InTech-Scheduling Algorithm and Bandwidth Allocation in Wimax
Bandwidth Allocation in Networks with Multiple Interferences
description
Transcript of Bandwidth Allocation in Networks with Multiple Interferences
Bandwidth Allocation in Networks with Multiple
InterferencesReuven Bar-Yehuda
Gleb PolevoyDror Rawitz
Technion
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Multiple interference
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(1 )ii
1 1 2 1 2 31 (1 ) (1 )(1 ) (1 )(1 )(1 )
1
1 2 3
1 1 2 1 2 3:1 (1 ) (1 ) (1 ) 1 ii
Additive we can approximate to For small interferences
Interval selection with multiple interference
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Base stations B={1,2,…,i,…,n}Interferences i <1Users U={1,2,…,j,…,m}Times {1,2,…,t,…,f}User j has a set of time interval
requests from base station i: Rij={Iij1,…,Iijk,….}
Each request ijk has a profit Pijk >0
Optimization problem: Allocating subsets of time intervals with maximum profit s.t:
• At most one interval per user• All intervals satisfied by a base
station are independent.
Rijj
'' : ( ')
(1 )ijk ijk
iI S t Ii i t alloc i
t
2
1
i
n
Main result: 7-approximation
This is achieved by getting:
k+1- approximation for strong interferences
-approximation for weak interferences
For k=2 it gives: 3+4=7 (will be shown)
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1 ki i
1 ki i
131k
Interval selection with multiple interference
Linearization & Normalization
We can transform:
To:
Where:
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'' : ( ')
1ijk ijk
iI S t I i i t alloc i
w
'' : ( ')
(1 )ijk ijk
iI S t Ii i t alloc i
''
log(1 )log
iiw
Maximize
s.t.
II SP
Common time &One req/user One req/base station
( { }) 1I Sw S I
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Riii
t
2
1
i
n
R11
R22
Rnn
t0
Maximize
s.t.
Bad news: NP-Hard (add width-less expensive box)
Good news: FPTAS (Dynamic programming approach)
Generalization to many base stations: the bipartite is a forest.
II SP
Open knapsack
( { }) 1I Sw S I
1Not feasible
( { }) 1w S I
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8
1
2
1
1
2
j
i
j
m n
uu
Baseu
uu
Baseu
Base u
Use open knapsack constraintsat interval’s right endpoints
maxI
I SP
( { }) 1Iw S Right II S
s.t. S contains at most one interval from a user contains at most one interval per base stationIS RightI S
Same user
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1
2
1
1
2
j
i
j
m n
uu
Baseu
uu
Baseu
Base u
Strong interferences: w > 1/k
Same user
Same time
Let Î be an interval that ends first; 1 if I in conflict with Î
For all intervals I define: p1 (I) = 0 else
For every feasible x: p1 ·x k+1 Every Î-maximal solution is k+1 approximation . For every Î-maximal x: p1 ·x 1
Î
Algorithm MaxIS( R, p )If R = Φ then return Φ ;If I S p(I) 0 then return MaxIS( R - {I}, p);Let Î R that ends first;
p(Î) if I in conflict with ÎI S define: p1 (I) =
0 elseIS = MaxIS( R, p- p1 ) ;If IS is Î-maximal then return IS else return IS {Î};
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Strong interferences: w > 1/kThe k+1 approximation algorithm
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1
2
1
1
2
j
i
j
m n
uu
Baseu
uu
Baseu
Base u
Weak interferences: w ≤ 1/k
Same user
Interference conflict
Let Î be an interval that ends first; 0 if I not in any conflict with Î
For all intervals I define: p1 (I) = 1-1/k else if I same base or same user as Î w(I) else if I in interference conflict with Î
For every feasible x: p1 ·x 3-2/k
Every Î-maximal is For every Î-maximal x: p1 ·x 1-1/k
Î Same base station
13 approximation1k
1/7-approximationR9
R8 w½ > R7 w > ½ w½ >
R6R5 w½ >
R4R3 w > ½ w½ >
R2R1 w > ½ w > ½ w½ >
Algorithm:GRAY = Find 1/3-approximation for gray (w>1/2) intervals;COLORED = Find 1/4-approximation for colored intervalsReturn the one with the larger profitAnalysis:If GRAY* 3/7OPT then GRAY 1/3(3/7OPT)=1/7OPT elseCOLORED* 4/7OPT thus COLORED 1/4(4/7OPT)=1/7OPT
Interval selection with multiple interference
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Base stations B={1,2,…,i,…,n}Interferences i <1Users U={1,2,…,j,…,m}Times {1,2,…,t,…,f}User j has a set of time interval
requests from base station i: Rij={Iij1,…,Iijk,….}
Each request ijk has a profit Pijk >0
Optimization problem: Allocating subsets of time intervals with maximum profit s.t:
• At most one interval per user• All intervals satisfied by a base
station are independent.
Riji
j
11 : ( 1)
(1 )ijk ijk
iI S t Ii i t alloc i
t
Frequency allocation with multiple interference
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Base stations B={1,2,…,i,…,n}Interferences i <1Users U={1,2,…,j,…,m}Frequencies {1,2,…,t,…,f}User j has a set of bandwidth
demands from base station i: Rij={dij1,…,dijk,….}
Each demand dijk has a profit pijk >0
Optimization problem: Allocating demands with maximum profit s.t:
• At most one demand satisfied per user
• All demands satisfied by a base station are independent.
• |alloc(ijk)|= dijk
Riji
j
{1,2,... }: ( )
(1 )ijkt fijk t alloc ijk
t
Main result: 12-approximation
This is achieved by getting:
- approximation for strong interferences
-approximation for weak interferences
For k=2 it gives: 5+7=12
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1 ki i
1 ki i
2 1k
Frequency allocation with multiple interference
251k
Thank you !
The Local-Ratio Technique: Basic definitions
Given a profit [penalty] vector p.
Maximize[Minimize] p·x Subject to: feasibility constraints F(x)
x is r-approximation if F(x) and p·x [] r · p·x*
An algorithm is r-approximation if for any p, Fit returns an r-approximation
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The Local-Ratio Theorem:
x is an r-approximation with respect to p1
x is an r-approximation with respect to p- p1
x is an r-approximation with respect to p
Proof: (For maximization) p1 · x r × p1*
p2 · x r × p2*
p · x r × ( p1*+ p2*) r × ( p1 + p2 )*
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Special case: Optimization is 1-approximation
x is an optimum with respect to p1
x is an optimum with respect to p- p1
x is an optimum with respect to p
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A Local-Ratio Schema for Maximization[Minimization] problems:
Algorithm r-ApproxMax[Min]( Set, p )
If Set = Φ then return Φ ;If I Set p(I) 0 then return r-ApproxMax( Set-{I}, p ) ;[If I Set p(I)=0 then return {I} r-ApproxMin( Set-{I}, p ) ;]
Define “good” p1 ;
REC = r-ApproxMax[Min]( S, p- p1 ) ;If REC is not an r-approximation w.r.t. p1 then “fix it”;
return REC; 20
The Local-Ratio Theorem: Applications
Applications to some optimization algorithms (r = 1) : ( MST )Minimum Spanning Tree (Kruskal)
( SHORTEST-PATH )s-t Shortest Path (Dijkstra) )LONGEST-PATH( s-t DAG Longest Path (Can be done with dynamic programming)
)INTERVAL-IS( Independents-Set in Interval Graphs Usually done with dynamic programming ) )LONG-SEQ( Longest (weighted) monotone subsequence (Can be done with dynamic programming)
( MIN_CUT )Minimum Capacity s,t Cut (e.g. Ford, Dinitz) Applications to some 2-Approximation algorithms: (r = 2)
( VC )Minimum Vertex Cover (Bar-Yehuda and Even) ( FVS )Vertex Feedback Set (Becker and Geiger)
( GSF )Generalized Steiner Forest (Williamson, Goemans, Mihail, and Vazirani) ( Min 2SAT )Minimum Two-Satisfibility (Gusfield and Pitt)
( 2VIP )Two Variable Integer Programming (Bar-Yehuda and Rawitz) ( PVC )Partial Vertex Cover (Bar-Yehuda)
( GVC )Generalized Vertex Cover (Bar-Yehuda and Rawitz) Applications to some other Approximations :
( SC )Minimum Set Cover (Bar-Yehuda and Even) ( PSC )Partial Set Cover (Bar-Yehuda)
( MSP )Maximum Set Packing (Arkin and Hasin)
Applications Resource Allocation and Scheduling :.…
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Single request to Single base stationI19I18I17I16I15I14I12I12I11
Maximize s.t: For each instance I:
For each freq. t:
I
IxIp )(
}1,0{Ix
)()(:
1IetIsIIx
R1j = {I1j}j
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Single base station: How to select P1 to get optimization?
I19I18I17I16I15I14I13I12I11
Î time
Let Î be an interval that ends first; 1 if I in conflict with Î
For all intervals I define: p1 (I) = 0 else
For every feasible x: p1 ·x 1 Every Î-
maximal is optimal. For every Î-maximal x: p1 ·x 1
P1=1
P1=1
P1=1
P1=1
P1=0
P1=0
P1=0
P1=0
P1=0
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Single base station: An Optimization Algorithm
I19I18I17I16I15I14I13I12I11 Î
time
Algorithm MaxIS( S, p )If S = Φ then return Φ ;If I S p(I) 0 then return MaxIS( S - {I}, p);Let Î S that ends first;
p(Î) if I in conflict with ÎI S define: p1 (I) =
0 elseIS = MaxIS( S, p- p1 ) ;If IS is Î-maximal then return IS else return IS {Î};
P1=0
P1=0
P1=0
P1=0
P1=0
P1=P(Î )
P1=P(Î )
P1=P(Î )
P1=P(Î )
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Single base station: Running Example
P)I1( = 5 -5
P)I4( = 9 -5 -4
P)I3( = 5 -5
P)I2( = 3 -5
P)I6( = 6 -4 -2
P)I5( = 3 -4
-5 -4 -2
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Approximation for weak interferencesFA-Weak(R, p)
If R= return (, )Let be minimum in R
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:Frequency allocation with multiple interference
1 2 1
2
ˆ1 ifˆDefine ( , , ) 2 elseif and define
2else
(1 )ˆ ˆˆ̂ ˆ ˆDefine { : ( , , ) 0}
( , ) ( , )ˆˆ̂return Augume
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e
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ak
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i ip i j k d j j p p p
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