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Directed Virtual Path LayoutsIn ATM networks
Jean-Claude Bermond Nausica Marlin
David PelegStepane Perennes
Presented by Boris Mudrik
2
Contents
1. Introduction
1.1 Definitions
Problem
2. The Model
Examples
3. The Cycle Cn
3.1 General Case
3.2 Case c =1
4. The Path Pn
3
1. Introduction
The transfer of data in ATM is based on packets of fixed length, termed cells.
Each cell is routed independently, based on two routing fields at the cell header, called virtual channel
identifier (VCI) and virtual path identifier (VPI).
This method effectively creates two types of predetermined simple routes in the network, namely, routes which are based on VPIs (called virtual paths or VPs) and routes based on VCIs and VPIs (called virtual channels or VCs).
4
1. Introduction (cont.)
VCs are used for connecting network users.
VPs are used for simplifying network management -
routing of VCs in particular.
Route of a VC may be viewed as a concatenation of
complete VPs.
A major problem in this framework is the one of
defining the set of VPs in such a way that some good
properties are achieved.
5
1. Introduction (cont.)
More formally, given a communication network, the VPs form a virtual directed graph (digraph) on the top of the physical one, with the same set of vertices but with a different set of arcs.
Specifically, a VP from u to v is represented by an arc from u to v in the virtual digraph.
This virtual digraph provides a Directed Virtual Path
Layout (DVPL) for the physical graph.
Each VC can be viewed as a simple dipath in the virtual digraph.
6
1.1 Definitions
A virtual path (VP) is a simple path in the graph (network)
A virtual channel (VC) of length k, connecting vertices u
and v, is a sequence p1,p2,…,pk of VPs such that pi
begins at vertex u, pk ends at vertex v and the
beginning of pi+1 coincides with the end of pi, for i < k.
A virtual path layout in the network is a collection of VPs, such that every pair of vertices is connected by a VC composed of VPs
7
1.1 Definitions
A hop count in the VPL is the maximum number of VPs
among a virtual channels, connecting every pair of
vertices. A capacity of an arc is the maximum number of VPs
can share this arc.
A load of an arc is the number of VPs sharing this arc (assuming that capacity of each VP is 1).
8
Problem
In this article, we consider the following problem:
Given a capacity on each physical arc, minimize the
diameter of an admissible virtual graph (a virtual
digraph that doesn't load an arc more than its
capacity)
9
The physical network is presented by a strongly connected directed graph (digraph) G=(V,E,c).
|V|=n. The vertex set V represents the networks switches and end-users.
The arc set E represents the set of physical directed arcs.
The parameter c is the capacity function, assigning to each arc e its capacity c(e) (the amount of data arc can carry) .
For simplicity in this paper eE, c(e)=c0.
2. The Model
10
The network formed by the VPs is represented by a strongly connected digraph H=(V,E’) and a function P assigning to each arc e’=(x,y)E’ a simple directedpath (dipath) P(e’) connecting x to y in G.
In our terminology, the pair (H,P) is a virtual digraph on G.
An arc of H is a virtual arc.
The dipath P(e’) in G associated with a virtual arc e’ is a virtual dipath (VP).
2. The Model (cont.)
11
The load of an arc e of G is the number of virtual dipaths containing the arc e, that is,
A virtual digraph (H,P) on G satisfying the requirement
is referred to as a c-admissible Directed Virtual Paths Layout of G, shortly denoted c-DVPL of G.
The aim is to design c-DVPL of G with minimum hop-count, i.e, to find a virtual digraph with minimum diameter.
2. The Model (cont.)
eP e Eeel
eceE, le
12
For any digraph F, dF(x,y) denotes the distance from x
to y in F, and DF denotes diameter of F (maximum
dF(x,y) for all x,y).
The virtual diameter, , of the digraph G with
respect to the capacity c, is the minimum of DH over
all the c-DVPL H of G.
2. The Model (cont.)
G,cD~
13
2. The Model (example)
1
2 3
4
5
6
P(5,2)=(5,6),(6,1),(1,2)
l(4,5)=l(2,3)=2l(6,3)=l(5,1)=1
G=(V,E,c)
eE, c(e)=2
V={1,2,3,4,5,6}
E={(1,2), (1,5), (2,3), (3,4), (4,5), (5,6), (5,1), (6,1), (6,3)}
DG=dG(1,6)=5
H=(V,E’)
E’={(1,3), (2,3), (3,4), (3,5), (4,6), (5,1), (5,2), (6,3)}
DH=dH(4,1)=dH(4,2)=4
14
In figure, G consists of the symmetric directed cycle Cn.
The virtual graph H consists of arcs (i,i+1) in the
clockwise direction and arcs (ip, (i-1)p) in the opposite
direction (assuming that p divides n). The load of every
arc of Cn is 1.
2. The Model (example)
15
3. The Cycle Cn
In this section the physical digraph G is Cn, the
symmetric directed cycle of length n. We choose
arbitrarily a direction on Cn. For concreteness, consider
as positive, or forward (resp., negative or backward)
the clockwise (resp., counterclockwise) direction. We
assume that eE, c(e)=c if e is a forward arc and c(e)=c
if e is a backward arc, for some constant nonnegative
integers c, c.
16
3. The Cycle Cn (cont.)
It turns out that our bounds can be expressed as
functions of = c+ c. It is then convenient to define
ubC(n,) (resp., lbC(n,)) as an upper bound (resp., lower
bound) for valid if c satisfies c+ c= .
By the definition, lbC(n,) ubC(n,).
G,cD~
G,cD~
17
3.1 General Case
In this section, we show the following upper and
lower bounds on the virtual diameter for the cycle.
The bounds are both proved by induction from the
next two lemmas.
12
2122
2~
2
111
nn,cCD
nn
18
Lemma 3.1
Proof.
Let H be an optimal c-DVPL of Cn.
Let [x1,y1]+ be the dipath consisting of all the vertices of
Cn between x1 and y1 in the positive direction.
Let d+(x1,y1) denote the number of arcs in [x1,y1]+.
1,,
2maxmin, plb
p
nnlb CNpC
19
Lemma 3.1 (cont.)
We say that [x1,y1]+ is covered by H if (the VP
corresponding to) some virtual arc e’ contains [x1,y1]+.
Abusively we say that [x1,y1]+ is covered by e’.
20
Lemma 3.1 (cont.)
First we prove that if [x1,y1]+ is covered by e’ then
For this, we shorten the cycle by identifying all the
nodes in [y1,x1]+ with x1, obtaining a cycle C’ of length d+
(x1,y1). Virtual arcs are transformed like in figure.
1,, 11 yxdlbD CH
21
Lemma 3.1 (cont.)
A virtual arc from x[x1,y1]+ to y[x1,y1]+ is left
unchanged.
A virtual arc from x[x1,y1]+ to y[y1,x1]+ is transformed
into the arc (x,x1).
Note that the virtual arc containing the positive arcs
of [x1,y1]+ is transformed into a loop.
We also remove loops or multiple virtual dipaths in
order to get a simple DVPL on C’.
22
Lemma 3.1 (cont.)
This transformation does not increase the load of any arc.
The virtual arc e’ that contained [x1,y1]+ disappears,
so the congestion of any positive arc decreases. Our transformation does not increase the virtual
diameter. Consequently, we obtain a c’-DVPL of C’ (a cycle of
length d+(x1,y1)) with c’++c’=1, and diameter at
most DH. It follows that
(1) 1,, 11 yxdlbD CH
23
Lemma 3.1 (cont.)
Now we argue that there exist vertices u and v with large d+(u,v) such that [u,v]+ is covered.
Let P be the shortest dipath in H from 0 to n/2, and assume w.l.o.g. that P contains the arcs of [0,n/2]+.
Let S denote the set of vertices of P between x and y in the positive direction.
Then S DH + 1, and therefore there exist vertices
u and v such that [u,v]+ is covered and with
(2) .
2,
HD
nvud
24
Lemma 3.1 (cont.)
Let
From (2) we have
And from (1) it follows that
covered is vuvudp ,,max
p
nDH 2
1, plbD CH
25
Lemma 3.2
Proof.
Let us construct a c-DVPL on Cn.
W.l.o.g. suppose that c+ c, so c+ 0.
Let pN+, we proceed as follows.
1,12min,
p
nubpnub CNpC
26
Lemma 3.2 (cont.)
Use n virtual arcs (i,i+1)i[0..n-1] of dilation 1 in positive direction.
Let S be the set of vertices ,
and note that vertices of S form a cycle .
Use an optimal c’-DVPL for with c’= c1, and
c’= c, that is c’+ c’= 1.
p
p
npp 1,,2,,0
p
nC
p
nC
27
Lemma 3.2 (cont.)
By construction, the diameter DS of the set S (i.e., the maximal distance of two vertices in S) is at most
For any vertex x, we have d(S,x) p1 and d(x,S) p1. Hence
1,
p
nubC
1,12,
p
nubpDy,SdxSdx,yd CS
28
Proposition 3.3
Proof.
First we consider the lower bound.
We prove by induction on that
12
2122
2~
2
111
nn,cCD
nn
.2
1,
1
nnlbC
29
Proposition 3.3 (cont.)
For the initial case we have lbC(n,1) = n – 1 n/2.
Now to go from – 1 to we use lemma3.1 witch
states that
Hence lbC(n,1) = n – 1 n/2 and the proof is completed.
1
1
2
1,
2maxmin, p
p
nnlb
NpC
.21
21
,2
max1
11
1
1
npnp
pn
for attained
30
Proposition 3.3 (cont.)
Now, we prove the upper bound.
First we show by induction on that for n = 2a, aN
122122
2,1
an
nubC
31
Proposition 3.3 (cont.)
For = 1 , ubC(n,1) n-1 is true.
For the inductive step from – 1 to , we apply
lemma 3.2 with p = a, getting
ubC(n, ) 2(a1) + ubC(2a1, 1)
By induction,
ubC(2a1, 1) 2( 1)a – 2( 1) + 1
So we get the expected result.
32
Proof of Proposition 3.3 (cont.)
For other values of n, the claim is proved as follows.
a is such that n 2a. As ubC is increasing function on n, we obtain
As , this implies
.2
1
n
a Let
122
21221
σ
nσσσan,σub
σ
C
12
1
n
a .12
2,1
nnubC
33
Corollary 3.4
If c+ = c= c then
12
4~
2
2
12
1
c
n
c nc,cCD
n
34
3.2 Case c =1
The upper bound is the one of proposition 3.3. The lower bound proof requires some care so we
first give some definitions.
Let H be an optimal virtual digraph on G with respect
to the capacity 1.
The following definitions are given for the positive
direction, but similar notions apply for the negative
direction as well.
12232
41,~
122
n
nCDOn n
35
Definition 3.5
The forward successor of a vertex x is denoted x+
[x,y]+ denotes the dipath from x to y in Cn in the
positive direction
A path Q=(e’1,…,e’q) from x to y in H is said to be of
type + if [x,y]+W(Q).
Where W(Q) is the route in Cn associated to the
dipath Q in H.
36
Definition 3.6
A circuit-bracelet of size n is a digraph A of order n constructed as follows (see the Figure): The digraph is made of a set of cycles Ci, iI
directed in a clockwise manner. For any i, Ci and C(i+1) mod |I| share a unique vertex
v (i+1) mod |I| .
The length of the dipath in Ci from vi-1 to vi is
denoted pi and is called the positive length of Ci.
The length of the dipath in Ci, from vi to vi-1 is
denoted ni and is called the negative length of Ci.
The successor of vi in Ci by wi, and the ancestor of
vi+1 in Ci by zi.
37
Lemma 3.7
f(n) is the minimal value of DA, where A is any circuit-bracelet of size n.Proof.
If an arc e of Cn is not used by a virtual dipath P(e’)
with e’E’, we add a virtual arc e’ such that P(e’)=(e). This transformation can only decrease the diameter
of H, which is of no consequence since we only seek for a lower bound on the virtual diameter.
Using this manipulation eE, e’E’ s.t. eP(e’). This implies
(3)
Where w(e’) is the dilation of a VP e’, i.e. the length of P(e’).
1,~
nCDnf
.newew type of e type of e
38
Lemma 3.7 (cont.)
Show that if e’=(x,y)E' is an arc of type + of dilation w(e’) 3 then all the arcs of type – between y– and x+ are of dilation 1.
Since eE, c(e)=1, and there is already a virtual arc of type + between x and y, there is no virtual arc of type + ending at any vertex between x+ and y–.
Since H(V,E’) is strongly connected, there is at least one arc ending at each one of this vertices. These arcs are of type –. For the same reasons of capacity and connectivity, these arcs are of dilation 1.
39
A circuit-bracelet
Due to this property it is easy to see that there exists a digraph isomorphism between H and a circuit-bracelet of size n. (see the figure).
C1w1
z1
C7 v7 C6
C3
z0
–
w5
C5 +v6
v5
z5
C2v2 v3
w2 = z2
H =
+
vowo
C0
v1
C4
v4
u1
u2
u3
u4
u5
u1=vi
u2
u3
u4
u5=vi+1
40
Lemma 3.8
and the total number of circuits in an optimal circuit-bracelet is .
Proof.
By the proposition 3.3, there exists a regular circuit-bracelet with diameter at most , so
The size of any circuit in an optimal circuit-bracelet is at most , otherwise the distance from
wi to the second neighbor of vi on the bigger cycle Ci
is more than f(n).
nnf n
122 n
nOnf
nOnf 2
41
Lemma 3.8 (cont.)
Hence there are at least circuits.
Moreover the total number of circuits is less than
, otherwise there exist two vertices at
distance more than f(n).
Thus and the lemma follows.
n
nOnf 2
nnf
42
Definitions
We prove proposition 3.10 for the special case of
regular circuit-bracelet satisfying i, ni = 1.
The circuits of regular circuit-bracelet all consist of a
single arc of type – and pi arcs of type +.
Remark that pi is then the length of Ci.
Let g(n) denote the minimal value of DA where A is
any regular circuit-bracelet of size n.
vi
vi+1
pi
43
Lemma 3.9
Proof. We assume that n is sufficiently large. Let p be an integer and D the diameter of the
considered circuit-bracelet. Call a circuit big if its size is greater than D/p, small
otherwise. Recall that the size of any circuit is less than D + 2. Let b be the number of big circuits Let s be the number of small circuits.
.122 Onng
44
Lemma 3.9 (cont.)
We have
(4)
Suppose that big circuits are ordered cyclically according to the circuit-bracelet structure:
as shown on the figure. Let k{0,1,…,b–1} and consider dipaths from In the positive direction the cost is exactly
.22 DbsDbp
Dsn and
110,,,
biii CCC
2
,
pkkj
ik jpd
.pkk ii zw
to
45
Lemma 3.9 (cont.)
These circuits are big Hence
So we must use the negative direction.
The length is then
where – the number of vertices in the all small circuits.
p
Dp
ji
22
1 DpDD
p
pdk
if
.3 Diisbppd kpkiik pkk
.321
0
bDbsbpsbbndbk
kk
46
Lemma 3.9 (cont.)
so
Note now that , so
(5)
If the coefficient of s in (5) is positive then the left
factor of (5) is greater than which is
greater than
In turn, the coefficient of s is positive if
.3
2D
b
sppsb
b
n
p
Ds
.32
12
Dpb
p
bp
Dsb
b
n
32
pbb
n
.322 pn
.2
pp
Db
47
Lemma 3.9 (cont.)
(4) implies
Using that fact that But
if , and the latter inequality is true if p
33 and n is large enough.
It is follows that
.2
222
p
D
D
nbbD
p
Dn so and
.8
162,22
p
pnbnD obtain we
pp
D
p
pn
2
8
162
.3622 nng
02
48
2
n
pp
48
Proposition 3.10
Proof.
The upper bound is the one given for the general
case. We conjecture that this bound is tight. It would
be desirable to obtain a simpler argument that could
extend to higher capacities.
Recall that Consider a circuit-bracelet, and
recall that ni+pi D+1, so that we can find an integer
k such that
12232
41,~
122
n
nCDOn n
.nD
.12,1,1
npnDpnki iiki ii with
49
Proposition 3.10 (cont.)
Consider the shortest dipath from v1 to vk+1 and suppose
that it uses the positive direction. So
It follows that
So, the dipath from vk to v1 cannot use the negative
direction, and must use the positive one.
It follows that
Globally,
If we remove this vertices we obtain a regular circuit-bracelet with lesser diameter. It follows that
.,1
Dpki i
.,1
Dnki i
.,1
Dpki i
.2 nDpi n
.1221
122
n
nnnngnf
50
4. The Path Pn
In this section the physical digraph G is the n-vertex
symmetric directed path Pn. Our bounds are valid for
any capacity function c such that positive (resp.,
negative) arcs have capacity c+ (resp., c–) and the
additional requirement c+ 1, c– 1.
Let = c++ c–.
51
Proposition 4.1
Proof.
Let us first prove the lower bound.
Let H be a c-DVPL of Pn.
We say that a sub-path [x,y] is covered by H if the
dipaths from x to y and from y to x are both contained
in (the VP corresponding to) some virtual arc.
422
112,
~2
1
11
1
n
cPDn
n
52
Proposition 4.1 (cont.)
First we show that if [x,y] is covered then
DH > lbC ( d(x,y) , - 2 ) .
Indeed if [x,y] is covered we identify x and y and
collapse the path into a cycle of length d(x,y).
We ignore the virtual paths covering [x,y] (see the
proof of lemma 3.1 for details).
We obtain a c’-DVPL for Cd(x,y) with c’++c’–= –2.
53
Proposition 4.1 (cont.)
Now, consider two shortest dipaths in H, one from 0
to n–1 and the second from n–1 to 0.
There are at most 2DH intermediate points (including
0 and n–1) on these two dipaths.
Hence we can find two consecutive intermediate
vertices x and y, with [x,y] covered, such that
If m = max{d(x,y) | [x,y] is covered}, we have
But due to the covering property DH lbC(m, – 2).
.2
,HD
nyxd
.2mn
DH
54
Proposition 4.1 (cont.)
Hence
Using the lower bound on lbC(m,) given in proposition 3.3, and maximizing in m, completes the lower bound proof.
.
2,,
2max, mlb
m
nmlb CP
55
Proposition 4.1 (cont.)
To prove the upper bound, we construct a VPL based
on the best VPL we know on the cycle Cn with c’+ = c+
and c’– = c– – 1.
In this VPL, no VP passes over vertex 0. So we cut
the cycle at vertex 0 and consider it as the path Pn.
On the negative direction, we add a VP of dilation n
from n – 1 to 0 (See Fig. 4). The added VP is used at
most once in a path on H. The bound is the one for
the cycle Cn–1, sum of capacities – 1 plus 1.
56
Figure 4: Pn, c=2