume that drops are not permitted and consider the eect of restricting the delivery of each product to be done consecutively. We show a bound of r 2 for an undirected graph in which the delivery route is a path. (This implies the same bound for a cycle delivery route, since a cycle is a special case of a path in which the start and end nodes are the same.) The bound is tight when the number of products is not bounded, as is shown by Example 3.3. The bound for directed graphs is given by Observation 1.1, and is shown to be asymptotically tight in Example 3.6. Theorem 3.2 For the delivery problem of products, each available at a single warehouse, in an undirected graph with a delivery route which is a path with a given start node s and end node t, and drops are not permitted, the ratio of the optimal solution to a restricted problem to the optimal solution of an unrestricted problem, is at most 2, i.e., r 2. 7

8 Proof: Consider the directed graph D = (W; A), representing an optimal solution to the unrestricted problem: Its nodes are the warehouses W, and the start and end nodes s and t. We have an arc (w i ; w j ) 2 A if in the unrestricted optimal solution warehouse w j is visited by the delivery vehicle immediately after warehouse w i. Also, there is an arc from s to the rst warehouse visited, and an arc from the last warehouse visited to t. Clearly, D is a directed graph containing an Euler path. A simple case is if D also contains a Hamiltonian path from s to t. Let s; w 1 ; w 2 ; : : : w n ; t be a Hamiltonian path (if one exists). For each i, the set of edges of G in the unrestricted optimal solution, in which the vehicle carries product i, form a connected graph G i containing node w i. Also, add to G i non-carrying edges from the unrestricted optimal solution such that G i contains a path from w i to w i+1. Clearly the total lengths of the edges of G i 's is no more than the length of the unrestricted optimal solution. Finally we construct a feasible solution to the restricted problem in this case as follows: Start at s and go to w 1 then traverse G 1 at most twice, and end at w 2. Repeat for w i, i = 2; : : : n. The length of this solution is at most twice the length of the G i 's and hence at most twice the unrestricted optimal solution. For the general case in which D is not necessarily Hamiltonian, we use a similar idea. Since D contains an Euler path from s to t, the directed graph obtained by adding the edge (t; s) to the graph D is strongly connected. We apply Theorem 4.2 below to obtain B, a subgraph of D with one arc leaving each warehouse vertex, w i, (except possibly t), such that the undirected version of B is connected. Construct D 0 a directed graph whose nodes are the nodes of D, and whose arcs are the arcs of B and their reverse arcs, except for the reverse arcs that would constitute a directed simple path from t to s. We claim that D 0 contains an Euler path from s to t, since the indegree of each node, other than s and t is equal to its out degree, by denition. The indegree of s (resp. t) is one less (resp. more) than the outdegree of s (resp. t), again by denition. Furthermore, the undirected version of the graph D 0 is clearly connected. We dene E 0 to be the edges of G in which product i is being carried, except for the edges i of G corresponding to paths that dene arcs of B. We dene G 0 to be the subgraph of G i induced by edges Ei. 0 Note that G 0 is a connected graph, possibly containing only the single i warehouse vertex w i. Finally, we construct a feasible solution to the restricted problem, by combining D 0 and the G 0 's as follows: Traverse the paths in G corresponding to an Euler i path in D 0, but before leaving a warehouse vertex w i along an arc of B (recall, there is a unique such arc for all nodes of D except possibly t), traverse over G 0 by doubling the edges i Ei. 0 If t is a warehouse, say w k, and it does not have an arc in B leaving it, then, at the end of the Euler path in D 0 by traversing the edges of G 0 at most twice, we deliver also product k. k This yields a delivery route in which all the deliveries of a product i are done consecutively. The optimal solution to the unrestricted problem contains all the paths corresponding to arcs of B and also all the arcs of G 0 for every i. The restricted solution we generated i contains these arcs twice, and hence the ratio r is bounded by 2. To see that this bound is tight, even in the simple case in which D is Hamiltonian we have the following example for a delivery route that is a cycle. This implies a tight bound also for the path case, since a cycle is a special case of a path in which the given start node is the same as the given nal node. 8

9 Example 3.3 We construct an undirected graph that has p > 2 warehouse nodes, placed on a straight line, where w i is on the integer point i. All edges of the graph have length which is the Euclidean distance between the points on the line. There are two demand nodes for product i, at point i? 1 and i + 1, except for products 1 and p that have a single demand node at point 2 and p? 1 respectively. An optimal solution to the unrestricted problem is of length 2(p? 1). An optimal solution to the restricted problem is of length 4(p? 2) + 2 = 4(p? 1)? 2, since it traverses all but one of the line segments [i; i + 1] 4 times, and the remaining such segment twice. Thus, the ratio r = 2? (1=(p? 1)) which tends to 2 as p tends to innity. In the example above, the ratio of 2 is \achieved" for a large number of products. We conjecture that the example above is the worst possible, namely that the ratio, r, is bounded by 2? (1=(p? 1)). In the case of p = 2, we show a matching upper and lower bound of r = 3=2: Theorem 3.4 For the delivery problem of 2 products, each available at a single warehouse, in an undirected graph with a delivery route which is a cycle, and drops are not permitted, the ratio of the optimal solution to a restricted problem to the optimal solution of an unrestricted problem, is at most 3=2, i.e., r 3=2, and this bound is tight. Proof: Consider an optimal solution to the unrestricted problem. It contains walks from warehouse w 1 to warehouse w 2 along which product 1 is delivered, and then back to warehouse w 1 delivering product 2, and so on, say k times. Now, if k is odd, we can combine pairs of paths along which product 1 is delivered, and end up at warehouse w 2, having delivered all of product 1, without any additional travel. Then by combining pairs of paths along which product 2 is delivered, we can deliver all of product 2, and end up at warehouse w 1. Thus if k is odd, we get a restricted solution of the same length as the unrestricted solution. Now for k 2 even, a restricted solution can be constructed as before, except that the delivery vehicle will have to travel one additional time from w 1 to w 2 and back, carrying no product (deadheading). Clearly such travel can be done along a shortest path between w 1 and w 2. This extra travel has length at most half the length of the unrestricted optimal solution ND, since the unrestricted optimal solution travels at least twice back and forth between w 1 and w 2 (k 2). We have thus generated a feasible solution to the restricted problem whose length RST (3=2)ND, and thus the ratio r is bounded above by 3=2. An example showing that this bound is tight is easily constructed, as in the upper bound proof: Let w 1 ; w 2 be the two warehouses, and they are connected by four paths, each path of length 1. Demands are spread along each of the paths, on two paths there are demands for product 1 and on the other two paths demands for product 2. All other lengths are induced by the triangle inequality. Clearly, an unrestricted solution with no drops has length 4, while an optimal solution to the restricted problem has length 6. For directed graphs with a path delivery route, the ratio between the restricted and no-drops optimal solutions can be unbounded if a product is available at more than one warehouse. (Such examples are easy to construct.) However, in the case we are considering, where each product is available at a single warehouse, the ratio is (surprisingly) bounded by p, the number of products. Example 3.6 below shows this bound is tight. 9

10 Theorem 3.5 For the delivery problem of products, each available at a single warehouse, in a directed graph with a delivery route which is a path from s to t, and drops are not permitted, the ratio of the optimal solution to a restricted problem to the optimal solution of an unrestricted problem, is at most p, the number of products, i.e., r p. Proof: Given an optimal solution to the unrestricted problem without drops, denoted ND, let w 1 be the rst warehouse visited by ND, after s. We dene other w i+1 inductively, once w i is dened: w i+1 is the rst warehouse visited by ND, just after leaving w i for the last time. For some warehouse w q, once it is left for the last time no further warehouses are visited, and ND proceeds to t. Note that q p, namely every warehouse appears at most once in this ordering along ND. We consider rst the easier case, in which q = p, namely, all warehouses appear in the ordering given. Let i, for i = 1; : : : ; p, be the path in ND from the last visit to w i to the \rst" visit of w i+1 (the rst visit of of w i+1 after w i is last visited). q is the path in ND from the last visit to w q to t, and s is the path from s to w 1 in ND. Let C i for i = 1; : : : ; q be the closed walk that starts at the rst visit to w i and ends at the last visit to w i. Note that along C i, ND may visit other warehouses and distribute products other than i. Furthermore, since p = q in this case, we have that ND is the union of the closed walks C i and the paths i. We construct a solution to the restricted problem on the q products, RST(q), as follows: If a warehouse w i is visited only once, then the delivery of product i in ND is consecutive (recall that ND does not permit drops), so we include in RST the path i along which all the deliveries of product i are done. Otherwise, we include C i and then i, and deliver along them only product i. (Arcs along C i in which other products were delivered in ND, will be non carrying arcs in this part of RST.) Since p = q, RST(q) is a feasible solution to the restricted problem. Finally, note that every arc of ND appears in at most one i and at most q? 1 dierent C i 's, and at most once in each i and each C i. Thus every arc of ND appears in RST at most p times, implying that the length of RST is at most q = p times the length of ND. Now assume that q < p. Dene the path from s to t to be the concatenation of all the paths i, including s, i.e., = ( s ; 1 ; : : : ; q ). By our assumption that q < p, we have that NDn contains p? q warehouses, and is comprised of one or more connected components, each of which is a directed closed walk starting (and ending) at a warehouse node on. Furthermore, each of these additional p? q warehouses appears in exactly one connected component. We now use Observation 1.1 to construct a separate solution to a restricted problem on each connected component, in which the products to be delivered correspond to those of the additional p? q warehouses present in that connected component. The length of the union of these individual solutions is no more than p? q times the length of NDn. Finally, we combine these individual solutions with a solution to the restricted problem for products 1; : : : ; q, RST(q), obtained as in the simple case above, whose length is at most q times the length of ND. Clearly the length of this combined solution is at most p times the length of ND. The combination is done as follows: Walk from s to w 1 along RST(q), carrying and delivering no products. If there is a connected component of NDn that contains w 1, walk along the individual solution for that component, delivering all products of that component except for w 1. Next continue along RST(q) to w 2, delivering product 1. 10

11 In general, upon reaching a warehouse w i on RST(q), check whether there is a connected component containing w i, and if so, whether the restricted solution corresponding to it has already been traversed. Traverse the restricted solution when necessary, and proceed along RST(q) to w i+1 (or to t if i = q). The following example shows that for a directed graph, the bound obtained by Observation 1.1 is asymptotically tight for cycle delivery route, and hence also for paths, even when compared to ND, the optimal solution to the problem with drops not permitted: Example 3.6 We construct a directed graph with p warehouse nodes w 1 ; : : : w p. From each warehouse node w i, i = 1; : : : p there are k \parallel" paths to w i+1, each path of length 1 (where w p+1 is w 1 ). On each such path from w i to w i+1 there is a demand node for product i. All other arcs in the graph have length induced by the triangle inequality. The unrestricted optimal solution has length k p as it goes around the cycle w 1 ; w 2 : : : ; w p ; w 1 k times (and does not benet from the use drops even if they are permitted). A restricted optimal solution will traverse the length of this cycle (k? 1)p + 1 times, and thus be of length (k? 1)p 2 + p. The ratio r is equal to ((k? 1)p + 1)=k which tends to p, the number of products, as k tends to innity. We now turn to the case in which drops are allowed in the unrestricted problem. As noted before, drops cannot improve the quality of a solution to a restricted problem. Observe that Example 3.6 shows that for directed graphs the general bound p of Observation 1.1 is asymptotically tight for the \drop" case as well. Theorem 3.7 For the delivery problem of products, each available at a single warehouse, in an undirected graph with a delivery route which is a cycle, and drops are permitted, the ratio of the optimal solution to a restricted problem to the optimal solution of an unrestricted problem, is at most 3, i.e., r 3. Proof: Without loss of generality, assume that the order by which warehouses are visited for the rst time in the unrestricted optimal solution, OPT, is w 1 ; w 2 ; : : : w p. Let E i be the edges of G in which product i is carried in OPT, and G i be the subgraph of G induced by edges E i. Note that G i is a connected graph. We generate a feasible solution to the restricted problem as follows: Double the graph G i for i = 1; 2 : : : p and traverse it, then continue as in OPT from w i to w i+1 (where w p+1 is dened to be w 1 ). Clearly, the length of this solution is bounded by 3OPT. An immediate corollary of this theorem, is the following: Corollary 3.8 Consider the delivery problem of products, each available at a single warehouse, in an undirected graph with a delivery route which is a cycle. Let UNR be an optimal solution to an unrestricted problem with drops permitted, and ND an optimal solution to the same unrestricted problem with drops not permitted, then ND=UNR 3. The following example shows that both bounds given by Theorem 3.7 and Corollary 3.8 are asymptotically tight, by giving an example in which ND=UNR tends to 3. 11

12 3 2 1 n Figure 1: Example 3.9 Example 3.9 There are two types of products, primary and secondary. There are n warehouses for primary products w 1 ; w 2 ; : : : w n. Demand for primary product 1 is at each of the n? 1 warehouses w 2 ; : : : w n. There is a special node, 0, in which there is demand for all other primary products, 2; : : : ; n. Between every primary warehouse w i and node 0 there are two paths, each of length 1, with \many" secondary warehouse nodes spread along each path. There is one demand node for each secondary product, at a very small distance from its warehouse. All other edge lengths are induced by the triangle inequality. See Figure 1, in which the warehouses w 1 ; : : : ; w n are depicted as squares, and node 0 is the circle in the center. The unrestricted optimal solution (with drops permitted) traverses each of the paths once, with small excursions for delivery of secondary products, and thus has length 2n. In this solution primary product 1 is stored at node 0, to meet the demands in warehouses w 3 ; : : : w n. An optimal solution to the unrestricted problem forbidding drops must traverse the equivalent of all but 2 of the paths just for the delivery of primary product 1, thereby traversing length 2n? 2. To deliver all other primary products, it must traverse the same length again, but along the way for each i = 2; : : : ; n the secondary demands on one of the two paths connecting node 0 to w i, but not both(!) can also be met. Finally, supplying the remaining secondary demands again requires an additional 2n? 2. The total length of an optimal restricted delivery route is thus at least 6n? 6. Thus r here is at least 3? 3=n, which tends to 3 as n tends to innity. Remark 3.10 The same bounds of 3 (both upper and lower) apply also to the case in which the delivery route is a path, and the graph is undirected. For a directed graph in which the delivery route is a path from s to t, the following example shows that the optimal solution to the problem without drops (whether delivery must be consecutive or not) can be innitely longer than an optimal solution to the problem with drops, even for the case of 2 products: Example 3.11 There are four nodes in a directed graph: node 1 is the start node, and is also w 1, node 2 is w 2, node 3 is d 2, and node 4 is d 1, which is also the nal node. The 12

13 distances along the directed arcs are 1 for arcs (i; j), with i < j, and M very large for arcs (i; j) with i > j, except for arc (3; 2) (which is the arc from d 2 to w 2 ) which is also of length 1. An optimal solution to a problem permitting drops carries product 1 to node 2, drops it there, carries product 2 from node 2 to node 3, returns to node 2 to pick up product 1 and carries it to its destination, node 4. The length of such a path is 5. If drops are not permitted, any feasible delivery route must traverse at least one arc of length M, and thus the ratio of the lengths tends to innity as M tends to innity. We summarize the bounds obtained in the following table. The rst entry is the upper bound, and the second entry is the lower bound achieved by a given example. We let UNR (resp. ND, RST) denote the value of an optimal solution to an unrestricted problem with drops permitted (resp. no drops permitted, and restricted problems). p is the number of products, n, the number of primary products ( p), and k an arbitrary number. UNR / ND ND / RST UNR / RST directed graph cycle p ; p(k? 1)=k p ; p(k? 1)=k p ; p(k? 1)=k directed graph path 1 p ; p 1 undirected graph path/cycle 3 ; 3? (3=n) 2 ; 2? (1=(p? 1)) 3 ; 3? (3=n) Remark 3.12 Let R CTSP be the factor by which a Clustered TSP with xed start points in each cluster, can be approximated. Then the unrestricted single warehouse per product delivery problem can be approximated by a factor of r R CTSP in polynomial time: Dene a cluster for each product i, containing the demand nodes and single warehouse node of product i. The designated start node for the cluster is the warehouse node. In Section 5 we give a factor of 7/2 approximation algorithm for The Clustered TSP with given start nodes in each cluster, for undirected graphs. Thus, the single warehouse per product problem on undirected graphs with no drops permitted (resp. with drops permitted) can be approximated by a factor of 7 (resp. 3 (7=2)). A better approximation algorithm can be derived for the case of 2 products and no drops allowed. Let R TSPT (resp. R TSPP ) be the factor by a shortest tour (resp. path with given start and end nodes) can be approximated. Theorem 3.13 The unrestricted delivery problem of 2 products, each available at a single warehouse, on an undirected graph with cycle delivery route,with no drops allowed, can be approximated by a factor of maxfr TSPT ; R TSPP g in polynomial time. Proof: In every feasible delivery route of 2 products, the delivery vehicle must switch the product being transported at least once. We claim that there is an optimal solution to the unrestricted problem, in which the delivery vehicle switches the product carried at most twice. Such a delivery route can be constructed at no extra length, from any given delivery route which is a cycle, as described in Lemma 3.4, by combining pairs of routes delivering the same product. Let the number of walks from warehouse w 1 along which product 1 is delivered to warehouse w 2, (which must be equal to the number of walks from w 2 back to w 1 delivering product 2) be k. Now, if k is odd, we can combine pairs of paths along which product 1 is delivered, and end up at warehouse w 2, having delivered all of product 1, without 13

14 any additional travel. Then, switch to product 2 and deliver it back and forth along all k paths carrying product 2, and ending at w 1. If k is even, we carry product 1 on k? 1 of the paths carrying product 1, ending at w 2, then carry product 2 along all k paths, and switch back to product 1, for the last time, returning to w 1. Let OPT 1 and OPT 2 be the length of an optimal delivery route with the additional restriction that products carried by the vehicle are switched exactly once or twice, respectively. By the above discussion, OPT = minfopt 1 ; OPT 2 g. We now describe two algorithms that result in two feasible solutions to the unrestricted problem, APX 1 and APX 2, with one and two product switches respectively, which approximate OPT 1 and OPT 2. Algorithm 1: 1. Find P 1, an approximate shortest path starting at w 1, ending at w 2, and visiting all demand nodes for product Find P 2, an approximate shortest path starting at w 2, ending at w 1, and visiting all demand nodes for product Traverse P 1 distributing product 1, then P 2, distributing product 2. Algorithm 2: 1. Find T 1, an approximate shortest tour on w 1, w 2, and all demand nodes for product 1. The tour is comprised of two paths between w 1 and w 2, call them T 1 1 and T Find T 2, an approximate shortest tour on w 1, w 2, and all demand nodes for product 2. The tour is comprised of two paths between w 1 and w 2, call them T 1 2 and T Traverse T 1 1 delivering product 1, then T 1 2 and T 2 2 delivering product 2, and nally T 2 1 delivering product 1. Let APX 1 and APX 2 be the lengths of the solutions generated by Algorithms 1 and 2, respectively. We get that APX 1 R TSPP OPT 1, and APX 2 R TSPT OPT 2. Thus we obtain a maxfr TSPT ; R TSPP g approximation algorithm for our problem. Using Christodes' algorithm for steps 1 and 2 in both algorithms, we have that R TSPP = 5=3 [12], and R TSPT = 3=2 [6, 14], which yield a 5=3 factor approximation to our problem. 4 Proof of Graph Theoretic Theorems Theorem 4.1 Let D = (N; A) be a strongly connected directed graph, and let A i be the set of arcs leaving node i 2 N. Let G = (N; E) be the undirected version of D, in which for each directed arc (i; j) we dene its undirected version, an edge fi; jg. Let E i be the undirected versions of A i. One can select one edge from each set E i, such that the resulting collection of edges contains a spanning tree of G. 14

15 v 1 v 2 v 3 v 1 v 2 v 3 v r v r B 1 before path exchange v r+1 B 1 after path exchange v r+1 Figure 2: Proof of Theorem 4.1 Proof: Let B be a collection of jnj arcs selected arbitrarily, one from each A i, and B 0 its undirected version. There are two possible cases: Case (i) B 0 is a connected subgraph of G: Since B 0 contains jnj edges and is connected, it clearly contains a spanning tree of G. Case (ii) B 0 is not connected: Let B 0 be a connected component of 1 B0, and B 1 its directed version. We claim that B 1 contains a directed cycle, and all other arcs in B 1 are directed into the cycle. This holds since every node in B 1 has a single arc leaving it in B 1, thus the number of arcs is equal to the number of nodes, and hence B 1 contains a cycle. This cycle must be directed, since it can have no node of out degree greater than 1 in B 1, by denition. Denote this cycle by C. Since B 0 is connected, every node in B 1 1 n C is connected to C by a path. Since no node in C can have an arc in B 1 directed out of C, this path must be directed toward C. We now show how to replace some edges of B by other edges from E, such that we maintain the property that we have exactly one edge from every E i, and the number of connected components is reduced by 1. Clearly, a repeated application of this replacement procedure, completes the proof of the theorem. Let P = fv 1 ; v 2 ; : : : v r ; v r+1 g be a directed path in D starting at a node v 1 2 C B 1, such that v 2 ; : : : v r 2 B 1 nc and v r+1 2 B2, 0 where B 0 2 is another connected component of B, and B 2 its directed version. Replace the arcs of B 1 leaving nodes v 1 ; : : : v r by the arcs of P. See Figure 2 for an illustration. The property that each node of D has one selected arc leaving it, is clearly maintained. Furthermore, B 0 1 remains connected: Consider an arc (v j ; k) that is removed from B (and replaced by arc (v j ; v j+1 )). Node k is still connected to the nodes of C by our claim above that stated that nodes of B 1 not in C are connected to C by arcs directed towards C. Finally, including the arc (v r ; v r+1 ) in the modied B creates a single connected component containing the nodes of B 0 1 [ B2, 0 and thus the number of connected components of B 0 has been reduced. Theorem 4.2 Let D = (N; A) be a directed graph, such that D [ (t; s) is strongly connected for some s; t 2 N, and let A i be the set of arcs in D leaving node i 2 N. One can select one arc from each set A i, to obtain B, whose undirected version, B 0 is connected. (If A t = ;, B 15

16 does not include an arc from A t.) Proof: Let ^D be the graph D with an additional arc from t to s, and ^A t = A t [ (t; s). Apply Theorem 4.1 to ^D to obtain ^B whose undirected version is connected. Now if ^B does not contain the arc (t; s), we let B = ^B, and we are done. So assume (t; s) 2 ^B. Remove arc (t; s) from ^B, and add in an arc from At (if possible, namely, if A t is not empty). If the undirected version of the new ^B is connected, we are also done. So assume the new ^B consists of two connected components B 1 containing s and B 2 containing t. We now continue as in the proof of Theorem 4.1: B 1 consists of a directed cycle C and paths directed towards C. Find a path P from C to a node in B 2, and replace some of the arcs of B by P. Note that since the graph D had an Euler path from s to t, there is a directed path from every node in D to t, so we can choose a path P as required. 5 Approximating The Clustered TSP In this section we describe a constant factor approximation algorithm for The Clustered Traveling Salesman Problem. In Clustered TSP the nodes of an undirected graph are partitioned into clusters that must be traversed consecutively by a salesman. There are two versions of Clustered TSP, one in which a start node in each cluster is specied, and the other in which the start point can be any point of the cluster. The problem with unspecied start nodes was studied by [5, 19, 13], who describe (exponential time) algorithms for its solution. The algorithm in [5] is based on the following idea: Add a large constant to the length of all edges connecting nodes of dierent clusters. This ensures that a \usual TSP" tour will have the desired structure, namely traverse each cluster consecutively, since the edges of large costs are avoided as much as possible. Note that this transformation does not preserve the error ratio and thus approximation algorithms for TSP are not helpful (see [9] for an elaboration of this issue). Gendreau, Hertz and Laporte describe a 1.5 approximation for the problem with two clusters and possibly an additional node (the depot). [10] provide a factor of 2 approximation algorithm for the clustered TSP where the clusters must be visited in a xed sequence. Here we describe an approximation algorithm for Clustered TSP with designated start nodes, which clearly, is also an approximation algorithm for the problem with unspecied start nodes. Before describing our approximation algorithm, we dene some notation. Let OPT be the length of an optimal solution to Clustered TSP. We denote by MST(G) the length of a minimum spanning tree on graph G, and by TOUR(G) the length of some constructed tour on graph G. The input to Clustered TSP is an undirected graph G = (N; E) with lengths on the edges satisfying the triangle inequality. Let N 1 ; N 2 ; : : : ; N p be the nodes of the clusters, i.e., [ i N i = N and N i \ N j = ; whenever i 6= j. Denote by G i the subgraph of G induced by the nodes of cluster i, N i. For the problem with designated start nodes, we are also given a node in each cluster w i 2 N i, such that the salesman must start his visit to cluster i by visiting node w i. 16