I have the optimization problem
\begin{align}
\: \max \: \sum_{i = 1}^{M} \log_2\left(1 + \frac{S_{i}}{N_i}\right) \\
\text {Subject to} \: \sum_{i = 1}^{M} S_{i} \leq P_T
\end{align}.
$N_i$ are variables given in the problem. I have to determine the $S_i$
I tried using Lagrange multiplier method, which gives me:
$H(S_1, S_2, …, S_M) = B\sum_{i = 1}^{M} \log_2\left(1 + \frac{S_{i}}{N_i}\right) – \lambda \left(\sum_{i = 1}^{M} S_{i} – P_t\right)$.
Using the Lagrange multipliers technique means that we will solve the system:
$\forall S_i,\: 0 = \frac{\partial H}{\partial S_i} = \frac{1}{N_i + S_i} – \lambda \\$
$0 = \frac{\partial H}{\partial \lambda} = \sum_{i = 1}^{n} S_{i} – P_T$
This leads to $(N_1 + S_1) = (N_2 + S_2) = \dots = (N_M + S_M)$ with $\sum_{i = 1}^{n} S_{i} = P_T$.
Best Answer
Since $\ S_i=x-N_i\ $ for all $\ i\ $ ($\ x\ $ must actually be $\ \frac{1}{\lambda}\ $), and $\ \displaystyle\sum_{j=1}^M S_j=P_T\ $, then you get $$ P_T= \sum_{j=1}^M\left(x+N_j\right)=Mx+ \sum_{j=1}^MN_j\ , $$ and hence \begin{align} x&=\frac{\displaystyle P_T-\sum_{j=1}^MN_j}{M}\ \text{ , and}\\ S_i&= \frac{\displaystyle P_T-\sum_{j=1}^MN_j}{M}-N_i\ . \end{align}