A 4096 byte file is transferred over a connection at 512 bits per second. Determine the size of the file (in Kbytes) that can be transferred in the same time over another connection at a speed of 2048 bits per second. Please provide one number in your answer – the file size in Kbytes.
Size of the transferred file = transfer time · transfer rate. Note that the transmission rate in the second case is 2048/512 = 4 times the rate in the first case. Since the file transfer times are the same, the file size that can be transferred in the second case is also 4 times larger. It will be 4 * 4 KB = 16 KB.
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