We apply capacity region to K-user scenario.

With NOMA system, the K-user capacity region is described by $2^K -1$ constraints, one for each possible non-empty subset $\mathcal{S}$ 􏸙of users:

$$\sum _ { k \in \mathcal{S} } R _ { k } < \operatorname { log_{2} } \left( 1 + \frac { \sum _ { k \in \mathcal{S} } P _ { k } } { N _ { 0 } } \right) \quad \text { for all } \mathcal{S} \subset \{ 1 , \ldots , K \}$$ The right hand side corresponds to the maximum sum rate that can be achieved by a single transmitter with the total power of the users in 􏸙$\mathcal{S}$ and with no other users in the system. The sum capacity is $$C_{sum} = \log_{2}\left(1+\frac{\sum_{k=1}^K P_k}{N_0}\right)$$ It can be shown that there are exactly $K!$ corner points, each one corresponding to a successive cancellation order among the users.

The equal received power case ( $P_1$ = $\ldots$ = $P_K$ = $P$ ) is particularly simple. The sum capacity is

$$C_{sum} = \log_{2}\left(1+\frac{KP}{N_0}\right)$$

With FDMA, as is clearly shown in slide 3, the sum capacity of all users can achive the same as NOMA does, when the degree of freedom are equally allocated to each user.

With TDMA, the sum capacity of all users with equal power $P$ is

$$C_{sum} = \log_{2}\left(1+\frac{P}{N_0}\right)$$ .

With conventional CDMA, also called as Treat Interference as Noise(TIN), the sum capacity of all users with equal power $P$ is

$$C_{sum} = K \cdot \operatorname { log } _ { 2 } \left( 1 + \frac { P } { ( K - 1 ) P + N _ { 0 } } \right)$$ , which approaches $$K \cdot \frac { P } { \left( K - 1 \right) P + N _ { 0 } } \operatorname { log } _ { 2 } e \approx \operatorname { log } _ { 2 } e = 1.442 bits / s / Hz$$ as $K \rightarrow \infty$