Introduction to Information Storage and Data Management
The introductory domain tests whether you can describe storage from first principles: how data becomes information, the three access methods (block, file, object), the host-to-storage I/O path, and the management services that turn raw capacity into a data platform.
Data, information and access methods
- Data is raw bits; information is data placed in context. Retention, classification and protection policies apply to information, not to blocks.
- Block storage exposes fixed-size logical blocks (LBAs) as a LUN or volume over FC, iSCSI or NVMe-oF; the host file system or database imposes structure.
- File storage exposes a hierarchical namespace with directory and file permissions over NFS or SMB. Roughly 80% of enterprise capacity is unstructured file and object data.
- Object storage uses a flat namespace where an object is data plus rich metadata addressed by a unique ID through an HTTP REST API such as S3; there is no partial in-place update.
The I/O path
- Application, file system or volume manager, multipathing, HBA/initiator, fabric, array front-end port, cache, back-end, media.
- Media sets the ceiling: SSD has no seek or rotational delay, 15K SAS pays seek plus rotational latency, 7.2K NL-SAS trades performance for capacity.
- Metrics are IOPS, bandwidth (MB/s is approximately IOPS times block size), latency and queue depth. A drive rated at 180 IOPS is not improved by adding cache.
Management services
Availability (RAID, multipathing, replication); protection (snapshots, backup, air gap); efficiency (thin provisioning, deduplication, compression, tiering); governance (retention, WORM, classification).
Worked example
An application needs 9,000 random write IOPS, each drive delivers 180 IOPS, and the volume sits on RAID 6. A parity update costs two reads plus two writes, so the RAID 6 write penalty is 6: 9,000 times 6 equals 54,000 back-end IOPS, and 54,000 divided by 180 equals 300 drives. On RAID 10 with a penalty of 2 the same workload needs 18,000 divided by 180, which is 100 drives, but only half of raw capacity is usable. The decision is a capacity-versus-drive-count trade-off.
Traps that cost marks
- Treating IOPS and throughput as interchangeable: 10,000 IOPS is 40 MB/s at 4 KB blocks but 2.5 GB/s at 256 KB blocks.
- Assuming battery-backed write cache removes the write penalty. Cache absorbs bursts; sustained load still reaches the back end.
- Confusing thin provisioning (delayed allocation) with deduplication (eliminating duplicate blocks).