DMA %CPU blocked equation

Question

% CPU blocked or consumed = (x/y)   (in case of cycle stealing)

where x= data preparation time and y= data transfer time..

 I read questions and answers using this formula but i cant understand how this formula is formulated .. how we are calculating the % Cpu is blocked based on this x and y terms , I mean that in cycle stealing CPU is blocked while DMAC setup and during corresponding interrupt...CPU is not blocked during data preparation time ... also during data transfer time (y) CPU may be blocked or may not be ... Also how this percentage is calculated here.... Does it mean that out of ' data transfer time ' , CPU is blocked for a period equal 'data preparation time'?

Please someone help me with equation

Answer 1

https://gateoverflow.in/?qa=blob&qa_blobid=13639721099184478126

Comments

@ Satyajeet Singh  

In cycle steal : formula is x/y

In burst mode , formula is : x/x+y 

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

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 .

1. First read this best answer https://gateoverflow.in/134178/dma-concept
2.  Next read this best answer  https://gateoverflow.in/116524/coa-dma-doubt

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@ Satyajeet Singh  

Can you please explain what is included in data preparation time and transfer time? Do data preparation time include time to buffer the IO data or does it also include setup timings?

I am giving you an example and tell you what data preparation time means .

Suppose Question is given :

A device 1 MBPS is operating in a cycle mode of DMA. Whenever 16 B word is available it is transferred into memory in 4 microsecond.What is the ℅ of time the processor is blocked due to DMA ?

Ans:  It says Cycle Steal Mode .

First of all , Data is transferred word by word from DMA to memory .

Here total data size is 16 Bytes . And 1 word = 16 Bytes ( given )

so for 16 B data we need to transfer it  in 1 word means 1 time bus access is require by DMA. 

each cycle time is 4 micro second as given,  so to transfer 16 Bytes data it takes 1 * 4 =4 micro seconds time .

so Data preparation time is 8 microseconds . [ data preparation time is total cycle time , means how many cycles are there to transfer and how much time it takes to transfer in those cycles , as each cycle time is given generlly .]

Now, second part, what is  transfer time?

Now, hard disk transfer rate is 1 MBPS = 10^6 Bytes/sec as given .

so (10^6 * 8 ) bits is sending in 1 sec 

1  bit is sending 1/(10^6 * 8 ) sec

16 * 8 bits data is sending in 16 * 8 /(10^6 * 8 ) seconds = ( 16 * 8 /10^6 * 8 ) * 10^6 micro secs = 16 micro secs 

so, to transfer 16 Bytes data hard disk needs 16 micro secs

This is transfer time , 16 microsecs .

Then, % of CPU time consume for DMA operation is : ( 4/16 ) * 100 = 0.25 * 100 = 25% [ by using formula x/y ]

Hope you get it now.

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@Bikram sir word preparation time should be 4 us.