Question

For this problem we are distributing a file of size F = 2 Gbits to each of 6 peers on a peer to peer network. Suppose the server has an upload rate of u = 56 Mbps, and that the 6 peers have upload rates of: Un = 27 Mbps, uz = 21 Mbps, uz = 24 Mbps, 44 = 29 Mbps, u5 = 29 Mbps, 46 = 21 Mbps, and download rates of: dy = 23 Mbps, d2 = 23 Mbps, dz = 31 Mbps, d4 = 12 Mbps, d5 = 24 Mbps, do = 19 Mbps, Answer the following questions: What is the minimum time needed to distribute this file using peer-to-peer download? What is the root case of this specific minimum time?

Answer 1

Solution

1)

The formula for finding minimum distribution time is

D_cs = max{NF/u_s , F/d_min}

N=6

F=2 Gbits

us=56 Mbps

N F/us

=6(2000)/56

=12000/56

=214.28

F/dmin

=2000/12

=166.66

214.28>166.66

=166.66 sec

So the minimum time needed to distribute the file using P2P download

166.66 secs

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2)

N=6

F=2 Gbits

us=56 Mbps

D_P2P = max{F/u_s , F/d_min , NF/(u_s + ∑u_i)}

F/us

2000/56=35.71

F/dmin

2000/12=166.66

NF/(u_s + ∑u_i)

(6*2000)/(56+(27+21+24+29+29+21))

=57.97

=max{35.71,166.66,57.97}

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all the best