Localization
Paul Tseng
Mathematics, University of Washington Seattle
Chinese University of Hong Kong February 2008
(Ongoing work with Ting Kei Pong) Abstract
This is a talk given at CUHK, 2008.
Talk Outline
• Sensor network localization and SDP, ESDP relaxations
• Properties of SDP, ESDP
• A robust version of ESDP for the noisy case
• Conclusion & Ongoing work
Sensor Network Localization
Basic Problem
:• n pts in <2.
• Know last n − m pts (‘anchors’) xm+1, ..., xn and Eucl. dist. estimate for pairs of ‘neighboring’ pts
dij ≥ 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, ...
Optimization Problem Formulation
υopt := min
x1,...,xm
X
(i,j)∈A
kxi − xjk22 − d2ij
• Objective function is nonconvex. m can be large (m > 1000).
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• Problem is NP-hard (reduction from PARTITION).
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• Use a convex (SDP, SOCP) relaxation. High soln accuracy unnecessary.
• Seek “simple” distributed methods (important for practical implementation).
SDP Relaxation
Let X := [x1 · · · xm], A := [xm+1 · · · xn].
SDP relaxation (Biswas,Ye ’03):
υsdp := min
Z
X
(i,j)∈A,j>m
Yii − 2xTj xi + kxjk22 − d2ij
+ X
(i,j)∈A,j≤m
Yii − 2Yij + Yjj − d2ij s.t. Z = Y XT
X I
0
Adding the nonconvex constraint rankZ = 2 yields original problem.
υsdp ≤ υopt.
But SDP relaxation is still expensive to solve for m large..
SOCP Relaxation
υopt = min
x1,...,xm,yij
X
(i,j)∈A
yij − d2ij
s.t. yij = kxi − xjk22 ∀(i, j) ∈ A
Relax “=” to “≥” constraint (Doherty,Pister,El Ghaoui ’03):
υsocp := min
x1,...,xm,yij
X
(i,j)∈A
yij − d2ij
s.t. yij ≥ kxi − xjk22 ∀(i, j) ∈ A
υsocp ≤ υsdp.
SOCP is much easier to solve than SDP relaxation (T ’07), but can be much weaker.
ESDP Relaxation
ESDP relaxation (Wang, Zheng, Boyd, Ye ’06):
υesdp := min
Z
X
(i,j)∈A,j>m
Yii − 2xTj xi + kxjk22 − d2ij
+ X
(i,j)∈A,j≤m
Yii − 2Yij + Yjj − d2ij s.t. Z = Y XT
X I
Yii Yij xTi Yij Yjj xTj xi xj I
0 ∀(i, j) ∈ A, j ≤ m
Yii xTi xi I
0 ∀i ≤ m
υsocp ≤ υesdp ≤ υsdp. In simulation, ESDP is nearly as strong as SDP, and solvable much faster by IP method.
An Example
n = 3, m = 1, d12 = d13 = 2
Problem
:0 = min
x1∈<2
|kx1 − (1, 0)k22 − 4| + |kx1 − (−1, 0)k22 − 4|
SDP/ESDP Relaxation
:0 = min
x1=(α,β)∈<2 Y11∈<
|Y11 − 2α − 3| + |Y11 + 2α − 3|
s.t.
Y11 α β
α 1 0
β 0 1
0
If solve SDP/ESDP by IP method, then likely get analy. center.
SOCP Relaxation
:0 = min
x1∈<2 y12,y13∈<
|y12 − 4| + |y13 − 4|
s.t. y12 ≥ kx1 − (1, 0)k22 y13 ≥ kx1 − (−1, 0)k22
If solve SOCP by IP method, then likely get analy. center.
SDP Relaxation: a larger example with noise
:n = 64, m = 60. Anchors at (+.45, +.45) (“◦”). Sensors uniformly distributed on [−.5, .5]2 (“∗”).
(i, j) ∈ A whenever kxtruei − xtruej k2 ≤ 0.3
Normally distributed noise: dij = dtrueij · max{0, 1 + .2ν}, ν ∼ N (0, 1).
The SDP soln found by SeDuMi 1.05 is shown (“·”) joined to its true position (“∗”) by a line.
−0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.5
−0.5
−0.4
−0.3
−0.2
−0.1 0 0.1 0.2 0.3 0.4 0.5
Properties of SDP & ESDP Relaxations
Fact 0
: Sol(SDP) and Sol(ESDP) are nonempty, closed, convex, and bounded if each i ≤ m is conn. to some j > m in the graph ({1, ..., n}, A).tri[Z] := Yii − kxik22, i = 1, ..., m. “ith trace”
Fact 1
(Biswas,Ye ’03, T ’07, Wang et al ’06): For each i,tri[Z] = 0 ∃Z ∈ ri(Sol(ESDP)) =⇒ xi is invariant over Sol(ESDP).
Still true with “ESDP” changed to “SDP”.
Fact 2
(Pong, T ’08): Suppose υopt = 0. For each i,tri[Z] = 0 ∀Z ∈ Sol(ESDP) ⇐⇒ xi is invariant over Sol(ESDP).
Proof is by induction, starting from sensors that neighbor anchors.
(Q: True for SDP?)
Proof sketch for Fact 2
:1. For (i, j) ∈ A, j > m, if xi is invariant over Sol(ESDP), then tri(Z) = 0 for all Z ∈ Sol(ESDP).
Why: υopt = 0 and xi invariant over Sol(ESDP) imply, for any Z ∈ Sol(ESDP), Yii − 2xTj xi + kxjk22 = d2ij, kxi − xjk22 = d2ij
So tri(Z) = Yii − kxik22 = d2ij − kxi − xjk22 = 0.
2. For (i, j) ∈ A, j ≤ m, if xi is invariant over Sol(ESDP), then tri(Z) = trj(Z) for all Z ∈ Sol(ESDP).
Why? υopt = 0 and xi invariant over Sol(ESDP) imply, for any Z ∈ Sol(ESDP), Yii − 2Yij + Yjj = d2ij, kxi − xjk22 = d2ij
So Yij − xTi xj = 12(tri(Z) + trj(Z)).
Then
Yii Yij xTi Yij Yjj xTj xi xj I
= tri(Z)
1 12 0 0
1
2 0 0 0
0 0 0 0 0 0 0 0
+ trj(Z)
0 12 0 0
1
2 1 0 0
0 0 0 0 0 0 0 0
+
xTi xTj I
[ xi xj I ]
This is psd, which implies ...that tri(Z) = trj(Z).
When there is measurement noise, does tri[Z] = 0 (with Z ∈ ri(Sol(ESDP))) imply xi is near the true position of sensor i?
Let
d2ij = ¯d2ij + δij ∀(i, j) ∈ A,
where d¯ij := kxtruei − xtruej k2 (xtruei = xi ∀ i > m). δ := max¯
(i,j)∈A
|δij|.
Fact 3
(Pong, T ’08): For δ ≈ 0¯ and for each i,tri[Z] = 0 ∃Z ∈ ri(Sol(ESDP)) 6=⇒ kxi − xtruei k2 ≈ 0.
Still true with “ESDP” changed to “SDP”.
Proof is by counter-example.
An example of sensitivity of SDP/ESDP solns to measurement noise:
x1
x2
−1 1 2
a3 a
4
a2
a1 1
−1
x1
x2
−1 1 2
−1
a3 a
4
a2
a1 1
Thus, even when Z ∈ Sol(SDP/ESDP) is unique, tri[Z] = 0 certifies accuracy of xi only in the noiseless case!
Robust ESDP
Fix ρ > ¯δ.
Sol(ρESDP) denotes the set of Z = Y XT
X I
satisfying
Yii Yij xTi Yij Yjj xTj xi xj I
0 ∀(i, j) ∈ A, j ≤ m
Yii xTi xi I
0 ∀i ≤ m
|Yii − 2xTj xi + kxjk2 − d2ij| ≤ ρ ∀(i, j) ∈ A, j > m
|Yii − 2Yij + Yjj − d2ij| ≤ ρ ∀(i, j) ∈ A, j ≤ m
Note: Xtrue I T
Xtrue I ∈ Sol(ρESDP).
Let
Zρ := arg min
Z∈Sol(ρESDP)
− X
(i,j)∈A,j≤m
ln det
Yii Yij xTi Yij Yjj xTj xi xj I
− X
i≤m
ln det Yii xTi xi I
Fact 4
(Pong, T ’08): ∃¯ρ > ¯δ and τ > 0 such that, for δ < ρ ≤ ¯¯ ρ and for each i, xi is invariant over Sol(ESDP|¯dij) ⇐⇒ tri[Zρ] < τ=⇒ kxρi − xtruei k2 ≤ p2|A| + n (tri[Zρ])1/2
Conclusion & Ongoing work
SDP and ESDP are stronger relaxations, but inherit the soln instability relative to measurement noise. Lack soln accuracy certificate.
SOCP and ρESDP are weaker relaxations, but have more stable solns. Have soln accuracy certificate. Is ρESDP better?
• Distributed method to compute Zρ?
• Simulation and numerical testing?
Thanks for coming!
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