88 Hard Disk ManagementHard Disk ManagementThis Image Servers employ an advanced Serial-ATA controller that manages four high-capacity hard disks operating in a RAID-5 configuration. This design increases systemperformance, 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 earliershared-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 entirestorage system down.By contrast, the Image Server uses a non-blocking switched architecture to isolate the drivesfrom one another. Any drive failure makes that drive unavailable and the rest of the storage systemremains undisturbed. In addition, the Image Server uses Advanced Data Protection features, whereall drive commands are checked to ensure that no command corruption has taken place over theentire 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 panelAbout RAID 5A RAID 5 configuration features the data striping of RAID 0, combined with the parity benefits ofRAID 4. Using a parity (exclusive OR) function, RAID 5 can tolerate the loss of one drive. Parityinformation is distributed across all drives rather than being concentrated on a single disk. Thisavoids throughput loss due to contention for the parity drive. Spare drives can be used to rebuildan array after a drive is replaced.RAID-5 capacity equals the size of drive times (number of drives -1). In addition, the array’sstorage efficiency increases with the number of disks; from 66.7 % for 3 drives to 75 % for 4drives: storage efficiency = (number of drives -1) X (number of drives).