# JOURNAL OF NETWORKS, VOL. 4, NO. 7, SEPTEMBER

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3 JOURNAL OF NETWORKS, VOL. 4, NO. 7, SEPTEMBER is a special case of our result presented here. III. MODEL A network is a graph G =(V,E), wherev is the set of nodes and E is the set of direct communication lines between nodes, called edges. The number of nodes and edges are respectively denoted by V and E. Observe that in an Internet Service Provider (ISP) network consisting of a single Open Shortest Path Protocol (OSPF) area, each node v in the network forms a shortest path routing tree to route to all other nodes in the network. A shortest path between nodes s and t is denoted by P s,t and the length (i.e. number of edges) in this path is denoted by d(s, t). Clearly, between any two nodes there exist possibly more than one shortest path. However, as the name suggests all these paths have the same length d(s, t). For the purposes of our discussion we assume that graph G representing the network is connected. That is, there is a path between any two nodes in V. Consider node v. A shortest path tree T v rooted at v is a spanning tree of G which, for any node v i, contains a shortest path between v and v i. We assume that the graph is unweighted. That is, all links (i.e. edges) in the network have the same weight. This can occur, for instance, when OC-48 connections are used as the ISP backbone, all the links have the same capacity, or all the links have delay below an appropriate threshold. Let G beagraphandv V. We call an edge e = (a, b) E horizontal with respect to v if d(v, a) =d(v, b) in G. The following observation states that a SP-tree rooted at a node v cannot cover any edge that is horizontal with respect to v. Observation 1: Let G =(V,E) beagraphandv be a node of G. Any edge of G which is horizontal with respect to v cannot belong to any SP-tree T v rooted at v. Proof: Suppose that e = (a, b) is a horizontal edge with respect to v but nevertheless there is a SP-tree T v containing e. Since the shortest distances from v to a and b are the same, we obtain that there is a loop in T v,which is a contradiction. We call an edge e =(a, b) E vertical with respect to v if d(v, a) =d(v, b) +1 or d(v, b) =d(v, a) +1. Let e =(a, b) be a vertical edge with respect to v and assume that d(v, a) =d(v, b)+1 holds. Then, e is called an unavoidable edge by v if any shortest path between v and a includes Observe that if edge e = (a, b) is unavoidable with respect to v, then e = (a, b) is also vertical edge with respect to v. However, the opposite is not necessarily correct. To illustrate, in the network depicted on Fig. 1, edge (2, 3) is unavoidable and edge (2, n +2) is vertical (but not unavoidable) with respect to node 1. The following observation holds. Observation 2: Let G =(V,E) be a graph and v be a node of G. LetS v be the set of all possible shortest path trees rooted at node v. Edge e of G is unavoidable by v if and only if any tree T v S v contains e. Proof: Clearly, if any tree T v S v contains e =(a, b), then e is vertical with respect to v and any shortest path from a to v includes b if d(v, a) =d(v, b)+1. Suppose now, that there is a tree T v in S v that does not contain e = (a, b) and e is an unavoidable edge by v. Assume without loss of generality that d(v, a) =d(v, b)+1.sincet v is a SP-tree, every node of G must be a node of the tree. Consequently, both a and b are in T v but the edge (a, b) is not in T v. On the other hand, since (a, b) is unavoidable by v, any shortest path from a to v contains node b. A contradiction obtained completes the proof. We are interested in the following three problems and describe their applications in practice. E-Problem ( Exist -Problem): Given a graph G = (V,E), select a minimum set of nodes R V and for each node v R atreet v S v such that the union of selected trees covers all edges of G. A-Problem ( Any -Problem): Given a graph G = (V,E), select a minimum set of nodes R V such that, regardless which tree T v S v is selected for a node v R, the union of those trees cover all edges of G. BR-Problem ( Bejerano/Rastogi -Problem): Given agraphg =(V,E) and a SP-tree T v for each v V, select a minimum set of nodes R V such that the union of T v, v R, covers all edges of G. First we demonstrate that optimal solutions for each of the first two problems are considerably different. (Clearly, the size of the optimal solution to the third problem depends on the choice of the family {T v : v V }.) Consider, for example a rectilinear grid of size n n depicted on Fig. 1. We number nodes of the grid from 1 to n row-wise (ith row nodes are (i 1) n+1, (i 1) n+2,...,i n). Figure 1. A-Problem vs E-Problem. On the rectilinear grid of size n n, optimal solution to the E-Problem consists of two trees while optimal solution to the A-Problem consists of n trees. One can select two SP-trees, one rooted at node 1 and another rooted at node n, such that their union covers all the edges of G. Indeed, one can consider a tree rooted at node 1 which is formed by edges ((i 1) n+1,i n+1), (j, j +1) and (i n + j, i n + j +1) (socalled first column and all rows -tree) and a tree rooted at node n which is formed by edges (n n + j, n n + j +1), (i n, (i+1) n) and ((i 1) n+j, i n+j) (so called last row and all columns -tree), where 1 i n 1, 1 j n 1. It is easy to see that both trees are SP-trees. Thus, a solution to the E-problem consists of only two SP-trees. In view of Observation 2, it is rather

4 660 JOURNAL OF NETWORKS, VOL. 4, NO. 7, SEPTEMBER 2009 simple to show also that an optimal solution to the A- problem consists of n roots (SP-trees). Indeed, since edges of column j (j =1,..., n) are unavoidable only by nodes of this column, at least one SP-tree rooted at a node of column j must be present in an optimal solution. Any SP-tree rooted at this node will cover all edges of column j, and for any node x not from column j and any edge e from this column, there exists a SP-tree rooted at x which does not contain the edge e. Analogously, at least one SP-tree rooted at a node of row i (i =1,..., n) must be present in an optimal solution. Thus, an optimal solution to the A-problem on a rectilinear grid of size n n must consist of n SP-trees with roots one per each column and each row. It is easy to see that selecting the nodes on a diagonal of the grid generates a set of SP-trees that completely cover all edges of the grid. IV. HARDNESS RESULTS In this subsection, we prove the hardness of A- Problem, E-Problem, andbr-problem on unweighted graphs. We first prove that the A-Problem is NP-hard on unweighted graphs by reducing the set cover problem to it. As a byproduct we get also the NP-hardness of the BR-problem even on unweighted graphs. For this one needs to refine substantially the construction of [1]. Theorem 1: The A-Problem is NP-hard even for unweighted graphs. Proof: We show that the A-problem is NP-hard by reducing a set cover problem (SC) to it. Consider an instance of SC problem I(Z, Q), wherez is a set of m elements and Q is a set of n subsets of these elements. We construct the following unweighted graph G =(V,E). For each element z i Z (1 i m), V contains six nodes u i, w i, v i and a i, b i, c i,ande contains seven edges (u i,w i ), (u i,v i ), (v i,w i ), (a i,b i ), (a i,c i ), (b i,c i ) and (c i,v i ) (i.e., one has two triangles joined with a bridge edge). For each set q j Q, V contains a node labelled by s j. For each z i q j, E contains edge (s j,u i ). For each z i / q j, E contains edge (s j,v i ). In addition, G contains two cliques of size 3 (called anchor cliques) clique 1 and clique 2. Clique 1 has nodes r 1, s 1,andt 1,andclique 2 has nodes r 2, s 2, and t 2. Also, we have an auxiliary node y which is connected to t 1 and to nodes a i and b i (1 i m), and m auxiliary nodes d i (1 i m) connected to t 1 and v i. Finally, node t 2 is connected to nodes u i, v i,andw i (1 i m). An example of such a graph is depicted in Fig. 2 for the SC problem with Z = {z 1, z 2, z 3, z 4 } and four subsets q 1 = {z 1, z 2, z 3 }, q 2 = {z 2, z 3, z 4 }, q 3 = {z 1, z 3, z 4 },andq 4 = {z 1, z 2, z 4 }. We claim that there is a solution of size k to the given SC problem if and only if there is a solution of size k + m + 2to the A-problem. Let a solution of the SC problem consist of the subsets q j1,...,q jk.we show that any set S of k + m +2 SP-trees rooted at r 1, r 2, a i (1 i m), and s j1,..., s jk covers all edges of G. Indeed, any SP-tree T r1 rooted at r 1 evidently covers edges (r 1,s 1 ) and (r 1,t 1 ), (t 1,y), (t 1,d i ), (d i,v i ), (y, a i ),and(y, b i ) (i =1,...,m). Since edges (t 2,r 2 ) Figure 2. The graph 1 G for an instance of the SC problem with Z = {z 1, z 2, z 3, z 4 } and four subsets q 1 = {z 1, z 2, z 3 }, q 2 = {z 2, z 3, z 4 }, q 3 = {z 1, z 3, z 4 },andq 4 = {z 1, z 2, z 4 }. and (t 2,s 2 ) are unavoidable by r 1 they are also covered by T r1. Any SP-tree T r2 rooted at r 2 covers edges (r 2,s 2 ), (t 2,u i ), (t 2,v i ), (t 2,w i ), (v i,c i ), (v i,d i ), (c i,b i ), and (c i,a i ) (i = 1,...,m). Any SP-tree T ai rooted at a i (1 i m) covers edges (a i,b i ), (v i,u i ), (v i,w i ) and either (v i,s j ) (j {1,...,n}) if this edge exists or (u i,s j ) otherwise. Note that both v i and u i cannot be connected to the same s j. Finally, for every t = i 1,...,i k, any SP-tree T st rooted at s t covers edges (u i,w i ) for each i {1,...,m} such that z i q t. Since subsets q j1,...,q jk cover all elements of Z, each edge (u i,w i ) (1 i m) will be covered. Thus, all graph edges will be covered by at least one of these k + m +2 SP-trees. Next, we show that if there is a set of k + m + 2 SPtrees whose union covers all edges of G, then there is a solution for the SC problem of size k. Note that, among those k + m +2 SP-trees, there needs to be a SP-tree rooted at node a i or b i to cover the edge (a i,b i ),and this holds for any i {1,...,m}. Also, there must exist, among those trees, a SP-tree rooted at r 1 or s 1 to cover the edge (r 1,s 1 ) and a SP-tree rooted at r 2 or s 2 to cover the edge (r 2,s 2 ). Without loss of generality, suppose that the selected roots are r 1, r 2,anda 1,...,a m.weshow that none of these SP-trees covers edges (u i,w i ), 1 i m. SP-trees rooted at r 1 and r 2 cannot cover edges (u i,w i ) as these edges are horizontal with respect to both r 1 and r 2 (as well as to s 1,t 1,s 2,t 2 ). Similarly, for any i = 1,...,m, edge (u i,w i ) is horizontal with respect to a i (aswellastob i and c i ). Also, any SP-tree T ai rooted at a i cannot cover edges (u l,w l ) for any l i because d(a i,u l )=d(a i,w l )=4,whichmeans(u l,w l ) is horizontal with respect to a i (distances d(a i,u l ) and d(a i,w l ) are realized via node t 2 ). It is easy to see also that edges (u l,w l ) (l =1,...,m) are horizontal with respect to nodes b i,c i,d i,v i and y, too. From the previous discussion we see that the remaining k SP-trees must be rooted at nodes u i, w i,ands j for some

5 JOURNAL OF NETWORKS, VOL. 4, NO. 7, SEPTEMBER Figure 3. The unavoidable edges of SP-trees T ai, T sj, T r2 and T r1. subset q j containing element z i. Also, any SP-tree rooted at u i or w i covers only edge (u i,w i ) for element z i and cannot cover any other edge (u l,w l ) for any l i as they are horizontal to both u i and w i. Let S Q be a collection of subsets defined as follows. For every SP-tree (out of k remaining SP-trees) rooted at node s j,weaddthesetq j Q to S, and for every SP-tree rooted at either node u i or w i,weaddtos an arbitrary set q j Q such that z i q j. Since the set of k remaining SP-trees covers all element edges (u i,w i ), i =1,...,m, the collection S covers all the elements of Z, andisa solution to the SC problem of size at most k. Thus, the theorem is proven. Using the technique developed in [1] to prove that the BR-problem is NP-hard on weighted graphs, one can easily show that the E-Problem is NP-hard even for unweighted graphs. The freedom to choose an arbitrary tree T v from S v for a node v, makes the weight 1+ɛ, used in the reduction of [1], obsolete. The NP-hardness of the E-problem on unweighted graphs immediately follows from the construction given in [1]. It is easy to see also that this result can be directly derived also from the proof of Theorem 1 as shown below. Theorem 2: The E-Problem is NP-hard even for unweighted graphs. Proof: The argument of this proof is very similar to the one used to proof Theorem 1. To prove the hardness of E-Problem on unweighted graphs, we use the same construction G =(V,E) as in the proof of Theorem 1. We show that if there is a set of k + m + 2 SP-trees such that each edge e G is unavoidable with respect to at least one of these k + m + 2 SP-trees, then there is a solution for the SC problem of size k. Note that, among those k + m +2 SP-trees, there needs to be a SP-tree rooted at node a i or b i to cover the edge (a i,b i ),and this holds for any i {1,...,m}. Also, there must exist, among those trees, a SP-tree rooted at r 1 or s 1 to cover the edge (r 1,s 1 ) and a SP-tree rooted at r 2 or s 2 to cover the edge (r 2,s 2 ). Without loss of generality, suppose that the selected roots are r 1, r 2,anda 1,...,a m.thesetof unavoidable edges for each of these nodes is depicted in Fig. 3. Observe that none of these SP-trees covers edges (u i,w i ), 1 i m. Thus, the remaining k SP-trees must be rooted at nodes u i, w i,ands j for some subset q j containing element z i. Also, any SP-tree rooted at u i or w i covers only edge (u i,w i ) for element z i and cannot cover any other edge (u l,w l ) for any l i as they are horizontal to both u i and w i. Let S Q be a collection of subsets defined as follows. For every SP-tree (out of k remaining SP-trees) rooted at node s j,weaddthesetq j Q to S, and for every SP-tree rooted at either node u i or w i,weaddtos an arbitrary set q j Q such that z i q j. Since the set of k remaining SP-trees must cover all element edges (u i,w i ), i =1,...,m, the collection S covers all the elements of Z, and is a solution to the SC problem of size at most k. Thus, the theorem is proven. Moreover, the same proof proves the NP-hardness of the BR-problem even on unweighted graphs as shown below. Theorem 3: The BR-Problem is NP-hard even on unweighted graphs. Proof: Recall that in the BR-Problem, a SP-tree T v is given for each node v V. However, each given SPtree T v must include the edges that are unavoidable with respect to node v. Since the unavoidable edges of SP-trees rooted at nodes r 1, r 2, a i,ands j,(i =1,...,m), (j = 1,...,n) cover all edges of G, using the same argument as in the proof of Theorem 2, we find that there is a set of k + m + 2 SP-trees such that each edge e G is unavoidable with respect to at least one of these k + m + 2 SP-trees, if and only if there is a solution for the SC problem of size k. Thus, the theorem is proven. V. BEST FACTOR APPROXIMATION FOR THE A-Problem Now we provide a heuristic for the A-problem and point out that our heuristic is the best factor approximation algorithm for the problem. For each node v of graph G =(V,E) we construct asetu v of unavoidable edges by v in G. Itiseasyto see that for a given v, thesetu v can be obtained in time O( E ). Consider now an instance of the set cover problem (E,{U v : v V }), wheree is the universe of elements and {U v : v V } is the collection of subsets. We have the following lemma.

6 662 JOURNAL OF NETWORKS, VOL. 4, NO. 7, SEPTEMBER 2009 Lemma 5.1: AsetR V is an optimal solution to the A-Problem on a graph G =(V,E) if and only if {U v : v R} is an optimal solution to the corresponding set cover problem. Proof: Note that, by Observation 2, any SP-tree T v rooted at a node v must contain all edges of U v. Therefore, {Uv : v R} = E if and only if R is a solution to the A-Problem. The well-known greedy algorithm for the set cover problem translates into a greedy algorithm, depicted in Fig. 5, for the A-Problem. According to [5], the greedy algorithm is a (ln( ) + 1)-approximation algorithm for the set cover problem, where is the size of the biggest subset. Since in our case, for any v V, U v cannot contain more edges than a SP-tree rooted at v has, we have the following result. Lemma 5.2: The Greedy algorithm computes a (ln( V )+1)-approximation for the A-Problem. Note that the worst-case time complexity of the Greedy algorithm can be shown to be O( V E ). The reduction from the set cover problem, presented in the proof of Theorem 1, can be extended to derive a lower bound for the best approximation ratio achievable by any polynomial time algorithm (see for details [1], [2] where a similar result was proven for the BR-problem). Lemma 5.3: The lower bound of any polynomial time approximation algorithm for the A-Problem as well as for the E-Problem is ln( V ). VI. BEST FACTOR APPROXIMATION FOR THE A-Problem Now we provide a heuristic for the A-problem and point out that our heuristic is the best factor approximation algorithm for the problem. For each node v of graph G =(V,E) we construct asetu v of unavoidable edges by v in G. It is easy to see that for a given v, thesetu v can be obtained in time O( E ). Consider now an instance of the set cover problem (E,{U v : v V }), wheree is the universe of elements and {U v : v V } is the collection of subsets. We have the following lemma. Lemma 6.1: AsetR V is an optimal solution to the A-Problem on a graph G =(V,E) if and only if {U v : v R} is an optimal solution to the corresponding set cover problem. Proof: Note that, by Observation 2, any SP-tree T v rooted at a node v must contain all edges of U v. Therefore, {Uv : v R} = E if and only if R is a solution to the A-Problem. The well-known greedy algorithm for the set cover problem translates into a greedy algorithm, depicted in Fig. 5, for the A-Problem. According to [5], the greedy algorithm is a (ln( ) + 1)-approximation algorithm for the set cover problem, where is the size of the biggest subset. Since in our case, for any v V, U v cannot contain more edges than a SP-tree rooted at v has, we have the following result. Lemma 6.2: The Greedy algorithm computes a (ln( V )+1)-approximation for the A-Problem. Note that the worst-case time complexity of the Greedy algorithm can be shown to be O( V E ). The reduction from the set cover problem, presented in the proof of Theorem 1, can be extended to derive a lower bound for the best approximation ratio achievable by any polynomial time algorithm (see for details [1], [2] where a similar result was proven for the BR-problem). Lemma 6.3: The lower bound of any polynomial time approximation algorithm for the A-Problem as well as for the E-Problem is ln( V ). VII. HEURISTICS FOR THE E-Problem In this section we provide several algorithms to find a solution to the E-problem. A natural greedy algorithm for the E-Problem would be a procedure where at each step a SP-tree (and hence a root) is chosen which covers the maximum number of not yet covered edges of G. We call this tree a current best SP-tree. To find a current best SP-tree, one should not iterate over all possible SP-trees (the number of which could be exponential). Instead, one can do the following. Iterate over all not considered yet nodes of G, say nodes of S V, building for each node v of S a best possible SP-tree rooted at v, i.e., a SP-tree T v which contains the maximum number of uncovered yet edges of G (a so called current best SP-tree rooted at v). And then, among those trees {T x : x S}, choose a tree T v which covers the maximum number of uncovered edges. To find a current best SP-tree rooted at a node v one can use a function given in Fig. 6. Clearly, this function works in linear time. Now we can give a formal description of the greedy strategy described above for the E-Problem (see Fig. 7). We call it Max New Edges algorithm. It is easy to see that the runtime of this algorithm is O( R V E ). A rather standard technique can be used to show that the Max New Edges is an O(ln V )-approximation algorithm for the E-Problem. Theorem 4: The Max New Edges algorithm computes a O(ln V )-approximation for the E-Problem. Proof: Let c denote the minimum number of SP-trees needed to cover all edges of a graph G =(V,E), and let g denote the number of SP-trees produced by the Max New Edges algorithm. We will show that (g 1)/c ln E 2 ln V. Let m := E. Initially, there are m 0 = m edges left to be covered, and we know that there are c SP-trees that cover all edges of G. Therefore, by the pigeon-hole principle, there must be at least one SP-tree that covers at least m 0 /c edges. Since the Max New Edges algorithm selects a SP-tree that covers maximum number of uncovered yet edges, it will select a SP-tree that covers at least this many edges. The number of edges that remain to be covered is at most m 1 = m 0 m 0 /c = m 0 (1 1/c). Applying the argument again, we know that we can cover these edges with c SP-trees (the optimal cover), and hence there

7 JOURNAL OF NETWORKS, VOL. 4, NO. 7, SEPTEMBER Input: A graph G =(V,E) Output: AsetR V of roots set R := ; for each v V compute set U v of edges unavoidable by v; while E do choose a node v V \ R such that U v E is maximum; (break ties randomly) set R := R {v}, E := E \ U v; return R; Figure 4. A formal description of the Greedy algorithm for the A-Problem (called A-Heuristic). Input: A graph G =(V,E) Output: AsetR V of roots set R := ; for each v V compute set U v of edges unavoidable by v; while E do choose a node v V \ R such that U v E is maximum; (break ties randomly) set R := R {v}, E := E \ U v; return R; Figure 5. A formal description of the Greedy algorithm for the A-Problem (called A-Heuristic). Input: A graph G =(V,E), a node v of G, and a subset E E of uncovered yet edges Output: A current best SP-tree T v rooted at v set U := and q := max{d(u, v) :u V }; compute the layers L i(v) :={u V : d(u, v) =i}, i =1,...,q,ofG with respect to v; for each u V \{v} do let u belong to the layer L i(v); if there exists an edge (u, x) in E such that x L i 1(v) then add such an edge (u, x) to U; else add to U an arbitrary edge (u, x) with x L i 1(v); return tree T v := (V,U); Figure 6. A function current best SP-tree(G, v, E ) which, given a graph G =(V, E), a node v and a set of uncovered yet edges E E, returns a SP-tree T v rooted at v which contains the maximum number of edges from E. Input: A graph G =(V,E) Output: AsetR V of roots and a family T = {T v : v R} of R SP-trees set R :=, T := and E := E; while E do for each v V \ R compute T v :=current best SP-tree(G, v, E ); among the trees {T v : v V \ R} computed, choose a tree T x which contains the maximum number of edges from E ; (break ties randomly) set R := R {x}, E := E \{the edge set of T x} and T := T {T x} ; return R and T ; Figure 7. A formal description of the Max New Edges algorithm for the E-Problem. exists a SP-tree that covers at least m 1 /c uncovered edges, edges uncovered. If we apply this argument g 1 times, leaving at most m 2 = m 1 m 1 /c = m 0 (1 1/c) 2 each time we succeed in covering at least a fraction of

8 664 JOURNAL OF NETWORKS, VOL. 4, NO. 7, SEPTEMBER 2009 (1 1/c) of the remaining edges. Then the number of edges, that remain uncovered after g 1 SP-trees have been chosen by the Max New Edges heuristic, is at most m g 1 = m 0 (1 1/c) g 1. We are interested in the largest value of g such that 1 m(1 1/c) g 1. We can rewrite this as 1 m((1 1/c) c ) (g 1)/c, and, using the fact that for all c>0, (1 1/c) c 1/e (where e is the base of the natural logarithm), obtain 1 m(1/e) (g 1)/c.Thatis,e (g 1)/c m and therefore, (g 1)/c ln m 2 ln V. Our next algorithm makes use of the notion of unavoidable edges. In this method, the number of unavoidable edges is calculated with respect to each node in the graph. Then, a node v with the maximum number of uncovered unavoidable edges is selected. If there are more than one such nodes, we break ties by selecting a node arbitrarily. Finally, using a functiongivenin Fig. 6, a current best SPtree rooted at node v is calculated, and the edges of this tree are declared covered. The process is repeated until all edges of the graph are covered. We call this heuristic Max Unavoidables. Its formal description is given in Fig. 8. Clearly, it runs also in time O( R V E ). We consider also the following two naive but natural heuristics. Each of these heuristics selects a new root v using different strategy but both of them construct a current best SP-tree rooted at node v using function current best SP-tree. Heuristic Max Degree chooses v to be a node from V \ R with the maximum number of uncovered incident edges, while heuristic Random Root chooses v randomly from V \ R. In the remaining part of this section we demonstrate that the difference between the optimal solution and the one returned by Max Degree heuristic (and therefore, by Random Root heuristic) can be significantly different for agivengraphg. As an example, we construct a (5n+4)- node graph as follows. We consider two sets of n nodes labelled a 1, a 2,..., a n, e 1, e 2,..., e n,twosetsofn+1 nodes labelled b 1, b 2,..., b n, b n+1, d 1, d 2,..., d n+1,anda set of n+2 nodes labelled c 0, c 1, c 2,..., c n+1. We connect these nodes as follows. Each node a i (i<n) is connected to nodes c i and a i+1. Similarly, each node e i (i<n)is connected to nodes c i and e i+1. Each node b i (i n) is connected to nodes b i+1, c i 1,andc i. Similarly, each node d i (i n) is connected to nodes d i+1, c i 1,andc i. An example of such a graph with 5*5+4 nodes is depicted in Fig. 9. Heuristic Max Degree will return O( V ) SPtrees rooted at black nodes c 1, c 2,..., c n, while it is easy to see that there are three SP-trees rooted at grey nodes b 1, d 1 and e 1 which cover all edges of the graph. As our experimental results show, even these naive Max Degree and Random Root heuristics outperform the one proposed in [12], [13] in minimizing network monitoring overhead in Power-Low graphs. VIII. COMPARISON In this section, we compare the performance of our algorithms with other methods in the literature that ad- Figure 9. A graph G for which there are three SP-trees covering all edges of G, but heuristic Max Degree will return O( V ) SP-trees. dress similar problem. Namely, we compare the method suggested in [12] and [13], which is called the Beacon Minimization Problem (BMP) method, with the method we proposed for the A-Problem. Then, we compare the Max New Edges, Max Unavoidables with the method proposed for the BMP problem. We first present theoretical worst-case comparison between the method proposed for the BMP and our methods and then we discuss experimental results showing average cases. In the following subsections, we refer to the method proposed in [12] and [13] as BMP method. A. Theoretical Comparison The authors in [12] compared their approach with the one proposed in [10] and showed that their algorithm reduced the number of beacons required by [10] by more than 50%. We show now that comparing to BMP our methods may decrease the number of monitors by factor of n,wheren is the number of network nodes. Indeed, consider again the network depicted in Fig. 1. The BMP algorithm finds a uncovered edge in the network and selects a node that is incident to that edge. Then, it adds the set of edges that are incident to the newly selected beacon to the set of covered (i.e. monitored) edges. Note that since there is more than one path between each two nodes, every beacon selected by the BMP algorithm will cover only the links that are incident to the selected beacon. Assume that the BMP algorithm finds a uncovered link (i, i+1) or (i, i + n), i= 1,..., n-1, and then selects node i as beacon. In this case, n-1 nodes are required by the BMP algorithm to monitor all links. However, as we showed previously, applying the A-Heuristic, only n beacons are required to monitor all links. Furthermore, all Max Unavoidables, Max New Edges and Max Degree will monitor all links by only two beacons. B. Experimental Environment Now we discuss our experiment environment and results. To compare the performance of the method proposed in [12] with the Max Unavoidables, we generated 100-, 500- and 1000-node power-law-based networks using BRITE [6]. Each node had degree at least three and

10 666 JOURNAL OF NETWORKS, VOL. 4, NO. 7, SEPTEMBER 2009 Figure 10. Number of beacons required by the A Heuristic and BMP methods to monitor all network links when the network size was: (a) 100 nodes, (b) 500 nodes, and (c) 1000 nodes. Figure 11. Number of beacons required by the Max Unavoidables, Max New Edges, Random Root and BMP methods to monitor all network links when the network size was: (a) 100 nodes, (b) 500 nodes, and (c) 1000 nodes. [8] M. GONEN, Y.SHAVITT, AΘ(logn)-approximation for the set cover problem with set ownership, Information Processing Letters, (2009), [9] A. GUBIN, W.YURCIK, L.BRUMBAUGH, PingTV: A Case Study in Visual Network Monitoring, In Proceedings of the Conference on Visualization, (2001), [10] J. HORTON, A.LOPEZ-ORTIZ, On the Number of Distributed Measurement Points for Network Tomography, In Proceedings of the Conference on Internet Measurement Conference, (2003), [11] S. JAMIN, C. JIN, Y. JIN, D. RAZ, Y. SHAVITT, L. ZHANG, On the Placement of Internet Instrumentation, In Proceedings of IEEE INFOCOM, (2000), [12] R. KUMAR AND J. KAUR, Practical Beacon Placement for Link Monitoring Using Network Tomography, In Proceedings of Internet Measurement Conference, (2004). [13] R. KUMAR AND J. KAUR, Practical Beacon Placement for Link Monitoring Using Network Tomography, In IEEE Journal on Selected Areas in Communication (J-SAC), special issue on Sampling the Internet: Techniques and Applications, (Dec 2006). [14] J. MOULIERAC AND M. MOLNAR, Active monitoring of delays with asymmetric routes, IRISA Institut de recherche en informatique et systemes aleatoires, (July, 2005), 16 pages. [15] K. SUH, Y.NG GUOY, J. KUROSE AND D. TOWSLEY, Locating network monitors: complexity, heuristics, and coverage, UMass Computer Science Techincal Report , (July, 2004). [16] J. WALZ, B. LEVINE, A Hierarchical Multicast Monitoring Scheme, In Proceedings of NGC on Networked Group Communication, (2000),

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