Question

[Analytical Question] P2P. We are distributing a file of size F=20Gbits to N peers. The server has an upload rate of u_s=20Mbps, and each peer has a download rate of d_i=1Mbps and an upload rate of u. For N=100 and u_i=2Mbps, What is the chart providing minimum distribution time for both client-server and P2P distributions.

Group of answer choices

1.

D_cs = Max {NF/u_s, F/d_min} = Max {100x20x10⁹/20x10⁶ , 20x10⁹/10⁶} = 100000s
D_P2P = Max{F/u_s, F/d_min, NF/(u_s+ ∑ u)
D_P2P= Max{20x10⁹/20x10⁶ , 20x10⁹/10⁶, 100x20x10⁹/(20x10⁶+100x2x10⁶)} = 20000s

2.

D_cs = Max {NF/u_s, F/d_min} = Max {20x100x10⁹/20x10⁶ , 20x10⁹/10⁶} = 100000s
D_P2P = Max{NF/u_s, F/d_min, NF/(u_s+ ∑ u)
D_P2P= Max{100x20x10⁹/20x10⁶ , 20x10⁹/10⁶, 100x10⁹/20x10⁶+100x10⁶} = 909s

3.

D_cs = Max {F/u_s, F/d_min} = Max {20x100x10⁹/20x10⁶ , 20x10⁹/10⁶} = 1000s
D_P2P = Max{F/u_s, NF/d_min, NF/(u_s+ N∑ u)
D_P2P= Max{20x10⁹/20x10⁶ , 20x10⁹/10⁶, 100x20x10⁹/20x10⁶+100x10⁶} = 1000s

4.

D_cs = Max {F/u_s, F/d_min} = Max {20x100x10⁹/20x10⁶ , 20x10⁹/10⁶} = 1000s
D_P2P = Max{NF/u_s, F/d_min, NF/(u_s+ ∑ u)
D_P2P= Max{20x10⁹/20x10⁶ , 20x10⁹/10⁶, 100x20x10⁹/20x10⁶+100x10⁶} = 10000s

Answer 1

Option 1 is Correct

Explanation:

Given :

F=20Gbits = 20 * bits

u_s=20Mbps = 20 * bps

d_i=1Mbps = 1 * bps

u_i=2Mbps = 20 * bps

N=100

Formula to calculate minimum distribution time for client-server distribution is:

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

Substituting the values in the formula we get:

D_cs = Max {100x20x10⁹/20x10⁶ , 20x10⁹/10⁶}

D_cs = Max { , 2 * }

D_cs = Max(100000 , 20000}

D_cs = 100000s

Formula to calculate Peer to Peer distribution:

D_P2P = Max{F/u_s, F/d_min, NF/(u_s + ∑ u)

D_P2P = Max{20x10⁹/20x10⁶ , 20x10⁹/10⁶, 100x20x10⁹/(20x10⁶+100x2x10⁶)}

D_P2P = 20000s