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Variations on the Theme of Network Coding by gegeshandong

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									   Network Coding Tomography for
          Network Failures


Sidharth Jaggi       Hongyi Yao
Minghua Chen
Computerized Axial
Tomography (CAT Scan)




                        1
Tomography


  Heart




             2
Network Tomography [V96]…


                                               Objects:
                                                   •Topology estimation
            @#$%&*
             001001                                •Failure localization




Failure type:
     •Adversarial error: The corrupted packets are carefully chosen by
     the enemies for specific reasons.
     •Random error: The network packets are randomly polluted.

                                                                           3
          Tomography type
   Active tomography[RMGR04,CAS06]:
       All network nodes work cooperatively for tomography.
       Probe packets from the sources are required.
       Heavy overhead on computation & throughput.
   Passive tomography [RMGR04, CA05, Ho05, This work]:
       Tomography is done during normal communications.
       Zero overhead on computation & throughput.



                                                               4
         Network coding
                                                            S

   Network coding suffices to achieve to         m1            m2

    the optimal throughput for
                                                       m1   m2
    multicast[RNSY00].

                                                        m1+m2
                                                       am1+bm2

                                                 m1              m2
   Random linear network coding suffices,
    in addition to its distributed feature and
    low design complexity[TMJMD03].                   r1         r2




                                                                      5
        Random Linear Network Coding
   Source: Sends packets. Organized as:

                          X        I
                                                v1     v2
   Internal Nodes: Random linear coding
        v1
                                                 a1v1+a2v2
                              a1v1+a2v2
        v2

                                           Information T: Recover
   Sink gets Y:                           Topology [Sharma08]


       Y=T            X        I       =   TX               T
                                                                6
                                                                                    back




             Network Coding Aids Tomography
    
         Networkscheme scheme by used by u:x(e), x(e4)=x(e2). 2),
          Routing coding is used is u: x(e3)=x(e1 3)=x(e1)+2x(e
          x(e4)=x(e1)+x(e2).
Probe messages:          x=2
                         x
                                        3+2 . 2x
  M=[1, 2]
                 1
                      e1       3       3    e3
                                                    x
                                                    3
                                                    7
                                                   2x                    =[7, 2]
                                                                       YE=[3, 5]
            s                                            r               =[5,3]
                                                                       YM=[1,2]
                 2             2   u   2            x
                                                    0
                                                    2
                                                    5
                         x                 3+2 x             x[2,1]
                                                             x[1,0]
                                                             x[0,1]
                                                             x[1,1]           =[2,2]
                                                                       E=YE-YM=[2,0]

                      e2                     e4                        e1          e3

        Network coding scheme is enough for r to locate error edge e1.

        Routing scheme is not enough for r to locate error edge e1.
                                                                                           7
    Summary of Contribution
    Passive tomography for random linear network coding

                          WHY?
       turns out Topology estimation Failure localization
    It Failure type that the idea underlying the example
    holds even the coding is done in a random fashion.
                                         Exponential
                        No result        [HLCWK05]
        Adversary
                                  Hardness advantages.
    Random linear network coding has greatproof
        error
                    Exponential
                       [This work]        [This work]
                                          Exponential
    Passive = low      No result
                    overhead.
      Random                          [FM05,HLCWK05]
          error        Polynomial         Polynomial
                      [This work]         [This work]

                                                             8
        Core Concept: IRV
                                                                              0           0

  Edge Impulse Response Vector (IRV):                            2 1




                                                                 [
                                                                 9e1
The linear transform from the edge to the receiver.              6




                                                                 ]
Using IRVs we can estimate topology and locate




                                                                              [
                                                                            0
failures.                                                            3
                                                         2               1 3
                                                                         e3 2
                                                                          3               1




                                                                              ]
                                                                                              1
                                                                                              3
1. Relation between IRVs and network                             2                3                   4
structure:                                                       1                    3
                                                                                      9
                                                                                                  2
IRV(e1) is in the linear space spanned by IRV(e2) and IRV(e3).                                    2
                                                                                                  6




                                                                         [
                                                                         1
                                                                         0   e2
                                                                         0
                                                                                  2 9
                                                                                  1 0         6
                                                                                              0




                                                                         ]
2. Unique mapping from edge to IRV:
For random linear network coding, two independent edges
has independent IRVs with high probability.
                                                                                                          9
    Network tomography by IRVs
   The concept of IRV significantly aids
    network tomography:
       The relations between IRVs and network
        structure is used to estimate network
        topology.
       The unique mapping between network
        edge and its IRV is used to locate network
        failures.
        Topology Estimation for Random Errors

   Why study random failures:

       For network without errors, the only information about
        the network is the transform matrix T. Thus recovering
        network topology is hard [SS08].

       Surprisingly, for network with random failures (errors,
        or packet loss), the IRV of the failure edge will be
        exposed, letting us recovering network topology
        efficiently.
  Topology Estimation for Random Errors

     Stage 1: Collect IRVs
                                       [2,1]
      4 , 2                   0 , 0
                                                  [1,3]
E1=   27 , 15       E2=       3 , 3




                                           [
      18 , 10                 6 , 14        0
                                            3
                                          [3,2]
                                         [1,1]
                                            2




                                           ]
                          [


                          0
      <E1>      <E2>= <   3   >
                          2
                          ]




                                                          10
      Topology Estimation for Random Errors

   Stage 2: Recover topology




                                               [




                                                           [
                                               2           0
                                               9           0
                                               6           4



                               [

                                   [

                                        [
    IRVs from Stage 1:         0    2   0




                                               ]




                                                           ]
                               3    9   0




                                                   [
                                                   0
                               2    6   4          3




                                                           [
                               ]

                                   ]

                                        ]
                                                           0
                                                   2
                                                           0




                                                   ]
                                                           2




                                                           ]
According to: IRV(e1) is in the linear space




                                                   [

                                                       [

                                                               [
                                                   1   0       0
                                                   0   1       0
spanned by IRV(e2) and IRV(e3).                    0   0       1




                                                   ]

                                                       ]

                                                               ]
                  e1

             e2    e3

                                                                   11
    Random Failure Localization                                                           Exp

   Preliminaries: The Impulse Response Vector (IRV) of each edge.
   As long as the topology is given, we can do error localization.




                                                          [
                                                              [
                                                          4 2
                                                          27 15




                                             [
                 [

                     [

                         [

                               [

                                    [

                                         [
                 2   0     0    0   0    0   1




                                                                                  [

                                                                                      [
                                                   in <   18 10 >?




                                                                  [
IRVs:                                                           2 [2,1]           2   0
                 9   3     0    0   1    0   0




                                                          ]
                                                              ]
                                                                  9               9   3
                 6   2     2    4   0    1   0                    6
                                                                                  6   2
                                             ]
                 ]

                     ]

                         ]

                               ]

                                    ]

                                         ]




                                                                  ]




                                                                              [
                                                                                  ]

                                                                                      ]
                                                                              0
                                                                              3
                                                                      [3,2]
        Locating random failures:                                             2




                                                                              ]
             [




                           [




             2             0             4 , 2
        E=   9   [2,1] +   3   [3,2] =   27 , 15
             6             2             18 , 10
             ]




                           ]




                                                                                            12
Summary of our contribution

  Failure type   Topology estimation   Failure localization

                                          Exponential
                     No result            [HLCWK05]
    Adversary
                     Exponential         Hardness proof
       error
                     [This work]          [This work]
                                          Exponential
                     No result
    Random                             [FM05,HLCWK05]
      error         Polynomial            Polynomial
                   [This work]            [This work]
Future direction
   Current work: From existing good network
    codes to tomography algorithms.
   Another direction: From some criteria to new
    network codes.
   For instance, network Reed-Solomon
    code[HS10], satisfies:
       Optimal multicast throughput
       Low complexity and distributed designing.
       Significantly aids tomography:
            Failure localization without centralized topology
             information.
            Adversary localization can be done in polynomial time.
Related works
            Network Coding Tomography for
                   Network Failures

         Thanks!



         Questions?


Details in: Hongyi Yao and Sidharth Jaggi and Minghua Chen, Network
Tomography for Network Failures, under submission to IEEE Trans. on
Information Theory, and arxiv: 0908-0711


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