# How can $N_0$ be zero in this example of calculating throughput in non-orthogonal multiple access?

I am very new to the area and I have a question. I came across this example, I'd appreciate it if anyone could clarify this for me. In the paper it says:

For example, we assume a multiple access scenario with 4 activated users $$\mathcal{X} = \{1,2,3,4\}$$. The equivalent transmission powers of 4 users are assumed as $$P_1 = 20 \text{dBm}$$, $$P_2 = 15 \text{dBm}$$, $$P_3 = 10 \text{dBm}$$, $$P_4 = 5 \text{dBm}$$ and the power of AWGN is assumed as $$N_0 = 0 \text{dBm}$$. Hence the theoretical throughput achieved via direct superposition scheme can be calculated as: $$\sum\limits_{i=1}^{4} R_i \leq \log_2{\left( 1+ \dfrac{\displaystyle\sum_{i=1}^4 P_i}{N_0} \right)} = 7.19 \text{bits/s/Hz}$$

My question is how does this equation hold? In paper, additive white gaussian noise (AWGN) is given as $$N(0,N_0)$$.

Here is the link for the paper (page 2, Equation 3).

That seems to work out. If I do the calculation in Python:

import numpy as np

N0 = 0.001
# P1=20dBm, P2=15dBm, P3=10dBm, P4=5dBm

P1 = np.power(10.0, 20/10)*N0
P2 = np.power(10.0, 15/10)*N0
P3 = np.power(10.0, 10/10)*N0
P4 = np.power(10.0, 5/10)*N0

sum = np.log2(1 + P1/N0 + P2/N0 + P3/N0 + P4/N0)
print(sum)


then I get:

7.187699013365945

Which tallies with the number you get to three significant figures.