Question

3. Consider a system employing interrupt-driven I/O for a particular device that transfers data at an average of 10 KB/s on a continuous basis. a) Assume that interrupt processing takes about 80 us (i.e., the time to jump to the interrupt service routine (ISR), execute it, and return to the main program). Determine how much processor time is consumed per second by this I/O device if it interrupts for every byte. b) Now assume that the device has two 32-byte buffers and interrupts the processor when one of the buffers is full. Naturally, interrupt processing takes longer, because the ISR must transfer 32 bytes. While executing the ISR, the processor takes about 10 us for the transfer of each byte. Determine how much processor time is consumed by this I/O device in this case.

Answer 1

a) Data transfer rate = 10KB/s = 10 * 1024 = 10240 B/s

if it interrupts for every byte,

Time to transfer 1 byte = 80 us.

So, time to transfer 10240 B = 10240 * 80 = 819200 us or .8192 seconds.

means around 0.8 seconds of processor are wasting in this ISR itself.

b) Data transfer rate is same ie 10240 B/s

First processor must transfer 8 * 4 *2 bytes

For 1 ISR Processor is transferring first 2 bytes = 80+10*4+10*4 = 160 us.

Then it will transfer 32 byte per interrupt because whenever one of the buffer is empty it will interrupt the processor,

Time for 32 bytes transfer = 80 + 40 = 120 us

So time taken for transfer of 10240 Bytes = 160 + (10240-64)/32*120 =   38320 us =0.3832 seconds

Which is far less than previous Question.