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I have following optimization problem

$$ \arg \max_{x \in \mathbb{R}_{++}} \frac{ \ln(1+ax) + \ln(1+bx) }{x+c} $$

Where $ x $ can have only positive values and $ a, b,c $ are positive constants.

Is there a close form expression possible for optimal $ x $ in the above problem?

I know that the objective function of the problem is a quasi concave function however I wonder if it admits a close form solution.
Any help in this regard will be much appreciated. Thanks in advance.

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  • $\begingroup$ I think you Need a numerical mathod to solve this Problem. Do you have concrete values fot $a,b,c$? $\endgroup$ – Dr. Sonnhard Graubner Apr 9 '18 at 8:49
  • $\begingroup$ @Dr.SonnhardGraubner no i do not have known values of $a,b,c$. But I can solve the following problem in closed form $$\underbrace{\max}_{x}\frac{\ln(1+ax+bx)}{x+c}$$ and I think the close form solution of $x^*$ that I will get for this lower bound of the original problem in my post will be a lower bound on the optimal value of $x$ in my original problem. What do you say? If this is right then at least I can reduce the range of $x$ over which Bisection method is to be used. What your take on this? $\endgroup$ – Frank Moses Apr 9 '18 at 8:53
  • $\begingroup$ You Can also use that $$\log(a)+\log(b)=\log(ab)$$ to simplify your objective function $\endgroup$ – Dr. Sonnhard Graubner Apr 9 '18 at 8:56
  • $\begingroup$ @Dr.SonnhardGraubner yes I used that to obtain the lower bound of the numerator that I used in the comment. $\endgroup$ – Frank Moses Apr 9 '18 at 8:58
  • $\begingroup$ But the first derivative containes the Log-function, this make the difficulties to get the solution of $$f'(x)=0$$ $\endgroup$ – Dr. Sonnhard Graubner Apr 9 '18 at 9:01
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I have tried Dinkelbach's and Schaible's method, but have not found a solution. I still provide my notes so anyone can check them and maybe provide useful comments.

Denote the problem as $$\max_{x \in \mathbb{R}^n_{++}} \frac{\ln(1+ax) + \ln(1+bx)}{x+c}$$ and define: $$h(\lambda) = \max_{x \in \mathbb{R}^n_{++}} \{ \ln(1+ax) + \ln(1+bx)-\lambda (x+c) \}$$ Dinkelbach (1967) showed that $h(\lambda) \geq 0$ if and only if: $$\max_{x \in \mathbb{R}^n_{++}} \frac{\ln(1+ax) + \ln(1+bx)}{x+c} \geq \lambda.$$ The problem now is to find the largest $\lambda$ such that $h(\lambda) \geq 0$. Since $h$ is concave, its maximum can be found with the first order condition $a/(1+ax) + b/(1+bx) - \lambda = 0$, which simplifies to: $$\lambda abx^2 + (2ab - 2a\lambda - 2b\lambda) x + (a+b+\lambda) = 0.$$ Plugging this into $h(\lambda)$ and solving for $h(\lambda)=0$ does not give me a closed form solution.


Schaible (1974) showed that you can substitute $y = x / (x+c)$ and $t = 1/(x+c)$ to get an equivalent but convex optimization problem: $$\max_{y \in \mathbb{R}^n_{++}} \{ t \ln(1+ay/t) + t \ln(1+by/t) : y+ct\geq 1 \}.$$ The KKT conditions are: $$\begin{align} \frac{at}{t+ay} + \frac{bt}{t+by} - \lambda = 0 \\ \ln(1+ay/t) + \ln(1+by/t) - \frac{a}{t+ay} - \frac{b}{t+ay} - c\lambda = 0 \\ \lambda(1-y-ct) = 0 \\ y+ct \geq 1 \\ \lambda \geq 0 \end{align}$$ I do not find a closed form solution for this either.

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