1. Mobile Assisted Localization in Wireless Sensor Networks N.B. Priyantha, H. Balakrishnan, E.D. Demaine, S. Teller MIT Computer Science Presenters: Puneet Gupta Sol Lederer

2. Case for Mobile Assisted Localization Obstructions, especially in indoor environments Sparse node deployments Geometric dilution of precision (GDOP) Hence, finding 4 reference points for each node for localization is difficult

3. Overview of scheme Initially no nodes know their location Mobile node finds cluster of nearby nodes Explores “visibility region” and measures distance # of measurements required is linear in the # of nodes Virtual nodes are discarded

4. Theorem 1 A graph is globally rigid if it is formed by starting from a clique of 4 non- coplanar nodes and repeatedly adding a node connected to at least 4 nodes.

5. MAL: Distance Measurement First case: Two nodes, n 0 and n 1, single unknown ||n 0 - n 1 || Adding mobile node, m, introduces 3 unknowns (m x, m y, m z ), making problem more difficult Necessary condition: # deg of freedom (unknowns – knowns) ≤ 0. Solution: Use three mobile locations along the same line in a plane containing n 0 and n 1

6. Case of 2 nodes solved 6 constraints from measurements of ||n i – m j || for I = 0,1 and j = 0,1,2 Extra constraint obtained from colinearity of mobile points unknowns – knowns = 0 Solve system of polynomial equations

7. Case of 3 nodes Three nodes, n 0 n 1 n 2, three unknowns, ||n 0 - n 1 || ||n 1 - n 2 || ||n 0 - n 2 || Each mobile position gives #unknowns (m x, m y, m z ) = 3 #constraints (||m – n i ||, i = 0,1,2) = 3 Three additional constraints needed

8. Case of 3 nodes  Solution Restriction: All mobile positions lie in a common plane k mobile locations  k-3 additional co- planarity constraints Solution: k = 6, geometry of n 0, n 1, n 2 above the plane containing 6 coplanar points m 0, m 1, m 2, m 3, m 4, m 5 no three of which are collinear, determined by the distances ||m i - n j ||, i = 0…5 & j = 0...2

9. Case of 4 or More Number of nodes = j ≥ 4 Initially: Number of unknowns = (3j – 5) 3 coordinates per node Minus 3 deg of translational motion Minus 2 deg of rotational motion Each mobile node adds (j – 3) deg of freedom (j distances – 3 coordinates of mobile position) j – 3 >= 1

10. Case of 4 or more  Solution Require at least (3j – 5)/(j – 3) mobile positions E.g. for j = 4, required mobile positions to uniquely determine the geometry = 7 But, no 4 of the 11 nodes (4 + 7) may be coplanar

11. MAL: Movement Strategy Initialize: Find 4 nodes that can all be seen from a common location Move the mobile to 7 nearby locations & measure distances Compute pair-wise distances Loop: Pick a localized stationary node (not yet considered by this loop) Move mobile in perimeter of this node, searching for positions to hear a non-localized node Localize this node

12. AFL: Anchor-free localization Elect five nodes as shown Get crude coordinates based on hop count to anchors

13. AFL Use non-linear optimization algorithm to minimize sum-squared energy E Coordinate assignments satisfy all 1- hop node distances when E = 0

14. Graph from running AFL—using RF connectivity information Graph obtained by MAL

15. Performance Layout of nodes in test scenario

16. Estimate error

17. Critique Pros: Innovative stategy Cons: In a cumbersome terrain (e.g. forest) it may not be feasible to deploy a roving node.

18. The End