SOCP Relaxation of Sensor Network Localization

Transcription

1 SOCP Relaxation of Sensor Network Localization Paul Tseng Mathematics, University of Washington Seattle University of Vienna/Wien June 19, 2006 Abstract This is a talk given at Univ. Vienna, 2006.

2 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 1 Talk Outline Problem description SDP and SOCP relaxations Properties of SDP and SOCP relaxations Performance of SOCP relaxation and efficient solution methods Conclusions & Future Directions

3 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 2 Sensor Network Localization Basic Problem: n pts in R d (d = 1, 2, 3). Know last n m pts ( anchors ) x m+1,..., x n pairs of neighboring pts and Eucl. dist. estimate for d ij 0 (i, j) A with A {(i, j) : 1 i < j n}. Estimate first m pts ( sensors ). History? Graph realization, position estimation in wireless sensor network, determining protein structures,...

4 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 3 Optimization Problem Formulation υ opt := min x 1,...,x m xi x j 2 d 2 ij Objective function is nonconvex... Problem is NP-hard (reduction from PARTITION)... Use a convex (SDP, SOCP) relaxation.

5 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 4 NP-hardness PARTITION: Given positive integers a 1, a 2,..., a n, a partition I 1, I 2 of {1,..., n} with i I 1 a i = i I 2 a i? (NP-complete) Reduction to our problem (d = 1) (Saxe 79): Let m = n 1, x n = 0, d n1 = a 1, d 12 = a 2,..., d n 1,n = a n [This extends to d 2] PARTITION yes υ opt = 0

6 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 5 SDP Relaxation Let X := [x 1 x m ], A := [x m+1 x n ]. Then (Biswas,Ye 03) x i x j 2 = tr ( b ij b T ij [ X T X X T X I d ]) with b ij := [ ] Im 0 (e 0 A i e j ). Fact: [ Y X T X I d ] 0 has rank d Y = X T X

7 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 6 Thus υ opt = Drop low-rank constraint: min X,Y s.t. Z = ( tr bij b T ijz ) d 2 ij [ Y X T X I d ] 0, rankz = d υ sdp := min X,Y s.t. Z = tr ( b ij b T ijz ) d 2 ij [ Y X T X I d ] 0 Biswas and Ye gave probabilistic interpretation of SDP soln, and proposed a distributed (domain partitioning) method for solving SDP when n > 100.

8 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 7 SOCP Relaxation Second-order cone program (SOCP) is easier to solve than SDP. Q: Is SOCP relaxation a good approximation? Or a mixed SDP-SOCP relaxation? Q: How to efficiently solve SOCP relaxation?

9 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 8 SOCP Relaxation υ opt = min x 1,...,x m,y ij yij d 2 ij s.t. y ij = x i x j 2 (i, j) A Relax = to constraint: υ socp = min x 1,...,x m,y ij yij d 2 ij s.t. y ij x i x j 2 (i, j) A y x 2 y + 1 (y 1, 2x) (also Doherty,Pister,El Ghaoui 03)

10 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 9 Properties of SDP, SOCP Relaxations d = 2, n = 3, m = 1, d 12 = d 13 = 2 Problem: 0 = min x 1 =(α,β) R 2 (1 α) 2 + β ( 1 α) 2 + β 2 4

11 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 10 SDP Relaxation: 0 = min y 2α 3 + y + 2α 3 x 1 =(α,β) R 2 y R s.t. y α β α β 0 1 If solve SDP by IP method, then likely get analy. center.

12 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 11 SOCP Relaxation: 0 = min x 1 =(α,β) R 2 y,z R y 4 + z 4 s.t. y (1 α) 2 + β 2 z ( 1 α) 2 + β 2 If solve SOCP by IP method, then likely get analy. center.

13 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 12 Properties of SDP & SOCP Relaxations Fact 1: υ socp υ sdp. If υ socp = υ sdp, then {SOCP (x 1,..., x m ) solns} {SDP (x 1,..., x m ) solns}. Fact 2: If (x 1,..., x m, y ij ) is the analytic center soln of SOCP, then x i conv {x j } j N (i) i m with N (i) := {j : (i, j) A}.

14 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 13 Opt soln (m = 900, n = 1000, nhbrs if dist<.06) SOCP soln found by IP method (SeDuMi)

15 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 14 Fact 3: If X = [x 1 x m ], Y is a relative-interior SDP soln (e.g., analytic center), then for each i, x i 2 = Y ii = x i appears in every SDP soln. Fact 4: If (x 1,..., x m, y ij ) is a relative-interior SOCP soln (e.g., analytic center), then for each i, x i x j 2 = y ij for some j N (i) x i appears in every SOCP soln.

16 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 15 Error Bounds What if distances have errors? d 2 ij = d 2 ij + δ ij, where δ ij R and d ij := x true i x true j (x true i = x i i > m). Fact 5: If (x 1,..., x m, y ij ) is a relative-interior SOCP soln corresp. (d ij ) and δ ij δ, then for each i, x i x j 2 = y ij for some j N (i) = x i x true i = O( δ ij ). Fact 6: As δ ij 0, (analytic center SOCP soln corresp. (d ij ) ) (analytic center SOCP soln corresp. ( d ij ) ).

17 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 16 Error bounds for the analytic center SOCP soln when distances have small errors.

18 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 17 Solving SOCP Relaxation I: IP Method min x 1,...,x m,y ij yij d 2 ij s.t. y ij x i x j 2 (i, j) A Put into conic form: min u ij + v ij s.t. x i x j w ij = 0 (i, j) A y ij u ij + v ij = d 2 ij (i, j) A α ij = 1 2 (i, j) A u ij 0, v ij 0, (α ij, y ij, w ij ) Rcone d+2 (i, j) A with Rcone d+2 := {(α, y, w) R R R d : w 2 /2 αy}. Solve by an IP method, e.g., SeDuMi 1.05 (Sturm 01).

19 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 18 Solving SOCP Relaxation II: Smoothing + Coordinate Gradient Descent So SOCP relaxation: min y z y d2 = max{0, z d 2 } min x 1,...,x m max{0, x i x j 2 d 2 ij} This is an unconstrained nonsmooth convex program. Smooth approximation: max{0, t} µh(t/µ) (µ > 0) h smooth convex, lim t h(t) = lim t h(t) t = 0. We use h(t) = ((t 2 + 4) 1/2 + t)/2 (CHKS).

20 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 19 SOCP approximation: min f µ (x 1,.., x m ) := ( ) xi x j 2 d 2 ij µh µ Add a smoothed log-barrier term µ ( ( )) d 2 ij x i x j 2 log µh µ Solve the smooth approximation using coordinate gradient descent (SCGD): If xi f µ = Ω(µ), then update x i by moving it along the Newton direction [ 2 x i x i f µ ] 1 xi f µ, with Armijo stepsize rule, and re-iterate. Decrease µ when xi f µ = O(µ) i. µ init = 1e 5. µ end = 1e 9. Decrease µ by a factor of 10. Code in Fortran. Computation easily distributes.

21 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 20 Simulation Results Uniformly generate x true 1,..., x true n in [0, 1] 2, m =.9n, two pts are nhbrs if dist< radiorange. Set d ij = x true i ɛ ij N(0, 1) (Biswas, Ye 03) Solve SOCP using SeDuMi 1.05 or SCGD. x true j max{0, 1 + ɛ ij nf}, Sensor i is uniquely positioned if x i x j 2 y ij 10 7 d ij for some j N (i).

22 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 21 SeDuMi SCGD n m nf cpu/m up /Err up cpu/m up /Err up /402/7.2e-4.4/365/3.7e /473/1.8e-3 3.3/451/1.5e /554/1.5e-2 2.2/518/1.1e /1534/4.3e-4 1.3/1541/3.3e /1464/3.6e-3 6.8/1466/3.6e /1710/5.1e-2 3.7/1710/5.1e /2851/4.0e-4 2.5/2864/3.2e /2938/3.2e /2900/3.0e /3073/1.0e /3033/9.1e-3 Table 1: radiorange =.06(.035) for n = 1000, 2000(4000) cpu (sec) times are on a HP DL360 workstation, running Linux 3.5. m up := number of uniquely positioned sensors. Err up := max i uniq. pos. x i x true i.

23 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 22 True positions of sensors (dots) and anchors (circles) (m = 900, n = 1000) SOCP soln found by SeDuMi and SCGD

24 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 23 Mixed SDP-SOCP Relaxation Choose 0 l m. Let B := {(i, j) A : i l, j l}. min x 1,...,x m,y ij,y (i,j) B s.t. Z = tr ( b ij b T ijz ) d 2 ij + [ Y X T X I d ] 0 y ij d 2 ij \B X = [x 1 x l ] y ij x i x j 2 (i, j) A \ B with b ij := [ ] Il 0 0 (e 0 0 A i e j ). Easier to solve than SDP? As good a relaxation? Properties?

25 SOCP RELAXATION OF SENSOR NETWORK LOCALIZATION 24 Conclusions & Future Directions SOCP relaxation may be a good pre-processor. Faster methods for solving SOCP? Exploiting network structures of SOCP? (For d = 1, solvable by ɛ-relaxation method (Bertsekas,Polymenakos,T 97)) Error bound for SDP relaxation? Additional (convex) constraints? x i x j d ij 2? Other objective functions, e.g., Replace 2-norm by a p-norm (1 p )? p-order cone relaxation? Q: For any x 1,..., x n R d, does arg min x (1 < p < ) A: Yes for d 2. No for d 3. n x x i p p conv {x 1,..., x n }? i=1