Giai đoạn 2: ti - 1, cho nó downto 0 Sieve (i ), Giai đoạn 3: lặp lại Sieve (0); mãi mãi Bây giờ chúng ta quay về với nhiệm vụ đánh giá số lượng khe thời gian nó có giao thức chấm dứt. Trong giai đoạn 1, một khi tình trạng của kênh là NULL thoát ra giao thức cho vòng lặp. | NONUNIFORM LEADER ELECTION PROTOCOL 235 With this preamble out of the way we are now in a position to spell out the details of our leader election protocol. Protocol Nonuniform-election initially all the stations are active Phase 1 for z 0 to 00 sieve z exit for-loop if the status of the channel is NULL Phase. te-i- 1 for i 1 downto 0 Si eve z Pha-e 3 repeat s ieve o forever wp now turn to the task of evaluating the number of time slots it takes the protocol to terminate. In Phase 1 once the status of the channel is NULL the protocol exits the for loop. Itius thoremost exist an integer t such that the status of the channel is r SINGLE or COLLISION in Sieve O sieve l sieve 2 . . . sieve - 1 NULL in s i eve i ló t Pt 1 be an arbitrary real number. Write u Iiog1og 4n 1 Equation guaraneee- that 21s Vgf. Assnmitha Sgave O sieve l . . sieve are performed in Phase 1 and let A be the random variable denoting the number of stations that transmitted msievc t. SuppoerthatweCarert most n ectivertationt. and sieve e isrerformed. Let X denote the number of stations that transmits in sieve . Clearly the expected value EIA ofAfis _ n n 1 V 2 4 ị 4 Using the SIork4v inequality and wecga vgute v A i ptyA 4 4 v p . v . . Equation 15 guarantees that with probability at least 1 - 1 4 the status of the channel in sieve is NULL. In particular this mew that 236 LEADER ELECTION PROTOCOLS FOR RADIO NETWORKS . 1 t s holds with probability at least 1 - and therefore Phase 1 terminates in t 1 s 1 Flog log 4n 1 log log n O log log time tum thrsimpliesthatPhase2 in log log n O log log time slots. Thus we have the following result. . . . e. . Lemma With probability exceeding 1 - 1 4 Phase 1 and Phase 2 combined take at most 2 log log n O log log time slots. Recall that Phase 2 involves t calls namely sieve - 1 sieve - 2 . . Sieve O . For convenience 0 1 the analysis we regardthelast call si eve of Phase 1 as .
