Exercises

With $N = 1$, the received signal at AP $m$ is scalar: $y_m = \sum_{k=1}^{K} g_{mk} \sqrt{p_k} s_k + w_m$. The Level 1 local estimate is $\hat{s}_{mk} = \hat{g}_{mk}^* y_m$.

$\hat{s}_k = \sum_{m=1}^{M} \hat{s}_{mk} = \sum_{m=1}^{M} \hat{g}_{mk}^* y_m$

This is equivalent to MRC with the stacked channel estimate $\hat{\mathbf{g}}_k = [\hat{g}_{1k}, \ldots, \hat{g}_{Mk}]^T$: $\hat{s}_k = \hat{\mathbf{g}}_k^H \mathbf{y}$.

Each AP forwards one complex scalar per served user per channel use: $|\mathcal{D}_m| = 8$ complex samples per channel use.

Each AP forwards its full received vector: $N = 4$ complex samples per channel use.

In this case, Levels 1–3 actually require more fronthaul than Level 4 ($8 > 4$). This happens when $|\mathcal{D}_m| > N$. However, the Level 1–3 estimates are lower-dimensional (scalars vs. vectors), can be more aggressively quantized, and do not require the CPU to perform $MN \times MN$ matrix inversion. The comparison is fairer when accounting for quantization: Level 4 needs higher precision (more bits per sample) because the CPU must reconstruct the signal space.

$\mathbb{E}[\hat{g}_{mk}^* g_{mk}] = \mathbb{E}[|\hat{g}_{mk}|^2] = \gamma_{mk}$ (since $g_{mk} = \hat{g}_{mk} + \tilde{g}_{mk}$ and $\hat{g}_{mk} \perp \tilde{g}_{mk}$).

Desired signal power: $p \left(\sum_{m=1}^{M} \gamma_{mk}\right)^2$.

For user $j \neq k$: $\text{Var}(\hat{g}_{mk}^* g_{mj}) = \mathbb{E}[|\hat{g}_{mk}|^2 |g_{mj}|^2] = \gamma_{mk} \beta_{mj}$ (independent channels across APs, so cross-AP terms vanish).

Total interference: $\sum_{j \neq k} p \sum_m \gamma_{mk} \beta_{mj}$.

The beamforming uncertainty for user $k$: $\text{Var}(\hat{g}_{mk}^* g_{mk}) = \gamma_{mk} \beta_{mk}$

(Note: $\mathbb{E}[|\hat{g}_{mk}|^2 |g_{mk}|^2] = \gamma_{mk}(\gamma_{mk} + \beta_{mk})$, minus $\gamma_{mk}^2$ for the mean.)

Noise amplification: $\sigma^2 \sum_m \gamma_{mk}$.

$\text{SINR}_k^{(1)} = \frac{p \left(\sum_m \gamma_{mk}\right)^2}{\sum_j p \sum_m \gamma_{mk} \beta_{mj} + \sigma^2 \sum_m \gamma_{mk}}$$where the sum over$j$in the denominator includes$j = k$(beamforming uncertainty is part of the effective noise in the UatF bound).$\blacksquare$

$\text{SINR}_k^{(3)} = p_k \frac{|\boldsymbol{\alpha}_k^H \mathbf{b}_k|^2}{\boldsymbol{\alpha}_k^H \mathbf{E}_k \boldsymbol{\alpha}_k} = p_k \frac{\boldsymbol{\alpha}_k^H (\mathbf{b}_k \mathbf{b}_k^H) \boldsymbol{\alpha}_k}{\boldsymbol{\alpha}_k^H \mathbf{E}_k \boldsymbol{\alpha}_k}$

This is a generalized Rayleigh quotient with rank-one numerator matrix. The maximum is $p_k \mathbf{b}_k^H \mathbf{E}_k^{-1} \mathbf{b}_k$, achieved by $\boldsymbol{\alpha}_k^{\star} = c \, \mathbf{E}_k^{-1} \mathbf{b}_k$.

The LSFD SINR equals the centralized MMSE SINR when the interference covariance matrix $\mathbf{E}_k$ is block-diagonal with the same block structure as the local combining. This occurs when: (a) There is no pilot contamination across APs (interference is AP-local), and (b) The local MMSE combiner at each AP is optimal given its own received signal.

In practice, pilot contamination introduces off-diagonal terms in $\mathbf{E}_k$ (correlation of interference across APs), creating a gap between Level 3 and Level 4. $\blacksquare$

$\gamma_{mk} = \frac{2 \beta_{mk}^{2}}{2 \beta_{mk} + 0.01}$

$\gamma_{11} = 2/2.01 = 0.995$, $\gamma_{12} = 0.02/0.21 = 0.095$ $\gamma_{21} = 0.5/1.01 = 0.495$, $\gamma_{22} = 0.5/1.01 = 0.495$ $\gamma_{31} = 0.02/0.21 = 0.095$, $\gamma_{32} = 2/2.01 = 0.995$

Numerator: $(\gamma_{11} + \gamma_{21} + \gamma_{31})^2 = (0.995 + 0.495 + 0.095)^2 = (1.585)^2 = 2.512$

Denominator: $\sum_j \sum_m \gamma_{m1} \beta_{mj} + \sigma^2 \sum_m \gamma_{m1}$

For $j = 1$: $0.995 \times 1 + 0.495 \times 0.5 + 0.095 \times 0.1 = 0.995 + 0.248 + 0.010 = 1.253$ For $j = 2$: $0.995 \times 0.1 + 0.495 \times 0.5 + 0.095 \times 1.0 = 0.100 + 0.248 + 0.095 = 0.443$ Noise: $0.01 \times 1.585 = 0.016$

$\text{SINR}_1^{(1)} = \frac{2.512}{1.253 + 0.443 + 0.016} = \frac{2.512}{1.712} = 1.467$ (1.66 dB)

$\mathbf{b}_1 = [0.995, 0.495, 0.095]^T$

$[\mathbf{E}_1]_{mm} = \sum_j \gamma_{m1} \beta_{mj} + \sigma^2 \gamma_{m1}$

$[\mathbf{E}_1]_{11} = 0.995(1.0 + 0.1 + 0.01) = 0.995 \times 1.11 = 1.104$ $[\mathbf{E}_1]_{22} = 0.495(0.5 + 0.5 + 0.01) = 0.495 \times 1.01 = 0.500$ $[\mathbf{E}_1]_{33} = 0.095(0.1 + 1.0 + 0.01) = 0.095 \times 1.11 = 0.105$

$\boldsymbol{\alpha}_1^{\star} = [0.995/1.104, \; 0.495/0.500, \; 0.095/0.105]^T = [0.901, \; 0.990, \; 0.905]^T$

LSFD SINR: $\mathbf{b}_1^H \mathbf{E}_1^{-1} \mathbf{b}_1 = 0.995^2/1.104 + 0.495^2/0.500 + 0.095^2/0.105 = 0.897 + 0.490 + 0.086 = 1.473$

The LSFD gain over Level 1 is $10\log_{10}(1.473/1.467) = 0.02$ dB — negligible in this small example because the interference is already low.

Level 1 uses $\mathbf{a}_{mk} = \hat{\mathbf{g}}_{mk}$ (MRC). Level 2 uses $\mathbf{a}_{mk} = \mathbf{Z}_m^{-1} \hat{\mathbf{g}}_{mk}$ (MMSE). The MMSE combiner maximizes the SINR among all local combiners, so the Level 2 local SINR is at least as large as the Level 1 local SINR at every AP. Since the CPU uses the same combining (equal weights) in both levels, the final SINR inherits the ordering.

Level 2 uses $\alpha_{mk} = 1$ for all $m$. Level 3 optimizes $\boldsymbol{\alpha}_k$ to maximize the SINR. Since $\boldsymbol{\alpha}_k = \mathbf{1}$ is a feasible choice for the Level 3 optimization, the optimal Level 3 SINR is at least as large as Level 2.

Level 3 restricts the combining vector to the form $\mathbf{v}_{k} = \sum_m \alpha_{mk} [\mathbf{0}, \ldots, \mathbf{a}_{mk}, \ldots, \mathbf{0}]^T$ — a block-structured vector determined by local combiners and scalar weights. Level 4 optimizes over all $\mathbf{v}_{k} \in \mathbb{C}^{MN}$ without this structural constraint. Since the Level 3 feasible set is a subset of the Level 4 feasible set, $\text{SINR}_k^{(3)} \leq \text{SINR}_k^{(4)}$. $\blacksquare$

$[\mathbf{b}_k]_m = \mathbb{E}[\hat{\mathbf{g}}_{mk}^H \mathbf{g}_{mk}]$

Since $\mathbf{g}_{mk} = \hat{\mathbf{g}}_{mk} + \tilde{\mathbf{g}}_{mk}$ and $\hat{\mathbf{g}}_{mk} \perp \tilde{\mathbf{g}}_{mk}$: $[\mathbf{b}_k]_m = \mathbb{E}[\|\hat{\mathbf{g}}_{mk}\|^2] = N \gamma_{mk}$

$[\mathbf{D}_{kj}]_{mm} = \mathbb{E}[|\hat{\mathbf{g}}_{mk}^H \mathbf{g}_{mj}|^2]$

For independent channels: $\hat{\mathbf{g}}_{mk}$ and $\mathbf{g}_{mj}$ are independent. $= \mathbb{E}[\|\hat{\mathbf{g}}_{mk}\|^2] \mathbb{E}[|(\hat{\mathbf{g}}_{mk}/\|\hat{\mathbf{g}}_{mk}\|)^H \mathbf{g}_{mj}|^2]$

The first factor is $N \gamma_{mk}$. The second is $\mathbb{E}[|\mathbf{u}^H \mathbf{g}_{mj}|^2] = \beta_{mj}$ for unit vector $\mathbf{u}$.

Thus $[\mathbf{D}_{kj}]_{mm} = N \gamma_{mk} \beta_{mj}$.

$[\mathbf{D}_{kk}]_{mm} = \mathbb{E}[|\hat{\mathbf{g}}_{mk}^H \mathbf{g}_{mk}|^2]$

$= \mathbb{E}[|\hat{\mathbf{g}}_{mk}^H (\hat{\mathbf{g}}_{mk} + \tilde{\mathbf{g}}_{mk})|^2]$ $= \mathbb{E}[|\|\hat{\mathbf{g}}_{mk}\|^2 + \hat{\mathbf{g}}_{mk}^H \tilde{\mathbf{g}}_{mk}|^2]$ $= \mathbb{E}[\|\hat{\mathbf{g}}_{mk}\|^4] + \mathbb{E}[\|\hat{\mathbf{g}}_{mk}\|^2] \mathbb{E}[|\mathbf{u}^H \tilde{\mathbf{g}}_{mk}|^2]$

For i.i.d. $\hat{\mathbf{g}}_{mk} \sim \mathcal{CN}(\mathbf{0}, \gamma_{mk} \mathbf{I}_N)$: $\mathbb{E}[\|\hat{\mathbf{g}}_{mk}\|^4] = N(N+1) \gamma_{mk}^2$

$[\mathbf{D}_{kk}]_{mm} = N(N+1) \gamma_{mk}^2 + N \gamma_{mk} (\beta_{mk} - \gamma_{mk}) = N^2 \gamma_{mk}^2 + N \gamma_{mk} \beta_{mk}$ $\blacksquare$

Fronthaul must carry $2b \times |\mathcal{D}_m| \times B$ bits per second (Nyquist rate, real and imaginary parts): $2b \times 12 \times 20 \times 10^6 \leq 2 \times 10^9$ $b \leq 2 \times 10^9 / (480 \times 10^6) = 4.17$

Maximum $b = 4$ bits per real dimension.

Dynamic range: $6\sigma_s = 6\sqrt{0.5} \approx 4.24$. $\sigma^2_{q} = \frac{(4.24)^2}{12 \times 2^8} = \frac{18}{3072} = 5.86 \times 10^{-3}$ per real dimension.

Per complex sample: $2\sigma^2_{q} = 1.17 \times 10^{-2}$.

At 12 dB SINR, the effective noise floor is $\sigma_s^2 / 10^{1.2} = 0.5/15.85 = 0.0316$. Quantization noise ratio: $1.17 \times 10^{-2} / 0.0316 = 0.370$. SINR loss: $10 \log_{10}(1 + 0.370) \approx 1.37$ dB.

This is significant. With $b = 4$ bits, the quantization degrades the 12 dB SINR to approximately 10.6 dB. To reduce the loss below 0.5 dB, we would need $b = 6$ bits, which requires $C_{\text{fh}} \geq 2.88$ Gbit/s.

$S = K \times \bar{R} = 40 \times 50 = 2000$ Mbit/s = 2 Gbit/s.

$P = M(P_{\text{circuit}} + P_{\text{fh}}) + P_{\text{tx}} + P_{\text{CPU}}$ $= 100(0.2 + 0.5) + 2 + 10 = 70 + 2 + 10 = 82$ W.

$\text{EE} = 2 \times 10^9 / 82 = 24.4$ Mbit/Joule.

Note that the circuit and fronthaul power ($70$ W) dominates the transmit power ($2$ W).

The local MMSE combiner at AP $m$ for user $k$ is $a_{mk}^{(2)} = \left(\sum_{j \in \mathcal{D}_m} p_j |\hat{g}_{mj}|^2 + \sum_j p_j c_{mj} + \sigma^2\right)^{-1} \hat{g}_{mk}$

With $N = 1$, this is a scalar: $a_{mk}^{(2)} = \hat{g}_{mk} / z_m$

where $z_m$ is the same scalar for all $k$. The local estimate becomes: $\hat{s}_{mk} = (a_{mk}^{(2)})^* y_m = \hat{g}_{mk}^* y_m / z_m^*$

Since $z_m$ is a common scalar factor that cancels in the SINR ratio, the Level 2 SINR with $N = 1$ equals the Level 1 SINR.

The benefit of local MMSE over MRC comes from the AP's ability to spatially suppress interference, which requires $N \geq 2$ antennas. With a single antenna, the AP cannot distinguish between users based on spatial signatures. This implies that Level 2 provides no gain over Level 1 for single-antenna APs, and the jump from Level 1 to Level 3 (LSFD) provides all the intermediate performance improvement. $\blacksquare$

When $\mathbf{E}_k = \text{diag}(e_1, \ldots, e_M)$: $\mathbf{b}_k^H \mathbf{E}_k^{-1} \mathbf{b}_k = \sum_{m=1}^{M} \frac{|[\mathbf{b}_k]_m|^2}{e_m}$

Each term $p_k |[\mathbf{b}_k]_m|^2 / e_m$ is the contribution of AP $m$ to the overall SINR, which equals the SINR that AP $m$ would achieve if it were the only AP.

The total SINR is the sum of per-AP SINRs. This is the macro-diversity gain of cell-free massive MIMO: having $M$ distributed APs is equivalent (in SINR) to having $M$ independent receivers whose SINRs add up. This is selection diversity replaced by maximal-ratio diversity across APs.

The condition for diagonal $\mathbf{E}_k$ — no inter-AP interference correlation — holds when pilots are orthogonal (no pilot contamination) and channels across APs are independent. $\blacksquare$

Let $f(\rho) = \rho \bar{R}(\rho)$ (numerator up to constants) and $g(\rho) = P_{\text{circuit}} + c_{\text{fh}} \rho + \rho p / \eta$ (denominator up to constants). Then $\text{EE}(\rho) = B f(\rho) / g(\rho)$.

The superlevel set $\{\rho : f(\rho)/g(\rho) \geq \alpha\} = \{\rho : f(\rho) - \alpha g(\rho) \geq 0\}$. Since $f(\rho)$ is concave (product of linear and concave) and $g(\rho)$ is linear (hence convex), $f - \alpha g$ is concave. Superlevel sets of concave functions are convex, so $\text{EE}$ is quasi-concave.

Using the quotient rule: $\frac{d}{d\rho} \frac{f(\rho)}{g(\rho)} = \frac{f'(\rho) g(\rho) - f(\rho) g'(\rho)}{g(\rho)^2} = 0$

Thus $f'(\rho^{\star}) g(\rho^{\star}) = f(\rho^{\star}) g'(\rho^{\star})$, or equivalently: $\frac{f'(\rho^{\star})}{f(\rho^{\star})} = \frac{g'(\rho^{\star})}{g(\rho^{\star})}$

This says the "elasticity" of the sum rate equals the "elasticity" of the power. Substituting: $\frac{\bar{R}(\rho^{\star}) + \rho^{\star} \bar{R}'(\rho^{\star})}{\rho^{\star} \bar{R}(\rho^{\star})} = \frac{c_{\text{fh}} + p/\eta}{P_{\text{circuit}} + (c_{\text{fh}} + p/\eta) \rho^{\star}}$.

Since $\text{EE}(\rho)$ is quasi-concave with $\text{EE}(0) = 0$ and $\text{EE}(\rho) \to 0$ as $\rho \to \infty$:

1. Set $\rho_{\text{lo}} = 0$, $\rho_{\text{hi}} = \rho_{\max}$, tolerance $\epsilon$.
2. While $\rho_{\text{hi}} - \rho_{\text{lo}} > \epsilon$: a. $\rho_{\text{mid}} = (\rho_{\text{lo}} + \rho_{\text{hi}}) / 2$ b. Evaluate $\text{EE}'(\rho_{\text{mid}})$ c. If $\text{EE}' > 0$: $\rho_{\text{lo}} = \rho_{\text{mid}}$; else: $\rho_{\text{hi}} = \rho_{\text{mid}}$
3. Return $\rho^{\star} = (\rho_{\text{lo}} + \rho_{\text{hi}}) / 2$.

Convergence: $O(\log_2(1/\epsilon))$ iterations. $\blacksquare$

Level 4 requires forwarding $\mathbf{y}_m \in \mathbb{C}^N$: $R_{\text{fh}} = 2 \times 8 \times 8 \times 100 \times 10^6 = 12.8$ Gbit/s.

$12.8 > 10$ Gbit/s. No, the fronthaul link cannot support Level 4 with $b = 8$. Maximum bits: $b = \lfloor 10 \times 10^9 / (16 \times 10^8) \rfloor = \lfloor 6.25 \rfloor = 6$ bits. Alternatively, the AP could use Level 3 (LSFD) which requires only $2 \times 8 \times |\mathcal{D}_m| \times 100 \times 10^6$ bit/s. For $|\mathcal{D}_m| = 10$: $16$ Gbit/s (still too high with $b = 8$). With $b = 5$: $10$ Gbit/s — just feasible.

Level 4: $R_4 = 2bNf_s$ bit/s per AP. Level 3: $R_3 = 2b|\mathcal{D}_m|f_s$ bit/s per AP. Ratio: $R_4/R_3 = N / |\mathcal{D}_m|$.

Level 4 has lower fronthaul when $N < |\mathcal{D}_m|$, i.e., when each AP serves more users than it has antennas. This occurs in dense user deployments with few-antenna APs.

For example, with $N = 1$ (single-antenna APs) and $|\mathcal{D}_m| = 10$ users: Level 4 requires $1/10$ of Level 3's fronthaul. This is the regime where centralized processing is both optimal in SINR and efficient in fronthaul.

Conversely, with $N = 8$ and $|\mathcal{D}_m| = 4$: Level 4 requires $2\times$ the fronthaul of Level 3. Here, distributed processing is the clear choice. $\blacksquare$

An AP at distance $d_{mk}$ serves user $k$ if $d_{mk} \leq r_{\max} = d_0 \beta_{\text{thr}}^{-1/\alpha}$. With APs on a square grid with spacing $d$, the number of APs within a circle of radius $r_{\max}$ is approximately $|\mathcal{M}_k| \approx \pi r_{\max}^2 / d^2 = \pi d_0^2 \beta_{\text{thr}}^{-2/\alpha} / d^2$.

For $\alpha = 4$: $|\mathcal{M}_k| \approx \pi d_0^2 / (d^2 \sqrt{\beta_{\text{thr}}})$.

In the noise-limited regime (interference negligible), $c_m \approx \sigma^2 \gamma_{mk}$ and the SINR becomes $\text{SINR}_k \approx (p / \sigma^2) \sum_{m \in \mathcal{M}_k} \gamma_{mk}$.

For APs uniformly distributed within the cluster: $\sum_m \gamma_{mk} \propto |\mathcal{M}_k| \bar{\gamma}$

where $\bar{\gamma}$ is the average estimate quality. Thus the SINR scales approximately linearly with $|\mathcal{M}_k|$ — classic macro-diversity gain.

Define the per-user utility: $U(|\mathcal{M}_k|) = B \log_2(1 + a |\mathcal{M}_k|) - c |\mathcal{M}_k|$ where $a$ captures the average per-AP SINR contribution and $c$ is the fronthaul cost per AP-user pair.

Taking the derivative: $\frac{dU}{d|\mathcal{M}_k|} = \frac{aB}{(1 + a|\mathcal{M}_k|) \ln 2} - c = 0$

$|\mathcal{M}_k|^{\star} = \frac{1}{a}\left(\frac{aB}{c \ln 2} - 1\right) = \frac{B}{c \ln 2} - \frac{1}{a}$

The optimal cluster size grows with the bandwidth $B$ and decreases with the fronthaul cost $c$. When fronthaul is cheap ($c \to 0$), the optimal cluster size diverges — use all available APs. $\blacksquare$

Sum rate: $S_a = 60 \times \bar{R}$ (to be normalized). Power: $P_a = 200(0.15 + 0.3) + 60 \times 0.1/0.3 + 15 = 90 + 20 + 15 = 125$ W. $\text{EE}_a = S_a / 125$.

Sum rate: $S_b = 0.92 S_a$ (8% reduction). Power: $P_b = 120(0.15 + 0.3) + 20 + 15 = 54 + 20 + 15 = 89$ W. $\text{EE}_b = 0.92 S_a / 89$.

$\text{EE}_b / \text{EE}_a = 0.92 \times 125 / 89 = 1.29$.

AP switching improves the energy efficiency by 29% despite an 8% rate loss. The power savings from turning off 80 APs (36 W in circuit + fronthaul) far outweigh the rate reduction.

The Bussgang theorem states that for a Gaussian input $x$ passed through a memoryless nonlinearity $Q(\cdot)$, the LMMSE decomposition is $Q(x) = \alpha_B x + e_B$ where $\alpha_B$ minimizes $\mathbb{E}[|Q(x) - \alpha x|^2]$.

Taking the derivative: $\alpha_B = \mathbb{E}[x^* Q(x)] / \mathbb{E}[|x|^2] = \mathbb{E}[x^* Q(x)] / \sigma_x^2$.

For the real part: $Q_R(x_R) = \sigma_q \, \text{sign}(x_R)$ where $\sigma_q$ is the output level. $\mathbb{E}[x_R Q_R(x_R)] = \sigma_q \mathbb{E}[|x_R|] = \sigma_q \sqrt{2\sigma_{x,R}^2/\pi} = \sigma_q \sigma_x \sqrt{1/\pi}$.

With unit output ($\sigma_q = 1$): $\alpha_B = \sqrt{2/(\pi \sigma_x^2)}$ per real dimension.

For the complex case: $\alpha_B = \sqrt{2/\pi} / \sigma_x$ (where $\sigma_x^2$ is the variance per real dimension).

$\sigma_e^2 = \mathbb{E}[|Q(x)|^2] - |\alpha_B|^2 \sigma_x^2$

For $b = 1$: $\mathbb{E}[|Q(x)|^2] = 2\sigma_q^2 = 2$ (unit levels, real + imaginary). $\sigma_e^2 = 2 - 2/\pi \cdot \sigma_x^2 / \sigma_x^2 \cdot 2 = 2(1 - 2/\pi) \approx 0.727$.

In general: $\sigma_e^2 = \mathbb{E}[|Q(x)|^2] - \alpha_B^2 \sigma_x^2$, which can be computed numerically for arbitrary $b$. For large $b$: $\alpha_B \to 1$ and $\sigma_e^2 \to \sigma_x^2 2^{-2b}/3$ (matching the additive noise model). $\blacksquare$

Level 1: Local estimates $\{\hat{s}_{mk}\}$ — just complex scalars. The CPU simply sums them. No channel knowledge needed at the CPU.

Level 2: Same as Level 1 — local estimates $\{\hat{s}_{mk}\}$.

Level 3: Local estimates $\{\hat{s}_{mk}\}$ plus large-scale fading statistics $\{\beta_{mk}\}$ (for computing LSFD weights). The statistics change slowly.

Level 4: Full received signals $\{\mathbf{y}_m\}$ plus channel estimates $\{\hat{\mathbf{g}}_{mk}\}$ for all $m, k$. The CPU needs instantaneous CSI.

Levels 1 and 2 require the least CPU-side information: only the scalar local estimates. Level 1 is the simplest because the APs also require the least information (only $\hat{g}_{mk}$ for the target user).

After the KLT, the components $\tilde{s}_{m,i}$ are independent with variances $\lambda_i$. The total distortion is $D = \sum_i d_i$ where $d_i$ is the per-component distortion.

Minimize $R = \sum_i \frac{1}{2} \log_2(\lambda_i / d_i)$ subject to $\sum_i d_i = D$ and $0 < d_i \leq \lambda_i$.

By KKT conditions: $d_i = \min(\lambda_i, \theta)$ where $\theta$ is chosen so that $\sum_i \min(\lambda_i, \theta) = D$.

Components with $\lambda_i \leq \theta$ get zero rate ($d_i = \lambda_i$); the rest get $R_i = \frac{1}{2}\log_2(\lambda_i / \theta)$ bits.

Scalar quantization allocates $b$ bits uniformly: $d_i = \lambda_i 2^{-2b}$, giving $D_{\text{scalar}} = 2^{-2b} \sum_i \lambda_i$ and $R_{\text{scalar}} = b |\mathcal{D}_m|$.

For the same distortion $D$, vector quantization achieves rate $R_{\text{vector}} = \sum_i \frac{1}{2} \log_2(\lambda_i / \theta)$ (summing only active components).