cs dept, city univ.1 the complexity of connectivity in wireless networks presented by luo hongbo

CS Dept, City Univ. 1

The Complexity of Connectivity in Wireless

Networks

Presented by LUO Hongbo


Outline

Introduction Scheduling Algorithm Scheduling Complexity Analysis


Introduction- Interference Models

Protocol Model Physical Model (SINR)

A message transmitted from a node xs is successfully received by a node xr if

}\{ ),(

),(

si xxri

i

rs

s

xxdP

N

xxdP

)1,2(


Introduction- Scheduling Complexity Some notations

d(xi,xj) : Euclidean distance between two nodes xi and xj

: The distance between the endpoints for a directed link fij = (xi,xj)

B(xi,r) : The ball of radius r around xi containing all nodes xj for which d(xi,xj) r

: The power level of nodes xi in time-slot t

Scheduling Complexity

Minimal number of time slots to schedule all the links

Power assignment schemes

(linear)

)( ijf

)( it x

dPx


Introduction- Limitations of Uniform power assignment

Theorem 1: Assume that every node vi has the same transmission power. The scheduling complexity is at least

Proof: Assume for contradiction that there are nodes sending successfully in the same time-slot. The SINR at xr is at most

)(2

nn

22

L

2

2

1

2

)),(2()1(

),(L

xxdP

LN

xxdP

rs

rs


Introduction- Limitations of Linear power assignment

Theorem 2: Assume that every node xi that intends to send a message over a link of length transmits with power .

The scheduling complexity is at least

Proof: Let xi be a transmitting node in a time-slot t, and it transmits with the power . Therefore, all nodes xj, j<i face an interference of at least

Now let xs be the left-most node that sends a message in time-slot t, and let xr be its receiver, the SINR at every xr

)(}2/,1min{ nn

sP

2)),(2(

),()(

1

1

ii

iiij

xxd

xxdxI

),( 1 iii xxdP

2

2

2

2

),(

),(

max1

RRNRN

xxd

xxd

ii

rs


Scheduling Algorithm –Main body


Scheduling Algorithm –Subroutine


Scheduling Algorithm - Phases

The algorithm is composed of many phases, and each phase contains the following three steps: Topology construction

A nearest neighborhood forest is formed Classification

All the links in the forest are classified into different length classes Scheduling

The links in all these length classes are scheduled in different time-slots

A directed spanning tree towards a single node is formed Strong connectivity is satisfied in a single additional time-slot

by this node


Scheduling Algorithm – Step 1 Topology Construction

Active node

Non-active node


Active node

Non-active node




Active node

Non-active node

),(

),()(

)(

max

ri

ri

it

r

xxNd

N

N

xxdx

xSINR

max)( Nxit

ix


Classify the links into different length classes

Scheduling Algorithm – Step 2 Classification

02))log((,...,, max110 LLLL

Group the links in L to be scheduledThere are groups, and each group consists of length classes

,...)4(2 nLogL

n

np

4log

12 322212 kk2

,)4( nLogL ,0L

0L

n4log

)4(..., nLognL

1L 2L kL

,...1)4(2 nLogL ,1)4( nLogL ,1L 1)4(..., nLognL

,...1)4(3 nLogL ,1)4(2 nLogL ,1)4log( nL 1..., nL


ObjectiveGiven , we want to get the scheduleE = {E1,E2,…,Et,…}.

Scheduling Algorithm – Step 3 Scheduling

}|{,,...,, ''1'

0'' LffFLLLL uvuvp

Schedule the links in F

t:=1;while( ) do

Et = ScheduleLinks(F,t);F:=F \ Et ; t:=t + 1;

end while

F


Scheduling Algorithm – Step 3 ScheduleLinks (F, t)

)()4()(

)()(

)()4()(

*1

2

*1

*)(

fnCr

fCr

fnx

sk

sxst

1C

1f

sx

sy

2f

3f

*f

4f

5f

2C


Scheduling Complexity Analysis-Goals

Theorem 3: The scheduling complexity of strong connectivity in wireless networks is at most To prove the theorem, there are two goals to achieve:

All scheduled transmissions are received successfully by the intended receivers.

For every network, the algorithm produces a correct schedule S that induces a strongly connected sub-graph. Furthermore, the length of the schedule is

)(log4 n

)(log)( 4 nST


Scheduling Complexity Analysis-Goals (1)

Theorem 4: Consider an arbitrary time-slot t. All scheduled transmissions Et in t are received successfully by the intended receivers.

Proof: Consider a link fx = (xs,xr), from Theorem 5 we know that the total interference faced at xr is at most

Hence, by defining , we get

)(1)(0 )4(4

3ss xx

rrrr nIIII )(1)( )4(: ss xx n

31

4

43

)4(4

3)(

)()4(

)()(1)(

)(

NnN

f

fn

xSINRss

s

xx

x

xx

r



Theorem 5: Consider a scheduled link fx =(xs,xr). The total interference experienced at receiver xr that was caused by simultaneously scheduled links from smaller, the same, or larger length classes respectively is bounded by

Smaller length classes:

Larger length classes :

The same length class:

)(1)( )4(4

ss xx n

)()(: sir xyI

)()(:0sir xyI

)()(: sir xyI



First we prove for any transmitting node yi

with We begin by showing that the interference Ir(yi) at xr caused by yi is at most

Since the two links are scheduled in the same time slot,

Then we can get

So the total interference

)(1)( )4(4

ss xxr nI

)()( si xy

1)()(

)4(),(

)()4(

),()( s

i

x

ri

yy

ri

iir n

xyd

fn

xyd

PyI

1)()4()( sxir nyI

1)()(1

)4()()4()(),(

siis xy

yyri nfnfxyd

)(1)(1)(

)()(:

0 )4(4

)4()( sss

sii

xxx

xyyirr nnnyII


Scheduling Complexity Analysis-Goals (1)-

iy

sx

rx

yf

xf

)()( si xy



Then we prove for any transmitting node yi

with Since every sender has a link to its closest neighbor, for all links

fy with Intended transmitter yi. The interference at xr caused by yi is at most

By summing up all nodes, the total interference

)(1)( )4(4

ss xxr nI

)()( si xy

1)()()(

)4()4()(

)()4(

),()( si

i

xy

y

yy

ri

iir nn

f

fn

xyd

PyI

),()( riy xydf

)(1)(1)(

)()(:

)4(4

)4()( sss

sii

xxx

xyyirr nnnyII


Scheduling Complexity Analysis-Goals (1)-

iy

sx rx

yf

xf

)()( si xy



Finally we prove for any transmitting node yi

with For each link fi, , with transmitting node yi and , it holds that

According to the algorithm, around each transmitting node yi, there can be no

other scheduled sender yj from the same length class within the distance at least

This means that disks Di of radius centered at all transmitting nodes yi

from the same length class do not overlap. The area of each such disk is

)(1)(0 )4(4

ss xxr nI

)()( si xy

)(2

1)()(2 xix fff

xi ff

)(4

2xf

)()( si xy

)(2

2)()2()()( xiji ffff

2))(4

2()( xi fDA



jf

ix

rxif

)( if

jf

ix

rxif

)()( ji ff



)()3)(1(2

1),()()3(

2

1xrix fkxydfk

Rk

R0

Xs

Extended ring

Xr

)(4

2xf


Scheduling Complexity Analysis-Goals (1)The area of the “extended” ring

Each transmitter yi in Rk has distance at least from xr and sends with

power at most , then

Summing up the interferences over all rings, we get

)3(

2)4(24

))()3(21

(

))(2()4(

)(

)()()(

1

2)()(0

k

n

fk

fn

DA

RAyIRI

si

ki

x

x

xy

Ry i

kirkr

22

4

)()2(

2

)()3(

4

)()2(

2

)()3)(1()( xxxx

k

ffkffkRA

)()3(2

1xfk

22 )()2)(2

1( xfk

))(2()4( )(x

y fn i

)(1)(

11

1

2)(00 )4(

4

1

)3(

2)4(24)( ss

sxx

kk

x

krr nk

nRII


Scheduling Complexity Analysis-Goals (1)-Counterexample

sx rxxf

)( xf

sy

ry

yf)( yf



Lemma 1: Consider any time-slot t and Let F be the set of links remain to be scheduled at the beginning of t. It holds that for some constant

Lemma 2: Let Ap denote the set of active nodes at the beginning of phase p. For each p, it holds that

n

FEt

log

||||

2/|||| 1 pp AA

,0



Theorem 6: For every network, The length of the schedule produced by Algorithm 1 is

Proof: Let m denote the total number of links that are to be scheduled during

a subroutine call, i.e., |F|=m n. After the first time-slot, at least

nodes have been scheduled. Then after the kth time-slot, for

The number of links that have not been scheduled is at most

The algorithm’s scheduling complexity is

1)log

1( lnlog

mk

nk ememn

m

/logln nmk

)(log)( 4 nST

)(loglog)4log(logln)( 4 nnnnmST

n

m

log



Proof of Lemma 1Proof: For every selected link f*, we bound the number of dropped links

that are in the same length class as f*,denoted by P0(f*), and the number

of dropped links in higher length class than f*,denoted by P+(f*). P0(f*) P+(f*) )2(log n

2)3(4



Based on this, for every link that is selected in the ScheduleLinks() of theschedule subroutine for scheduling in a time-slot t, the dropped links are atmost

Since it holds that,

the number of communication links |Et| that are scheduled in time-slot t is atleast

)2(log)3(4)()()( 2**0* nfPfPfP

||||)(|| * FEfPE tt

n

F

n

FEt

log

||

1)2(log)3(4

||||

2



We start with P0(f*). For each dropped link fuv with ,it holds

that . Consider a disk Du of radius around

Its transmitter xu for every fuv, disk Du do not overlap. The area of each disk is

According to the algorithm, the transmitting node xu must be located within

distance within of xs. Hence, all disks Du are entirely contained

in a disk D* centered at xs with radius

Thus,

)()()(2 ** fff uv )()( *

uvff

2

2*

2*2*0 )3(4

)(41

)()3()(

f

ffP

2*2 )(4

1)(

4

1)( ffDA uvu

)()( *uvff

)()3()(23)( ** fff uv

)(21

uvf



2C

)(2

1)(

2

1)(

)()3()(

)()2()(

)()(

*

*3

*2

*1

ffDr

fCr

fCr

fCr

uvu

1Csx

rx*f

3C



Lemma 3: Let fxy and fuv be the two links that are considered in the same subroutine call, and let ,Then, it holds that

Proof: According to the algorithm, only links in the length classes

are considered in the same subroutine.

It follows that

)()( ux

xuxu nfff xyn

xyuv )4(

2

1)(2)()( 1)4log(

,...,, )4log()4log( jnjnj LLL

)()4(2

1)( xyuv fnf xu



Now We turn to P+(f*). By the definition of the algorithm, a link fi is droppedif and only if . Then for satisfying ,

for a dropped link fi with , the length of the link must be

at least . Now consider the disks Cj and the ring Rj,

)()4(,( *1

fnxBrsi

si

sisi 1

)()4( *fnrj

j

)()4(2

1)( *fnfi

)()4(),( *fnrxd is

)()4(),()()4( *1

* fnxxdfnj

is

j

1C

sx

jC

3C2C

1jC

jR



Lemma 4: Consider a disk C with radius rc, and disks Di with centers ci and radius ri, ri rc for all i. Let be the maximal number of such disks Di such that both of the following properties hold:

Every Di overlaps with C in at least one point; No disk Di contains a center cj for Then, it holds that

Lemma 5: At most links with receiver in C3 are dropped from Ft. Lemma 6: For any k 3, there can be at most dropped receivers in

rings .

2

ji

)1(,..., kk RR

12



2c

1c

2r

1r

cr



Proof of Lemma 5We want to bound the links with receiver in C3. There are two cases:

1) Consider all links fi for which ,

According to Lemma 3, it holds that

Since fi was dropped, its receiver must be within the disk C of radius

around xs. For and , it holds that

Consider the disk C and disks Di of radius around each sender si.

2) Consider the remaining links fi for which ,

)()()4()4(21 1 Crfnn i 1

)()()4(2

1)()4(),(

)()4(2

1)(

*2*3

*2

iis

i

ffnfnrxd

fnf

)()4(2

1)( *fnfi

)()4()( *1 fnCr

1si

2

)( if2si



ir if

C

sx

rx*f

3C

)(Cr



Proof of Lemma 6On one hand, every dropped link with receiver in rings

must be of length

On the other hand, the distance between a receiver in these rings and xs

It follows by Lemma 3 that there can be at most dropped links with receiver

in rings

)()4()()4(),( *1

1*

1)1(

fnfnrxdk

k

is

)()4(2

1)( *fnf k

i )1(,..., kk RR

)1(,..., kk RR

)()()4(2

1 *i

k ffn



)1( kRkR

sx

rx*f

1)1( kC

kC



Based on the Lemma 5 and Lemma 6, we can bound the total numberof dropped links. Rings (Circles) The number of dropped links

j times

Based on the analysis, we can get

:,...,, 243 RRR

njn jj

h

hjj log2

1

1

:,...., 1 jjRR

2:3C

:,...,, 222212 RRR

)2(log)( * nfP


Thanks!

cs dept, city univ.1 the complexity of connectivity in wireless networks presented by luo hongbo

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