**I need help with Part B. Please write clearly!***

A server has to distribute a file of size F = 8 Gbits to each of 10 peers. Suppose the server has an upload rate of u = 77 Mbps, and that the 10 peers have upload rates of:

u1 = 17 Mbps, u2 = 18 Mbps, u3 = 18 Mbps, u4 = 18 Mbps, u5 = 12 Mbps, u6 = 20 Mbps, u7 = 16 Mbps, u8 = 12 Mbps, u9 = 17 Mbps, u10 = 12 Mbps,

and download rates of:

d1 = 19 Mbps, d2 = 10 Mbps, d3 = 14 Mbps, d4 = 20 Mbps, d5 = 17 Mbps, d6 = 20 Mbps, d7 = 20 Mbps, d8 = 32 Mbps, d9 = 16 Mbps, d10 = 15 Mbps.

Peer 1 Peer 2 Peer 3 Peer 4 Peer 5 Upload Rate 17 Mops 18 Mbps 18 Mbps 18 Mbps 12 Mbps 20 Mbps 16 Mbps 12 Mbps 17 Mbps 12 Mbps Download Rate 19 Mbps 10 Mbps 14 Mbps 20 Mbps 17 Mbps 20 Mbps 20 Mbps 32 Mbps 16 Mbps 15 Mbps Peer 6 Peer 7 Peer 8 Peer 9 Peer 10 a. What is the minimum time needed to distribute this file from the central server to the 10 peers using the client-server model? Size of File = F = 8 Ghita 8* 10 bits Server needs 10 copies of the file total upload size = 10F bits Server upload rate = u = 77 Mbps ~ 77 * 10 bps Minimum time needed to upload the files = 10F/u = (10*8*109)/(17* 109) = 1038.961039 seconds = 1039 seconds Slowest server's receive rate = duin = min{d1d2.d3.d4 d5,06,07,08,09,110} = 10 Mbps + 10 * 10 bps Minimum time needed = Fdoia=(8 * 109/10 * 10) = 800 seconds Minimum time needed to distribute the file using Client-server model=max(10F/u, F du = max(1039s, 800s) = 1039 seconds b. What is the minimum time needed to distribute this file using peer-to-peer download? Peer 1 Peer 2 Peer 3 Peer 4 Peer 5 Upload Rate 17 Mops 18 Mbps 18 Mbps 18 Mbps 12 Mbps 20 Mbps 16 Mbps 12 Mbps 17 Mbps 12 Mbps Download Rate 19 Mbps 10 Mbps 14 Mbps 20 Mbps 17 Mbps 20 Mbps 20 Mbps 32 Mbps 16 Mbps 15 Mbps Peer 6 Peer 7 Peer 8 Peer 9 Peer 10 a. What is the minimum time needed to distribute this file from the central server to the 10 peers using the client-server model? Size of File = F = 8 Ghita 8* 10 bits Server needs 10 copies of the file total upload size = 10F bits Server upload rate = u = 77 Mbps ~ 77 * 10 bps Minimum time needed to upload the files = 10F/u = (10*8*109)/(17* 109) = 1038.961039 seconds = 1039 seconds Slowest server's receive rate = duin = min{d1d2.d3.d4 d5,06,07,08,09,110} = 10 Mbps + 10 * 10 bps Minimum time needed = Fdoia=(8 * 109/10 * 10) = 800 seconds Minimum time needed to distribute the file using Client-server model=max(10F/u, F du = max(1039s, 800s) = 1039 seconds b. What is the minimum time needed to distribute this file using peer-to-peer download