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Bandwidth Allocation in Networks with Multiple Interferences

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### Bandwidth Allocation in Networks with Multiple Interferences

Reuven Bar-Yehuda

GlebPolevoy

DrorRawitz

Technion

The Model

Base Stations {1,2,…,i,…,n}

- Interferences (i) <1

Users {1,2,…,j,…,m}

Frequencies {1,2,…,t,…,f}

User j has a set of bandwidth requests from base station i: Rij={Iij1,…,Iijk,….}

- Each request ijk has a profit Pijk>0

Optimization problem: Allocating subsets of demands with maximum profit s.t:

- At most one demand per user
- All demands satisfied by a base station are independent.
- If t allocated by base i to user j then

i

Rij

j

The Local-Ratio Technique: Basic definitions

Given a profit [penalty] vector p.

Maximize[Minimize]p·x

Subject to: feasibility constraints F(x)

x isr-approximationif F(x) andp·x[]r·p·x*

An algorithm is r-approximationif for any p, F

it returns an r-approximation

The Local-Ratio Theorem:

xis an r-approximation with respect to p1

xis an r-approximation with respect to p- p1

xis 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 )*

Special case: Optimization is 1-approximation

xis an optimum with respect to p1

xis an optimum with respect to p- p1

xis an optimum with respect to p

A Local-Ratio Schema for Maximization[Minimization] problems:

Algorithm r-ApproxMax[Min]( Set, p )

If Set = Φ then returnΦ ;

If I Setp(I) 0 then returnr-ApproxMax( Set-{I}, p ) ;

[If I Setp(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;

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: ….

Fatal interference, one request per user

I99

I88

I77

I66

I55

I44

I33

I22

I11

Maximize

s.t:

For each instance I:

For each time t:

Rij = {Iij}

i

j

Fatal interference, one request per user : How to select P1 to get optimization?

P1=0

P1=1

P1=0

Activity9

Activity8

Activity7

Activity6

Activity5

Activity4

Activity3

Activity2

Activity1

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=0

P1=0

P1=1

P1=0

P1=1

P1=1

Fatal interference, one request per user: An Optimization Algorithm

P1=P(Î )

P1=0

Activity9

Activity8

Activity7

Activity6

Activity5

Activity4

Activity3

Activity2

Activity1 Î

time

Algorithm MaxIS( S, p )

If S = Φ then returnΦ ;

If ISp(I) 0 then returnMaxIS( S - {I}, p);

Let ÎS that ends first;

IS define: p1(I) = p(Î) (I in conflict with Î) ;

IS = MaxIS( S, p- p1) ;

If IS isÎ-maximal then returnIS else return IS {Î};

P1=0

P1=0

P1=0

P1=P(Î )

P1=0

P1=P(Î )

P1=P(Î )

Fatal interference, one request per user :Running Example

P(I5) = 3 -4

P(I6) = 6 -4 -2

P(I3) = 5 -5

P(I2) = 3 -5

P(I1) = 5 -5

P(I4) = 9 -5 -4

-4

-5

-2

Single Machine Scheduling :

Bar-Noy, Guha, Naor and Schieber STOC 99: 1/2 LP

Berman, DasGupta, STOC 00: 1/2

This Talk, STOC 00(Independent) 1/2

- Activity9
- Activity8
- Activity7
- Activity6
- Activity5
- Activity4
- Activity3
- Activity2
- Activity1 ?????????????
- time
- Maximize s.t.For each instance I:
- For each time t:
- For each activity A:

Single Machine Scheduling: How to select P1 to get ½-approximation ?

P1=1

P1=0

P1=0

P1=1

P1=0

P1=0

Activity9

Activity8

Activity7

Activity6

Activity5

Activity4

Activity3

Activity2

Activity 1

Î time

Let Î be an interval that ends first; 1if I in conflict with Î

1 if I in conflict with Î

For all intervals I define: p1(I) =

0 else

For every feasible x: p1 ·x 2

Every Î-maximal is 1/2-approximation

For every Î-maximal x: p1 ·x 1

P1=0

P1=0

P1=0

P1=0

P1=1

P1=0

P1=1

P1=0

P1=0

P1=1

P1=1

P1=1

P1=1

P1=1

The ½-approximation Algorithm

Activity9

Activity8

Activity7

Activity6

Activity5

Activity4

Activity3

Activity2

Activity1 Î

time

Algorithm MaxIS( S, p )

If S = Φ then returnΦ ;

If ISp(I) 0 then returnMaxIS( S - {I}, p);

Let ÎS that ends first;

IS define: p1(I) = p(Î) (I in conflict with Î) ;

IS = MaxIS( S, p- p1) ;

If IS isÎ-maximal then returnIS else return IS {Î};

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