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Unformatted text preview: Transmit time = (500 bytes)(8 B/bit)/10 Mbps = 4000 bits/10 Mbps = 400 usec Cycle time = 30,000 usec + 400 usec = 30,400 usec. Only one segment can be sent per cycle time, So data rate is 4000 bits/30,400 usec < 132 kbps 2. GBN ARQ requires a send window no greater than 2 n-1, where n is the number of bits in the sequence numbers, and a receive window of size 1. The number of segments needed to fill the pipeline is then Cycle time/transmit time 30,400/400 = 76 So n >=ceil( log 2 (76+1)) = 7 bits....
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This note was uploaded on 01/17/2012 for the course CNT 5106c taught by Professor Helmy during the Spring '09 term at University of Florida.
- Spring '09