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TCP: maximum throughput achievable for end-user machine

An end-user machine running TCP sends full windows of 65,535 bytes over a 1-Gbps channel that has a 10 msec one-way delay. What is the maximum throughput achievable?

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Round Trip Time = 2 * one-way delay = 2 * 10 msec = 20 msec

Which ...

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Solution shows all the steps of computation along with needed explanation.