SOCP Relaxation of Sensor Network Localization

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.

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

Sensor Network Localization Basic Problem

:

• n pts in <^d (d = 1, 2, 3).

• Know last n − m pts (‘anchors’) x_m+1, ..., x_n and Eucl. dist. estimate for pairs of ‘neighboring’ pts

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, ...

Optimization Problem Formulation

υ_opt := min

x₁,...,x_m

X

(i,j)∈A

kx_i − x_jk² − d²_ij 

• Objective function is nonconvex.

6. .

_

• Problem is NP-hard (reduction from PARTITION).

6. .

_

• Use a convex (SDP, SOCP) relaxation.

NP-hardness

PARTITION: Given positive integers a₁, a₂, ..., a_n, ∃ a partition I₁, I₂ of {1, ..., n} with P

i∈I₁ a_i = P

i∈I₂ a_i? (NP-complete)

Reduction to our problem

(d = 1) (Saxe ’79):

Let m = n − 1, x_n = 0, d_n1 = a₁, d₁₂ = a₂, ..., d_n−1,n = a_n

PARTITION ‘yes’ ⇐⇒ υ_opt = 0 [This extends to d ≥ 2]

SDP Relaxation

Let X := [x₁ · · · x_m], A := [x_m+1 · · · x_n].

Then (Biswas,Ye ’03)

kx_i − x_jk² = tr



b_ijb^T_ij  X^TX X^T

X I_d



with b_ij :=  I_m 0

0 A



(e_i − e_j).

Fact

:  Y X^T X I_d



 0 has rank d ⇐⇒ Y = X^TX

Thus

υ_opt = min

X,Y

X

(i,j)∈A

tr b_ijb^T_ijZ
− d²_ij  s.t. Z =  Y X^T

X I_d



 0, rankZ = d Drop low-rank constraint:

υ_sdp := min

X,Y

X

(i,j)∈A

tr b_ijb^T_ijZ
− d²_ij  s.t. Z =  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.

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?

SOCP Relaxation

υ_opt = min

x₁,...,x_m,y_ij

X

(i,j)∈A

y_ij − d²_ij 

s.t. y_ij = kx_i − x_jk² ∀(i, j) ∈ A

Relax “=” to “≥” constraint:

υ_socp = min

x₁,...,x_m,y_ij

X

(i,j)∈A



y_ij − d²_ij 

s.t. y_ij ≥ kx_i − x_jk² ∀(i, j) ∈ A

y ≥ kxk² ⇐⇒ y + 1 ≥ k(y − 1, 2x)k

(also Doherty,Pister,El Ghaoui ’03)

Properties of SDP, SOCP Relaxations

d = 2, n = 3, m = 1, d₁₂ = d₁₃ = 2

Problem

:

0 = min

x₁=(α,β)∈<²

|(1 − α)² + β² − 4| + |(−1 − α)² + β² − 4|

SDP Relaxation

:

0 = min

x1=(α,β)∈<2 y∈<

|y − 2α − 3| + |y + 2α − 3|

s.t.





y α β

α 1 0

β 0 1



  0

If solve SDP by IP method, then likely get analy. center.

SOCP Relaxation

:

0 = min

x1=(α,β)∈<2 y,z∈<

|y − 4| + |z − 4|

s.t. y ≥ (1 − α)² + β² z ≥ (−1 − α)² + β²

If solve SOCP by IP method, then likely get analy. center.

Properties of SDP & SOCP Relaxations Fact 1

: υ_socp ≤ υ_sdp. If υ_socp = υ_sdp, then

{SOCP (x₁, ..., x_m) solns} ⊇ {SDP (x₁, ..., x_m) solns}.

Fact 2

: If (x₁, ..., x_m, y_ij)_(i,j)∈A 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}.

Opt soln (m = 900, n = 1000, nhbrs if dist< .06)

SOCP soln found by IP method (SeDuMi)

Fact 3

: If X = [x₁ · · · x_m], Y is a relative-interior SDP soln (e.g., analytic center), then for each i,

kx_ik² = Y_ii =⇒ x_i appears in every SDP soln.

Fact 4

: If (x₁, ..., x_m, y_ij)_(i,j)∈A is a relative-interior SOCP soln (e.g., analytic center), then for each i,

kx_i − x_jk² = y_ij for some j ∈ N (i) ⇐⇒ x_i appears in every SOCP soln.

Error Bounds

What if distances have errors?

d²_ij = ¯d²_ij + δ_ij,

where δ_ij ∈ < and d¯_ij := kx^true_i − x^true_j k (x^true_i = x_i ∀ i > m).

Fact 5

: If (x₁, ..., x_m, y_ij)_(i,j)∈A is a relative-interior SOCP soln corresp.

(d_ij)_(i,j)∈A and X

(i,j)∈A

|δ_ij| ≤ δ, then for each i,

kx_i − x_jk² = y_ij for some j ∈ N (i) =⇒ kx_i − x^true_i k = O(

s X

(i,j)∈A

|δ_ij|).

Fact 6

: As X

(i,j)∈A

|δ_ij| → 0, (analytic center SOCP soln corresp. (d_ij)_(i,j)∈A)

→ (analytic center SOCP soln corresp. ( ¯d_ij)_(i,j)∈A).

Error bounds for the analytic center SOCP soln when distances have small errors.

Solving SOCP Relaxation I: IP Method

x₁,...,xmin_m,y_ij

X

(i,j)∈A

y_ij − d²_ij 

s.t. y_ij ≥ kx_i − x_jk² ∀(i, j) ∈ A

Put into conic form:

min X

(i,j)∈A

u_ij + v_ij

s.t. x_i − x_j − w_ij = 0 ∀(i, j) ∈ A

y_ij − u_ij + v_ij = d²_ij ∀(i, j) ∈ A

α_ij = ¹₂ ∀(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) ∈ < × < × <^d : kwk²/2 ≤ αy}.

Solve by an IP method, e.g., SeDuMi 1.05 (Sturm ’01).

Solving SOCP Relaxation II: Smoothing + Coordinate Gradient Descent

miny≥z |y − d²| = max{0, z − d²} So SOCP relaxation:

x₁min,...,x_m

X

(i,j)∈A

max{0, kx_i − x_jk² − d²_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) = ((t² + 4)^1/2 + t)/2 (CHKS).

SOCP approximation:

min f_µ(x₁, .., x_m) := X

(i,j)∈A

µh kx_i − x_jk² − d²_ij µ

!

Add a smoothed log-barrier term −µ X

(i,j)∈A

log µh d²_ij − kx_i − x_jk² µ

!!

Solve the smooth approximation using coordinate gradient descent (SCGD):

• If k∇_xif_µk = Ω(µ), then update x_i by moving it along the Newton direction

−[∇²_x

ix_if_µ]⁻¹∇_xif_µ, with Armijo stepsize rule, and re-iterate.

• Decrease µ when k∇_xif_µk = O(µ) ∀i.

µ^init = 1e − 5. µ^end = 1e − 9. Decrease µ by a factor of 10.

Code in Fortran. Computation easily distributes.

Simulation Results

• Uniformly generate x^true₁ , ..., x^true_n in [0, 1]², m = .9n, two pts are nhbrs if dist< radiorange.

Set

d_ij = kx^true_i − x^true_j k · max{0, 1 + _ij · nf },

_ij ∼ N (0, 1) (Biswas, Ye ’03)

• Solve SOCP using SeDuMi 1.05 or SCGD.

• Sensor i is uniquely positioned if 

  

kx_i − x_jk² − y_ij    

≤ 10⁻⁷d_ij for some j ∈ N (i).

SeDuMi SCGD n m nf cpu/m_up/Err_up cpu/m_up/Err_up 1000 900 0 5.5/402/7.2e-4 .4/365/3.7e-5 1000 900 .001 5.4/473/1.8e-3 3.3/451/1.5e-3 1000 900 .01 5.6/554/1.5e-2 2.2/518/1.1e-2 2000 1800 0 209.6/1534/4.3e-4 1.3/1541/3.3e-4 2000 1800 .001 230.1/1464/3.6e-3 6.8/1466/3.6e-3 2000 1800 .01 176.6/1710/5.1e-2 3.7/1710/5.1e-2 4000 3600 0 203.1/2851/4.0e-4 2.5/2864/3.2e-4 4000 3600 .001 205.2/2938/3.2e-3 15.1/2900/3.0e-3 4000 3600 .01 201.3/3073/1.0e-2 23.2/3033/9.1e-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 := maxi uniq. pos.kx_i − x^true_i k.

True positions of sensors (dots) and anchors (circles) (m = 900, n = 1000)

SOCP soln found by SeDuMi and SCGD

Mixed SDP-SOCP Relaxation

Choose 0 ≤ ` ≤ m. Let B := {(i, j) ∈ A : i ≤ `, j ≤ `}.

x₁,...,xmin_m,y_ij,Y

X

(i,j)∈B



tr b_ijb^T_ijZ
− d²_ij

 + X

(i,j)∈A\B



y_ij − d²_ij 

s.t. Z =  Y X^T

X I_d



 0

X = [x₁ · · · x_`]

y_ij ≥ kx_i − x_jk² ∀(i, j) ∈ A \ B

with b_ij :=  I_` 0 0

0 0 A



(e_i − e_j).

Easier to solve than SDP? As good a relaxation? Properties?

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 -relaxation method (Bertsekas,Polymenakos,T ’97))

• Error bound for SDP relaxation?

• Additional (convex) constraints? Other objective functions, e.g., X

(i,j)∈A

|kx_i − x_jk − d_ij|² ?

• Replace 2-norm by a p-norm (1 ≤ p ≤ ∞)? p-order cone relaxation?

• Q: For any x₁, ..., x_n ∈ <^d, does arg min

x

n

X

i=1

kx − x_ik^p_p ∈ conv {x₁, ..., x_n}?

(1 < p < ∞)

A: Yes for d ≤ 2. No for d ≥ 3.