Question

What is the minimum time needed to distribute this file from the central server to the 10 peers using the client-server model?

Answer 1

The answer is "
`0.143 * 10^3 \ s`"

Step-by-step explanation:

Please find the complete question in the attachment.

The size of the file
`= F = 2 \ Gbits = 2 * 10^9 \ bits`

8 file copies have had to be uploaded from Server. So uploading size of to all

Rate of Server upload
`= u = 83 Mbps = 83 * 10^6 \ bps`

therefore, the minimum time for the server to upload the file:

`= 8 (F)/(u) \\\\= ((8 * 2 * 10^9))/((83 * 10^6)) \ s \\\\= 0.193 * 10^3 \ s`

Receive rate of the slower server:

`=dmin \\\\= min\{d_1,d_2,d_3,d_4,d_5,d_6,d_7,d_8\} \\\\= 14\ Mbps \\\\= 14* 10^6\ bps`

The minimum time necessary to get the file:

`= (F)/(dmin)\\\\ = ((2 * 10^9))/((14 * 10^6))\ s \\\\= 0.143 * 10^3 \ s`