D. ANALYSISThere are several possib

D. ANALYSISThere are several possible reasons for a neighboring node B with higher ranking not to transmit in its corresponding time slot, as observed by node A. Most commonly, B may not have the copy of message due to packet loss. Next, B could retransmit but the packet may not be received by A. B may not be located where assumed by A, causing even disconnection from A. Finally, since each node ranks its neighbors based on its local knowledge, diversity in neighboring knowledge among nodes may lead to inconsistency among nodes may lead to inconsistency among local rankings.For instance, in nodeA’s local ranking(calculated based on A’s neighbor set), A ranks 2nd and B ranks 1st. But in node B’s local ranking, B itself may not rank 1st because it may have different neighbor set. This loop waiting may lead to loss of efficiency since no node devotes itself to transmit in the first time slot after reception.In this theoretical analysis, we assume that no packet loss occurs during transmission, and ignore the mobility of nodes. First, we discuss the loop waiting issue. ReC is loop waiting free, meaning that in a set of forwarder candidates {F 1 ,F 2 ,...,F k }, there exists at least one node F i who ranks itself 1st in its local ranking, and will retransmit immediately after reception of the message. Here we provide a proof for the simplest case of two nodes. For two nodes A and B (both received the message m), we will show that it is not possible that B ranks higher than A in A’s local ranking, while A ranks higher than B in B’s local ranking.