Download presentation

Presentation is loading. Please wait.

Published byKole Deakins Modified over 2 years ago

1
~1~ Infocom’04 Mar. 10th. 2004 On Finding Disjoint Paths in Single and Dual Link Cost Networks Chunming Qiao* LANDER, CSE Department SUNY at Buffalo *Collaborators: Dahai Xu, Yang Chen, Yizhi Xiong and Xin He

2
~2~ Infocom’04 Mar. 10th. 2004 Outline The Min-Min Problem Motivation and Definition Existing and Proposed Heuristics Application and Performance Evaluation Summary

3
~3~ Infocom’04 Mar. 10th. 2004 Finding Disjoint Path Pairs Basic and important problem in survivable routing The Min-Min Problem Definition: Finding a link (node) disjoint path pair such that the length of the shorter path is minimized. Applications Encrypted data on the shorter path, and decryption key on the longer path Shared Path Protection (use the shorter path as AP) Counterpart problems Min-Max Min-Sum

4
~4~ Infocom’04 Mar. 10th. 2004 Computational Complexities Min-Sum (P) [Suurballe-74] Min-Max (NP Complete) [Li-90] Min-Min (P or NP Hard?) NP Complete! proved by Xu et. al. in INFOCOM’04] Reduction from a well-known NPC problem 3SAT We also proved that it is NP-hard to obtain a k- approximation to the optimal solution for any k > 1

5
~5~ Infocom’04 Mar. 10th. 2004 Solving The Min-Min Problem Active Path First (APF) Heuristic Finds a shortest path for use as AP, followed by searching a disjoint BP. It may fail to find such a BP even though a disjoint path pair does exist. K Shortest Path (KSP) Heuristic First K shortest paths are found and tested in the increasing order of their costs (path lengths) to see if a disjoint BP exists. Could be time-consuming

6
~6~ Infocom’04 Mar. 10th. 2004 Inefficiency of KSP Any path from s to d consists of two sub-paths in domain E1 and E2 respectively. Links in E1 is much shorter than those in E2. The number of all possible sub-paths in E1 is very large 1st 2nd

7
~7~ Infocom’04 Mar. 10th. 2004 Proposed Approach Find a shortest AP first (as in APF) If the AP doesn’t have a disjoint BP, determine the “conflicting link set” that are causing the problem Try another AP without using these problematic links

8
~8~ Infocom’04 Mar. 10th. 2004 Conflicting Link Set Definition A minimal subset of the links on AP such that no path using ALL these “problematic” links can find a disjoint counterpart, e.g., e1 and e2 in the following example. The Min-Min problem can be solved much faster by avoiding using at least one link in the conflicting link set for the next shortest AP.

9
~9~ Infocom’04 Mar. 10th. 2004 Divide and Conquer Let P(I, O) be the problem of finding a disjoint path pair where AP must use the links in set I (Inclusion) but not the links in O (the Exclusion set). Denote the original Min-Min problem by P( , ) Find a shortest AP; If no disjoint BP, find the Conflicting Link Set Divide P( , ) into sub-problems based on the conflicting link set, e.g., P( , {e1}) and P({e1}, ) in the previous example. The same procedure may be applied recursively on these sub-problems, e.g., P({e1}, ) can be further divided into P({e1},{e2}) and P({e1,e2}, ). The definition of conflicting link set means that we do not need to try to solve P({e1,e2}, ).

10
~10~ Infocom’04 Mar. 10th. 2004 The Proposed Conflicting Link Exclusion (COLE) Heuristic An algorithm to find the conflicting link set (to be discussed) Usually has fewer links than the half of the links on AP Fewer sub-problems than KSP “Divide and Conquer” based on the conflicting link set (rather than all the links on AP as in KSP) Then pick a best solution (with a shorter AP) among those for the sub-problems. Find a optimal or near-optimal result for each sub- problem Each sub-problem may be solved recursively using Divide-and-Conquer

11
~11~ Infocom’04 Mar. 10th. 2004 Solving the Sub-problem Finding a shortest path consisting of certain links (e.g. set I) is itself NP-Hard Approximation method to speed up the computation. [Xu et al. OFC’04]

12
~12~ Infocom’04 Mar. 10th. 2004 Finding Conflicting Link Set Finding a link-disjoint path pair between nodes s and d in graph G=(V, E) = Finding two unit-flows in a flow network where each link's capacity is set to 1 unit Assume that the network is symmetrical For the chosen AP, construct a new graph G 0 G 0 uses the same V and E of G The capacity of the links in AP is set to 1 The capacity of the reverse links in AP is set to 0. The capacity of all other links with non-zero capacity in G (except those in AP) is set to |AP|+1 (or a larger value).

13
~13~ Infocom’04 Mar. 10th. 2004 Finding Conflicting Link Set (II) Let Φ 0 =(S, D) be a min-cut of G 0, S={s, 3, 7} D={1, 2, 4, 5, 6, d} The set of negative links (from D to S) on AP of Φ 0 is a Conflicting Link Set {e1, e2}

14
~14~ Infocom’04 Mar. 10th. 2004 Reason for Not Using an Ordinary Min-Cut Divide and conquer based on Ordinary Min-Cut might not help reducing the computational complexity. AP 0 is the shortest path (and no link-disjoint BP exists) AP opt is the shortest path with a link-disjoint BP The min-cut: The partition S = {s}, positive links are a and b Divide the original Min-Min problem into P( , {a}) and P({a}, {b}) (no solution in P({a, b}, ) ) Solving P( , {a}) leads to a non-optimal solution, and trying to solve P({a}, {b}) will again yield AP 0

15
~15~ Infocom’04 Mar. 10th. 2004 Shared Path Protection Two BPs can share backup bandwidth on a common link as long as their APs are disjoint (with a single failure)

16
~16~ Infocom’04 Mar. 10th. 2004 Performance Evaluation Solution to Min-Min problem (Single Link Cost Networks) COLE will stop iteration after finding optimal result. KSP can find the optimal result with a large enough K but has a longer running time than COLE In both algorithms, the time for each invocation of the Dijkstra Algorithm to find the (next) shortest path dominates Application to shared path protection (Dual Link Cost Networks) COLE is compared with the optimal shared Min-Sum and optimal shared Min-Max solutions (based on ILP) Tradeoffs between bandwidth overhead and recovery time.

17
~17~ Infocom’04 Mar. 10th. 2004 Number of Dijkstra Invocations (Min-Min) Net 1 (46 Nodes, 76 edges), Net 2 (79 Nodes, 108 edges), Net 3 (119Nodes, 190 edges) KSP calls the sub-routine significantly more times than COLE, especially for large networks

18
~18~ Infocom’04 Mar. 10th. 2004 Performance in Shared Path Protection Min-Sum Min-Min Min-Max Bandwidth Overhead: Percentage increase in the total bandwidth (active + backup) required over the standard active bandwidth

19
~19~ Infocom’04 Mar. 10th. 2004 Summary The Min-Min problem is formulated and applied to shared path protection The concept of Conflict Link Set is defined, which helps to solve the Min-Min problem fast A novel heuristic algorithm COLE capable of solving the Min-Min problem faster than KSP is proposed COLE is also found to be competitive in providing shared path protection.

Similar presentations

OK

The Byzantine Generals Strike Again Danny Dolev. Introduction We’ll build on the LSP presentation. Prove a necessary and sufficient condition on the network.

The Byzantine Generals Strike Again Danny Dolev. Introduction We’ll build on the LSP presentation. Prove a necessary and sufficient condition on the network.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on instant messaging and chat Ppt on 21st century skills for 21st Ppt on agriculture in india pdf Ppt on trade fairs Ppt on coalition government in uk Ppt on international financial markets Ppt on x ray diffraction Ppt on shell scripting for dummies Maldi ms ppt online Ppt on cse related topics in economics