Disk Throughput Calculator
Calculate sequential read/write throughput in MB/s from block size and IOPS.
About Disk Throughput Calculator
Throughput is the product of how much data moves per operation (block size) and how many operations complete per second (IOPS): a device doing 8,000 IOPS at a 128KB block size is moving roughly 1000MB/s. Real-world sustained throughput on a given link or interface rarely reaches the full theoretical figure because of protocol overhead, contention and controller limits, so an achievable link-utilization percentage is applied on top to produce a realistic effective figure.
Formulas
Theoretical throughput
throughput = block_size × IOPS
- B
- block size
- I
- IOPS
Effective throughput
effective = theoretical × utilization%
- u
- achievable utilization as a fraction
Worked example
With the default values below, Disk Throughput Calculator returns Effective Throughput of 1,000.00. Change any input above and every figure here is recalculated in your browser.
Inputs used
- Block Sizeminimum 0.5 KB
- 128 KB
- IOPSminimum 0
- 8,000
- Link/Interface Utilizationaccepts 1–100 %
- 100 %
Results
- Effective Throughput
- 1,000.00 MB/s
- Theoretical Throughput
- 1,000.00 MB/s
- Effective Throughput
- 0.9766 GB/s
- Data Movable per Hour
- 3,515.63 GB
Step by step
- 1Values used
- 2Theoretical throughput
- 3Effective throughput
- 4Effective Throughput= 1,000.00 MB/s
- 5Theoretical Throughput= 1,000.00 MB/s
- 6Effective Throughput= 0.9766 GB/s
- 7Data Movable per Hour= 3,515.63 GB
Assumptions and limits
- The result is derived only from the 3 inputs listed above. Anything not among them is not part of the calculation.
- In particular, this calculator does not take inflation, tax, fees or charges or compounding frequency into account.
- Figures are rounded for display to 4 decimal places, so totals built from several rounded parts can differ slightly from the unrounded arithmetic.
How these calculators are built and tested is described on the Methodology page. If a figure looks wrong, please report it.
Frequently Asked Questions
Why do larger block sizes generally yield higher throughput?
Larger blocks amortize the fixed per-operation overhead (command issue, seek/latency, protocol framing) over more data, so sequential workloads using large blocks (128KB-1MB) typically report much higher MB/s than random 4KB workloads even on the same device, despite the device's IOPS ceiling actually being higher for small blocks.
Why is IOPS often higher for small blocks but throughput higher for large blocks?
IOPS and throughput trade off against each other: small blocks let a device complete more discrete operations per second (higher IOPS), but each operation moves less data, so total MB/s can be lower than a workload using fewer, larger operations that saturate the same underlying bandwidth.
What utilization percentage should I assume for planning?
For well-tuned sequential workloads on modern NVMe SSDs, 85-95% of theoretical throughput is common. Network-attached or shared storage, older SATA/SAS interfaces, and highly random workloads typically see larger gaps between theoretical and effective throughput, often 50-70%.