Consider a disk pack with 16 surfaces, 128 tracks per surface and 256 sectors per track. 512 bytes of data... Capacity and bits for address?
- A. 256 Mbyte, 19 bits
- B. 256 Mbyte, 28 bits
- C. 512 Mbyte, 20 bits
- D. 64 Gbyte, 28 bits
Answer: A. 256 Mbyte, 19 bits
Explanation: Capacity = 16 * 128 * 256 * 512 = 256 MB. Address bits = log2(16)+log2(128)+log2(256) = 4+7+8 = 19 bits.