Answer:
<u>t = 0.620 seconds</u>
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,
<u>18 Mbps means 18 x 10^6 bits are transmitted in 1 second</u>.
Now we need to convert the file size of FileX into bits because the transmission rate is given in bits per second.
<u>1 MiB = 1.049 x 10^6 B</u> and <u>1 B (Byte) = 8 b (bits)</u>
So,
1 MiB = 1.049 x 10^6 x 8 b
= <u>8392000 bits
</u>
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 = <u>0.466 seconds</u>
t1 = 0.466 seconds
Now Computer Y will start transmitting FileY (340 KiB). Converting the file size from Kibibytes to bits:
<u>1 KiB = 1024 B</u> so, 340 KiB = 340 x 1024 = 348160 B
Now convert Bytes into bits:
348160 x 8 = <u>2785280 bits</u>
Sending these bits over the 18Mbps connection will require:
2785280/18 x 10^6 = <u>0.154 seconds</u>.
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
