GATE CSE 2005 | Question: 69

Question

A device with data transfer rate $10$ KB/sec is connected to a CPU. Data is transferred byte-wise. Let the interrupt overhead be $4\mu$sec. The byte transfer time between the device interface register and CPU or memory is negligible. What is the minimum performance gain of operating the device under interrupt mode over operating it under program-controlled mode?

• A. $15$
• B. $25$
• C. $35$
• D. $45$

Comments

I think this question is a bit ambiguous. WHY?

Because suppose data to be transferred is just 1B, then programmed i/o needs 1B of status register data transfer and then 1B of user data. Whereas interrupt doesn’t require any status bits.

WHEREAS if we consider 10KB data then this 1B status register data seems negligible and can be ignored, and hence the answer changes. THEY should have mentioned the amount of data to be transferred.

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They should have asked max performance gain, it would be if programmed was sending data (1B in 100 micro sec) and interrupt was waiting for generation of data, so only the interrupt overhead (4 micro second) if interrupt also starts sending data, then it's time would be morr here and no speedup

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Ans

Answer 1

In Programmed I/O, the CPU issues a command and waits for I/O operations to complete.

So here, CPU will wait for $1\text{ sec}$ to transfer $10\ KB$ of data.

The minimum performance gain for interrupt mode happens for the smallest unit of data transfer – which here is $1$ byte.

Time to transfer $1$ byte of data in programmed I/O mode $=\dfrac{1}{10\; KBps} = 100 \mu s$

In Interrupt mode, to transfer $1$ byte of data, overhead is $4 \times 10^{-6}s = 4\mu s$
Performance gain  $=\dfrac{100}{4}= 25$

Thus, (b) is correct answer.

Comments

there is a doubt, In interrupt driven IO, CPU doesnt have to wait for the entire IO to complete??

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Can one exceed the data transfer rate of a device using interrupt mode IO over programmed IO?

Isn't data transfer rate the maximum under all circumstances?

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Why are you taking time for 10 KB transfer ,when question says that data is transferred byte wise.

Although answer remains same.

As you mentioned that in interrupt driven data is transferred word by word.Can you give some reference to that?I think it depends on the buffer capacity.It may not be one word always.

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As you mentioned that in interrupt driven data is transferred word by word.Can you give some reference to that?

@rahul sharma 5 ji,

It is mentioned in the question.So it is assumed. 

I think it depends on the buffer capacity.

Yes. Correct.

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@sushmita ji,

there is a doubt, In interrupt driven IO, CPU doesn't have to wait for the entire IO to complete??

Not in general. Otherwise what is the purpose of interrupt.

But yes CPU can wait in some special cases if there is some data dependency or I/O operation is Synchronous Refer --> https://gateoverflow.in/487/gate2008-64-isro2009-13

Now even if it waits nothing much is specified so i think it could be assumed it is included in interrupt overhead time. 

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Can one exceed the data transfer rate of a device using interrupt mode IO over programmed IO?

@Kai ji,

I did not get you could please explain little bit more what exactly you want to ask ?

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@Chhotu can u give some reference to study this concept of programmed and interrupt driven i o 

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hI @Chhotu

I have one basic doubt , i interrupt i/o , it will also take 1 sec to transfer 10KB apart from 4*10$^{-2}$ sec.

But processor will not devote 1 sec , processor will keep on doing other work.

Right?

 

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@Chhotu

I don't think CPU is waiting in case of any synchronous I/O. It's the process that is waiting for the I/O operation to be completed. CPU is executing some other process in the meantime.

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Can you tell me if this approach is correct?

For Interrupt mode overhead = 4 micro sec

For programmed I/O mode : For 10*1024 B time needed = 1sec

For 1 B (since, minimum asked) time needed = 10^6/(10*1024) micro sec = 97.65 micro sec

Performance gain = 97.65/4 = (nearly 25)

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Same question
A device with data transfer rate of 20 KBps is connected to a CPU byte wise. Assume byte transfer time between the device and CPU is negligible and interrupt overhead is 20 μsec. The performance gain under interrupt mode over program controlled mode is ________. (Upto 1 decimal place)

Can any explain what the meaning of this line
"What is the minimum performance gain of operating the device under interrupt mode over operating it under program-controlled mode?"
It bit confusing
Is it mean  interrupt mode/program controlled mode

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Not in general. Otherwise what is the purpose of interrupt.

But yes CPU can wait in some special cases if there is some data dependency

@Chhotu How can the CPU go on doing other work when it is busy servicing the interrupt. Can you please explain? I think the CPU will wait in all cases. Does here wait refer to waiting for the non interrupt instructions to execute? Please explain. Thank you

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I think entire data tranfer will take place  in both the cases..and then status of i/o module is made to ready...but in programmed i/o cpu is busy checking ready status of i/o module  hence for 1 sec it will in busy waiting..unlike

in interrupt i/o cpu will not busy wait ..and interrupt overhead is 4 micro sec per  byte .But in both cases data will be transferred word wise….Correct me if wrong

Reference:

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CPU never waits like this……. It schedules some other task for that duration, and when the interrupt is finished executing IO, it sends signal back to CPU, then CPU, accesses it.

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Isn’t the formula you are using is of speedup and not performance gain?

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Maximun performance gain depends on buffersize???

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is that 1B for flag ?

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Performance gain and speed up are same thing

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@sushmita For Interrupt IO shouldn't it be 4 μs +1 sec(total data transfer time )? Why we only considered Overhead time ?

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.

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Performance checks for CPU spent time on respective device

In interrupt mode CPU will spend time when interrupt occurs remaining time it will do it's work it won't service the interrupt mode.

So perfomence time is 4mue sec 

While data transferring CPU won't interfere so we don't calculate data transfer time for performance

While in programmed IO unitl data tranfer completes CPU serves for IO so for 1 byte 1/100mue sec ( In programmed IO interrupt time is 0)

Perfomance gain of interrupt over programmed is 100mue / 4mue = 25

Hope you get it 

https://youtu.be/Tb7KEdrRIPg?si=csLOSZWAbOSS5Kwo&t=8121

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Thank you for the video link.

Answer 2

In programmed I/O, CPU does continuous polling,

To transfer 1B CPU polls for 10^(-4) sec or 100 micro-sec of processing (since transfer speed is 10KBps)

In interrupt mode CPU is interrupted on completion of i\o,

To transfer 1B CPU does 4 micro-sec of processing(since transfer time between other components is negligible).

Gain = 100 / 4 = 25

Answer 3

Just calculate the time taken for 1 Byte transfer for both I/o and interrupt.

For I/O it is given 10kb/sec so 1 byte transfer will take 10^-4 sec or 0.1 milli sec for each byte.

For interrupt they have given 4 micro sec as overhead meaning 4 micro sec for 1 byte transfer.

Now they are asking about performance. They are asking performance of interrupt in comparison to I/O so just do simple division  

performance = 0.1 millisec / 4 micro sec

                     = 100/4 = 25

I don’t know why there is so much confusion over this here.Its pretty simple.

Comments

I agree with you.

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@Vishal_kumar98  @yashhoskere @Chhotu 

For I/O it is given 10kb/sec so 1 byte transfer will take 10^-4 sec or 0.1 milli sec for each byte.

while CPU is waiting and checking the status bit in status reg, I/O device controller is actually reading the 1 byte of data and setting the status flag in control register is 1 ??

My Question is in this time period what is the actual work is done by I/O device??

Answer 4

$Programmed\quad IO:$

1) CPU polls continuously while the device prepares data in the buffer.

2) CPU transfers the data to main memory.

 

$Interrupt\quad driven\quad IO:$

1) Device generates an interrupt when data is ready.

2) CPU handles the interrupt (interrupt overhead).

3) CPU transfers the data to main memory.

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$Given:$

Device transfer rate = 10 KB/sec = 10 * 1000 = 10,000 Bytes/sec

Time per byte = $10^{-4}$ sec = $100 × 10^{-6} $ sec = 100 microseconds

$Programmed\quad IO:$

Since CPU polls continuously and the CPU to main memory transfer time is negligible (given)

CPU time per byte = device time per byte

CPU time per byte = 100 microseconds

 

$Interrupt\quad driven\quad IO:$

Interrupt overhead per byte = $4 * 10^{-6}$ seconds

CPU to main memory transfer time is negligible

 

Performance gain = Old/New

CPU time in Programmed IO = $100 * 10^{-6}$ seconds

CPU time in Interrupt driven IO = $4 * 10^{-6}$ seconds

Performance gain = $(100 * 10^{-6}) / (4 * 10^{-6})$ = 25

Answer 5

ans is 25

Comments

Can you please explain what do you mean by 
in programed mode
100/(100+4)=100/104 ?

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If i take in programmed control the time of  data  transfer of 1 byte that comes out to be 100 microsecond and divide it by interupt control overhead for 1 byte too then

100/4 = 25 speed up is this approach right...?

Answer 6

Programmed i/o time=status register rad time+ data transfer time

Given data trasfer time is negligible =0

status register time(defaultly taken of one byte)

10kb---1sec

1b=1/10ksec

1b=0.1milisecond

Programmed i/o time=0.1milisecond+0=0.1milisecond=100microsecond

 

interrupt driven time=data transfer time+interrupt overhead=0+4microecond=4 microsecond

performance gain=100microsecond/4microsecond=25