CompTIA Server+ (SK0-005)Server Hardware Installation and ManagementMedium
A server has four 1.8 TB SAS drives. The administrator wants to configure a RAID array that provides both high performance and fault tolerance, capable of surviving the failure of two drives without data loss. What is the maximum usable storage capacity for this array?
- A1.8 TB
- B5.4 TB
- C7.2 TB
- D3.6 TB
Show answer & explanationAnswer & explanation
Correct answer: D. 3.6 TB
To survive the failure of two drives, RAID 6 is the appropriate choice. RAID 6 requires a minimum of four drives and dedicates two drives' worth of capacity for parity information. With four 1.8 TB drives, the usable capacity is (4 - 2) * 1.8 TB = 2 * 1.8 TB = 3.6 TB.
Why the other options are wrong
- A. 1.8 TB would be the capacity if only one drive was used, or if it was RAID 10 with two drives.
- B. 5.4 TB would imply 3 drives usable, which is not possible with RAID 6 and 4 drives.
- C. 7.2 TB is the raw capacity (4 * 1.8 TB) without any RAID overhead.
RAID 6
A RAID level that provides fault tolerance against two simultaneous drive failures by using two independent parity blocks distributed across all drives in the array.
- Minimum 4 drives
- N-2 usable capacity
- High fault tolerance, good performance
Memory trick: RAID 0 is fast, RAID 1 mirrors past, RAID 5 one parity, RAID 6 two drives last.