DMA Concept

Question

reference to this pdf :https://gateoverflow.in/?qa=blob&qa_blobid=13639721099184478126

Can any one define in simple terms

X μsec =data transfer time or preparation time (words/block) ?

Y μsec =memory cycle time or cycle time or transfer time (words/block) ?

Answer 1

 x=  Data preparation time or CPU time   and y=  Data transfer time 

When to use which formula: % of cpu time consume   , that depends on the exact wordings in question .

 Case 1: In Burst Mode  -

In this case we use ( x/x+y  )                  

 example : burst mode in both cases 

• Gate question https://gateoverflow.in/2777/gate1996_25 
• https://gateoverflow.in/1062/gate2004-68?

Case 2:  In Cycle Stealing Mode -

In this case we use  ( x/y ) 
example -  explicitly mention it is in cycle stealing mode

• https://gateoverflow.in/18941/drdo-q32 
• https://gateoverflow.in/1393/gate2005-70 

In cycle stealing mode we always follow pipelining concept that means when one byte is getting transferred then at the same time Device is parallel way preparing the next byte.So we use concept of x/y .

And default is burst mode, when no mode mention explicitly , we use Burst mode like  1996 and 2004 Gate questions.

We use cycle stealing formula (x/y) only when the term "cycle stealing" is mention explicitly . In cycle stealing 1 byte data is transferred at a time in theoretically.

Now, coming to x =  preparation time, it is given in question,say " A hard disk with a transfer rate of 10 Mbytes/second is constantly transferring data to memory using DMA. The processor runs at 600 MHz, and takes 300 and 900 clock cycles to initiate and complete DMA transfer respectively. If the size of the transfer is 20 Kbytes,"

Clock cycle time = 600 MHz =1/ (600 x 10⁶ ) seconds

total cycle time for processor is (300+900)=1200 and each cycle time is  1/ (600 x 10⁶  ) seconds

So data preparation time = x , which is 1200 * 1/ (600 x 10⁶ ) * 10⁶  microsecs = 2 microsec.  This 2 microsec is x .

Now coming to y = data transfer time

here data transfer rate from disk is 10 MB/sec , total data to transfer is 20 KB . 

so, Disk Transfer data at a rate of 10 Mbytes/sec

 1 byte transfer in 1/10⁷ sec

 20 Kbytes transfer in 2 milisec = 2000 micro sec .. this 2000 microsec is data transfer time of y .

now here nothing is mention so we use formula for burst that is x/x+y 

so put x and y value and get the answer.

here if cycle stealing mode explicitly mention like " assume disk is using cycle stealing mode" then we should use x/y formula..

hope you get it now .

Comments

Thanks for so big answer but i got above thing previously

My doubt was and are :->

what is meant by data transfer time  ..is it means transfer of data between i/o to processor through dma mode?

and what is meant by preparation time?

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let there is a device transfer rate is  x bytes/sec .

"A hard disk with a transfer rate of 10 Mbytes/second is constantly transferring data to memory using DMA" --- here device transfer rate is 10 MB. now what time device take to transfer total  20 KB, is known as transfer time.

Yes,  it means time taken to transfer data between i/o device to memory through dma mode .

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There are cpu cycles, some cycle need to set up DMA and some cycle need to complete that DMA transfer . each cycle took some amount time.

so total cycle time is ( number of cycles * each cycle time ), this time is called as Data preparation time.

 If processor can transfer y bytes(i.e bus length) in 1 cycle(assume 1 cycle of 1 sec ) then bus speed is y bytes/sec.

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please help but how ????? 600 MHz =(1/600) x 10⁶

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@ learner_geek 

1 Hz = 1/seconds

1 seconds = 10⁶ microseconds .

1 MHz = 1 x 10⁶ Hz = 10^-6 seconds

600 MHz = 600 * 10⁶ Hz

600 Mhz  = 1/ (600 * 10⁶  )  * 10⁶ microseconds = 1/600 microseconds 

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@Bikram sir,

Please correct me if wrong

ex - If we are given that memory cycle time is 80 nsec. And cycle stealing mode is used 1 word is transferred when bus is aquired .

Lets take data transfer time for 1 word is 2 microsecond ,

data preparation time ==> we need cycles to aquire and leave bus( i.e BR and BG ) .assuming BG and BR can be done in 1 machine cycle ..hence for 1word we need 1 cycle to get bus one time therefore data preparation time is 80 nsec

during total 2 microseconds of transfer time for 80 nsec we need cpu for BR and BG

hence CPU consume / blocked = 80nsec / 2 microsec

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@Bikram sir, i don't understand how x/y tells us about the CPU idle time?Can you explain this a bit.