Question

Given a link with a maximum transmission rate of 18 Mbps. Only two computers, X and Y, wish to transmit starting at time t = 0 seconds. Computer X sends fileX (1 MiB) and computer Y sends fileY (340 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 FileY finish transmitting? Give answer in seconds, without units, and round to two decimal places.

Answer 1

Total time taken = 0.62 seconds

Step-by-step explanation:

Transmission rate = 18 Mbps

The size of file X winch will go fast= 11.73 MiB (Mebibyte) = 8.38 Mb(Megabit)

The time it will take to transfer = File size / Transmission rate

= 8.38 / 18 seconds

= 0.46 seconds

Mb in 1 KiB = 116.415

The file Y that will go next has size = 340 Kib

= 340 / 116.41 Mb

= 2.92 Mb

The time it will take to transfer = 2.92 / 18 seconds

= 0.16

Total time = Time taken for file X + Time taken for file Y

= 0.46 + 0.16

= 0.62