Analysis of the Batch Algorithm Sample Clauses

Analysis of the Batch Algorithm. Since the actual per- formance of the Batch algorithm depends on the membership leave positions whenever L > 0, we consider only the expected E[RQueue bat h℄ > (N N=2 ) >8 0; if J = 0, L = N h 1 l performance measures rather than the exact ones, although n the exact and expected results are the same when L = 0. Our analysis adopts the convention that the combination equals = ><> Pl=0 2 1 L > h 1 l L
Sample 1
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Analysis of the Batch Algorithm. In analyzing the performance of the Batch algorithm, we consider the following five cases. Note that when , the performance metrics will depend on the membership leave positions and exact metrics cannot be obtained. Therefore, whenever , we derive the expected performance mea- sures. We also define and be the two perfor- xxxxx measures under condition . We also adopt the con- vention that the combination equals 0 if , or . Due to limited space, readers can refer to [5] for detailed mathematical derivation and results.
Sample 1
Analysis of the Batch Algorithm. Since the actual per- formance of the Batch algorithm depends on the membership leave positions whenever L > 0, we consider only the expected E[RQueue bat h℄ >
Sample 1
Analysis of the Batch Algorithm. When , the performance metrics will depend on the membership leave positions and exact metrics cannot be obtained. Therefore, whenever , we derive the ex- pected performance measures. Due to limited space, read- ers can refer to [5] for detailed mathematical derivation and results.
Sample 1
Analysis of the Batch Algorithm. Since the actual per- formance of the Batch algorithm depends on the membership E[RQueue−batch] leave positions whenever L > 0, we consider only the expected performance measures rather than the exact ones, although r  0,  h 1 2l  Σ − Σ (N−N/2l) Σ L L − L, if J = 0, 0 ≤ L< N if J = 0, L = N . Σ the exact and expected results are the same when L = 0. Our analysis adopts the convention that the combination n equals 0 if n < 0, r < 0, or n < r. = l=0 1 − L l=0 2 1 − Σ  Σh−1 l Σ (N) (N−N/2l) L (N) − (L − 1), otherwise. (8) We first derive the expected number of renewed nodes. Consider a node v at level l. In a completely balanced tree, node v has N/2l descendants. At a rekeying instance, node v can be in one of the three distinct states: no-change, pruned, and renewed. Node v can be in the “no-change” state if none of its N/2l descendants wish to leave. Thus, the probability of node v being in the “no-change” state is . Σ .N −N/2lΣ N
Sample 1
Analysis of the Batch Algorithm. Since the actual per- formance of the Batch algorithm depends on the membership E[RQueue−batch] r leave positions whenever L > 0, we consider only the expected performance measures rather than the exact ones, although the exact and expected results are the same when L = 0. Our analysis adopts the convention that the combination n equals 0 if n < 0, r < 0, or n < r.  0,  h 1 2l =  Σl=−0 l=0 2  Σh−1 l 1 − 1 − (N −N/2l) (N ) L L L L (N −N/2l) (N ) if J = 0, L = N − L, if J = 0, 0 ≤ L<N − (L − 1), otherwise.
Sample 1

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