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2011_Midterm_1_sol

# In the end all 5 data segments have been correctly

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Unformatted text preview: and 2 Mbps, prepare a chart giving the minimum distribution time for each of the combination of N and u for both client ­server distribution and P2P distribution. Solution: For calculating the minimum distribution time for client-server distribution, we use the following formula: Dcs = max {NF/us, F/dmin} Similarly, for calculating the minimum distribution time for P2P distribution, we use the following formula: Where, F = 15 Gbits = 15 * 1024 Mbits us = 30 Mbps dmin = di = 2 Mbps Note, 300Kbps = 300/1024 Mbps. Client Server u 300 Kbps 700 Kbps 2 Mbps 10 7680 7680 7680 N 100 51200 51200 51200 1000 512000 512000 512000 10 7680 7680 7680 N 100 25904 15616 7680 1000 47559 21525 7680 Peer to Peer u 300 Kbps 700 Kbps 2 Mbps Problem 5: Compare GBN, SR, and TCP (no delayed ACK). Assume that the timeout values for all three protocols are sufficiently long such that 5 consecutive data segments and their corresponding ACKs can be received (if not lost in the channel) by the receiving host (Host B) and the sending host (Host A) respectively. Suppose Host A...
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