Creator & media

RAID capacity and rebuild window calculator

Estimate conventional smallest-drive RAID capacity, explicit spare reservations and a one-member full-device rebuild window from entered throughput.

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Explore the result

Reserve the declared number of largest installed drives as spares. Equalize active members to the smallest capacity, apply the chosen parity/mirror convention and entered overhead, then model one full-member rebuild separately.

Worked example comparison
ExampleOutputWhat to notice
Worked exampleModeled usable capacity22.8 TB decimalFour active 8 TB members in RAID 5 have 24 TB before overhead and 22.8 TB after a 5% reduction. A full 8 TB write at 100 MB/s reduced by 20% takes about 27.778 hours.
RAID 6Modeled usable capacity15.2 TB decimalTwo member capacities are reserved for parity.
No throughput reductionModeled usable capacity22.8 TB decimalThe rebuild uses the full entered MB/s rate.

Check your understanding

Does this implement Synology SHR?

No, only conventional equalized RAID layouts.

Does rebuild use only occupied filesystem space?

No, it models one full smallest-capacity device.

Behind the numbers

How this calculation works

Reserve the declared number of largest installed drives as spares. Equalize active members to the smallest capacity, apply the chosen parity/mirror convention and entered overhead, then model one full-member rebuild separately.

THE FORMULARAID 5 raw usable TB = (active drives − 1) × smallest drive TB; rebuild seconds = one member bytes / effective bytes per second

A worked example

Four active 8 TB members in RAID 5 have 24 TB before overhead and 22.8 TB after a 5% reduction. A full 8 TB write at 100 MB/s reduced by 20% takes about 27.778 hours.

Good to know

Conventional equalized RAID only, not SHR or a backup/recovery guarantee. All-member RAID 1 mirror convention. Rebuild uses one full smallest member, not filesystem-used data or a failure probability model.

Reserve spares before applying the RAID formula

The model reserves the largest declared drives so they can cover a smallest-capacity member. All active members are then modeled at the smallest installed capacity.

Extra space on larger active drives is not automatically reused. Hybrid allocation schemes have different rules and are outside this conventional equalized-capacity calculation.

Capacity and rebuild time are separate questions

Parity and mirrors reduce usable capacity; the entered overhead reduces it further. Those capacity deductions do not directly tell how fast a replacement member can be rebuilt.

Rebuild time uses a full smallest-member capacity and an effective write rate after the entered load reduction. A controller that rebuilds only allocated regions may behave differently, as may read bottlenecks and concurrent activity.

Redundancy is not recovery assurance

The tool enforces basic member-count constraints, including even active members for RAID 10. Passing that arithmetic check does not predict successful recovery, error tolerance in a particular failure sequence or backup quality.

RAID 0 has no redundancy-based rebuild estimate. The remaining levels still need separate operational checks and backups; this page does not estimate disk failure probabilities.

Use the result with the next measurement

The PCM audio file size and WAV storage calculator estimates one kind of stored payload. Array capacity and backup/rebuild planning also need container overhead, retention and redundancy assumptions that the audio byte count does not include.

Conventional equalized RAID only, not SHR or a backup/recovery guarantee. All-member RAID 1 mirror convention. Rebuild uses one full smallest member, not filesystem-used data or a failure probability model.

A few questions, answered

Does this implement Synology SHR?

No, only conventional equalized RAID layouts.

Does rebuild use only occupied filesystem space?

No, it models one full smallest-capacity device.

Is RAID 0 rebuild time reported?

No, RAID 0 has no redundancy-based rebuild here.

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