1. Quickest path and Quickest routing: A dynamic routing method Research Topic: Jiang, XidongMS candidate in computer science at California State University, Chico, CA 95929-0410 USA (e-mail: joe588@yahoo.com)

2. Adenda Introduction The quickest path problem overview Related work Quickest path problem for all M value date Local optimal quickest path solution The conflict free optimal quickest path algorithm Further work Conclusion

3. Introduction QP was initialed 1990 It was quickly noticed and have variety topic It is different with shortest path protocol which has been widely used

4. Quickest path review Given a network N, denote N = (V, E, c, l, s, t), where G = (V, A) is simple connected directed graph. V is node set of G and E is edge set of G. c(u, v) ≥ 0 is the capacity of an edge (u, v) E. l (u, v) ≥ 0 is a lead time or time delay of travel time of a packet passing though an edge (u, v) E.

5. Quickest path review (cont.) denoted as p = (s= u1, 2,…, uk = t),p Ƥ = { p | p is the path from s to t}. The lead time of traversal delay of path p is l(p) = The capacity of path p s c(p) = min {c(u,v)|(u,v) E } To any m unit data requires transmission time from s to t is T. T = l(p) + m/c(p)

6. Quickest path review (cont.) Observation 1: to any graph G, p is QP of G to transmit m volume data, the sub path of p could not maintain the property of sub QP of G. Observation 2: to any graph G, the QP is function of f(m, c, l). In a c and l are given graph G, QP is m dependent.

7. Quickest path review (cont.)

8. QP is unit value dependent

9. Quickest path review (cont.) QP sub path is not QP The QP abef sub path ade is not QP for 60 units data. ace is QP for 60 units data

10. Related work Chen and Chin first named QP problem Marta Pascoal etc. survey of PQ is a good reference Herminia I. Calvete’ paper discussed batch constraint QP

11. QP for all M value of data QP is units value data dependent Find all paths for different m value of data For certain value of data could have multiple path be selected

12. Local optimal QP Split entire network by Layers from top to bottom Each layer has limited hops Layers linked each other Apply QP in layer

13. Conflict free optimal QP Algorithm From Chan and Chin we can get paths for different value of data From Herminia I. Calvete, divide whole data set into a few small portions. Utilize the different QP transfer data. However if the k paths have common link the algorithm will cause contention. The goal of QP protocol is to automatically distribute traffic, erase network congestion. If route conflict, congestion will occur. In order to resolve such problem we provide splitting data transfer protocol to eliminate the path collision.

14. Conflict free optimal QP Algorithm Let are the h quickest paths., (i = 1,…h, 1<n = 1,2,,,.). For different unit value of data (h = 1, 2,….), not lost generality let p i is the QP for m i unit value of data. The T i (i = 1,…h) is the corresponding transmission time used. c i (i=1,2…h) is the capacity of path p i. l(p i ) is the lead time of p i.

15. Conflict free optimal QP Algorithm Lemma 1 [1]: If the amount of data to be transmitted is more then mj, then the quickest path must be one of those paths. On the contrary, if the amount of data to be transmitted is less than mj then the quickest path must be one of those paths.

16. The conflict free optimal QP Definition: q is a QP path from s to s. The road capacity links are those link (u,v) E(q), that c(u,v) = c (q). Preposition 1: if q1, q2 are QP for different units value data, then q1, q2 will not have comment links are route capacity link. Proof: if the q1, q2 have share the same links to be capacity link, they have the same c(q), then either q1=q2 or to any units value of data T(q1)>T(q2) or contrary. #

17. Conflict free optimal QP Algorithm Preposition 1: if q1, q2 are QP for different units value data, then q1, q2 will not have comment links are route capacity link. Proof: if the q1, q2 have share the same links to be capacity link, they have the same c(q), then either q1=q2 or to any units value of data T(q1)>T(q2) or contrary. #

18. Conflict free optimal QP Algorithm OQP algorithm: Step 1: For mi value of data, default path is pi. We searching path. If they do not have shared links, split mi value of data to small section which corresponding path pi-1 and pi-2. From lemma we knew for small amount of data p(k<i) paths are the split mi value of data QP. If they have intersection do step 2.

19. Conflict free optimal QP Algorithm OQP algorithm (cont.) Step 2: If find shared links then it is means that they could cause route contention. The two paths have the same source and destination. As step 1 we split data and send in both paths at such situation. From preposition 1 we knew that the shared links is not c(p) links, so those links capacity larger then both c(p i -1) and c(p i -2), so the data pass along those links will use less time then pass both c(p i -1) and c(p i -2) links. Data will be split to pass those non shared links of p i -1 and p i -2.

20. Conflict free optimal QP Algorithm Theory 1: Denote T(m) as the time used for default QP for m unit value of data, then T < T(m). Proof: Omit (It is obvious by combine the information algorithm, lemma and preposition 1).

21. Discussion OQP algorithm is consistence with the choice of different path for different value of data. Network traffic analog is complicate. From above mention there always an extra capacities were waste during data transmission. OQP is one of solutions. QP algorithm combines storage and forwarding routing strategy, it is possible to maximize to utilize the network resources.

22. Conclusion In this paper provides solutions for QP path in real world. One is layer routing protocol. The other is OQP which is based on previous researches. It is a splitting data method of conflict free QP algorithm. It is use a lower unit value of data QP for higher units value data transfer to increased traffic and less time consume.

23. Acknowledgement Dr. Ben Juliano, professor of computer science department, CSU, Chico. Without him I was probably could not finished.