1 / 26

Bandwidth Allocation in Networks with Multiple Interferences

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}

giza
Download Presentation

Bandwidth Allocation in Networks with Multiple Interferences

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Bandwidth Allocation in Networks with Multiple Interferences Reuven Bar-Yehuda GlebPolevoy DrorRawitz Technion

  2. Multiple interference we can approximate to For small interferences

  3. 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

  4. 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)

  5. Linearization & Normalization We can transform: To: Where:

  6. Common time • & One req/userOne req/base station R11 Maximize s.t. R22 Rii i Rnn t t0

  7. 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

  8. 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

  9. Strong interferences: w >1/k 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

  10. Strong interferences: w >1/k • The k+1 approximation algorithm 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};

  11. 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

  12. 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

  13. 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

  14. 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

  15. 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

  16. Thankyou!

  17. 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

  18. 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 )*

  19. 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

  20. 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;

  21. 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: ….

  22. 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

  23. 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

  24. 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 )

  25. 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

  26. :Frequency allocation with multiple interference Approximation for weak interferences FA-Weak(R, p) If R= return (, ) Let be minimum in R

More Related