Network Link Capacity Calculator — Shannon-Hartley, Link Budget & Throughput
Compute channel capacity from bandwidth and SNR with Shannon-Hartley, build a free-space or fiber link budget, and find the physical, transport, and hardware throughput ceilings.
[ Data Throughput and Bandwidth ]
A clear path from inputs to results.
Estimate the real data rate of a network link from its physics — the Shannon-Hartley capacity set by bandwidth and signal-to-noise ratio, built up from a free-space or fiber link budget. It compares terrestrial fiber, a low-Earth-orbit satellite hop, and a deep-space link, then layers on round-trip time, bandwidth-delay product, and transport/hardware ceilings to show what actually limits throughput.
How to use
- Pick a link preset: terrestrial fiber core, LEO satellite (Starlink-like), or deep-space network (Mars backhaul). Each loads a distance, bandwidth, line rate, transmit power, antenna gains, noise figure, and losses.
- The engine builds the link budget: for radio links it uses ITU-R free-space path loss from frequency and distance; for fiber it uses attenuation per km. Received power is transmit power plus antenna gains minus path loss and misc losses.
- It computes the thermal noise floor from bandwidth and noise figure, forms the SNR, and applies the Shannon-Hartley limit, capping the result at the link’s line rate.
- Choose a transport protocol (CUBIC or BBRv3) and execution layer (Linux sockets, eBPF, or DPDK) to set the hardware/transport ceilings alongside the physical capacity.
- Adjust link stress (offered load) and burstiness (Hurst parameter) to see queueing delay, loss probability, round-trip time, and bandwidth-delay product respond.
- Read the headline capacity, SNR, RTT, queue delay, and loss, plus the physical, transport, and hardware ceilings that bound achievable throughput.
The formula
Reference: Shannon-Hartley channel-capacity theorem (Shannon, 1948); free-space path loss per ITU-R P.525; thermal-noise floor −10·log₁₀(kT₀) = −174 dBm/Hz at T₀ = 290 K (Boltzmann constant, Johnson-Nyquist noise).
Worked example
Deep-space network capacity (Mars backhaul)
- Deep-space preset: carrier frequency f = 8400 MHz, distance d = 221,000,000 km, bandwidth B = 20 MHz, line rate 50 Mbps, transmit power 43 dBm (~20 W), transmit gain 48 dB, receive gain 74 dB, noise figure 1.5 dB, misc loss 3.0 dB.
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Free-space path loss is FSPL = 32.44 + 20·log₁₀(8400) + 20·log₁₀(221×10⁶) = 32.44 + 78.49 + 166.89 = 277.81 dB. Received power is P_rx = 43 + 48 + 74 − 277.81 − 3.0 = −115.81 dBm. The thermal noise floor is N = −174 + 10·log₁₀(20×10⁶) + 1.5 = −174 + 73.01 + 1.5 = −99.49 dBm. So SNR = −115.81 − (−99.49) = −16.32 dB, i.e. a linear SNR of 10^(−1.632) = 0.0233. The Shannon capacity is C = B·log₂(1 + SNR) = 20×10⁶ × log₂(1.0233) = 20×10⁶ × 0.03324 = 664,700 bps. Because 665 kbps is far below the 50 Mbps line rate, the Shannon limit binds and sets the capacity.
SNR ≈ −16.32 dB and Shannon capacity ≈ 665 kbps — even with large dish gains, the enormous path loss over 221 million km makes the channel noise-limited, so capacity is set by Shannon, not the 50 Mbps line rate.
Interpreting the result
What limits a link depends on the regime. On the deep-space link, path loss is so large that SNR is negative in decibels and Shannon-Hartley caps capacity at a few hundred kbps regardless of the hardware line rate. On fiber and LEO links the SNR is high, so the Shannon value exceeds the line rate and capacity clamps to the line rate instead — the hardware, not the channel, is the bottleneck. Beyond the physical capacity, achievable throughput is further bounded by the transport protocol filling the bandwidth-delay product and by CPU/PCIe/execution-layer ceilings; the smallest of these is the real limit.
Assumptions & limitations
- Shannon-Hartley gives the theoretical capacity of an ideal channel; real modulation and coding achieve a fraction of it, so treat the capacity as an upper bound.
- Radio links use the ITU-R free-space path-loss model (no atmospheric, rain, or pointing losses beyond the entered misc-loss term); fiber uses a constant attenuation per km.
- The noise floor assumes a reference temperature of 290 K; actual system noise temperature (especially for cryogenically cooled deep-space receivers) can differ, shifting SNR.
- Antenna gains, powers, and losses are the preset or entered link-budget values and are treated as fixed; no adaptive coding, beamforming, or antenna-tracking dynamics are modeled.
- Queueing delay and loss use simplified utilization-based models; the animated protocol dynamics (CUBIC/BBR window evolution) are illustrative and are not part of the capacity/link-budget calculation.
Frequently asked questions
What is the Shannon-Hartley limit and why does it matter?
It is the maximum error-free data rate of a channel: C = B·log₂(1 + SNR), where B is bandwidth and SNR is the linear signal-to-noise ratio. It sets a hard ceiling no modulation or coding can beat. On the deep-space preset, B = 20 MHz and SNR ≈ 0.023 give C ≈ 665 kbps — the channel, not the hardware, is the bottleneck.
When does the line rate bind instead of Shannon?
When SNR is high enough that B·log₂(1 + SNR) exceeds the hardware line rate, the engine clamps capacity to the line rate. That is the case for the fiber and LEO presets, where the channel is clean and the electronics/optics are the limit — the opposite of the deep-space regime.
What is the bandwidth-delay product and why does it appear?
The BDP is capacity × round-trip time — the amount of data “in flight” a link can hold. A transport protocol must keep roughly one BDP of unacknowledged data outstanding to fill the pipe. On high-latency links (LEO, and especially deep space) the BDP is large, so small windows or buffers leave throughput far below the physical capacity.
Why is the deep-space link capacity so low despite 20 MHz of bandwidth?
Because capacity grows only logarithmically with SNR, and the SNR here is negative in dB. Free-space loss over 221 million km is about 277.8 dB, leaving received power at −115.8 dBm against a −99.5 dBm noise floor — an SNR of −16.3 dB. log₂(1 + 0.023) ≈ 0.033 bits/Hz, so 20 MHz yields only ~665 kbps.
How is the noise floor computed?
From thermal (Johnson-Nyquist) noise: N(dBm) = −174 + 10·log₁₀(B) + NF, where −174 dBm/Hz is kT₀ at 290 K, B is bandwidth in Hz, and NF is the receiver noise figure. For 20 MHz and a 1.5 dB noise figure that gives −99.49 dBm.
Do the protocol and execution-layer choices change the capacity?
They do not change the physical Shannon/line-rate capacity, but they set the transport and hardware ceilings. The execution layer (Linux sockets, eBPF, DPDK) applies an efficiency factor, and CPU/PCIe limits cap the hardware rate. Achievable throughput is the minimum of the physical, transport, and hardware ceilings.
Methodology last reviewed 2026-10-06. Results are informational and not financial or engineering advice — see our disclaimer.