Question

: A system using interrupt-driven I/O that transfers data at an average of 8 KB/s on a continuous basis is connected to a storage device a. The interrupt processing time is about 100 ms (which includes the time to jump to the (ISR), execute it, and return to the main program). Consider if the interrupt is for every byte, what portion of processor time is used up by this I/O device. b. Consider that the device has two buffers of 16-byte and it interrupts the microprocessor when one of the two buffers is full. Logically, interrupt takes longer to process, as 16 bytes must be transferred by the ISR. The processor takes about 8 ms for the transfer of each byte while executing the ISR,. Calculate what percentage of processor time is consumed by this I/O device in this case. c. Assume now that a block transfer I/O instruction is available in the processor. This permits the associated ISR to transfer each byte of a block in only 2 ms. Calculate the percentage of processor time consumed by this I/O device in this case.

Answer 1

Answer:

Compute fraction of the processor time:

Given that,

The data transfer rate is 8KB/s, so per one second the total number of bytes that can be transferred is calculated as .

• Since 1 Kilo represents 1024 in terms of data.

The time interval for an interrupt is given as follows:

Time required for interrupt processing =

The fraction of the processor time is calculated as follows:

Therefore, the fraction of the processor used for input- output device (I/O device) is