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RAID Capacity, Mixed-Workload IOPS and Annual Failure Probability

Compare RAID capacity and derive host IOPS from backend operation costs. Explore a one-year drive-failure probability under an explicit constant-hazard model.

Capacity before storage-system overhead

Use equal-size active drives, excluding hot spares. RAID0 uses all drive capacity; RAID5 and RAID6 reserve the equivalent of one and two drives for distributed parity, not dedicated parity drives. This exercise models RAID1 as a two-drive mirror and RAID10 as pairs of mirrored drives with an even count of at least four. Controller-specific configurations can differ.

Capacity is in decimal TB before metadata, formatting, snapshots and other reservations. Different-size drives generally require a separate layout calculation. These are planning calculations, not guarantees of usable filesystem space.

=IF(COUNT(B3:B4)<>2,"Enter numeric drive inputs",IF(OR(B3<1,B3<>INT(B3),B4<=0),"Invalid drive inputs",SWITCH(UPPER(TRIM(B2)),"RAID0",B3*B4,"RAID1",IF(B3<>2,"Use two drives for this RAID1 model",B4),"RAID5",IF(B3<3,"Need >=3 drives",(B3-1)*B4),"RAID6",IF(B3<4,"Need >=4 drives",(B3-2)*B4),"RAID10",IF(OR(B3<4,MOD(B3,2)<>0),"Need even >=4 drives",B3/2*B4),"Unknown RAID")))

Host IOPS from a host read/write mix

B5 is the available backend operations-per-second budget. B6 is the fraction of HOST operations that are reads, from 0 to 1. B7 is backend operations per host write. Assume one backend operation per host read. Each host operation therefore costs r + (1-r)p backend operations on average, so divide the backend budget by this cost.

For 4000 backend IOPS, 70% host reads and penalty 6, the estimate is 4000 / 2.5 = 1600 host IOPS. Those comprise 1120 reads and 480 writes, costing 1120 + 480*6 = 4000 backend operations. With penalty 2, the estimate is approximately 3076.923 host IOPS.

B7 is an independent workload assumption: conventional small random read-modify-write examples use 4 for RAID5 and 6 for RAID6, while two-copy writes use 2. Changing B2 does not automatically change B7. Cache, full-stripe writes, block size, latency, queue depth, rebuilds and hardware may substantially change observed throughput. This model is not a measured maximum.

=IF(COUNT(B5:B7)<>3,"Enter numeric IOPS inputs",IF(OR(B5<0,B6<0,B6>1,B7<1),"Invalid IOPS inputs",B5/(B6+(1-B6)*B7)))

Annual probability under a constant failure hazard

B8 is positive MTBF in hours. E4 calculates 1 - EXP(-8760/MTBF), a fraction, for one non-leap year under an exponential lifetime model. E5 converts it to percentage points. At 1.2 million hours it gives about 0.727342%.

This modeled probability is different from an observed fleet annualized failure rate based on failures per drive-year, or the small-rate approximation 8760/MTBF. MTBF is not an individual drive lifetime or warranty. Vendor conditions and population assumptions matter. This is not array data-loss probability; redundancy, correlated failures, rebuilds and backups need separate treatment.

=IF(NOT(ISNUMBER(B8)),"Enter numeric MTBF",IF(B8<=0,"MTBF must be positive",1-EXP(-8760/B8)))

Try the example

Use the original values or change them to test your understanding.

Further reading