NOMA Downlink: Two-User Power Allocation, Sum Rate and Fairness

Calculate two-user NOMA rates, power allocation and Jain fairness. Compare manual or equal-rate allocation with OMA and explore residual SIC interference.

#5G #downlink #Jain fairness #NOMA #OMA #power allocation #Shannon capacity #SIC

This NOMA downlink calculator helps communications engineers, students and researchers explore power allocation for two users sharing one time–frequency resource. It calculates each user's achievable rate, the sum rate and Jain's fairness index. Use it to compare a fixed power split with an equal-rate allocation, investigate imperfect successive interference cancellation (SIC), or check a link-level simulation.

Enter the total transmit power assigned to the shared resource, its bandwidth and the integrated receiver noise power. Then enter the two channel power gains as linear ratios. These gains include propagation losses and the fading realization you want to examine; they are not distances, amplitudes or decibel values. For example, −80 dB corresponds to a power gain of 10⁻⁸. The strong user must have a gain at least as large as the weak user's because both receivers use the same noise power in this model.

In manual mode, α is the fraction of total transmit power assigned to the weak user's message. The remaining fraction goes to the strong user. Start with the default 80/20 split, then move α to see how the rates change. Equal-rate mode chooses the split that maximizes the smaller of the two modeled rates. The reported equal-rate fraction is also available as a suggestion while you explore manual allocation. Giving both messages equal power generally does not give their users equal rates.

The weak user decodes its message while treating the strong user's layer as interference. The strong receiver first decodes the weak layer, cancels it and then decodes its own message. Set the residual SIC power fraction to zero for ideal cancellation. A value of 0.01 leaves one percent of the canceled layer's received power as interference. This is a sensitivity model, not a bit-error-rate or decoder simulation.

The power-split charts show the throughput–fairness tradeoff across the complete allocation range. Jain's index ranges from 0.5 when only one of two users receives a nonzero rate to 1 when the rates are equal. A high index alone does not imply high throughput. The rate-tradeoff chart compares NOMA with time sharing. The OMA baseline gives each user half the time, full bandwidth and full transmit power during its slot, preserving average total power.

Results use Shannon rates with Gaussian signaling, without coding gaps, pilots, guard intervals, retransmissions or scheduling overhead. They represent one channel snapshot, not an average over Rayleigh fading or an outage probability. Keep receiver noise consistent when changing bandwidth; the calculator holds the entered integrated noise fixed. It does not design antennas, optimize beamforming, pair many users, model inter-cell interference or guarantee a practical modulation and coding scheme. Use measured channel/noise data and a detailed link simulation for deployment decisions.

How it is calculated

For a channel snapshot, let gs=∣hs∣2≥gw=∣hw∣2g_s=|h_s|^2\ge g_w=|h_w|^2 and let NN be the equal noise power integrated over bandwidth BB. The full-power received SNRs are ρk=Pgk/N\rho_k=P g_k/N. The selected weak-user fraction is α\alpha and the strong-user fraction is 1−α1-\alpha.

The two-user Shannon model follows Islam et al., Power-Domain NOMA in 5G Systems: Potentials and Challenges, arXiv manuscript pp. 7–9, and the two WiredWhite articles linked below:

γw=αρw1+(1−α)ρw,γw→s=αρs1+(1−α)ρs.\gamma_w=\frac{\alpha\rho_w}{1+(1-\alpha)\rho_w},\qquad \gamma_{w\to s}=\frac{\alpha\rho_s}{1+(1-\alpha)\rho_s}.

For sensitivity to imperfect cancellation, the strong-user SINR is modeled as

γs=(1−α)ρs1+εαρs.\gamma_s=\frac{(1-\alpha)\rho_s}{1+\varepsilon\alpha\rho_s}.

Here ε\varepsilon is a power residual fraction. This extension treats the residual as additional uncorrelated interference; it does not simulate SIC decoding errors.

User spectral efficiencies are Ck=log⁡2(1+γk)C_k=\log_2(1+\gamma_k) in bit/s/Hz. Rates are Rk=BCkR_k=B C_k with BB in Hz. Because the channels are ordered and the noise powers are equal, γw→s≥γw\gamma_{w\to s}\ge\gamma_w; the strong receiver supports the weak-layer coding rate. The displayed headroom is Cw→s−CwC_{w\to s}-C_w.

Equal-rate allocation solves γs=γw\gamma_s=\gamma_w, which maximizes the smaller rate because one rate falls and the other rises with α\alpha. With t=1−αt=1-\alpha, define q=ρs−1+ρw−1+2εq=\rho_s^{-1}+\rho_w^{-1}+2\varepsilon and c=ρs−1+εc=\rho_s^{-1}+\varepsilon. The derived quadratic is (1−ε)t2+qt−c=0(1-\varepsilon)t^2+qt-c=0, evaluated stably as

t=2cq+q2+4(1−ε)c.t=\frac{2c}{q+\sqrt{q^2+4(1-\varepsilon)c}}.

At perfect SIC this agrees with Taricco (2023), eqs. (12)–(16). The residual-SIC form is derived here.

Jain fairness is J=(Cw+Cs)2/[2(Cw2+Cs2)]J=(C_w+C_s)^2/[2(C_w^2+C_s^2)]. The OMA reference gives each user half the time at full power and bandwidth: Ck,OMA=12log⁡2(1+ρk)C_{k,\mathrm{OMA}}=\tfrac12\log_2(1+\rho_k). Its average transmit power is the same as NOMA. The rate-tradeoff chart varies OMA time sharing. A short small-SINR series evaluates the logarithm without cancellation at very low SNR.

Chart sampling combines a uniform grid with logarithmically spaced samples near both allocation endpoints and a dense neighborhood of the equal-rate split. The exact equal-rate and selected splits are included. Complementary power fractions are retained separately to preserve precision near an endpoint.

Assumptions

  • Two users share one downlink resource with a common effective channel for both superposed layers at each receiver. Gains are linear power ratios, include the chosen channel snapshot, and remain fixed during the calculation.
  • Both receivers have equal integrated noise power. The strong user is ordered by channel-to-noise quality, not by received power of its own layer.
  • Gaussian signaling, Shannon rate adaptation, perfect channel knowledge and synchronization. No implementation or coding gap is applied.
  • Residual SIC is modeled as a fixed fraction of canceled received weak-layer power. The weak layer is decoded before the strong user's own message.
  • All fractions from 0 to 1 are allowed to show single-user limits. A practical finite-constellation receiver may impose additional power-ratio and decoding constraints; a weak fraction above 0.5 alone does not prove reliable SIC.
  • Bandwidth changes hold the entered total noise power fixed. Update that input if the noise spectral density is held fixed instead.

Where it applies

This is an instantaneous two-user rate and allocation model. It does not calculate a fading average, an outage probability, BER/BLER, finite-blocklength performance, practical MCS selection, antenna patterns, beamforming or multi-cell scheduling. Under residual SIC, the plotted rate tradeoff is the stated interference model, not a claim about the broadcast-channel capacity region.

Equal rates maximize Jain's index and the smaller modeled rate; they do not generally maximize sum rate. NOMA is not guaranteed to exceed the equal-time OMA reference at every allocation or SIC residual. Coding, pilot and control overhead reduce delivered payload throughput.

Reference audit: Abu Mahady (2025), PDF p. 31 (printed p. 11), reports 4.53 bit/s NOMA and 3.38 bit/s OMA at 0/20 dB, which disagree with its own equations. Those totals are not used as expected values. The 10/30 dB example is retained as a published approximate benchmark. Its reported NOMA sum 9.52 bit/s is compatible with truncating the correct 9.525521 bit/s. Its reported OMA sum 6.72 bit/s contains a small arithmetic discrepancy: the equations give 6.713329 bit/s, which rounds or truncates to 6.71. The absolute reference tolerance of 0.01 bit/s explicitly accommodates the OMA discrepancy of 0.006671 bit/s (about 0.10%) and the NOMA reporting precision. The correct formulas are kept. Taricco's Fig. 2 labels the strong SNR as 13 dB and reports a strong-user allocation of 0.1608; exact 13 dB gives about 0.16106. The check allows the figure's SNR rounding.

Source integrity: attachment s4 was attached with an incorrect title; its actual subject is holographic superconductors, and it is excluded from this calculator. The verified NOMA manuscript is attachment s6. Only the author can remove the unrelated attachment.

References

  • W. Javaid, Power Allocation Scheme for Throughput of NOMA Systems in 5G Networks, WiredWhite, 22 July 2025 (post 30884). https://wiredwhite.com/power-allocation-noma-5g/
  • W. Javaid, Performance Analysis of Power-Domain Non-Orthogonal Multiple Access in 5G Networks Using MATLAB, WiredWhite, 28 April 2026 (post 47987). https://wiredwhite.com/power-domain-noma-5g-matlab/
  • S. M. R. Islam, N. Avazov, O. A. Dobre and K.-S. Kwak, Power-Domain Non-Orthogonal Multiple Access (NOMA) in 5G Systems: Potentials and Challenges, IEEE Communications Surveys & Tutorials 19(2), 2017, pp. 721–742. DOI: 10.1109/COMST.2016.2621116. Verified open manuscript: https://arxiv.org/abs/1609.06261 (attachment s6), pp. 7–9.
  • G. Taricco, Fair Power Allocation Policies for Power-Domain Non-Orthogonal Multiple Access Transmission With Complete or Limited Successive Interference Cancellation, IEEE Access 11, 2023, pp. 46793–46803. DOI: 10.1109/ACCESS.2023.3274470, eqs. (12)–(16) and Fig. 2. https://doi.org/10.1109/ACCESS.2023.3274470
  • I. Abu Mahady, Improving Performance of NOMA, Lakehead University PhD thesis, 2025, PDF pp. 30–32, especially p. 31, eqs. (2.2)–(2.7). Open repository: https://thesis.lakeheadu.ca/bitstream/handle/2453/5484/AbuMahadyI2025m-2b.pdf?isAllowed=y&sequence=1
  • R. Jain, D.-M. Chiu and W. Hawe, A Quantitative Measure of Fairness and Discrimination for Resource Allocation in Shared Computer Systems, DEC-TR-301, 1984. https://arxiv.org/abs/cs/9809099

Inputs

InputSymbolUnitDefault
Total transmit power
Power assigned to this two-user resource block, before channel attenuation.
P_txW1
Shared bandwidth
Both NOMA users occupy this bandwidth simultaneously. Noise below is integrated over this bandwidth.
B_chMHz1
Receiver noise power
Equal integrated noise at both receivers, including receiver noise figure. If bandwidth changes, update noise consistently.
N_rxW1e-14
Strong-user channel power gain
Linear received/transmitted power ratio, including path loss and the chosen fading realization. Power gain, not amplitude gain or decibels.
g_s1e-12
Weak-user channel power gain
Use a gain no greater than the strong-user gain. A power gain of −90 dB is 10⁻⁹.
g_w1e-13
Power allocation policy
Equal-rate mode maximizes the smaller of the two modeled rates over all power splits from 0 to 1.
allocationManual weak-user fraction
Weak-user power fraction α
0.8 assigns 80% of transmit power to the weak user. Endpoints are single-user limiting cases.
a_manual0.8
Residual SIC power fraction ε
Fraction of the canceled weak-user received power left as interference at the strong receiver: 0 = perfect cancellation; 0.01 = 1% residual power.
epsilon_sic0

Results

With the default inputs above, the calc gives:

ResultValue
Weak-user rate1.874 Mb/s
Strong-user rate4.392 Mb/s
NOMA sum rate6.267 Mb/s
NOMA sum spectral efficiency6.267 b/s/Hz
Jain fairness0.861
Equal-time OMA sum rate5.059 Mb/s
NOMA / OMA sum-rate ratio1.239
Applied weak-user power fraction0.8
Equal-rate weak-user fraction0.9409
Weak-layer SIC decoding headroom0.391 b/s/Hz

Checked against sources

Each case below was calculated by this calc and agrees with the value from its source (within the tolerance where one is shown).

Taricco Fig. 2: equal-rate allocation at 13/10 dB

Source: G. Taricco (2023), IEEE Access 11, Fig. 2, p. 46797, DOI 10.1109/ACCESS.2023.3274470. Published optimum strong-user fraction 0.1608 at labeled SNRs 13/10 dB. Exact 13 dB gives 0.161061; 0.1608 corresponds to rounding the linear SNR to 20 (13.0103 dB). Absolute tolerance 0.0005 covers the figure's rounded SNR. No formula is adjusted.

Inputs: Receiver noise power = 1 × 10⁻¹⁰ W; Strong-user channel power gain = 1.995 × 10⁻⁹; Weak-user channel power gain = 1 × 10⁻⁹; Power allocation policy = Equal rates / max-min fairness; the others as above.

ResultSourceThis calc
Strong-user fraction for equal rates0.1608 ± 5 × 10⁻⁴0.1611

Published 10/30 dB, 80/20 split: NOMA and OMA totals

Source: I. Abu Mahady (2025), Improving Performance of NOMA, Lakehead thesis, PDF p. 31 / printed p. 11: weak/strong full-power SNRs 10/30 dB, weak/strong shares 0.8/0.2, 1 Hz bandwidth. Published NOMA sum 9.52 and OMA sum 6.72 bit/s are retained as reference values. The correct equations give NOMA 9.525521 bit/s (compatible with two-decimal truncation) and OMA 6.713329 bit/s, which rounds or truncates to 6.71, not 6.72. The published OMA total contains an arithmetic discrepancy of 0.006671 bit/s (about 0.10%). Absolute tolerance 0.01 bit/s explicitly accommodates that discrepancy and the NOMA reporting precision; it is not a claim that 6.72 is a correctly rounded result. The formulas remain unchanged. The preceding 0/20 dB case has larger inconsistent totals (4.53/3.38) and is not used. URL: https://thesis.lakeheadu.ca/bitstream/handle/2453/5484/AbuMahadyI2025m-2b.pdf?isAllowed=y&sequence=1

Inputs: Receiver noise power = 1 × 10⁻¹⁰ W; Strong-user channel power gain = 1 × 10⁻⁷; Weak-user channel power gain = 1 × 10⁻⁹; Shared bandwidth = 1 × 10⁻⁶ MHz; the others as above.

ResultSourceThis calc
NOMA sum spectral efficiency9.52 ± 0.01 b/s/Hz9.526 b/s/Hz
Equal-time OMA sum spectral efficiency6.72 ± 0.01 b/s/Hz6.713 b/s/Hz

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