MadeEasy CO Test q

Question

consider a system that uses interrupt driven I/O for a particular device which has an average data transfer rate of 10KBps.The processing of the interrupt which includes the time to jum to ISR, its execution and returning to the main program is 250 us. what fraction of processor time is consumed by IO device, if IO dvice interrupts for every 2 byte (in %)?

a)80

b)40

c)12

Answer 1

Time taken for the transfer of 2 bytes=>

2B/10KB => 200 us

Processor cycle time=> 250 us

Time for which processor remains idle => (200/250)*100 => 80%

Comments

i think 250 micro sec includes the 200 micro sec. Then the solution seems correct.

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The 250us is interrupt processing time and 200us is interrupt generation time. Interrupt processing time is taken by the CPU whereas interrupt generation time is taken by any I/O device. So, how can 200us be included in 250us? As per my understanding, in interrupt driven I/O, when a device issues an interrupt, the CPU starts servicing the interrupt. In this case suppose a device issues interrupt at t=0us, the CPU finishes ISR at t=250us but the device has already issued another interrupt at t=200us i.e. even before the finishing of ISR.

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since the question reads “...the time to jump to ISR, its execution and returning”. So i think it means that ISR was followed and data transfer took place and then returned back to the main program.