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Consider CSMA/CD LAN with bandwidth 10 Mbps and propagation delay of 5 μsec. Repeaters are not allowed in this system. Data frames are 512 bits long, including 32 bits of header, checksum and other overhead. The first bit slot after a successful transmission is reserved for the receiver to capture the channel in order to send a 32 bit acknowledgment frame. The effective data rate of the network is ________ (in Mbps). (Excluding overhead, assuming that there are no collisions)
in Computer Networks by Active (4.5k points) | 221 views
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Solution: 6.45 Mbps

Use the formula  effective data rate  ( or throughput ) = BW * efficiency

where efficiency =              (Tt for payload)   /  ( Channel Seize time+ Tt entire packet + Tt ackn + Tp entire packet + Tp ack)

                                            where Channel Seize time = 2*Tp

0
7.45 mbps it should be?

2 Answers

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by Veteran (74.4k points)
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5.68 Mbps.
ago by (383 points)
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