GATE CS 2024 Set 2 Q1 — I/O Interface: Interrupt and DMA Mode

To find the data transfer rate of the DMA controller in bits per second, we can break down the calculation into the following steps:

1. Determine the total processor cycles per second:

   $$\text{Processor frequency} = 4 \text{ MHz} = 4 \times 10^6 \text{ cycles/second}$$

2. Calculate the cycles stolen by the DMA controller per second:

   Since $1\%$ of the processor cycles are used for DMA, the number of cycles available for DMA per second is:

   $$\text{DMA cycles per second} = 1\% \text{ of } (4 \times 10^6) = 0.01 \times 4,000,000 = 40,000 \text{ cycles/second}$$

3. Calculate the data transferred in bytes per second:

   The DMA controller transfers $8$ bytes in each stolen cycle:

   $$\text{Data transfer rate (in bytes/sec)} = 40,000 \text{ cycles/second} \times 8 \text{ bytes/cycle} = 320,000 \text{ bytes/second}$$

4. Convert the data transfer rate to bits per second:

   Since $1 \text{ byte} = 8 \text{ bits}$:

   $$\text{Data transfer rate (in bits/sec)} = 320,000 \times 8 = 2,560,000 \text{ bits/second}$$

Using the Indian numbering system representation, $2,560,000$ is written as $25,60,000$.

Therefore, the correct option is C.