HW 3 Sol - ECE 445 Computer Networks and Telecommunications...

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ECE 445: Computer Networks and Telecommunications Fall 2007 HW Solution 3 — Due: October 16 Professor: Stephen Wicker TA: Prapun Suksompong 4.2 [5 pt] Recall that the throughput of pure ALOHA is 1 2 e . Hence, the effective data rate is 56 × 1 2 e = 10300 . 6 [bps]. Each station requires 1000 100 = 10 [bps]. So, the maximum value of N is ± 10300 . 6 10 ² = 1030 . 4.3 [5 pt] Pure ALOHA has less delay. With pure ALOHA, transmission can start in- stantly. At low load, no collisions are expected so the transmission is likely to be successful. With slotted ALOHA, it has to wait for the next slot. This introduces half a slot time of delay. 4.5 [10 pt] See p. 254–255 of the textbook. Here, G = 50 × 40 × 10 - 3 = 2 requests per slot time. Then, the number of requests X for each slot is a Poisson random variable with parameter G . Consider the transmission of a test frame. (a) The probability that a test frame will avoid a collision is the probability that all the other users are silent in that slot which is
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This homework help was uploaded on 02/20/2008 for the course ECE 4450 taught by Professor Wicker during the Fall '07 term at Cornell.

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HW 3 Sol - ECE 445 Computer Networks and Telecommunications...

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