I want to maximize $$\sum_{i=1}^{k}[\frac{n_i}{T_i}\ln \lambda_i-\lambda_i]T_i,~~~n_i, T_i>0 $$ subject to $\lambda_1\geq \lambda_2 \geq \ldots \geq \lambda_k>0.$ How can I show that the solution $\hat{\lambda}_1, \ldots, \hat{\lambda}_k$ of this extremum problem satisfies $$\sum_{i=1}^{k}(\frac{n_i}{T_i}-\hat{\lambda}_i)T_i=0~?$$
How can I maximize the following function?
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$\begingroup$
functions
optimization
maxima-minima
2 Answers
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If $\hat{\lambda}_1\geq\ldots\geq \hat{\lambda}_k$ and if $b>0$ then also $b\hat{\lambda}_1\geq\ldots\geq b\hat{\lambda}_k.$ Thus $$\sum_{i=1}^{k}[\frac{n_i}{T_i}ln( b \hat{\lambda}_i)-b\hat{\lambda_i}]T_i $$achieves its maximum as a function of $b$ at $b=1.$ On setting its derivatives at $b=1$ equal to zero, we have $$\sum_{i=1}^{k}(\frac{n_i}{T_i}-\hat{\lambda}_i)T_i=0.$$
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The function is concave, meaning that a KKT point corresponds to a maximum. The Lagrangian is: $$L(\lambda,y,z) = \sum_{i=1}^k \left[ \frac{n_i}{T_i} - \lambda_i\right]T_i + \sum_{i=1}^{k-1} y_i(\lambda_i - \lambda_{i+1}) + \sum_{i=1}^k z_i \lambda_i $$ The KKT conditions are: $$\sum_{i=1}^k \left[ \frac{n_i}{T_i \lambda_i} - 1\right]T_i + y_1 -y_k= 0$$ $$y_i(\lambda_i - \lambda_{i+1}) = 0$$ $$\lambda_1 \geq \lambda_2 \geq \ldots \geq \lambda_k > 0$$ $$y \geq 0$$ This is not exactly what you got.