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Addressing and routing in hexagonal networks with applications in location update and c onnection rerouting in mobile ph

Addressing and routing in hexagonal networks with applications in location update and c onnection rerouting in mobile phone network. Fabian Garcia Nocetti Ivan Stojmenovic Jingyuan Zhang. Hexagonal networks. Honeycomb mesh -interconnection network Stojmenovic 1998 Cellular phone network

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Addressing and routing in hexagonal networks with applications in location update and c onnection rerouting in mobile ph

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  1. Addressing and routing in hexagonal networks with applications in location update and connection rerouting in mobile phone network Fabian Garcia Nocetti Ivan Stojmenovic Jingyuan Zhang

  2. Hexagonal networks Honeycomb mesh -interconnection network Stojmenovic 1998 Cellular phone network Base stations at centers of hexagons Benzenoid hydrocarbons

  3. Triangular networks Hexagonal mesh -interconnection network – Chen, Shin, Kandlur 1990 Centers of cells in cellular network and connections with neighbors

  4. Dual triangular-hexagonal networks BS= base stations = centers of hexagons Cells in cellular network = hexagons

  5. Location management in cellular networks Mobile phones update periodically current cell position with their home location server Time, movement and distance based methods Bar-Noy, Kessler, Sidi 1994 Distance from last updated BS=? BS mobile phone

  6. Connection rerouting in cellular nets Wired network path modification or Wireless path extension How to shorten extended wireless path?

  7. Addressing in ‘hexagonal’ networks (0,-2,1) i+j+k=0 i (a,b,c)= ai + bj + ck x k j z (0,0,0) (a’,b’,c’)= (a”,b”,c”)  a’i+b’j+c’k= a”i+b”j+c”k y (-1,0,1) (-1,0,0)=(0,1,1) (-1,0,2)

  8. Shortest path representation Lemma: D-S=(a,b,c) shortest path from S to D  abc=0 Proof:assume abc0 , a>0, b>0 D-S= ai+bj+ck= ai+bj+ck –(i+j+k)= (a-1)i+ (b-1)k +(c-1)k, a=|a-1|+1, b=|b-1|+1, |a|+|b|+|c|= |a-1|+|b-1|+2+|c| > |a-1|+|b-1|+|c-1| contradiction.

  9. Length of shortest path Corollary: c=0  ab 0. Assume ab>0 a>0 b>0ai+bj=ai+bj-(i+j+k)= (a-1)i+(b-1)j - k |a|+|b|=|a-1|+1+|b-1|+1> |a-1|+|b-1|+|-1| contradiction. Lemma: D-S= ai+bj  |D-S|=min(|a|+|b|, |a-b|+|b|, |a-b|+|a|) Proof: D-S=ai+bj=(a-b)i+bk=(b-a)j+ak

  10. Routing and connection rerouting D-S= ai+bjandab 0. Route |a| times in direction sign(a)x (=ai) Route |b| times in direction sign(b)y (=bj) C(|a|+|b|,|a|) shortest paths (choice may depend on load) Distance determined by three integers (one is 0) Routing corresponds to GEDIR = distance based routing: Route to neighbor that is closest to destination

  11. Fault tolerance in ‘hexagonal’ networks Hexagonal network is planar graph Apply GFG = GEDIR-FACE-GEDIR for guaranteed delivery GFG also works in other planar graphs: square mesh, honeycomb mesh, semi-regular tessellations GEDIR works for other networks e.g. hypercubes

  12. Extension – BS at arbitrary positions BS determine Delaunay triangulation = dual Voronoi diagram GEDIR guarantees delivery in Delaunay triangulations (Bose, Morin, 2000) Faulty BSs: apply GFG (DT is planar graph)

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