IIT m november 2016

Question

A DMA needs 5 clock cycle to transfer 8 byte data.how many clock needed to transfer 1024 byte of data ,if burst transfer is not used

Answer 1

We can consider 8 byte data as a word ..

2 popular modes of DMA transfer are burst mode DMA and cycle stealing DMA..

CPU remains less inactive in cycle stealing DMA as compared to burst mode DMA ..

In cycle stealing, transfer takes word by word..

Given 

No of cycles to transmit 8 byte data  =  5 cycles

So no of cycles to transmit 1024 bytes data  =  (1024 / 8) * 5

                                                                 =   128 * 5

                                                                 = 640 cycles

Hence 640 cycles should be the correct answer..

Comments

can you explain what happens in burst mode?

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In  burst mode transfer takes place in terms of memory  blocks whereas in cycle stealing mode word by word transfer takes place..

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The number of cycles required to obtain bus for DMA is not given then in this question how burst mode and cycle stealing mode are different? Also what will be answer for burst mode?

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The burst mode case can be calculated if we are given either block size or word count .Word count is mentioned in DMA initialisation phase which means no of words in a single memory block..

But here nothing given and 8 bytes is a too small size to be considered for a block..Modern processors are 64 bit (8 bytes) which mean nothing but word size of 8 bytes..In other words data size that CPU can process at a time,