VMS / NVR Server
Enter the number of cameras, bitrate and retention and I compute the recording bandwidth, storage (TB), write IOPS, server RAM and how many servers you need by their throughput/channel limit.
Why that many servers?
| Criterion | Calculation | Servers |
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Storage uses decimal TB (1 TB = 10¹² bytes): BW(Mbps)/8 = MB/s, × 86400 s/day × days ÷ 10⁶. Add headroom for the operating system, indexes, RAID and growth. IOPS are low because recording is large-file sequential writing (≈ one stream per camera).
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How it works
The calculator sizes the recording server of a CCTV system from the number of cameras, the bitrate per camera and the retention days. Recording bandwidth is Mbps = cameras × bitrate and storage is computed in decimal TB (1 TB = 10¹² bytes): TB = (Mbps/8) × 86400 × days ÷ 10⁶. Write IOPS are estimated as one sequential stream per camera, because recording video is sequential writing of large files, not random I/O.
It also splits the load across servers according to two configurable limits (advanced section): maximum throughput per server (300 Mbps by default, typical for a VMS on standard hardware) and maximum channels per license or software (128 by default). The server count is max(⌈total Mbps / Mbps limit⌉, ⌈cameras / channel limit⌉), and RAM is estimated as base + cameras × RAM per camera (8 GB + 0.25 GB per channel by default).
Example: 64 cameras at 4 Mbps with 30 days of retention
- Bandwidth:
64 × 4 = 256 Mbpsof sustained recording. - Storage:
256/8 = 32 MB/s × 86400 × 30 ÷ 10⁶ ≈ 82.9 TB. - Servers: by throughput
⌈256/300⌉ = 1and by channels⌈64/128⌉ = 1→ 1 server is enough. - Estimated RAM:
8 + 64 × 0.25 = 24 GB, with about 64 sequential write IOPS.