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CS Dept, City Univ. 3
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(
CS Dept, City Univ. 4
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
CS Dept, City Univ. 5
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
CS Dept, City Univ. 6
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
CS Dept, City Univ. 9
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
CS Dept, City Univ. 10
Scheduling Algorithm – Step 1 Topology Construction
Active node
Non-active node
CS Dept, City Univ. 11
Active node
Non-active node
Scheduling Algorithm – Step 1 Topology Construction
CS Dept, City Univ. 12
Scheduling Algorithm – Step 1 Topology Construction
Active node
Non-active node
CS Dept, City Univ. 13
Scheduling Algorithm – Step 1 Topology Construction
Active node
Non-active node
),(
),()(
)(
max
ri
ri
it
r
xxNd
N
N
xxdx
xSINR
max)( Nxit
ix
CS Dept, City Univ. 14
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
CS Dept, City Univ. 15
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
CS Dept, City Univ. 16
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
CS Dept, City Univ. 17
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
CS Dept, City Univ. 18
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
CS Dept, City Univ. 19
Scheduling Complexity Analysis-Goals (1)
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
CS Dept, City Univ. 20
Scheduling Complexity Analysis-Goals (1)
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
CS Dept, City Univ. 22
Scheduling Complexity Analysis-Goals (1)
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
CS Dept, City Univ. 24
Scheduling Complexity Analysis-Goals (1)
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
CS Dept, City Univ. 25
Scheduling Complexity Analysis-Goals (1)
jf
ix
rxif
)( if
jf
ix
rxif
)()( ji ff
CS Dept, City Univ. 26
Scheduling Complexity Analysis-Goals (1)
)()3)(1(2
1),()()3(
2
1xrix fkxydfk
Rk
R0
Xs
Extended ring
Xr
)(4
2xf
CS Dept, City Univ. 27
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
CS Dept, City Univ. 28
Scheduling Complexity Analysis-Goals (1)-Counterexample
sx rxxf
)( xf
sy
ry
yf)( yf
CS Dept, City Univ. 29
Scheduling Complexity Analysis-Goals (2)
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
CS Dept, City Univ. 30
Scheduling Complexity Analysis-Goals (2)
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
CS Dept, City Univ. 31
Scheduling Complexity Analysis-Goals (2)
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
CS Dept, City Univ. 32
Scheduling Complexity Analysis-Goals (2)
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
CS Dept, City Univ. 33
Scheduling Complexity Analysis-Goals (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
CS Dept, City Univ. 34
Scheduling Complexity Analysis-Goals (2)
2C
)(2
1)(
2
1)(
)()3()(
)()2()(
)()(
*
*3
*2
*1
ffDr
fCr
fCr
fCr
uvu
1Csx
rx*f
3C
CS Dept, City Univ. 35
Scheduling Complexity Analysis-Goals (2)
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
CS Dept, City Univ. 36
Scheduling Complexity Analysis-Goals (2)
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
CS Dept, City Univ. 37
Scheduling Complexity Analysis-Goals (2)
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
CS Dept, City Univ. 39
Scheduling Complexity Analysis-Goals (2)
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
CS Dept, City Univ. 41
Scheduling Complexity Analysis-Goals (2)
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
CS Dept, City Univ. 43
Scheduling Complexity Analysis-Goals (2)
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