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v2009.01.01 - Convex Optimization

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290 CHAPTER 4. SEMIDEFINITE PROGRAMMING<br />

direction matrix W<br />

Denote by Z ⋆ an optimal composite matrix from semidefinite program (678).<br />

Then for Z ⋆ ∈ S N+n whose eigenvalues λ(Z ⋆ )∈ R N+n are arranged in<br />

nonincreasing order, (Ky Fan)<br />

N+n<br />

∑<br />

λ(Z ⋆ ) i<br />

i=n+1<br />

= minimize<br />

W ∈ S N+n 〈Z ⋆ , W 〉<br />

subject to 0 ≼ W ≼ I<br />

trW = N<br />

(1581a)<br />

which has an optimal solution that is known in closed form. This eigenvalue<br />

sum is zero when Z ⋆ has rank n or less.<br />

Foreknowledge of optimal Z ⋆ , to make possible this search for W , implies<br />

iteration; id est, semidefinite program (678) is solved for Z ⋆ initializing<br />

W = I or W = 0. Once found, Z ⋆ becomes constant in semidefinite<br />

program (1581a) where a new normal direction W is found as its optimal<br />

solution. Then the cycle (678) (1581a) iterates until convergence. When<br />

rankZ ⋆ = n , solution via this convex iteration is optimal for sensor-network<br />

localization problem (672) and its equivalent (677).<br />

numerical solution<br />

In all examples to follow, number of anchors<br />

m = √ N (680)<br />

equals square root of cardinality N of list X . Indices set I identifying all<br />

measurable distances • is ascertained from connectivity matrix (673), (674),<br />

(675), or (676). We solve iteration (678) (1581a) in dimension n = 2 for each<br />

respective example illustrated in Figure 71 through Figure 74.<br />

In presence of negligible noise, true position is reliably localized for every<br />

standardized example; noteworthy in so far as each example represents an<br />

incomplete graph. This means the set of all solutions having lowest rank is<br />

a single point, to within a rigid transformation.<br />

To make the examples interesting and consistent with previous work, we<br />

randomize each range of distance-square that bounds 〈G, (e i −e j )(e i −e j ) T 〉<br />

in (678); id est, for each and every (i,j)∈ I<br />

d ij = d ij (1 + √ 3η χ l<br />

) 2<br />

d ij = d ij (1 − √ 3η χ l+1<br />

) 2 (681)

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