1 Internet Networking Spring 2002 Tutorial 6 Network Cost of Minimum Spanning Tree

2 Motivation Effective implementation of Multicast at the Internet Understanding cost of performing multicast at the network Finding optimal multicast tree (Steiner tree) is NP-hard computational problem We will see the heuristics for finding a multicast tree that is least than 2 times worse than the optimal Steiner tree

3 Motivation (cont.) S C C C C C i1i2 C+ε 1 C+ε 2 Routing by Shortest Path Tree gives a cost of 5C comparing to C+ε Finding optimal tree is a hard problem: Should we select i1 or i2 (or neither)

4 Definitions Steiner tree problem: Let G(V,E) be a weighted graph, a set D  V and the source node s  V\D. |D|=m, m=1 – unicast, m=|V|-1 – broadcast. R is a tree that it is a sub graph of G consist of (at least) S and all vertexes of D (R is a multicast tree). P* is Steiner Tree if it is a tree with a minimal sum of edge weights between all possible R trees Remark: Finding Steiner Tree in the general graph in NP-hard problem

5 Heuristic We build a graph I(G), that it’s nodes will be: the node S and all nodes of D This will be a clique graph (each node has an edge with each other node) Weight of an edge e i,j will be a cost of the shortest path in the graph G between i and j We will find the Minimum Spanning Tree in the graph I(G) and will route according to it –This algorithm can be calculated in the polynomial time

6 Heuristic (cont.) a b c S G(V,E) S c b a I(G) S c b a T

7 Heuristic (remark) It’s possible that different paths in I(G) could contain identical parts in G(V,E) Simple circle of nodes in I(G) is also the minimum cost circle of the nodes in the same order in the graph G G(V,E) S a I(G) S b a b

8 Heuristic – proving upper bound by 2 S a d c b Diagram of the optimal Steiner tree for a given graph P 1 a circle that passes through the nodes s->a->b->c->d->s. It uses each edge of P* twice. Therefore: C(P 1 ) = 2C(P*) P 2 is a simple circle in graph I(G) that pass through all the nodes in the same order as P 1.

9 Heuristic – proving upper bound by 2 C(P 1 )≥C(P 2 ) because that weights of edges in I(G) represents shortest path weights in G –We saw that P* is optimal Steiner tree –We saw that P 1 passes twice in the path of P* and we got that: C(P 1 ) = 2C(P*) –We saw that a circle P 2 passes through nodes of D and s in the same order as P 1 and we got: C(P 1 )≥C(P 2 )

10 Heuristic – proving upper bound by 2 Let drop from P 2 the maximal weight edge. We got a tree. Let sign it P 3. It is clear that C(P 2 )≥C(P 3 ) (we dropped an edge) and because we dropped the edge with the maximal weight, its weight should be less than the average: (m+1) -1 C(P 2 ) (reminding |D|=m) Therefore: m(m+1) -1 C(P 2 ) ≥ C(P 3 ) In the heuristic we search for the minimum spanning tree in I(G), we call it T. It’s clear that C(P 3 ) ≥ C(T), because a T is minimum spanning tree in I(G) and P 3 is a just a one of the trees in I(G).

11 Heuristic – proving upper bound by 2 Therefore we got: 2m(m+1) -1 C(P*)  m(m+1) -1 C(P 2 )  C(P 3 )  C(T) and: 2 > 2m/(m+1)  C(T)/C(P*)

12 Bad case of the heuristic Steiner tree is a star around I and connected to S, with the cost: 5+5ε. A three that selects the heuristic is a direct connection of the nodes to S with the cost of 8. This holds the upper bound. (It can be seen that it is 2m/(m+1). If we put m=4, then we get 8/5). In the general case, the worst result that the heuristic can give is 2m/m+1 worse than the optimal S 1+ε 2 I