Hop-Congestion Trade-Offs For ATM Networks

Evangelos KranakisDanny KrizancAndrzej Pelc

Presented by Eugenia Bouts

2

Abstract & Introduction

The model we use in this paper is based on the Virtual Path Layout model introduced by Gerstel and Zaks [4, 5]. Messages may be transmitted through arbitrarily long virtual paths. Packets are routed along those paths by maintaining a routing field whose subfields determine intermediary destinations of the packet, i.e. endpoints of virtual paths on its way to the final destination

In such a network it is important to construct path layouts that minimize the hop number (i.e. the number of virtual paths used to travel between any two nodes) as a function of edge congestion (i.e. the number of virtual paths passing through a link).

3

Notation and definitions

N - arbitrary connected network VP - virtual path in N is a simple chain in this

network (v1 , … , vk) VC - virtual channel of length k, joining

vertices u and v, is a sequence of k VP's P –layout in the network N, is a collection of

virtual paths in N , such that every pair of vertices u ,v of N is joined by a VC composed of VP's from P.

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Notation and definitions (cont.)

We assume that traversing any VP is made in a single hop. HopsN(P, u, v) is the minimum number of hops using VP's

in P to go from u to v. HopsN(P) = MAX{HopsN(P, u, v) | u,v Є N}

We are interested in the hop number for layouts of bounded congestion.

CN(P)- congestion of given layout P as the maximum number of VP's from P passing through any link of N.

HopsN(c)= MIN{HopsN (P) | P : CN(P)=c} (for any c ≥ 1)

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A General Lower Bound

Theorem 1: For any network N on n vertices, with maximal degree d, and for any c ≥ 1 we have the following lower bound:

HopsN(c) ≥ 1)log(

)log(

dc

n

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A General Lower Bound (cont.)

Proof: Let H= HopsN(c). Choose any vertex of the network as a root, say r. Since the congestion is c and maximum degree of a node is ≤ d, at most dc nodes can be reached from r with one hop.More generally, with at most h hops at most dc + (dc)2 + … + (dc)h vertices can be reached from r.Put h=H and it follows that n ≤ (dc)H+1, which implies that H+1 ≥ □

)log(

)log(

dc

n

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Chain Graphs

The results in this section are stated for the chain Ln with vertices 0,1,...,n.

Lower Bounds

Theorem 2: HopsLn(c) ≥ ½·n1/c

for any c ≥ 1

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Notions

Notions: P - a layout for Ln

I = [a,b] Ln ( any segment of the chain ) for J P such that Define the layout PI induced by the layout P on the

segment I as the set

Observation: if P has congestion c then PI has congestion <c.

IJ IJJ )(

}{\}:)({ IPJJ

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Figure 1- Induced Layout

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Lemma

Lemma: For any layout P in Ln with congestion c,

there is a vertex u such that HopsLn(P,0,u) ≥ ½·n1/c

HopsLn(P,u,n) ≥ ½·n1/c

Proof: We prove the statement by induction on c. For c = 1 the result is easy; take as u the midpoint of Ln. Assume the lemma is true for c-1.Let P be an arbitrary layout of Ln with congestion c.

Let I=[a,b] be the largest virtual path in layout P. We consider two cases. |I| ≥ n(c-1)/c |I| ≤ n(c-1)/c

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Lemma (proof)

Case 1: |I| ≤ n(c-1)/c In this case at least n1/c hops are necessary to reach one end-point of the chain Ln and the inequality of the lemma is clearly satisfied.

Case 2: |I| ≥ n(c-1)/c Take u be a mid-point of I. To prove the first inequality assume not and let C=(p1, … ,pk) be a virtual channel in P, of length k< ½·n1/c, joining 1 with u.Let (h0, … ,hk) be the end-points of consecutive VP’s in C.Let hr be the last vertex such hr ≤ a or hr ≥ b. Without loss of generality we may assume that hr ≤ a.Let qr+1 denote the VP joining a with hr+1. thus C’=(qr+1,pr+2, … ,pk) is a virtual channel in the layout PI induced by P on I. By the inductive hypotesis, the length of C’ is at least½|I|1/(c-1) ≥ ½n1/c. Hence, k ≥ ½n1/c, contradiction. □

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Lower Bound

Theorem 2: HopsLn(c) ≥ ½·n1/c for any c ≥ 1

Proof: By Lemma there is a vertex u such that HopsLn(P,0,u) ≥ ½·n1/c

HopsLn(P,u,n) ≥ ½·n1/c

This completes the proof of the theorem. □

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Asymptotically Optimal Path Layouts

Theorem 3: For any positive integer c, HopsLn(c) ≤c·n1/c

Proof: Let c be a positive integer. For simplicity assume that n1/c is an integer. The construction can be easily modified in the general case. We construct the layout consisting of nested virtual paths of lengths n(c-1)/c, n(c-2)/c,…,n1/c,1. VP’s of each length form a virtual channel joining both ends of Ln (Figure 2).The layout consisting of all those VP’s has congestion c and hop-number at most cn1/c. □

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Figure 2

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Theorem 4

Corollary: For constant c we have

Theorem 4: For any integer k we have HopsLn(kn1/k) ≤ 2k

)( (c)Hops /1Ln

cn

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Theorem 4 - proof

The layout proving this upper bound is given in Figure3.

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Theorem 4 – proof (cont.)

Construct VP’s with left end-point 0, of lengthsn(k-1)/k,2n(k-1)/k,… ,n1/k,n(k-1)/k.

From their right end-points of construct VP’s going right, of lengthsn(k-2)/k,2n(k-2)/k,… ,n1/k,n(k-2)/k.

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Theorem 4 – proof (cont.)

Continue in this way with non-overlapping VP's of sizes n(k-j)/k,2n(k-j)/k,… ,n1/k,n(k-j)/k for j=1,…,k.

The layout consisting of all those VP’s has congestion kn1/k and hop-number at most 2k □

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Asymptotically Tight Bound

As a corollary we obtain an asymptotically tight bound for congestion log 2 n/log log n.

Corollary: HopsLn (log2n/ loglog n) Θ(log n/loglog n).

Proof:The upper bound is obtained from theorem 4 for k = log n/log log n. Then n1/k= log n and kn1/k= log2n/ loglog n. The lower bound follows immediately from theorem 1. □

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Theorem 5

In case of congestion 2 we have a more precise result: we give a layout for the chain which is optimal up to an additive constant.

Theorem 5: For the chain Ln we have

22)2(52 nHopsn Ln

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Theorem 5 - proof

Proof: We first prove the upper bound. Let t=We create 2 virtual channels C1,C2 consisting of non –overlapping paths whose lengths form arithmetic progressions 2,4,6,…,2i,…2t left to right and right to left respectively. Verticles t(t+1) and n-t(t+1) are joined by a third virtual channel C3 consisting of three non-overlapping Vp’s of lengths differing by at most 1.

12

n

C3

2 24 42t 2t

C1C2

… …

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Theorem 5 – proof (cont.)

We construct the layout P for Ln consisting of all VP’s of length 1 (links of Ln) and all VP’s in chains C1,C2 and C3.

Since all VP’s in the above chains are non-overlapping, layout P has congestion 2.

The distance between end points t(t+1) and n-t(t+1) of channel C3 is

Thus each VP in this channel has length at most

42

612

22

2)1(2

nnnnttnD

22

3

nDE

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Theorem 5 – proof (cont.)

Consider any pair of vertices u and v in Ln

Case 1: u is inside a VP ( length 2i ) of channel C1 and v is inside a VP ( length 2j ) of channel C2 we have :

Case 2: u is inside a VP ( length 2i ) of channel C1 and v is inside a VP of channel C3 we have :

All other cases when u and v are in VP’s from different channels are similar.

It remains to consider the situation when u and v are in VP’s from the same channel. If this is channel C1 or C2 HopsLn(P,u,v) is less than in case 1.

1232)(3)(),,( ntjjtitivuPHopsnL

122

2)(),,(

n

EitivuPHops

nL

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Theorem 5 – proof (cont.)

Case 3: u and v are inside VP’s of channel C3. The worst case occures when u and v in different

external VP’s of C3.We have

It follows that for any u,v

and consequently

This implies the upper bound.

22),,( nvuPHopsnL

22)( nPHopsnL

2222

12

),,(

nE

EEvuPHops

nL

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Theorem 5 – proof (cont.)

In order to prove the lower bound, consider any layout P in Ln, with congestion at most 2.Let Q be the set of VP’s in P of length at least 2 and let q be the size of Q.

The overlap of any VP’s from Q can have length at most 1 ,hence a left-right ordering of these VP’s is well defined.

Definitions: A1,A2,…Ak first k VP’s ordered from left to right

Bk,Bk-1,…B1 last k VP’s in this ordering

If Q is odd let C0 be the remaining VP in Q

Let ai,bi,c0 be the length of Ai,Bi,C0 respectively If Q is even c0=0

2qk

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Theorem 5 – proof (cont.)

We shall prove that HopsLn(P)>X= Suppose not. If u is the mid-point of C0 and v is the end–point of this

VP we get

If u and v are mid-points of Ak and Bk we get

If u and v are mid-points of Ak-1 and Bk-1 we get

If u and v are mid-points of Ak-i and Bk-i we get

52 n

XvuPHopscnL ),,(2

0

XvuPHopsba

nLkk

),,(1

2

XvuPHopsba

nLkk

),,(222

11

XvuPHopsiba

nLikik

),,(222

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Theorem 5 – proof (cont.)

If u and v are mid-points of A1 and B1 we get

Adding all the above inequalities we get

Since all Vp’s in Q cover Ln, this implies

and hence

XvuPHopskba

nL

),,()1(222

11

XkkXkk

cbbbaaa kk 2422

)1(4)...()...( 02121

02121 )...()...( cbbbaaan kk

kXkkXn 6222 2

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Theorem 5 – proof (cont.)

Let Substituting in the latter inequality we get

This is a contradiction because is negative for any t.

It follows that HopsLn(P)>X □

tnk 2/52 nX

)1042( 2 ttnn

)1042( 2 tt

22)2(52 nHopsn Ln

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References

[1] B. Awerbuch, A. BarNoy, N. Linial and D. Peleg, “Improved Routing with Succinct Tables”

[2] J. Y. Le Boudec, “The Asynchronous Transfer Mode: A Tutorial”

[3] I. Cidon, O. Gerstel and S. Zaks, ”A Scalable Approach to Routing in ATM Networks”

[4] O. Gerstel and S. Zaks,”The Virtual Path Layout Problem in Fast Networks”

[5] O. Gerstel and S. Zaks, “The Virtual Path Layout Problem in ATM Ring and Mesh Networks”

[6] D. E. McDysan and D. L. Spohn,”ATM: Theory and Applications”

[7] M. de Prycker, “Asynchronous Transfer Mode: Solutions for Broadband ISDN

[8] N. Santoro and R. Khatib,”Labeling and Implicit Routing in Networks”