# Difference of two optimization problem's optimal value

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 $$|\alpha_1 -\alpha_2|$$?

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 $$|\alpha_1 - \alpha_2| \leq \sum_{m=1}^M (\log(1+\beta_m) - \log(\beta_m)) < \infty$$.
• Thank you for your comment. You are correct. I add the $f_m$. Sep 15, 2020 at 20:51
• You should further assume that $\beta_{m,t} > 0$, otherwise (if $A = 0$) $\alpha_1 = \alpha_2 = \infty$. Sep 15, 2020 at 21:01