Exercises

(a) $\mathbb{E}[M_{10}] = H_{10} = 1 + 1/2 + \cdots + 1/10 \approx 2.93$.

(b) SNR (linear) $= 10^{1.5} \approx 31.6$.

$C_{\text{sum}}(10) \approx \log_2(1 + 31.6 \times 2.93) = \log_2(93.6) \approx 6.55$ bits/s/Hz.

(c) $C_1 = \mathbb{E}[\log_2(1 + 31.6\,|h|^2)]$. With $\mathbb{E}[|h|^2] = 1$: $C_1 \approx \log_2(32.6) \approx 5.03$ bits/s/Hz.

Multiuser diversity gain: $6.55 - 5.03 = 1.52$ bits/s/Hz (30% improvement with 10 users). $\blacksquare$

(a) Rayleigh ($\kappa = 0$): $\text{Var} = 1/1 = 1$.

Rician ($\kappa = 10$): $\text{Var} = 21/121 = 0.174$.

The Rician channel has 5.7$\times$ less variance.

(b) Multiuser diversity exploits the fact that different users have different instantaneous channel gains. When fading variance is low (strong LoS), all users have similar gains close to the mean, so $\max_k |h_k|^2 \approx \mathbb{E}[|h|^2]$ regardless of $K$. The opportunistic scheduler finds no "peaks" to exploit.

(c) 5G mmWave channels often have strong LoS components (high $\kappa$), reducing the multiuser diversity gain. Schedulers in mmWave systems rely more on spatial multiplexing (beamforming to multiple users) than temporal opportunism. $\blacksquare$

(a) Expanding:

$$C_{\text{sum}}(K) = K\sum_{j=0}^{K-1}\binom{K-1}{j}(-1)^j \int_0^{\infty}\log_2(1+\text{SNR}\,x)\,e^{-(j+1)x}\,dx$$

Each integral has the form:

$$\int_0^{\infty}\log_2(1+sx)\,\lambda e^{-\lambda x}\,dx = \frac{e^{\lambda/s}}{\ln 2}\,\text{E}_1(\lambda/s)$$

where $\text{E}_1(z) = \int_z^{\infty}e^{-t}/t\,dt$.

With SNR $= 31.6$ (10 dB):

(c) The approximation overestimates by 0.5--0.7 bits/s/Hz (8--10%). The gap arises because $\log_2(1+\text{SNR}\,x)$ is concave in $x$, so Jensen's inequality gives $\mathbb{E}[\log_2(1+\text{SNR}\,M_K)] < \log_2(1+\text{SNR}\,\mathbb{E}[M_K])$. $\blacksquare$

(a) Max-rate: $\arg\max_k R_k =$ User 1 ($R_1 = 8.0$).

(b) PF metrics: $m_1 = 8/6 = 1.33$, $m_2 = 3/1.5 = 2.00$, $m_3 = 5/4 = 1.25$, $m_4 = 1/0.5 = 2.00$.

Tie between Users 2 and 4. Break ties arbitrarily (e.g., User 2).

(c) Max-min: $\arg\min_k \bar{T}_k =$ User 4 ($\bar{T}_4 = 0.5$). Serve the worst-off user. $\blacksquare$

At the PF fixed point, $\bar{T}_1^{\star} = \bar{T}_2^{\star}$ by symmetry (both users are statistically identical).

With equal average throughputs, the PF metric reduces to $R_k/\bar{T}^{\star}$, which selects $\arg\max_k R_k$.

Since $R_1$ and $R_2$ are i.i.d. continuous random variables, $\Pr[R_1 > R_2] = 1/2$ by symmetry.

At the fixed point: $\bar{T}_k^{\star} = \mathbb{E}[\mathbf{1}_{k=k^{\star}} R_k]$.

By symmetry, $\Pr[k^{\star} = 1] = \Pr[k^{\star} = 2] = 1/2$.

Therefore each user is scheduled in exactly half the slots and $\bar{T}_1^{\star} = \bar{T}_2^{\star} = \mathbb{E}[R_1 \mid R_1 > R_2]/2$, confirming the symmetric equilibrium. $\blacksquare$

The $\alpha$-fair scheduler maximises $\sum_k U_{\alpha}(\bar{T}_k)$. By the gradient argument:

$$k^{\star}[t] = \arg\max_k \; R_k[t] \cdot \bar{T}_k^{-\alpha}$$

(a) $\alpha = 0$: $U_0(x) = x$. Metric: $R_k \cdot 1 = R_k$. This is the max-rate rule.

(b) $\alpha = 1$: $U_1(x) = \ln x$. Metric: $R_k / \bar{T}_k$. This is PF.

(c) $\alpha \to \infty$: Metric $R_k \cdot \bar{T}_k^{-\alpha}$. As $\alpha \to \infty$, the term $\bar{T}_k^{-\alpha}$ dominates for the user with the smallest $\bar{T}_k$, so the scheduler always serves the user with the lowest average throughput (regardless of instantaneous rate). This is max-min fairness. $\blacksquare$

(a) With $t_c = 1$: $\bar{T}_k[t+1] = R_k[t]$ if scheduled, and $\bar{T}_k[t+1] = 0$ otherwise. The PF metric becomes $R_k[t]/R_k[t-1]$ for the last-scheduled user and $R_k[t]/0 = \infty$ for unscheduled users. This degenerates to round-robin (every unscheduled user has infinite priority).

(b) With $t_c \to \infty$, $\bar{T}_k$ converges to the true ergodic average and never changes. The PF metric $R_k/\bar{T}_k$ becomes proportional to $R_k/\bar{T}_k^{\infty}$, a fixed scaling of each user's instantaneous rate. The scheduler becomes a channel-aware weighted max-rate scheme.

(c) A good rule of thumb is $t_c \in [10\,T_{\text{coh}},\, 100\,T_{\text{coh}}]$, so $t_c \in [100, 1000]$ slots. This ensures that $\bar{T}_k$ averages over many independent channel realisations (capturing the ergodic mean) while still adapting to slow changes in user geometry. $\blacksquare$

(a) Subcarriers 1--4 to User 1 (gain 4 > 1). Subcarriers 5--8 to User 2 (gain 4 > 1).

(b) Each user gets 4 subcarriers with gain 4 and power 1:

$R_k = 4 \times \log_2(1 + 4) = 4 \times 2.32 = 9.29$ bits/s/Hz. Both users achieve the same rate.

(c) Yes — by the symmetry of the channel, the throughput-optimal assignment is also perfectly fair. This is a special case; in general, sum-rate maximisation and fairness conflict. $\blacksquare$

If $g_n = g$ for all $n = 1, \ldots, N$, then:

$p_n = \mu - 1/g$ for all $n$ (assuming all are active).

From the power constraint: $N(\mu - 1/g) = P$, so $\mu = P/N + 1/g$ and $p_n = P/N$ for all $n$.

Water-filling degenerates to equal power allocation when all channels are identical.

Water-filling redistributes power from weak to strong subcarriers. When all subcarriers are equally strong, there is nothing to redistribute, and equal power is optimal. $\blacksquare$

(a) The subcarrier assignment with per-user rate constraints is a generalised assignment problem (GAP). GAP is known to be NP-hard (Martello and Toth, 1990). Specifically, even the feasibility question — does there exist an assignment satisfying all $R_{\min}$ constraints? — reduces to bin packing.

(b) Constraint-first heuristic:

1. For each user $k$, find the subcarrier $n_k^{\star}$ with the highest gain $g_{k,n}$ and tentatively assign it.
2. Allocate minimum power to each assigned subcarrier to meet $R_{\min}$ constraints.
3. Assign remaining subcarriers greedily (best-gain user).
4. Distribute remaining power via water-filling.

(c) In the worst case, the heuristic can be arbitrarily suboptimal (the GAP has no constant-factor polynomial-time approximation unless P = NP). However, for typical wireless channel statistics, the heuristic achieves within 5--10% of the optimum (verified by simulation in the literature, e.g., Jang and Lee, 2003). $\blacksquare$

(a) $\eta = 0.332 \times \log_2(64) = 0.332 \times 6 = 1.99$ bits/s/Hz.

(b) Total bandwidth: $10 \times 180 = 1800$ kHz $= 1.8$ MHz. Number of OFDM symbols per 0.5 ms slot: 7 (normal CP). Approximate RE count: $10 \times 12 \times 7 = 840$ REs. (Subtract $\sim$20% for reference signals: 672 data REs.)

TBS $\approx 1.99 \times 672 \approx 1337$ bits.

(c) With 2 slots per TTI (1 ms): Peak rate $= 1337 \times 2 / 10^{-3} \approx 2.67$ Mbps. $\blacksquare$

(a) Zero-drift condition: $(1-0.1) \times \delta_{\text{up}} = 0.1 \times \delta_{\text{down}}$

$\delta_{\text{down}}/\delta_{\text{up}} = 0.9/0.1 = 9$.

(b) $\delta_{\text{down}} = 9 \times 0.1 = 0.9$ dB.

(c) $\Delta[0] = 0$ ACK: $\Delta[1] = 0 + 0.1 = 0.1$ dB ACK: $\Delta[2] = 0.1 + 0.1 = 0.2$ dB NACK: $\Delta[3] = 0.2 - 0.9 = -0.7$ dB ACK: $\Delta[4] = -0.7 + 0.1 = -0.6$ dB ACK: $\Delta[5] = -0.6 + 0.1 = -0.5$ dB

The NACK caused a large downward correction (conservative shift), which is gradually recovered by successive ACKs. $\blacksquare$

(a) $\text{SINR}^{(1)} = 3^2/6 = 9/6 = 1.5$ (1.76 dB).

(b) $\text{SINR}^{(3)} = 3^2 \times 3^2/6 = 81/6 = 13.5$ (11.3 dB).

(c) $G_3 = 3^{4/2} = 9$ (9.54 dB improvement). Cell-edge SINR improves from 1.76 dB to 11.3 dB. $\blacksquare$

(a) A natural split proportional to user fraction: Inner zone: 35 RBs (shared by all cells, serves 70% of users). Outer zone: 15 RBs $\div$ 3 = 5 RBs per cell (reuse-3).

(b) Each cell has 5 exclusive RBs for edge users. With 30% of users ($= 0.3K$) on the cell edge, each edge user competes for 5 RBs. If $K = 20$: 6 edge users share 5 RBs.

(c) Reuse-1 edge rate per RB: $\log_2(1 + 10^{0/10}) = \log_2(2) = 1.0$ bit/s/Hz. Per-user rate (sharing 15 RBs among 6 users): $1.0 \times 15/6 = 2.5$ bits/s/Hz total.

FFR edge rate per RB: $\log_2(1 + 10) = 3.46$ bits/s/Hz. Per-user rate (sharing 5 RBs among 6 users): $3.46 \times 5/6 = 2.88$ bits/s/Hz total.

FFR improves edge user rate by 15% despite using 1/3 the bandwidth. $\blacksquare$

(a) With users uniformly distributed in a cell of radius $R$:

$K_{\text{in}}(r_{\text{th}}) = K \cdot (r_{\text{th}}/R)^2$ (fraction of cell area inside $r_{\text{th}}$).

Sum rate:

$$\mathcal{R}(r_{\text{th}}, W_{\text{in}}) = K_{\text{in}} \cdot W_{\text{in}} \cdot \bar{\eta}_{\text{in}}(\text{SINR}_{\text{in}}) + K_{\text{out}} \cdot W_{\text{out}} \cdot \bar{\eta}_{\text{out}}(\text{SINR}_{\text{out}})$$

where $W_{\text{out}} = (W - W_{\text{in}})/\Delta$ and $\bar{\eta}$ are the average spectral efficiencies.

(b) $K_{\text{in}}$ is quadratic in $r_{\text{th}}$, while the SINR functions depend nonlinearly on the zone boundary and interfering power levels. The product of these terms is in general non-convex.

(c) Since the problem has effectively one degree of freedom ($r_{\text{th}}$, with $W_{\text{in}}$ determined by the user fraction), a simple line search over $r_{\text{th}} \in [0, R]$ with step $\delta r$ suffices:

For each $r_{\text{th}}$, compute the sum rate analytically and pick the maximum. Complexity: $O(R/\delta r)$. $\blacksquare$

(a) $m_1 = 3.32/3.0 = 1.107$, $m_2 = 1.18/0.8 = 1.475$, $m_3 = 2.41/2.5 = 0.964$.

User 2 is scheduled ($m_2 = 1.475$ is highest).

(b) If User 2 is scheduled but HARQ NACK occurs, the effective rate is 0. Two approaches:

Approach 1 (count the NACK): $\bar{T}_2 = 0.99 \times 0.8 + 0.01 \times 0 = 0.792$.

Approach 2 (retransmission counts later): Defer the EWMA update until the HARQ process resolves (after combining). This is the approach used in LTE/NR.

(c) HARQ retransmissions are typically given highest priority in the scheduler (they are not subject to PF competition), because abandoning a partially decoded transport block wastes the initial transmission's resources. The PF metric is updated only after final ACK/NACK resolution. $\blacksquare$

The gradient of $\sum_k \ln \bar{T}_k$ with respect to the rate allocated on subcarrier $n$ to user $k$ is $1/\bar{T}_k$. The marginal contribution of assigning subcarrier $n$ to user $k$ is $\eta_k(n) / \bar{T}_k$.

Maximising the sum of marginal contributions over all subcarriers gives:

$$\pi^{\star}(n) = \arg\max_k \frac{\eta_k(n)}{\bar{T}_k}$$

This is the natural frequency-domain extension of PF: each subcarrier is independently assigned to the user with the highest PF metric on that subcarrier. When $N = 1$, it reduces to the standard time-domain PF rule. In LTE/NR, this per-RB PF scheduling is the standard approach. $\blacksquare$

(a) The SINR of user $k$ in cell $j$ on subcarrier $n$ is:

$$\gamma_{k,n}^{(j)} = \frac{p_n^{(j)} g_{k,n}^{(j,j)}} {p_n^{(\bar{j})} g_{k,n}^{(\bar{j},j)} + \sigma^2}$$

where $\bar{j}$ is the other cell. The rate depends on whether the other cell also transmits on subcarrier $n$.

(b) Even for a single cell with per-user constraints, OFDMA assignment is NP-hard. The two-cell coupling (through interference) makes the problem strictly harder.

(c) Best-response iteration: Each cell $j$ solves its own PF allocation problem treating the other cell's assignment as fixed interference, then updates. This is a best-response dynamic in a game-theoretic framework.

(d) If the interference coupling is weak (cell-centre users dominate), the best-response dynamics can be shown to converge via the theory of supermodular games or exact potential games. Under strong coupling (cell-edge users dominate), convergence is not guaranteed, and oscillations may occur. Practical implementations use damping (partial updates) or penalty terms to ensure convergence. $\blacksquare$