I'm trying to solve the following optimization problem: \begin{equation}\end{equation} \begin{align} \text{argmax}_{\left\{x_i\right\},\left\{y_j\right\}} & \frac{1}{2}\left( \sum_{i=1}^{N} \log_2\left(1+\alpha_i x_i\right) + \sum_{j=1}^{M} \log_2\left(1+ \beta_j y_j\right) \right) \\[4pt] \nonumber \text{subject to: }& \sum_{i=1}^N x_i + \sum_{j=1}^M y_j \leq A \\ & \sum_{j=1}^{M} \log_2\left(1+ \beta_j y_j\right) \leq B \end{align}

To this end, I tried to use the Lagrangian multipliers method. I defined the Lagrangian function \begin{align} \mathcal{L}(x_1,\ldots,x_N,y_1,\ldots,y_M,\lambda_1,\lambda_2) &= \left( \sum_{i=1}^{N} \log_2\left(1+\alpha_i x_i\right) + \sum_{j=1}^{M} \log_2\left(1+ \beta_j y_j\right) \right) \\ & - \lambda_1 \left(\sum_{i=1}^N x_i + \sum_{j=1}^M y_j - A\right) \\ & -\lambda_2 \left(\sum_{j=1}^{M} \log_2\left(1+ \beta_j y_j\right) -B\right) \end{align} and I computed the partial derivatives of the Lagrangian function with respect to generic $x_i$ and $y_j$. Setting them to zero led to \begin{align} & \frac{\partial \mathcal{L}}{\partial x_i} = \frac{\alpha_i}{\alpha_i x_i \log(2) + \log(2)} - \lambda_1 = 0, \\ & \frac{\partial \mathcal{L}}{\partial y_j} = \frac{\beta_j}{\beta_j y_j \log(2) + \log(2)} - \lambda_1 -\lambda_2 \frac{\beta_j}{\beta_j y_j \log(2) + \log(2)} = 0, \end{align} which resulted in \begin{align} & x_i = \frac{1}{\log(2) \lambda_1} - \frac{1}{\alpha_i}, \\ & y_j = \frac{1-\lambda_2}{\log(2) \lambda_1} - \frac{1}{\beta_j}. \end{align} I don't know how to proceed from here onwards. Any help?

  • $\begingroup$ Well first I would manipulate the problem so you have equality constraints. $\endgroup$ – mattos Mar 20 at 14:45
  • $\begingroup$ If you show us your work in computing the partial derivatives, we could help you debug that process. If you're asking what to do after computing the partial derivatives, that's a further question. $\endgroup$ – Brian Borchers Mar 20 at 16:09
  • $\begingroup$ @BrianBorchers partial derivatives added in the OP. I'm not sure what I should do next $\endgroup$ – cubicle01 Mar 20 at 16:21
  • $\begingroup$ When you substitute those values of $x$ and $y$ into the Lagrangian, you'll obtain the dual function $g(\lambda)$. $\endgroup$ – Brian Borchers Mar 20 at 16:49
  • $\begingroup$ @BrianBorchers I still don't understand how I should proceed after having substituted $x$ and $y$ in the Lagrangian, as the dual function contains sereval sums of logarithms. Would you mind explaining the concept a bit further, please? $\endgroup$ – cubicle01 Mar 20 at 17:15


I will use $\ln$ instead to facilitate the notation issues. First you should transform your inequalities into equations with the introduction of some slack variables $s_1, s_2$ so the Lagrangian reads

\begin{align} \mathcal{L}(x_1,\ldots,x_N,y_1,\ldots,y_M,\lambda_1,\lambda_2,s_1,s_2) &= \frac 12\left( \sum_{i=1}^{N} \ln\left(1+\alpha_i x_i\right) + \sum_{j=1}^{M} \ln\left(1+ \beta_j y_j\right) \right) \\ & - \lambda_1 \left(\sum_{i=1}^N x_i + \sum_{j=1}^M y_j - A+s_1^2\right) \\ & -\lambda_2 \left(\sum_{j=1}^{M} \ln\left(1+ \beta_j y_j\right) -B+s_2^2\right) \end{align}

now, the stationary points are the solutions for

\begin{align} & x_i = \frac{1}{2 \lambda_1} - \frac{1}{\alpha_i}, \\ & y_j = \frac{1-2\lambda_2}{2\lambda_1} - \frac{1}{\beta_j},\\ & \sum_{i=1}^N x_i + \sum_{j=1}^M y_j - A+s_1^2= 0,\\ & \sum_{j=1}^{M} \ln\left(1+ \beta_j y_j\right) -B+s_2^2=0,\\ & \lambda_1s_1 = 0,\\ & \lambda_2 s_2 = 0, \end{align}

We can proceed by substituting the values found to $x_i,y_j$ into the restrictions giving

\begin{align} &\frac{N}{2 \lambda_1} - \sum_{i=1}^N\frac{1}{\alpha_i}+M\frac{1-2\lambda_2}{2\lambda_1}- \sum_{j=1}^M\frac{1}{\beta_j}-A+s_1^2=0,\\ &\left(\frac{1-2\lambda_2}{2\lambda_1}\right)^M\prod_{j=1}^M\beta_j=2^Me^{B-s_2^2} \end{align}

Now using $\lambda_is_i=0$ we can (?) calculate $\lambda_i$ and finally to have the stationary points. The solutions obtained with $s_i = 0$ are solutions at the feasible region boundary.

  • $\begingroup$ Thank you. Can we direcly assume that $s_1=s_2 = 0$ if the values of $x_i$ and $y_j$ have to be finite? $\endgroup$ – cubicle01 Mar 20 at 20:41
  • $\begingroup$ It seems true if $\alpha_i,\ \beta_j$ are non negative. $\endgroup$ – Cesareo Mar 20 at 20:50

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