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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 clientserver 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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This test prep was uploaded on 03/06/2014 for the course ECE 374 taught by Professor Ganz during the Spring '08 term at UMass (Amherst).
 Spring '08
 GANZ

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