Computer Organization And Architecture

RAID 3 requires a single redundant disk (Correct). In this RAID configuration, a single disk stores the parity information for the entire array. RAID 4 requires a dedicated parity disk, not multiple blocks or disks storing parity calculations (Incorrect). RAID 5 distributes the parity strips across all disks (Correct). RAID 5 uses striping with distributed parity, which avoids having a single parity disk. RAID 2 uses Hamming code for error correction, where the number of redundant disks is proportional to the log of the number of data disks (Correct). Thus, the correct answer is: (A), (C) and (D) only.

To calculate the disk capacity: Capacity = Surfaces × Tracks per surface × Sectors per track × Bytes per sector. Substitute the values: Capacity = 16 × 128 × 256 × 512 bytes. Convert to MB: Capacity = 16 × 128 × 256 × 512 / 1024² MB = 512 MB.

To calculate the average access time, we use the formula: Average Access Time = (Hit Ratio) × (Cache Access Time) + (1 − Hit Ratio) × (Main Memory Time). Given: Hit Ratio = 0.98, Cache Access Time = 12 nsec, Main Memory Access Time = 1.5 msec = 1.5 × 10^6 nsec. Substitute the values into the formula: Average Access Time = (0.98) × (12) + (0.02) × (1.5 × 10^6). Average Access Time = 11.76 + 30,000 = 30,011.76 nsec. The average access time is approximately 41.76 nsec.

During a typical fetch operation: 1. PC (Program Counter): Holds the address of the instruction to be fetched. 2. MAR (Memory Address Register): Receives the address from the PC. 3. Memory: The instruction is fetched from the memory location pointed to by MAR. 4. MDR (Memory Data Register): Holds the fetched instruction. 5. IR (Instruction Register): Stores the instruction for execution.