Capacity design method (2) : Physical capacity from logical capacity (for HPEC 4D+2P)

Virtual Storage Platform One SDS Block Storage Administrator Guide

Version

1.17.x

Audience

anonymous

Part Number

MK-24VSP1SDS002-04

The following is assumed:

Each storage node (or drive) in the storage cluster has the same valid physical capacity.
The storage cluster does not have faulty storage nodes or faulty drives.
The configuration, the capacity of each storage node, and the number of drives meet the system requirements.

For details about storage node requirements, see Storage node requirements in the VSP One SDS Block Setup and Configuration Guide for the model that you are using.

In the following example procedure, you calculate the physical capacity required per storage node when a logical capacity of 50 TiB is allocated to six storage nodes (20 device slots). The rebuild capacity policy is set to "Fixed", and the number of tolerable drive failures is set to 1.

Since the logical capacity can be increased in 595728-MiB increments (section Capacity design principles (2) : Logical capacity (for HPEC 4D+2P)), the valid physical capacity required in each storage node is:
```
RoundUp(((50 × 1024²) / 6 + 4200) × (893592 / 595728)) + 893592
    = 14007092[MiB/Node]
    RoundUp(numerical value)
        : A numerical value is rounded up to an integer.
```
Calculate the valid physical capacity of the used drives.

For example, the valid physical capacity of a drive (1.6 TB) is 1340388 [MiB/device].

Because the capacity for one drive is allocated as the rebuild capacity, the number of necessary drives can be calculated by using the following expression:
```
RoundUp(14007092 / 1340388) + 1 = 12[Drives/Node]
　　RoundUp(value): Result of rounding value up to an integer
```
Through the preceding calculation, it is found that 50-TiB logical capacity is available by constructing a storage cluster consisting of six storage nodes with twelve 1.6-TB drives.