Ferkans — Interactive Telecom Tutor

Fairness Through Power Control

The achievable rates derived in Section 18.3 depend critically on the large-scale fading coefficients $\beta_k$ . Without power control, cell-edge users with small $\beta_k$ achieve much lower rates than cell-centre users with large $\beta_k$ , creating severe unfairness. In the massive MIMO regime, the effective channels harden to deterministic values, making power control particularly effective: the optimisation is over the slowly-varying large-scale fading coefficients rather than the rapidly-fluctuating instantaneous channels. Max-min fairness power control equalises the worst-user rate, ensuring uniform quality of service across the cell.

Definition:
Max-Min Fairness Power Control

The max-min fairness power control problem for the massive MIMO uplink with MR combining is:

$\max_{\{p_k\}} \;\min_{k=1,\ldots,K} \; R_k(\{p_j\})$

subject to $0 \leq p_k \leq P_{\max}$ for all $k$ , where:

$R_k = \log_2\!\left(1 + \text{SINR}_k\right)$

Since $\log_2(1+x)$ is monotone increasing, this is equivalent to:

$\max_{\{p_k\}} \;\min_{k} \;\text{SINR}_k(p_1, \ldots, p_K)$

For MR combining in the large- $M$ regime with perfect CSI:

$\text{SINR}_k \approx \frac{p_k M \beta_k} {\sum_{j \neq k} p_j \beta_j + \sigma^2}$

The max-min problem seeks the power allocation that makes all users' SINRs equal while satisfying the power constraints.

,

Theorem: Optimal Max-Min Power Allocation with MR Processing

For the massive MIMO uplink with MR combining in the large- $M$ regime, the max-min optimal power allocation is:

$p_k^{\star} = \frac{\eta}{\beta_k}$

where $\eta > 0$ is chosen such that $\max_k p_k^{\star} = P_{\max}$ , i.e.:

$\eta = P_{\max} \cdot \min_k \beta_k$

The resulting equal SINR for all users is:

$\text{SINR}^{\star} = \frac{\eta M} {(K-1)\eta + \sigma^2} = \frac{P_{\max}\beta_{\min} M} {(K-1)P_{\max}\beta_{\min} + \sigma^2}$

where $\beta_{\min} = \min_k \beta_k$ .

The optimal strategy inverts the channel: users with worse channels (smaller $\beta_k$ ) transmit with more power. This is precisely the channel inversion strategy. The equal SINR is determined by the weakest user (the one with the smallest $\beta_k$ , which transmits at $P_{\max}$ ). All other users reduce their power so that their received power at the BS matches the weakest user's received power: $p_k\beta_k = p_j\beta_j = \eta$ for all $k, j$ .

Proof

Symmetry of the max-min solution

At the max-min optimum, all users must achieve the same SINR (otherwise, reducing the power of a higher-SINR user would decrease interference and increase the minimum SINR).

Setting $\text{SINR}_k = \gamma^{\star}$ for all $k$ :

$p_k M\beta_k = \gamma^{\star}\!\left(\sum_{j \neq k}p_j\beta_j + \sigma^2\right)$

Solving for the power allocation

Let $q_k = p_k\beta_k$ (received power from user $k$ ). The constraint becomes:

$q_k M = \gamma^{\star}\!\left(\sum_{j \neq k}q_j + \sigma^2\right)$

For all users to have the same SINR, we need $\sum_{j \neq k}q_j$ to be the same for all $k$ , which requires $q_k = q$ for all $k$ (equal received powers).

Then: $qM = \gamma^{\star}((K-1)q + \sigma^2)$ , giving $\gamma^{\star} = qM/((K-1)q + \sigma^2)$ .

Optimal power and equal received power

Equal received power $q = p_k\beta_k$ means $p_k = q/\beta_k$ . The user with the smallest $\beta_k$ uses the most power. Setting $p_{\min_k\beta_k} = P_{\max}$ :

$q = P_{\max}\beta_{\min}, \quad p_k = \frac{P_{\max}\beta_{\min}}{\beta_k}$

This is the channel inversion power control with $\eta = P_{\max}\beta_{\min}$ . $\blacksquare$

Max-Min Power Control for Massive MIMO

Complexity:

O(K)

— a single pass over the users

Input: Large-scale fading

\{\beta_k\}_{k=1}^{K}

, max power

P_{\max}

,

noise variance

\sigma^2

, number of antennas

M

Output: Power allocation

\{p_k^{\star}\}_{k=1}^{K}

, common SINR

\gamma^{\star}

1.

\beta_{\min} \leftarrow \min_{k=1}^{K} \beta_k

2.

\eta \leftarrow P_{\max} \cdot \beta_{\min}

3. for

k = 1, \ldots, K

do

4.

\quad p_k^{\star} \leftarrow \eta / \beta_k

5. end for

6.

\gamma^{\star} \leftarrow \dfrac{\eta M}{(K-1)\eta + \sigma^2}

7. return

\{p_k^{\star}\}

,

\gamma^{\star}

This closed-form solution is specific to MR combining with i.i.d. Rayleigh fading. For ZF or MMSE combining, or with correlated channels, the max-min problem generally requires iterative algorithms (e.g., bisection on the target SINR combined with linear feasibility checks).

Power Control Effect on User Rates

Compare the per-user rate distribution with and without max-min power control. Without power control, cell-edge users suffer; with power control, all users achieve the same (equalised) rate. Observe how the 5th-percentile rate improves dramatically.

Parameters

M

(BS antennas)100

K

(number of users)10

SNR [dB]0

Example: Power Control with Two Users

A massive MIMO base station with $M = 100$ antennas serves $K = 2$ users with $\beta_1 = 1$ (cell centre) and $\beta_2 = 0.01$ (cell edge). The noise variance is $\sigma^2 = 1$ and $P_{\max} = 10$ .

(a) Without power control ( $p_1 = p_2 = P_{\max}$ ), compute each user's rate. (b) With max-min power control, compute the equalised rate.

Solution

Without power control

$\text{SINR}_1 = \frac{10 \times 100 \times 1}{10 \times 0.01 + 1} = \frac{1000}{1.1} \approx 909KATEXPLACEHOLDER0ENDR_1 = \log_2(910) \approx 9.83\;\text{bits/s/Hz}KATEXPLACEHOLDER1END\text{SINR}_2 = \frac{10 \times 100 \times 0.01}{10 \times 1 + 1} = \frac{10}{11} \approx 0.91KATEXPLACEHOLDER2ENDR_2 = \log_2(1.91) \approx 0.93\;\text{bits/s/Hz}$ $

The cell-edge user achieves only 9.5% of the cell-centre rate.

With max-min power control

$\eta = P_{\max}\beta_{\min} = 10 \times 0.01 = 0.1KATEXPLACEHOLDER0ENDp_1^{\star} = 0.1/1 = 0.1, \quad p_2^{\star} = 0.1/0.01 = 10KATEXPLACEHOLDER1END\text{SINR}^{\star} = \frac{0.1 \times 100}{0.1 + 1} = \frac{10}{1.1} \approx 9.09KATEXPLACEHOLDER2ENDR^{\star} = \log_2(10.09) \approx 3.33\;\text{bits/s/Hz}$ $

Comparison

Metric	No PC	Max-min PC
$R_1$	9.83	3.33
$R_2$	0.93	3.33
Min rate	0.93	3.33
Sum rate	10.76	6.67

Max-min power control increases the minimum rate by $3.6\times$ at the cost of reduced sum rate. The cell-edge user's rate improves from 0.93 to 3.33 bits/s/Hz. $\blacksquare$

Quick Check

In max-min fairness power control for massive MIMO with MR combining, the optimal transmit power for user $k$ is proportional to:

$\beta_k$

$1/\beta_k$

$\beta_k^2$

$\sqrt{\beta_k}$

Correction:

1/\beta_k

The channel inversion strategy $p_k \propto 1/\beta_k$ equalises the received power $p_k\beta_k$ across all users, achieving max-min fairness.

Max-Min Fairness

A resource allocation criterion that maximises the minimum user rate (or SINR): $\max \min_k R_k$ . In massive MIMO with MR combining, the optimal solution is channel inversion $p_k \propto 1/\beta_k$ , equalising all users' rates.

Channel Inversion Power Control

A power control strategy where each user's transmit power is set inversely proportional to its channel gain: $p_k = \eta/\beta_k$ , so that the received power $p_k\beta_k = \eta$ is the same for all users. Optimal for max-min fairness with MR combining in massive MIMO.

Related: Max-Min Fairness, Massive MIMO

Power Control for Massive MIMO