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 File Y finish transmitting?

Answer 1

t = 0.620 seconds

Step-by-step explanation:

The files are being sent as a stream so first FileX will transmit over the 18Mbps connection and then FileY will follow.

Computer X transmits FileX (1 MiB) over a 18Mbps connection. So,

18 Mbps means 18 x 10^6 bits are transmitted in 1 second.

Now we need to convert the file size of FileX into bits because the transmission rate is given in bits per second.

1 MiB = 1.049 x 10^6 B and 1 B (Byte) = 8 b (bits)

So,

1 MiB = 1.049 x 10^6 x 8 b

= 8392000 bits

Calculating the time taken for FileX to transmit:

18 x 10^6 bits take 1s to transmit so

8392000 bits will take: 8392000/18x10^6 = 0.466 seconds

t1 = 0.466 seconds

Now Computer Y will start transmitting FileY (340 KiB). Converting the file size from Kibibytes to bits:

1 KiB = 1024 B so, 340 KiB = 340 x 1024 = 348160 B

Now convert Bytes into bits:

348160 x 8 = 2785280 bits

Sending these bits over the 18Mbps connection will require:

2785280/18 x 10^6 = 0.154 seconds.

t2 = 0.154 seconds

Computer Y has finished transmission. The total time taken can be calculated by adding both the transmission times t1 and t2.

t = t1 + t2

=0.466 + 0.154

t = 0.620 seconds