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Tracking Mobile Users in Wiewless Communications NetworksPowerPoint Presentation

Tracking Mobile Users in Wiewless Communications Networks

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### Tracking Mobile Users in Wiewless Communications Networks

Amotz Bar-Noy and Ilan Kessler

IEEE Trans. On Info. Theo. VOL.39,NO.6,Nov 1993,p1877-1886

Speaker : Cheng-Chung,Li

Outline

- Introduction
- Problem Defined
- Weighted Graph
- Lines
- Trees
- Arbitrary Graphs(Approx.Algorithms)

- Discussion

Introduction

- A important issue in wireless networks is the design and analysis of tracking mobile users
- The users are mobile and could be anywhere within network area

- In this paper , the issue considered is the cost of utilizing the wireless links for the tracking mobile users in cellular networks
- Update
- Find

- Two extreme strategies
- Always-Update
- Never-Update

- But how to minimized the total (update+find) cost ?
- Increasing one cost leads to a decrease in the other one

- We will construct a model called Mobility Graph and defined the problem called reporting center problem , and try to solve above question

Problem Defined the problem called reporting center problem , and try to solve above question

- The mobility graph G of the network is the graph in which each vertex corresponds to a different cell , and two vertices are connected by an edge if and only if the corresponding cells overlap
- Each vertex i of the mobility graph has a weight wi > 0

1 the problem called reporting center problem , and try to solve above question

2

3

cell

Mobility graph

6

5

4

9

7

8

2,4,6,8 are reporting centers

- Let I be a set of vertices , referred to as centers . The the problem called reporting center problem , and try to solve above questionvicinity of center v is the set of all vertices not in I that are reachable from v by a path containing no centers
- By definition , the vicinity of center v includes v
- The weight of I is w(I)=iIwi,and the size of the largest vicinity in the graph is denoted by z(I)

- The the problem called reporting center problem , and try to solve above questionReporting Centers Problem-C(G,Z):Given a weighted graph G and integer Z , select a set of centers S such that z(S)Z and w(S)w(S’) for all S’ such that z(S’)Z
- We are so greedy ! We want to find the min.(update+find) solution

- An import special case of C(G,Z) is the case un which all the weighted are equal to one – But we just concert weighted graph in this report
- C(G,Z) is an NP-Complete problem for any Z2

Weighted Graph-Lines the weighted are equal to one

- Given an integer 1<Z<n , the goal is to find a set of centers S such that the following hold
- (a)The largest vicinity contains at most Z vertices
- (b)w(S)=minI{1,-,n}{w(I)|(a)holds for I}

- We denote this problem by C(n,Z)

1

2

n

wn

w1

w2

- The modified problem is to find for a given integer 0 the weighted are equal to one kZ a set of centers Sk , such that the following hold
- (a)the set Sk contains the vertex n-k and does not contain the vertices n-k+1,…,n
- (b)the largest vicinity contains at most Z vertices
- (c)w(Sk)= minI{1,-,n}{w(I)|(a)and(b)holds for I}.

- We denote the modified problem by Ck(N,Z)

1

2

n-k

n

n-k+1

- Clearly , at least one of the sets {S the weighted are equal to one k,k=0,1,…,Z-1} is a solution to C(n,Z)
- Let k’ be an index for which w(Sk’)=min0k<Zw(Sk) . Then Sk’ is a solution to C(n,Z)

- For every i=0, the weighted are equal to one …,Z-1 and j=1,…,n-1 , letSi(j) to be a solution to Ci(j,Z)
- For every 0k<Z , let rk be an index for which w(Srk(n-k-1))=min0r<Z-kw(Sr(n-k-1))
- Then Sk(n)=(n-k)Srk(n-k-1) is a solution to Ck(n,Z)

2

1

n-k

n

n-Z

n-k-1

n-k+1

- For all 0 the weighted are equal to one iZ-1 and Z-i <jn-iw(Si(i+j))=min{wj+w(SZ-i-1(j-1)),w(Si+1(j+i+1))}
- The above algorithms can solve C(n,Z) in O(nZ) time

Z values

1

2

j

j+i

j-Z+1

j-1

j+1

Weighted Graph-Trees the weighted are equal to one

- We describe the simple binary tree T first
- For any set of vertices I in the tree T , consider the connected components that are obtained when all vertices of I are removed from T
- We denote by a(I) the size of the connected component that contains the root of T , i.e. a(i)=0 if I contains the root

- The modified problem is to find for given nonnegative integers k and l such that k+lZ-1 , a set of centers Slk such that the following hold
- a.a(Slk)=k
- b.the largest vicinity contains at most Z vertices
- c.the largest vicinity that contains the root has at most Z-l vertices (l:external vertices)
- D.w(Slk)=minIT{w(T)|a,b,c hold for I}

- a.For every l integers k and l such that k+lZ-1 , let i’ and j’ be indexes for which w(S0i’(TL))+w(S0j’(TR))=mini+j+lZ-1 w(S0i(TL))+w(S0j(TR))Then Sl0(T)={r}S0i’(TL)S0j’(TR) is a solution to Cl0(T,Z)
- b.For every k>0 and l 0 such that k+lZ-1 , let i’ and j’ be indexes for which w(S1+l+j’i’(TL))+w(S1+l+i’j’(TR))=mini+j+1=kw(S1+l+ji(TL))+wS1+l+ij(TR))Then Slk(T)=S1+l+j’i’(TL) S1+l+i’j’(TR) is a solution to Clk(T,Z)

- Each min operation in the equation a. can be done by O(Z integers k and l such that k+l2) operations , and it computed for at most Z values of k
- Each min operation in the equation b. can be done by O(Z) operations , and it computed for at most Z2 values of k and l
- So the above solution for C(T,Z) is O(nZ3)
- If T is an arbitrary tree , the solutions of T are computed after computing the solutions for all subtrees of T , time complexity is sill O(nZ3)

Weighted Graph-Arbitrary Graphs integers k and l such that k+l

- Initially, all vertices are designated as centers
- Then the centers are checked in an order of decreasing weights , and if making a center a noncenter vertex does not create a vicinity larger than Z , then this center is made a noncenter vertex

Approx.Algorithms integers k and l such that k+l

- Let the vertices of the graph be denoted by 1,…,n and w.l.og. Assume that w1w2…wn
- 1.S={1,…,n}
- 2.x=1
- 3.If z(S-x)Z then S=S-{x}
- 4.x=x+1
- 5.If xn the go to step 3
- 6.Return the set S

- Two centers are said to be integers k and l such that k+lsiblings if their vicinities are not disjoint
- Let is the maximum degree of the graph
- Each center has at most Z sibilings

- For any greedy center x integers k and l such that k+lS, there exists an opt. center yR such that wywx , and x is either in the vicinity of y or in the vicinity of a sibling of y (where the vicinities are with respect tp the opt. set R)
- We can prove it by contradiction

- integers k and l such that k+lxSwxZ2yRwy
- Let x be a greedy center and let y(x) to be an opt. center such that wy(x)x and x is either in the vicinity of y(x) or x is in the vicinity of a sibling of y(x)
- So xSwxxSwy(x)
- But the largest vicinity has at most Z vertices , that each opt. center appears at most Z2 times
- Therefore xSwxZ2yRwy

Discussion integers k and l such that k+l

- As for weighted graphs , important special cases other than trees could be almost-trees and planar graphs
- Also we believe there exists a better approx. algorithm for arbitrary graphs

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