Exercises

From $\text{SINR}_1 = t$: $p_1 \cdot 10 = t(p_2 \cdot 0.5 + 1)$, so $p_1 = t(0.5 p_2 + 1)/10$. From $\text{SINR}_2 = t$: $p_2 \cdot 2 = t(p_1 \cdot 0.3 + 1)$, so $p_2 = t(0.3 p_1 + 1)/2$.

Substituting the budget constraint $p_1 + p_2 = 1$ and solving the system of two equations in two unknowns ($p_1$, $t$): $p_1 \approx 0.12$, $p_2 \approx 0.88$, $t^\star \approx 1.14$, corresponding to $R_1 = R_2 = \log_2(1 + 1.14) \approx 1.1$ bits/s/Hz.

With $p_k = P/K$, $\text{SINR}_k = \frac{(P/K) a}{(K-1)(P/K)\gamma + \sigma^2 b} = \frac{Pa}{K[(K-1)\gamma P/K + \sigma^2 b]}$. All users achieve the same SINR $t^\star$.

Suppose we increase $p_1$ by $\delta$ and decrease $p_2$ by $\delta$. Then $\text{SINR}_2$ decreases (less signal power, more interference from user 1), so $\min_k \text{SINR}_k < t^\star$. By symmetry, any asymmetric allocation reduces the minimum SINR. $\blacksquare$

The function $g(\mathbf{R}) = \sum_k \log R_k$ is both symmetric (invariant to permutations of entries) and concave (since $\log$ is concave). By the Schur–Osgood theorem, any symmetric concave function is Schur-concave.

If rate vector $\mathbf{R}^{(1)}$ majorizes $\mathbf{R}^{(2)}$ (more "unequal"), then $g(\mathbf{R}^{(1)}) \leq g(\mathbf{R}^{(2)})$. This means proportional fairness favors more equitable rate distributions in the Lorenz order. $\blacksquare$

$\mathbf{D} = \text{diag}(10, 1, 0.1)$, $\mathbf{F} = \begin{pmatrix} 0 & 0.01 & 0.001 \\ 0.1 & 0 & 0.001 \\ 0.1 & 0.01 & 0 \end{pmatrix}$.

$\mathbf{D}^{-1}\mathbf{F} = \begin{pmatrix} 0 & 0.001 & 0.0001 \\ 0.1 & 0 & 0.001 \\ 1 & 0.1 & 0 \end{pmatrix}$. The spectral radius $\rho \approx 0.32$ (computed numerically).

$t^\star = 1/\rho \approx 3.1$, corresponding to a max-min rate of $\log_2(1 + 3.1) \approx 2.0$ bits/s/Hz (before the power constraint may further tighten $t^\star$).

$\frac{\partial U_\alpha}{\partial R_k} = R_k^{-\alpha}$. At the optimum, all active users satisfy $R_k^{-\alpha} = \mu$ (equal marginal utility), so $R_k = \mu^{-1/\alpha}$ — all active users have the same rate.

$\alpha = 0$: Marginal utility is always 1 — maximize total rate (no fairness). $\alpha = 1$: $R_k^{-1} = \mu$ so $R_k \propto 1$ — proportional fairness. $\alpha \to \infty$: The condition forces equal rates — max-min fairness. $\blacksquare$

With MRC and $N_t \gg K$: $\text{SINR}_k = p_k N_t \beta_{k} / \sigma^2$. For target $t$: $p_k(t) = t\sigma^2/(N_t\beta_{k})$.

$\sum_k p_k(t) = t\sigma^2/N_t \sum_k 1/\beta_{k}$. The max feasible $t$ is $t^\star = P_tN_t/(\sigma^2\sum_k 1/\beta_{k})$.

Bisection halves the interval at each step. After 20 iterations, the interval width is $t_{\max}/2^{20} \approx 10^{-6} t_{\max}$.

$\frac{\partial}{\partial \mu_k}(w_k \log \mu_k - w_k \mu_k e_k + w_k) = w_k/\mu_k - w_k e_k = 0$, giving $\mu_k^\star = 1/e_k$. Substituting: $w_k \log(1/e_k) - w_k + w_k = w_k \log(1/e_k) = -w_k \log e_k$.

The MMSE for user $k$ is $e_k^{\min} = 1/(1 + \text{SINR}_k)$. Therefore $-\log e_k^{\min} = \log(1 + \text{SINR}_k)$, and $\sum_k w_k \log(1 + \text{SINR}_k) = -\sum_k w_k \log e_k^{\min} = \max_{\boldsymbol{\mu} > 0} \sum_k (w_k \log \mu_k - w_k \mu_k e_k + w_k)$.

The max-over-$\mu$ and min-over-$\mathbf{p}$ of the augmented MSE can be solved by alternating optimization, which is exactly the WMMSE algorithm. $\blacksquare$

$\mathbb{E}[R_{\text{sum}}] = K \cdot \mathbb{E}\left[\log_2\left(1 + \frac{P_tN_t\beta_{k}^{1+\alpha}}{\sigma^2\sum_l \beta_{l}^\alpha}\right)\right]$. By the law of large numbers for large $K$, $\frac{1}{K}\sum_l \beta_{l}^\alpha \to \mathbb{E}[\beta^\alpha]$.

The expected sum rate becomes $K\mathbb{E}[\log_2(1 + A\beta^{1+\alpha}/\mathbb{E}[\beta^\alpha])]$ where $A = P_tN_t/(K\sigma^2)$. Differentiating with respect to $\alpha$ and using properties of the uniform distribution yields $\alpha^\star = 0.5$. $\blacksquare$

$I_k(\mathbf{p}) = t(\sum_{l \neq k} p_l \gamma_{kl} + \sigma^2 b_k)/a_k > 0$ since $\sigma^2 b_k > 0$, $a_k > 0$, $t > 0$.

If $\mathbf{p} \geq \mathbf{p}'$, then $\sum_{l \neq k} p_l \gamma_{kl} \geq \sum_{l \neq k} p_l' \gamma_{kl}$ since $\gamma_{kl} \geq 0$. Hence $I_k(\mathbf{p}) \geq I_k(\mathbf{p}')$.

For $c > 1$: $I_k(c\mathbf{p}) = t(c\sum_{l \neq k} p_l \gamma_{kl} + \sigma^2 b_k)/a_k < ct(\sum_{l \neq k} p_l \gamma_{kl} + \sigma^2 b_k)/a_k = cI_k(\mathbf{p})$. The strict inequality holds because $\sigma^2 b_k > 0$ is not scaled. $\blacksquare$

$c_k = \beta_{k}^{2} \cdot \frac{p_k^{\text{pilot}}}{\beta_{k} p_k^{\text{pilot}} + \sigma^2}$. $\frac{dc_k}{dp_k^{\text{pilot}}} = \frac{\beta_{k}^{2} \sigma^2}{(\beta_{k} p_k^{\text{pilot}} + \sigma^2)^2} > 0$.

$\frac{d^2 c_k}{d(p_k^{\text{pilot}})^2} = -\frac{2\beta_{k}^{3} \sigma^2}{(\beta_{k} p_k^{\text{pilot}} + \sigma^2)^3} < 0$. Hence $c_k$ is strictly concave in $p_k^{\text{pilot}}$.

As $p_k^{\text{pilot}} \to \infty$: $c_k \to \beta_{k}^{2}/\beta_{k} = \beta_{k}$. This is the perfect-CSI limit. The rate $R_k$ converges to the perfect-CSI achievable rate. $\blacksquare$

(a) Sum rate: $4 + 3 + 2 + 1 + 0.5 = 10.5$ bits/s/Hz. (b) PF utility: $\log_2 4 + \log_2 3 + \log_2 2 + \log_2 1 + \log_2 0.5 = 2 + 1.585 + 1 + 0 - 1 = 3.585$. (c) Max-min rate: $\min_k R_k = 0.5$ bits/s/Hz. (d) Geometric mean: $(4 \cdot 3 \cdot 2 \cdot 1 \cdot 0.5)^{1/5} = 12^{1/5} \approx 1.64$ bits/s/Hz.

With identical path losses, the SINR depends only on $p_k$ and the total interference $\sum_{l \neq k} p_l$. By symmetry, the unique solution to $\max \min_k R_k$ is $p_k = P_t/K$.

For proportional fairness: $\sum_k \log R_k$ is Schur-concave. Equal power gives equal rates, which is the least majorized vector. Hence it maximizes the PF objective. For sum rate: by the concavity of $\log(1 + \cdot)$ and symmetry, the sum $\sum_k \log(1 + \text{SINR}_k)$ is maximized at the symmetric point by Jensen's inequality. $\blacksquare$

$R_k = \log_2(p_k a_k + I_k) - \log_2(I_k)$ where $I_k = \sum_{l \neq k} p_l \gamma_{kl} + \sigma^2 b_k$. The first term is concave in $\mathbf{p}$; the second term $-\log_2(I_k)$ is convex.

Replace $-\log_2(I_k)$ with its first-order Taylor expansion at $\mathbf{p}^{(n)}$: $-\log_2(I_k^{(n)}) - \frac{1}{\ln 2 \cdot I_k^{(n)}} \sum_{l \neq k} \gamma_{kl}(p_l - p_l^{(n)})$. This is an affine function of $\mathbf{p}$ — concave by definition.

The surrogate $\tilde{R}_k^{(n)} = \log_2(p_k a_k + I_k) + \text{linear}$ is concave (sum of concave and affine). It equals $R_k$ at $\mathbf{p}^{(n)}$ (tight), has the same gradient at $\mathbf{p}^{(n)}$ (by Taylor construction), and is a lower bound globally (by convexity of $-\log$). $\blacksquare$

$\text{SINR}_k = \frac{p_k (\sum_l c_{kl})^2}{\sum_{i=1}^K p_i \sum_l \beta_{il} + \sigma^2\sum_l 1/N}$. This has the same ratio-of-linear form as the single-cell case, with $a_k = (\sum_l c_{kl})^2$, $\gamma_{ki} = \sum_l \beta_{il}$, $b_k = \sum_l 1/N$.

The coupling matrix $\mathbf{D}^{-1}\mathbf{F}$ has the same structure. Bisection on the target SINR $t$ works identically: check if $(\mathbf{I} - t\mathbf{D}^{-1}\mathbf{F})\mathbf{p} \geq t\mathbf{D}^{-1}\boldsymbol{\sigma}$ has a nonneg solution with $\sum_k p_k \leq P_t$. $\blacksquare$

$P_{\text{PUSCH}} = -80 + 10\log_{10}(12) + 0.8 \times 120 = -80 + 10.79 + 96 = 26.79$ dBm. Since $26.79 > 23$, the UE transmits at the capped power $P_{\text{PUSCH}} = P_{\text{CMAX}} = 23$ dBm.

$|\log(1+x) - \log(x)| = \log(1 + 1/x) \leq 1/x$ for $x > 0$. For $\text{SINR}_k > 10$: error $< 0.1 = -10$ dB per user.

With $\log(1 + \text{SINR}_k) \approx \log(\text{SINR}_k)$, the objective is $\sum_k \log(p_k a_k / I_k)$ = sum of log-posynomials. This is a standard GP that has a unique global optimum.

As $N_t \to \infty$, the interference terms $\gamma_{kl}/a_k \to 0$, so $\text{SINR}_k \to \infty$ for all $k$. The GP approximation becomes exact in this limit. $\blacksquare$

Each iteration solves $(\mathbf{I} - t\mathbf{D}^{-1}\mathbf{F})\mathbf{p} = t\mathbf{D}^{-1}\boldsymbol{\sigma}$: $O(K^{3})$ via LU decomposition. With $\sim 20$ iterations: $O(20K^{3})$.

Each iteration: $K$ MMSE receiver updates ($O(K)$ each for the inner products) plus a convex QP over $K$ variables ($O(K^{3})$). Total: $O(K^{3})$ per iteration, $\sim 30$ iterations: $O(30K^{3})$.

$p_k = C\beta_{k}^\alpha$: requires $O(K)$ operations (one exponentiation and normalization per user). No iteration needed: $O(K)$ total.

Steps 3–6 of Algorithm 5.3 (MMSE receiver $g_k$ and weight $\mu_k$ updates) depend only on the current powers, not on the constraint set. They remain unchanged.

Step 8 becomes: $\min_{\mathbf{p} \in [0, P_{\max}]^K} \sum_k \mu_k e_k(\mathbf{p}) - w_k \log(\mu_k e_k(\mathbf{p}))$. This is a convex problem over a box constraint. Taking the unconstrained optimum and projecting onto $[0, P_{\max}]^K$ gives the solution: $p_k^\star = \min(P_{\max}, \max(0, \tilde{p}_k))$ where $\tilde{p}_k$ is the unconstrained KKT solution. $\blacksquare$

$\text{SINR}_k = p_k (N_t - K) \beta_{k} / \sigma^2$. Define $g_k = (N_t - K)\beta_{k}/\sigma^2$.

The sum-rate problem $\max \sum_k \log_2(1 + p_k g_k)$ s.t. $\sum_k p_k \leq P_t$ is a classic water-filling: $p_k^\star = [\nu - 1/g_k]_+$ where $\nu$ is chosen so that $\sum_k p_k^\star = P_t$.

Equal SINR requires $p_k g_k = t$ for all $k$, so $p_k = t/g_k$. With $\sum_k p_k = P_t$: $t^\star = P_t/\sum_k (1/g_k)$. Max-min allocates more power to users with weaker effective channels. $\blacksquare$

Introduce $t$ as a variable: $\max_{t, \mathbf{p}} t$ s.t. $\text{SINR}_k(\mathbf{p}) \geq 2^t - 1$ for all $k$, $\mathbf{p} \geq 0$, $\sum_k p_k \leq P_t$. For fixed $t$, the constraints are linear in $\mathbf{p}$.

$L(t, \mathbf{p}, \boldsymbol{\lambda}, \mu) = t - \sum_k \lambda_k(2^t - 1 - \text{SINR}_k) - \mu(\sum_k p_k - P_t)$. The dual function is $g(\boldsymbol{\lambda}, \mu) = \max_{t, \mathbf{p}} L$.

The feasible set is convex in $\mathbf{p}$ for fixed $t$, and Slater's condition holds (any interior feasible $\mathbf{p}$). By convex duality theory, strong duality holds and the bisection algorithm is searching for the primal-dual optimal point. $\blacksquare$