Let we have two following optimization problems: \begin{align} \text{(P1)}\quad \alpha_1 = \max_{x_1,\ldots,x_M} &\quad \sum_{m=1}^{M}\log(1+f_m(x_1,\ldots,x_M))\\ \textrm{s.t.} &\quad \boldsymbol{A}\boldsymbol{x}\leq\mathbf{1}\\ &\quad x_m\geq0,\forall m. \end{align}

\begin{align} \text{(P2)}\quad \alpha_2 =\max_{x_1,\ldots,x_M} &\quad \sum_{m=1}^{M}\log(f_m(x_1,\ldots,x_M))\\ \textrm{s.t.} &\quad \boldsymbol{A}\boldsymbol{x}\leq\mathbf{1}\\ &\quad x_m\geq0,\forall m, \end{align} where $\boldsymbol{x}=[x_1,\ldots,x_M]^{\mathrm{T}}$, $\boldsymbol{A}$ is an arbitrary $N\times M$ matrix with positive entries and $\mathbf{1}$ is an all-one $N\times 1$ vector. Moreover, assume that $f_m$ is as follows \begin{align} f_m(x_1,\ldots,x_M)=\frac{x_m}{1+\sum_{t=1}^M \beta_{m,t}x_t}, \end{align} where $\beta_{m,t}$ is an arbitrary positive coefficients. Is there any bound for $|\alpha_1 -\alpha_2|$?


1 Answer 1


A necessary assumption will be that $f_m > 0$. Further, assume that each $f_m$ is constant. Then there is nothing to optimize and you have to compare $\alpha_2$, which may go to $-\infty$ and $\alpha_1 > 0$. So without any essential additional assumptions there is no bound. Of course, if you assume that each $f_m > \beta_m$ (not necessarily constant) for some $\beta_m > 0$ then you have $|\alpha_1 - \alpha_2| \leq \sum_{m=1}^M (\log(1+\beta_m) - \log(\beta_m)) < \infty$.

  • $\begingroup$ Thank you for your comment. You are correct. I add the $f_m$. $\endgroup$
    – Math_Y
    Sep 15, 2020 at 20:51
  • $\begingroup$ You should further assume that $\beta_{m,t} > 0$, otherwise (if $A = 0$) $\alpha_1 = \alpha_2 = \infty$. $\endgroup$ Sep 15, 2020 at 21:01
  • $\begingroup$ Yes. I change the non-negativity to positivity. $\endgroup$
    – Math_Y
    Sep 15, 2020 at 21:55

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