Edge-based semidefinite programming relaxation of sensor network localization with lower bound constraints

Abstract

In this paper, we strengthen the edge-based semidefinite programming relaxation (ESDP) recently proposed by Wang, Zheng, Boyd, and Ye (SIAM J. Optim. 19:655–673, 2008) by adding lower bound constraints. We show that, when distances are exact, zero individual trace is necessary and sufficient for a sensor to be correctly positioned by an interior solution. To extend this characterization of accurately positioned sensors to the noisy case, we propose a noise-aware version of ESDP_lb (ρ-ESDP_lb) and show that, for small noise, a small individual trace is equivalent to the sensor being accurately positioned by a certain analytic center solution. We then propose a postprocessing heuristic based on ρ-ESDP_lb and a distributed algorithm to solve it. Our computational results show that, when applied to a solution obtained by solving ρ-ESDP proposed of Pong and Tseng (Math. Program. doi:10.1007/s10107-009-0338-x), this heuristics usually improves the RMSD by at least 10%. Furthermore, it provides a certificate for identifying accurately positioned sensors in the refined solution, which is not common for existing refinement heuristics.

Appendices

6 Unrealistic lower bound estimates

In this appendix, we give two counterexamples for the trace test under unrealistic lower bound estimate. Then we remark on some possible heuristics in picking realistic lower bound estimates in applications.

The first example shows that Theorem 1 can fail if r is not realistic, even when there is no noise in distance measurement.

Example 2

Consider n=4, m=1, $x_{1}^{\mathrm {true}}=(0,1)^{T}$, x ₂=(1,0)^T, x ₃=(−1,0)^T, x ₄=(0,−1)^T, x ₅=(0,1)^T, $d_{12}=d_{13}=\sqrt{2}$, d ₁₄ and d ₁₅ are not known, and $r_{14}=r_{15}=\sqrt{2}$. Then the problem (4) becomes

(12)

We claim that is the unique solution. First, note that Z ₀ is an optimal solution (12) since it is feasible with objective value zero. Thus, for any optimal solutions, $|y_{11}-2x_{2}^{T}x_{1}+\|x_{2}\|^{2}-2|+|y_{11}-2x_{3}^{T}x_{1}+\|x_{3}\|^{2}-2|=0$, and hence y ₁₁=1 and x ₁=(0,t)^T for some t∈[−1,1]. Next, we have from $y_{11}-2x_{4}^{T}x_{1}+\|x_{4}\|^{2}\ge2$ that

$$1+2t+1\ge2\quad \Rightarrow\quad t\ge0.$$

Similarly, $y_{11}-2x_{5}^{T}x_{1}+\|x_{5}\|^{2}\ge2$ implies t≤0. Hence, x ₁=(0,0). This shows that Z ₀ is the unique solution to (12). However, tr₁(Z ₀)=1>0. Note that the lower bound estimate is not realistic since

Moreover, in unrealistic cases, zero individual trace for some relative interior solution can fail to imply that sensor i is accurately positioned in the noiseless case; though it is true that x _i is invariant.

Example 3

Consider n=3, m=1, $x_{1}^{\mathrm {true}}=(0,1)^{T}$, x ₂=(1,0)^T, x ₃=(−1,0)^T, x ₄=(0,2)^T, $d_{12}=d_{13}=\sqrt{2}$, d ₁₄ is not known and r ₁₄=3. Then the problem (4) becomes

(13)

We claim that

is the unique solution. Again, Z ₀ is an optimal solution of (13) since it is feasible with objective value zero. Thus, by setting the objective function zero, we see that any optimal solutions must satisfy y ₁₁=1 and x ₁=(0,t)^T for some t∈[−1,1]. Next, we have from $y_{11}-2x_{4}^{T}x_{1}+\|x_{4}\|^{2}\ge9$ that

$$1-4t+4\ge9\quad \Rightarrow\quad -1\ge t.$$

Hence, x ₁=(0,−1). This shows that Z ₀ is the unique solution to (12). Moreover, tr₁(Z ₀)=0 but $x_{1}\neq x_{1}^{\mathrm {true}}$. Note that the lower bound estimate is not realistic since

In view of the above examples, we see that if the lower bound estimates are not realistic, the trace test does not provide theoretical guarantee about the true position of sensors. To pick a realistic lower bound estimate r, one heuristic is to choose r as in (9): since $\hat{d}^{\mathrm{true}}(i):=\max_{(i,j)\in {\mathcal {A}}}d^{\mathrm{true}}_{ij}$ is likely strictly less than the radio range, such an r is likely realistic when noise is small. A more conservative heuristic in the case of Gaussian noise (8) is to set $r_{ij} = \max\{\hat{d}(i),\hat{d}(j)\}/(1+2\hat{\sigma})$, where $\hat{\sigma}$ is an estimate of σ and $\hat{d}(i)$ as defined in Sect. 4.1, following the similar consideration that gives (10). A way of choosing tighter realistic lower bound estimates will conceivably give more accurate solutions.

7 Proof of Theorem 1

The idea of the proof of Theorem 1 is similar to that of [20, Theorem 1], except that we need Lemma 3 to deal with the lower bound constraints in (4) corresponding to ${\mathcal {B}}$. Below, we also repeat some essential arguments of the first part of the proof of [20, Theorem 1] for the ease of readers.

Fix any $\bar{i}\in {\mathcal {I}}_{\mathrm{esdp}}^{0,r}$ and $\bar{Z}\in\mathrm{ri}( {\mathcal {S}}_{\mathrm{esdp}}^{0,r})$. Suppose to the contrary that $\mathrm{tr}_{\bar{i}}(\bar{Z})>0$. Let ${\bar{{\mathcal {I}}}}$ be the collection of all nodes in ${\mathcal {I}}_{\mathrm{esdp}}^{0,r}$ such that it is connected to $\bar{i}$ by a path with arcs in the following set:

Hence, ${\mathcal {N}}({\bar{{\mathcal {I}}}})\not=\emptyset$ by Assumption 1. It follows from Lemma 2(a) and definition that $\mathrm{tr}_{i}(\bar{Z})>0$ for all $i\in {\bar{{\mathcal {I}}}}$. Hence, Lemma 2(b) implies that

$$ {\mathcal {N}}({\bar{{\mathcal {I}}}})\subseteq\{1,\dots,m\}\setminus {\mathcal {I}}_{\mathrm{esdp}}^{0,r}.$$

(14)

We will derive a contradiction below.

First, note that there exists a $Z\in\mathrm{ri}( {\mathcal {S}}_{\mathrm{esdp}}^{0,r})$ such that $x_{j}\neq x_{j}^{\mathrm {true}}$ for all $j\notin {\mathcal {I}}_{\mathrm{esdp}}^{0,r}$, and hence, by (14), for all $j\in {\mathcal {N}}({\bar{{\mathcal {I}}}})$. Since $Z^{\mathrm{true}}\in {\mathcal {S}}_{\mathrm{esdp}}^{0,r}$, by convexity, we have

$$Z^\alpha:= \alpha Z^{\mathrm{true}}+(1-\alpha)Z\in\mathrm{ri}( {\mathcal {S}}_{\mathrm{esdp}}^{0,r}) \quad\forall 0\le\alpha< 1.$$

Next, consider any $(i,j)\in {\mathcal {A}}({\bar{{\mathcal {I}}}})$ with $i\in {\bar{{\mathcal {I}}}}$. It follows from (14) that j≤m and hence $(i,j)\in {\mathcal {A}}^{s}$. Applying [20, Lemma 4] with $\bar{A}=(x_{i}^{\mathrm {true}} \ x_{j}^{\mathrm {true}})$, A=(x _i x _j ), $\bar{B}=Z^{\mathrm{true}}_{\{i,j\}}$, $B=Z^{\mathrm{true}}_{\{i,j\}}$ and making use of the fact that $x_{i}=x^{\mathrm{true}}_{i}$, we see that

Since $\mathrm{tr}_{i}(\bar{Z})>0$ implies tr _i (Z)>0 and also $x_{j}\neq x^{\mathrm{true}}_{j}$, we conclude as in [20, Theorem 1] that $Z^{\alpha}_{\{i,j,m^{+}\}}$ is nonsingular for all 0<α<1 sufficiently small.

On the other hand, let $(i,j)\in {\mathcal {B}}^{s}({\bar{{\mathcal {I}}}})$ with $j\notin {\mathcal {I}}_{\mathrm{esdp}}^{0,r}$. Then, by arguing as in the previous paragraph, we have that $Z^{\alpha}_{\{i,j,m^{+}\}}$ is nonsingular for all 0<α<1 sufficiently small.

Choose a 0<α<1 such that $Z^{\alpha}_{\{i,j,m^{+}\}}$ is nonsingular for all $(i,j)\in {\mathcal {A}}({\bar{{\mathcal {I}}}})\cup {\mathcal {B}}^{s}({\bar{{\mathcal {I}}}})$ with $j\notin {\mathcal {I}}_{\mathrm{esdp}}^{0,r}$. By translating all points in the same direction if necessary, we assume without loss of generality that $x^{\alpha}_{i}\neq0$ for all $i\in {\bar{{\mathcal {I}}}}$. Then, for any sufficiently small θ>0, we have

(15)

where U _θ ∈ℝ^d×d be an orthogonal matrix with 0<∥U _θ −I _d ∥ _F =O(θ), ∥⋅∥ _F is the Fröbenius norm. We now show that θ>0 can be further chosen such that the lower bound constraints are satisfied.

For any $(i,j)\in {\mathcal {B}}^{s}({\bar{{\mathcal {I}}}})$ with $j\notin {\mathcal {I}}_{\mathrm{esdp}}^{0,r}$, the corresponding positive semidefinite constraint in (4) is satisfied. Hence, with $y_{ij}^{\alpha}$ unchanged, we see that the corresponding lower bound constraint is also satisfied. This places no further restrictions on the choice of θ. On the other hand, consider any $(i,j)\in {\mathcal {B}}^{s}({\bar{{\mathcal {I}}}})$ with $j\in {\mathcal {I}}_{\mathrm{esdp}}^{0,r}\backslash {\bar{{\mathcal {I}}}}$. We shall alter $y_{ij}^{\alpha}$ in such a way that both the lower bound constraint and the positive semidefinite constraint are satisfied. To this end, note that tr _j (Z ^α)=0 by definition of ${\bar{{\mathcal {I}}}}$ and that

$$ \mathrm{tr}_i(Z^\alpha)\ge\alpha\,\mathrm{tr}_i(Z^{\mathrm{true}})+(1-\alpha)\mathrm{tr}_i(Z)>0.$$

(16)

These imply $\ell_{ij}(Z^{\alpha})>r_{ij}^{2}$ by Lemma 3(b). Moreover, by the third constraint in (4) and tr _j (Z ^α)=0, we have $y_{ij}^{\alpha}=(x_{i}^{\alpha})^{T}x_{j}^{\alpha}$. Hence, for sufficiently small θ>0, we have that

(17)

and by (16) that

(18)

The relationship (18) is the same as

Next, consider any $(i,j)\in {\mathcal {B}}^{as}$ with $i\in {\bar{{\mathcal {I}}}}$. Since $Z^{\alpha}\in\mathrm{ri}( {\mathcal {S}}_{\mathrm{esdp}}^{0,r})$ and tr _i (Z ^α)>0, it follows from Lemma 3(a) that $\ell_{ij}(Z^{\alpha})>r_{ij}^{2}$. Hence, for θ>0 sufficiently small, we have that

$$ y^\alpha_{ii}-2x_j^TU_\theta x^\alpha_i+\|x_j\|^2>r_{ij}^2\quad \forall(i,j)\in {\mathcal {B}}^{as},\ i\in {\bar{{\mathcal {I}}}}, j>m.$$

(19)

Fix any θ>0 such that (15), (17) and (19) hold. For each $(i,j)\in {\mathcal {A}}^{s}\cup {\mathcal {B}}^{s}$ with $i,j\in {\bar{{\mathcal {I}}}}$, we have from $Z^{\alpha}\in {\mathcal {F}}_{\mathrm{esdp}}^{r}$ that

from which it follows that

Thus, replacing $x^{\alpha}_{i}$ in Z ^α by $U_{\theta}x^{\alpha}_{i}$ for all $i\in {\bar{{\mathcal {I}}}}$, and $y_{ij}^{\alpha}$ by $(U_{\theta}x_{i}^{\alpha})^{T}x_{j}^{\alpha}$ for all $(i,j)\in {\mathcal {B}}^{s}({\bar{{\mathcal {I}}}})$ with $j\in {\mathcal {I}}_{\mathrm{esdp}}^{0,r}\backslash {\bar{{\mathcal {I}}}}$ yields a $\tilde{Z}^{\alpha}$ feasible for (4). Moreover, $\tilde{Z}^{\alpha}$ is an optimal solution of (4) since the objective function of (4) does not depend on x _i for $i\in {\bar{{\mathcal {I}}}}\subseteq {\bar{{\mathcal {I}}}}\cup {\mathcal {N}}({\bar{{\mathcal {I}}}})\subseteq\{1,\dots,m\}$, nor on y _ij for $(i,j)\in {\mathcal {B}}^{s}$. Thus $\tilde{Z}^{\alpha}\in {\mathcal {S}}_{\mathrm{esdp}}^{0,r}$. However, its $U_{\theta}x^{\alpha}_{i}$ component is different from the $x^{\alpha}_{i}$ component of Z ^α for all $i\in {\bar{{\mathcal {I}}}}$, which contradicts the definition of ${\bar{{\mathcal {I}}}}$.