Datacenter

IOPS with RAID Penalty

Define the disks, RAID level and read/write mix and I compute the functional IOPS applying the write penalty (RAID5=4, RAID6=6, RAID1/10=2) over the raw IOPS.

redzilla.cl — iops

Functional IOPS

—

application IOPS

Raw IOPS

—

physical array ceiling

Penalty

—

disk I/O per write

Functional —

Raw —

Backend W —

Disk I/O split

Reads to disk —

Writes to disk (with penalty) —

At the operating point, the array's physical I/O (—) splits between reads and writes; each application write costs — disk I/O.

The same workload on other RAID levels

RAID	Penalty	Functional IOPS	Backend W

With 8 disks of 180 IOPS and 70% read.

How it is computed · IOPS and the write penalty

Raw raw IOPS = number of disks × IOPS per disk. It is the physical ceiling of the array (all I/O to the backend).

Penalty each application write triggers several disk I/Os: RAID 0 = 1, RAID 1/10 = 2, RAID 5 = 4, RAID 6 = 6.

Functional functional IOPS = raw ÷ (read fraction + write fraction × penalty). These are the IOPS the application sees with that workload mix.

Backend W disk I/O from writes = functional IOPS × write fraction × penalty. Added to the reads it equals the raw IOPS.

Array-sizing model: it estimates how many disks you need for a workload at a given write percentage. The per-disk IOPS are reference values; measure your own for exact figures.

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How it works

The calculator estimates the functional IOPS of a disk array: the IOPS the application actually sees once the cost of RAID redundancy is subtracted. You start from three inputs: number of disks, IOPS per disk (with reference profiles: 7.2k HDD≈80, 10k≈140, 15k≈180, SATA SSD≈20,000, NVMe≈100,000) and the read/write mix of the workload.

The model is the standard storage-sizing formula: raw IOPS are disks × IOPS per disk, and every application write triggers several physical I/Os depending on the RAID level (write penalty: RAID 0 = 1, RAID 1/10 = 2, RAID 5 = 4, RAID 6 = 6). Functional IOPS come from raw ÷ (read fraction + write fraction × penalty). The tool also shows the write backend, the split of physical I/O and a comparison table with the same workload on all five RAID levels.

Example: 8 × 15k HDDs in RAID 10 with a 70/30 mix

Raw IOPS: 8 × 180 = 1,440 (the physical ceiling of the array).
RAID 10 has a 2× write penalty; with 70 % reads and 30 % writes the divisor is 0.7 + 0.3 × 2 = 1.3.
Functional IOPS: 1,440 ÷ 1.3 ≈ 1,108. The same array in RAID 5 (4× penalty) would give 1,440 ÷ 1.9 ≈ 758.
Write backend: 1,108 × 0.3 × 2 ≈ 665 disk I/Os, which added to the reads complete the 1,440 raw IOPS.

Frequently asked questions

Why does RAID 5 have a write penalty of 4 and RAID 6 a penalty of 6?

Because of the parity read-modify-write cycle. In RAID 5 every small write must read the data block and the parity block, then write both back: 4 disk I/Os. RAID 6 keeps two parity blocks, so it takes 2 reads and 4 writes: 6 I/Os in total.

How many IOPS does an HDD deliver compared with an SSD?

A 7,200 rpm HDD sits around 80 IOPS, a 10k drive around 140 and a 15k drive around 180, because they are limited by arm and platter mechanics. A SATA SSD delivers on the order of 20,000 IOPS and an NVMe drive can exceed 100,000. These are reference values: for precise sizing you should measure your actual drives.

What happens if my workload has more writes than I estimated?

Functional IOPS drop, and they drop faster the higher the RAID write penalty. With 8 disks of 180 IOPS in RAID 5, moving from 70/30 to 50/50 lowers functional IOPS from about 758 to about 576. That is why write-heavy workloads (databases, VDI) are usually sized on RAID 10 rather than RAID 5/6.

Does the difference between raw and functional IOPS matter when deciding how many disks to buy?

Yes, it is the whole point: the application only sees functional IOPS. If you need 2,000 IOPS with 30 % writes on RAID 5, the array must deliver 2,000 × (0.7 + 0.3 × 4) = 3,800 raw IOPS, which means almost twice the disks you would estimate from the raw figure alone.

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