DMA i/o devices

Question

Please give detailed explanation.

Answer 1

For Cycle Steal Mode :

Data is transferred word by word from DMA to memory .

Here total data size is 4 Bytes . And 1 word = 4 Bytes ( given )

so for 4 B data we need to transfer in 1 word means 1 time bus access is require by DMA. 

each cycle time is 50 micro second, to transfer 16 Bytes it takes 1 * 50 = 50 micro seconds to transfer 4 Bytes data .

so Data preparation time is 50 microseconds .

Now, hard disk transfer rate is 10 KBPS = 10 * 10³ Bytes/sec

so (10⁴ * 8 ) bits is sending in 1 sec 

1  bit is sending 1/(10⁴ * 8 ) sec

4 * 8 bits data is sending in 4 * 8 /(10⁴ * 8 ) seconds = ( 4 * 8 /10⁴ * 8 ) * 10⁶ micro secs = 400 micro secs 

so, to transfer 4 Bytes data hard disk needs 400 micro secs == Transfer time.

Then, % of CPU time consume for DMA operation is : ( 50/400 ) * 100 = 0.125 * 100 = 12.5%

So, Ans is 12.5% CPU time is consumed in Cycle Steal Mode .

Comments

50 micro second is transfer time when word is ready to be transferred? why have you used it in the preparation time?

Answer 2

In cycle stealing mode transfer takes place word by word..

Here word size given = 4 B

Here time CPU busy which is also known as preparation time  =  Data size / I/O Device transfer rate

                                                   In   1 s , data transferred  =    10⁴ B

                                                   So   time take for 4 B  =  10^-4 s * 4

                                                                                     =  400 microsec

 And time CPU is blocked ( also known as transfer time )    =   50 microsec

So % time CPU is blocked                                                =   (50 / 400) * 100

                                                                                       =   12.50 % [In cycle stealing mode , pipelining is there]

So by this amount only CPU activity is reduced..

Hence 12.50 % should be the correct answer..

                                 

Comments

It means answer should be (50/400)*100=12.5

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Thanks @Bikram sir , @learner_geek for rectification..Edited my answer..

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@Habibkhan, what if it is asked ...

1)% of time CPU consumed

2)% of time CPU blocked(or held up)

I think 1) is 0 (bcz no interrupt time is given)

 and 2) is 12.5%