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Consider a typical disk that rotates at 150000 Rotations Per Minute (RPM) and has a transfer rate of 50 × 106 bytes/sec. If the average seek time of the disk is twice the average rotation delay and the controller's transfer time is 10 times the disk transfer time, the average time (in milliseconds) to read or write a 512-byte sector of the disk is _____.
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- 6 ms
- 111 ms
- 16.1 ms
- 6.1 ms
Correct Option: D
60sec → 15000 rotations
| 1 sector → | = 4 ms ← 1 rotation | |
| 15000 |
| ∴ Average rotational delay = | × 4 ms = 2 ms | |
| 2 |
⇒ As per question, average seek time = 2 × Avg. rotational delay
= 2 × 2 = 4ms
⇒ As per question, controller's transfer time is = 10 × 0.01 ms = 0.1 ms
∴ Avg. Time = 4ms + 0.1 ms + 2ms = 6.1 ms
