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Bandwidth Allocation in Networks with Multiple InterferencesPowerPoint Presentation

Bandwidth Allocation in Networks with Multiple Interferences

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### Bandwidth Allocation in Networks with Multiple Interferences

Reuven Bar-Yehuda

GlebPolevoy

DrorRawitz

Technion

Multiple interference

we can approximate to For small interferences

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

Interval selection with multiple interference

Main result: 7-approximationThis 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)

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

- Strong interferences: w >1/k

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

- Strong interferences: w >1/k
- The k+1 approximation algorithm

Algorithm MaxIS( R, p )

If R = Φ then returnΦ ;

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

Let ÎRthat ends first;

p(Î) if I in conflict with Î

IS define: p1(I) =

0 else

IS= MaxIS( R, p- p1) ;

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

- Weak interferences: w ≤1/k

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

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 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

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

Frequency allocation with multiple interference

Main result: 12-approximationThis 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

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: ….

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 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

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

Let ÎS that ends first;

p(Î) if I in conflict with Î

IS 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

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

:Frequency allocation with multiple interference

Approximation for weak interferencesFA-Weak(R, p)

If R= return (, )

Let be minimum in R

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