Presentation is loading. Please wait.

Presentation is loading. Please wait.

1 Internet Networking Spring 2004 Tutorial 6 Network Cost of Minimum Spanning Tree.

Similar presentations


Presentation on theme: "1 Internet Networking Spring 2004 Tutorial 6 Network Cost of Minimum Spanning Tree."— Presentation transcript:

1 1 Internet Networking Spring 2004 Tutorial 6 Network Cost of Minimum Spanning Tree

2 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 at most 2 times worse than the optimal Steiner tree

3 3 Motivation (cont.) 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)? 12 4 3 5 S C C C C C i1i2 C+ε 1 C+ε 2 00 0 0 0 0 0 00 0

4 4 Definitions The Steiner tree problem: Let G(V,E) be a weighted graph, D  V is the destination group, and s  V\D is the source node. R is a multicast tree if it consists of s and all vertexes of D. A Steiner Tree: a tree with minimal sum of edge weights among all possible R trees. Namely, P* is a Steiner tree if: C(P*) = min R {C(R)} Finding a Steiner Tree in a general graph is an NP- hard problem

5 5 A 2-approx algorithm We build graph I(G), whose nodes are: {s} ∪ D This graph will be a clique: each node has edge connecting it with each other node The weight of an edge e i,j is the cost of the shortest path in the graph G between nodes i and j We find the Minimum Spanning Tree in I(G) and route according to it

6 6 A 2-approx algorithm (cont.) a b c S 1 2 1 1 1 2 4 3 4 2 2 3 5 3 3 G(V,E) S c b a 4 3 4 1 6 5 I(G) S c b a 3 4 1 T

7 7 A 2-approx algorithm (remark) It’s possible that different paths in I(G) will 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 8 A 2-approx algorithm – proving upper bound by 2 P 1 let be a circle that passes through nodes s->a->b->c->d->s. It uses each edge of P* twice. Therefore: C(P 1 ) = 2C(P*) P 2 let be a simple circle in I(G) that passes through all nodes in the same order as P 1 does. C(P1)≥C(P2) because weights of edges in I(G) represents shortest path weights in G S a d c b Consider the Steiner tree for a given s, D instance

9 9 A 2-approx algorithm – proving upper bound by 2 (cont.) Let us drop from P 2 the maximal weight edge. We got a tree, called P 3. It is clear that C(P 2 )≥C(P 3 ), since we dropped an edge. Since we dropped the edge with the maximal weight, its weight is higher than the average: C(P 2 )/(m+1), (where |D|=m). Therefore: C(P 2 ) – C(P 2 )/m+1 = m(m+1) -1 C(P 2 ) ≥ C(P 3 ) Namely: C(P 3 ) ≤ m(m+1) -1 C(P 2 ) ≤ m(m+1) -1 C(P 1 ) = 2C(P*)m(m+1) -1 In the heuristic we searched for the minimum spanning tree in I(G), we call it T. It ’ s clear that C(P 3 ) ≥ C(T), because T is minimum spanning tree in I(G) and P 3 is just a one of the trees in I(G). Therefore: C(T) ≤ 2C(P*)m/m+1

10 10 A 2-approx algorithm – proving upper bound by 2 (cont.) Therefore we got: C(T) ≤ C(P 3 ) ≤ m(m+1) -1 C(P 2 ) ≤ m(m+1) -1 C(P 1 ) = 2m(m+1) -1 C(P*) and: C(T)/C(P*) ≤ 2m/(m+1) < 2

11 11 Worst case of the heuristic Steiner tree is a star around I and connected to S, with the cost: 5+5ε. A tree selected by 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. 12 43 S 1+ε 2 I 2 2 2 2 2

12 12 Worst case of the heuristic Steiner tree is a star around I and connected to S, with the cost: 5+5ε. A tree selected by 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. 12 43 S 1+ε 2 I 2 2 2 2 2


Download ppt "1 Internet Networking Spring 2004 Tutorial 6 Network Cost of Minimum Spanning Tree."

Similar presentations


Ads by Google