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​​​Consider a computer with a $4 \mathrm{MHz}$ processor. Its $\text{DMA}$ controller can transfer $8$ bytes in $1$ cycle from a device to main memory through cycle stealing at regular intervals. Which one of the following is the data transfer rate (in bits per second) of the $\text{DMA}$ controller if $1 \%$ of the processor cycles are used for $\text{DMA}$?

  1. $2,56,000$
  2. $3,200$
  3. $25,60,000$
  4. $32,000$
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1 Answer

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Cycle time=1/4MHZ=0.25us

Transfer time=ty=1 cycle=0.25 us

Cycle stealing mode (given)

% cpu block=(ty/tx)*100

1=ty/tx *100

tx=ty*100

tx=25us

In 25 us 8B data is transferred

In 25us 8*8 bits data is transferred

In 1 sec  64/25us data is transferred

So data transfer rate =64/25us

                                        =2.56*10^6

 So answer is option C
Answer:

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