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I have the optimization problem as below:

$$\begin{align} &\text{maximize } \sum_{k=1}^{M} \alpha_k {R}_k\\ &\text{subject to: } \exp \left[ - (2^{{R}_k } -1) \left(\frac{\tilde{Z} g_{k} p_{\max} + \sigma^2}{g_{pu} p_{pu}} \right) \right] \leq q \end{align}$$

$\alpha_k$ is the weight factor associated to the $R_k$.

This is a nonlinear optimization. But I am completely new to optimization. So any help is highly appreciated.

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  • $\begingroup$ Welcome to SciComp! I don't understand your notation with the colons. $\endgroup$ Commented Aug 14, 2015 at 8:35
  • $\begingroup$ The only variables are the $R_{k}$, right? Is the constant factor within the exponential positive, negative, or could it be of either sign? $\endgroup$ Commented Aug 14, 2015 at 12:52
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    $\begingroup$ Cross-posted verbatim and answered at Math.StackExchnage.com. Please don't cross post. $\endgroup$
    – horchler
    Commented Aug 14, 2015 at 15:52
  • $\begingroup$ @BrianBorchers,, yap $R_k$ is the variable and rest of the notations are positives. $\endgroup$
    – jhon_wick
    Commented Aug 14, 2015 at 16:45
  • $\begingroup$ Presuming all the terms are positive enough, take the natural log then the log base 2? $\endgroup$
    – Bill Barth
    Commented Aug 15, 2015 at 0:23

1 Answer 1

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If the $R_k$ are your only optimization variables, then the constraint $$ \exp \left[ - (2^{{R}_k } -1) \left(\frac{\tilde{Z} g_{k} p_{\max} + \sigma^2}{g_{pu} p_{pu}} \right) \right] \leq q $$ is equivalent to $$ - (2^{{R}_k } -1) \left(\frac{\tilde{Z} g_{k} p_{\max} + \sigma^2}{g_{pu} p_{pu}} \right) \leq \ln q $$ which is equivalent to $$ (2^{{R}_k } -1) \leq -(\ln q) \left(\frac{g_{pu} p_{pu}}{\tilde{Z} g_{k} p_{\max} + \sigma^2} \right) $$ which is equivalent to $$ 2^{{R}_k } \leq -(\ln q) \left(\frac{g_{pu} p_{pu}}{\tilde{Z} g_{k} p_{\max} + \sigma^2} \right)+1 $$ which is equivalent to $$ {R}_k \leq \log_2 \left\{ -(\ln q) \left(\frac{g_{pu} p_{pu}}{\tilde{Z} g_{k} p_{\max} + \sigma^2} \right)+1\right\}. $$ This is a linear constraint, and your objective function is already linear. So you ended with a standard linear program, for which there is plenty of software (and if you don't want to use any of that, plenty of simple to implement methods).

The transformation above shows an important principle: one can often transform complicated-looking problems into much simpler problems if one just looks at them closely enough.

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