For a K-user MAC, the capacity region has 2^K - 1 constraints (one for each non-empty subset). For K=3, how many rate constraints define the region (not counting the trivial R_i >= 0)?
CFour constraints: individual rates and sum rate only
DTen constraints from all possible combinations
For K users, the MAC capacity region is a polymatroid with 2^K - 1 constraints. For K=3, this gives 7 constraints: three individual (R_1 <= I(X_1;Y|X_2,X_3), etc.), three pairwise sums (R_i+R_j <= I(X_i,X_j;Y|X_k), etc.), and one three-way sum (R_1+R_2+R_3 <= I(X_1,X_2,X_3;Y)). The constraint for subset S is sum_{i in S} R_i <= I(X_S; Y | X_{S^c}), which is the rate of users in S given perfect knowledge at the receiver of users outside S (as if they were decoded and subtracted). This polymatroidal structure generalizes beyond MACs to network information theory.
Question 2 True / False
The sum-capacity of a Gaussian K-user MAC is log2(1 + (sum_i P_i) / N). This rate is achievable by TDMA (time division), so TDMA is optimal for Gaussian MACs.
TTrue
FFalse
Answer: False
TDMA (each user gets 1/K of the time) achieves sum rate (1/K) * K * (1/2) log2(1 + K*P/N) = (1/2) log2(1 + K*P/N) for equal power P. With successive interference cancellation (SIC), the sum rate is (1/2) log2(1 + (K*P)/N), which is strictly larger than TDMA when K > 1. The SIC-achieving coding scheme allows all K users to transmit simultaneously (non-orthogonal access), and the receiver separates them via SIC instead of dividing time. The power factor grows from K*P (TDMA) to (sum of all powers) without time division, making SIC more efficient.
Question 3 Short Answer
In successive interference cancellation (SIC), the decoding order matters for the individual rates R_i but not the sum-rate boundary. Explain why the sum rate is the same regardless of SIC order.
Think about your answer, then reveal below.
Model answer: The sum-rate constraint is R_1 + R_2 + ... + R_K <= I(X_1, X_2, ..., X_K; Y), which depends only on the joint information between all inputs and the output, independent of how the receiver processes them. Decoding order affects individual rates (which user decodes first, which second, etc.), yielding different rate tuples, but these tuples all lie on the same hyperplane in rate space. Different decoding orders correspond to different corner points of the capacity region's dominant face. Time-sharing (randomizing the decoding order) traces the full dominant face. Any point on the dominant face is achievable; points deeper in the region (below the dominant face) use probabilistic time-sharing between SIC orders.
This is a consequence of the polymatroidal structure: all rate points with the same sum lie on the same hyperplane, and the union of achievable rate points forms a convex polytope. The dominant face is the sum-rate boundary where all constraints are tight. Individual rates vary, but the aggregate (sum) is fixed once we commit to the sum-rate constraint.
Question 4 Multiple Choice
In a Gaussian 2-user MAC with users having power constraints P_1 = 10, P_2 = 1, and noise power N = 1. User 1 decodes first (treating user 2 as noise). What is user 1's rate R_1 (in bits)?
If user 1 decodes first, they must treat user 2's power P_2 = 1 as additional noise. The signal-to-interference-and-noise ratio is SNR = P_1/(P_2+N) = 10/(1+1) = 5, so R_1 = (1/2)*log2(1+5) = (1/2)*log2(6) ≈ 1.29 bits. If user 2 decoded first instead, user 2 would get R_2 = (1/2)*log2(1+1/1) = 0.5 bits, then user 1 would get R_1 = (1/2)*log2(1+10/1) ≈ 1.79 bits after SIC removes user 2. Different orders yield different rate tuples on the capacity region boundary.