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<p>The logarithm of a negative number is undefined, so it does not make sense to evaluate the function for x<=0. So I'd suggest to lower bound x to zero or a small number, assuming that x is an optimisation variable. The function should then not be evaluated
for x<=0.<br>
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<p>Filip <br>
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<div id="divRplyFwdMsg" dir="ltr"><font style="font-size:11pt" color="#000000" face="Calibri, sans-serif"><b>Van:</b> Ipopt <ipopt-bounces@coin-or.org> namens Maxime Boulay <mboulay@flogen.com><br>
<b>Verzonden:</b> dinsdag 28 november 2017 22:26<br>
<b>Aan:</b> Ipopt@list.coin-or.org<br>
<b>CC:</b> RMerdjani@flogen.com<br>
<b>Onderwerp:</b> [Ipopt] Question concerning presence of indefinite terms in function</font>
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<p class="MsoNormal">Hello,</p>
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<p class="MsoNormal">I am using ipopt to solve an objective function which contains an expression of the form x*log(x/(a+x)). This results in a derivative of the form log(x/(a+x)) and a second derivative of the form a/(x*(x+a)) when considering other terms
in the function. Since this will lead to problems whenever x = 0 or x < 0, I am not sure about the best way to treat those cases. I have tried equating the expression to 0 whenever those cases are met but this seems wrong since the expression doesn’t tend
to 0 when approaching the critical values. This also seems to slow down the time taken to find a solution whenever the initial values given or the solution include an x=0.
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<p class="MsoNormal">Given the fact that many of these may pop out since the function includes a summation, what would be the most efficient and correct way to deal with this?</p>
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<p class="MsoNormal">Thank you.</p>
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