Consider distributing a file of to peers. The server has an upload rate of mu_s=

Question

Consider distributing a file of to peers. The server has an upload rate of mu_s=30 Mbps, and each peer has a download rate of d_i=2 Mbps and an upload rate of mu. For N=10,100,1000 and mu=300 Kbps, 700 Kbps, 2 Mbps, prepare a chart giving the minimum distribution time for each of the combinations of N and mu for both client-server distribution and P2P distribution.

I need help with finding the minimum distribution time for P2P. For D_P2P, I need help understanding how to find NF/(u_s + summation of u_i with N and i = 1. I found an answer on Chegg, which shows me the answer to the problem. However, it does not explain how they got the summation value. For example, for N = 10, they have the equation as (15*1024)(10)/(30 + 3). If this is right, how did they get 3? I've found many other sites with answer keys to this type of question, but none of them show how they got the summation value. Therefore, I'm now on here hoping to find out.

Explanation / Answer

SOLUTION: Given:

     Server upload rate = 30 Mbps

                      Download rate of peer d_i = 2 mbps

The given upload rate of mu. Used for n=10, 100 and 1000 and mu=300 Kbps, 700 Kbps, 2 Mbps

The formula for finding the min/max division time for user/server division

                                                  Division= Maximum {NF/Us, F/d_i}

User server:

N= upper limit {(20*1024)(10)/30,(20)(1024)/2}

     =upper limit {6827, 10240}

10 =, upper limit {(20*1024)(100)/30,(20)(1024)/2}

        =upper limit {68277, 10240}

Here is no 3 in the equation comes. In the above this the result of n= 10 .

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.