Questions: Multiple Access Channel

Question 1

Two users share a Gaussian MAC with power constraints P_1 and P_2 and noise power N. The sum capacity is log2(1 + (P_1+P_2)/N). How does this compare to giving each user a separate channel with half the bandwidth?

• A. The MAC sum rate is always lower because of interference
• B. They are always equal because the total power and noise are the same
• C. The MAC sum rate is higher — treating interference as part of the signal (via SIC) is more efficient than splitting the channel
• D. The comparison depends on which user has more power

Answer: (C)

With orthogonal access (each user gets half the bandwidth), the sum rate is (1/2)log(1+2P_1/N) + (1/2)log(1+2P_2/N), which is strictly less than log(1+(P_1+P_2)/N) by the concavity of the log function. The MAC allows both users to transmit simultaneously across the full bandwidth, and SIC at the receiver separates them — this is more efficient than dividing the channel. Orthogonal multiple access (TDMA, FDMA) is suboptimal from an information-theoretic perspective.

Question 2

In successive interference cancellation, the decoding order affects the individual rates R_1 and R_2 but not the sum rate R_1 + R_2.

• T. True
• F. False

Answer: True

Decoding user 1 first (treating user 2 as noise) gives R_1 <= log(1 + P_1/(P_2+N)), then subtracting user 1 and decoding user 2 gives R_2 <= log(1 + P_2/N). Reversing the order gives R_1 <= log(1 + P_1/N) and R_2 <= log(1 + P_2/(P_1+N)). In both cases, R_1 + R_2 <= log(1 + (P_1+P_2)/N). The sum rate boundary is the same; the decoding order just selects different corner points of the capacity region. Time-sharing between decoding orders traces out the full dominant face.

Question 3

Explain why the MAC capacity region is a polygon (for two users) and how its corner points correspond to different decoding strategies.

Model answer: The two-user MAC capacity region is defined by R_1 <= I(X_1;Y|X_2), R_2 <= I(X_2;Y|X_1), and R_1+R_2 <= I(X_1,X_2;Y). The first constraint is the rate if user 2's signal were removed (decoded first); the second is the rate if user 1's signal were removed. The sum constraint bounds the total throughput. The resulting region is a pentagon. The two corner points on the dominant face correspond to the two SIC decoding orders: (I(X_1;Y|X_2), I(X_1,X_2;Y) - I(X_1;Y|X_2)) and (I(X_1,X_2;Y) - I(X_2;Y|X_1), I(X_2;Y|X_1)). Time-sharing between corners achieves any point on the dominant face. Each corner point gives one user the best possible rate while the other user gets whatever remains.

This pentagon structure generalizes to K users as a polymatroidal region. The number of constraints grows exponentially (2^K - 1 subset constraints), but the structure remains clean. The fact that the full region is achievable — not just corner points — is a remarkable result that relies on the flexibility of random coding.