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Proof for Absence of a Dispute Cycle when system follows Guideline A and export Policies define by Gao and Rexford Kumaran Vijayasankar Erik Jonsson School of Engineering and Computer Science The University of Texas at Dallas E-mails: [email protected]

Abstract— For any sysytem with customer, service provider and peer relations with Acyclic Customer-Service Provider graph, the Guideline and export Policies suggested by Gao and Rexford in [1], assures a stable path when using BGP. In [2] it is proved that any system with stable path will have an Acyclic Dispute Path, that is there will not be any Dispute Cycle. In this work, it is proved that the method suggested by Gao and Rexford does indeed provide a system with no dispute cycle. .

3210(Pc) 3 210 (Pe) 2

I. I NTRODUCTION The Guideline suggested is that a Customer provided path will be given more preference than a pathe provided by service provider or a peer. The Export policy restricts from exporting a path that is obtained from a service provider or a peer to other peer or service provider. A dispute cycle is defined as a path with policy disputes occurence of which indicates the presence of a stable path problem [2].

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II. T YPES OF PATH Customer Path (Pc) : If next hop along the path is a customer. Service Provide Path(Ps) : If next Hop along the path is a service provider. Peer Path (Pe) : If next hop along the path is a peer.

III. A RCS THAT CANNOT APPEAR A. No transmission or Dispute Arc from a Pe to Pc In Fig. 1, since 2 does not export the path 210 to 3 as it got the path from a peer and 3 is a service provider. In Generel a Peer path is not exported a service provide so : Transmission arc from Pe to Pc cannot exist. Since the peer path is of a lower local preference than a Customer path, Dispute arc from Pe to Pc cannot exist. B. No transmission or Dispute Arc from a Ps to Pc In Fig. 2, since 2 does not export the path 210 to 3 as 3 is a service provider. In Generel a service provider path is not exported a service provide so : Transmission arc from Ps to Pc cannot exist. Since the service provider path is of a lower local preference than a Customer path, Dispute arc from Ps to Pc cannot exist.

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0 Fig. 1. Absence of Pe to Pc Arcs

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Fig. 2. Absence of Ps to Pc Arcs

C. No transmission or Dispute Arc from a Ps to Pe In Fig. 3, since 4 does not export the path 420 to 5 as 420 is a service provider path and 5 is a peer. In Generel a service provider path is not exported a peer so : Transmission arc from Ps to Pe cannot exist. Since the service provider path is of equal local preference to a peer path, Dispute arc from Ps to Pe cannot exist. D. No transmission or Dispute Arc from a Pe to Pe In Fig. 4, since 2 does not export the path 210 to 3 as 3 is a peer In Generel a peer path is not exported a peer so : Transmission arc from Pe to Pe cannot exist.

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Pe cannot go to Pc or Pe by III-A, III-D respectively. Only possibility is for the path to go to Ps by III-C Ps cannot go to Pe. From IV-A, IV-B, IV-C It can be infered that no matter what path we start we cannot end in a path that is even of the same type as the one we started from. Thus it proves the absence of a dispute cycle. R EFERENCES

Fig. 3. Absence of Ps to Pe Arcs

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C. Assume path begins from Pe

[1] L. Gao and J. Rexford, “Stable internet routing without global coordination,” IEEE/ACM Transactions on Networking, 2001. [2] G. W. Timothy G. Griffin, F. Bruce Shepherd, “The stable paths problem and interdomain routing,” IEEE/ACM Transactions on Networking, 2002.

1 10 (Pc)

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Since both are peer paths, they have same local preference and so : Dispute arc from Pe to Pe cannot exist. IV. A BSENCE OF D ISPUTE C YCLE In order for a dispute cycle to exist, a arc must start from a path and some other path in this route should end in this path. ARGUMENT : If a path gives an arc then no path in htis route will give an arc to even a path of the same type This argument if proved will rule out the existence of a dispute cycle. The proof is developed by considering all possible cases.

A. Assume path begins from Pc 1) Pc gives arc eventually to Pe : Pe cannot give an arc to pc by III-A. If Pe gives an arc to Ps. Ps cannot give an arc to Pc by III-B. Ps cannot return to Pe by III-C. Pe cannot go to itself by III-D. so the path cannot end in Pc. 2) Pc gives arc only to Pc : Pc will not eventually end in another Pc cause if it did it will violate the ”ACYCLIC Customer - Service Provider graph” requirement.

B. Assume path begins from Ps By III-B, III-C Ps cannot give go to Pc or Pe. Ps cannot eventually end in Ps as it would violate the ”ACYCLIC Customer - Service Provider graph” requirement.