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ECE578-Solution08-2007 - ECE 478/578 Fundamentals of...

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Solution to Assignment #8 P1. (a) The advertised window should be large enough to keep the pipe full; delay (RTT) × bandwidth here is 100 ms × 100 Mbps = 10 Mb = 1.25 MB of data. This requires 21 bits (221 = 2, 097, 152) for the AdvertisedWindow field. The sequence number field must not wrap around in the maximum segment lifetime. In 60 seconds, 750 MB can be transmitted. 30 bits allows a sequence space of 1024 MB, and so will not wrap in 60 seconds. (b) The bandwidth is straightforward from the hardware; the RTT is also a precise measurement but will be a ected by any future change in the size of the network. The MSL is perhaps the least certain value, depending as it does on such things as the size and complexity of the network, and on how long it takes routing loops to be resolved. P2. If every other packet is lost, we transmit each packet twice. (a) Let E 1 be the value for EstimatedRTT, and T = 2 × E be the value for TimeOut. We lose the first packet and back o TimeOut to 2 × T . Then, when the packet arrives, we resume with EstimatedRTT =
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