EBP: THE ENERGY BALANCE PROTOCOL in Visual Studio .NET

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EBP: THE ENERGY BALANCE PROTOCOL
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Proof. Let i,j an indicator random variable that is equal to 1 if area Ti forwards the message j to the area Ti 1 and 0 otherwise. Thus i,j = 1 0 with probability pi with probability 1 pi
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Clearly, i,j depends only on i, but we add j for counting purposes. Obviously, E[ i,j ] = pi . It is
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Similarly to the proof of Lemma 9, we get
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hi+1 j=0
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P{h i+1,j hi+1 = n I i+1 = n}
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E[fi ] =
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and the proof is completed. Recall that, according to De nition 15.5.8, to achieve the same on the average energy dissipation per area unit (and thus per sensor) in the network area, the following equality should hold:
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i, j {1, . . . , n} (15.5)
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E k=1 Si
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that is, the average energy consumption per sensor should be equal in any two ring sectors. By induction, it suf ces to guarantee this for any two adjacent sectors. In what follows, we guarantee the above balance property, requiring a certain recurrence relation to hold. This recurrence basically relates three successive terms of the E[fi ] sequence (the E[gi ] terms depend only on i and on input parameters). Theorem 15.5.1. To achieve energy balance in the network, the following recurrency equation should hold: ai+1 E[fi+1 ] (di + ai )E[fi ] + di 1 E[fi 1 ] = ai E[gi ] ai+1 E[gi+1 ]
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where ai =
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di =
(i+1)2 1 2i+1
Proof. For the case j = i + 1 of Eq. (15.5) and using Lemmas 7 and 9, we have E[hi ]E[ i,j ] E[hi+1 ]E[ i+1,j ] = Si Si+1 E[hi ][i2 pi (i2 1)] E[hi+1 ]{(i + 1)2 pi+1 [(i + 1)2 1]} = (2i 1) (2i + 1)
i2 i2 1 (i + 1)2 1 (i + 1)2 E[hi ] pi E[hi ] = E[hi+1 ] pi+1 E[hi+1 ] 2i 1 2i 1 2i + 1 2i + 1 Let ai , and di be as de ned in the theorem statement above. By Lemma 11 we know that pi E[hi ] = E[fi 1 ], and by Lemma 10 it is E[hi ] = E[gi ] + E[fi ]; thus the last equation becomes ai+1 E[fi+1 ] (di + ai )E[fi ] + di 1 E[fi 1 ] = ai E[gi ] ai+1 E[gi+1 ] To solve the above recurrency we must compute E[gi ]. Lemma 12. If N is the total number of events that are generated in the network, the mean value of gi is given by the following relationship: E[gi ] = N i Proof. Because the position of each event is independent of other events and because for each sector i, probability i is the same, clearly gi is binomial with parameters N and i . In order to have a simpler recurrence involving only two (successive in fact) terms of the E[fi ] sequence, we will transform the recurrency relation of Theorem 15.5.1. into the following (easier to solve) relation: Lemma 13. The recurrency relation ti ti 1 = ai E[fi ] ai+1 E[fi+1 ] t0 = a1 E[f1 ] has as a solution the function for i = 1, . . . , n 1 and
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ti =
aj E[gj ] aj+1 E[gj+1 ] + a1 E[f1 ]
Proof. The proof is done by induction on i. For i = 0, it is obviously true. Let it be true for i 1. For i we have ti = ti 1 + ai E[gi ] ai+1 E[gi+1 ] By the induction hypothesis, we get the solution
ti =
aj E[gj ] aj+1 E[gj+1 ] + a1 E[f1 ]
Now the recurrency relation of Theorem 15.5.1. is simpli ed: ai+1 E[fi+1 ] di E[fi ] = ti , i = 1, . . . , n 1
Thus, we get a recurrence for sequence E[fi ] involving only two successive terms of the sequence. Theorem 15.5.2. The recurrency relation ai+1 E[fi+1 ] di E[fi ] = ti , i = 1, . . . , n 1
where ti is de ned in Lemma 13, has the following solution:
E[fn i ] =