GATE CS 2005 Q70 — I/O Interface: Interrupt and DMA Mode

To find the percentage of time the CPU gets blocked, we first determine the data transfer rate of the disk drive.

1. Rotational Speed: $$\text{Rotations per second} = \frac{3000 \text{ rpm}}{60} = 50 \text{ rps}$$
2. Track Capacity: $$\text{Capacity per track} = 512 \text{ sectors} \times 1 \text{ KB/sector} = 512 \text{ KB}$$ Using $1 \text{ KB} = 1024 \text{ bytes}$: $$\text{Capacity per track} = 512 \times 1024 = 524,288 \text{ bytes}$$
3. Data Transfer Rate (DTR): $$\text{DTR} = 50 \text{ rps} \times 524,288 \text{ bytes} = 26,214,400 \text{ bytes/sec}$$
4. Time to prepare 1 word (4 bytes): $$T_{\text{prep}} = \frac{4 \text{ bytes}}{26,214,400 \text{ bytes/sec}} \approx 152.59 \text{ nsec}$$

In cycle stealing mode, the DMA controller steals one memory cycle ($40 \text{ nsec}$) to transfer each $4$-byte word. During this cycle, the CPU is blocked.

$$\text{Percentage of time CPU is blocked} = \frac{\text{Memory Cycle Time}}{T_{\text{prep}}} \times 100\%$$

$$\text{Percentage blocked} = \frac{40 \text{ nsec}}{152.59 \text{ nsec}} \times 100\% \approx 26.21\%$$

This is closest to 25% (Option B). (Using $1 \text{ KB} = 1000 \text{ bytes}$ yields $25.6\%$, which also rounds to $25\%$).