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 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 $K \rightarrow \infty$