An 8K file is transferred over the connection at 4096 bits per second.

An 8K file is transferred over the connection at 4096 bits per second. Determine the size of the file (in bytes) that can be transferred over the same time over another connection at a rate of 256 bits per second.

I = 8 kB = 8 * 2 ^ 10 bytes = 23 * 2 ^ 13 bits = 2 ^ 16 bits.

v = 4096 bps = 2 ^ 12 bps.

Let’s determine the time during which the first file was transferred.

t = I / v = 2 ^ 16 bits / 2 ^ 12 bits / s = 2 ^ (16-12) s = 24 s = 16 s.

Then the size of the file that can be transmitted in 16 seconds at a speed of 256 bps is I = 16 s * 256 bps = 24 * 28 bits = 24 + 8 bits = 212 bits = 212/23 bytes = 29 bytes = 512 bytes …

Answer: the information volume of the file is 512 bytes.