Exercises

$a^2 P_2 = 9 \times 10 = 90 \geq P_1 + \sigma^2 = 10 + 1 = 11$. Yes, very strong.

$R_{1} \leq \frac{1}{2}\log(1 + 10) = \frac{1}{2}\log 11 \approx 1.73$ bits. $R_{2} \leq \frac{1}{2}\log(1 + 10) = 1.73$ bits. Both users achieve interference-free capacity simultaneously.

$R_{\text{TIN}} = \frac{1}{2}\log(1 + 100/(1+10)) = \frac{1}{2}\log(1 + 100/11) = \frac{1}{2}\log(10.09) \approx 1.67$ bits.

TIN sum: $2 \times 1.67 = 3.34$ bits. Interference-free sum: $2 \times \frac{1}{2}\log(101) = \log(101) \approx 6.66$ bits. TIN achieves about 50% of the interference-free sum rate.

$a^2 \geq 1$, independent of SNR.

$a^2 P \geq P + \sigma^2$, i.e., $a^2 \geq 1 + 1/\text{SNR}$. At high SNR, this approaches $a^2 \geq 1$ — the gap between strong and very strong vanishes.

For $a^2 \in [1, 1 + 1/\text{SNR})$: strong but not very strong. Joint decoding is optimal but interference-free rates are not achievable. For $a^2 \geq 1 + 1/\text{SNR}$: very strong, interference-free rates achieved.

At Rx 1: $R_{1} + R_{2} \leq \frac{1}{2}\log(1 + P + a^2 P) = \frac{1}{2}\log(1 + (1+a^2)P)$. At Rx 2 (symmetric): same bound. Individual: $R_{k} \leq \frac{1}{2}\log(1 + P)$.

$C_{\text{sum}} = \min\{\frac{1}{2}\log(1+(1+a^2)P), \; 2 \times \frac{1}{2}\log(1+P)\}$. When $a^2$ is large: the MAC bound $\frac{1}{2}\log(1+(1+a^2)P)$ exceeds $\log(1+P)$, and the individual bounds are tight. When $a^2$ is close to 1: the MAC bound can be the active constraint.

$R_{\text{TIN}} = \frac{1}{2}\log\left(1 + \frac{P}{1 + P^{\alpha}}\right) = \frac{1}{2}\log\left(\frac{1 + P + P^{\alpha}}{1 + P^{\alpha}}\right)$$

For $\alpha < 1$ and large $P$: $1 + P + P^{\alpha} \approx P$ and $1 + P^{\alpha} \approx P^{\alpha}$. So: $$R_{\text{TIN}} \approx \frac{1}{2}\log(P / P^{\alpha}) = \frac{1}{2}\log(P^{1-\alpha}) = \frac{1-\alpha}{2}\log P$$

$d(\alpha) = \lim_{P \to \infty} \frac{R_{\text{TIN}}}{\frac{1}{2}\log P} = 1 - \alpha$$ confirming the GDoF formula for the TIN-optimal regime.

With TDMA, user $k$ transmits for a fraction $\tau_k$ of the time with $\sum_k \tau_k = 1$. User $k$'s rate: $\tau_k \cdot \frac{1}{2}\log(1 + \text{SNR}_{k} / \tau_k)$ (power concentration). The DoF per user: $\tau_k$. Sum DoF: $\sum_k \tau_k = 1$.

With IA, each user achieves DoF = $1/2$. Sum DoF: $K/2$. For $K = 3$: IA gives 3/2, TDMA gives 1. Improvement: 50%. For $K = 10$: IA gives 5, TDMA gives 1. Improvement: 400%.

This improvement is at infinite SNR. At finite SNR, the noise enhancement from IA (using many channel extensions with imperfect alignment) erodes the gains. For practical $K$ and moderate SNR, the IA advantage may not materialize.

Since Rx 2 sees no interference: $R_{2} \leq \frac{1}{2}\log(1 + P_2/\sigma^2)$.

User 1 sees $Y_1 = X_1 + aX_2 + Z_1$. This is a channel with state $aX_2$ known at Tx 2 but not at Tx 1 or Rx 1. User 1 treats $aX_2$ as noise: $R_{1} \leq \frac{1}{2}\log(1 + P_1/(a^2 P_2 + \sigma^2))$.

Tx 2 knows $X_2$ and could use DPC to help Rx 1 — but Tx 2 has its own message to send and no incentive to help. The capacity region is: $$R_{1} \leq \frac{1}{2}\log\left(1 + \frac{P_1}{a^2 P_2 + \sigma^2}\right), \quad R_{2} \leq \frac{1}{2}\log(1 + P_2/\sigma^2)$$ This is the TIN region, and it equals the capacity for the Z-IC.

Each user's entire message is the common message. Both receivers decode both messages jointly. This gives the MAC rate region at each receiver: $R_{1} + R_{2} \leq I(X_1, X_2; Y_k)$ for $k = 1, 2$. This is exactly the strong interference capacity region.

Each receiver decodes only its own message, treating the other user's entire signal as noise. The rate region is: $R_{k} \leq I(X_k; Y_k)$ where the mutual information is computed with interference treated as noise. This is exactly TIN.

For intermediate $\alpha \in (0, 1)$, HK smoothly interpolates between these two extremes, allowing each receiver to decode part of the interference (the common portion) while tolerating the rest (the private portion).

With genie side information, $R_{1} \leq I(X_1; Y_1, S_1) \leq \frac{1}{2}\log(1 + \text{SNR}_{1})$ since the genie allows Rx 1 to subtract interference perfectly.

$n(R_{1} + R_{2}) \leq I(X_1^n; Y_1^n, S_1^n) + I(X_2^n; Y_2^n)KATEXPLACEHOLDER0END\leq n \cdot \frac{1}{2}\log(1 + \text{SNR}_{1}) + n \cdot \frac{1}{2}\log\left(1 + \frac{\text{INR}_{21}}{1 + \text{SNR}_{1}}\right) + n\epsilon_n$$The second term arises from the correlation between$Y_2$and$S_1$via$X_2$, combined with the entropy power inequality.

The bound says: user 1 gets at most the interference-free rate, and the sum rate is further constrained by how much "excess" information Rx 2 can extract beyond what the genie already provided to Rx 1.

$\text{SNR} = 1000$ (linear), $\text{INR} = 0.3 \times 1000 = 300$.

Each user: $R_{k} = \frac{1}{3} \cdot \frac{1}{2}\log(1 + 3000) \approx \frac{1}{3} \times 5.79 = 1.93$ bits. Sum: $3 \times 1.93 = 5.79$ bits.

Each user: $R_{k} = \frac{1}{2}\log(1 + 1000/(1+600)) = \frac{1}{2}\log(2.66) \approx 0.71$ bits. Sum: $3 \times 0.71 = 2.12$ bits.

Sum rate $\approx \frac{3}{2} \times \frac{1}{2}\log(1000) = \frac{3}{2} \times 4.99 = 7.48$ bits. IA wins at this high SNR. But note that TIN performs very poorly due to the relatively strong interference ($\text{INR}/\text{SNR} = 0.3$). TDMA is the practical middle ground.

At Rx 1: $Y_1 = X_1 + X_2 + Z_1$. This is a 2-user MAC with equal-power Gaussian inputs. Sum-rate: $\frac{1}{2}\log(1 + 2P/\sigma^2)$.

$R_{k} \leq \frac{1}{2}\log(1 + P/(P + \sigma^2))$ when the other user is treated as noise (SIC first decodes the other user). Alternatively, $R_{k} \leq \frac{1}{2}\log(1 + P/\sigma^2)$ when the other user is decoded and removed. The sum capacity is $\frac{1}{2}\log(1 + 2P/\sigma^2)$.

• $a^2 \geq 1 + 1/\text{SNR}$: very strong, exact capacity known.
• $1 \leq a^2 < 1 + 1/\text{SNR}$: strong, exact capacity known.
• $1/(1+\text{SNR}) \leq a^2 < 1$: moderate, HK within 1 bit.
• $a^2 < 1/(1+\text{SNR})$: noisy, TIN within 0.5 bits.

The largest gap between inner and outer bounds occurs in the "moderate" regime where $a^2$ is close to 1 from below. Here, TIN is suboptimal and full decoding is not possible. The HK scheme with ETW split is within 1 bit, but the exact capacity within that 1-bit window remains unknown.

The difficulty stems from the non-convexity of the IC capacity optimization and the lack of a degradation structure. Unlike the BC (where channel enhancement works) or the MAC (where SIC is optimal), the IC has no structural property that enables tight single-letter characterization in the moderate regime.