Question

Given a link with a maximum transmission rate of 52.3 Mbps. Only two computers, X and Y, wish to transmit starting at time t = 0 seconds. Computer X sends fileX (7 MiB) and computer Y sends fileY (341 KiB), both starting at time t = 0. Computer X gets the transmission medium first, so Computer Y must wait. For the following calculations, assume maximum transmission rate during transmission. Suppose that entire files are sent as a stream (no packets, no multiplexing). At what time (t = ?) would FileX finish transmitting? Give answer in seconds, without units, and round to two decimal places (e.g. for an answer of 12.4567 seconds you would enter "12.46" without the quotes)

Answer 1

Approximately 1.12 seconds

Step-by-step explanation:

The calculation units are different, as bits and bytes are different. So, we must convert one of the measures to keep calculation simple. In this case, we'll convert MiB (mebibytes) to Mb (megabits), so we can calculate all with megabits.

1 MiB (mebibyte) ≈ 8.39 Mb (megabits)

fileX = 7 MiB * 8.39 ≈ 58.73 Mb

Maximum transmission rate (x) = 52.3 Mbps

So we get this:

`(58.73)/(52.3)` ≈ 1.12 seconds = t