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Page 2 Lecture Notes in Computer Science 2865 Edited by G. Goos ...

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42 L. Q<strong>in</strong> and T. KunzP = Prdlog( ) + X σ )d 010( 10β2PG G λ dt t r0 β*10= * ( ) * 10r 0 2 2(4π) d L d0− (4)Because the mean of X σ is zero, if we set P rto the threshold P s, we will f<strong>in</strong>d a meantransmission range for different β values with Equation (4). Accord<strong>in</strong>g to the parametersused <strong>in</strong> NS2, the radio frequency is 914MHz, P t=0.281838 W, L=1, and thresholdP s=3.652x10 -10 W. The reference distance d 0is 1 m. The result<strong>in</strong>g average transmissionranges are shown <strong>in</strong> Table 2.Table 2 shows that for a transmission range equivalent to the ideal model, which is250 m, β=2.385. β values less than these correspond to a longer mean transmissionrange than ideal model, bigger values correspond to a shorter mean transmissionrange. To fairly compare the performance, we should conduct simulations under thesame or similar conditions. If we still use the same number of nodes <strong>in</strong> the same areafor higher β values, some nodes may be isolated. Bettstetter [13] proposed an algorithmfor obta<strong>in</strong><strong>in</strong>g the m<strong>in</strong>imum radio transmission range for a homogeneous nodedensity so that every node <strong>in</strong> the network is connected. But <strong>in</strong> NS2 simulations, arandom waypo<strong>in</strong>t model is used, and this model does not result <strong>in</strong> a uniform nodedistribution, so this algorithm cannot be applied. We apply a simple rule for fair comparison:for two different simulations with different transmission range, the total nodecoverage for a certa<strong>in</strong> area should be equal, i.e:πrn1w l211 1= n2πrw lwhere r 1,r 2are average radio transmission ranges, w 1, w 2and l 1,l 2are width and lengthof the simulation area, and n 1,n 2are number of mobile nodes respectively. Based onour simulations with an ideal radio model with 50 nodes <strong>in</strong> a 1500x300 m area withtransmission range of 250m, we calculate the required number of nodes for different βvalues <strong>in</strong> Table 2.Table 2. Mean Transmission Range and Required Nodes for Different Beta Values.β 2.0 2.1 2.2 2.3 2.385 2.4 2.5 2.6 2.7 2.8 2.9 3.0r(m)725 530 398 307 250 242 194 158 131 110 93 80#nodes 6 12 20 34 50 54 84 126 183 259 362 489222 20.1Xσ(5)3.3.2 Improv<strong>in</strong>g Packet Delivery RatioIn the shadow<strong>in</strong>g model, signal power strength fluctuates and exist<strong>in</strong>g routes maybecome unstable if some l<strong>in</strong>ks of a route are on the edge of the radio transmissionrange. The results <strong>in</strong> the previous sections also demonstrate that even a route discovereda very short time ago still might be assumed broken when the Route Reply isgo<strong>in</strong>g to be sent. So if nodes can have more stable routes that resist to the fluctuationthe route will live longer and a source node does not have to f<strong>in</strong>d a new route frequently,which <strong>in</strong> turn will <strong>in</strong>crease the chance to successfully deliver data packets.We can divide the stable route requirement <strong>in</strong>to two parts. First we try to f<strong>in</strong>d somestable routes, second we have to ma<strong>in</strong>ta<strong>in</strong> these routes because nodes are mov<strong>in</strong>grandomly. In this paper, we solve the first part and will give some suggestion for the

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