ENC3250 Solutions pg 43

# ENC3250 Solutions pg 43 - M = log2 (N ) 8.25 a. ••• 0...

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Unformatted text preview: M = log2 (N ) 8.25 a. ••• 0 1 2 3 4 5 6 7 0 ••• ••• 0 1 2 3 4 5 6 7 0 ••• ••• 0 1 2 3 4 5 6 7 0 ••• b. c. 8.26 Let t1 = time to transmit a single frame t1 = 1024 bits = 1. 024 m sec 106 bps The transmitting station can send 7 frames without an acknowledgment. From the beginning of the transmission of the first frame, the time to receive the acknowledgment of that frame is: t2 = 270 + t1 + 270 = 541.024 msec During the time t2 , 7 frames are sent. Data per frame = 1024 – 48 = 976 7 × 976 bits Throughput = = 12. 6 kbps 541. 024 × 10− 3 sec -43- ...
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## This note was uploaded on 02/20/2012 for the course ENC 3250 taught by Professor Dr.timothycerner during the Fall '11 term at UNF.

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