# Load Balancing in Ad Hoc Networks: Single-path Routing vs. Multi-path Routing

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3 1.8 B.6.4. W W..4 ig. 4. Region C B ig. 3. The rectangle model.. The Rectangle Model Consider nodes and B in the network. If the network is dense, the shortest path between these two nodes will be very close to the line segment B. Under multi-path routing, the first K shortest paths will be used. We can expect that these paths form K parallel lines from to B. To see this effect we run some simulations. igure 3 depicts an example. We can see that the set of nodes used in these paths approximately from a rectangle. This assumption is more accurate for a dense network. The length of this rectangle is the same as distance between and B. However, its width mainly depends on number of paths, node density, and the way paths are selected. We denote the width of this rectangle by W, and assume that it is independent of the position of nodes and B. In Section IV-D, the selection of this parameter is discussed in more details. W ig. 5. Region. B. Locus Problem Consider two fixed points and. We want to find the set of all points B such that point is inside the rectangle created by points and B, as described above. This set of points corresponds to the nodes that under multi-path routing policy, part of their traffic destined to node, passes through node. We need to consider two separate cases. Let d denote the distance between points and. 1) irst Case: d >W: ll points B should satisfy two conditions. Distance from point to the line segment B should be less than W and the projection of point on line B should lie between and B. These two conditions leads to the following two constraints: 1) Consider a circle with radius W and center.wedraw the two tangent lines from point to this circle. These two lines form the shaded region which is shown in igure 4. ll points B should lie inside this region. To show that this is a necessary condition, consider a point B outside this region. Clearly the distance from point to the line B will be larger than W, and ig. 6. Locus of all points B for the first case. therefore, point can not be inside the rectangle formed by points and B. ) Consider a circle with line segment as its diameter. ll points B should be outside this circle. They should also be inside the half space created by the line normal to. This is the shaded region shown in igure 5. Consider an arbitrary point B, and let C denote the point where line B intersects with the circle. The angle C ˆ is 9 degrees (since is the diameter), so C is the projection of on line B. IfB is outside the circle, then C lies between and B. Combining the above two regions, we find the locus of all points B which is shown in igure 6. ) Second Case: d W : When d is less than W, point will lie inside the circle with center and radius W, so we can not draw the tangent lines. It is easy to see that, in /4/\$ (C) 4 IEEE IEEE INOCOM 4

4 S(,) Traffic λ K R π π δs(, ) δr dφ dr, (1) where (r,φ ) shows the position of point in polar system. Recall that λ represents the amount of traffic generated by each flow. We note that, since K paths are used to transmit each flow, in the above equation, λ is first normalized by K. We use numerical methods to compute the above expression. D. How to choose W? s mentioned before, W is a very important parameter. Clearly this parameter depends on the method used to find paths. irst, we consider the case where only the shortest path between the nodes is computed, and we find an approximate expression for W. Then we generalize it to the case of multiple paths. Since it is assumed that the nodes are distributed uniformly and their positions are independent of each other, the number of nodes in a fixed region with area will have a Poisson distribution [5] with mean δ: S(,) ig. 7. Definition of S(, ). this case, the locus of points B is same as Region shown in igure 5. It is formed by excluding all points inside the circle with diameter from the half space created by a line perpendicular to line. C. Definition of S(, ) and Traffic Computation Consider two fixed points and which are now restricted to be inside a circle with radius R. Then, find the locus of points B as described in previous section and intersect it with the circle. Two examples corresponding to the two different cases, described in the previous section, are shown in igure 7. Later, we use the area of this region to compute the traffic load on different nodes. We denote this area by S(, ). Note that S(, ) depends on position of point and and parameter W. This function can be computed using analytical methods. Unfortunately, the resulting answer has a very complex closed form expression. In the ppendix, we describe an algorithmic method to compute this area. If we multiply this area by the density of nodes (δ), we get the number of nodes that have the property that part of their flow destined to a node at point, will pass through a node at point. Thus, to compute the total traffic passing through node, we need to consider all different positions of point, and sum up δs(, ) computed for each position of point. Mathematically, we can express this as follows: δ (δ)k P (number nodes in area = k) =e. () k! Consider two points and B, such that their distance is T + ɛ, where ɛ is very small positive number. Recall that T denotes the transmission range of each node. Consider the rectangle over points and B with width W. If the shortest pate lies within this rectangle, then we need to have at least one point in it. The average number of points in the rectangle is δwt. If this average is in the order of 1, (e.g., we choose 5) then the probability of not getting any node will be very small: P(No node) =e δwt (= e 5 =67), (3) and therefore W should be: W = O(1) δt. (4) This shows that W is inversely proportional to density of nodes, δ, and the communication range, T. If for the single shortest path, W 1 = W is chosen, we assume that for multipath case when K shortest paths are used, W K KW.This assumption is justified by simulation results. The value of W also depends on the path discovery algorithm. or example, in case of node-disjoint paths, simulations shows that W is typically larger than the case where linkdisjoint paths are used. However, the assumption that W scales linearly with the number of paths and is independent of the distance between two points, is valid for different rout discovery methods /4/\$ (C) 4 IEEE IEEE INOCOM 4

6 R S y C ig. 1. Computation of S(, ). [7] N.S.V. Rao and S.G. Batsell. QoS routing via multiple paths using bandwidth reservation. In IEEE INOCOM 98, volume 1, pages IEEE, [8] R. Rom, I. Cidon, and Y. Shavitt. nalysis of multi-path routing. IEEE/CM Transactions on Networking, 7(6): , [9]. Tsirigos and Z.J. Haas. Multi-path routing in the presence of frequent topological changes. IEEE Communications Magazine, November 1. [1] L. Wang et al.. Multi-path source routing in wireless ad hoc networks. In Canadian Conf. Elec. Comp. Eng., volume 1, pages ,. PPENDIX In this section, we describe an algorithmic method for computation of S(, ). Let us assume that and are respectively located at (x,y ) and (x,y ). Then: B S BC d = (x x ) +(y y ). (5) irst, consider the case in which W d. The angle α = BC shown in igure 1 is given by: ( ) W α =sin 1, (6) d and the slope of line is: m = y y. (7) x x We need to compute the coordinates of points B and C. Note that the slope of B and C can be obtained from: x m B = m + tan ( ) α 1 m tan ( α ), (8) m C = m tan ( ) α 1+m tan ( α ). (9) The coordinates of point B are given by: B = ( 1+m B) R (m B x y ), (1) ( ) m x B = B x m B y ± B, (11) 1+m B y B = m B (x B x )+y. (1) The same set of equations can be used to compute the coordinates of point C. Note that in (11) the plus or minus sign depends on the position of point and. We should make sure that the correct sign is used. Since the position of the three points,b, and C are known, we can compute the area of the triangle BC: S BC = 1 [x (y B y C ) x B (y y C ) (13) + x C (y y B )]. (14) To compute S(, ) we also need the area of the region shown in figure as S BC. To find that we need β = BOC: ( ) β =sin 1 dbc. (15) R Then S BC is: S BC = R (β sin(β)). (16) The only other parameter needed is the area S : ( ) d S = (α +sinα). (17) Combining all these results we obtain: S(, )=S BC + S BC S. (18) or the case d <W, we can just set α = π. Note that S BC will be zero in this case /4/\$ (C) 4 IEEE IEEE INOCOM 4

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