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I am trying to solve the following optimization problem. I do not care about the particular values of the $x$'s, I just care about the optimal value of the function.

\begin{equation} \min_{x\geq0} z_{1}\left(x_{1}+x_{2}+x_{3}\right)^{\frac{1}{\eta}}+z_{2}\left(x_{4}+x_{5}+x_{6}\right)^{\frac{1}{\eta}} -\mu_{1}\left(\frac{x_{1}}{a_{1}}+\frac{x_{4}}{a_{4}}\right) -\mu_{2}\left(\frac{x_{2}}{a_{2}}+\frac{x_{5}}{a_{5}}\right) -\mu_{3}\left(\frac{x_{3}}{a_{3}}+\frac{x_{6}}{a_{6}}\right) \end{equation}

where the vectors $z, a, \mu$ are positive parameters of the problem and the variables are the $x$, which are non-negative. Also, $1>\eta$.

The solution I am getting is the following

Potential answer (please let me know if I am wrong):

Let's take $x_{1}$, the first order condition is given by

$$ \frac{z_{1}}{\eta}(x_{1}+x_{2}+x_{3})^{\frac{1-\eta}{\eta}} - \frac{\mu_{1}}{a_{1}} \geq 0 $$

From the first order conditions one can show that

\begin{align} x_{1}+x_{2}+x_{3} = \left( \frac{\eta}{z_{1}} \max\left\{\frac{\mu_{1}}{a_{1}},\frac{\mu_{2}}{a_{2}},\frac{\mu_{3}}{a_{3}}\right\} \right)^{\frac{\eta}{1-\eta}} \\ x_{4}+x_{5}+x_{6} = \left( \frac{\eta}{z_{1}} \max\left\{\frac{\mu_{1}}{a_{4}},\frac{\mu_{2}}{a_{5}},\frac{\mu_{3}}{a_{6}}\right\} \right)^{\frac{\eta}{1-\eta}} \end{align}

By having these conditions we know what is the value of the first two terms of the function. However, can we say something about the last three terms?

Also, we could plug in these conditions in the first order constraints for each $x_{i}$, for example for x_{1} we would get that

$$ \max\left\{\frac{\mu_{1}}{a_{1}},\frac{\mu_{2}}{a_{2}},\frac{\mu_{3}}{a_{3}}\right\} \geq \frac{\mu_{1}}{a_{1}} $$

which should hold with equality if $x_{1}>0$.

Also, given values of $\mu$ and the $a$ one can show that some of the x cannot be simultaneously positive. After finding out which $x$ are $0$ any combination of positive $x$ for which the previous condition holds is a solution.

However, I am not sure at all whether this is progress toward the correct solution.

Any help would be very much appreciated.

EDIT: More context and work after a month thinking about this.

econ_ugrad
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1 Answers1

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The minimum is given by $$z_1 \left(\frac{c_1\eta}{z_1}\right)^{1/(1 - \eta)} - c_1 \left(\frac{c_1\eta}{z_1}\right)^{\eta/(1 - \eta)} + z_2 \left(\frac{c_2\eta}{z_2}\right)^{1/(1 - \eta)} - c_2 \left(\frac{c_2\eta}{z_2}\right)^{\eta/(1 - \eta)}$$ where $$c_1 = \max\left(\frac{\mu_1}{a_1}, \frac{\mu_2}{a_2}, \frac{\mu_3}{a_3}\right), \quad c_2 = \max\left(\frac{\mu_1}{a_4}, \frac{\mu_2}{a_5}, \frac{\mu_3}{a_6}\right).$$

$\phantom{2}$



Proof:

First of all, we give the following auxiliary result.

Fact 1: Let $z > 0, c > 0, r > 1$ be given. The minimum of $f(x) = z x^r - c x$ on $x \ge 0$ is given by $$f_{\min} = z \left(\frac{c}{zr}\right)^{r/(r - 1)} - c \left(\frac{c}{zr}\right)^{1/(r - 1)}$$ which is attained at $x = \left(\frac{c}{zr}\right)^{1/(r - 1)}$.
(The proof is easy and thus omitted.)

$\phantom{2}$

Now, let \begin{align*} g(x_1, x_2, x_3) &= z_1(x_1 + x_2 + x_3)^{1/\eta} - \left(\frac{\mu_1}{a_1}x_1 + \frac{\mu_2}{a_2}x_2 + \frac{\mu_3}{a_3}x_3\right), \\ h(x_4, x_5, x_6) &= z_2(x_4 + x_5 + x_6)^{1/\eta} - \left(\frac{\mu_1}{a_4}x_4 + \frac{\mu_2}{a_5}x_5 + \frac{\mu_3}{a_6}x_6\right). \end{align*}

Clearly, we have, for all $x_1, x_2, x_3 \ge 0$, \begin{align*} \frac{\mu_1}{a_1}x_1 + \frac{\mu_2}{a_2}x_2 + \frac{\mu_3}{a_3}x_3 &\le (x_1 + x_2 + x_3)\max\left(\frac{\mu_1}{a_1}, \frac{\mu_2}{a_2}, \frac{\mu_3}{a_3}\right) \end{align*} with equality if $$\left[\max\left(\frac{\mu_1}{a_1}, \frac{\mu_2}{a_2}, \frac{\mu_3}{a_3}\right) - \frac{\mu_i}{a_i}\right] x_i = 0, \quad i = 1, 2, 3.$$

Thus, using Fact 1, we have \begin{align*} g(x_1, x_2, x_3) &\ge z_1(x_1 + x_2 + x_3)^{1/\eta} - (x_1 + x_2 + x_3)\max\left(\frac{\mu_1}{a_1}, \frac{\mu_2}{a_2}, \frac{\mu_3}{a_3}\right)\\ &\ge z_1 \left(\frac{c_1\eta}{z_1}\right)^{1/(1 - \eta)} - c_1 \left(\frac{c_1\eta}{z_1}\right)^{\eta/(1 - \eta)} \end{align*} with equality if \begin{align*} &x_1 + x_2 + x_3 = \left(\frac{c_1\eta}{z_1}\right)^{\eta/(1 - \eta)}, \\ &\left(c_1 - \frac{\mu_i}{a_i}\right) x_i = 0, \quad i = 1, 2, 3. \end{align*}

Similarly, we have \begin{align*} h(x_4, x_5, x_6) &\ge z_2(x_4 + x_5 + x_6)^{1/\eta} - (x_4 + x_5 + x_6)\max\left(\frac{\mu_1}{a_4}, \frac{\mu_2}{a_5}, \frac{\mu_3}{a_6}\right)\\ &\ge z_2 \left(\frac{c_2\eta}{z_2}\right)^{1/(1 - \eta)} - c_2 \left(\frac{c_2\eta}{z_2}\right)^{\eta/(1 - \eta)} \end{align*} with equality if \begin{align*} &x_4 + x_5 + x_6 = \left(\frac{c_2\eta}{z_2}\right)^{\eta/(1 - \eta)}, \\ &\left(c_2 - \frac{\mu_i}{a_{i + 3}}\right) x_{i + 3} = 0, \quad i = 1, 2, 3. \end{align*}

Finally, we have \begin{align*} &\min_{x_i \ge 0, \forall i} [g(x_1, x_2, x_3) + h(x_4, x_5, x_6)]\\ =\,& z_1 \left(\frac{c_1\eta}{z_1}\right)^{1/(1 - \eta)} - c_1 \left(\frac{c_1\eta}{z_1}\right)^{\eta/(1 - \eta)} + z_2 \left(\frac{c_2\eta}{z_2}\right)^{1/(1 - \eta)} - c_2 \left(\frac{c_2\eta}{z_2}\right)^{\eta/(1 - \eta)}. \end{align*}

We are done.

River Li
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  • Thanks so much! This is a great and elegant way of finding the solution. If you have time, can you take a look at this problem https://math.stackexchange.com/questions/4394750/non-linear-system-of-equations-involving-max-and-min – econ_ugrad Mar 08 '22 at 15:14
  • @econ_ugrad You are welcome. I will try. – River Li Mar 08 '22 at 15:18