Question

1) A computer is writing on an external hard disk drive via a USB cable. The USB connection speed is assumed to be 5 Gb/sec which is the nominal transmission rate of the current USB 3 standard. How many bytes is the computer writing on the external device in one minute? Compare your answer with a realistic transfer speed that you encounter when you use USB flash devices and hard disks. Any explanation why the USB 3 standard appears to be slower than expected?

2) The standard int type in C language has 32 bits per integer on most machines. Its range is from -2147483648 up to 2147483647. Suppose that a machine defines a new size of 48 bits for integers. Then these 48-bit integers range in the interval

Answer 1

(B=Byte, b=bit, 1Byte = 8bit)

1. Theoretically, the computer should be able to write
5Gb * 3600 = 18000Gb = 2250GB

in an hour. But, since the USB 3.0 is just an interface, and the data being written to either a flash drive of a disk drive, the respective write speeds of the mediums become the bottleneck.

For example, if we are writing to a USB 3.0 compatible flash drive, and the drive's maximum speed is 100 Mbps, then in an hour, the computer will be able to write

100Mb * 3600 = 360000Mb = 45000MB = 45GB

of data to the flash drive.

2. for the bit width of 'n', the range of an integer is given by

-2ⁿ to (2ⁿ - 1)

hence, if n is 48, the the range will be:[ -2⁴⁸ to (2⁴⁸ - 1) ]

from -281474976710656 to 281474976710655

Please provide your feedback for the answer.