I have following optimization problems. The first problem is $$\underbrace{\max}_{x_1,x_2} \frac{\ln(1+a_1x_1)+\ln(1+a_2x_2)}{x_1+x_2+c}\\ \text{s.t. } 0\leq x_1\leq \bar{x}\\ 0\leq x_2\leq \bar{x}$$ and the second problem is $$\frac{\underbrace{\max}_{x_1,x_2}\ln(1+a_1x_1)+\ln(1+a_2x_2)}{x_1+x_2+c} \\ \\ \text{s.t.} \\0\leq x_1\leq \bar{x}\\ 0\leq x_2\leq \bar{x}$$ where $a_1,a_2$ and $c$ are some positive constants. $\bar{x}$ is some positive value which $x_1$ and $x_2$ cannot exceed. Which of the problem will have a higher optimized value? First or second? Any help in this regard will be much appreciated. Thanks in advance.

  • $\begingroup$ What is $\bar{x}$? The mean of $x_1$ and $x_2$? $\endgroup$ – YukiJ Apr 10 '18 at 6:25
  • $\begingroup$ @YukiJ no its a bound on the maximum value of $x_1$ and $x_2$. I have also added it in my question. $\endgroup$ – Frank Moses Apr 10 '18 at 6:27
  • $\begingroup$ @OnceUponACrinoid you mean in the same expression for $x_1$ i use different values of $x_1$ in the numerator and different value for $x_1$ in the denominator? Is it possible? $\endgroup$ – Frank Moses Apr 10 '18 at 6:43
  • $\begingroup$ What is the value of $x$ in the denominator? $\endgroup$ – LinAlg Apr 10 '18 at 16:03

I am presuming that $x_1$ and $x_2$ in the denominator of the 2nd problem are the respective argmaxes, $x_1^*$ and $x_2^*$,of the first problem, and therefore are input data, not optimization variables, for the 2nd problem.. Let $x_1^{**}$ and $x_2^{**}$ be the argmaxes of the 2nd problem. Then because the constraints are identical, obviously $$\frac{\ln(1+a_1x_1^*)+\ln(1+a_2x_2^*)}{x_1^*+x_2^*+c} \ge \frac{\ln(1+a_1x_1^{**})+\ln(1+a_2x_2^{**)}}{x_1^*+x_2^*+c}$$ becauee the 2nd problem has in effect been suboptimized.

  • $\begingroup$ thank you for your answer. What if we put $x_1=x_1^{**}$ and $x_2=x_2^{**}$ in the second problem?(please comment) Actually we have to pick $x_1$ and $x_2$ in the second problem such that the numerator is maximized in the second problem. While in the first problem we have to pick $x_1,x_2$ such that the ratio is maximized. $\endgroup$ – Frank Moses Apr 10 '18 at 23:20
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    $\begingroup$ If the denominator in the 2nd problem has x's which are optimized only for the numerator, then similarly to the case in my answer, the resulting ratio will be no better, and may be worse, than the first problem, which is optimized for the ratio. That is just another way of suboptimizing. $\endgroup$ – Mark L. Stone Apr 10 '18 at 23:42
  • $\begingroup$ @Frank Moses Does that answer your questions on this matter? $\endgroup$ – Mark L. Stone Apr 11 '18 at 1:01
  • $\begingroup$ I have marked it correct $\endgroup$ – Frank Moses Apr 11 '18 at 1:11

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