GATE CS 2004 Q68 — I/O Interface: Interrupt and DMA Mode

To find the percentage of processor time consumed during the DMA transfer, we compare the time the processor is busy with the total time taken for the transfer.

1. Processor Busy Time:

   The processor is active only during the initiation and completion of the DMA transfer.

   $$\text{Total cycles consumed} = 300 + 900 = 1200 \text{ cycles}$$ $$\text{Processor busy time} = \frac{1200 \text{ cycles}}{600 \times 10^6 \text{ Hz}} = 2 \times 10^{-6} \text{ seconds} = 2 \ \mu\text{s}$$
2. Total Transfer Time:

   The hard disk transfers $20 \text{ KB}$ of data at a rate of $10 \text{ MB/s}$.

   $$\text{Total transfer time} = \frac{20 \times 10^3 \text{ bytes}}{10 \times 10^6 \text{ bytes/second}} = 2 \times 10^{-3} \text{ seconds} = 2 \text{ ms}$$
3. Percentage of Processor Time Consumed: $$\text{Percentage} = \left( \frac{\text{Processor busy time}}{\text{Total transfer time}} \right) \times 100 = \left( \frac{2 \ \mu\text{s}}{2 \text{ ms}} \right) \times 100 = 0.1\%$$

Thus, the processor is consumed for $0.1\%$ of the transfer time.