GATE CS 1996 Q25 — I/O Interface: Interrupt and DMA Mode

To find the fraction of processor time consumed by the disk, we calculate the processor overhead incurred by the DMA transfers per second.

Step 1: Calculate the processor overhead per DMA transfer
$$\text{Total overhead per transfer} = \text{Setup time} + \text{Interrupt handling time}$$ $$\text{Total overhead per transfer} = 1000 + 500 = 1500 \text{ clock cycles}$$

Step 2: Calculate the number of transfers per second
Depending on how the unit "K" is interpreted, we have two standard cases:

• Case 1 (Decimal interpretation, $1\text{ KB} = 1000\text{ bytes}$):
  $$\text{Transfers per second} = \frac{2000 \times 10^3 \text{ bytes/sec}}{4 \times 10^3 \text{ bytes}} = 500 \text{ transfers/sec}$$
• Case 2 (Binary interpretation for block size, $1\text{ KB} = 1024\text{ bytes}$):
  $$\text{Transfers per second} = \frac{2000 \times 10^3 \text{ bytes/sec}}{4 \times 1024 \text{ bytes}} \approx 488.28 \text{ transfers/sec}$$

Step 3: Calculate the fraction of processor time consumed
$$\text{Total processor clock cycles per second} = 50 \text{ MHz} = 50 \times 10^6 \text{ cycles/sec}$$

• For Case 1:
  $$\text{Processor cycles used per second} = 500 \times 1500 = 7.5 \times 10^5 \text{ cycles/sec}$$ $$\text{Fraction consumed} = \frac{7.5 \times 10^5}{50 \times 10^6} = 0.015 \text{ (or } 1.5\%\text{)}$$
• For Case 2:
  $$\text{Processor cycles used per second} = 488.28 \times 1500 \approx 732,422 \text{ cycles/sec}$$ $$\text{Fraction consumed} = \frac{732,422}{50 \times 10^6} \approx 0.0146 \text{ (or } 1.46\%\text{)}$$

Thus, the fraction of processor time consumed is in the range of $1.4\%$ to $1.5\%$ (or $0.014$ to $0.015$).