360 Systems MAXX-1000SD - Manual (Page 96)

Page 96 MAXX Owner’s Manual
Hard Disk Management
Image Servers employ an advanced Serial-ATA controller that manages four high-capacity hard
disks operating in a RAID-5 configuration. This design increases system performance, and by
storing parity data, helps prevent data loss should a drive fail.
The Image Server brings a high level of reliability to Serial-ATA RAID through a new, switched
architecture that exceeds the reliability of SCSI shared-bus storage systems. The earlier shared-bus
architecture of SCSI has inherent performance limitations due to arbitration latency, since only one
drive may use the bus at a time. Further, a single drive failure can bring the entire storage system
down.
By contrast, the Image Server uses a non-blocking switched architecture to isolate the drives from
one another. Any drive failure makes that drive unavailable and the rest of the storage system
remains undisturbed. In addition, the Image Server uses Advanced Data Protection features, where
all drive commands are checked to ensure that no command corruption has taken place over the
entire data path.
Some of the Image Server RAID implementation features and benefits include:
• Non-blocking switch technology with RAID 5 parity
• On-board processor minimizes host CPU overhead
• Easy-to-configure arrays
• Dynamic sector repair for robust data protection
• Accelerated RAID-5 writes
• Easy drive replacement from front panel
About RAID 5
A RAID 5 configuration features the data striping of RAID 0, combined with the parity benefits of
RAID 4. Using a parity (exclusive OR) function, RAID 5 can tolerate the loss of one drive. Parity
information is distributed across all drives rather than being concentrated on a single disk. This
avoids throughput loss due to contention for the parity drive. Spare drives can be used to rebuild
an array after a drive is replaced.
RAID-5 capacity equals the size of drive times (number of drives -1). In addition, the array’s
storage efficiency increases with the number of disks; from 66.7 % for 3 drives to 75 % for 4
drives: storage efficiency = (number of drives -1) X (number of drives).