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Bandwidth Allocation in Networks with Multiple Interferences. Reuven Bar- Yehuda Gleb Polevoy Dror Rawitz Technion. Multiple interference. w e can approximate to For small interferences. Interval selection with multiple interference. Base stations B={1,2,…, i ,…,n}

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Bandwidth allocation in networks with multiple interferences

Bandwidth Allocation in Networks with Multiple Interferences

Reuven Bar-Yehuda

GlebPolevoy

DrorRawitz

Technion


Multiple interference
Multiple interference

we can approximate to For small interferences


Interval selection with multiple interference
Interval selection with multiple interference

Base stations B={1,2,…,i,…,n}

  • Interferences i <1

    Users 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.

Rij

j

t


Main result 7 approximation

Interval selection with multiple interference

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)


Linearization normalization
Linearization & Normalization

We can transform:

To:

Where:


Common time

  • & One req/userOne req/base station

R11

Maximize

s.t.

R22

Rii

i

Rnn

t

t0


Open knapsack

Not feasible

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.

1


Use open knapsack constraints

at interval’s right endpoints

  • s.t. S contains at most one interval from a user

contains at most one interval per base station


Same user

Same user

Î

Same user

Same time

LetÎbe an interval that ends first;

1 if I in conflict withÎ

For all intervals I define: p1(I) =

0 else

For every feasible x: p1 ·x  k+1

 Every Î-maximalsolution is k+1 approximation .

For everyÎ-maximal x: p1·x  1


Algorithm MaxIS( R, p )

If R = Φ then returnΦ ;

If ISp(I) 0 then returnMaxIS( R- {I}, p);

Let ÎRthat ends first;

p(Î) if I in conflict with Î

IS define: p1(I) =

0 else

IS= MaxIS( R, p- p1) ;

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


Same base station

Î

Same user

  • Interference conflict

LetÎbe an interval that ends first;

0 ifInot in any conflict with Î

For all intervals I define: p1(I) = 1-1/kelse 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-maximalis

For everyÎ-maximalx: p1·x  1-1/k


1 7 approximation
1/7-approximation

R9

R8 w > ½

R7 w > ½ w > ½

R6

R5 w > ½

R4

R3 w > ½ w > ½

R2

R1 w > ½w > ½ w > ½

Algorithm:

GRAY = Find 1/3-approximation for gray (w>1/2) intervals;

COLORED = Find 1/4-approximation for colored intervals

Return the one with the larger profit

Analysis:

If GRAY* 3/7OPTthen GRAY 1/3(3/7OPT)=1/7OPTelse

COLORED* 4/7OPTthus COLORED 1/4(4/7OPT)=1/7OPT


Interval selection with multiple interference1
Interval selection with multiple interference

Base stations B={1,2,…,i,…,n}

  • Interferences i <1

    Users 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.

i

Rij

j

t


Frequency allocation with multiple interference
Frequency allocation with multiple interference

Base stations B={1,2,…,i,…,n}

  • Interferences i<1

    Users U={1,2,…,j,…,m}

    Frequencies {1,2,…,t,…,f}

    User j has a set of bandwidth demands from base stationi: 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

i

Rij

j

t


Main result 12 approximation

Frequency allocation with multiple interference

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


Thankyou!


The local ratio technique basic definitions
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
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
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
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
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: ….


Single request to single base station
Single request to Single base station

I19

I18

I17

I16

I15

I14

I12

I12

I11

Maximize

s.t:

For each instance I:

For each freq. t:

R1j = {I1j}

j


Single base station how to select p 1 to get optimization
Single base station: How to select P1 to get optimization?

P1=0

P1=1

P1=0

I19

I18

I17

I16

I15

I14

I13

I12

I11

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


Single base station an optimization algorithm
Single base station: An Optimization Algorithm

P1=P(Î )

P1=0

P1=0

I19

I18

I17

I16

I15

I14

I13

I12

I11 Î

time

Algorithm MaxIS( S, p )

If S = Φ then returnΦ ;

If ISp(I) 0 then returnMaxIS( S - {I}, p);

Let ÎS that ends first;

p(Î) if I in conflict with Î

IS define: p1(I) =

0 else

IS= MaxIS( S, p- p1) ;

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

P1=0

P1=0

P1=P(Î )

P1=0

P1=P(Î )

P1=P(Î )


Single base station running example
Single base station: 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


Approximation for weak interferences

:Frequency allocation with multiple interference

Approximation for weak interferences

FA-Weak(R, p)

If R= return (, )

Let be minimum in R


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