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Question

Assume that interrupt processing takes 75us. Interrupt driven I/O for a device that transfer data an average of 8KB/sec on a continuous basis.Find percentage of the processor time is consumed by the I/O device if it interrupts for every byte?

a. 20

b. 40

c.60

d.80

Comments

b. 40%  ??

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Correct answer is 60%

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ohh , yes, right

i calculated wrong....

Answer 1

Key issue here is "Interrupt driven I/O for a device that transfer data an average of 8KB/sec on a continuous basis"

that means I/O interrupt will occur whenever one character is ready which is 125 micro sec. So time interval between two consecutive interrupt is 125 micro sec (not more than that).

So, % of CPU time for interrupt processing is 75/125 = 60 % 

Comments

@Debashish couldn't understand how you comeup with 122usec, can you please explain?

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@debashish

Here,  125 is total data transfer time and 75 is data preparation time , as it said " device that transfer data an average of 8KB/sec on a continuous basis " 

8 * 10³ Byte transfer in 1 sec

1 Byte  transfer in ( 1/ 8 * 10³ ) * 10⁶ microsec =  0.000125  * 10⁶ =125 microsec 

as it is interrupt driven I/O , formula used x/y , x= data preparation time and y = data transfer time

so, percentage of the processor time is consumed by the I/O device if it interrupts for every byte

=( 75/125 ) * 100 = 0.6 * 100 = 60 

so answer is  60% , Option C .

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Interrupt driven IO comes into picture, when the device is slow to process interrupt , so there are delays to process interrupt. The delays are generally called the overhead

See here interrupt request comes in,. So, processing the interrupt there are we need to stop current instruction and process the interrupt

Here the timing diagram shows, to service an interrupt in IO , which delays are involved?

1) Interrupt Latency Time

2) Saving Main Context

3) Identifying device (done by vectored interrupt(where a particular device do a work which is known by CPU) or by non vectored interrupt)

4) Interrupt Service - that is the main work, where interrupt is processed

5) Restore main context again

Here (2),(3),(5) are overheads of interrupt service

Answer 2

here my answer

The device generates 8 × 1024 = 8192 bytes per second or a rate of one every 122 µs. If each interrupt consumes 100 µs, then the fraction of processor time consumed is 100/122 = 0.82The device generate 8*1024 bytes or rate of one every 122 us If interrupt is 75us, then the fraction of processor time will be 75/122 = 0.61

 thus it should be c as that's the closest

Comments

in every 122ns I/O device dumps some data inside the controller and trigger the interrupt PIN. On the other hand in 75ns CPU transfer  that data to required location using with the help of ISR (executing some specified code segment) and so, this 75 ns time is called interrupt service time. Why CPU transfer so fast compared to the device (75ns < 122ns) ? because the internal bus is very fast. 

Yes, we have to match the in:out . Otherwise, controller buffer overflows.

Please tell is there anything wrong ?

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No, by "everything matches" means everything is now in place :)

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Thanks :)